sm ch (1)

Problem P1.2-11 (1-4 in text) Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with kA = 1 W/m-K and B with kB = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A; the heating element provides q ′′ = 5000 W/m 2 of heat. The surface of the wall at x = 2L is exposed to fluid at Tf,in = 300 K with heat transfer coefficient hin = 100 W/m2-K. 2 q ′′ = 5000 W/m

insulated

material A kA = 1 W/m-K L = 1 cm

x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

Figure P1.2-11(a): Composite wall with a heater.

You may neglect radiation and contact resistance for parts (a) through (c) of this problem. a.) Draw a resistance network to represent this problem; clearly indicate what each resistance represents and calculate the value of each resistance. The input parameters are entered in EES: “P1.2-11: Heater" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q_flux=100 [W/m^2] L = 1.0 [cm]*convert(cm,m) k_A=1.0 [W/m-K] k_B=5.0 [W/m-K] T_f_in=300 [K] h_in=100 [W/m^2-K] A=1 [m^2]

"heat flux provided by the heater" "thickness of each layer" "conductivity of material A" "conductivity of material B" "fluid temperature at inside surface" "heat transfer on inside surface" "per unit area"

The resistance network that represents the problem shown in Figure 2 is:

Figure 2: Resistance network.

The resistances due to conduction through materials A and B are:

RA =

RB =

L

(1)

kA A

L

(2)

kB A

where A is the area of the wall, taken to be 1 m2 in order to carry out the analysis on a per unit area basis. The resistance due to convection is:

Rconv ,in =

1

(3)

hin A

"part (a)" R_A=L/(k_A*A) R_B=L/(k_B*A) R_conv_in=1/(h_in*A) "resistance to convection on inner surface"

"resistance to conduction through A" "resistance to conduction through B"

which leads to RA = 0.01 K/W, RB = 0.002 K/W, and Rconv,in = 0.01 K/W. b.) Use your resistance network from (a) to determine the temperature of the heating element. The resistance network for this problem is simple; the temperature drop across each resistor is equal to the product of the heat transferred through the resistor and its resistance. In this simple case, all of the heat provided by the heater must pass through materials A, B, and into the fluid by convection so these resistances are in series. The heater temperature (Thtr) is therefore:

Thtr = T f ,in + ( RA + RB + Rconv ,in ) q ′′ A

(4)

T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"

which leads to Thtr = 410 K. c.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (b). The temperatures at x = L and x = 2L can be computed according to:

Tx = L = T f ,in + ( RB + Rconv ,in ) q ′′ A

(5)

Tx = 2 L = T f ,in + Rconv ,in q ′′ A

(6)

T_L=T_f_in+(R_B+R_conv_in)*q_flux*A T_2L=T_f_in+R_conv_in*q_flux*A

"temperature at x=L" "temperature at x=2L"

which leads to Tx=L = 360 K and Tx=2L = 350 K. The temperature distribution is sketched on the axes in Figure 3.

Figure 3: Sketch of temperature distribution.

Notice that the temperature drop through the two larger resistances (RA and RB) are much larger than the temperature drop across the small resistance, RB. Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm. material C kC = 2 W/m-K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

insulated x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B L = 1 cm k = 5 W/m-K B

Figure P1.2-11(b): Composite wall with Material C.

Neglect radiation and contact resistance for parts (d) through (f) of this problem. d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly indicate what each resistance represents and calculate the value of each resistance. There is an additional resistor corresponding to conduction through material C, RC, as shown below:

Notice that the boundary condition at the end of RC corresponds to the insulated wall; that is, no heat can be transferred through this resistance. The resistance to conduction through material C is:

RC = "part (b)" k_C=2.0 [W/m-K] R_C=L/(k_C*A)

L

(7)

kC A

"conductivity of material C" "resistance to conduction through C"

which leads to RC = 0.005 K/W. e.) Use your resistance network from (d) to determine the temperature of the heating element. Because there is no heat transferred through RC, all of the heat must still go through materials A and B and be convected from the inner surface of the wall. Therefore, the answer is not changed from part (b), Thtr = 410 K. f.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (e). The answer is unchanged from part (c) except that there is material to the left of the heater. However, no heat is transferred through material C and therefore there is no temperature gradient in the material.

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a contact resistance between materials A and B, Rc′′ = 0.01 K-m 2 /W , and the surface of the wall at

x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient hout = 10 W/m2-K. material C kC = 2 W/m-K T f ,out = 400 K 2 hout = 10 W/m -K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

x

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

L = 1 cm L = 1 cm

Rc′′ = 0.01 K-m /W 2

Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.

Neglect radiation for parts (g) through (i) of this problem. g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly indicate what each resistance represents and calculate the value of each resistance. The additional resistances associated with contact resistance and convection to the fluid at the outer surface are indicated. Notice that the boundary condition has changed; heat provided by the heater has two paths ( qout and qin ) and so the problem is not as easy to solve.

The additional resistances are computed according to: Rconv ,out =

1

(8)

hout A

Rcontact =

Rc′′ A

"part (c)" R``_c=0.01 [K-m^2/W] h_out=10 [W/m^2-K] T_f_out=400 [K] R_contact=R``_c/A R_conv_out=1/(h_out*A) "convection resistance on outer surface"

(9)

"area specific contact resistance" "heat transfer coefficient" "fluid temperature on outside surface" "contact resistance"

which leads to Rcontact = 0.01 K/W and Rconv,out = 0.1 K/W. h.) Use your resistance network from (j) to determine the temperature of the heating element. It is necessary to carry out an energy balance on the heater: q ′′ A = qin + qout

(10)

The heat transfer rates can be related to Thtr according to: qin =

(T

htr

− T f ,in )

RA + Rcontact + RB + Rconv ,in

qout =

(T

htr

− T f ,out )

(11)

(12)

RC + Rconv ,out

These are 3 equations in 3 unknowns, Thtr, qout and qin , and therefore can be solved simultaneously in EES (note that the previous temperature calculations from part (b) must be commented out): {T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A T_L=T_f_in+(R_B+R_conv_in)*q_flux*A

"heater temperature" "temperature at x=L"

T_2L=T_f_in+R_conv_in*q_flux*A q_flux*A=q_dot_in+q_dot_out q_dot_in=(T_htr-T_f_in)/(R_A+R_contact+R_B+R_conv_in) q_dot_out=(T_htr-T_f_out)/(R_C+R_conv_out)

"temperature at x=2L"} "energy balance on the heater" "heat flow to inner fluid" "heat flow to outer fluid"

which leads to Thtr = 446 K. The other intermediate temperatures shown on the resistance diagram can be computed: Tx = L − = Thtr − RA qin

(13)

Tx = L + = Thtr − ( RA + Rcontact ) qin

(14)

Tx = 2 L = Thtr − ( RA + Rcontact + RB ) qin

(15)

Tx =− L = Thtr − RC qout

(16)

"intermediate temperatures" T_Lm=T_htr-R_A*q_dot_in T_Lp=T_htr-(R_A+R_contact)*q_dot_in T_2L=T_htr-(R_A+R_contact+R_B)*q_dot_in T_mL=T_htr-R_C*q_dot_out

which leads to Tx=L- = 400.4 K, Tx=L+ = 354.7 K, Tx=2L = 345.6 K, and Tx=-L = 443.8 K. i.) Sketch the temperature distribution on the axes provided below.

P1.1-1: Viscosity of a dilute gas Momentum transfer occurs in a fluid due to interactions between molecules that results in a transfer of momentum. This process is characterized by viscosity, which relates the shear stress to a velocity gradient in the same way Fourier’s Law relates heat flux to a temperature gradient. It is not surprising, then, that the viscosity and thermal conductivity of an ideal gas are analogous transport properties. a.) Using reasoning similar to that provided in Section 1.1.2 for thermal conductivity, show that the viscosity of an ideal gas can be estimated according to μ ∝ T MW / σ 2 . Consider momentum transfer through a fluid in which a velocity gradient has been established in the x-direction, as shown in Figure 1. We can evaluate the net rate of momentum transferred through a plane that is located at position x. The flux of molecules passing through the plane from left-to-right (i.e., in the positive x-direction) is proportional to the number density of the molecules (nms) and their mean velocity (vms). The molecules that are moving in the positive xdirection experienced their last interaction at x–Lms (on average), where Lms is the distance between molecular interactions. The rate of momentum associated with these molecules per unit area is the product of the rate of molecules passing through the plane (nms vms) momentum and the momentum per molecule; the momentum per molecule is the product of the mass of the molecule (M) and its x-velocity at the point where it experienced its last collision, x-Lms ( M u x − Lms ). Therefore, the rate of momentum passing through the plane from left-to-right ( M x′′+ ) is given approximately by: M x′′+ ≈ nms vms M ums , x − Lms

(1)

Similarly, the momentum per unit area passing through the plane from right-to-left ( M x′′− ) is given by: M x′′+ ≈ nms vms M u x + Lms Velocity

(2)

M x′′+ M x′′− Lms

x-Lms

x+Lms x

Position

Figure 1: Momentum flows through a plane in a material.

The net rate of momentum flux passing through the plane per unit area in the positive x-direction ( M ′′ ) is the difference between M x′′+ and M x′′− ,

(

)

(3)

) ≈ −2n

(4)

M ′′ ≈ nms vms M u x − Lms − u x + Lms which can be rearranged to yield: M ′′ ≈ −2 nms vms M Lms

(u

− u x − Lms

x + Lms

Lms  

∂u vms M Lms  

∂x ms

∝μ

∂u ∂x

Comparing Eq. (4) with the definition of viscosity shows that the viscosity is proportional to the product of the number of molecules per unit volume, their average velocity, the mass of each molecule, and the mean distance between their interactions.

μ ∝ nms vms M Lms

(5)

The mass of a molecule is the molecular weight, MW. As noted in Eq. (1-14), kinetic theory indicates that Runiv T MW

vms ∝

(6)

where Runiv is the universal gas constant and T is the absolute temperature. The distance between molecular interactions was derived in Eq. (1-17) is

Lms =

1 nms π σ 2

(7)

where σ is the equivalent radius of the molecule. Substituting Eqs. (6) and (7) into Eq. (5) shows that:

μ∝

1

σ2

T MW

which is identical to Eq. (1-18) for conductivity if the specific heat capacity is removed.

(8)

P1.1-2 (1-1 in text): Conductivity of a dilute gas Section 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at ideal gas conditions. Test the validity of this approximation by comparing the conductivity estimated using Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., low pressure argon) provided by the internal function in EES. Note that the molecular radius, σ, is provided in EES by the Lennard-Jones potential using the function sigma_LJ. a.) What is the value and units of the proportionality constant required to make Eq. (1-18) an equality? Equation (1-18) is repeated below: k∝

cv

σ

T MW

2

(1)

Equation (1) is written as an equality by including a constant of proportionality (Ck): k = Ck

cv

σ

2

T MW

(2)

MW T

(3)

Solving for Ck leads to:

Ck =

kσ 2 cv

which indicates that Ck has units m-kg1.5/s-kgmol05-K0.5. The inputs are entered in EES for Argon at relatively low pressure (0.1 MPa) and 300 K. "Problem 1.1-2" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in T=300 [K] F$='Argon' P_MPa=0.1 [MPa] P=P_MPa*convert(MPa, Pa)

"temperature" "fluid" "pressure, in MPa" "pressure"

The conductivity, specific heat capacity, Lennard-Jones potential, and molecular weight of Argon (k, cv, σ, and MW) are evaluated using EES' built-in funcions. Equation (3) is used to evaluate the proportionality constant. k=conductivity(F$,T=T,P=P) cv=cv(F$,T=T,P=P) MW=molarMass(F$) sigma=sigma_LJ(F$) C_k=k*sigma^2*sqrt(MW/T)/cv

"conductivity" "specific heat capacity at constant volume" "molecular weight" "Lennard-Jones potential" "constant of proportionality"

which leads to Ck = 2.619x10-24 m-kg1.5/s-kgmol0.5-K0.5. b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and 100 MPa on a semi-log plot with pressure on the log scale. At what pressure does the approximation given in Eq. (1-18) begin to fail at 300 K for argon? Figure 1 illustrates the constant of proportionality as a function of pressure for argon at 300 K. The approximation provided by Eq. (1-18) breaks down at approximately 1 MPa. 8x10-24

-K

5x10-24

Ck (m-kg

1.5

0.5

6x10-24

/s-kgmol

0.5

)

7x10-24

4x10-24 3x10-24 2x10-24 10-24 0x100 0.001

0.01

0.1

1

10

100

Pressure (MPa)

Figure 1: Constant of proportionality in Eq. (3) as a function of pressure for argon at 300 K.

P1.1-3: Conductivity of a polyatomic gas Equation (1-18) cannot be used to understand the thermal conductivity of a polyatomic ideal gas, such as low pressure oxygen, because the ideal gas thermal conductivity is the sum of two terms corresponding to translational and internal contributions. k = ktrans + kint

(1)

Equation (1-18) only considers the translatonal contribution. Because thermal conductivity and viscosity are analagous transport properties, the translation term for the thermal conductivity of a dilute gas can be estimated as a function of the viscosity (μ) of the gas according to:

ktrans =

15Runiv μ 4 MW

(2)

where Runiv is the universal gas constant and MW is the molar mass of the of the gas. The internal contribution for a polyatomic molecule results from the transfer of energy associated with rotational and vibrational degrees of freedom. An estimate of the internal contribution is provided by the Eucken1 correlation

kint ≈

μ ⎡

cp − MW ⎢⎣

5Runiv ⎤ 2 ⎥⎦

(3)

where the viscosity is in units of Pa-s and the constant pressure specific heat and gas constant are in units of J/kmol-K. The internal contribution is zero for a monotonic gas. Choose a gas and use the EES viscosity function to determine its viscosity as a function of pressure and temperature. Then calculate and plot the thermal conductivity as a function of pressure at several temperatures. Compare the values you obtain from the dilute gas theory described above with the values provided at the same conditions obtained from the EES conductivity function. Use your program to answer the following questions. a.) The thermal conductivity of an ideal gas should only depend on temperature. At what pressure does this requirement fail for the temperature and gas you have selected? Hydrogen is selected as the gas and the inputs are entered in EES: "Problem 1.1-3" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in T=300 [K] F$='Hydrogen' P_MPa=0.1 [MPa] P=P_MPa*convert(MPa, Pa)

1

"temperature" "fluid" "pressure, in MPa" "pressure"

Hirschfelder, J.L., Curtiss, C.F, and Bird, R,B., “Molecular Theory of Gases and Liquids”, John Wiley and Sons, 1967

The viscosity, specific heat capacity at constant pressure, and molecular weight of the gas (μ, cp, and MW) are obtained using EES' built-in property function: mu=viscosity(F$,T=T,P=P) MW=MolarMass(F$) cP=cp(F$,T=T,P=P)

"viscosity" "molecular weight" "specific heat capacity"

The translation term in the thermal conductivity is estimated using Eq. (2): k_trans=15*R#*mu/(4*MW)

"translational contribution"

The internal term in the thermal conductivity is estimated using Eq. (3): cP_molar=cP*MW k_int=(mu/MW)*(cP_molar-5*R#/2)

"specific heat capacity on a molar basis" "internal contribution"

The dilute gas estimate of the thermal conductivity (kdilute) is obtained from Eq. (1) and compared to the value obtained from EES (k): k_dilute=k_trans+k_int k=conductivity(F$,T=T,P=P)

"dilute gas estimate of the thermal conductivity" "conductivity from EES' internal function"

Figure 1 illustrates the conductivity of hydrogen and the dilute gas estimate as a function of pressure at several values of temperature. It appears that the conductivity is independent of pressure up to about 1 MPa for hydrogen, although this value decreases with reduced temperature.

Thermal conductivity (W/m-K)

0.2

300 K

0.18 0.16

200 K

0.14 0.12 0.1 100 K

0.08 0.06 0.04

dilute gas theory EES function

0.02 0 0.001

0.01

0.1

1

10

40

Pressure (MPa)

Figure 1: Thermal conductivity as a function of pressure estimated by the dilute gas theory and using EES internal property routines for several temperatures.

b.) How does thermal conductivity vary with temperature? What causes this behavior?

Thermal conductivity increases with temperature. This is due to higher molecular velocities (primarily) but also due to more modes of energy storage being activated with temperature. c.) How does thermal conductivity vary with the choice of gas. Is there are relationship between the thermal conductivity and the number of atoms per molecule? Figure 2 illustrates the conductivity of 6 different gases at 300 K and 100 kPa. There does not appear to be a clear correlation between conductivity and the number of atoms per molecule. 0.036

Conductivity (W/m-K)

0.032 0.028 0.024 0.02 0.016

an e nbu t

io xi d nD C

ar bo

M et ha ne

e

n xy ge O

itr og en N

Ar go n

0.012

Figure 2: Thermal conductivity for several gases at 300 K and 100 kPa.

PROBLEM 1.2-1: Composite Wall A plane wall is a composite of a low conductivity material (with thickness L1 and conductivity k1) and a high conductivity material (with thickness L2 = L1 and conductivity k2). The edge of the wall at x = 0 is at temperature T1 and the edge at x = L1 + L2 has temperature T2, as shown in Figure P1.2-1(a). T1 is greater than T2. The wall is at steady-state and the temperature distribution in the wall is one-dimensional in x. x T1

k1

L1

x

x

k2

L2

k1

T2

k1

k2

q ′′

k2

T T1

0

L1

T2 x L1+L2 0

L1

x L1+L2

(a) (b) Figure P1.2-1: (a) Composite wall with k1 < k2, and (b) sketch of heat flux and temperature.

a.) Sketch the heat flux ( q ′′ ) and temperature (T) as a function of position within the wall on the axes in Fig. 1.2-1(b). Make sure that your sketch reflects the fact that (1) the wall is at steady state, and (2) k1 < k2. If the process is at steady state, then I can draw a control volume that extends from one surface to any location x in the material, as shown in Figure 2. T1

q0′′ A

q′′x A

x Figure 2: Control volume for solution

An energy balance on the control volume leads to: q0′′ A = q ′′x A

(1)

Equation (1) shows that the heat fux at any location x must be constant. The heat flux associated with conduction is governed by Fourier’s law:

q ′′x = −k

dT dx

Solving Eq. (2) for the temperature gradient leads to:

(2)

q ′′ dT =− x dx k

(3)

The numerator of Eq. (3), the heat flux, is constant while the denominator changes depending on whether you are in material 1 or material 2. In the low conductivity material 1, the temperature gradient will be higher than in the high conductivity material 2. Within each material, the temperature gradient must be constant (i.e., the temperature must be linear with x). The solution is shown in Figure 3. k1

k2

k1

x

k2

x T

q′′ T1 q′′x

T2 x L1

L1+L2

x L1

L1+L2

(a) (b) Figure 3: (a) Heat transfer rate and (b) temperature as a function of position within wall.

Problem P1.2-2: Conduction Through a Shape with Varying Cross-sectional Area The temperature distribution for the shape shown in Figure P1.2-2 can be assumed to be 1-D in the coordinate s. The problem is at steady state and the area available for conduction changes with s according to an arbitrary function, A(s). The temperatures of the two ends of the shape are specified; TH at s1 and TC at s2. s2 s1 s TH

TC

adiabatic T TH

TC

s Figure P1.2-2: Conduction through a shape in which the cross-sectional area varies according to A(s).

a.) Sketch the temperature distribution through the shape on the axes below the figure. The rate of conductive heat transfer ( q ) at any position s is given by Fourier’s law:

q = − k A

dT ds

(1)

At steady state, the heat transfer rate must be constant with position and therefore the temperature gradient is inversely proportional to area:

dT q =− ds kA The temperature gradient will be steepest where the area is smallest, as shown in Figure 2.

(2)

TH

TC

s1

s2

T TH

TC

s

Figure 2: Temperature distribution.

b.) Derive the governing differential equation for the problem; the governing differential equation should include only temperature T and its derivatives with respect to s as well as the area and its derivatives with respect to s. A differentially small control volume is defined, as shown in Figure 3.

TH s1

qs + ds

qs

TC

s2

Figure 3: Differential control volume.

An energy balance on the control volume leads to:

q s = q s + ds

(3)

Expanding the s+ds term in Eq. (3) leads to:

qs = qs +

dq ds ds

(4)

which can be simplified:

dq =0 ds

(5)

Substituting in Fourier’s law into Eq. (5) leads to: d ⎡ dT ⎤ −k A ⎥ = 0 ⎢ ds ⎣ ds ⎦

(6)

You can divide through by -k to get the governing differential equation:

d ⎡ dT ⎤ dA dT d 2T or A = 0 + A =0 ds ds ds 2 ds ⎢⎣ ds ⎥⎦

(7)

Problem 1.2-3 (1-2 in text): Conduction through a Wall Figure P1.2-3 illustrates a plane wall made of a very thin (thw = 0.001 m) and conductive (k = 100 W/m-K) material that separates two fluids, A and fluid B. Fluid A is at TA = 100°C and the heat transfer coefficient between the fluid and the wall is hA = 10 W/m2-K while fluid B is at TB = 0°C with hB = 100 W/m2-K. thw = 0.001 m TA = 100°C

TB = 0°C 2 hB = 100 W/m -K

hA = 10 W/m -K 2

k = 100 W/m-K Figure P1.2-3: Plane wall separating two fluids

a.) Draw a resistance network that represents this situation and calculate the value of each resistor (assuming a unit area for the wall, A = 1 m2). Heat flowing from fluid A to fluid B must pass through a fluid A-to-wall convective resistance (Rconv,A), a resistance to conduction through the wall (Rcond), and a wall-to-fluid B convective resistance (Rconv,B). These resistors are in series. The network and values of the resistors are shown in Figure 2. 0.1

K W

0.0001

K W

0.01

K W

TA = 100°C

TB = 0°C Rconv , A =

1 hA A

Rcond =

tw kA

Rcond , B =

1 hB A

Figure 2: Thermal resistance network representing the wall.

b.) If you wanted to predict the heat transfer rate from fluid A to B very accurately, then which parameter (e.g., thw, k, etc.) would you try to understand/measure very carefully and which parameters are not very important? Justify your answer. The largest resistance in a series network will control the heat transfer. For the wall above, the largest resistance is Rconv,A. Therefore, I would focus on predicting this resistance accurately. This would suggest that hA is the most important parameter and the others do not matter much.

Problem 1.2-4 (1-4 in text): Resistance Network Figure P1.2-4 illustrates a plane wall that is composed of two materials, A and B. The interface between the materials is characterized by a contact resistance. The left surface of material A is held at TH and the right surface of material B radiates to surroundings at TC and is also exposed to convection to a fluid at TC. material A

material B

TH

convection and radiation to TC

contact resistance

Figure P1.2-4: Composite wall with contact resistance, convection and radiation

The resistance network that represents the situation in Figure P1.2-4 should include five thermal resistors; their values are provided below: Rcond,A = 0.05 K/W, resistance to conduction through material A Rcontact = 0.01 K/W, contact resistance Rcond,B = 0.05 K/W, resistance to conduction through material B Rconv = 1.0 K/W, resistance to convection Rrad = 10.0 K/W, resistance to radiation a.) Draw a resistance network that represents the situation in Figure P1.2-4. Each resistance in the network should be labeled according to Rcond,A, Rcontact, Rcond,B, Rconv, and Rrad. Show where the temperatures TH and TC appear on your network.

Figure 2: Resistance network that represents Figure P1.2-4.

b.) What is the most important resistor in the network? That is, the heat transfer from TH to TC is most sensitive to which of the five resistances? The most important resistor in a series combination is the largest. The largest resistance is the parallel combination of Rconv and Rrad. The most important resistance in a parallel combination is the smallest; the smallest of Rconv and Rrad is Rconv. Thus, Rconv is the most important resistance. c.) What is the least important resistor in the network? The least important resistance is the contact resistance; it is the smallest in a series of resistors that are themselves unimportant relative to convection and radiation.

Problem 1.2-5 Figure P1.2-5 illustrates a wafer that is being developed in an optical lithography process. T∞ = 20°C 2 h = 15 W/m -K

q = 2 W

thp = 0.5 cm

chuck posts -4 2 Rc′′ = 5x10 K-m /W f = 0.1 ε = 0.7 wafer

thch = 1.5 cm Dw = 4 inch Tb = 20°C

chuck base kch = 25 W/m-K

Figure P1.2-5: Wafer being developed in an optical lithography process.

The energy required to develop the resist is deposited at a rate of q = 2 W near the center of the upper side of the wafer. The wafer has diameter Dw = 4 inch and is made of a conductive material; therefore, you may assume that the wafer is isothermal. The wafer is cooled by convection and radiation to the surroundings at T∞ as well as conduction to the chuck. The surrounding air is at T∞ = 20ºC and the heat transfer coefficient is h = 15 W/m2-K. The emissivity of the wafer surface is ε = 0.7. The chuck is made out of a single piece of material with conductivity kch = 25 W/m-K and consists of a base that is thch = 1.5 cm thick and an array of posts that are thp = 0.5 cm tall. The area of the base of the chuck is the same as the area of the wafer. The posts occupy f = 10% of the chuck area and the wafer rests on the top of the posts. There is an area specific contact resistance of Rc′′ = 5x10-4 K-m2/W between the bottom of the wafer and the top of the posts. The bottom surface of the chuck base is maintained at Tb = 20 ºC. a.) What is the temperature of the wafer at steady-state? The inputs are entered in EES: "Problem 1.2-5" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" D_w=4.0 [inch]*convert(inch,m) e=0.7 [-] h_bar=15 [W/m^2-K] q_dot=2 [W] th_ch=1.5 [cm]*convert(cm,m) k_ch=25 [W/m-K] R``_c=5e-4 [K-m^2/W] th_p=0.5 [cm]*convert(cm,m) f = 0.1 [-] T_infinity_C=20[C] T_infinity=converttemp(C,K,T_infinity_C) T_b_C=20 [C]

"diameter of wafer" "emissivity of wafer" "heat transfer coefficient" "power" "chuck base thickness" "chuck conductivity" "contact resistance" "post height" "fraction of post coverage" "ambient temperature in C" "ambient temperature" "chuck base temperature in C"

T_b=converttemp(C,K,T_b_C)

"chuck base temperature"

Note that the inputs are converted to base SI units and the units for each variable are set in the Variables Information window. The resistance network used to represent this problem is shown in Figure P1.2-5-2: Rcond , p

Rrad = 30.58

K = 0.247 W

K W

q = 2 W

Tb = 20°C

q2

Tp,b

T∞ = 20°C

q1 Tw

Rcond ,ch

K = 0.074 W Rc = 0.617

K W

Rconv = 8.223

K W

The resistances include: Rcond,ch = conduction through chuck base Rcond,p = conduction through posts Rc = contact resistance Rrad = radiation resistance Rconv = convection resistance

Figure P1.2-5-2: Resistance network.

In order to compute the resistance to radiation, it is necessary to guess a value of the wafer temperature (Tw) and subsequently comment out this guess in order to close up the solution. A reasonable value is chosen: T_w=300 [K]

"guess for wafer temperature - will be commented out"

The cross-sectional area of the wafer is: Aw =

π Dw2

(1)

4

The resistance to convection from the top surface of the wafer is:

Rconv = A_w=pi*D_w^2/4 R_conv=1/(A_w*h_bar)

1 Aw h

(2) "wafer area" "convection resistance"

The equations should be solved and the units set as you move through the problem (rather than at the end); this prevents the accumulation of small errors that are difficult to debug. The resistance to radiation is:

Rrad =

1 Aw ε (T + T∞2 ) (Tw + T∞ ) 2 w

(3)

R_rad=1/(A_w*sigma#*e*(T_w^2+T_infinity^2)*(T_w+T_infinity)) "radiation resistance"

The contact resistance is: Rc =

Rc′′ Aw f

(4)

Notice that the factor f in the denominator accounts for the contact area between the posts and the wafer. R_c=R``_c/(A_w*f)

"contact resistance"

The resistance to conduction through the posts is: Rcond , p =

thp

(5)

kch Aw f

and the resistance to conduction through the base is:

Rcond ,ch = R_cond_p=th_p/(k_ch*A_w*f) R_cond_ch=th_ch/(k_ch*A_w)

thch kch Aw

(6)

"resistance to conduction through posts" "resistance to conduction through chuck"

The rate of heat transfer by radiation and convection ( q1 ) and through the chuck ( q2 ) are computed:

q1 =

q2 =

(Tw − T∞ ) ⎛ 1 1 ⎞ + ⎜ ⎟ ⎝ Rconv Rrad ⎠

−1

(Tw − Tb ) Rc + Rcond , p + Rcond ,ch

q_dot_1=(T_w-T_infinity)/(1/R_conv+1/R_rad)^(-1) q_dot_2=(T_w-T_b)/(R_c+R_cond_p+R_cond_ch)

(7)

(8)

"rate of heat transfer by convection and radiation" "rate of heat transfer to chuck"

Because we guessed a value for Tw, it is not likely that q1 and q2 sum to the applied power to the wafer, as required by an energy balance:

q = q1 + q2

(9)

In order to finish the solution it is necessary to vary Tw until an energy balance is satisfied. EES automates this process; however, it will work best if it starts from a good set of guess values. Therefore, select Update Guesses from the Calculate menu. Then comment out the assumed value of Tw: {T_w=300 [K]}

"guess for wafer temperature - will be commented out"

and enter the energy balance: q_dot=q_dot_1+q_dot_2 T_w_C=converttemp(K,C,T_w)

"energy balance" "wafer temperature in C"

which leads to Tw = 294.8 K (21.64ºC). b.) Prepare a plot showing the wafer temperature as a function of the applied power, q . 29 28

Temperature (°C)

27 26 25 24 23 22 21 20 0

1

2

3

4

5

6

7

8

9

10

Heat transfer (W)

Figure P1.2-5-3: Wafer temperature as a function of applied power.

c.) What are the dominant heat transfer mechanisms for this problem? What aspects of the problem are least important? The values of the resistances at the nominal conditions given in the problem statement are shown in Figure P1.2-5-2. The value of the radiation and convection resistances are both large relative to the sum of resistances between Tw and Tb and therefore these mechanisms are not likely to play an important role in the problem. The resistance to conduction through the base of the chuck is small relative to the resistance to conduction through the posts and the contact resistance; therefore, conduction through the chuck base is not very important. The dominant

resistance in the problem is the contact resistance and the resistance to conduction through the posts is also important. d.) Radiation between the underside of the wafer and the top of the chuck base was ignored in the analysis; is this an important mechanism for heat transfer? Assume that the chuck surface is black and justify your answer. The resistance network, modified to include the resistance to radiation from the bottom of the wafer to the top of the chuck, is shown in Figure P1.2-5-4.

Rcond , p = 0.247

Rc = 0.617

K W

Rrad = 30.58

K W

K W

q = 2 W

Tb = 20°C Tp,b

q2

T∞ = 20°C

q1 Tw

Rcond ,ch

K = 0.074 W Rrad , wc = 33.96

K W

Rconv = 8.223

K W

The resistances include: Rcond,ch = conduction through chuck base Rcond,p = conduction through posts Rc = contact resistance Rrad = radiation resistance Rconv = convection resistance Rrad,wc = radiation resistance from top of chuck to bottom of wafer

Figure P1.2-5-4: Resistance network, including radiation from the wafer bottom.

The temperature of the top of the chuck is estimated using our previous solution:

Tp ,b = Tw − q1 ( Rc + Rcond , p )

(10)

and used to estimate the resistance to radiation from the top of the chuck to the bottom of the wafer:

Rrad , wc =

1 (1 − f ) Aw ε (T + Tp2,b ) (Tw + Tp ,b ) 2 w

(11)

T_p_b=T_w-q_dot_2*(R_c+R_cond_p) "temperature of the top surface of chuck" R_rad_wc=1/(A_w*(1-f)*sigma#*e*(T_w^2+T_p_b^2)*(T_w+T_p_b)) "radiation resistance between bottom of wafer and top of chuck"

which leads to Rrad,wc = 33.96 K/W. Because Rrad,wc is in series with Rc and Rcond,p and much larger than the sum of these resistances it is not very important to the problem.

e.) In an effort to maintain the wafer temperature at Tw= 20ºC, you decide to try to reduce and control the chuck base temperature, Tb. What temperature do you need to reduce Tb to in order that Tw= 20ºC? If you can only control Tb to within ±0.5 K then how well can you control Tw? The specified chuck temperature is commented out and instead the wafer temperature is specified: {T_b_C=20 [C]} T_w_C=20 [C]

"chuck base temperature in C" "specified wafer temperature"

which leads to Tb = 291.3 K (18.13ºC). In order to evaluate the impact of a ±0.5 K fluctuation of Tb on Tw, the required value of Tb is specified and the value of Tw is again commented out: T_b_C=18.13 [C] {T_w_C=20 [C]

"chuck base temperature in C" "specified wafer temperature"}

which leads to Tw = 293.2 K (20ºC), as expected. Now the value of Tb is elevated by 0.5 K in order to determine the impact on Tw: T_b_C=18.13 [C] + 0.5 [K]

"chuck base temperature in C"

which leads to Tw = 293.6 K (20.44ºC). Therefore, the ±0.5 K uncertainty in Tb leads to a ±0.44 K uncertainty in Tw. f.) Perform the same analysis you carried out in (e), but this time evaluate the merit of controlling the surrounding temperature, T∞, rather than the chuck temperature. What are the advantages and disadvantages associated with controlling T∞? The chuck temperature is returned to 20ºC: T_b_C=20 [C]

"chuck base temperature in C"

The specified surrounding temperature is commented out and instead the wafer temperature is specified: {T_infinity_C=20[C]} T_w_C=20 [C]

"ambient temperature in C" "specified wafer temperature"

which leads to T∞ = 280.0 K (6.835ºC); clearly the ambient temperature would need to be reduced by much more than the chuck temperature due to the weaker interaction between the wafer and the surroundings. This is a disadvantage of using the ambient temperature to control the wafer temperature. In order to evaluate the impact of a ±0.5 K fluctuation of T∞ on Tw, the required value of T∞ is specified and the value of Tw is again commented out: T_infinity_C=6.835 [C] "ambient temperature in C"

{T_w_C=20 [C]

"specified wafer temperature"}

which leads to Tw = 293.2 K (20ºC), as expected. Now the value of T∞ is elevated by 0.5 K in order to determine the impact on Tw: T_infinity_C=6.835 [C]+0.5 [K]

"ambient temperature in C"

which leads to Tw = 293.2 K (20.06ºC). Therefore, the ±0.5 K uncertainty in T∞ leads to a ±0.06 K uncertainty in Tw. This is an advantage of using T∞ to control the wafer temperature and is also related to the relatively weak thermal interaction between T∞ and Tw.

P1.2-6: Freezer Wall You have designed a wall for a freezer. A cross-section of your freezer wall is shown in Figure P1.2-6. The wall separates the freezer air at Tf = -10°C from air within the room at Tr = 20°C. The heat transfer coefficient between the freezer air and the inner wall of the freezer is h f = 10 W/m2-K and the heat transfer coefficient between the room air and the outer wall of the freezer is hr = 10 W/m2-K. The wall is composed of a thb = 1.0 cm thick layer of fiberglass blanket sandwiched between two thw = 5.0 mm sheets of stainless steel. The thermal conductivity of fiberglass and stainless steel are kb = 0.06 W/m-K and kw = 15 W/m-K, respectively. Assume that the cross-sectional area of the wall is Ac = 1 m2. Neglect radiation from either the inner or outer walls. thw = 5 mm

Tr = 20°C 2 hr = 10 W/m -K

thb = 1 cm thw = 5 mm

T f = −10°C 2 h f = 10 W/m -K

stainless steel, kw = 15 W/m-K

fiberglass blanket, kb = 0.06 W/m-K Figure P1.2-6: Freezer wall.

a.) Draw a resistance network to illustrate this problem. Be sure to label the resistances in your network so that it is clear what each resistance is meant to represent. There are five resistances associated with the problem; convection to the room and the freezer, Rconv,r and Rconv,f, and conduction through each of the stainless steel walls and the fiberglass blanket, Rcond,w and Rcond,f. These are placed in series since the heat transfer must pass through all of them, as shown in Figure P1.2-6-2.

Figure P1.2-6-2: Thermal resistance network.

b.) Enter all of the inputs in the problem into an EES program. Convert each input into the corresponding base SI unit (i.e., m, kg, K, W, N, etc.) and set the unit for each variable using the Variable Information window. Using comments, indicate what each variable means. Make sure that you set and check units of each variable that you use in the remainder of the solution process.

The inputs are entered in EES and converted to base SI: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" t_w = 5.0 [mm]*convert(mm,m) t_b = 1.0 [cm]*convert(cm,m) T_r = converttemp(C,K,20) h_r = 10 [W/m^2-K] "room air to outer wall heat transfer coefficient" k_w = 15 [W/m-K] k_b = 0.06 [W/m-K] h_f = 10 [W/m^2-K] "freezer air to inner wall heat transfer coefficient" T_f_C=-10 [C] T_f = converttemp(C,K,T_f_C) A = 1 [m^2]

"SS wall thickness" "fiberglass thickness" "room air temperature"

"SS conductivity" "fiberglass conductivity"

"freezer temperature in C" "freezer air temperature in K" "freezer area"

The units for each variable are set in the Variable Information window (see Figure P1.2-6-3).

Figure P1.2-6-3: Variable Information window

c.) Calculate the net heat transfer to the freezer (W). The values of each of the resistances in Figure P1.2-6-2 are calculated. The convection resistances between the room air and the outer wall of the freezer and the freezer air and the inner wall are:

Rconv ,r =

1 hr A

(1)

Rconv , f =

1

(2)

hf A

R_conv_r = 1/(h_r*A) R_conv_f=1/(h_f*A)

"convection resistance with room air" "convection resistance with freezer air"

The units of the two resistances are set in the Variable Information window (to K/W) and the units are checked to ensure that the equations entered are dimensionally consistent. The two conduction resistances are: Rcond , w =

Rcond ,b = R_cond_w=t_w/(k_w*A) R_cond_b=t_b/(k_b*A)

tw kw A tb kb A

(3)

(4)

"conduction resistance through SS wall" "conduction resistance through fiberglass wall"

The total heat transfer through the wall ( q ) is:

q =

(T

r

− Tf

)

Rconv ,r + 2 Rcond , w + Rcond ,b + Rconv , f

(5)

q_dot=(T_r-T_f)/(R_conv_r+R_cond_w+R_cond_b+R_cond_w+R_conv_f) "net heat transfer to freezer"

The Solution Window is shown in Figure P1.2-6-4, the heat load on the freezer is 81.7 W per m2 of wall area.

Figure P1.2-6-4: Solution window.

d.) Your boss wants to make a more energy efficient freezer by reducing the rate of heat transfer to the freezer. He suggests that you increase the thickness of the stainless steel wall panels in order to accomplish this. Is this a good idea? Justify your answer briefly. The value of the resistances are highlighted in Figure P1.2-6.4. Notice that Rcond,w is approximately 1000x less than the others. Your boss’ idea is not so good because in a series combination of resistances, it is the large resistances that dominate the problem. The wall is not important from a heat transfer standpoint. e.) Prepare a plot showing the heat transfer to the freezer as a function of the thickness of the stainless steel walls. Prepare a second plot showing the heat transfer to the freezer as a function of the thickness of the fiberglass. Make sure that your plots are clear (axes are labeled, etc.) A parametric table must be created to vary the thickness of the steel walls. Select New Parametric Table from the Tables menu (Figure P1.2-6-5) and place the variables q_dot and t_w in the table (highlight these variables from the list in the left hand box and select Add, then hit OK).

Figure P1.2-6-5: New Parametric Table dialog

Vary the thickness of the stainless steel walls from 0 to 2.0 cm (which corresponds to an extremely heavy freezer); right-click on the column of the parametric table that contains the variable t_w and select Alter Values (Figure P1.2-6-6).

Figure P1.2-6-6: Alter values of t_w to carry out the parametric investigation.

A dialog window will open asking what range you would like to vary t_w over; select 0 to 0.02 m (Figure P1.2-6-7) and hit OK.

Figure P1.2-6-7: Vary t_w from 0 to 0.02 m.

The entries in the t_w column will be automatically filled in. Each time one row of the Table is solved, the corresponding value of t_w will be used in the Equations Window; therefore, it is necessary to remove the value of t_w from the Equations Window. In order to do this temporarily (you will want to go back to the value in the problem statement), you should highlight the section of the code that specifies the value and right click. Select Comment to temporarily remove the code (Figure P1.6-2-8); subsequently performing the same operation and selecting Undo Comment will remove the comment indicators and “reactivate” the assignment.

Figure P1.2-6-8: Comment out the assignment of t_w in the Equation window

Solve the table by selecting Solve Table from the Calculate menu; the corresponding value of q_dot will be entered in each row of the parametric table (Figure P1.2-6-9).

Figure P1.2-6-9: Parametric table with solution

The solution can be plotted by selecting New Plot Window from the Plots menu and then X-Y plot to bring up the dialog shown in Figure P1.2-6-10. Select the source of the data (there is only one source in your EES file which is the single parametric table that exists) and specify that t_w will be on the x-axis and q_dot on the y-axis.

Figure P1.2-6-10: New Plot Setup window.

Select OK to create the plot and then edit it so that it looks good (include axes with descriptive names and units, grid line, etc.); the result should be similar to Figure P1.2-6-11.

Figure P1.2-6-11: Heat transfer to the freezer as a function of the freezer wall thickness.

Follow the same steps to generate Figure P1.2-6-12, which shows the freezer load as a function of the fiberglass thickness. Note that you will need to un-comment the line in the code where you specify the wall thickness.

Figure P1.2-6-12: Heat transfer to the freezer as a function of the fiberglass thickness.

f.) What design change to your wall would you suggest in order to improve the energy efficiency of the freezer. The largest resistance in Figure P1.2-6-4 is the conduction resistance through the fiberglass; I suggest that the thickness be increased. g.) One of your design requirements is that no condensation must form on the external surface of your freezer wall, even if the relative humidity in the room reaches 75%. This implies that the temperature of the external surface of the freezer wall must be greater than 15°C. Does your freezer wall satisfy this requirement? Calculate the external surface temperature (°C). The temperature at the surface of the freezer wall (Ts) corresponds to the node between Rconv,r and Rcond,w in Figure P1.2-6-2; the value of this temperature can be calculated according to: Ts = Tr − q Rconv ,r T_s = T_r-q_dot*(R_conv_r+R_cond_w) T_s_C=converttemp(K,C,T_s)

(6)

"surface temperature" "surface temperature in C"

The solution indicates that Ts =11.8°C which is less than 15°C and therefore condensation on the outside of the freezer is likely. h.) In order to prevent condensation, you suggest placing a heater between the outer stainless steel wall and the fiberglass. How much heat would be required to keep condensation from forming? Assume that the heater is very thin and conductive. The addition of the heater provides an additional heat input ( qw ) to the resistance network that enters between Rcond,w and Rcond,b on the air-side of the circuit, as shown in Figure P1.2-6-13.

Figure P1.2-6-13: Heater power added to the resistance network.

The required surface temperature is Ts,rq = 15°C. Therefore, the heat transfer through Rconv,r ( q1 ) is: q1 = "With the heater added" T_s_rq = converttemp(C,K,15) q_dot_1=(T_r-T_s_rq)/R_conv_r

(T

r

− Ts , rq )

(7)

Rconv ,r

"required surface temperature" "heat transfer from the room"

The heater temperature (Thtr) is therefore: Thtr = Ts ,rq − q1 Rcond , w

(8)

and the heat transfer to the freezer space ( q2 ) is: q2 =

(T

htr

− Tf

)

Rcond ,b + Rcond , w + Rconv , f

(9)

The heat transfer required by the heater ( qhtr ) is obtained by an energy balance on the heater node: qhtr = q2 − q1 T_htr=T_s_rq-q_dot_1*R_cond_w q_dot_2=(T_htr-T_f)/(R_cond_b+R_cond_w+R_conv_f) q_dot_htr=q_dot_2-q_dot_1

(10) "heater temperature" "heat transfer to freezer space" "heater power"

The solution indicates that qhtr = 43.6 W. i.) Prepare a plot showing the amount of heat required by the heater as a function of the freezer air temperature.

The plot is generated following essentially the same steps discussed in part (e) and shown in Figure P1.2-6-14.

Figure P1.2-6-14: Heater power as a function of the freezer air temperature.

Problem 1.2-7: Measuring Contact Resistance You have designed the experimental apparatus shown in Figure P1.2-7 to measure contact resistance. Four thermocouples (labeled TC1 through TC4) are embedded in two sample blocks at precise locations. The thermocouples are placed L1 = 0.25 inch from the edges of the sample blocks and L2 = 1.0 inch apart, as shown. Heat is applied to the top of the apparatus and removed from the bottom using a flow of coolant. The sides of the sample blocks are insulated. The sample blocks are fabricated from an alloy with a precisely-known and nearly constant thermal conductivity, ks = 2.5 W/m-K. The apparatus is activated and allowed to reach steady state. The temperatures recorded by the thermocouples are TC1 = 53.3°C, TC2 = 43.1°C, TC3 = 22.6°C, and TC4 = 12.3°C. The contact resistance of interest is the interface between the sample blocks. heat sample block, ks = 2.5 W/m-K

δks = 0.4 W/m-K interface insulation

L1 = 0.25 inch

δL = 0.01 inch TC1 TC2 TC3 TC4

L2 = 1 inch

δL = 0.01 inch L1 = 0.25 inch

δL = 0.01 inch

cooled block Figure P1.2-7: Experimental device to measure contact resistance.

a.) Use the data provided above to compute the measured heat flux in the upper and lower sample blocks. The input parameters are entered in EES: "P1.2-7 " $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in k_s=2.5 [W/m-K] L_1=0.25 [inch]*convert(inch,m) L_2=1.0 [inch]*convert(inch,m) TC_1=converttemp(C,K,53.3) TC_2=converttemp(C,K,43.1) TC_3=converttemp(C,K,22.6) TC_4=converttemp(C,K,12.3)

"conductivity" "distance between sensor and interface" "distance between sensors" "thermocouple 1 measurement" "thermocouple 2 measurement" "thermocouple 3 measurement" "thermocouple 4 measurement"

The heat transfer through the sample blocks is one-dimensional, steady state conduction through a constant cross-sectional area and therefore the heat flux through the upper and lower sample blocks are given by:

q1′′ = k s

q2′′ = k s

(TC1 − TC2 )

(1)

L2

(TC3 − TC4 )

(2)

L2

q_flux_1=(TC_1-TC_2)*k_s/L_2 q_flux_2=(TC_3-TC_4)*k_s/L_2

"heat flux in hot block" "heat flux in cold block"

The heat flux measurements are q1′′ =1004 W/m2 and q2′′ =1014 W/m2. Note that these values should be the same but are different due to measurement uncertainty or heat loss through the insulation. b.) Use the data to compute the temperature on the hot and cold sides of the interface. Figure 2 illustrates the measured temperatures as a function of position; the temperatures on the hot and cold sides of the interface (Th and Tc) can be obtained by extrapolating the temperature gradient to the interface, as shown in Figure 2.

Figure 2: Measured temperatures as a function of position and extrapolated temperatures at the interface.

The temperatures at the hot and cold sides of the interface are estimated according to:

Th = TC2 − (TC1 − TC2 )

L1 L2

(3)

Tc = TC3 + (TC3 − TC4 ) T_h=TC_2-(TC_1-TC_2)*L_1/L_2 T_c=TC_3+(TC_3-TC_4)*L_1/L_2

L1 L2

(4)

"extrapolated temperature at the hot interface" "extrapolated temperature at the cold interface"

The extrapolated temperatures at the interface are Th = 313.7 K and Tc = 298.3 K. c.) Use the data to compute the measured contact resistance. The average of the two heat flux measurements is:

( q1′′ + q2′′ )

q ′′ =

(5)

2

The measured value of the contact resistance is therefore: Rc′′ =

(Th − Tc )

(6)

q ′′

q_flux=(q_flux_1+q_flux_2)/2 R_contact=(T_h-T_c)/q_flux

"average of heat flux calculations" "measured contact resistance"

The measured contact resistance is Rc′′ = 0.0152 K-m2/W. It is important to estimate the uncertainty in your measurement. The uncertainty in the distance measurements is δL= 0.01 inch, the uncertainty in the conductivity of the sample blocks is δks = 0.4 W/m-K, and the uncertainty in the temperature measurements is δT = 0.5 K. d.) Estimate the uncertainty in the measurement of the heat flux in the upper sample block, the answer for (a), manually; that is carry out the uncertainty propagation calculations explicitly. The uncertainties are entered in EES: dk_s=0.1 [W/m-K] dL=0.01 [inch]*convert(inch,m) dT=0.5 [K]

"uncertainty in conductivity" "uncertainty in position measurements" "uncertainty in temperature measurement"

The uncertainty in q1′′ is related to the uncertainty in the measured quantities used to calculate q1′′ :

q1′′ = k s

(TC1 − TC2 ) L2

′′ 1 ) is obtained according to: The uncertainty in q1′′ due to TC1 ( δ q1,TC

(7)

δ q1,′′TC = 1

∂q1′′ δT δ T = ks ∂TC1 L2

(8)

′′ 2 ) is also: and the uncertainty in q1′′ due to TC2 ( δ q1,TC

δ q1,′′TC = ks 2

δT

(9)

L2

The uncertainty in q1′′ due to ks ( δ q1,′′ks ) is:

δ q1,′′k = (TC1 − TC2 )

δ ks

s

(10)

L2

′′ 2 ) is: and the uncertainty in q1′′ due to L2 ( δ q1,L

δ q1,′′L = ks (TC1 − TC2 ) 2

δL

(11)

L22

The total uncertainty in the heat flux is obtained by combining these contributions using the rootsum-square (RSS) technique: 2 2 + δ q1,′′TC + δ q1,′′k2 + δ q1,′′L2 δ q1′′ = δ q1,′′TC 1

2

s

2

(12)

"Manual calculation of the uncertainty" dq_flux_1_TC_1=dT*k_s/L_2 "uncertainty in heat flux 1 due to TC_1" dq_flux_1_TC_2=dT*k_s/L_2 "uncertainty in heat flux 1 due to TC_2" dq_flux_1_k_s=(TC_1-TC_2)*dk_s/L_2 "uncertainty in heat flux 1 due to k_s" dq_flux_1_L_2=(TC_1-TC_2)*k_s*dL/L_2^2 "uncertainty in heat flux 1 due to L_2" dq_flux_1=sqrt(dq_flux_1_TC_1^2+dq_flux_1_TC_2^2+dq_flux_1_k_s^2+dq_flux_1_L_2^2) "uncertainty in heat flux 1 measurement"

The total uncertainty in the heat flux is δ q1′′ = 81.0 W/m2. e.) Verify that EES' uncertainty propagation function provides the same answer obtained in (d). The uncertainty propagation capability of EES is accessed by selecting Uncertainty Propagation from the Calculate menu (Figure 3).

Figure 3: Uncertainty Propagation Window.

The calculated variable of interest is q_flux_1 and this should be selected from the Calculated variable list. The measured variables with uncertainty include the variables k_s, L_1, L_2, TC_1, TC_2, TC_3, and TC_4; these should be selected from the Measured variable list. The uncertainty associated with these measured variables can be specified by selecting Set uncertainties (Figure 4).

Figure 4: Uncertainties of Measured Variables Window.

The absolute uncertainties of each of the measured variables are assigned using the corresponding variable names (Figure 4). Select OK twice to see the results of the uncertainty propagation calculation (Figure 5).

Figure 5: Uncertainty Results Window.

Notice that the heat flux uncertainty calculated by EES is also 81.0 W/m2. The Uncertainty Results Window also delineates the sources of the uncertainty. f.) Use EES' uncertainty propagation function to determine the uncertainty in the measured value of the contact resistance. What is the % uncertainty in your measurement? Rather than the variable q_flux_1, the variable R_contact is selected in the Uncertainty of Measured Variables Window. The result of the calculation is shown in Figure 6.

Figure 6: Uncertainty Results Window.

The uncertainty in the contact resistance is 0.0017 K-m2/W or 11%. g.) Which of the fundamental measurements that are required by your test facility should be improved in order to improve your measurement of the contact resistance? That is, would you focus your attention on reducing δks, δL, or δT? Justify your answer. Examination of Figure 6 suggests that the uncertainty in the contact resistance is due almost entirely to the temperature measurements. Therefore, I would focus my attention on reducing δT.

Problem 1.2-8 (1-3 in text): Frozen Gutters You have a problem with your house. Every spring at some point the snow immediately adjacent to your roof melts and runs along the roof line until it reaches the gutter. The water in the gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutter and causing water to run into your attic. The situation is shown in Figure P1.2-8. snow melts at this surface 2 Tout , hout = 15 W/m -K Ls = 2.5 inch

snow, ks = 0.08 W/m-K insulation, kins = 0.05 W/m-K 2 Tin = 22°C, hin = 10 W/m -K Lins = 3 inch plywood, L p = 0.5 inch, k p = 0.2 W/m-K

Figure P1.2-8: Roof of your house.

The air in the attic is at Tin = 22°C and the heat transfer coefficient between the inside air and the inner surface of the roof is hin = 10 W/m2-K. The roof is composed of a Lins = 3.0 inch thick piece of insulation with conductivity kins = 0.05 W/m-K that is sandwiched between two Lp = 0.5 inch thick pieces of plywood with conductivity kp = 0.2 W/m-K. There is an Ls = 2.5 inch thick layer of snow on the roof with conductivity ks = 0.08 W/m-K. The heat transfer coefficient between the outside air at temperature Tout and the surface of the snow is hout = 15 W/m2-K. Neglect radiation and contact resistances for part (a) of this problem. a.) What is the range of outdoor air temperatures where you should be concerned that your gutters will become blocked by ice? The input parameters are entered in EES and converted to base SI units (N, m, J, K) in order to eliminate any unit conversion errors; note that units should still be checked as you work the problem but that this is actually a check on the unit consistency of the equations. "P1.2-8: Frozen Gutters" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in T_in=converttemp(C,K,22) "temperature in your attic" L_ins=3 [inch]*convert(inch,m) "insulation thickness" L_p=0.5 [inch]*convert(inch,m) "plywood thickness" k_ins=0.05 [W/m-K] "insulation conductivity" k_p=0.2 [W/m-K] "plywood conductivity" k_s=0.08 [W/m-K] "snow conductivity" L_s=2.5 [inch]*convert(inch,m) "snow thickness" h_in=10 [W/m^2-K] "heat transfer coefficient between attic air and inner surface of roof" h_out=15 [W/m^2-K] "heat transfer coefficient between outside air and snow" A=1 [m^2] "per unit area"

The problem may be represented by the resistance network shown in Figure 2.

Figure 2: Resistance network representing the roof of your house.

The network includes resistances that correspond to convection with the inside and outside air:

Rconv ,out =

Rconv ,in =

1

(1)

hout A 1

(2)

hin A

where A is 1 m2 in order to accomplish the problem on a per unit area basis. There are also conduction resistances associated with the insulation, plywood and snow: Rins =

Lins kins A

Rp =

Lp

Rs = R_conv_out=1/(h_out*A) R_s=L_s/(k_s*A) R_p=L_p/(k_p*A) R_ins=L_ins/(k_ins*A) R_conv_in=1/(h_in*A)

(3)

(4)

kp A Ls ks A

(5) "outer convection resistance" "snow resistance" "plywood resistance" "insulation resistance" "inner convection resistance"

Which leads to Rconv,out = 0.07 K/W, Rs = 0.79 K/W, Rp = 0.06 K/W, Rins = 1.52 K/W and Rconv,in = 0.10 K/W. Therefore, the dominant effects for this problem are conduction through the insulation and the snow; the other effects (convection and the plywood conduction) are not terribly important since the largest resistances will dominate in a series network. If the snow at the surface of the room is melting then the temperature at the connection between Rs and Rp must be Ts = 0°C (see Figure 2). Therefore, the heat transferred through the roof ( q in Figure 2) must be:

q =

(Tin − Ts )

(6)

Rconv ,in + 2 R p + Rins

The temperature of the outside air must therefore be:

Tout = Ts − q ( Rs + Rconv ,out )

(7)

T_s=converttemp(C,K,0) "roof-to-snow interface temperature must be melting point of water" q_dot=(T_in-T_s)/(R_conv_in+2*R_p+R_ins) "heat transfer from the attic to the snow when melting point is reached" T_out=T_s-q_dot*(R_s+R_conv_out) "outside temperature required to reach melting point at roof surface" T_out_C=converttemp(K,C,T_out) "outside temperature in C"

which leads to Tout = -10.8°C. If the temperature is below this then the roof temperature will be below freezing and the snow will not melt. If the temperature is above 0°C then the water will not refreeze upon hitting the gutter. Therefore, the range of temperatures of concern are -10.8°C < Tout < 0°C. b.) Would your answer change much if you considered radiation from the outside surface of the snow to surroundings at Tout? Assume that the emissivity of snow is εs = 0.82. The modified resistance network that includes radiation is shown in Figure 3.

Figure 3: Resistance network representing the roof of your house and including radiation.

The additional resistance for radiation is in parallel with convection from the surface of the snow as heat is transferred from the surface by both mechanisms. The radiation resistance can be calculated approximately according to:

Rrad =

1

(8)

4T ε s σ A 3

where T is the average temperature of the surroundings and the snow surface. In order to get a quick idea of the magnitude of this resistance we can approximate T with its largest possible value (which will result in the largest possible amount of radiation); the maximum temperature of the snow is 0°C: e_s=0.82 [-]

"emissivity of snow"

R_rad=1/(4*T_s^3*e_s*sigma#*A)

"radiation resistance"

which leads to Rrad = 0.26 K/W. Notice that Rrad is much larger than Rconv,out; the smallest resistance in a parallel combination dominates and therefore the impact of radiation will be minimal. Furthermore, Rconv,out is not even a very important resistance in the original series circuit shown in Figure 2.

Problem 1.2-9: Computer Chip Cooling Computer chips tend to work better if they are kept cold. You are examining the feasibility of maintaining the processor of a personal computer at the sub-ambient temperature of Tchip = 0°F. Assume that the operation of the computer chip itself generates qchip =10 W of power. Model the processor unit as a box that is a = 2 inch x b = 6 inch x c = 4 inch. Assume that all six sides of the box is exposed to air at Tair = 70°F with a convection heat transfer coefficient of h =10 W/m2-K. The box experiences a radiation heat transfer with surroundings that are at Tsur = 70°F. The emissivity of the processor surface is ε =0.7 and all six sides experience the radiation heat transfer. You are asked to size the refrigeration system required to maintain the temperature of the processor. a.) What is the refrigeration load that your refrigeration system must be able to remove to maintain the processor at a steady-state temperature (W)? The input parameters are entered in EES; notice that the units of each parameter are immediately converted into SI and the units of the associated variables are set (by you) in the Variable Information Window (Figure 2). $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "INPUTS" T_chip = converttemp(F,K,0) q_dot_chip = 10 [W] a = 2 [inch]*convert(inch,m) b = 6 [inch]*convert(inch,m) c = 4 [inch]*convert(inch,m) h = 10 [W/m^2-K] T_air=converttemp(F,K,70) T_sur=converttemp(F,K,70) e = 0.7

"chip temperature" "chip generation" "dimensions of processor"

"heat transfer coefficient" "air temperature" "temperature of surroundings" "emissivity of surface"

Figure 2:Variable Information window showing the units for each variable set.

A control volume encompasses just the processor and includes the internal generation from operating the chip ( qchip ) as well as convection ( qconv ) and radiation ( qrad ) and the heat transfer removed by the refrigeration system ( qload ). The energy balance is:

qchip + qconv + qrad = qload

(1)

The convection and radiation heat transfer rates may be evaluated using the associated rate equations:

qconv = h As (Tair − Tchip )

(2)

4 4 qrad = σ ε As (Tsur − Tchip )

(3)

where σ is Stefan-Boltzmann’s constant and As is the surface area of the processor: As = 2 ( a b + b c + a c )

(4)

These equations are programmed in EES: "part (a)" A_s=2*(a*b+b*c+a*c) q_dot_conv=h*A_s*(T_air-T_chip) q_dot_rad=sigma#*e*A_s*(T_sur^4-T_chip^4) q_dot_chip+q_dot_conv+q_dot_rad=q_dot_load

"surface area of processor" "convective heat transfer" "radiation heat transfer" "energy balance"

The units of the variables that have been added are also entered in the Variable Information window (Figure 3).

Figure 3: Variable Information window with additional units entered.

You can check that your solution is dimensionally consistent by selecting Check Units from the Calculate menu (Figure 4).

Figure 4: Check Units message

Solving the problem (Solve from the Calculate menu) will bring up the Solution Window (Figure 5) and shows that the refrigeration load is 39.4 W.

Figure 5: Solution Window

b.) If the coefficient of performance (COP) of the refrigeration system is nominally 3.5, then how much heat must be rejected to the ambient air (W)? Recall that COP is the ratio of the amount of refrigeration provided to the amount of input power required. The definition of COP is: COP =

qload w ref

(5)

which is programmed in EES: "part (b)" COP = 3.5 COP = q_dot_load/w_dot_ref

"specified COP" "refrigeration power"

and solved to show that the refrigeration power will be 11.3 W. c.) If electricity costs 12¢/kW-hr, how much does it cost to run the refrigeration system for a year, assuming that the computer is never shut off. The cost of electricity and time of operation are both converted to SI units and used to evaluate the cost per year. "part (c)" ecost = 12 [cents/kW-hr]*convert(cents/kW-hr,$/J) time=1 [year]*convert(year,s) cost=time*ecost*w_dot_ref

"cost of electricity" "time of operation" "cost of operating system for 1 year"

The cost of operating the system for 1 year is $11.8.

Problem 1.2-10: Insulation Conductivity Test You have been contracted by ASHRAE (the American Society of Heating, Refrigeration, and Air-Conditioning Engineers) to measure the thermal conductivity of various, new materials for insulating pipes. Your contract specifies that you will measure the thermal conductivity to within 10%. Your initial design for the test setup is shown in Figure P1.2-10. The test facility consists of a pipe (with conductivity kpipe = 120 W/m-K) with inner diameter, Di,pipe = 6.0 inch and thickness thpipe = 0.5 inch that carries a flow of chilled water, Twater = 10°C. The heat transfer coefficient between the water and the internal surface of the pipe is hwater = 300 W/m2-K. The pipe is covered by a thins = 2.0 inch thick layer of the insulation (with conductivity kins) that is being tested. Two thermocouples are embedded in the insulation, one connected to the outer surface (Tins,out) and the other to the inner surface (Tins,in). The insulation material is surrounded by a thm = 3.0 inch thick layer of a material with a well-known thermal conductivity, km = 2.0 W/m-K. Two thermocouples are embedded in the material at its inner and outer surface (Tm,in and Tm,out, respectively). Finally, a band heater is wrapped around the outer surface of the material. Assume that the thickness of the band heater is negligibly small. The band heater provides qband = 3 kW/m. The outer surface of the band heater is exposed to ambient air at Tair = 20°C and has a heat transfer coefficient, hair = 10 W/m2-K and emissivity ε = 0.5. A contact resistance of Rc′′ =1x10-4 m2-K/W is present at all 3 interfaces in the problem (i.e., between the pipe and the insulation, the insulation and the material, and the material and the band heater). Tair = 20°C 2 hair = 10 W/m -K ε = 0.5 Tm,out Tm,in Tins,out Tins,in

band heater qband = 3 kW Twater = 10°C 2 hwater = 300 W/m -K pipe, kpipe = 120 W/m-K thpipe = 0.5 inch Rc′′ = 1x10 K-m /W -4

2

thins = 2 inch thm = 3 inch Di,pipe = 6 inch

insulation being tested, kins material with known conductivity, km = 2.0 W/m-K

Figure P1.2-10: Test facility for measuring pipe insulation

You may assume that the problem is 1-D (i.e., there are no variations along the length or circumference of the pipe) and do the problem on a per unit length of pipe (L=1 m) basis. a.) Draw a resistance network that represents the test facility. Clearly label each resistance and indicate what it represents. Be sure to indicate where in the network the heat input from the band heater will be applied and also the location of the thermocouples mentioned in the problem statement.

The resistance network is shown in Figure 2 and includes convection with the water and the air (Rconv,w and Rair), conduction through the pipe, insulation, and material (Rpipe, Rcond,ins, and Rcond,m), contact resistances between the pipe and insulation (Rc,1), the insulation and material (Rc,2), and the material and the band heater (Rc,3), and radiation (Rrad).

Figure 2: Resistance network representing the test facility.

b.) If the coefficient of performance (COP) of the refrigeration system is nominally 3.5, then how much heat must be rejected to the ambient air (W)? Recall that COP is the ratio of the amount of refrigeration provided to the amount of input power required. The known information is entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" k_pipe=120 [W/m-K] D_i_pipe=6.0 [inch]*convert(inch,m) th_pipe=0.5 [inch]*convert(inch,m) T_water=converttemp(C,K,10 [C]) h_water=300 [W/m^2-K] th_ins=2.0 [inch]*convert(inch,m) k_m=2.0 [W/m-K] th_m=3.0 [inch]*convert(inch,m) T_air=converttemp(C,K,20 [C]) h_air=10 [W/m^2-K] e=0.5 [-] R``_c=1e-4 [m^2-K/W] L=1 [m] q_dot_htr=3 [kW]*convert(kW,W) k_ins=1.0 [W/m-K]

"pipe conductivity" "pipe inner diameter" "pipe thickness" "water temperature" "water to pipe heat transfer coefficient" "insulation thickness" "material thermal conductivity" "material thickness" "air temperature" "air to heater heat transfer coefficient" "emissivity of band heater surface" "contact resistance" "length of pipe" "heater power" "insulation conductivity"

The values of the convection resistances are computed: 1 hwater π L Di , pipe

(1)

1 hair π L ( Di , pipe + 2 thpipe + 2 thins + 2 thm )

(2)

Rconv , w =

Rconv ,air =

R_conv_w=1/(h_water*pi*L*D_i_pipe) R_conv_air=1/(h_air*pi*L*(D_i_pipe+2*th_pipe+2*th_ins+2*th_m))

"pipe-to-water convection" "heater to air convection"

The conduction resistances are calculated according to:

R pipe

Rcond ,ins

Rcond ,m

⎛D + 2 thpipe ⎞ ln ⎜ i , pipe ⎟⎟ ⎜ Di , pipe ⎝ ⎠ = 2 π L k pipe

⎛D + 2 thpipe + 2 thins ln ⎜ i , pipe ⎜ Di , pipe + 2 thpipe = ⎝ 2 π L kins

(3) ⎞ ⎟⎟ ⎠

⎛D + 2 thpipe + 2 thins + 2 thm ⎞ ln ⎜ i , pipe ⎟⎟ ⎜ Di , pipe + 2 thpipe + 2 thins ⎝ ⎠ = 2 π L km

(4)

(5)

R_pipe=ln((D_i_pipe/2+th_pipe)/(D_i_pipe/2))/(2*pi*L*k_pipe) "pipe conduction resistance" R_cond_ins=ln((D_i_pipe/2+th_pipe+th_ins)/(D_i_pipe/2+th_pipe))/(2*pi*L*k_ins) "insulation conduction resistance" R_cond_m=ln((D_i_pipe/2+th_pipe+th_ins+th_m)/(D_i_pipe/2+th_pipe+th_ins))/(2*pi*L*k_m) "material conduction resistance"

The contact resistances are calculated according to: Rc′′ π L ( Di , pipe + 2 thpipe )

(6)

Rc′′ π L ( Di , pipe + 2 thpipe + 2 thins )

(7)

Rc′′ π L ( Di , pipe + 2 thpipe + 2 thins + 2 thm )

(8)

Rc ,1 =

Rc ,2 =

Rc ,3 =

R_c_1=R``_c/(pi*(D_i_pipe+2*th_pipe)*L) "pipe-to-insulation contact resistance" R_c_2=R``_c/(pi*(D_i_pipe+2*th_pipe+2*th_ins)*L) "insulation-to-material contact resistance" R_c_3=R``_c/(pi*(D_i_pipe+2*th_pipe+2*th_ins+2*th_m)*L) "material-to-heater contact resistance"

Finally, the radiation resistance is calculated according to:

Rrad =

1 π L ( Di , pipe + 2 thpipe + 2 thins + 2 thm ) σ ε (Thtr2 + Tair2 )(Thtr + Tair )

(9)

but Thtr is not known in Eq. (9); therefore, a guess value of Thtr must be used to allow the calculation of the resistance. This guess value will be removed once a solution is obtained. A reasonable guess value for the heater temperature is something higher than the ambient temperature. T_htr_g=500 [K] "this is a guess for the heater temperature - it allows me to calculate the radiation resistance" "this guess will be removed to complete the solution" R_rad=1/(pi*(D_i_pipe+2*th_pipe+2*th_ins+2*th_m)*L*sigma#*e*(T_htr_g^2+T_air^2)*(T_htr_g+T_air)) "radiation resistance"

The heat transferred to the heater must either pass inwards to the water or outwards to the ambient air. qhtr =

(Thtr − Twater ) Rconv , w + R pipe + Rc ,1 + Rcond ,ins + Rc ,2 + Rcond ,m + Rc ,3

+

(Thtr − Tair ) ⎛ 1 1 + ⎜⎜ ⎝ Rconv , air Rrad

⎞ ⎟⎟ ⎠

−1

(10)

q_dot_htr=(T_htr-T_water)/(R_conv_w+R_pipe+R_c_1+R_cond_ins+R_c_2+R_cond_m+R_c_3)+(T_htrT_air)/((1/R_conv_air+1/R_rad)^(-1)) "heater power"

The calculated and guessed values of Thtr will not be the same (unless you are very lucky); update the guess values for the calculation (select Update Guesses from the Calculate menu) and then specify that T_htr_g must equal T_htr. You will need to comment the assignment of T_htr_g to avoid over-specifying the problem. {T_htr_g=500 [K] temperature - it allows me to calculate the radiation resistance" "this guess will be removed to complete the solution"} T_htr=T_htr_g

"this is a guess for the heater

The heater temperature is found to be 394.6 K (about 120°C which is too hot to touch). The heat transferred through the insulation is: qins =

(Thtr − Twater ) Rconv , w + R pipe + Rc ,1 + Rcond ,ins + Rc ,2 + Rcond ,m + Rc ,3

q_dot_ins=(T_htr-T_water)/(R_conv_w+R_pipe+R_c_1+R_cond_ins+R_c_2+R_cond_m+R_c_3) "heat transferred through insulation"

which leads to qins = 976.3 W (most of the heat is transferred to the surrounding air).

(11)

When the test facility is operated, the heater power is not measured nor are the contact resistances, emissivity, or heat transfer coefficients known with any precision. Also, the insulation thermal conductivity is not known, but rather must be calculated from the measured temperatures. The heat transferred through the material with known thermal conductivity is the same as the heat transferred through the insulation that is being measured. Therefore: qins =

Tm ,out − Tm ,in Rcond ,m

=

Tins ,out − Tins ,in Rcond ,ins

(12)

and so the resistance of the insulation can be calculated based on the ratio of the temperature differences: Rcond ,ins =

(T (T

ins , out m , out

− Tins ,in ) − Tm ,in )

Rcond ,m

(13)

Equation (13) indicates that the test facility relies on accurately measuring the temperature differences across the insulation and the temperature difference across the material. c.) Using your model, predict the temperature difference across the insulation ( ΔTins = Tins ,out − Tins ,in ) and the material ( ΔTm = Tm ,out − Tm,in ). Using Eq. (12), the two temperature differences are: ΔTins = qins Rcond ,ins

(14)

ΔTm = qins Rcond ,m

(15)

DeltaT_ins=q_dot_ins*R_cond_ins DeltaT_m=q_dot_ins*R_cond_m

"insulation temperature difference" "material temperature difference"

which leads to ΔTins= 70.2 K and ΔTm = 33.8 K. d.) Prepare a plot showing ΔTm as a function of the material thickness (thm) for thicknesses ranging from 5.0 mm to 50 cm. Explain the shape of your plot. The plot requested by the problem statement is generated using a parametric table that include the variables th_m and DeltaT_m. The result is shown in Figure 3.

Figure 3: Temperature difference across the material as a function of the material thickness.

Notice that at low thm the value of ΔTm is small because the resistance of the material is small. At high values of thm the resistance of the material is large but the heat transferred through the material becomes small (more of the energy is transferred to the air) and so the value of ΔTm begins to decrease. e.) Based on your plot from part (d), what is a reasonable value for thm? Remember that you need to measure the temperature difference and therefore you would like it to be large. A value of thm around 10 cm provides a large value of ΔTm; further increases are probably not justified. A similar plot and design point could be obtained for ΔTins by varying thins.

Problem P1.2-11: Heater Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with kA = 1 W/m-K and B with kB = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A; the heating element provides q ′′ = 5000 W/m 2 of heat. The surface of the wall at x = 2L is exposed to fluid at Tf,in = 300 K with heat transfer coefficient hin = 100 W/m2-K. 2 q ′′ = 5000 W/m

insulated

material A kA = 1 W/m-K L = 1 cm

x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

Figure P1.2-11(a): Composite wall with a heater.

You may neglect radiation and contact resistance for parts (a) through (c) of this problem. a.) Draw a resistance network to represent this problem; clearly indicate what each resistance represents and calculate the value of each resistance. The input parameters are entered in EES: “P1.2-11: Heater" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q_flux=100 [W/m^2] L = 1.0 [cm]*convert(cm,m) k_A=1.0 [W/m-K] k_B=5.0 [W/m-K] T_f_in=300 [K] h_in=100 [W/m^2-K] A=1 [m^2]

"heat flux provided by the heater" "thickness of each layer" "conductivity of material A" "conductivity of material B" "fluid temperature at inside surface" "heat transfer on inside surface" "per unit area"

The resistance network that represents the problem shown in Figure 2 is:

Figure 2: Resistance network.

The resistances due to conduction through materials A and B are:

RA =

RB =

L

(1)

kA A

L

(2)

kB A

where A is the area of the wall, taken to be 1 m2 in order to carry out the analysis on a per unit area basis. The resistance due to convection is:

Rconv ,in =

1

(3)

hin A

"part (a)" R_A=L/(k_A*A) R_B=L/(k_B*A) R_conv_in=1/(h_in*A) "resistance to convection on inner surface"

"resistance to conduction through A" "resistance to conduction through B"

which leads to RA = 0.01 K/W, RB = 0.002 K/W, and Rconv,in = 0.01 K/W. b.) Use your resistance network from (a) to determine the temperature of the heating element. The resistance network for this problem is simple; the temperature drop across each resistor is equal to the product of the heat transferred through the resistor and its resistance. In this simple case, all of the heat provided by the heater must pass through materials A, B, and into the fluid by convection so these resistances are in series. The heater temperature (Thtr) is therefore:

Thtr = T f ,in + ( RA + RB + Rconv ,in ) q ′′ A

(4)

T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"

which leads to Thtr = 410 K. c.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (b). The temperatures at x = L and x = 2L can be computed according to:

Tx = L = T f ,in + ( RB + Rconv ,in ) q ′′ A

(5)

Tx = 2 L = T f ,in + Rconv ,in q ′′ A

(6)

T_L=T_f_in+(R_B+R_conv_in)*q_flux*A T_2L=T_f_in+R_conv_in*q_flux*A

"temperature at x=L" "temperature at x=2L"

which leads to Tx=L = 360 K and Tx=2L = 350 K. The temperature distribution is sketched on the axes in Figure 3.

Figure 3: Sketch of temperature distribution.

Notice that the temperature drop through the two larger resistances (RA and RB) are much larger than the temperature drop across the small resistance, RB. Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm. material C kC = 2 W/m-K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

insulated x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B L = 1 cm k = 5 W/m-K B

Figure P1.2-11(b): Composite wall with Material C.

Neglect radiation and contact resistance for parts (d) through (f) of this problem. d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly indicate what each resistance represents and calculate the value of each resistance. There is an additional resistor corresponding to conduction through material C, RC, as shown below:

Notice that the boundary condition at the end of RC corresponds to the insulated wall; that is, no heat can be transferred through this resistance. The resistance to conduction through material C is:

RC = "part (b)" k_C=2.0 [W/m-K] R_C=L/(k_C*A)

L

(7)

kC A

"conductivity of material C" "resistance to conduction through C"

which leads to RC = 0.005 K/W. e.) Use your resistance network from (d) to determine the temperature of the heating element. Because there is no heat transferred through RC, all of the heat must still go through materials A and B and be convected from the inner surface of the wall. Therefore, the answer is not changed from part (b), Thtr = 410 K. f.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (e). The answer is unchanged from part (c) except that there is material to the left of the heater. However, no heat is transferred through material C and therefore there is no temperature gradient in the material.

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a contact resistance between materials A and B, Rc′′ = 0.01 K-m 2 /W , and the surface of the wall at

x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient hout = 10 W/m2-K. material C kC = 2 W/m-K T f ,out = 400 K 2 hout = 10 W/m -K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

x

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

L = 1 cm L = 1 cm

Rc′′ = 0.01 K-m /W 2

Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.

Neglect radiation for parts (g) through (i) of this problem. g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly indicate what each resistance represents and calculate the value of each resistance. The additional resistances associated with contact resistance and convection to the fluid at the outer surface are indicated. Notice that the boundary condition has changed; heat provided by the heater has two paths ( qout and qin ) and so the problem is not as easy to solve.

The additional resistances are computed according to: Rconv ,out =

1

(8)

hout A

Rcontact =

Rc′′ A

"part (c)" R``_c=0.01 [K-m^2/W] h_out=10 [W/m^2-K] T_f_out=400 [K] R_contact=R``_c/A R_conv_out=1/(h_out*A) "convection resistance on outer surface"

(9)

"area specific contact resistance" "heat transfer coefficient" "fluid temperature on outside surface" "contact resistance"

which leads to Rcontact = 0.01 K/W and Rconv,out = 0.1 K/W. h.) Use your resistance network from (j) to determine the temperature of the heating element. It is necessary to carry out an energy balance on the heater: q ′′ A = qin + qout

(10)

The heat transfer rates can be related to Thtr according to: qin =

(T

htr

− T f ,in )

RA + Rcontact + RB + Rconv ,in

qout =

(T

htr

− T f ,out )

(11)

(12)

RC + Rconv ,out

These are 3 equations in 3 unknowns, Thtr, qout and qin , and therefore can be solved simultaneously in EES (note that the previous temperature calculations from part (b) must be commented out): {T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A T_L=T_f_in+(R_B+R_conv_in)*q_flux*A

"heater temperature" "temperature at x=L"

T_2L=T_f_in+R_conv_in*q_flux*A q_flux*A=q_dot_in+q_dot_out q_dot_in=(T_htr-T_f_in)/(R_A+R_contact+R_B+R_conv_in) q_dot_out=(T_htr-T_f_out)/(R_C+R_conv_out)

"temperature at x=2L"} "energy balance on the heater" "heat flow to inner fluid" "heat flow to outer fluid"

which leads to Thtr = 446 K. The other intermediate temperatures shown on the resistance diagram can be computed: Tx = L − = Thtr − RA qin

(13)

Tx = L + = Thtr − ( RA + Rcontact ) qin

(14)

Tx = 2 L = Thtr − ( RA + Rcontact + RB ) qin

(15)

Tx =− L = Thtr − RC qout

(16)

"intermediate temperatures" T_Lm=T_htr-R_A*q_dot_in T_Lp=T_htr-(R_A+R_contact)*q_dot_in T_2L=T_htr-(R_A+R_contact+R_B)*q_dot_in T_mL=T_htr-R_C*q_dot_out

which leads to Tx=L- = 400.4 K, Tx=L+ = 354.7 K, Tx=2L = 345.6 K, and Tx=-L = 443.8 K. i.) Sketch the temperature distribution on the axes provided below.

Problems 1.2-12 (1-5 in text): Floor Heater You have decided to install a strip heater under the linoleum in your bathroom in order to keep your feet warm on cold winter mornings. Figure P1.2-12 illustrates a cross-section of the bathroom floor. The bathroom is located on the first story of your house and is W = 2.5 m wide by L = 2.5 m long. The linoleum thickness is thL = 5 mm and has conductivity kL = 0.05 W/m-K. The strip heater under the linoleum is negligibly thin. Beneath the heater is a piece of plywood with thickness thp = 5 mm and conductivity kp = 0.4 W/m-K. The plywood is supported by ths = 6 cm thick studs that are Ws = 4 cm wide with thermal conductivity ks = 0.4 W/m-K. The centerto-center distance between studs is ps = 25.0 cm. Between each stud are pockets of air that can be considered to be stagnant with conductivity kair = 0.025 W/m-K. A sheet of drywall is nailed to the bottom of the studs. The thickness of the drywall is thd = 9.0 mm and the conductivity of drywall is kd = 0.1 W/m-K. The air above in the bathroom is at Tair,1 = 15°C while the air in the basement is at Tair,2 = 5°C. The heat transfer coefficient on both sides of the floor is h = 15 W/m2-K. You may neglect radiation and contact resistance for this problem. 2 Tair ,1 = 15°C, h = 15 W/m -K

strip heater thp = 5 mm

ps = 25 cm

linoleum, kL = 0.05 W/m-K plywood, kp = 0.4 W/m-K thL = 5 mm

ths = 6 cm Ws = 4 cm studs, ks = 0.4 W/m-K drywall, kd = 0.1 W/m-K air, ka = 0.025 W/m-K

thd = 9 mm 2 Tair ,2 = 5°C, h = 15 W/m -K

Figure P1.2-12: Bathroom floor with heater.

a.) Draw a thermal resistance network that can be used to represent this situation. Be sure to label the temperatures of the air above and below the floor (Tair,1 and Tair,2), the temperature at the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat input to the strip heater ( qh ) on your diagram. The resistance diagram corresponding to this problem is shown in Figure 2.

Figure 2: Resistance diagram representing the bathroom floor.

Starting at the left-hand side of Figure 2 (i.e., from the basement air), the resistances correspond to convection between the air in the basement and the drywall (Rconv), conduction through the drywall (Rd), conduction through the air (Rair) and studs (Rs) in parallel, conduction through the plywood (Rp), conduction through the linoleum (RL), and convection between the air in the bathroom and the linoleum (Rconv). b.) Compute the value of each of the resistances from part (a). The known values for the problem are entered in EES and converted to base SI units: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" W=2.5 [m] L=2.5 [m] T_air_1=converttemp(C,K,15) T_air_2=converttemp(C,K,5) h=15 [W/m^2-K] th_L=5.0 [mm]*convert(mm,m) k_L=0.05 [W/m-K] th_P=5.0 [mm]*convert(mm,m) k_P=0.4 [W/m-K] th_s=6.0 [cm]*convert(cm,m) W_s=4.0 [cm]*convert(cm,m) k_s=0.4 [W/m-K] p_s=25 [cm]*convert(cm,m) k_air=0.025 [W/m-K] th_d=9.0 [mm]*convert(mm,m) k_d=0.1 [W/m-K]

"width of bathroom" "length of bathroom" "air temperature in the bathroom" "air temperature in the basement" "heat transfer coefficient" "linoleum thickness" "linoleum thermal conductivity" "plywood thickness" "plywood thermal conductivity" "stud thickness" "stud width" "stud conductivity" "stud center-to-center distance" "air conductivity" "drywall thickness" "drywall conductivity"

The units for each of these variables is set in the Variable Information window (select Variable Information from the Options menu), Figure 3. The units of each of the additional variables that are added as you solve the problem should be set in the Variable Information window.

Figure 3: Variable Information window.

The area of the floor is:

A = LW

(1)

The convection resistance is computed according to:

Rconv =

1 hA

(2)

The conduction resistances of the linoleum, plywood, and drywall are computed: RL =

thL kL A

Rp =

thp

Rd = A=L*W R_conv=1/(h*A) R_L=th_L/(k_L*A) R_P=th_P/(k_P*A) R_d=th_d/(k_d*A)

(3)

(4)

kp A thd kd A

(5) "area for conduction through floor" "convection resistance" "linoleum resistance" "plywood resistance" "drywall resistance"

The conduction resistance of the studs is computed according to:

Rs =

ths ⎛W ⎞ ks A ⎜ s ⎟ ⎝ ps ⎠

(6)

Note that the area for conduction is the product of the area of the floor and the fraction of the floor occupied by the studs. The conduction resistance of the air is: ths

Rair = kair

(7)

⎛ p − Ws ⎞ A⎜ s ⎟ ⎝ ps ⎠

R_s=th_s/(k_s*A*W_s/p_s) R_air=th_s/(k_air*A*(p_s-W_s)/p_s)

"stud resistance" "air resistance"

These calculations lead to Rconv = 0.011 K/W, RL = 0.016 K/W, Rp = 0.002 K/W, Rd = 0.014 K/W, Rs = 0.15 K/W, and Rair = 0.46 K/W. c.) How much heat must be added by the heater to raise the temperature of the floor to a comfortable 20°C? If Ts in Figure 2 is 20°C then the heat transferred to the bathroom ( q1 ) is: q1 =

TL − Tair ,1

(8)

Rconv

T_L=converttemp(C,K,20) q_dot_1=(T_L-T_air_1)/R_conv

"linoleum temperature" "heat transferred to bathroom"

which leads to q1 = 469 W. The temperature of the heater is therefore: Th = TL + q1 RL

(9)

T_h=T_L+q_dot_1*R_L

"strip heater temperature"

which leads to a heater temperature, Th = 300.7 K. The heater must provide q2 (the heat transferred to the bathroom) as well as q1 (the heat transferred to the basement). q1 =

and the total heater power is:

Th − Tair ,2 −1

⎡1 1 ⎤ Rconv + Rd + ⎢ + ⎥ + Rp ⎣ Rs Rair ⎦

(10)

qh = q1 + q2

(11)

q_dot_2=(T_h-T_air_2)/(R_conv+R_d+(1/R_s+1/R_air)^(-1)+R_P) "heat lost to lower story" q_dot_h=q_dot_1+q_dot_2 "total heater power"

which leads to q2 = 161 W and qh = 630 W. d.) What physical quantities are most important to your analysis? What physical quantities are unimportant to your analysis? Figure 4 illustrates the values of the resistances on the resistance diagram.

Figure 4: Resistance values.

Examination of Figure 4 shows that Rp, Rd, and Rconv are unimportant relative to the amount of heat transferred to the basement; these resistances are small in a series combination. Therefore, you can expect that the conductivity of the drywall and plywood as well as their thickness are not very important. Furthermore, the resistance of the air is larger than the resistance of the studs; in a parallel combination, the larger resistance is not important. Therefore, the conductivity of air is likely not very important. The important quantities include the conductivity of the studs and their size as well as the thickness and conductivity of the linoleum and its thickness. The heat transfer coefficient is also important. e.) Discuss at least one technique that could be used to substantially reduce the amount of heater power required while still maintaining the floor at 20°C. Note that you have no control over Tair,1 or h. The heat transferred to the bathroom is given by Eq. (8); you cannot change h and therefore the value of Rconv is fixed. The only way for you can reduce the heater power is to reduce the amount of heat transferred to the basement. This can be done most effectively by increasing the resistance of the studs, perhaps by increasing their thickness or reducing their width.

Problem 1.2-13: Burner An electric burner for a stove is formed by taking a cylindrical piece of metal that is D = 0.32 inch in diameter and L = 36 inch long and winding it into a spiral shape. The burner consumes electrical power at a rate of q = 900 W. The burner surface has an emissivity of ε = 0.80. The heat transfer coefficient between the burner and the surrounding air ( h ) depends on the surface temperature of the burner (Ts) according to:

⎡ W ⎤ ⎡ W ⎤ h = 10.7 ⎢ 2 ⎥ + 0.0048 ⎢ 2 2 ⎥ Ts ⎣ m -K ⎦ ⎣ m -K ⎦

(1)

where h is in [W/m2-K] and Ts is in [K]. The surroundings and the surrounding air temperature are at Tsur = 20°C. a.) Determine the steady state surface temperature of the burner. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" D=0.32 [inch]*convert(inch,m) L=36 [inch]*convert(inch,m) q_dot=900 [W] e=0.80 [-] T_sur=converttemp(C,K,20)

"diameter of burner" "length of burner" "burner power consumption" "burner emissivity" "surrounding temperature"

In order to move logically through the problem solution it is best to initially assume a surface temperature, calculate the heat transfer rates, and finally adjust the surface temperature until the heat transfer rates are consistent with the problem statement. Therefore, an initial and reasonable guess for the surface temperature is made and used to compute the heat transfer coefficient with Eq. (1). T_s=1000 [K] "initial guess for the surface temperature" h=10.7 [W/m^2-K]+0.0048 [W/m^2-K^2]*T_s

"heat transfer coefficient"

The surface area of the burner is: As = π D L

(2)

The rate of convective heat transfer from the burner to the air is: qconv = h As ( Ts − Tsur )

and the rate of radiative heat transfer from the burner to the surroundings is:

(3)

qrad = σ ε As (Ts − Tsur )

(4)

where σ is Stefan-Boltzmann’s constant. A_s=pi*D*L q_dot_conv=A_s*h*(T_s-T_sur) q_dot_rad=A_s*e*sigma#*(T_s^4-T_sur^4)

"surface area" "convection" "radiation"

At this point, it is necessary to adjust the surface temperature assumption so that the sum of the radiative and convective heat transfer rates are equal to q , the energy provided to the burner. This could be done manually, adjust the variable T_s up or down as necessary. However, EES allows you to automate this process by solving the nonlinear set of equations; like any equation solver, EES begins this process from an initial set of values (guess values) for each of the variables and iteratively solves the equations over and over to minimize the error (the residual). The advantage of the solution approach provided above is that you have a “good” starting point defined (a good set of guess values); therefore, select Update Guesses from the Calculate menu in order to lock in these guess values. Then, remove the initially assumed value of the variable T_s (just comment it out – that is, highlight the line and right click on it, select Comment {} frorm the menu that will appear) and specify instead that the heat transfer rates must sum to q . q = qrad + qconv {T_s=1000 [K] "initial guess for the surface temperature"} q_dot=q_dot_conv+q_dot_rad

(5)

"energy balance on the burner"

Solving the EES program will lead to Ts = 900 K. b.) Prepare a plot showing the surface temperature as a function of the burner input power. This is done most easily using a Parametric Table. Comment out the specified value of the burner input power: {q_dot=900 [W]

"burner power consumption"}

and prepare a Parametric Table that includes the variables q_dot and T_s. Vary the value of for example from 250 W to 1500 W, and solve the table. Plot the results to obtain Figure 2. q_dot,

Figure 2: Burner surface temperature as a function of burner power

Problem 1.2-14: Ice Rink Ice for an ice skating rink is formed by running refrigerant at Tr = -30°C through a series of cast iron pipes that are embedded in concrete, as shown in Figure P1.2-14. The cast iron pipes have an outer diameter of Do,p = 4 cm and an inner diameter of Di,p = 3 cm. The pipes are spaced Lptp = 8.0 cm apart. The heat transfer coefficient between the refrigerant and the pipe surface is hr = 100 W/m2-K. The concrete slab is Lc = 8 cm thick and the pipes are in the center of the slab. The bottom of the slab is insulated (assume perfectly). The thermal conductivity of concrete and iron are kc = 4.5 W/m-K and kiron = 51 W/m-K, respectively. An Lfill = 1 cm thick layer of water is placed on the top of the concrete slab. The refrigerant cools the top of the slab and the water turns to ice slowly. Assume that the water is stagnant and can be treated as a solid. The conductivity of ice and water are kice = 2.2 W/m-K and kw = 0.6 W/m-K, respectively. The heat transfer coefficient between the top of the water layer and the surrounding air at Ta = 15°C is ha = 10.0 W/m2-K. The top of the water surface has an emissivity of ε = 0.90 and radiates to surroundings at Tsur = 15°C. Ta = 15°C 2 ha = 10 W/m -K Lfill = 1 cm

ε = 0.9

Tsur = 15°C water, kw = 0.6 W/m-K ice, kice = 2.2 W/m-K

Lice

Lptp = 8 cm Lc = 8 cm Lp = 4 cm

cast iron tubes kiron = 51 W/m-K Do,p = 4 cm Di,p = 3 cm

insulation

concrete, kc = 4.5 W/m-K Tr = −30°C 2 hr = 10 W/m -K

Figure P1.2-14: Schematic of ice rink

a.) Draw a network that represents this situation using 1-D resistances. (Some of the resistances must be approximate since it is not possible to exactly calculate a 1-D resistance to the conduction heat flow in the concrete). Include an energy term related to the energy that is added to the system by the generation of ice. Clearly label the resistors. The resistance network is shown in Figure 1.

Rrad Tsur

Rw Tice

Rice

Rc

Rp

Rconv,r Tr

Rconv,a Ta heat removed from water to make ice The resistors include: Rrad = radiation resistance Rconv,a = convection resistance to air Rw = conduction through water Rice= conduction through ice Rc = conduction through concrete Rp = conduction through pipe Rconv,r = convection resistance to refrigerant Figure 1: Resistance network representing the ice rink

The resistance network interacts with the air temperature (Ta), the surroundings (Ts), and the refrigerant (Tr). There is a heat transfer rate at the interface between the ice and water related to the heat removed from the water in order to form more ice. This heat is accepted because there is more energy removed by the refrigerant than is provided from the surroundings. b.) Estimate the magnitude of each of the resistances in your network when the ice is 0.5 cm thick (i.e., Lice = 0.5 cm). We will deal with a unit cell of the sub-floor structure, as shown in Figure 2:

Figure 2: Unit cell of the sub-floor structure

The solution process will be described in the context of EES. It is assumed that you have already been exposed to the EES software by carrying out the self-guided tutorial contained in Appendix A. The first step in preparing a successful solution to any problem with EES is to enter the inputs to the problem and set their units. Experience has shown that it is generally best to work exclusively in SI units (m, J, K, kg, Pa, etc.). This unit system is entirely self-consistent. If the problem statement includes parameters in other units they are converted to SI units within the “Inputs” section of the code. The upper section of your EES code should look something like: $UnitSystem SI MASS RAD PA K J

$Tabstops 0.2 0.4 0.6 3.5 in "Inputs" T_r=converttemp(C,K,-30 [C]) h_r=100 [W/m^-K] D_i_p=3.0*convert(cm,m) D_o_p=4.0*convert(cm,m) k_iron=51 [W/m-K] L_ptp=8.0*convert(cm,m) k_c=4.5 [W/m-K] L_c=8.0*convert(cm,m) L_p=4.0*convert(cm,m) L_fill=1.0*convert(cm,m) k_ice=2.2 [W/m-K] k_w=0.6 [W/m-K] h_a=10 [W/m^2-K] T_a=converttemp(C,K,15 [C]) T_sur=converttemp(C,K,15 [C]) e=0.9 [-] W=1 [m] L_ice=0.5*convert(cm,m)

"refrigerant temperature (K)" "heat transfer coefficient between refrigerant and pipe" "pipe inner diameter (m)" "pipe outer diameter (m)" "pipe conductivity (W/m-K)" "pipe-to-pipe distance (m)" "concrete thermal conductivity (W/m-K)" "thickness of concrete (m)" "center of pipe to upper surface of concrete thickness (m)" "thickness of water layer on concrete (m)" "conductivity of ice (W/m-K)" "conductivity of water (W/m-K)" "water-to-air heat transfer coefficient (W/m^2-K)" "temperature of air on top of slab (K)" "temperature of radiation surrounding on top (K)" "emissivity of water surface (-)" "width of surface (m)" "thickness of ice (m)"

The radiation resistance is:

Rrad =

1 4 σ ε T 3 As

(1)

Tsur + Ts 2

(2)

where

T =

and Ts is the temperature at the surface of the water. The area of water in the unit cell that is exposed to air is: As = Lptp W

(3)

where W is the width of the unit cell into the page (here, W = 1 m for a solution per unit length of the floor). The surface temperature cannot be known until the problem is solved and yet it must be used to calculate the resistance to radiation, Rrad. One of the nice things about using the Engineering Equation Solver (EES) software to solve this problem is that the software can deal with this type of nonlinearity and provide the solution to the implicit equation. It is this capability that simultaneously makes EES so powerful and yet sometimes, ironically, difficult to use. EES should be able to solve equations regardless of the order in which they are entered. However, you should enter equations in a sequence that allows you to solve them as you enter them; this is exactly what you would be forced to do if you were to solve the problem using a typical programming language (e.g., MATLAB, FORTRAN, etc.). This technique of entering your

equations in a systematic order provides you with the opportunity to debug each subset of equations as you move along rather than waiting until you have entered all of your equations and tried to solve only to find that there are multiple problems. Another benefit of approaching a problem in this manner is that you can consistently update your guess values associated with the variables in your problem; EES solves your equations using a nonlinear relaxation technique and therefore the closer your variables are to “reasonable” values the better this process will go. To proceed with the solution to this ice rink problem using EES, it is a helpful idea to assume initially a reasonable surface temperature (e.g., 273 K) so that it is possible to estimate the radiation resistance and continue with the solution. The next few lines in your EES code should look something like: "Resistances" "Radiation" A_s=W*L_ptp T_s=273.2

"surface area" "this is a guess for the surface temperature - eventually we will comment this out" T_bar=(T_s+T_sur)/2 "average temperature" R_rad=1/(4*e*sigma#*T_bar^3*A_s) "radiation resistance"

If you solve the equations that have been entered (Calculate/Solve) you can check that your answers make sense and you can verify that your equations have a consistent set of units. It would be good to do this and then update your guess values (Calculate/Update Guesses); this operation sets the guess values for each of your variables to their current value and therefore helps EES iterate to the correct solution. Finally, you should set the units for each of your variables. The best way to do this is to go to the Variable Information window (Options/Variable Info) and enter the unit for each variable in the Units column. Once you have done that you can check units (Calculate/Check Units) in order to make sure that all of the units you set are consistent with the equations that you’ve entered. The convection resistance to the air is: Rconv ,a =

1 ha As

(4)

The conduction resistances through the water and ice are: Rw =

(L

Rw =

fill

− Lice )

kw A Lice kice A

(5)

(6)

We will learn how to calculate the resistances of the concrete and pipe more exactly when we get to extended surfaces and 2-D conduction. For now we will estimate them very approximately using the concept of an effective length and cross-sectional area for conduction. The length for

conduction will be taken to be the distance that the pipe is submerged beneath the surface and the area will be taken to be the area of the unit cell: Rc =

Lp

(7)

kc As

The pipe resistance is taken to be the resistance of half a cylinder (in fact, the bottom of the pipe probably also participates if the pipe is very conductive): ⎛D ⎞ ln ⎜ o , p ⎟ ⎜ Di , p ⎟ ⎠ Rp = ⎝ π kiron W

(8)

Finally, the convection resistance to the refrigerant is: Rconv ,r =

2 hr π Do , p W

(9)

The resulting EES code is shown below: R_conv_a=1/(h_a*A_s) R_w=(L_fill-L_ice)/(k_w*A_s) R_ice=L_ice/(k_ice*A_s) R_c=L_p/(k_c*A_s) R_p=2*ln(D_o_p/D_i_p)/(2*pi*W*k_iron) R_conv_r=1/(h_r*W*pi*D_i_p/2)

"air convection resistance" "resistance of water layer" "resistance of ice layer" "concrete resistance (approximate)" "pipe resistance" "refrigerant convection resistance"

EES will calculate the resistances in the network (although the radiation resistance continues to be only approximate). These resistances are placed on the network in Figure 3. As h fs Rrad = 2.77

K W

Tsur = 15°C

Tice = 0°C

dLice dt

Rc = 0.11

K W

Rconv ,r = 0.21

K W Tr = -30°C

Rw = 0.10

Ta = 15°C Rconv ,a =1.25

K K Rice = 0.03 W W

R p = 0.002

K W

K W Figure 3: Resistances calculated for ice rink

Resistance networks often provide substantial intuition relative to a problem. For example, Figure 3 shows that the resistances associated with convection and radiation from the surface of the water are of the same order of magnitude and large relative to others in the circuit; therefore,

both radiation and convection is important for this problem. If the radiation resistance had been much larger than the convection resistance (as is often the case in forced convection problems where the convection coefficient is much larger) then radiation could be neglected; the smallest resistance in a parallel network will dominate the problem because most of the thermal energy will tend to flow through that resistance. In a series resistance network, the larger resistors dominate the problem and the smaller ones can be neglected. Therefore, we could safely neglect the conduction resistance through the water as it is small relative to the parallel combination of the radiation and convection resistances. Similarly, conduction through the ice and the pipe are not important to this problem. It is almost always a good idea to estimate the size of the resistances in a heat transfer problem prior to solving it; often it is possible to simplify the problem considerably and the size of the resistances can certainly be used to guide your efforts. For the ice rink problem, a detailed analysis of conduction through the pipe would be a misguided use of your time whereas a more accurate simulation of the conduction through the concrete would be very important. c.) Calculate the rate of change in the thickness of the ice when the ice thickness is 0.5 cm. The heat transfer from the ice/water interface to the refrigerant ( qr ) is higher than the heat transfer from the air and surroundings to the ice/water interface ( qa ) and therefore ice will be formed. These heat transfer rates can be estimated according to: qr =

qa =

(Tice − Tr ) Rice + Rc + R p + Rconv ,r

(Ta − Tice ) ⎡ 1 1 ⎤ + Rw + ⎢ ⎥ ⎣ Rrad Rconv ,a ⎦

−1

(10)

(11)

An energy balance at the interface leads to: qr − qa = A h fs

where hfs is the latent heat of fusion for ice and

dLice dt

dLice is the rate of ice formation. The additional dt

EES code needed to solve this problem is: "Rate of ice formation" h_fs=Enthalpy_fusion(Water) rho_ice=1000 T_ice=convertTemp(C,K,0)

(12)

"enthalpy of fusion of ice" "density of ice" "temperature at which water freezes"

q_dot_r=(T_ice-T_r)/(R_conv_r+R_p+R_c+R_ice) "heat transfer to refrigerant" q_dot_a=(T_a-T_ice)/(R_w+(1/R_conv_a+1/R_rad)^(-1)) "heat transfer from air"

dLicedt=(q_dot_r-q_dot_a)/(A_s*rho_ice*h_fs)

"rate of change of ice layer"

At this point, we can use the heat transfer rates to recalculate the water surface temperature (as opposed to assuming it). Ts = Tice + qa Rw

(13)

It is necessary to comment out or delete the equation that provided the assumed surface temperature and instead calculate the surface temperature correctly. {T_s=273.2} "this is a guess for the surface temperature - eventually we will comment this out" T_s=T_ice+q_dot_a*R_w "recalculate the surface temperature to make radiation resistance exact"

The rate of formation of ice is 2.6e-6 m/s (or 0.94 cm/hr). Note that it will take about 1 hr to freeze all of the water on the rink based on this answer; the rate of ice formation will not be significantly affected by the amount of ice because the conduction resistances of the ice and water were found to be relatively insignificant. Once the ice is completely frozen, the surface temperature of the ice will drop until qr is balanced by qa .

Problem 1.2-15 (1-6 in text): The super ice-auger You are a fan of ice fishing but don't enjoy the process of augering out your fishing hole in the ice. Therefore, you want to build a device, the super ice-auger, that melts a hole in the ice. The device is shown in Figure P1.2-15.

2

h = 50 W/m -K T∞ = 5 ° C

ε = 0.9 insulation, kins = 2.2 W/m-K thp = 0.75 inch

heater, activated with V = 12 V and I = 150 A plate, kp = 10 W/m-K D = 10 inch thins = 0.5 inch thice = 5 inch

ρice = 920 kg/m3 Δifus = 333.6 kJ/kg

Figure P1.2-15: The super ice-auger.

A heater is attached to the back of a D = 10 inch plate and electrically activated by your truck battery, which is capable of providing V = 12 V and I = 150 A. The plate is thp = 0.75 inch thick and has conductivity kp = 10 W/m-K. The back of the heater is insulated; the thickness of the insulation is thins = 0.5 inch and the insulation has conductivity kins = 2.2 W/m-K. The surface of the insulation experiences convection with surrounding air at T∞ = 5°C and radiation with surroundings also at T∞ = 5°C. The emissivity of the surface of the insulation is ε = 0.9 and the heat transfer coefficient between the surface and the air is h = 50 W/m2-K. The super ice-auger is placed on the ice and activated, resulting in a heat transfer to the plate-ice interface that causes the ice to melt. Assume that the water under the ice is at Tice = 0°C so that no heat is conducted away from the plate-ice interface; all of the energy transferred to the plate-ice interface goes into melting the ice. The thickness of the ice is thice = 5 inch and the ice has density ρice = 920 kg/m3. The latent heat of fusion for the ice is Δifus = 333.6 kJ/kg. a.) Determine the heat transfer rate to the plate-ice interface. The inputs are entered in EES: "P1.2-15" $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 D=10 [inch]*convert(inch,m) th_ins=0.5 [inch]*convert(inch,m) k_ins=2.2 [W/m-K] th_p=0.75 [inch]*convert(inch,m) k_p=10 [W/m-K] e=0.9 [-] h_bar=50 [W/m^2-K] T_infinity=converttemp(C,K,5 [C]) V=12 [V] I=150 [A]

"diameter of ice fishing hole" "insulation thickness" "insulation conductivity" "plate thickness" "conductivity of plate" "emissivity" "air heat transfer coefficient" "ambient temperature" "battery voltage" "current"

T_ice=converttemp(C,K,0 [C]) th_ice=5 [inch]*convert(inch,m) rho_ice=920 [kg/m^3] DELTAi_fus=333.6 [kJ/kg]*convert(kJ/kg,J/kg)

"temperature of ice-water interface" "thickness of ice" "density of ice" "enthalpy of fusion of ice"

The power provided to the heater is the product of the voltage and current:

q = V I q_dot=V*I

(1) "power to melting plate"

A resistance network that can be used to represent this problem is shown in Figure P1.2-15-2. Rrad = 3.56 K/W q q2

q1

Tice = 0°C

T∞ = 5°C

Ttop

Rcond , p = 0.038 K/W Rcond ,ins = 0.114 K/W

Rconv = 0.395 K/W

The resistances include: Rcond,p = conduction through plate Rcond,ins = conduction through insulation Rrad = radiation resistance Rconv = convection resistance

Figure P1.2-15-2: The super ice-auger.

In order to compute the radiation resistance required to solve the problem, it is necessary to assume a value of Ttop, the temperature at the top of the insulation (this value will eventually be commented out in order to complete the problem): T_top=360 [K]

"guess for top surface temperature"

The cross-sectional area of the plate is computed:

Ac =

π D2

(2)

4

and the radiation resistance is computed according to:

Rrad =

1 ε Ac σ (T + T∞2 ) (Ttop + T∞ )

(3)

2 top

A_c=pi*D^2/4 R_rad=1/(e*A_c*sigma#*(T_top^2+T_infinity^2)*(T_top+T_infinity))

"area of hole" "radiation resistance"

Note that the equations should be entered, the units set, and the EES code solved line by line in order to debug the code in small segments. The convection resistance is computed according to:

Rconv =

1 Ac h

(4)

and the conduction resistances are computed according to:

Rcond ,ins =

Rcond , p =

thins Ac kins

(5)

thp

(6)

Ac k p

R_conv=1/(A_c*h_bar) R_cond_ins=th_ins/(k_ins*A_c) R_cond_p=th_p/(k_p*A_c)

"air convection resistance" "conduction resistance of insulation" "conduction resistance of plate"

The heat transfer from the heater to the ambient surroundings ( q1 in Figure P1.2-15-2) is:

q1 =

(Th − T∞ ) ⎛ 1 1 ⎞ Rcond ,ins + ⎜ + ⎟ ⎝ Rrad Rconv ⎠

−1

(7)

and the heat transfer to the ice is:

q2 =

(Th − Tice ) Rcond , p

(8)

where Th is the heater temperature. An energy balance on the heater leads to:

q = q1 + q2

(9)

Equations (7) through (9) are 3 equations in 3 unknowns ( q , q1 , and q2 ) and can be solved using EES: q_dot_1=(T_h-T_infinity)/(R_cond_ins+(1/R_rad+1/R_conv)^(-1)) "heat transfer to ambient" q_dot_2=(T_h-T_ice)/R_cond_p "heat transfer to ice" q_dot=q_dot_1+q_dot_2 "energy balance"

The temperature at the top of the plate can be computed based on the solution. Update the guess values for the problem (select Update Guess Values from the Calculate menu) and comment out the guessed value for Ttop: {T_top=360 [K]}

"guess for top surface temperature"

and calculate Ttop according to the resistance network:

Ttop = Th − q1 Rcond ,ins T_top=T_h-q_dot_1*R_cond_ins

(10) "recalculate top temperature"

The result is q2 = 1676 W. The values of the resistances are shown in Figure P1.2-15-2; notice that radiation does not play an important role in the problem because it is a large resistance in parallel with a much smaller one. The resistance to conduction through the plate is also unimportant since it is so small. The resistance to conduction through the insulation and convection are dominant. b.) How long will it take to melt a hole in the ice? An energy balance on the ice-to-plate interface leads to:

q2 = Ac Δi fus ρice

dthice dt

(11)

dthice is the rate at which the thickness of the ice is reduced. Because there is no energy dt lost to the water, the rate of ice melting is constant with ice thickness. Therefore the time required to melt the ice is estimated according to:

where

time

dthice = thice dt

q_dot_2=A_c*DELTAi_fus*dth_icedt*rho_ice dth_icedt*time=th_ice time_min=time*convert(s,min)

(12) "energy balance on ice interface" "time to melt ice" "in min"

which leads to time = 1178 s (19.6 min). c.) What is the efficiency of the melting process? The efficiency is defined as the ratio of the energy provided to the plate-to-ice interface to the energy provided to the heater:

η=

q2 q

eta=q_dot_2/q_dot

(13) "efficiency of process"

which leads to η = 0.93. d.) If your battery is rated at 100 amp-hr at 12 V then what fraction of the battery's charge is depleted by running the super ice-auger? The total amount of energy required to melt a hole in the ice is:

Q = q time

(14)

The energy stored in the battery (Ebattery) is the product of the voltage and the amp-hr rating. The fraction of the battery charge required is: f = Q=q_dot*time E_battery=100 [amp-hr]*V*convert(A-V-hr,J) f=Q/E_battery

which leads to f = 0.491.

Q Ebattery "total energy required" "car battery energy" "fraction of car battery energy used"

(15)

Problem 1.2-16 The temperature distribution across a L = 0.3 m thick wall at a certain instant of time is given by T = 900 − 900 x − 500 x 2 where T is the temperature in the wall in °C at position x in m. The density, specific heat capacity and thermal conductivity of the wall are ρ = 2050 kg/m3, c = 0.96 kJ/kg-K and k = 1.13 W/m-K, respectively. a) Calculate the rate of change of the average wall temperature. The inputs are entered in EES: "P1.2-16" $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "known" k=1.13 [W/m-K] rho=2050 [kg/m^3] c=0.96 [kJ/kg-K]*convert(kJ/kg-K,J/kg-K) L=0.3 [m]

"conductivity" "density" "specific heat capacity" "thickness of the wall"

A control volume that includes the wall is shown in Figure 1.

dU dt

q x =0

q x = L

x

Figure 1: Energy balance on the wall.

The energy balance on the control volume is:

q x =0 =

dU + q x = L dt

(1)

Fourier's law is used to compute the conduction heat transfer at each edge: q x =0 = − k A

q x = L = − k A

The energy storage is given by:

dT dx

x =0

dT dx

x= L

(2)

(3)

dU dT = ALρ c dt dt

(4)

where T is the average temperature in the wall. Substituting Eqs. (2), (3), and (4) into Eq. (1) leads to: −k A

dT dx

= ALρ c x =0

dT dT −k A dt dx

(5) x= L

Solving for the rate of change of the average temperature leads to: dT k ⎛ dT = ⎜ dt L ρ c ⎝ dx

− x=L

dT dx

⎞ ⎟ x =0 ⎠

(6)

where the temperature gradients can be obtained by taking the derivative of the temperature distribution given in the problem statement: dTdx0=-900 [C/m] dTdxL=-900 [C/m] - 2*500 [C/m^2]*L dTbardt=k*(dTdxL-dTdx0)/(L*rho*c)

which leads to

"temperature gradient at x=0" "temperature gradient at x=L" "rate of change of the average temperature"

dT = -0.000574 ºC/s. dt

b) The left surface of the wall (at x = 0) is exposed to air at T∞ = 1000°C. Determine the average convection coefficient between the air and the wall surface at x = 0 m. Newton's law of cooling defines the heat transfer coefficient according to: h A (T∞ − Tx =0 ) = − k A

∂T ∂x

(7) x =0

solving for the heat transfer coefficient: ∂T ∂x x =0 h= (T∞ − Tx =0 ) −k

T_infinity=1000 [C] T0=900[C] h_bar=-k*dTdx0/(T_infinity-T0)

(8) "ambient temperature at x=0" "temperature at x=0" "heat transfer coefficient"

Problem 1.2-17 Figure P1.2-17 illustrates the temperature distribution in a plane wall at a particular instant of time. T

x Figure P1.2-17: Temperature distribution in a plane wall at a certain instant in time.

Select the correct statement from those listed below and justify your answer briefly. • The heat transfer at the left-hand face of the wall (i.e., at x = 0) is into the wall (in the positive x direction), • The heat transfer at the left-hand face of the wall is out of the wall (in the negative x direction), • It is not possible to tell the direction of the heat transfer at the left-hand face of the wall. Fourier’s law states that conduction is proportional to the negative of the temperature gradient. At the left-hand face of the wall the temperature gradient is positive; therefore, the heat transfer must be in the negative x-direction or out of the wall.

Problem 1.2-18 Urea formaldehyde foam with conductivity kfoam = 0.020 Btu/hr-ft-F is commonly used as an insulation material in building walls. The major advantage of foam insulation is that it can be installed in existing walls by injection through a small hole. In a particular case, foam is to be installed in a wall consisting of a thplaster = 5/8 inch thick sheet of plaster board with conductivity kplaster = 0.028 Btu/hr-ft-F, a thas = 3.5 inch air space and a thbrick = 2.5 inch layer of brick with conductivity kbrick = 0.038 Btu/hr-ft-F. Before the air gap is filled with foam, there is natural convection associated with buoyancy induced air motion. The equivalent thermal resistance of ′′ = 0.95 hr-ft2-F/Btu. The the air gap on a per unit area basis due to the natural convection is Rag convection coefficients for the inner and outer surface of the wall are hi = 1.5 Btu/hr-ft2-F and

ho = 3.5 Btu/hr-ft2-F, respectively. a) The R-value of a wall is the resistance of the wall on a per unit area basis, expressed in units ft2-F-hr/Btu. Calculate the R-value of the wall before the foam insulation is applied. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "known" k_brick=0.038 [Btu/hr-ft-F]*convert(Btu/hr-ft-F,W/m-K) k_plaster=0.028 [Btu/hr-ft-F]*convert(Btu/hr-ft-F,W/m-K) k_foam=0.020 [Btu/hr-ft-F]*convert(Btu/hr-ft-F,W/m-K) th_plaster=(5/8) [in]*convert(in,m) th_as=3.5 [in]*convert(in,m) th_brick=2.5 [in]*convert(in,m) R``_ag=0.95 [hr-ft^2-F/Btu]*convert(hr-ft^2-F/Btu,K-m^2/W) "resistance of air gap without foam per unit area" h_bar_i=1.5 [Btu/hr-ft^2-F]*convert(Btu/hr-ft^2-F,W/m^2-K) h_bar_o=3.5 [Btu/hr-ft^2-F]*convert(Btu/hr-ft^2-F,W/m^2-K)

"brick conductivity" "plaster conductivity" "formaldehyde foam conductivity" "thickness of plaster" "thickness of air space" "thickness of brick"

"inside heat transfer coefficient" "outside heat transfer coefficient"

The resistance per unit area without foam is:

′′ ,nofoam = Rtotal

th 1 thplaster 1 ′′ + brick + + + Rag hi k plaster kbrick ho

(1)

"No foam" R``_conv_i=1/h_bar_i "resistance to convection on inner surface per unit area" R``_plaster=th_plaster/k_plaster "resistance to conduction through plaster per unit area" R``_brick=th_brick/k_brick "resistance to conduction through brick per unit area" R``_conv_o=1/h_bar_o "resistance to convection on outer surface per unit area" R``_total_nofoam=R``_conv_i+R``_plaster+R``_ag+R``_brick+R``_conv_o "total resistance per unit area" R_value_nofoam=R``_total_nofoam*convert(K-m^2/W,ft^2-F-hr/Btu)"R-value of wall without foam"

which leads to an R-value of 9.245 ft2-F-hr/Btu. b) Calculate the R-value of the wall after insulation is applied. Assume that the insulating foam completely fills the air gap.

The resistance per unit area with foam is: ′′ , foam = Rtotal

1 thplaster thas thbrick 1 + + + + hi k plaster k foam kbrick ho

(2)

"Foam, no shrinkage" R``_foam=th_as/k_foam "resistance to conduction through foam per unit area" R``_total_foam=R``_conv_i+R``_plaster+R``_foam+R``_brick+R``_conv_o "total resistance per unit area" R_value_foam=R``_total_foam*convert(K-m^2/W,ft^2-F-hr/Btu) "R-value of wall with foam"

which leads to an R-value of 22.88 ft2-F-hr/Btu. c) Foam insulation ordinarily shrinks after it is installed by an amount dependent upon conditions such as the outdoor temperature. Calculate the R-value of the wall assuming that the foam shrinks by 3%. Assume that the air in the gap between the foam and the plaster and the foam and the brick is stagnant. The resistance per unit area with the foam considering shrinkage is: ′′ , foam = Rtotal

th th th 1 thplaster 1 + + (1 − shrinkage ) as + shrinkage as + brick + hi k plaster k foam ka kbrick ho

(3)

where ka is the thermal conductivity of air (evaluated at 20°C): "Foam, shrinkage" Shrinkage=0.03 [-] "amount of shrinkage" R``_foam_s=(1-Shrinkage)*th_as/k_foam "resistance to conduction through foam per unit area" k_a=conductivity(Air,T=converttemp(C,K,20 [C])) "conductivity of air" R``_air_s=Shrinkage*th_as/k_a "resistance to conduction through air per unit area" R``_total_foam_s=R``_conv_i+R``_plaster+R``_foam_s+R``_air_s+R``_brick+R``_conv_o "total resistance per unit area" R_value_foam_s=R``_total_foam_s*convert(K-m^2/W,ft^2-F-hr/Btu) "R-value of wall with foam"

which leads to an R-value of 23.04 ft2-F-hr/Btu. Note that the R-value actually improved with shrinkage because the stagnant air is less conductive than the foam. However, if the shrinkage is more extreme then natural convection will cause the resistance of the air to drop and the R-value to be reduced.

Problem 1.2-19 Figure P1.2-19 illustrates a cross-section of a thermal protection suit that is being designed for an astronaut. thins = 4 cm

thliner = 1 cm

thext = 1 mm tissue

Tspace = 4 K

ε = 0.25

Tb = 37°C kliner = 0.06 W/m-K

heater

kext = 14.5 W/m-K kins = 0.06 W/m-K

Figure P1.2-19: Cross-section of thermal protection suit.

The suit consists of a liner that is immediately adjacent to the skin. The skin temperature is maintained at Tb = 37ºC by the flow of blood in the tissue. The liner is thliner = 1 cm thick and has conductivity kliner = 0.06 W/m-K. A thin heater is installed at the outer surface of the liner. Outside of the heater is a layer of insulation that is thins = 4 cm with conductivity kins = 0.06 W/m-K. Finally, the outer layer of the suit is thext = 1 mm thick with conductivity kext = 14.5 W/m-K. The outer surface of the external layer has emissivity ε = 0.25 and is exposed by radiation only to outer space at Tspace = 4 K. a.) You want to design the heater so that it completely eliminates any heat loss from the skin. What is the heat transfer per unit area required? The inputs are entered in EES and the units converted to base SI units: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" T_b=converttemp(C,K,37 [C]) th_ins=4 [cm]*convert(cm,m) k_ins=0.06 [W/m-K] th_ext=1 [mm]*convert(mm,m) k_ext=14.5 [W/m-K] emm=0.25 [-] T_space = 4 [K] th_liner=1 [cm]*convert(cm,m) k_liner=0.06 [W/m-K] A=1 [m^2]

"tissue temperature" "insulation thickness" "insulation conductivity" "exterior wall thickness" "conductivity of exterior wall" "emissivity of exterior wall" "temperature of space" "liner thickness" "liner conductivity" "do problem on a unit area basis"

The units of each variable are set by right-clicking on each variable in the Solution Window and setting the units in the Units dialog (Figure 2):

Figure 2: Set units for variables.

The units are checked by selecting Check Units from the Calculate menu. A resistance diagram that represents the suit is shown in Figure 3. qskin

q

qout Text

Tb Rliner

Rext

Rins

Tspace Rrad

Figure 3: Resistance network representation of space suit.

The resistance to conduction through the liner, insulation and external layer are computed according to:

Rliner =

thliner kliner A

(1)

Rins =

thins kins A

(2)

Rext =

thext kext A

(3)

where A is taken to be 1 m2 to do the problem on a per unit area basis. "part (a)" R_cond_liner=th_liner/(k_liner*A) R_cond_ins=th_ins/(k_ins*A) R_cond_ext=th_ext/(k_ext*A)

"resistance to conduction through liner" "resistance to conduction through the insulation" "resistance to conduction through exterior wall"

The resistance to radiation can be computed according to:

Rrad =

As σ ε (T + T 2 ext

1

2 space

)(T

ext

+ Tspace )

(4)

However, Text - the external surface of the space suit, is not known. Therefore, we will guess or assume this temperature and subsequently complete the problem by calculating this value and removing this assumption. A reasonable guess is Text = 250 K. An energy balance on the heater (recall that the heater power is to be selected so that qskin = 0 and therefore Thtr = Tb) leads to: qout =

(T

b

− Tspace )

Rins + Rext + Rrad

(5)

T_ext=250 [K] "guess for exterior wall outside temperature" R_rad=1/(emm*A*sigma#*(T_ext^2+T_space^2)*(T_ext+T_space)) "radiation resistance" q_dot_out=(T_b-T_space)/(R_cond_ins+R_cond_ext+R_rad) "rate of heat transfer"

The problem is solved and the guess values in EES updated (select Update Guesses from the Calculate menu). The assumed value of Text is commented out and Text is recalculated according to: Text = Tb − qout ( Rins + Rext ) {T_ext=250 [K]} T_ext=T_b-q_dot_out*(R_cond_ins+R_cond_ext)

(6)

"guess for exterior wall outside temperature" "recalculate exterior wall outside temperature"

which leads to qout = 69 W. b.) In order, rate the importance of the following design parameters to your result from (a): kins, kext, and ε. The magnitude of the thermal resistances that participate in the process are Rins = 0.67 K/W, Rext = 6.9x10-5 K/W, and Rrad = 3.77 K/W. In a series resistance circuit, the largest resistors dominate and therefore the most important parameters are those that dictate Rrad and the least important are those that determine Rext. In order, the most important parameters are ε and kins. The value of kext is almost completely unimportant. c.) Plot the heat transfer per unit area required to eliminate heat loss as a function of the emissivity, ε. A parametric table is created (select New Parametric Table from the Tables menu) and the parameters emm and q_dot_out are added (Figure 4).

Figure 4: New Parametric Table window.

The value of emm is varied from 0.1 to 1.0 by right-clicking on the column heading and selecting Alter Values (Figure 5)

Figure 5: Alter values to vary emissivity in the parametric table.

The specified value of emissivity is commented out in the program and the table is run (select Solve Table from the Calculate menu). Select New Plot (X-Y Plot) from the Plots menu to generate Figure 6.

2

Heat transfer per unit area (W/m )

140 120 100 80 60 40 20 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Emissivity Figure 6: Required heat transfer per unit area as a function of the suit emissivity.

While the average emissivity of the suit's external surface is ε = 0.25, you have found that this value can change substantially based on how dirty or polished the suit is. You are worried about these local variations causing the astronaut discomfort due to local hot and cold spots. d.) Assume that the heater power is kept at the value calculated in (a). Plot the rate of heat transfer from the skin as a function of the fractional change in the emissivity of the suit surface. The code for part (a) is commented out: {"part (a)" R_cond_liner=th_liner/(k_liner*A) "resistance to conduction through liner" R_cond_ins=th_ins/(k_ins*A) "resistance to conduction through the insulation" R_cond_ext=th_ext/(k_ext*A) "resistance to conduction through exterior wall" {T_ext=250 [K]} "guess for exterior wall outside temperature" R_rad=1/(emm*A*sigma#*(T_ext^2+T_space^2)*(T_ext+T_space)) "radiation resistance" q_dot_out=(T_b-T_space)/(R_cond_ins+R_cond_ext+R_rad) T_ext=T_b-q_dot_out*(R_cond_ins+R_cond_ext) "recalculate exterior wall outside temperature"}

"

and the heat transfer rate is set according to the result calculated in (a). The emissivity is varied from its nominal value by an amount fct - the fractional change:

ε dirty = fct ε "part (d)" fct=1.5 [-] emm_dirty=emm*fct q_dot_htr=69 [W]

"fractional change in the emissivity" "emissivity at a location where suit has gotten tarnished" "heat transfer rate calculated in (a)"

The resistance of the liner, insulation, and external layer are computed as before: R_cond_liner=th_liner/(k_liner*A)

"resistance to conduction through liner"

(7)

R_cond_ins=th_ins/(k_ins*A) R_cond_ext=th_ext/(k_ext*A)

"resistance to conduction through the insulation" "resistance to conduction through exterior wall"

The external surface temperature, Text, is again assumed and the assumed value is used to compute Rrad: T_ext=250 [K] "guess for exterior wall outside temperature" R_rad=1/(emm_dirty*A*sigma#*(T_ext^2+T_space^2)*(T_ext+T_space)) "radiation resistance"

The energy balance on the heater is:

qhtr =

(T

htr

− Tspace )

Rins + Rext + Rrad

+

(Thtr − Tb ) Rliner

(8)

q_dot_htr=(T_htr-T_space)/(R_cond_ins+R_cond_ext+R_rad)+(T_htr-T_b)/R_cond_liner "energy balance on heater"

The heat transfer rate to space is:

qout =

(T

htr

− Tspace )

(9)

Rins + Rext + Rrad

q_dot_out=(T_htr-T_space)/(R_cond_ins+R_cond_ext+R_rad)

"heat transfer to space"

The problem is solved and the guess values in EES updated (select Update Guesses from the Calculate menu). The assumed value of Text is commented out and Text is recalculated according to: Text = Thtr − qout ( Rins + Rext )

(10)

{T_ext=250 [K]} "guess for exterior wall outside temperature" T_ext=T_htr-q_dot_out*(R_cond_ins+R_cond_ext) "recalculate exterior wall outside temperature"

The heat transfer rate from the skin is:

qskin = q_dot_skin=(T_b-T_htr)/R_cond_liner

(Tb − Thtr ) Rliner

(11)

"heat transfer from tissue"

which leads to qskin = 15.6 W. Figure 7 illustrates the rate of heat transfer from the skin as a function of the fractional change in the emissivity of the suit surface.

2)

Heat transfer from skin per unit area (W/m

20 15 10 5 0 -5 -10 -15 -20 -25 0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

Fractional change in emissivity

Figure 7: Heat transfer from the skin as a function of the fractional change in emissivity.

Problem 1.2-20 Figure P1.2-20 illustrates a temperature sensor that is mounted in a pipe and used to measure the temperature of a flow of air. Tw = 20°C support Ac = 1x10-6 m2 k = 10 W/m-K

L = 0.01 m

2 h = 30 W/m -K g = 0.1 W T∞ = 50°C sensor As = 0.001 m2 Figure P1.2-20: Temperature sensor.

The operation of the sensor leads to the dissipation of g = 0.1 W of electrical power. This power is either convected to the air at T∞ = 50°C or conducted along the support to the wall at Tw = 20°C. Treat the support as conduction through a plane wall (i.e., neglect convection from the edges of the support). The heat transfer coefficient between the air and the sensor is h = 30 W/m2-K. The surface area of the sensor is As = 0.001 m2. The support has cross-sectional area Ac = 1x10-6 m2, length L = 0.01 m, and conductivity k = 10 W/m-K. a.) What is the temperature of the temperature sensor? A resistance network representation of this problem is shown in Figure 2. Rconv

Rcond

T∞

Tw

g Figure 2: Resistance network.

The resistance to conduction through the support is: Rcond =

L 0.01 m K-m K = = 1000 -6 2 k Ac 10 W 1x10 m W

(1)

The resistance to convection from the sensor surface is: Rconv

1 m 2 -K K = = = 33.3 2 W h As 30 W 0.001 m

An energy balance on the sensor leads to:

(2)

g =

(Ts − T∞ ) + (Ts − Tw ) Rconv

Rcond

(3)

Solving for Ts leads to: T T∞ 323.2 K 293.2 K + w 0.1W+ + Rconv Rcond 33.3K/W 1000 K/W = = 325.4 K 1 1 1 1 + + 33.3 K/W 1000 K/W Rconv Rcond

g + Ts =

(4)

b.) What is the error associated with the sensor measurement (i.e., what is the difference between the sensor and the air temperature)? Is the error primarily due to self-heating of the sensor associated with g or due to the thermal communication between the sensor and with the wall? Justify your answer. The error is Ts - T∞ = 2.26 K. The error is dominated by self-heating rather than mounting error. The mounting error would cause the sensor temperature to be less than the fluid temperature whereas the self-heating causes the temperature of the sensor to be elevated relative to the fluid temperature. c.) Radiation has been neglected for this problem. If the emissivity of the sensor surface is ε = 0.02, then assess whether radiation is truly negligible. The radiation resistance is: 1 As ε σ (T + Tw2 )(Ts + Tw ) m2 K 4 = 0.001 m 2 0.02 5.67x10-8 W ( 325.42 + 293.22 ) ( 325.4 + 293.2 ) K 3 K = 7432 W

Rrad =

2 s

(5)

The radiation occurs in parallel with convection and conduction and is large relative to either of these resistances, therefore it is probably negligible.

Problem 1.2-21 Figure P1.2-21 illustrates a plane wall made of a material with a temperature-dependent conductivity. The conductivity of the material is given by:

k = bT

(1)

where b = 1 W/m2-K2 and T is the temperature in K.

TH k = bT

TC x L Figure P1.2-21: Plane wall with temperature-dependent conductivity.

The thickness of the wall is L = 1 m. The left side of the wall (at x= 0) is maintained at TH = 500 K and the right side (at x= L) is kept at TC = 50 K. The problem is steady-state and 1-D. a.) Sketch the temperature distribution in the wall (i.e., sketch T as a function of x). Make sure that you get the qualitative features of your sketch right. The temperatures at x = 0 and x = L are specified. The temperature variation from 0 < x < L will not be linear. The rate of heat transfer will be constant with x for this problem. Fourier's law governs the rate of conduction:

q = − k Ac

dT dx

(2)

According to Eq. (1), in regions where the temperature is high, the conductivity will be high; therefore, the temperature gradient will be small. In regions where the temperature is low, the conductivity will be low and the temperature gradient higher. Figure 2 reflects these characteristics.

TH k = bT

TC x L T 500 K

50 K x

0

L

Figure 2: Sketch of temperature distribution.

b.) Derive the ordinary differential equation that governs this problem. The first step towards developing an analytical solution for this, or any, problem involves the definition of a differential control volume. The control volume must encompass material at a uniform temperature; therefore, in this case it must be differentially small in the x-direction (i.e., it has width dx, see Figure 3) but can extend across the entire cross-sectional area of the wall as there are no temperature gradients in the y- or z-directions.

Figure 3: Differential control volume.

Next, the energy transfers across the control surfaces must be defined as well as any thermal energy generation or storage terms. For the steady-state, 1-D case considered here, there are only two energy transfers, corresponding to the rate of conduction heat transfer into the left side (i.e., at position x, q x ) and out of the right side (i.e., at position x+dx, q x + dx ) of the control volume. A steady-state energy balance for the differential control volume is therefore:

q x = q x + dx

(3)

A Taylor series expansion of the term at x+dx leads to:

q x + dx

dq d 2 q dx 2 d 3 q dx3 = q x + dx + 2 + + ... dx dx 2! dx3 3!

(4)

The analytical solution proceeds by taking the limit as dx goes to zero so that the higher order terms in Eq. (4) can be neglected:

q x + dx = q x +

dq dx dx

(5)

Substituting Eq. (5) into Eq. (3) leads to:

q x = q x +

dq dx dx

(6)

or

dq =0 dx

(7)

Equation (7) indicates that the rate of conduction heat transfer is not a function of x. For the problem in Figure 1, there are no sources or sinks of energy and no energy storage within the wall; therefore, there is no reason for the rate of heat transfer to vary with position. The final step in the derivation of the governing equation is to substitute appropriate rate equations that relate energy transfer rates to temperatures. The rate equation for conduction is Fourier’s law: q = − k Ac

dT dx

(8)

Substituting Eq. (8) into Eq. (7) leads to:

d ⎡ dT ⎤ −k Ac =0 ⎢ dx ⎣ dx ⎥⎦

(9)

The area is constant and can be divided out of Eq. (9). The thermal conductivity is given by Eq. (1): d ⎡ dT ⎤ bT =0 dx ⎢⎣ dx ⎥⎦ c.) What are the boundary conditions for this problem? The boundary conditions are:

(10)

Tx =0 = TH

(11)

Tx = L = TC

(12)

and

d.) Solve the governing differential equation from (b) - you should end up with a solution that involves two unknown constants of integration. Equation (10) is separated: ⎡ dT ⎤ d ⎢bT =0 dx ⎥⎦ ⎣

(13)

and integrated: ⎡

dT ⎤

bT

dT = C1 dx

∫ d ⎢⎣bT dx ⎥⎦ = ∫ 0

(14)

which leads to: (15)

where C1 is a constant of integration. Equation (15) is separated: T dT =

C1 dx b

(16)

and integrated: C1

∫ T dT = b ∫ dx

(17)

T 2 C1 = x + C2 b 2

(18)

which leads to:

e.) Use the boundary conditions from (c) with the solution from (d) in order to obtain two equations in the two unknown constants. Equation (18) is substituted into Eqs. (11) and (12):

TH2 = C2 2

(19)

TC2 C1 = L + C2 b 2

(20)

f.) Type the inputs for the problem and the equations from (e) into EES in order to evaluate the undetermined constants. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" T_H=500 [K] T_C=50 [K] b=1 [W/m-K^2] L=1 [m]

"temperature at left side of wall" "temperature at right side of wall" "coefficient for conductivity function" "thickness of wall"

and Eqs. (19) and (20) are entered in EES: T_H^2/2=C_2 T_C^2/2=C_1*L/b+C_2

"boundary condition at x=0" "boundary condition at x=L"

which leads to C1 = -123750 W/m2 and C2 = 125000 K2. g.) Prepare a plot of the temperature as a function of position in the wall using EES. The solution, Eq. (18), is entered in EES. T^2/2=C_1*x/b+C_2

"solution"

and a plot is prepared using a parametric table that contains the variables x and T. The result is shown in Figure 4, which is qualitatively similar to the sketch in Figure 2.

500 450

Temperature (K)

400 350 300 250 200 150 100 50 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Position (m) Figure 4: Temperature as a function of position.

0.9

1

Problem 1.2-22 You are designing a cubical case that contains electronic components that drive remotely located instruments. You have been asked to estimate the maximum and minimum operating temperature limits that should be used to specify the components within the case. The case is W = 8 inch on a side. The emissivity of the paint used on the case is ε = 0.85. The operation of the electronic components within the case generates between q = 5 and q = 10 W due to ohmic heating, depending on the intensity of the operation. The top surface of the case is exposed to a solar flux q ′′ . All of the surfaces of the case convect (with average heat transfer coefficient h ) and radiate to surroundings at T∞. The case will be deployed in a variety of climates, ranging ′′ = 850 W/m2) to from very hot (T∞,max = 110°F) to very cold (T∞,max = -40°F), very sunny ( qmax ′′ = 0 W/m2), and very windy ( hmax = 100 W/m2-K) to still ( hmin = 5 W/m2-K). For the night ( qmin following questions, assume that the case is at a single, uniform temperature and at steady state. a.) Come up with an estimate for the maximum operating temperature limit. The case temperature will be highest when the case generation is maximum, the ambient temperature is maximum, the solar flux is maximum, and the heat transfer coefficient is minimum. These inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" W=8 [inch]*convert(inch,m) e=0.85 [-] q_dot=10 [W] q``=850 [W/m^2] h_bar=5 [W/m^2-K] T_infinity=converttemp(F,K,120 [F])

"side dimension" "emissivity" "dissipation in case" "solar flux" "heat transfer coefficient" "ambient temperature"

The surface area of the case is:

As = 6W 2

(1)

The resistance to convection from the case is: Rconv = A_s=6*W^2 R_conv=1/(h_bar*A_s)

1 h As

(2) "surface area" "convection resistance"

The radiation resistance cannot be calculated without knowing the surface temperature of the case, T. Therefore, a reasonable value of the surface temperature is assumed. The radiation resistance is:

Rrad =

1 ε σ As (T + T∞2 )(T + T∞ ) 2

T=350 [K] R_rad=1/(e*A_s*sigma#*(T^2+T_infinity^2)*(T+T_infinity))

(3)

"guess for the case temperature" "radiation resistance"

The guess values are updated and the assumed value of the case temperature is commented out. An energy balance on the case leads to: qs + q =

(T − T∞ ) + (T − T∞ )

(4)

qs = W 2 q′′

(5)

Rconv

Rrad

where qs is the absorbed solar irradiation.

{T=350 [K]} q``*W^2+q_dot=(T-T_infinity)/R_conv+(T-T_infinity)/R_rad T_F=converttemp(K,F,T)

"guess for the case temperature" "energy balance on case" "case temperature in F"

which leads to a maximum operating temperature limit of T = 147.5°F. b.) Plot the maximum operating temperature as a function of the case size, W. Explain the shape of your plot (why does the temperature go up or down with W? if there is an asymptotic limit, explain why it exists). The value of W is commented out and a parametric table is generated that includes W and T. Figure 1 illustrates the maximum operating temperature as a function of the size of the enclosure. As the size of the enclosure is reduced, the maximum operating temperature increases because the 10 W of dissipation must be rejected but the area available for convection and radiation is reduced. As the size is increased, the maximum operating temperature reaches an asymptote. The limiting value of T is higher than T∞ because the absorbed solar irradiation and the surface area for convection and radiation both increase in proportion to W2; therefore the limit is consistent with the situation where the solar flux is exactly balanced by the heat flux associated with radiation and convection (the dissipation becomes insignificant relative to the solar flux).

Maximum operating temperature (°F)

230 220 210 200 190 180 170 160 150 140 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Size of enclosure (m) Figure 1: Maximum operating temperature as a function of the size of the enclosure.

c.) Come up with an estimate for the minimum operating temperature limit (with W = 8 inch). The case temperature will be lowest when the case generation is minimum, the ambient temperature is minimum, the solar flux is minimum, and the heat transfer coefficient is maximum. These inputs are entered in EES: q_dot=5 [W] q``=0 [W/m^2] h_bar=100 [W/m^2-K] T_infinity=converttemp(F,K,-40 [F])

"dissipation in case" "solar flux" "heat transfer coefficient" "ambient temperature"

The solution is run again at the predicted temperature is T= -39.7°F d.) Do you feel that the emissivity of the case surface is very important for determining the minimum operating temperature? Justify your answer. The emissivity is not important because radiation is not important. To see this, look at the resistance to convection, Rconv= 0.040 K/W, and the resistance to radiation, Rrad= 1.65 K/W. These two heat transfer mechanisms occur in parallel; the largest resistance in a parallel network is not important - therefore, radiation is much less important than convection.

Problem 1.2-23 Figure P1.2-23 illustrates a cross-sectional view of a water heater. T∞ = 20°C 2 hout = 15 W/m -K thins = 0.5 inch

heater 2 q ′′ = 10,000 W/m ktube = 12 W/m-K Din = 0.75 inch Dout = 0.875 inch

Rc′′ = 1x10 K-m /W -4

2

ε = 0.5 kins = 0.5 W/m-K

T f = 50°C p f = 18 psi 2 hin = 250 W/m -K

Figure P1.2-23: Water heater.

The water flows through a tube with inner diameter Din= 0.75 inch and outer diameter Dout = 0.875 inch. The conductivity of the tube material is ktube = 12 W/m-K. The water in the tube is at mean temperature Tf = 50°C and pressure pf = 18 psi. The heat transfer coefficient between the water and the internal surface of the tube is hin = 250 W/m2-K. A very thin heater is wrapped around the outer surface of the tube. The heater provides a heat transfer rate of q ′′ = 10,000 W/m2. Insulation is wrapped around the heater. The thickness of the insulation is thins = 0.5 inch and the conductivity is kins= 0.5 W/m-K. There is a contact resistance between the heater and the tube and the heater and the insulation. The area specific contact resistance for both interfaces is Rc′′ = 1x10-4 K-m2/W. The outer surface of the insulation radiates and convects to surroundings at T∞ = 20°C. The heat transfer coefficient between the surface of the insulation and the air is hout = 15 W/m2-K. The emissivity of the outer surface of the insulation is ε = 0.5. a.) Draw a resistance network that represents this problem. Label each resistance and clearly indicate what it represents. Show where the heater power enters your network. The resistance network is shown in Figure 2.

Rrad

Rcond,tube

Tf Rconv,in

qin

T∞ = 20°C

Rcond,ins

Rc T Rc h qout

Text

q′′π Dout L

Rconv,out

The resistances include: Rconv,in = convection to water Rcond,tube = conduction through tube Rc = contact resistance Rcond,ins = conduction through insulation Rconv,out = convection resistance to air Rrad = radiation resistance Figure 2: Resistance network.

b.) Using EES, determine the temperature of the heater and the rate of heat transfer to the water. The inputs are entered in EES; note that the problem is done on a per unit length basis, L= 1 m. $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" D_in=0.75 [inch]*convert(inch,m) D_out=0.875 [inch]*convert(inch,m) k_tube=12 [W/m-K] T_f=converttemp(C,K,50 [C]) p_f=18 [psi]*convert(psi,Pa) T_infinity=converttemp(C,K,20 [C]) h_bar_out=15 [W/m^2-K] h_bar_in=250 [W/m^2-K] e=0.5 [-] k_ins=0.5 [W/m-K] th_ins_inch=0.5 [inch] th_ins=th_ins_inch*convert(inch,m) R``=1e-4 [K-m^2/W] q``=10000 [W/m^2] L=1 [m]

"inner diameter" "outer diameter" "conductivity of tube" "fluid temperature" "fluid pressure" "ambient temperature" "heat transfer coefficient to ambient air" "heat transfer coefficient to fluid" "emissivity of surface of insulation" "conductivity of insulation" "thickness of insulation, in inch" "thickness of insulation" "area specific contact resistance" "heat flux provided by heater" "per unit length basis"

The conduction resistance of the tube and the insulation is:

Rcond ,tube

⎛D ⎞ ln ⎜ out ⎟ D = ⎝ in ⎠ 2 π ktube L

(1)

⎛ D + 2 thins ⎞ ln ⎜ out ⎟ Dout ⎝ ⎠ = 2 π kins L

Rcond ,ins

(2)

The contact resistance is:

Rc =

R′′ π Dout L

(3)

The convection resistance between the water and the tube surface is:

1 hin π Din L

(4)

1 hout π ( Dout + 2 thins ) L

(5)

Rconv ,in =

Rconv ,out =

R_cond_tube=ln(D_out/D_in)/(2*pi*k_tube*L) R_cond_ins=ln((D_out+2*th_ins)/D_out)/(2*pi*k_ins*L) R_c=R``/(pi*D_out*L) R_conv_in=1/(h_bar_in*pi*D_in*L) R_conv_out=1/(h_bar_out*pi*(D_out+2*th_ins)*L)

"tube conduction resistance" "insulation conduction resistance" "contact resistance" "internal convection resistance" "external convection resistance"

In order to calculate the radiation resistance, the external surface temperature Text, is assumed. The radiation resistance is:

Rrad =

1 ε σ π ( Dout + 2 thins ) L (Text2 + T∞2 )(Text + T∞ )

(6)

T_ext=T_infinity "guess for external surface temperature" R_rad=1/(e*sigma#*pi*(D_out+2*th_ins)*L*(T_ext^2+T_infinity^2)*(T_ext+T_infinity)) "radiation resistance"

An energy balance on the heater is: q ′′ π Dout =

(Th − T∞ ) ⎡ 1 1 ⎤ Rc + Rcond ,ins + ⎢ + ⎥ ⎣ Rrad Rconv ,out ⎦

−1

+

(T

h

− Tf

)

Rc + Rcond ,tube + Rconv ,in

q``*pi*D_out*L=(T_h-T_infinity)/(R_c+R_cond_ins+(1/R_rad+1/R_conv_out)^(-1))+& (T_h-T_f)/(R_c+R_cond_tube+R_conv_in) "energy balance on heater"

(7)

The problem is solved and the guess values are updated. The assumed value of Text is commented out and then recalculated based on the solution. The rate of heat transfer to the ambient is: qout =

(Th − T∞ ) ⎡ 1 1 ⎤ Rc + Rcond ,ins + ⎢ + ⎥ ⎣ Rrad Rconv ,out ⎦

−1

(8)

and the external surface temperature is: Text = Th − qout ( Rc + Rcond ,ins )

(9)

{T_ext=T_infinity} "guess for external surface temperature" q_dot_out=(T_h-T_infinity)/(R_c+R_cond_ins+(1/R_rad+1/R_conv_out)^(-1)) "rate of heat transfer to ambient" T_ext=T_h-q_dot_out*(R_c+R_cond_ins) "recalculate external temperature" T_h_C=converttemp(K,C,T_h) "heater temperature, in C"

which leads to Th = 90.85ºC. The rate of heat transfer to the water is computed: qin =

(T

h

− Tf

)

Rc + Rcond ,tube + Rconv ,in

q_dot_in=(T_h-T_f)/(R_c+R_cond_tube+R_conv_in)

(10)

"rate of heat transfer to fluid"

which leads to qin = 581 W. c.) What is the efficiency of the heater (the ratio of the power provided to the water to the power provided to the heater)? The efficiency is defined as:

η= eta=q_dot_in/(q_dot_out+q_dot_in)

qin qin + qout

(11)

"efficiency of heater"

which leads to η = 83.2%. d.) The efficiency of the heater is less than 100% due to heat lost to the atmosphere. Rank the following parameters in terms of their relative importance with respect to limiting heat loss to the atmosphere: ε, Rc′′ , kins, hout . Justify your answers using your resistance network and a discussion of the magnitude of the relevant resistances.

The resistances separating the heater from the ambient include: Rrad = 1.89 K/W, Rc = 0.0014 K/W, Rcond,ins = 0.24 K/W, and Rconv,out = 0.45 K/W. This suggests that convection is more important than radiation and also more important than either conduction or contact resistance. Conduction is the next-most important resistance followed by radiation and finally contact resistance (which is absolutely unimportant). Therefore, the relative importance is: hout , kins, ε, and Rc′′ . e.) Plot the efficiency as a function of the insulation thickness for 0 inch < thins < 1.5 inch. Explain the shape of your plot. Figure 3 illustrates the efficiency as a function of the insulation thickness. Notice that initially as the thickness increases the efficiency actually drops; this is because the convection resistance (which, from (d), is the most important resistance) will decrease with insulation thickness since the area for convection increases. Eventually, the insulation conduction resistance increases to the point where it becomes important and the efficiency begins to increase. 0.852 0.848

Efficiency

0.844 0.84 0.836 0.832 0.828 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Thickness of insulation (inch) Figure 3: Efficiency as a function of the insulation thickness.

f.) The temperature on the internal surface of the tube must remain below the saturation temperature of the water in order to prevent any local boiling of the water. Based on this criteria, determine the maximum possible heat flux that can be applied to the heater (for thins = 0.5 inch). The internal surface temperature of the tube is computed: Ts ,in = Th − qin ( Rc + Rcond ,tube )

(12)

The saturation temperature of the water (Tsat) is obtained from EES' internal thermodynamic property routines. The prescribed value of the heat flux is cancelled out and, instead, Ts,in is set equal to Tsat.

{q``=10000 [W/m^2]} T_s_in=T_h-q_dot_in*(R_c+R_cond_tube) T_sat=temperature(Water,p=p_f,x=1 [-]) T_s_in=T_sat

which leads to q ′′ = 14,061 W/m2.

"heat flux provided by heater" "inner surface temperature of tube" "saturation temperature" "maximum allowable temperature"

Problem 1.3-1: Composite Wall with Generation A plane wall is composed of two materials, A and B, with the same thickness, L, as shown in Figure P1.3-1. The same, spatially uniform, volumetric rate of generation is present in both materials ( g ′′′ = g ′′′A = g ′′′B ) and the wall is at steady state. The conductivity of material B is twice that of material A (kB = 2 kA). The left side of the wall is adiabatic and the right side is maintained at a temperature Tx=2L = To. L

L

material A

material B

kA g ′′′A

kB = 2kA g ′′′B = g ′′′A

To

x

Figure P1.3-1: Plane wall composed of materials A and B.

a.) Sketch the rate of heat transfer as a function of position within the wall ranging from x = 0 (the left face of material A) to x = 2L (the right face of material B). Note that the sketch should be qualitatively correct, but cannot be quantitative as you have not been given any numbers for the problem. The heat transfer rate must increase linearly from zero at x=0; to see this, consider the energy balance on the control volume shown in Figure 2.

Figure 2: Control volume.

The heat transfer rate at any position x must be:

q = g ′′′ A x

(1)

Figure 3: Rate of heat transfer as a function of position in the wall

b.) Sketch the temperature as a function of position within the wall. Again, be sure that your sketch has the correct qualitative features. The temperature gradient is, according to Fourier’s law:

dT q =− dx k

(2)

Therefore, the temperature gradient will become increasingly negative as you move towards positive x; however, there will be a step change in the temperature gradient at the interface between materials A and B (i.e., at x=L). The conductivity of material B is twice that of A and so the temperature gradient at x=L+ will be half that of x=L-. T

To

0

L

2L

x

Figure 4: Temperature as a function of position in the wall

Problem 1.3-2: Composite Wall with Generation A plane wall is composed of two materials, A and B, with the same conductivity k and thickness L, as shown in Figure P1.3-2. The left side of material A is adiabatic (i.e., well insulated) and the right side of material B is held at a temperature TL. There is no volumetric generation in material A but material B experiences a uniform rate of volumetric generation of thermal energy, g ′′′ . L material A kA ′′′ g A = 0

L material B kB = kA = g ′′′ g ′′′ B

TL

x

Figure P1.3-2: Plane wall composed of materials A and B.

a.) Sketch the rate of heat transfer ( q ) as a function of position within the wall. Note that the sketch should be qualitatively correct but cannot be quantitative as you have not been given any numbers for the problem. The control volume shown in Figure 2 can be used to evaluate the heat transfer from 0 < x < L.

Figure 2: Control volume.

The heat transfer rate must be zero until x = L:

q = 0 for 0 < x < L

(1)

The control volume shown in Figure 3 can be used to evaluate the heat transfer from L < x < 2 L.

Figure 3: Control volume.

The heat transfer rate must increase linearly from x = L to x= 2 L: q = g ′′′ A ( x − L ) for L < x < 2 L

(2)

The heat transfer rate is sketched in Figure 4.

Figure 4: Rate of heat transfer as a function of position in the wall

b.) Sketch the temperature as a function of position within the wall. Again, be sure that your sketch has the correct qualitative features. The temperature gradient is given by Fourier’s law: dT q =− dx k

(3)

Therefore, the temperature gradient will be zero within material A and then become increasingly negative as you move towards positive x within material B. Figure 5 shows a sketch of the temperature distribution.

Figure 5: Temperature as a function of position in the wall

Problem 1.3-3 (1-7 in text): Critical Evaluation of a Solution One of the engineers that you supervise has been asked to simulate the heat transfer problem shown in Figure P1.3-3(a). This is a 1-D, plane wall problem (i.e., the temperature varies only in the x-direction and the area for conduction is constant with x). Material A (from 0 < x < L) has conductivity kA and experiences a uniform rate of volumetric thermal energy generation, g ′′′ . The left side of material A (at x = 0) is completely insulated. Material B (from L < x < 2L) has lower conductivity, kB < kA. The right side of material B (at x= 2L) experiences convection with fluid at room temperature (20°C). Based on the facts above, critically examine the solution that has been provided to you by the engineer and is shown in Figure P1.3-3(b). There should be a few characteristics of the solution that do not agree with your knowledge of heat transfer; list as many of these characteristics as you can identify and provide a clear reason why you think the engineer’s solution must be wrong. 250 200

L

L

material A

material B

kA

kB < kA

g ′′′A = g ′′′

g B′′′ = 0

x

h , T f = 20°C

Temperature (°C)

150 100 50 0 -50 Material A -100

0

Material B L

2L

Position (m)

(a) (b) Figure P1.3-3: (a) Heat transfer problem and (b) "solution" provided by the engineer.

1. The left side of material A is insulated; therefore, the temperature gradient should be zero. 2. Material A has a higher conductivity than material B; therefore, at x = L the temperature gradient should be larger in material B than in material A. 3. Heat is transferred to the fluid at 20°C; therefore the temperature at x = 2 L must be greater than 20°C.

Problem 1.3-4: Cylinder Boundary Conditions A cylinder with conductivity k experiences a uniform rate of volumetric generation g ′′′ , as shown in Figure P1.3-4. The cylinder experiences 1-D, steady state conduction heat transfer in the radial direction and therefore the general solution to the ordinary differential equation for temperature (T) is:

g ′′′ r 2 T =− + C1 ln ( r ) + C2 4k

(1)

where r is the radial location and C1 and C2 are undetermined constants. At the inner radius of the cylinder (r = rin), a heater applies a uniform rate of heat transfer, qin . At the outer radius of the cylinder (r = rout), the temperature is fixed at Tout. The length of the cylinder is L. Write the two algebraic equations that can be solved in order to obtain the constants C1 and C2. Your equations must contain only the following symbols in the problem statement: qin , Tout, k, rin, rout, L, g ′′′ , C1, and C2. Do not solve these equations.

qin L

k , g ′′′

Tout rin rout Figure P1.3-4: Cylinder with uniform volumetric generation.

At the outer surface, the temperature is specified and therefore the boundary condition is:

Tout = −

2 g ′′′ rout + C1 ln ( rout ) + C2 4k

(2)

At the inner surface, the temperature is not specified and therefore it is necessary to do an energy balance on this interface, as shown in Figure 2.

Figure 2: Interface balance at r = rin. The interface energy balance is:

qin = q@ r = rin

(3)

Substituting Fourier's law for q@ r = rin leads to:

qin = − k 2 π rin L

dT dr

(4) r = rin

Substituting the general solution, Eq. (1), into Eq. (4) leads to: ⎡ g ′′′ rin C1 ⎤ qin = −k 2 π rin L ⎢ − + ⎥ rin ⎦ ⎣ 2k

(5)

Problem 1.3-5: Windings Figure P1.3-5(a) illustrates a motor that is constructed using windings that surround laminated iron poles. You have been asked to estimate the maximum temperature that will occur within the windings. The windings and poles are both approximated as being cylindrical, as shown in Figure P1.3-5(b).

pole

rout = 2 cm rin = 1 cm

L = 2 cm r

windings Tpole = 50°C

Tair = 25°C 2 h = 25 W/m -K windings k = 1 W/m-K 6 3 g ′′′ = 1x10 W/m

(a) (b) Figure P1.3-5: (a) Concentrated winding for a permanent magnet motor, (b) cylindrical model of the windings

The windings are a complicated composite formed from copper conductor, insulation and air that fills the gaps between adjacent wires. However, in most models, the windings are represented by a solid with equivalent properties that account for this underlying structure. You can therefore consider the windings in Figure P1.3-5(b) to be a solid. The electrical current in the windings causes an ohmic dissipation that can be modeled as a uniform volumetric generation rate of g ′′′ =1x106 W/m3. The conductivity of the windings is k = 1.0 W/m-K. The inner radius of the windings is rin = 1.0 cm and the outer radius is rout = 2.0 cm. The windings are L = 2.0 cm long and the upper and lower surfaces may be assumed to be insulated so that the temperature in the windings varies only in the radial direction. The stator pole is conductive and cooled externally; therefore, you can assume that the stator tooth has a uniform temperature of Tpole = 50°C. Neglect any contact resistance between the inner radius of the winding and the pole; therefore, the temperature of the windings at r = rin is Tpole. The outer radius of the windings is exposed to air at Tair = 20°C with a heat transfer coefficient of h = 25 W/m2-K. a.) Derive the governing differential equation for the temperature within the windings (i.e., the differential equation that is valid from rin < r < rout). You should end up with an ordinary differential equation for T in terms of the symbols provided in the problem statement. Clearly show your steps, which should include: 1. define a differentially small control volume, 2. do an energy balance on your control volume, 3. expand the r + dr terms in your energy balance and take the limit as dr → 0, 4. substitute rate equations into your energy balance. A differential control volume is shown in Figure 3.

qr

g

qr + dr dr

r

Figure 3: A differentially small control volume.

The energy balance on the control volume is:

qr + g = qr + dr

(1)

or, after expanding the r + dr term:

qr + g = qr +

dqr d 2 q dr 2 d 3 qr dr 3 dr + 2r + + ... dr dr 2! dr 3 3!

(2)

Taking the limit as dr approaches zero leads to:

qr + g = qr +

dqr dr dr

(3)

or

g =

dqr dr dr

(4)

The parameter g is the rate of thermal energy generation within the control volume, which can be expressed as the product of the volume of the control volume and the volumetric rate of generation:

g = g ′′′ 2 π r L dr

(5)

and the conduction term is expressed using Fourier’s law: dT dr

(6)

d ⎡ dT ⎤ −k 2 π r L dr ⎢ dr ⎣ dr ⎥⎦

(7)

q = − k 2 π r L Substituting Eqs. (5) and (6) into Eq. (4) leads to:

g ′′′ 2 π r L dr =

or g ′′′ r = −k

d ⎡ dT ⎤ r dr ⎢⎣ dr ⎥⎦

(8)

b.) Specify the boundary conditions for your differential equation. This should be easy at the inner radius, where the temperature is specified, but you will need to carry out an interface energy balance at r = rout. The boundary condition at the inner radius is: Tr = rin = Tpole

(9)

An interface energy balance at the outer radius is shown in Figure 4.

qr = r

qconv

out

Figure 4: An interface energy balance at the outer radius.

The interface energy balance leads to: qr = rout = qconv

(10)

Substituting Fourier's law and Newton's law of cooling into Eq. (10) leads to: − k 2 π rout L

dT dr

r = rout

dT dr

r = rout

(

= 2 π rout L h Tr = rout − Tair

)

(11)

or −k

(

= h Tr = rout − Tair

)

(12)

c.) Solve the governing differential equation that you derived in part (a) by integrating twice. You should end up with a solution that involves two constants of integration, C1 and C2. Equation (8) is rearranged:

g ′′′ r ⎡ dT ⎤ d⎢r dr = − k ⎣ dr ⎥⎦

(13)

and integrated:

⎡ dT ⎤

⎛ g ′′′ r ⎞ ⎟ dr k ⎠

∫ d ⎣⎢ r dr ⎥⎦ = ∫ ⎝⎜ −

(14)

to obtain:

r

dT g ′′′ r 2 =− + C1 dr 2k

(15)

dT g ′′′ r C1 =− + dr 2k r

(16)

so the temperature gradient is:

Equation (16) is again rearranged: ⎛ g ′′′ r C1 ⎞ + ⎟ dr 2k r ⎠

∫ dT = ∫ ⎜⎝ −

(17)

to obtain: T =−

g ′′′ r 2 + C1 ln ( r ) + C2 4k

(18)

d.) Substitute your answer from part (c) into the boundary conditions specified in part (b) to obtain two equations for your two unknown constants of integration, C1 and C2. Substituting Eq. (18) into Eq. (9) leads to:

g ′′′ rin2 − + C1 ln ( rin ) + C2 = Tpole 4k

(19)

Substituting Eqs. (18) and (16) into Eq. (12) leads to: 2 ⎛ g ′′′ rout C1 ⎞ ⎛ g ′′′ rout ⎞ −k ⎜ − + + C1 ln ( rout ) + C2 − Tair ⎟ ⎟ = h⎜− rout ⎠ 2k 4k ⎝ ⎠ ⎝

(20)

e.) Implement your results from (c) and (d) in EES and prepare a plot of the temperature in the stator as a function of radius. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" k=1 [W/m-K] gv=1e6 [W/m^3] T_air=converttemp(C,K,25) h=25 [W/m^2-K] r_out=2.0 [cm]*convert(cm,m) r_in=1.0 [cm]*convert(cm,m) T_pole=converttemp(C,K,50) L=2.0 [cm]*convert(cm,m)

"winding conductivity" "winding volumetric generation" "air temperature" "heat transfer coefficient" "outer radius of windings" "inner radius of windings" "pole temperature" "length of windings"

The two algebraic equations for C1 and C2, Eqs. (19) and (20), are entered: "boundary conditions" -gv*r_in^2/(4*k)+C_1*ln(r_in)+C_2=T_pole -k*(-gv*r_out/(2*k)+C_1/r_out)=h*(-gv*r_out^2/(4*k)+C_1*ln(r_out)+C_2-T_air)

"at r=r_in" "at r=r_out"

and the solution is obtained using Eq. (18). "solution" T=-gv*r^2/(4*k)+C_1*ln(r)+C_2 r_cm=r*convert(m,cm) T_C=converttemp(K,C,T)

Figure 5 illustrates the temperature as a function of radial position within the windings. 95

Temperature (°C)

90 85 80 75 70 65 60 55 50 1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

Radius (cm) Figure 5: Temperature as a function of position within the windings.

2

PROBLEM 1.3-6: Heating a Pipe You need to transport water through a pipe from one building to another in an arctic environment, as shown in Figure P1.3-6. The water leaves the building very close to freezing, at Tw = 5°C, and is exposed to a high velocity, very cold wind. The temperature of the surrounding air is Ta = -35°C and the heat transfer coefficient between the outer surface of the pipe and the air is ha = 50 W/m2-K. The pipe has an inner radius of rh,in = 2 inch and an outer radius of rh,out = 4 inch and is made of a material with a conductivity kh = 5 W/m-K. The heat transfer coefficient between the water and the inside surface of the pipe is very large and therefore the inside surface of the pipe can be assumed to be at the water temperature. Neglect radiation from the external surface of the pipe. rh,out = 4 inch rh,in = 2 inch

Ta = −35°C 2 ha = 50 W/m -K

water Tw = 5°C

kh = 5 W/m-K Figure P1.3-6: Heated pipe transporting near freezing water in an arctic environment.

a.) Determine the rate of heat lost from the water to the air for a unit length, L=1 m, of pipe. The known information is converted to base SI units and entered in EES. The units for each variable are entered in the Variable Information window. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_h_in=2 [inch]*convert(inch,m) r_h_out=4.0 [inch]*convert(inch,m) T_a=converttemp(C,K,-35) T_w=converttemp(C,K,5) h_a=50 [W/m^2-K] k_h=5 [W/m-K] L=1 [m]

"inner radius of plastic pipe" "outer radius of heater" "air temperature" "water temperature" "air to heater heat transfer coefficient" "heater conductivity" "unit length of pipe"

The resistance network that represents the situation includes a conduction resistance through the pipe (Rcond) and a convection resistance from the outer surface of the pipe (Rconv).

Rcond

⎛r ⎞ ln ⎜⎜ h ,out ⎟⎟ rh ,in ⎠ = ⎝ 2 π kh L

(1)

and Rconv =

1 ha 2 π rh ,out L

(2)

and the total heat loss is:

q =

(Tw − Ta ) Rcond + Rconv

(3)

These equations are entered in EES: "Part a" R_cond=ln(r_h_out/r_h_in)/(2*pi*k_h*L) R_conv=1/(h_a*2*pi*r_h_out*L) q_dot=(T_w-T_a)/(R_cond+R_conv)

"conduction through the pipe" "convection from outer surface" "heat loss"

The heat loss is 750 W per m of pipe. b.) Plot the heat lost from the water as a function of the pipe outer radius from 0.06 m to 0.3 m (keep the same inner radius for this study). Explain the shape of your plot. The value of r_h_out in the input section is commented (highlight the variable and select comment). A parametric table is created that includes the variables r_h_out and q_dot; the value of r_h_out in the table is varied from 0.01 m to 0.3 m (right click on the r_h_out column header and select Alter Values). The results in the parametric table are plotted in Figure 2.

Figure 2: Heat transfer from the water as a function of the outer radius of the pipe.

The result can be understood by plotting the resistances R_conv, R_cond, and R_conv+R_cond as a function of the outer radius of the pipe (as shown in Figure 3). Not that you’ll need to define a new variable, R_total:

R_total=R_cond+R_conv

"total resistance"

in order to make the plot.

Figure 3: Thermal resistance as a function of the pipe outer radius.

Initially, the convection resistance drops more rapidly than the conduction resistance increases as the radius increases; therefore, initially the total resistance is reduced with radius and the heat loss increases. Eventually, the conduction resistance dominates the convection resistance and therefore the total resistance rises with outer radius and therefore the heat loss is reduced. In order to reduce the heat loss from the water and therefore prevent freezing, you run current through the pipe material so that it generates thermal energy at with a uniform volumetric rate, g ′′′ = 5x105 W/m3. c.) Develop an analytical model capable of predicting the temperature distribution within the pipe. Implement your model in EES. The general solution for the temperature distribution in a cylindrical system with a constant rate of generation is:

T =−

g ′′′ r 2 + C1 ln ( r ) + C2 4k

(4)

where C1 and C2 are constants of integration that must come from the boundary conditions. The boundary conditions are a set temperature at rh,in:

Tw = − and an interface energy balance at rh,out:

g ′′′ rh2,in 4 kh

+ C1 ln ( rh ,in ) + C2

(5)

− kh

dT dr

r = rh ,out

(

= ha Tr = rh ,out − Ta

)

(6)

The derivative of temperature with respect to radius can be found by manipulating Eq. (4) or in Table 3-1: dT g ′′′ r C1 =− + dr 2k r

(7)

Substituting Eq. (7) and Eq. (4) into Eq. (6) leads to: ⎡ g ′′′ rh ,out ⎛ g ′′′ rh2,out ⎞ C1 ⎤ − kh ⎢ − + + C1 ln ( rh ,out ) + C2 − Ta ⎟⎟ ⎥ = ha ⎜⎜ − rh ,out ⎥⎦ 2 kh 4 kh ⎢⎣ ⎝ ⎠

(8)

Equations (5) and (8) are programmed in EES to obtain the constants C1 and C2. g```_dot=5e5 [W/m^3] "volumetric generation rate" "boundary conditions" T_w=-g```_dot*r_h_in^2/(4*k_h)+C1*ln(r_h_in)+C2 -k_h*(-g```_dot*r_h_out/(2*k_h)+C1/r_h_out)=h_a*(-g```_dot*r_h_out^2/(4*k_h)+C1*ln(r_h_out)+C2-T_a)

Note that the units of the constant should be set to K (Figure 4):

Figure 4: Variable Information window.

However, if you check units now you will obtain a unit error (Figure 5).

Figure 5: Unit error.

because the argument of the natural logarithm has units m. This cannot be helped; if the algebra associated with explicitly solving for each constant was followed through and these constants were substituted back into Eq. (4) then you would find that the arguments of the natural logarithm can be expressed as the ratio of two radii. The solution, Eq. (4), is programmed in EES: T=-g```_dot*r^2/(4*k_h)+C1*ln(r)+C2

"solution"

d.) Prepare a plot showing the temperature as a function of position within the pipe. The radius is expressed in terms of a non-dimensional radius (r_bar) to facilitate making the parametric table (it is easier to vary r_bar from 0 to 1 than it is to vary r from r_h_in to r_h_out, particularly if you plan on varying r_h_in or r_h_out). r_bar=(r-r_h_in)/(r_h_out-r_h_in) {r_bar=0} T_C=converttemp(K,C,T)

"dimensionless radius used to make plots"

The temperature distribution is shown in Figure 6.

Figure 6: Temperature in pipe as a function of radius.

e.) Calculate the heat transfer from the water when you are heating the pipe. The heat transfer from the water is obtained by applying Fourier’s law at r = rh,in: qw = − kh 2 π rh ,in L

dT dr

(9) r = rh ,in

Substituting Eq. (7) into Eq. (9):

⎡ g ′′′ rh ,in C1 ⎤ qw = −kh 2 π rh ,in L ⎢ − + ⎥ rh ,in ⎥⎦ 2k ⎢⎣ q_dot_w=-k_h*(-g```_dot*r_h_in/(2*k_h)+C1/r_h_in)*2*pi*r_h_in*L

(10)

"heat transfer from the water"

The heat transfer rate from the water is -8336 W per m of pipe; that is, heat is transferred to the water, which is evident from the temperature distribution. While you don’t want the water to freeze, you probably also don’t want to heat it and therefore g ′′′ =5e5 W/m3 is probably too high. f.) Determine the volumetric generation rate that is required so that there is no heat transferred from the water. EES can provide this solution. Simply comment out the generation rate that you set and then specify that the heat transfer rate from the water is 0. {g```_dot=5e5 [W/m^3]} "volumetric generation rate" "boundary conditions" T_w=-g```_dot*r_h_in^2/(4*k_h)+C1*ln(r_h_in)+C2 -k_h*(-g```_dot*r_h_out/(2*k_h)+C1/r_h_out)=h_a*(-g```_dot*r_h_out^2/(4*k_h)+C1*ln(r_h_out)+C2) T=-g```_dot*r^2/(4*k_h)+C1*ln(r)+C2 r_bar=(r-r_h_in)/(r_h_out-r_h_in) plots" r_bar=0 T_C=converttemp(K,C,T)

"solution" "dimensionless radius used to make

q_dot_w=-k_h*(-g```_dot*r_h_in/(2*k_h)+C1/r_h_in)*2*pi*r_h_in*L "heat transfer from the water" q_dot_w=0

The volumetric generation rate that results in no heat transfer to the water is 4.12x104 W/m3.

Problem 1.3-7: Nuclear Fuel Element Figure P1.3-7 illustrates a spherical, nuclear fuel element which consists of a sphere of fissionable material (fuel) with radius rfuel = 5 cm and kfuel = 1 W/m-K that is surrounded by a spherical shell of metal cladding with outer radius rclad = 7 cm and kclad = 300 W/m-K. The outer surface of the cladding is exposed to helium gas that is being heated by the reactor. The convection coefficient between the gas and the cladding surface is hgas = 100 W/m2-K and the temperature of the gas is Tgas = 500ºC. Neglect radiation heat transfer from the surface. Inside the fuel element, fission fragments are produced which have high velocities. The products collide with the atoms of the material and provide the thermal energy for the reactor. This process can be modeled as a volumetric source of heat generation in the material that is not uniform throughout the fuel. The volumetric generation ( g ′′′ ) can be approximated by the function:

⎛ r g ′′′ = g e′′′ ⎜ ⎜ rfuel ⎝

⎞ ⎟⎟ ⎠

b

where g e′′′ = 5x105 W/m3 is the volumetric rate of heat generation at the edge of the sphere and b = 1.0; note that the parameter b is a dimensionless positive constant that characterizes how quickly the generation rate increases in the radial direction. fissionable material kfuel = 1 W/m-K rfuel = 5 cm rclad = 7 cm

g ′′′

2 hgas = 100 W/m -K Tgas = 500°C

cladding kclad = 300 W/m-K

Figure P1.3-7: Spherical fuel element surrounded by cladding

a.) Enter the problem inputs into EES; be sure to set the units appropriately. The inputs are entered according to: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_fuel=5[cm]*convert(cm,m) k_fuel=1 [W/m-K] r_clad=7[cm]*convert(cm,m) k_clad=300 [W/m-K] h_gas=100 [W/m^2-K] T_gas=converttemp(C,K,500)

"radius of fuel" "conductivity of fuel" "cladding radius" "cladding conductivity" "convection coefficient" "gas temperature"

gve=5e5 [W/m^3] b=1 [-]

"generation at the center" "decay constant"

b.) Determine the governing differential equation that applies within the sphere (i.e., your differential equation should be valid for 0 < r < rfuel). The differential equation should include only those symbols given in the problem statement. Clearly show your steps. A differential control volume is shown in Figure 2 and includes conduction at r and r+dr at the inner and outer surfaces of the spherical shell as well as generation within the enclosed volume.

Figure 2: Differential control volume

The energy balance suggested by Figure 2 is:

qr + g = qr + dr

(1)

The term at r + dr can be expanded:

qr + dr = qr +

dqr dr dr

(2)

dqr dr dr

(3)

and substituted into Eq. (1):

qr + g = qr + and simplified:

g =

dqr dr dr

(4)

The conduction is governed by Fourier’s Law:

qr = −k fuel 4 π r 2

dT dr

and the generation is the product of the volume and the local generation rate:

(5)

⎛ r g = 4 π r dr g e′′′⎜ ⎜r ⎝ fuel 2

⎞ ⎟⎟ ⎠

b

(6)

The rate equations are substituted into Eq. (4): ⎛ r 4 π r dr g e′′′⎜ ⎜r ⎝ fuel 2

b

⎞ d ⎡ 2 dT ⎤ dr ⎟⎟ = ⎢ −k fuel 4 π r dr ⎣ dr ⎥⎦ ⎠

(7)

which can be simplified:

d ⎡ 2 dT ⎤ 2 g e′′′ ⎛ r +r r ⎜ dr ⎢⎣ dr ⎥⎦ k fuel ⎜⎝ rfuel

b

⎞ ⎟⎟ = 0 ⎠

(8)

Notice that it is not possible to cancel the r2 term from each side of Eq. (7) because it appears within the differential on the right side. c.) Enter the governing differential equation into Maple and use Maple to obtain a solution that includes two constants of integration. The generation function and the governing differential equation are entered according to: > gen:=gve*(r/r_fuel)^b;

r ⎞ gen := gve ⎛⎜⎜ ⎟⎟ r_fuel ⎝ ⎠

b

> GDE:=diff(r^2*diff(T(r),r),r)+r^2*gen/k_fuel=0;

r r gve ⎛⎜⎜ 2 d d r_fuel ⎞ ⎛ ⎝ GDE := 2 r ⎛⎜⎜ T( r ) ⎞⎟⎟ + r 2 ⎜⎜ 2 T( r ) ⎟⎟ + d r k_fuel ⎝ ⎠ ⎝ dr ⎠ 2

b

⎞ ⎟⎟ ⎠ =0

and solved using the dsolve command: > Tr:=rhs(dsolve(GDE)); b

r ⎞ r 2 gve ⎛⎜⎜ ⎟⎟ _C1 r_fuel ⎝ ⎠ Tr := − − + _C2 r k_fuel ( b2 + 6 + 5 b )

Notice the two constants of integration that must be determined using the boundary conditions. d.) The boundary condition at the center of the sphere is that the temperature must remain finite; this should eliminate one of the constants of integration in your Maple solution. Which constant must be zero?

In order for the temperature to remain finite as r approaches 0, the constant C1 must be zero. e.) Determine a symbolic equation for the remaining boundary condition (the one at r = rfuel) in terms of the temperature and temperature gradient evaluated at r = rfuel. An interface energy balance at r = rfuel includes conduction from the fuel and conduction into the cladding, as shown in Figure 3.

Figure 3: Interface balance at r = rfuel

The energy balance suggested by Figure 3 is:

qr = rruel = qout

(9)

The conduction term on the left side of Eq. (9) is evaluated using Fourier’s law: 2 qr = rruel = − k fuel 4 π rfuel

dT dr

(10) r = r fuel

while the heat transfer out of the cladding is driven by the difference between the temperature at interface between the fuel and the cladding and the temperature of the surrounding gas. The heat transfer is resisted by the sum of the conduction resistance of the cladding (Rclad): Rclad =

1 4 π kclad

⎡ 1 1 ⎤ − ⎢ ⎥ ⎢⎣ rfuel rclad ⎥⎦

(11)

and the convection resistance (Rconv): Rconv =

1 4π r

2 clad

hgas

(12)

so that:

qout =

Trfuel − Tgas

(13)

Rclad + Rconv

Subsituting Eqs. (13) and (10) into Eq. (9) leads to: 2 − k fuel 4 π rfuel

dT dr

= r = r fuel

Trfuel − Tgas

(14)

Rclad + Rconv

Equation (14) provides a single equation for the unknown constant of integration, C2. f.) Use the expressions from Maple to determine the required constant of integration within EES. Copy the solution for the temperature in the cladding from Maple and paste it into EES; modify the expression as necessary for compatibility (remember to eliminate the C1 term) and use it to generate plot of temperature vs radius within the cladding. The solution in Maple is manipulated using the diff and eval commands: > dTdr_rfuel:=eval(diff(Tr,r),r=r_fuel);

dTdr_rfuel :=

_C1 2 r_fuel gve r_fuel gve b − − 2 2 r_fuel k_fuel ( b + 6 + 5 b ) k_fuel ( b2 + 6 + 5 b )

> T_rfuel=eval(Tr,r=r_fuel);

T_rfuel = −

_C1 r_fuel 2 gve − + _C2 r_fuel k_fuel ( b2 + 6 + 5 b )

The expressions are copied and pasted into EES; the constant C1 is eliminated and the expressions are modified to be compatible with EES (the := is replaced with = and the _C2 is replaced with C2). Also, the _C1 portion of the expression for T_rfuel is deleted. "Boundary condition expressions" dTdr_rfuel =-2*r_fuel*gve/k_fuel/(b^2+6+5*b)-r_fuel*gve/k_fuel/(b^2+6+5*b)*b T_rfuel =-r_fuel^2*gve/k_fuel/(b^2+6+5*b)+C2

"from Maple" "from Maple"

The two resistance values must be calculated: Rst_clad=(1/r_fuel-1/r_clad)/(4*pi*k_clad) Rst_conv=1/(4*pi*r_clad^2*h_gas)

"conduction resistance of cladding" "convection resistance"

Note that the use of R_clad to represent the cladding resistance would have resulted in problems because of the existence of the variable r_clad; EES is not case-sensitive. Finally, the boundary condition, Eq. (14), is programmed: -4*pi*r_fuel^2*k_fuel*dTdr_rfuel=(T_rfuel-T_gas)/(Rst_clad+Rst_conv)

"boundary condition"

Solving the problem should provide a solution for C2 = -1916 K; note that the units should also be set and checked for all of your variables. The Maple solution is cut and pasted into EES: "Solution" T=-r^2*gve/k_fuel/(b^2+6+5*b)*(r/r_fuel)^b+C2 "solution from Maple"

To facilitate plotting, the solution is converted from K to °C and the radial location is defined in terms of a dimensionless radial position (r_bar) that goes from 0 to 1 (therefore, if the radius of the fuel sphere changes in parametric studies it is not necessary to reset the parametric table). r_bar=r/r_fuel T_C=converttemp(K,C,T)

Figure 4 illustrates the temperature as a function of radius

Figure 4: Temperature distribution in the fuel

PROBLEM 1.3-8 (1-8 in text): Hay Temperature Freshly cut hay is not really dead; chemical reactions continue in the plant cells and therefore a small amount of heat is released within the hay bale. This is an example of the conversion of chemical to thermal energy and can be thought of as thermal energy generation. The amount of thermal energy generation within a hay bale depends on the moisture content of the hay when it is baled. Baled hay can become a fire hazard if the rate of volumetric generation is sufficiently high and the hay bale sufficiently large so that the interior temperature of the bale reaches 170°F, the temperature at which self-ignition can occur. Here, we will model a round hay bale that is wrapped in plastic to protect it from the rain. You may assume that the bale is at steady state and is sufficiently long that it can be treated as a one-dimensional, radial conduction problem. The radius of the hay bale is Rbale = 5 ft and the bale is wrapped in plastic that is tp = 0.045 inch thick with conductivity kp = 0.15 W/m-K. The bale is surrounded by air at T∞ = 20°C with h = 10 W/m2-K. You may neglect radiation. The conductivity of the hay is k = 0.04 W/m-K. a.) If the volumetric rate of thermal energy generation is constant and equal to g ′′′ = 2 W/m3 then determine the maximum temperature in the hay bale. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in R_bale=5 [ft]*convert(ft,m) t_p=0.045 [inch]*convert(inch,m) k_p= 0.15 [W/m-K] h=10 [W/m^2-K] T_infinity=converttemp(C,K,20 [C]) L=1 [m] k = 0.04 [W/m-K] g_dot_v=2 [W/m^3]

"hay bale radius" "plastic thickness" "plastic conductivity" "heat transfer coefficient" "ambient temperature" "per unit length of bale" "conductivity of hay" "volumetric heat generation"

This is an example of a one-dimensional steady conduction problem with constant volumetric generation and therefore the formulae provided in Table 1-3 can used directly. The general solution is:

T =−

g ′′′ r 2 + C1 ln ( r ) + C2 4k

(1)

where C1 and C2 are constants selected to enforce the boundary conditions. The boundary condition at the center of the bale is either that the temperature remain bounded or that the temperature gradient be zero; either will lead to C1 = 0. An energy balance at the outer edge of the hay bale leads to: − k 2 π Rbale L

dT dr

= r = Rbale

Tr = Rbale − T∞ R p + Rconv

(2)

where Rp and Rconv are the thermal resistances associated with conduction through the plastic and convection from the outer surface of the bale, respectively:

Rp =

tp

(3)

k p 2 π Rbale L

Rconv =

1 h 2 π Rbale L

(4)

where L = 1 m for a problem that is done on a unit length basis. The temperature gradient and temperature at the outer radius of the bale are obtained using Eq. (1) with C1 = 0: dT dr

=− r = Rbale

Tr = Rbale = −

g ′′′ Rbale 2k

(5)

2 g ′′′ Rbale + C2 4k

(6)

Placing equations (2) through (6) into EES allows the constants of integration to be determined: R_p=t_p/(k_p*2*pi*R_bale*L) "thermal resistance associated with conduction through plastic" R_conv=1/(2*pi*R_bale*L*h) "thermal resistance associated with convection" dTdr_Rbale=-g_dot_v*R_bale/(2*k) "temperature gradient at outer edge" T_Rbale=-g_dot_v*R_bale^2/(4*k)+C_2 "temperature at outer edge" -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv) "interface energy balance"

The maximum temperature in the bale occurs at the center; according to Eq. (1) with C1 = 0, this temperature is given by: g ′′′ r 2 T =− + C1 ln ( r ) + C2 4k T_max=C_2 T_max_F=converttemp(K,F,T_max)

(7)

"maximum bale temperature" "maximum bale temperature in F"

The maximum temperature in the hay bale is 322.3 K or 120.6°F. b.) Prepare a plot showing the maximum temperature in the hay bale as a function of the hay bale radius. How large can the hay bale be before there is a problem with self-ignition? A parametric table is generated that contains the variables T_max_F and R_bale and used to generate Figure 1.

Figure 1: Maximum temperature as a function of the bale radius.

Note that a hay bale larger than approximately 2.1 m will result in a hay fire. Prepare a model that can consider temperature-dependent volumetric generation. Increasing temperature tends to increase the rate of chemical reaction and therefore increases the rate of generation of thermal energy according to: g ′′′ = a + bT where a = -1 W/m3 and b = 0.01 W/m3-K and T is in K. c.) Enter the governing equation into Maple and obtain the general solution (i.e., a solution that includes two constants). The governing differential equation is obtained as discussed in Section 1.3: g ′′′ r =

d ⎛ dT ⎞ ⎜ −k r ⎟ dr ⎝ dr ⎠

This ordinary differential equation is entered in Maple: > restart; > ODE:=(a+b*T(r))*r=diff(-k*r*diff(T(r),r),r);

2 d ⎞ ⎛d ⎞ ⎛ ⎜ T ( r ) k r ODE := ( a + b T( r ) ) r = −k ⎜⎜ T( r ) ⎟⎟ ⎟⎟ − 2 ⎜ d r ⎠ ⎝ ⎝ dr ⎠

and solved: > Ts:=dsolve(ODE);

(8)

⎛ Ts := T( r ) = BesselJ⎜⎜ 0, ⎝

b k

⎞ ⎛ r ⎟⎟ _C2 + BesselY⎜⎜ 0, ⎠ ⎝

b k

a ⎞ r ⎟⎟ _C1 − b ⎠

Note that the solution is given in the form of Bessel functions; ⎛ ⎛ b ⎞ b ⎞ a T = −C2 BesselJ ⎜⎜ 0, r ⎟⎟ + C1 BesselY ⎜⎜ 0, r ⎟⎟ − k k ⎝ ⎠ ⎝ ⎠ b

(9)

Even though we have not yet learned about Bessel functions, we can manipulate this solution within Maple. d.) Use the boundary conditions to obtain values for the two constants in your general solution (hint: one of the two constants must be zero in order to keep the temperature at the center of the hay bale finite). You should obtain a symbolic expression for the boundary condition in Maple that can be evaluated in EES. In part (a) we could not take the natural logarithm of 0 in Eq. (7) and therefore C1 was zero. A similar thing happens with the Bessel functions. We can evaluate the limits of the two Bessel functions as r → 0: > limit(BesselJ(0,r),r=0);

1

> limit(BesselY(0,r),r=0);

−∞

The BesselY function becomes infinite and therefore C1 in Eq. (9) must be 0. > Ts:=subs(_C1=0,Ts);

⎛ Ts := T( r ) = BesselJ⎜⎜ 0, ⎝

b k

a ⎞ r ⎟⎟ _C2 − b ⎠

The boundary condition at the outer surface of the hay does not change; the temperature and temperature gradient at Rbale can be evaluated symbolically using Maple: > dTdr_Rbale:=eval(diff(rhs(Ts),r),r=R_bale);

⎛ dTdr_Rbale := −BesselJ⎜⎜ 1, ⎝

b ⎞ R_bale ⎟⎟ k ⎠

b _C2 k

> T_Rbale:=eval(rhs(Ts),r=R_bale);

⎛ T_Rbale := BesselJ⎜⎜ 0, ⎝

b a ⎞ R_bale ⎟⎟ _C2 − k b ⎠

These symbolic expressions are cut and paste into EES and used to replace Eqs. (5) and (6) and provide a new constant C2: {g_dot_v=2 [W/m^3] a=-1 [W/m^3] "coefficients for volumetric generation function" b=0.01 [W/m^3-K]

"volumetric heat generation"}

R_p=t_p/(k_p*2*pi*R_bale*L) "thermal resistance associated with conduction through plastic" R_conv=1/(2*pi*R_bale*L*h) "thermal resistance associated with convection" {dTdr_Rbale=-g_dot_v*R_bale/(2*k) "temperature gradient at outer edge" T_Rbale=-g_dot_v*R_bale^2/(4*k)+C_2 "temperature at outer edge"} dTdr_Rbale = -BesselJ(1,(b/k)^(1/2)*R_bale)*(b/k)^(1/2)*C_2 "symbolic expressions cut and paste from Maple" T_Rbale = BesselJ(0,(b/k)^(1/2)*R_bale)*C_2-1/b*a -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv) "interface energy balance"

The maximum temperature is the temperature at the center of the bale; this is evaluated using Maple: > T_max=eval(rhs(Ts),r=0);

T_max = _C2 −

a b

and copied and pasted into EES: {T_max=C_2 T_max = C_2-1/b*a "symbolic expression cut and paste from Maple" T_max_F=converttemp(K,F,T_max)

"maximum bale temperature"}

"maximum bale temperature in F"

e.) Overlay on your plot from part (b) a plot of the maximum temperature in the hay bale as a function of bale radius when the volumetric generation is a function of temperature. The result is shown in Figure 1.

Problem 1.3-9 (1-9 in text): Mass Flow Meter Figure P1.3-9 illustrates a simple mass flow meter for use in an industrial refinery. T∞ = 20°C 2 hout = 20 W/m -K rout = 1 inch rin = 0.75 inch

insulation kins = 1.5 W/m-K test section 7 3 g ′′′ = 1x10 W/m k = 10 W/m-K m = 0.75kg/s T f = 18°C

L = 3 inch

thins = 0.25

Figure P1.3-9: A simple mass flow meter.

A flow of liquid passes through a test section consisting of an L = 3 inch section of pipe with inner and outer radii, rin = 0.75 inch and rout = 1.0 inch, respectively. The test section is uniformly heated by electrical dissipation at a rate g ′′′ = 1x107 W/m3 and has conductivity k = 10 W/m-K. The pipe is surrounded with insulation that is thins = 0.25 inch thick and has conductivity kins = 1.5 W/m-K. The external surface of the insulation experiences convection with air at T∞ = 20°C. The heat transfer coefficient on the external surface is hout = 20 W/m2-K. A thermocouple is embedded at the center of the pipe wall. By measuring the temperature of the thermocouple, it is possible to infer the mass flow rate of fluid because the heat transfer coefficient on the inner surface of the pipe ( hin ) is strongly related to mass flow rate ( m ). Testing has shown that the heat transfer coefficient and mass flow rate are related according to: ⎛ m ⎞ hin = C ⎜⎜ ⎟⎟ ⎝ 1[ kg/s ] ⎠

0.8

where C= 2500 W/m2-K. Under nominal conditions, the mass flow rate through the meter is m = 0.75 kg/s and the fluid temperature is Tf = 18°C. Assume that the ends of the test section are insulated so that the problem is 1-D. Neglect radiation and assume that the problem is steadystate. a.) Develop an analytical model in EES that can predict the temperature distribution in the test section. Plot the temperature as a function of radial position for the nominal conditions. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_out=1.0 [inch]*convert(inch,m) r_in=0.75 [inch]*convert(inch,m) h_bar_out=20 [W/m^2-K]

"outer radius of measurement section" "inner radius of measurement section" "external convection coefficient"

T_infinity=converttemp(C,K,20 [C]) T_f=converttemp(C,K, 18 [C]) k=10 [W/m-K] g```=1e7 [W/m^3] m_dot=0.75 [kg/s] th_ins=0.25 [inch]*convert(inch,m) k_ins=1.5 [W/m-K] L= 3 [inch]*convert(inch,m)

"ambient temperature" "fluid temperature" "conductivity" "volumetric rate of thermal energy generation" "mass flow rate" "thickness of insulation" "insulation conductivity" "length of test section"

The heat transfer coefficient on the internal surface is computed according to the specified mass flow rate: C=2500 [W/m^2-K] h_bar_in=C*(m_dot/1 [kg/s])^0.8

"constant for convection relationship" "internal convection coefficient"

The general solution to a 1-D problem in cylindrical coordinates with constant volumetric thermal energy generation was provided in Table 1-3, to within the unknown constants C1 and C2:

g ′′′ r 2 T =− + C1 ln ( r ) + C2 4k

(1)

dT g ′′′ r C1 =− + dr 2k r

(2)

The boundary condition at the outer edge of the test section is:

(

)

Tr = rout − T∞ ⎛ dT ⎞ − k 2 π rout L ⎜ = ⎟ ⎝ dr ⎠ r = rout ( Rins + Rconv ,out )

(3)

where Rins is the thermal resistance to conduction through the insulation (provided in Table 1-2): ⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎦ Rins = ⎣ 2 π L kins

(4)

and Rconv,out is the resistance to convection from the outer surface of the insulation: Rconv ,out =

1 2 π ( rout + thins ) L hout

R_ins=ln((r_out+th_ins)/r_out)/(2*pi*L*k_ins) R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out) T_r_out=-g```*r_out^2/(4*k)+C_1*ln(r_out)+C_2 dTdr_r_out=-g```*r_out/(2*k)+C_1/r_out

"resistance to conduction through insulation" "resistance to convection from outer surface" "temperature at outer surface of section" "temperature gradient at outer surface of section"

(5)

-k*2*pi*r_out*L*dTdr_r_out=(T_r_out-T_infinity)/(R_ins+R_conv_out) "boundary condition at r=r_out"

The boundary condition at the inner edge of the test section is:

⎛ dT ⎞ hin 2 π rin L T f − Tr = rin = −k 2 π rin L ⎜ ⎟ ⎝ dr ⎠ r = rin

(

)

(6)

T_r_in=-g```*r_in^2/(4*k)+C_1*ln(r_in)+C_2 "temperature at inner surface of section" dTdr_r_in=-g```*r_in/(2*k)+C_1/r_in "temperature gradient at inner surface of section" h_bar_in*2*pi*r_in*L*(T_f-T_r_in)=-k*2*pi*r_in*L*dTdr_r_in "boundary condition at r=r_in"

The EES code will provide the solution to the constants C1 and C2; note that it is not possible to eliminate the unit warnings that are associated with the argument of the natural logarithm in Eq. (1). In fact, if sufficient algebra was carried out, the equations could be placed in a form where the natural logarithm had a dimensionless argument. The location at which to evaluate the temperature (r) is specified in terms of a dimensionless radial position ( r ) that goes from 0 at the inner surface of the test section to 1 at the outer surface. The temperature is evaluated using Eq. (1): r_bar=0.5 [-] r=r_in+r_bar*(r_out-r_in) T=-g```*r^2/(4*k)+C_1*ln(r)+C_2 T_C=converttemp(K,C,T)

"dimensionless radial position" "radial position" "temperature" "in C"

Figure P1.3-9-2 illustrates the temperature as a function of radial position. 80

Temperature (°C)

75 70 65 60 55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure 1.3-9-2: Temperature as a function of radius.

b.) Using your model, develop a calibration curve for the meter; that is, prepare a plot of the mass flow rate as a function of the measured temperature at the mid-point of the pipe. The range of the instrument is 0.2 kg/s to 2.0 kg/s.

The dimensionless radial position is set to r =0.5, corresponding to the temperature of the center of the test section. Figure 1.3-9-3 illustrates the mass flow rate through the meter as a function of the measured temperature. 2 1.8

Mass flow rate (kg/s)

1.6 1.4 1.2

Tf = 28°C

1

Tf = 18°C

0.8

Tf = 8°C

0.6 0.4 0.2 0 40

60

80

100

120

140

160

Temperature (°C)

Figure 1.3-9-3: Mass flow rate as a function of the temperature at the center of the pipe wall for several values of the fluid temperature.

The meter must be robust to changes in the fluid temperature. That is, the calibration curve developed in (b) must continue to be valid even as the fluid temperature changes by as much as 10°C. c.) Overlay on your plot from (b) the mass flow rate as a function of the measured temperature for Tf = 8°C and Tf = 28°C. Is your meter robust to changes in Tf? The calibration curves generated at Tf = 8°C and Tf = 28°C are also shown in Figure 1.3-9-3. Notice that the fluid temperature has a large effect on the device. For example, if the measured temperature is 80°C then the mass flow rate could be anywhere from 0.45 kg/s to 0.75 kg/s depending on the fluid temperature. The meter is not robust to changes in Tf. In order to improve the meters ability to operate over a range of fluid temperature, a temperature sensor is installed in the fluid in order to measure Tf during operation. d.) Using your model, develop a calibration curve for the meter in terms of the mass flow rate as a function of ΔT, the difference between the measured temperatures at the mid-point of the pipe wall and the fluid. The temperature difference is calculated according to:

ΔT = Tr =0.5 − T f DT=T-T_f

"measured temperature difference"

Figure 1.3-9-4 illustrates the mass flow rate as a function of the temperature difference:

(7)

Mass flow rate (kg/s)

2 1.8

Tf = 28°C

1.6

Tf = 18°C

1.4

Tf = 8°C

1.2 1 0.8 0.6 0.4 0.2 0

30

40

50

60 70 80 90 100 Temperature difference (K)

110

120

Figure 1.3-9-4: Mass flow rate as a function of the temperature difference between the measured temperature at the center of the pipe wall and the fluid temperature for several values of the fluid temperature.

e.) Overlay on your plot from (d) the mass flow rate as a function of the difference between the measured temperatures at the mid-point of the pipe wall and the fluid if the fluid temperature is Tf = 8°C and Tf = 28°C. Is the meter robust to changes in Tf? The calibration curves for Tf = 8°C and Tf = 28°C are also shown in Figure 1.3-9-4; notice that the fluid temperature has almost no effect on the calibration curves and so the meter is robust to changes in the fluid temperature. f.) If you can measure the temperature difference to within δΔT = 1 K then what is the uncertainty in the mass flow rate measurement? (Use your plot from part (d) to answer this question.) The uncertainty in the measured mass flow rate that corresponds to an uncertainty in the temperature difference is evaluated according to:

⎛ ∂m ⎝ ∂ΔT

δ m = ⎜

⎞ ⎟ δΔT ⎠

(8)

From Figure 1.3-9-4 we see that the partial derivative of mass flow rate with respect to temperature difference decreases with flow rate. At high flow rates (around 2 kg/s), the partial derivative is approximately 0.08 kg/s-K which leads to an uncertainty of 0.08 kg/s. At low flow rates (around 0.2 kg/s), the partial derivative is approximately 0.04 kg/s-K which leads to an uncertainty of 0.04 kg/s. You can use the built-in uncertainty propagation feature in EES to assess uncertainty automatically. g.) Set the temperature difference to the value you calculated at the nominal conditions and allow EES to calculate the associated mass flow rate. Now, select Uncertainty Propagation from the Calculate menu and specify that the mass flow rate is the calculated variable while the temperature difference is the measured variable. Set the uncertainty in the temperature difference to 1 K and verify that EES obtains an answer that is approximately consistent with part (f).

The temperature difference is set to 50 K corresponding to approximately the middle of the range of the device. The mass flow rate is commented out and EES is used to calculate the mass flow rate from the temperature difference: DT=50 [K] {m_dot=0.75 [kg/s]}

"mass flow rate"

Select Uncertainty Propagation from the Calculate menu (Figure P1.3-9-5) and select the variable m_dot as the calculated variable and the variable DT as the measured variable.

Figure P1.3-9-5: Determine Propagation of Uncertainty dialog.

Select Set uncertainties and indicate that the uncertainty of the measured temperature difference is 1 K (Figure P1.3-9-6).

Figure P1.3-9-6: Uncertainties of Measured Variables dialog.

Select OK and then then OK again to carry out the calculation. The results are displayed in the Uncertainty Results tab of the Solution window (Figure P1.3-9-7).

Figure P1.3-9-7: Uncertainties Results tab of the Solution window.

The uncertainty calculated by EES is δ m = 0.031 kg/s, which falls between the bounds identified in part (e). h.) The nice thing about using EES to determine the uncertainty is that it becomes easy to assess the impact of multiple sources of uncertainty. In addition to the uncertainty δΔT, the constant C has an uncertainty of δC = 5% and the conductivity of the material is only known to within δk = 3%. Use EES' built-in uncertainty propagation to assess the resulting uncertainty in the mass flow rate measurement. Which source of uncertainty is the most important? Select Uncertainty Propagation from the Calculate menu and select the variable m_dot as the calculated variable and the variables DT, C, and k as the measured variables. Set the uncertainty of each of the measured variables according to the problem statement (Figure P1.3-9-8).

Figure P1.3-9-8: Uncertainties of Measured Variables dialog.

The results of the uncertainty calculation are shown in Figure P1.3-9-9.

Figure P1.3-9-9: Uncertainties Results tab of the Solution window.

Notice that the uncertainty has increased to δ m = 0.062 kg/s and that the dominant source of the uncertainty is related to C. The effect of the uncertainty in the conductivity is small (only 5.8% of the total). i.) The meter must be used in areas where the ambient temperature and heat transfer coefficient may vary substantially. Prepare a plot showing the mass flow rate predicted by your model for ΔT = 50 K as a function of T∞ for various values of hout . If the operating range of your

meter must include -5°C < T∞ < 35°C then use your plot to determine the range of hout that can be tolerated without substantial loss of accuracy. Figure P1.3-9-10 illustrates the mass flow rate as a function of T∞ for various values of hout . 0.9

2

h = 5 W/m -K

Mass flow rate (kg/s)

2

0.85

10 W/m -K 2 20 W/m -K

0.8

50 W/m -K

2

2

100 W/m -K

0.75 0.7 0.65 0.6 -10 -5

0

5

10 15 20 25 30 35 40 45 50 55

Air temperature (°C)

Figure P1.3-9-10: Mass flow rate predicted with ΔT = 50 K as a function of ambient temperature for various values of the air heat transfer coefficient.

The shaded region in Figure P1.3-9-10 indicates the operating temperature range (in the xdirection) and the region of acceptable accuracy (based approximately on the results of part (e)). Figure P1.3-9-10 shows that 5 W/m2-K < hout < 50 W/m2-K will keep you within the shaded region and therefore this is, approximately, the range of hout that can be tolerated without substantial loss of accuracy.

Problem 1.3-10 A current of 100 amps passes through a bare stainless-steel wire of D = 1.0 mm diameter. The thermal conductivity and electrical resistance per unit length of the wire are k = 15 W/m-K and Re′ = 0.14 ohm/m, respectively. The wire is submerged in an oil that is maintained at T∞ = 30°C. The steady-state temperature at the center of the wire is measured to be 180°C, independent of axial position within the oil bath. a) What is the temperature at the outer surface of the wire? The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in k=15 [W/m-K] R\L=0.14 [ohm/m] i=100 [amp] D=1[mm]/2*convert(mm,m) T_c=converttemp(C,K,180 [C]) T_oil=converttemp(C,K,30 [C]) L=1 [m]

"conductivity of wire" "resistance per unit length of wire" "current" "diameter" "temperature at center of wire" "temperature of oil" "per unit length basis"

The rate of generation per unit volume is: g ′′′ =

I 2 Re′ L 2 L π rout

(1)

D 2

(2)

where rout is the outer radius of the wire:

rout = r_out=D/2 g_dot```*(pi*r_out^2*L)=i^2*R\L*L

"outer radius" "rate of volumetric generation"

The general solution for radial conduction with uniform volumetric generation is provided in Table 1-3: T =−

g ′′′ r 2 + C1 ln ( r ) + C2 4k

dT g ′′′ r C1 =− + dr 2k r

(3)

(4)

In order for the temperature to remain bounded at r = 0, it is necessary that C1 be zero. The temperature at the center is therefore:

Tc = C2

(5)

T_c=C_2

"solve for C_2"

The temperature at the surface is given by: Ts = −

2 g ′′′ rout + C1 ln ( rout ) + C2 4k

T_s=-g_dot```*r_out^2/(4*k)+C_2 T_s_C=converttemp(K,C,T_s)

(6)

"surface temperature" "in C"

which leads to Ts = 172.6ºC. b) Estimate the convection coefficient between the submerged wire and the oil. An interface energy balance at r = rout leads to: h (Toil − Ts ) = k

dT dr

(7) r = rout

where the temperature gradient is evaluated using Eq. (4): dT dr

=− r = rout

g ′′′ rout 2k

dT\dr_s=-g_dot```*r_out/(2*k) h_bar*(T_oil-T_s)=k*dT\dr_s

(8)

"surface temperature gradient" "interface energy balance"

which leads to h = 6250 W/m2-K. c) A plastic material (kp = 0.05 W/m-K) can be applied to the outer surface of the wire. Can the insulation result in a reduction of the center temperature? If so, what insulation thickness should be applied? The resistance to conduction through the insulation is: ⎛ r + thins ⎞ ln ⎜ out ⎟ rout ⎝ ⎠ Rins = 2π k p L

and the resistance to convection from the surface of the insulation is:

(9)

Rconv =

1 2 π rout L h

(10)

The total resistance between the surface of the wire and the oil is: Rtotal = Rconv + Rins k_p=0.05 [W/m-K] R_ins=ln((r_out+th_ins)/r_out)/(2*pi*k_p*L) R_conv=1/(h_bar*2*pi*L*(r_out+th_ins)) R_total=R_ins+R_conv

(11)

"conductivity of plastic" "resistance to conduction through plastic" "resistance to convection from external surface" "total resistance from surface of wire"

Figure 1 illustrates Rins, Rconv, and Rtotal as a function of the insulation thickness. Notice that increasing the insulation thickness reduces Rconv because there is more surface area for convection but increases Rcond because the length for conduction is longer. In some situations, the reduction in Rconv dominates the problem and therefore the total resistance may be reduced by adding insulation. However, in this case, there is no such region.

Thermal resistance (K/W)

10

1

total resistance

convection resistance 0.1

0.01

0.001 0.000001

resistance to conduction through insulation

0.00001

0.0001

0.001

0.01

Insulation thickness (m) Figure 5: Resistance to convection, resistance to conduction through insulation, and total resistance from wire surface as a function of the insulation thickness.

Problem 1.3-11: Nuclear Fuel Element Figure P1.3-11 illustrates a spherical, nuclear fuel element which consists of a sphere of fissionable material (fuel) with radius rfuel = 5 cm and kfuel = 2 W/m-K that is surrounded by a spherical shell of metal cladding with outer radius rclad = 7 cm and kclad = 0.25 W/m-K. The outer surface of the cladding is exposed to fluid that is being heated by the reactor. The convection coefficient between the fluid and the cladding surface is h = 50 W/m2-K and the temperature of the fluid is T∞ = 500ºC. Neglect radiation heat transfer from the surface. Inside the fuel element, thermal energy is being generated for the reactor. This process can be modeled as a volumetric source of heat generation in the material that is not uniform throughout the fuel. The volumetric generation ( g ′′′ ) can be approximated by the function:

g ′′′ =

β r

where β = 5x103 W/m2. fissionable material kfuel = 2 W/m-K rfuel = 5 cm rclad = 7 cm

g ′′′

2 h = 50 W/m -K T∞ = 500°C

cladding kclad = 0.25 W/m-K Figure P1.3-11: Spherical fuel element surrounded by cladding

a.) Determine an analytical solution for the temperature distribution within the fuel element. Implement your solution in EES and plot the temperature as a function of radius for 0 < r < rfuel. The inputs are entered according to: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" r_fuel=5 [cm]*convert(cm,m) k_fuel=2 [W/m-K] r_clad=7 [cm]*convert(cm,m) k_clad=0.25 [W/m-K] h_bar=50 [W/m^2-K] T_infinity=converttemp(C,K,500[C]) beta=5e3 [W/m^2]

"radius of fuel element" "conductivity of fuel element" "radius of cladding" "conductivity of cladding" "heat transfer coefficient" "temperature of fluid" "constant for volumetric generation"

A differential control volume is shown in Figure 2 and includes conduction at r and r+dr at the inner and outer surfaces of the spherical shell as well as generation within the enclosed volume.

Figure 2: Differential control volume

The energy balance suggested by Figure 2 is:

qr + g = qr + dr

(1)

The term at r + dr can be expanded:

qr + dr = qr +

dqr dr dr

(2)

dqr dr dr

(3)

and substituted into Eq. (1):

qr + g = qr + and simplified:

g =

dqr dr dr

(4)

The conduction is governed by Fourier’s Law:

qr = −k fuel 4 π r 2

dT dr

(5)

and the generation is the product of the volume and the local generation rate: g = 4 π r 2 dr g ′′′= 4 π r 2 dr

β r

The rate equations, Eqs. (5) and (6), are substituted into Eq. (4):

(6)

4 π r dr β =

d ⎡ dT ⎤ dr − k fuel 4 π r 2 ⎢ dr ⎣ dr ⎥⎦

(7)

which can be simplified: d ⎡ 2 dT ⎤ rβ r = − dr ⎢⎣ dr ⎥⎦ k fuel

(8)

Equation (8) can be separated and integrated: ⎡

∫ d ⎢⎣ r

2

dT ⎤ rβ dr = ∫− ⎥ dr ⎦ k fuel

(9)

which leads to:

r2

dT r2 β =− + C1 dr 2 k fuel

(10)

where C1 is a constant of integration. Equation (10) can be separated and integrated: ⎛

βo

∫ dT = ∫ ⎜⎜ − 2 k ⎝

+

fuel

C1 ⎞ ⎟ dr r 2 ⎟⎠

(11)

which leads to: T =−

β 2 k fuel

r−

C1 + C2 r

(12)

where C2 is the second constant of integration. The boundary condition at r = 0 requires that the temperature remain finite and therefore C1 = 0. T =−

βo 2 k fuel

r + C2

(13)

The boundary condition at r = rfuel is obtained using an interface balance, as show in Figure 3. The interface energy balance includes conduction from the fuel and heat transfer into the cladding.

Figure 3: Interface balance at r = rfuel

The energy balance suggested by Figure 3 is:

qr = rruel = qout

(14)

The conduction term on the left side of Eq. (14) is evaluated using Fourier’s law: 2 qr = rruel = − k fuel 4 π rfuel

dT dr

(15) r = r fuel

while the heat transfer out of the cladding is driven by the difference between the temperature at interface between the fuel and the cladding and the temperature of the surrounding gas. The heat transfer is resisted by the sum of the conduction resistance of the cladding (Rcond,clad):

Rcond ,clad =

1 4 π kclad

⎡ 1 1 ⎤ − ⎢ ⎥ ⎢⎣ rfuel rclad ⎥⎦

(16)

and the convection resistance (Rconv):

Rconv =

1 2 4 π rclad h

(17)

so that: qout =

Tr = rfuel − T∞ Rcond ,clad + Rconv

Subsituting Eqs. (18) and (15) into Eq. (14) leads to:

(18)

2 − k fuel 4 π rfuel

dT dr

= r = r fuel

(T

r = r fuel

− T∞

)

(19)

Rcond ,clad + Rconv

Equation (19) provides a single equation for the unknown constant of integration, C2. Substituting Eq. (13) into Eq. (19) leads to:

⎛ β 2 − k fuel 4 π rfuel ⎜⎜ − ⎝ 2 k fuel

⎛ ⎞ β rfuel + C2 − T∞ ⎟ ⎜⎜ − ⎟ ⎞ ⎝ 2 k fuel ⎠ ⎟⎟ = + R R cond , clad conv ⎠

(20)

The resistances are computed according to Eqs. (16) and (17) and the constant C2 is computed according to Eq. (20). R_cond_clad=(1/r_fuel-1/r_clad)/(4*pi*k_clad) "resistance to conduction through cladding" R_conv=1/(4*pi*r_clad^2*h_bar) "resistance to convection from cladding" k_fuel*4*pi*r_fuel^2*(beta/(2*k_fuel))=(-beta*r_fuel/(2*k_fuel)+C_2-T_infinity)/(R_cond_clad+R_conv) "boundary condition"

The temperature distribution is obtained using Eq. (13). r=0 [m] T=-beta*r/(2*k_fuel)+C_2 T_C=converttemp(K,C,T)

"radius" "temperature distribution" "in C"

Figure 4 illustrates the temperature in the sphere as a function of position. 740 730

Temperature (°C)

720 710 700 690 680 670 660 0

0.01

0.02

0.03

0.04

0.05

Radius (m) Figure 4: Temperature as a function of radius.

b.) The maximum allowable temperature in the fuel element is Tmax = 1100ºC. What is the maximum value of β that can be used? What is the associated total rate that heat is transferred to the gas?

The maximum temperature occurs at r = 0. According to Eq. (13), the temperature at r = 0 is:

Tr =0 = C2

(21)

The guess values are updated and the specified value of β is commented out. The temperature at the center is specified to be Tmax: {beta=5e3 [W/m^2]} T_max_s=converttemp(C,K,1100 [C]) T_max=C_2 T_max_s=T_max

"constant for volumetric generation" "maximum allowable temperature" "maximum temperature in fuel" "adjust beta so that center temperature is equal to T_max_s"

which leads to β = 1.3x104 W/m2. The rate of heat transfer is given by Eq. (18):

qout =

Tr = rfuel − T∞ Rcond ,clad + Rconv

− =

βo 2 k fuel

rfuel + C2 − T∞

Rcond ,clad + Rconv

(22)

q_dot=(-beta*r_fuel/(2*k_fuel)+C_2-T_infinity)/(R_cond_clad+R_conv) "heat transfer from fuel"

which leads to qout = 204.1 W. c.) You are designing the fuel elements. You can vary rfuel and β. The cladding must always be 2 cm thick (that is rclad = rfuel + 2 cm). The constraint is that the fuel temperature cannot exceed Tmax = 1100ºC and the design target (the figure of merit to be maximized) is the rate of heat transfer per unit volume of material (fuel and cladding). What values of rfuel and β are optimal? The volume of the fuel and the cladding is: 4 3 V = π rclad 3

(23)

and therefore the heat transfer per unit volume can be determined. V=4*pi*r_clad^3/3 q_dot\V=q_dot/V

"volume of fuel" "heat transfer per volume"

The cladding radius is specified based on the fuel radius: r_clad=r_fuel+2 [cm]*convert(cm,m) {r_clad=7 [cm]*convert(cm,m)

"radius of cladding" "radius of cladding"}

The fuel radius is varied in a parametric table and the heat transfer per unit volume as a function of fuel radius is shown in Figure 5.

3

Heat transfer per unit volume (W/m )

2.3x10 5 2.0x10 5 1.8x10 5 1.5x10 5 1.2x10 5 1.0x10 5 7.5x10 4 5.0x10 4 2.5x10 4 0

0.02

0.04

0.06

0.08

0.1

Fuel element radius (m) Figure 5: Heat transfer per unit volume as a function of fuel element radius.

Figure 5 shows that the optimal value of rfuel is approximately 1.6 cm. A more exact value can be obtained using the Min/Max feature from the Calculate menu. The optimal design is rfuel = 1.57 cm with β = 2.6x104 W/m2.

Problem 1.3-12 Figure P1.3-12 illustrates a plane wall. The temperature distribution in the wall is 1-D and the problem is steady state.

qL′′

h , T∞

x L

g ′′′ = a x, k Figure P1.3-12: Plane wall.

There is generation of thermal energy in the wall. The generation per unit volume is not uniform but rather depends on position according to:

g ′′′ = a x

(1)

where a is a constant and x is position. The left side of the wall experiences a specified heat flux, q ′′L . The right side of the wall experiences convection with heat transfer coefficient h to fluid at temperature T∞. The thickness of the wall is L and the conductivity of the wall material, k, is constant. a.) Derive the ordinary differential equation that governs this problem. Clearly show your steps. A differential control volume is shown in Figure 2 and leads to: q x + g = q x + dx

q x + dx

q x g x dx

Figure 2: Differential control volume with energy terms.

After expanding the x + dx term:

(2)

dq dx dx

q x + g = q x +

(3)

The rate of thermal energy generation within the control volume is:

g = g ′′′ Ac dx where Ac is the cross-sectional area of the wall. Fourier’s law:

q = − k Ac

(4)

The conduction term is expressed using

dT dx

(5)

Substituting Eqs. (5) and (4) into Eq. (3) results in

g ′′′ Ac dx =

d ⎛ dT ⎞ ⎜ − k Ac ⎟ dx dx ⎝ dx ⎠

(6)

which can be simplified:

d ⎛ dT ⎜ dx ⎝ dx

g ′′′ ⎞ ⎟=− k ⎠

(7)

Substituting the position dependent generation into Eq. (7) leads to:

d ⎛ dT ⎜ dx ⎝ dx

ax ⎞ ⎟=− k ⎠

(8)

b.) Solve the differential equation that you obtained in (a). Your solution should include two undetermined constants. Equation (7) is separated and integrated:

⎛ dT ⎞

∫ d ⎜⎝ dx ⎟⎠ = ∫ −

ax dx k

(9)

which leads to:

dT a 2 =− x + C1 dx 2k where C1 is a constant of integration. Equation (10) is integrated again: B

B

(10)

⎛

a

∫ dT = ∫ ⎜⎝ − 2 k x

2

⎞ + C1 ⎟ dx ⎠

(11)

which leads to:

T =−

a 3 x + C1 x + C2 6k

(12)

c.) Specify the boundary conditions for the differential equation that you derived in (a). An interface energy balance at x = 0 leads to: q ′′L = − k

dT dx

(13) x =0

An interface energy balance at x = L leads to: −k

dT dx

= h (Tx = L − T∞ )

(14)

x= L

d.) Use the results of (b) and (c) to obtain two equations that can be solved for the two undetermined constants. Substituting Eq. (10) into Eq. (13) leads to:

q L′′ = − k C1

(15)

Substituting Eqs. (10) and (12) into Eq. (14) leads to:

⎛ a 2 ⎞ ⎛ a 3 ⎞ L + C1 ⎟ = h ⎜ − L + C1 L + C2 − T∞ ⎟ −k ⎜ − ⎝ 2k ⎠ ⎝ 6k ⎠ Equations (15) and (16) can be solved for C1 and C2.

(16)

Problem 1.4-1: A 3-Node Numerical Solution Figure P1.4.1(a) illustrates a plane wall with thickness L and cross-sectional area A that has a specified temperature TH on the left side (at x = 0) and a specified temperature TC on the right side (at x = L). There is no volumetric generation in the wall. However, the conductivity of the wall material is a function of temperature such that: k = b + cT where a and b are constants. You would like to model the wall using a finite difference solution; a model with only 3 nodes is shown in Figure P1.4-1(b).

TC

TH

T1

L

T2 Δx

T3

k = b+cT (a) (b) Figure P1.4-1: (a) A plane wall and (b) a numerical model with 3 nodes.

The distance between adjacent nodes for the 3 node solution is: Δx = L/2. a.) Write down the system of equations that could be solved in order to obtain the temperatures at the three nodes. Your equations should include the temperature of the nodes (T1, T2, and T3) and the other parameters listed in the problem statement: TH, TC, Δx, A, b, and c. The equations for T1 and T3 are easy, their temperatures are specified:

T [1] = TH

(1)

T [3] = TC

(2)

Figure 2 illustrates the control volume for the 2nd node.

Figure 2: Control volume for node 2.

An energy balance for the control volume shown in Fig. 3 leads to: q RHS [2] + q LHS [2] = 0

(3)

The energy transfer rates must be approximated according to: q RHS [2] =

(T [3] − T [ 2]) A ⎡b + c ⎛ T [3] + T [ 2] ⎞⎤

(4)

q LHS [2] =

(T [1] − T [ 2]) A ⎡b + c ⎛ T [1] + T [ 2] ⎞ ⎤

(5)

⎢ ⎣⎢

Δx

⎜ ⎝

2

⎟⎥ ⎠ ⎦⎥

and

⎢ ⎢⎣

Δx

⎜ ⎝

2

⎟⎥ ⎠ ⎥⎦

Notice that the temperature differences agree with the sign convention used in Figure 3 and that the conductivity is evaluated at the temperature of the interface. Substituting Eqs. (4) and (5) into Eq. (3) leads to:

(T [3] − T [ 2]) A ⎡b + c ⎛ T [3] + T [ 2] ⎞⎤ + (T [1] − T [ 2]) A ⎡b + c ⎛ T [1] + T [ 2] ⎞ ⎤ = 0 Δx

⎢ ⎢⎣

⎜ ⎝

2

⎟⎥ ⎠ ⎥⎦

Δx

⎢ ⎢⎣

⎜ ⎝

2

⎟⎥ ⎠ ⎥⎦

(6)

Equations (1), (2), and (6) together represent a system of three equations in the three unknown temperatures.

PROBLEM 1.4-2 (1-10 in text): Mass Flow Meter (revisited) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). The conductivity of the material that is used to make the test section is not actually constant as was assumed in Problem 1-9 but rather depends on temperature according to:

k = 10

W ⎡ W ⎤ + 0.035 ⎢ T − 300 [ K ]) 2 ( m-K ⎣ m-K ⎥⎦

a.) Develop a numerical model of the mass flow meter using EES. Plot the temperature as a function of radial position for the conditions shown in Figure P1.3-9 (Figure P1-9 in the text) with the temperature-dependent conductivity. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_out=1.0 [inch]*convert(inch,m) r_in=0.75 [inch]*convert(inch,m) h_bar_out=10 [W/m^2-K] T_infinity_C=20 [C] T_infinity=converttemp(C,K,T_infinity_C) T_f=converttemp(C,K, 18 [C]) g```=1e7 [W/m^3] m_dot=0.75 [kg/s] th_ins=0.25 [inch]*convert(inch,m) k_ins=1.5 [W/m-K] L= 3 [inch]*convert(inch,m) C=2500 [W/m^2-K] h_bar_in=C*(m_dot/1 [kg/s])^0.8

"outer radius of measurement section" "inner radius of measurement section" "external convection coefficient" "ambient temperature in C" "ambient temperature" "fluid temperature" "volumetric rate of thermal energy generation" "mass flow rate" "thickness of insulation" "insulation conductivity" "length of test section" "constant for convection relationship" "internal convection coefficient"

A function is defined that returns the conductivity of the material: Function k_t(T) "This function returns the conductivity of the test section material as a function of temperature" k_t=10 [W/m-K]+0.035 [W/m-K^2]*(T-300 [K]) end

A uniform distribution of nodes is used, the radial location of each node (ri) is:

ri = rin +

( i − 1) r − r ( ) ( N − 1) out in

for i = 1..N

(1)

where N is the number of nodes. The radial distance between adjacent nodes (Δr) is: Δr =

( rout − rin ) ( N − 1)

(2)

N=51 [-] DELTAr=(r_out-r_in)/(N-1) "Set up nodes" duplicate i=1,N r[i]=r_in+(r_out-r_in)*(i-1)/(N-1) end

"number of nodes" "distance between adjacent nodes (m)" "this loop assigns the radial location to each node"

An energy balance is carried out on a control volume surrounding each node. For node 1, placed at the inner surface (Figure P1.4-2-1): qconv ,in + qouter + g = 0

qouter g1 qconv ,in

(3)

T2 T1

Figure P1.4-2-1: Control volume around node 1.

The rate equation for convection is: qconv ,in = hin 2 π rin L (T f − T1 )

(4)

The rate equation for conduction is: Δr ⎞ (T2 − T1 ) ⎛ qouter = kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(5)

The rate equation for generation is: g = 2 π rin

Δr L g ′′′ 2

"Node 1" q_dot_conv_in=h_bar_in*2*pi*r_in*L*(T_f-T[1]) g_dot[1]=2*pi*r_in*L*DELTAr*g```/2 q_dot_outer[1]=k_t((T[1]+T[2])/2)*2*pi*(r[1]+DELTAr/2)*L*(T[2]-T[1])/DELTAr q_dot_conv_in+q_dot_outer[1]+g_dot[1]=0

(6)

"convection from fluid" "generation" "conduction from node 2" "energy balance on node 1"

An energy balance on an internal node is shown in Figure P1.4-2-2: qinner + qouter + g = 0

(7)

qouter

Ti+1

g

Ti

qinner

Ti-1

Figure P1.4-2-2: Control volume around internal node i.

The rate equations for conduction are: Δr ⎞ (Ti +1 − Ti ) ⎛ qouter = kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L 2 ⎠ Δr ⎝

(8)

Δr ⎞ ( T − T ) ⎛ qinner = kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i 2 ⎠ Δr ⎝

(9)

The rate equation for generation is: g = 2 π ri Δr L g ′′′

(10)

"Internal nodes" duplicate i=2,(N-1) q_dot_inner[i]=k_t((T[i]+T[i-1])/2)*2*pi*(r[i]-DELTAr/2)*L*(T[i-1]-T[i])/DELTAr "conduction from inner node" q_dot_outer[i]=k_t((T[i]+T[i+1])/2)*2*pi*(r[i]+DELTAr/2)*L*(T[i+1]-T[i])/DELTAr "conduction from outer node" g_dot[i]=2*pi*r[i]*L*DELTAr*g``` "generation" q_dot_inner[i]+q_dot_outer[i]+g_dot[i]=0 "energy balance on node i" end

An energy balance on node N placed on the outer surface is shown in Figure P1.4-2-3: qinner + qair + g = 0

(11)

qair TN g qinner

TN-1

Figure P1.4-2-3: Control volume around internal node N.

The rate equation for the heat transfer with the air is: qair =

(R

(T∞ − TN )

ins

+ Rconv ,out )

(12)

where ⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎦ Rins = ⎣ 2 π L kins Rconv ,out =

(13)

1 2 π ( rout + thins ) L hout

(14)

The rate equation for conduction is: Δr ⎞ (T − T ) ⎛ qinner = kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N Δr 2 ⎠ ⎝

(15)

The rate equation for generation is: g = 2 π rout

Δr L g ′′′ 2

(16)

"Node N" R_ins=ln((r_out+th_ins)/r_out)/(2*pi*L*k_ins) "resistance to conduction through insulation" R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out) "resistance to convection from outer surface" q_dot_air=(T_infinity-T[N])/(R_ins+R_conv_out) "heat transfer from air" q_dot_inner[N]=k_t((T[N]+T[N-1])/2)*2*pi*(r_out-DELTAr/2)*L*(T[N-1]-T[N])/DELTAr "conduction from node N-1" g_dot[N]=2*pi*r_out*L*DELTAr*g```/2 "generation" q_dot_air+q_dot_inner[N]+g_dot[N]=0 "energy balance on node N"

The solution is converted to degrees Celsius: duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

The solution is illustrated in Figure P1.4-2-4.

"convert solution to deg. C"

75 72.5

Temperature (°C)

70 67.5 65 62.5 60 57.5 55 52.5 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.4-2-4: Temperature as a function of radius.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in the text) in the limit that the conductivity is constant. The conductivity function is modified temporarily so that it returns a constant value: Function k_t(T) "This function returns the conductivity of the test section material as a function of temperature" k_t=10 [W/m-K]{+0.035 [W/m-K^2]*(T-300 [K])} end

The analytical solution from P1.3-9 is programmed and used to compute the analytical solution at each node: "Analytical solution from P1.3-9" k=k_t(300 [K]) "conductivity to use in the solution" T_r_out=-g```*r_out^2/(4*k)+C_1*ln(r_out)+C_2 "temperature at outer surface of section" dTdr_r_out=-g```*r_out/(2*k)+C_1/r_out "temperature gradient at outer surface of section" -k*2*pi*r_out*L*dTdr_r_out=(T_r_out-T_infinity)/(R_ins+R_conv_out) "boundary condition at r=r_out" T_r_in=-g```*r_in^2/(4*k)+C_1*ln(r_in)+C_2 "temperature at inner surface of section" dTdr_r_in=-g```*r_in/(2*k)+C_1/r_in "temperature gradient at inner surface of section" h_bar_in*2*pi*r_in*L*(T_f-T_r_in)=-k*2*pi*r_in*L*dTdr_r_in "boundary condition at r=r_in" duplicate i=1,N T_an[i]=-g```*r[i]^2/(4*k)+C_1*ln(r[i])+C_2 T_an_C[i]=converttemp(K,C,T_an[i]) end

"temperature" "in C"

Figure P1.4-2-5 illustrates the temperature distribution predicted by the numerical and analytical solutions in the limit that k is constant.

80

Temperature (°C)

75 70 65 60

analytical model numerical model

55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.4-2-5: Temperature as a function of radius predicted by the analytical and numerical models in the limit that k is constant.

c.) What effect does the temperature dependent conductivity have on the calibration curve that you generated in part (d) of Problem 1.3-9 (1-9)? The quantity measured by the meter is the difference between the temperature at the center of the pipe wall (T[26] when 51 nodes are used) and the fluid temperature: DT=T[26]-T_f

"temperature difference"

Figure P1.4-2-6 illustrates the calibration curve (i.e., the relationship between the temperature difference and the mass flow rate) with and without the temperature dependent conductivity included. 2

Mass flow rate (kg/s)

1.8 1.6 1.4 without temperature dependent conductivity

1.2 1

with temperature dependent conductivity

0.8 0.6 0.4 0.2 0 30

40

50

60

70

80

90

100

110

120

Temperature difference (K)

Figure P1.4-2-6: Calibration curve generated with and without the temperature dependent conductivity included.

PROBLEM 1.4-3: Fuel sphere (revisited) Reconsider Problem 1.3-7 using a numerical model developed in EES. a.) Plot the temperature as a function of position within the fuel. The inputs are entered in EES and a function is defined to return the volumetric generation. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function gv(gve,r,r_fuel,b) " Inputs: gve - volumetric generation at the edge (W/m^3) r - radius (m) r_fuel - radius of fuel element (m) b - exponent (-) Outputs: gv - volumetric rate of thermal energy generation (W/m^3)" gv=gve*(r/r_fuel)^b end "Inputs" r_fuel=5[cm]*convert(cm,m) k_fuel=1 [W/m-K] r_clad=7[cm]*convert(cm,m) k_clad=300 [W/m-K] h_gas=100 [W/m^2-K] T_gas=converttemp(C,K,500) gve=5e5 [W/m^3] b=1 [-]

"radius of fuel" "conductivity of fuel" "cladding radius" "cladding conductivity" "convection coefficient" "gas temperature" "generation at the center" "decay constant"

The nodal positions are specified: N=11 [-] duplicate i=1,N r[i]=(i-1)*r_fuel/(N-1) end Dr=r_fuel/(N-1)

"number of nodes" "location of each node" "distance between nodes"

Energy balances on the internal nodes lead to: "internal nodes" duplicate i=2,(N-1) g_dot[i]=4*pi*r[i]^2*Dr*gv(gve,r[i],r_fuel,b) q_dot_LHS[i]=4*pi*(r[i]-Dr/2)^2*k_fuel*(T[i-1]-T[i])/Dr q_dot_RHS[i]=4*pi*(r[i]+Dr/2)^2*k_fuel*(T[i+1]-T[i])/Dr g_dot[i]+q_dot_LHS[i]+q_dot_RHS[i]=0 end

An energy balance on node N leads to:

"generation" "conduction from node i-1" "conduction from node i+1" "energy balance"

"surface node" Rst_clad=(1/r_fuel-1/r_clad)/(4*pi*k_clad) Rst_conv=1/(4*pi*r_clad^2*h_gas) g_dot[N]=4*pi*r[N]^2*Dr*gv(gve,r[N],r_fuel,b)/2 q_dot_LHS[N]=4*pi*(r[N]-Dr/2)^2*k_fuel*(T[N-1]-T[N])/Dr q_dot_RHS[N]=(T_gas-T[N])/(Rst_clad+Rst_conv) g_dot[N]+q_dot_LHS[N]+q_dot_RHS[N]=0

"conduction resistance of cladding" "convection resistance" "generation" "conduction from node N-1" "heat transfer from gas" "energy balance"

An energy balance on node 1 leads to: "inner node" g_dot[1]=4*pi*(Dr/2)^3*gv(gve,r[1],r_fuel,b)/2/3 q_dot_RHS[1]=4*pi*(r[1]+Dr/2)^2*k_fuel*(T[2]-T[1])/Dr g_dot[1]+q_dot_RHS[1]=0

"generation" "conduction from node 2" "energy balance"

The solution is converted to °C. duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

The temperature as a function of position is shown in Figure 1. 640

Temperature (°C)

620 600 580 560 numerical solution 540 520 0

analytical solution from P1.3-7

0.01

0.02

0.03

0.04

0.05

Radius (m) Figure 1: Temperature as a function of radial position, predicted by the numerical model and the analytical model derived in Problem 1.3-7.

b.) Verify that your answer agrees with the analytical solution obtained in Problem 1.3-7. The analytical solution from Problem 1.3-7 is evaluated at the same radial locations used in the numerical model: "Boundary condition expressions" dTdr_rfuel =-2*r_fuel*gve/k_fuel/(b^2+6+5*b)-r_fuel*gve/k_fuel/(b^2+6+5*b)*b "from Maple"

T_rfuel =-r_fuel^2*gve/k_fuel/(b^2+6+5*b)+C2 "from Maple" -4*pi*r_fuel^2*k_fuel*dTdr_rfuel=(T_rfuel-T_gas)/(Rst_clad+Rst_conv)"boundary condition" "Solution" duplicate i=1,N T_an[i]=-r[i]^2*gve/k_fuel/(b^2+6+5*b)*(r[i]/r_fuel)^b+C2 T_an_C[i]=converttemp(K,C,T_an[i]) end

"solution from Maple"

The analytical solution is overlaid onto the numerical result in Figure 1. c.) Plot some aspect of the solution as a function of the number of nodes used in the numerical model and determine the number of nodes required for an accurate solution. Figure 2 illustrates the maximum temperature in the fuel element as a function of the number of nodes and shows that at least 50 nodes is required to obtain an accurate solution. 640

Maximum temperature (°C)

630 620 610 600 590 580 570 560 1

10

100

500

Number of nodes Figure 2: Maximum temperature in the fuel element as a function of the number of nodes.

Problem 1.4-4: Storing Hay in a Barn If you bale hay without allowing it to dry sufficiently then the hay bales will contain a lot of water. Besides making the bales heavy and therefore difficult to put in the barn, the water in the hay bails causes an exothermic chemical reaction to occur within the bale (i.e., the hay is rotting). The chemical reaction proceeds at a rate that is related to temperature and the bales may be thermally isolated (they are placed in a barn and surrounded by other hay bales); as a result, the hay can become very hot and even start a barn fire. Figure P1.4-4 illustrates a cross-section of a barn wall with hay stacked against it. L=5m Ta ,in = 20°C 2 ha ,in = 15 W/m -K

Ta ,out = −5°C 2 ha ,out = 45 W/m -K

g ′′′

x

thw = 1 cm

barn wall kw = 0.11 W/m-K

hay kh = 0.05 W/m-K

Figure P1.4-4: Barn wall with hay.

The air within the barn is maintained at Ta,in = 20°C and the heat transfer coefficient between the air and the inner surface of the hay is ha ,in = 15 W/m2-K. The outside air is at Ta,out = -5°C with ha ,out = 45 W/m2-K. Neglect radiation from the surfaces in this problem. The barn wall is

composed of wood (kw = 0.11 W/m-K) and is thw = 1 cm thick. The hay has been stacked L = 5 m thick against the wall. Hay is a composite structure composed of plant fiber and air. However, hay can be modeled as a single material with an effective conductivity kh = 0.05 W/mK. The volumetric generation of the hay due to the chemical reaction is given by: ⎛ T ⎞⎤ ⎡W ⎤⎡ g ′′′ = 1.5 ⎢ 3 ⎥ ⎢exp ⎜⎜ ⎟⎟ ⎥ ⎣ m ⎦ ⎣⎢ ⎝ 320 [ K ] ⎠ ⎦⎥

0.5

where T is temperature in K. a.) Develop a numerical model that can predict the temperature distribution within the hay. The input information is entered in EES and a function is used to define the volumetric generation: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function gen(T) "volumetric heat generation in wall" "Input - T, temperature [K]" "Output - gen, volumetric rate of heat generation [W/m^3]"

gen=1.5 [W/m^3]*sqrt(exp(T/320 [K])) end "Inputs" T_a_in=converttemp(C,K,20) h_a_in=15 [W/m^2-K] T_a_out=converttemp(C,K,-5) h_a_out=45 [W/m^2-K] k_w=0.11 [W/m-K] th_w=1.0 [cm]*convert(cm,m) L=5.0 [m] k_h=0.05 [W/m-K] A = 1 [m^2]

"temperature of air within barn" "internal heat transfer coefficient" "temperature of air outside barn" "external heat transfer coefficient" "conductivity of barn wall" "barn wall thickness" "thickness of hay" "conductivity of hay" "per unit area of wall"

Nodes are distributed uniformly throughout the computational domain (which consists only of the hay, not the barn wall), the location of each node (xi) is: xi =

(i − 1) L i = 1..N ( N − 1)

(1)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δx) is:

Δx =

L N −1

"Setup grid" N=10 [-] duplicate i=1,N x[i]=(i-1)*L/(N-1) end Deltax=L/(N-1)

(2)

"number of nodes" "position of each node" "distance between adjacent nodes"

A control volume is defined around each node and an energy balance is written for each control volume. The control volume for an arbitrary, internal node (i.e., a node that is not placed on the edge of the hay) experiences conduction heat transfer passing through the internal surface ( q LHS ), conduction heat transfer passing through the external surface ( q RHS ), and heat generation within the control volume ( g ). A steady-state energy balance for the control volume is shown in Fig. 2 and leads to: q LHS + q RHS + g = 0

(3)

Figure 2: An internal control volume

Each of the terms in the energy balance in Eq. (3) must be modeled using a rate equation. Conduction through the inner surface is driven by the temperature difference between nodes i-1 and i through the material that lies between these nodes. q LHS =

kh A (Ti −1 − Ti ) Δx

(4)

where A is the area of the wall (assumed to be 1 m2, corresponding to doing the problem on a per unit area of wall basis). The conduction into the outer surface is: q RHS =

kh A (Ti +1 − Ti ) Δx

(5)

The generation is the product of the volume of the control volume and the volumetric generation rate, which is approximately:

g = g T′′′i A Δx

(6)

where g ′′′ must be evaluated at the nodal temperature Ti. Substituting Eqs. (4) through (6) into Eq. (3) leads to:

kh A (Ti −1 − Ti ) kh A (Ti +1 − Ti ) + + g T′′′i A Δx = 0 for i = 2...( N − 1) Δx Δx

(7)

Figure 3 illustrates the control volume associated with the node that is placed on the outer surface of the hay (i.e., node N).

Figure 3: Control volume for node N located on hay outer surface

The energy balance for the control volume associated with node N is:

q LHS + g + qout = 0

(8)

where the conduction term is:

q LHS =

kh A (TN −1 − TN ) , Δx

(9)

the generation term is:

g = g T′′′N A

Δx , 2

(10)

(note the factor of 2 corresponding to half the volume), and the heat transfer to the external air is:

qconv =

(T

a , out

− TN )

Rw + Rconv ,out

(11)

where Rw =

thw kw A

(12)

and Rconv ,out =

1 ha ,out A

Substituting Eqs. (9) through (11) into Eq. (8) leads to:

(13)

kh A (TN −1 − TN ) Δx (Ta ,out − TN ) + g T′′′N A + =0 Δx 2 Rw + Rconv ,out

(14)

A similar procedure applied to the control volume associated with node 1 leads to: kh A (T2 − T1 ) Δx + g T′′′1 A + ha ,in A (Ta ,in − T1 ) = 0 Δx 2

(15)

Equations (7), (14), and (15) represent N equations in an equal number of unknowns; the solution of these equations provides the numerical solution. "Internal control volumes" duplicate i=2,(N-1) k_h*A*(T[i-1]-T[i])/Deltax+k_h*A*(T[i+1]-T[i])/Deltax+gen(T[i])*A*Deltax=0 end R_w=th_w/(k_w*A) "conduction resistance of barn wall" R_conv_out=1/(h_a_out*A) "convection resistance to external air" k_h*A*(T[N-1]-T[N])/Deltax+gen(T[N])*A*Deltax/2+(T_a_out-T[N])/(R_w+R_conv_out)=0 "Node N" k_h*A*(T[2]-T[1])/Deltax+gen(T[1])*A*Deltax/2+h_a_in*A*(T_a_in-T[1])=0 "Node 1"

If the EES program is solved then the temperature distribution will be placed in the Arrays window. b.) Prepare a plot that shows the temperature distribution as a function of position in the hay. The information in the Arrays table is used to prepare the plot shown in Figure 4.

Figure 4: Temperature as a function of position in the wall.

c.) Prepare a plot that shows that you are using a sufficient number of nodes in your numerical solution. The most relevant result of the calculation is the maximum temperature within the wall. T_max=MAX(T[1..N])

"Maximum temperature in the wall"

Comment out the number of nodes assignment: {N=10 [-]}

and prepare a parametric table that contains N and T_max (Figure 5).

Figure 5: Parametric table

The information in the parametric table is used to create Figure 6 which shows the maximum temperature as a function of the number of nodes; Figure 6 suggests that 50 nodes should be used to obtain a numerically convergent solution.

Figure 6: Predicted maximum temperature as a function of the number of nodes.

d.) Verify that your solution is correct by comparing it with an analytical solution in an appropriate limit. Prepare a plot that overlays your numerical solution and the analytical solution in this limit. There are a few limits; the easiest one would be to turn the generation off (i.e., set it to zero). Alternatively, set the generation rate to a constant value (e.g., 1 W/m3) and obtain the analytical solution. Modify the function: function gen(T) "volumetric heat generation in wall" "Input - T, temperature [K]" "Output - gen, volumetric rate of heat generation [W/m^3]" gen=1.0 [W/m^3] {1.5 [W/m^3]*sqrt(exp(T/320 [K]))} end

The general solution for a plane wall subjected to a constant generation rate was provided in Table 3-1: T =−

g ′′′ 2 x + C1 x + C2 2 kh

(16)

The boundary condition at x = L is: − kh A

dT dx

= x=L

(T

x= L

− Ta ,out )

Rw + Rconv ,out

(17)

where the temperature gradient can also be obtained from Table 3-1: dT dx

x= L

⎡ g ′′′ ⎤ g ′′′ = ⎢− x + C1 ⎥ = − L + C1 kh ⎣ kh ⎦ x=L

(18)

and ⎡ g ′′′ 2 ⎤ g ′′′ 2 Tx = L = ⎢ − x + C1 x + C2 ⎥ = − L + C1 L + C2 k k 2 2 h h ⎣ ⎦ x= L "Analytical solution for constant generation" g```_dot=gen(300 [K]) -k_h*A*dTdx_L=(T_L-T_a_out)/(R_w+R_conv_out) dTdx_L=-g```_dot*L/k_h+C_1 T_L=-g```_dot*L^2/(2*k_h)+C_1*L+C_2

The boundary condition at x = 0 is:

"obtain the rate of generation" "boundary condition at x=L" "temperature gradient at x=L" "temperature at x=L"

(19)

ha ,in A (Ta ,in − Tx =0 ) = −kh A

dT dx

(20) x =0

where the temperature gradient can also be obtained from Table 3-1: dT dx

x =0

⎡ g ′′′ ⎤ = ⎢− x + C1 ⎥ = C1 ⎣ kh ⎦ x =0

(21)

and ⎡ g ′′′ 2 ⎤ Tx =0 = ⎢ − x + C1 x + C2 ⎥ = C2 ⎣ 2 kh ⎦ x =0 h_a_in*A*(T_a_in-T_0)=-k_h*A*dTdx_0 dTdx_0=C_1 T_0=C_2

(22)

"boundary condition at x=0" "temperature gradient at x=0" "temperature at x=0"

Solving the problem shows that C1 = 45.0 K/m and C2 = 293.3 K. The solution at each node is obtained: duplicate i=1,N T_an[i]=-g```_dot*x[i]^2/(2*k_h)+C_1*x[i]+C_2 end

Figure 7 illustrates the analytical and numerical solutions and shows that they agree.

Figure 7: Numerical and analytical solutions in the limit that g ′′′ is constant.

Temperatures above Tfire= 200°F are considered to be a fire hazard and temperatures above Td = 140°F will result in a degradation of the hay to the point where it is not usable. e.) What is the maximum allowable thickness of hay (Lmax) based on keeping the maximum temperature below Tfire? You can either manually adjust L until the variable Tmax is equal to Tfire or simply set Tmax and comment out the assignment of the variable L and let EES automatically determine the correct value (note that you need to return the generation function to its original state). {L=5.0 [m]} T_fire=converttemp(F,K,200) T_max=T_fire

"thickness of hay" "combustion temperature"

Which leads to Lmax = 3.615 m. f.) If L = Lmax from (e) then how much of the hay will remain usable (what percent of the hay is lost to heat degradation)? Figure 8 illustrates the temperature distribution for L = 3.615 m and shows the extent of the region of the heat damaged hay.

Figure 8: Temperature distribution for L = 3.615 m.

The region of usable hay extends from 0 to 0.63 m and from 2.73 m to 3.62 m. Therefore, only 43% of the hay will be useable when it is removed from the bar. Note that simple calculations like this can be done easily using the Calculator function in EES (select Calculator from the Windows menu). The calculator environment includes all of the variables from the last run of EES. Therefore, typing ?L returns 3.615 (Figure 9).

Figure 9: Calculator window.

To calculate the efficiency of the storage process, use the Calculator window as shown in Figure 10.

Figure 10: Calculator window.

Problem 1.4-5 Solve the problem stated in EXAMPLE 1.3-2 numerically rather than analytically. a.) Develop a numerical model that can predict the temperature distribution within the lens. Prepare a plot of the temperature as a function of position. The inputs are entered in EES: "Problem 1.4-5" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q``_rad=0.1 [W/cm^2]*convert(W/cm^2,W/m^2) L=1.0 [cm]*convert(cm,m) T_a=converttemp(C,K,20) h=20 [W/m^2-K] k=1.5 [W/m-K] alpha=0.1 [1/mm]*convert(1/mm,1/m) A=1 [m^2]

"radiation incident on the lens" "thickness of lens" "ambient temperature" "heat transfer coefficient" "conductivity of lens" "absorption coefficient" "per unit area"

Nodes are distributed uniformly throughout the computational domain; the distance between adjacent nodes is: Δx =

L ( N − 1)

(1)

where N is the number of nodes. The position of each node is: xi = Δx ( i − 1) for i = 1...N N=10 [-] Dx=L/(N-1) duplicate i=1,N x[i]=Dx*(i-1) end

"number of nodes" "distance between adjacent nodes" "position of each node"

An energy balance on an internal control volume is shown in Figure 1.

Figure 1: Energy balance on an internal control volume

(2)

The energy balance is:

qtop + qbottom + g = 0

(3)

Substituting rate equations into Eq. (3) leads to:

kA kA ′′ α exp ( −α xi ) = 0 (Ti −1 − Ti ) + (Ti +1 − Ti ) + A Δx qrad Δx Δx

(4)

"internal control volume energy balances" duplicate i=2,(N-1) k*A*(T[i-1]-T[i])/Dx+k*A*(T[i+1]-T[i])/Dx+A*Dx*q``_rad*alpha*exp(-alpha*x[i])=0 end

An energy balance on node 1 located at the upper surface is shown in Figure 2.

Figure 2: Energy balance on the upper edge control volume

The energy balance for node 1 is: qconv + qbottom + g = 0

(5)

Substituting rate equations into Eq. (5) leads to: h A (Ta − T1 ) +

kA A Δx ′′ α exp ( −α x1 ) = 0 qrad (T2 − T1 ) + 2 Δx

(6)

The corresponding energy balance for node N located at the lower surface is: h A (Ta − TN ) +

kA A Δx ′′ α exp ( −α xN ) = 0 qrad (TN −1 − TN ) + 2 Δx

"upper edge" h*A*(T_a-T[1])+k*A*(T[2]-T[1])/Dx+A*Dx*q``_rad*alpha*exp(-alpha*x[1])/2=0 "lower edge" h*A*(T_a-T[N])+k*A*(T[N-1]-T[N])/Dx+A*Dx*q``_rad*alpha*exp(-alpha*x[N])/2=0

(7)

The temperature distribution in the lens is shown in Figure 3.

Figure 3: Temperature as a function of position in the lens

b.) Plot some characteristic of your solution as a function of the number of nodes to show that you are using a sufficient number of nodes. The maximum temperature in the lens is obtained using the Max command in EES: T_max=MAX(T[1..N])

"maximum temperature in lens"

The maximum temperature and number of nodes are placed in a parametric table; the number of nodes is varied and the results are shown in Figure 4.

Figure 4: Maximum temperature as a function of the number of nodes

c.) Think of a sanity check that you can use to gain confidence in your model; that is, can you change some input parameter and show that the solution behaves as you would expect. Support your answer with a plot. As the lens conductivity becomes very large, the temperature rise within the lens should be reduced. Figure 3 illustrates the predicted result when the conductivity is increased by a factor of 10, to 15 W/m-K. d.) Plot the maximum lens temperature as a function of the heat transfer coefficient, h . Figure 5 illustrates the maximum temperature in the lens as a function of the heat transfer coefficient.

Figure 5: Maximum temperature as a function of the heat transfer coefficient

Problem 1.4-6 A current lead must be designed to carry current to a cryogenic superconducting magnet, as shown in Figure P1.4-6. TH = 300 K current lead D = 1 cm L = 20 cm Ic = 1000 amp

TC = 100 K

Figure P1.4-6: Current lead.

The current lead carries Ic = 1000 amp and therefore experiences substantial generation of thermal energy due to ohmic dissipation. The electrical resistivity of the lead material depends on temperature according to:

⎡ ohm-m ⎤ ⎣ K ⎥⎦

ρe = 17x10−9 [ ohm-m ] + (T − 300 [ K ]) 5x10−11 ⎢

(1)

The length of the current lead is L = 20 cm and the diameter is D = 1 cm. The hot end of the lead (at x = 0) is maintained at Tx=0 = TH = 300 K and the cold end (at x = L) is maintained at Tx=L = TC = 100 K. The conductivity of the lead material is k = 400 W/m-K. The lead is installed in a vacuum chamber and therefore you may assume that the external surfaces of the lead (the outer surface of the cylinder) are adiabatic. a.) Develop a numerical model in EES that can predict the temperature distribution within the current lead. Plot the temperature as a function of position. The input information is entered in EES and a function is used to define the electrical resistivity according to Eq. (1): $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function rho_e(T) "Input: T - temperature (K) Output: rho_e - electrical resistivity (ohm-m)" rho_e=17e-9 [ohm-m]+(T-300 [K])*5e-11 [ohm-m/K] end "Inputs" L=20 [cm]*convert(cm,m) T_H=300 [K] T_C=100 [K]

"length of current lead" "hot end temperature" "cold end temperature"

Ic=1000 [amp] k=400 [W/m-K] D_cm=1 [cm] D=D_cm*convert(cm,m)

"current" "conductivity of lead" "diameter of lead, in cm" "diameter of lead"

Nodes are distributed uniformly throughout the computational, the location of each node (xi) is: xi =

(i − 1) L i = 1..N ( N − 1)

(2)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δx) is:

Δx = N=21 [-] duplicate i=1,N x[i]=L*(i-1)/(N-1) end Dx=L/(N-1)

L N −1

(3)

"number of nodes" "axial position" "distance between adjacent nodes"

A control volume is defined around each node and an energy balance is written for each control volume. The control volume for an arbitrary, internal node (i.e., a node that is not placed on the edge of the hay) experiences conduction heat transfer passing through the top surface ( qtop ), conduction heat transfer passing through the bottom surface ( qbottom ), and heat generation within the control volume ( g ). A steady-state energy balance for an internal control volume:

q LHS + q RHS + g = 0

(4)

Each of the terms in the energy balance in Eq. (4) must be modeled using a rate equation. Conduction through the inner surface is driven by the temperature difference between nodes i-1 and i through the material that lies between these nodes.

qtop =

k Ac (Ti −1 − Ti ) Δx

(5)

where Ac is the cross-sectional area of the lead:

Ac = The conduction through the bottom surface is:

π D2 4

(6)

qbottom =

k Ac (Ti +1 − Ti ) Δx

(7)

The generation is the product of the electrial resistance of the material in the control volume and the current squared:

g = ρ e,T =Ti

Δx 2 Ic Ac

(8)

Substituting Eqs. (5), (7), and (8) into Eq. (4) leads to:

k Ac (Ti −1 − Ti ) k Ac (Ti +1 − Ti ) Δx 2 Ic = 0 for i = 2...( N − 1) + + ρ e,T =Ti Ac Δx Δx

(9)

A_c=pi*D^2/4 "cross-sectional area of lead" duplicate i=2,(N-1) k*A_c*(T[i-1]-T[i])/Dx+k*A_c*(T[i+1]-T[i])/Dx+rho_e(T[i])*Dx*Ic^2/A_c=0 "energy balance on internal nodes" end

The temperatures of nodes 1 and N are specified:

T[1]=T_H T[N]=T_C

T1 = TH

(10)

TN = TH

(11)

"hot end temperature" "cold end temperature"

Figure 2 illustrates the temperature as a function of position. 300 280

Temperature (K)

260 240 220 200 180 160 140 120 100 0

0.04

0.08

0.12

0.16

Position (m)

Figure 2: Temperature as a function of position.

0.2

b.) Determine the rate of energy transfer into the superconducting magnet at the cold end of the current lead. This parasitic must be removed in order to keep the magnet cold and therefore must be minimized in the design of the current lead. An energy balance on node N leads to:

qC =

k Ac (TN −1 − TN ) Δx 2 Ic + ρe ,T =TN Δx 2 Ac

q_dot=k*A_c*(T[N-1]-T[N])/Dx+rho_e(T[N])*Dx*Ic^2/(2*A_c)

(12)

"parasitic to cold end"

which leads to qC = 45.7 W. c.) Prepare a plot showing the rate of energy transfer into the magnet as a function of the number of nodes used in your model.

Rate of heat transfer to cold end (W)

Figure 3 illustrates qC as a function of N. 46 45 44 43 42 41 40 2

10

100

200

Number of nodes Figure 3: Rate of heat transfer to the cold end as a function of the number of nodes.

d.) Plot the rate of heat transfer to the cold end as a function of the diameter of the lead. You should see a minimum value and therefore an optimal diameter - explain why this occurs. Figure 4 illustrates the rate of heat transfer to the cold end as a function of the diameter of the lead. At very low diameters the ohmic dissipation is large because the electrical resistance is high and therefore the parasitic is large. At very large diameters, the thermal resistance of the lead is large therefore the parasitic is large. The optimal diameter is around 0.9 cm and balances these effects.

Rate of heat transfer to cold end (W)

300 250 200 150 100 50 0 0.5

0.75

1

1.25

1.5

1.75

2

Diameter of lead (cm) Figure 4: Rate of heat transfer to cold end as a function of the diameter of the lead.

e.) Prepare a plot showing the optimal diameter and minimized rate of heat transfer to the cold end as a function of the current that must be carried by the lead. You may want to use the Min/Max Table selection from the Calculate menu to accomplish this.

225

2

200

1.8

175 1.6 150 1.4

D

125 100

1.2

qC

75 1 50 0.8

25 0 500

1000

1500

2000

2500

3000

3500

4000

4500

Diameter of lead (cm)

Rate of heat transfer to the cold end (W)

Figure 5 illustrates the minimized rate of heat transfer to the cold end and the optimal diameter as function of the level of current.

0.6 5000

Current (amp)

Figure 5: Minimized rate of heat transfer and optimal lead diameter as a function of current.

Problem 1.4-7 Figure P1.4-7 illustrates a plane wall. The temperature distribution in the wall is 1-D and the problem is steady state.

h , T∞

node 1

node 2 node 3 qL′′ x L g ′′′ = a x, k Figure P1.4-7: Three-node model of a plane wall.

There is generation of thermal energy in the wall. The generation per unit volume is not uniform but rather depends on position according to:

g ′′′ = a x

(1)

where a is a constant and x is position. The left side of the wall experiences a specified heat flux, q ′′L . The right side of the wall experiences convection with heat transfer coefficient h to fluid at temperature T∞. The thickness of the wall is L and the conductivity of the wall material, k, is constant. You are going to develop a numerical model with 3 nodes, as shown in Figure P1.4-7. The nodes are distributed uniformly throughout the domain. Derive the three equations that must be solved in order to implement the numerical model. Do not solve these equations. An energy balance on node 1 leads to:

q ′′L Ac + g ′′′x =0 Ac

Δx k Ac + (T2 − T1 ) = 0 Δx 2

(2)

Substituting Eq. (1) into Eq. (2) and dividing through by Ac leads to: q L′′ +

k (T2 − T1 ) = 0 Δx

(3)

An energy balance on node 2 leads to: k Ac kA (T1 − T2 ) + g ′′′x= L / 2 Ac Δx + c (T3 − T2 ) = 0 Δx Δx Substituting Eq. (1) into Eq. (4) and dividing through by Ac leads to:

(4)

k L k (T1 − T2 ) + a Δx + (T3 − T2 ) = 0 Δx 2 Δx

(5)

An energy balance on node 3 leads to: k Ac Δx (T2 − T3 ) + g ′′′x= L Ac + h Ac (T∞ − T3 ) = 0 2 Δx

(6)

Substituting Eq. (1) into Eq. (6) and dividing through by Ac leads to: k Δx (T2 − T3 ) + a L + h (T∞ − T3 ) = 0 Δx 2 Equations (3), (5), and (7) can be solved to provide T1, T2, and T3.

(7)

PROBLEM 1.5-1 (1-11 in text): Hay Temperature (revisited) Reconsider Problem P1.3-8, but obtain a solution numerically using MATLAB. The description of the hay bale is provided in Problem P1.3-8. Prepare a model that can consider the effect of temperature on the volumetric generation. Increasing temperature tends to increase the rate of reaction and therefore increase the rate of generation of thermal energy; the volumetric rate of generation can be approximated by: g ′′′ = a + bT where a = -1 W/m3 and b = 0.01 W/m3-K. Note that at T = 300 K, the generation is 2 W/m3 but that the generation increases with temperature. a.) Prepare a numerical model of the hay bale using EES. Plot the temperature as a function of position within the hay bale. The input information is entered in EES and a function is used to define the volumetric generation: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function gen(T) "volumetric heat generation in wall" "Input - T, temperature [K]" "Output - gen, volumetric rate of heat generation [W/m^3]" a=-1 [W/m^3] b=0.01 [W/m^3-K] gen=a+b*T

"coefficients in generation function"

end "Inputs" L = 1 [m] R_bale= 5 [ft]*convert(ft,m) t_p=0.045 [inch]*convert(inch,m) k_p=0.15 [W/m-K] T_infinity=converttemp(C,K,20) h=10 [W/m^2-K] k=0.04 [W/m-K]

"per unit length of bale" "bale radius" "plastic thickness" "plastic conductivity" "ambient temperature" "heat transfer coefficient" "hay conductivity"

Nodes are distributed uniformly throughout the computational domain (which consists only of the hay, not the plastic), the location of each node (ri) is: ri =

(i − 1) R ( N − 1) bale

i = 1..N

(1)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δr) is: Δr =

Rbale ( N − 1)

(2)

"Setup grid" N=50 [-] duplicate i=1,N r[i]=(i-1)*R_bale/(N-1) end Deltar=R_bale/(N-1)

"number of nodes" "position of each node" "distance between adjacent nodes"

A control volume is defined around each node and an energy balance is written for each control volume. The control volume for an arbitrary, internal node (i.e., a node that is not placed on the edge or at the center of the hay) experiences conduction heat transfer passing through the internal surface ( q LHS ), conduction heat transfer passing through the external surface ( q RHS ), and heat generation within the control volume ( g ). A steady-state energy balance for the control volume is shown in Figure 1: q LHS + q RHS + g = 0

(3)

Figure 1: Internal node energy balance

Each of the terms in the energy balance in Eq. (3) must be modeled using a rate equation. Conduction through the inner surface is driven by the temperature difference between nodes i-1 and i through the material that lies between these nodes.

q LHS

Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ L 2 ⎠ ⎝ = (Ti −1 − Ti ) Δr

(4)

where L is the length of the bale (assumed to be 1 m, corresponding to doing the problem on a per unit length of bale basis). The conduction into the outer surface of the control volume is:

q RHS

Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟L 2 ⎠ ⎝ = (Ti +1 − Ti ) Δr

(5)

The generation is the product of the volume of the control volume and the volumetric generation rate, which is approximately: 2 2 ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ g = g ′′′ (Ti ) π L ⎢⎜ ri + r − − ⎟ ⎜ i ⎟ ⎥ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎢⎣⎝

(6)

where g ′′′ (Ti ) is the volumetric rate of generation evaluated at the nodal temperature Ti. Substituting Eqs. (4) through (6) into Eq. (3) leads to: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ L k 2 π ⎜ ri + 2 2 ⎟L ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ 2 ⎠ 2 ⎠ ⎝ ⎝ ′′′  π T T T T g T L r r − + − + + − − ( i −1 i ) ( i +1 i ) ( i ) ⎢⎜ i ⎟ ⎜ i ⎟ ⎥ = 0 (7) 2 ⎠ ⎝ 2 ⎠ ⎥⎦ Δr Δr ⎢⎣⎝ for i = 2...( N − 1) Figure 2 illustrates the control volume associated with the node that is placed on the outer surface of the hay (i.e., node N).

Figure 2: Control volume for node N located on hay outer surface

The energy balance for the control volume associated with node N is: q LHS + g = qout

(8)

where the conduction term is:

q LHS

Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎠ ⎝ = (TN −1 − TN ) , Δr

(9)

the generation term is: 2 ⎡ Δr ⎞ ⎤ ⎛ g = g ′′′ (TN ) π L ⎢ rN2 − ⎜ rN − ⎟ ⎥ , 2 ⎠ ⎦⎥ ⎝ ⎣⎢

(10)

Note that the volume in Eq. (10) is calculated differently from the volume in Eq. (6) because the control volume is half as wide radially. The heat transfer to the external air is: qout =

where

(TN − T∞ ) R p + Rconv

(11)

Rp =

thp

(12)

k p 2 π Rbale L

and

Rconv ,out =

1 h 2 π Rbale L

(13)

Substituting Eqs. (9) through (11) into Eq. (8) leads to: Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎡ 2 ⎛ Δr ⎞ ⎤ (TN − T∞ ) 2 ⎠ ⎝ ′′′  T T g T L r r π − + − − ( N −1 N ) ( N) ⎢ N ⎜ N ⎟ ⎥= Δr 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣

(14)

A similar procedure applied to the control volume associated with node 1 leads to: Δr ⎞ ⎛ k 2 π ⎜ r1 + ⎟L 2 ⎠ ⎝ (T2 − T1 ) + g ′′′ (T1 ) π Δr

Δr ⎞ ⎛ L ⎜ r1 + ⎟ =0 2 ⎠ ⎝ 2

(15)

Equations (7), (14), and (15) represent N equations in an equal number of unknowns; the solution of these equations provides the numerical solution. "Internal control volumes" duplicate i=2,(N-1) k*2*pi*(r[i]-Deltar/2)*L*(T[i-1]-T[i])/Deltar+k*2*pi*(r[i]+Deltar/2)*L*(T[i+1]T[i])/Deltar+gen(T[i])*pi*L*((r[i]+Deltar/2)^2-(r[i]-Deltar/2)^2)=0 end "node N" R_p=t_p/(k_p*2*pi*R_bale*L) "conduction resistance of plastic" R_conv=1/(h*2*pi*R_bale*L) "convection resistance" k*2*pi*(r[N]-Deltar/2)*L*(T[N-1]-T[N])/Deltar+gen(T[N])*pi*L*(r[N]^2-(r[N]-Deltar/2)^2)=(T[N]T_infinity)/(R_p+R_conv) "node 1" k*2*pi*(r[1]+Deltar/2)*L*(T[2]-T[1])/Deltar+gen(T[1])*pi*L*(r[1]+Deltar/2)^2=0

If the EES program is solved then the temperature distribution will be placed in the Arrays window. The temperature as a function of position is shown in Figure 3.

Figure 3: Temperature as a function of position within the bale

b.) Show that your model has numerically converged; that is, show some aspect of your solution as a function of the number of nodes in your solution and discuss an appropriate number of nodes to use. The maximum temperature (i.e., the temperature at the center of the bale) is shown in Figure 4 as a function of the number of nodes. The model is numerically converged after approximately N = 20.

Figure 4: Predicted maximum temperature as a function of the number of nodes

c.) Verify your numerical model by comparing your answer to an analytical solution in some, appropriate limit. The result of this step should be a plot which shows the temperature as a function of radius predicted by both your numerical solution and the analytical solution and demonstrates that they agree. The analytical solution derived in the problem 1.3-8 is used to compute the temperature at each nodal position: "Analytical solution from Problem 1.3-8" a=-1 [W/m^3] "coefficients for volumetric generation function" b=0.01 [W/m^3-K] dTdr_Rbale = -BesselJ(1,(b/k)^(1/2)*R_bale)*(b/k)^(1/2)*C_2 "symbolic expressions from Maple" T_Rbale = BesselJ(0,(b/k)^(1/2)*R_bale)*C_2-1/b*a -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv)"interface energy balance" duplicate i=1,N T_an[i]=BesselJ(0,sqrt(b/k)*r[i])*C_2-a/b end

Figure 3 illustrates the analytical solution overlaid on the numerical solution and demonstrates agreement. d.) Prepare a numerical model of the hay bale using MATLAB. Plot the temperature as a function of position within the hay bale. A new m-file is opened and formatted as a function with a single input (the number of nodes) and two outputs (vectors containing the radial position and temperature at each node). function[r,T]=P1p5_1(N) L = 1; R_bale= 1.524; t_p=0.00114; k_p=0.15; T_infinity=293.2; h=10; k=0.04;

%per unit length of bale (m) %bale radius (m) %plastic thickness (m) %plastic conductivity (W/m-K) %ambient temperature (K) %heat transfer coefficient (W/m^2-K) %hay conductivity (W/m-K)

end

A function is defined that returns the volumetric rate of generation as a function of temperature; the function is placed at the bottom of the same m-file so that it is accessible locally to P1p5_1. function[gv]=gen(T) %coefficients of function a=-1; %(W/m^3) b=0.01; %(W/m^3-K) gv=a+b*T; end

The radial position of each node is stored in the vector r. Deltar=R_bale/(N-1); %distance between adjacent nodes (m) for i=1:N r(i,1)=Deltar*(i-1); %radial location of each node (m) end

The problem is nonlinear because the generation rate depends on temperature; therefore, the method of successive substitution is used. An initial guess for the temperature distribution is stored in the vector T_g: %initial guess for temperature distribution for i=1:N T_g(i,1)=T_infinity; end

The guess values for temperature are used to setup the matrix A and vector b which contain the matrix formulation of the equations. The energy balance for node 1 is placed in row 1 of A. Δr ⎞ ⎛ k 2 π ⎜ r1 + ⎟L 2 ⎠ ⎝ (T2 − T1 ) + g ′′′ (T1* ) π Δr

Δr ⎞ ⎛ L ⎜ r1 + ⎟ =0 2 ⎠ ⎝ 2

(16)

where T1* is the guess value of the temperature or ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ 2 ⎢ k 2 π ⎜ r1 + 2 ⎟ L ⎥ ⎢ k 2 π ⎜ r1 + 2 ⎟ L ⎥ ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥ = − g ′′′ T * π L ⎛ r + Δr ⎞ T1 ⎢ − ( 1 ) ⎜⎝ 1 2 ⎟⎠ 2 Δr Δr ⎢ ⎥ ⎢ ⎥   

⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ b (1)  

 

A (1,1)

(17)

A (1,2)

The energy balances for the internal nodes are: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ L k 2 π ⎜ ri + 2 2 ⎟L ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ 2 ⎠ 2 ⎠ ⎝ ⎝ * (Ti −1 − Ti ) + (Ti +1 − Ti ) + g ′′′ (Ti ) π L ⎢⎜ ri + ⎟ − ⎜ ri − ⎟ ⎥ = 0 (18) 2 ⎠ ⎝ 2 ⎠ ⎥⎦ Δr Δr ⎢⎣⎝ for i = 2...( N − 1) or

⎡ Δr ⎞ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ L k 2 π ⎜ ri + 2 ⎟ L ⎥ ⎢ k 2 π ⎜ ri − 2 ⎟ L ⎥ ⎝ ⎠ − ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥+ Ti ⎢ − i −1 Δr Δr Δr ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥  

 

⎦ A ( i ,i −1)

A ( i ,i )

⎡ Δr ⎞ ⎤ ⎛ 2 2 ⎢ k 2 π ⎜ ri + 2 ⎟ L ⎥ ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ ⎝ ⎠ * ′′′  ⎢ ⎥ = − + Ti +1 g (Ti ) π L ⎢⎜ ri ⎟ − ⎜ ri − ⎟ ⎥ Δr 2 ⎠ ⎦⎥ 2 ⎠ ⎝ ⎢ ⎥ ⎢⎝ ⎣  

⎢⎣ ⎥⎦  b (i )  

(19)

A ( i ,i +1)

for i = 2 .. ( N − 1)

The energy balance for node N is: Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎡ 2 ⎛ Δr ⎞ ⎤ (TN − T∞ ) 2 ⎠ ⎝ * (TN −1 − TN ) + g ′′′ (TN ) π L ⎢ rN − ⎜ rN − ⎟ ⎥ = Δr 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣

(20)

or ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ k 2 π ⎜ rN − ⎟ L ⎥ ⎢ k 2 π ⎜ rN − 2 ⎟ L ⎥ ⎢ 1 2 ⎠ ⎝ ⎠ − ⎝ ⎥ + TN −1 ⎢ ⎥= TN ⎢ − Δr R p + Rconv ⎥ Δr ⎢ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥  

 

⎦ A ( N , N −1)

A( N , N )

(21)

2 ⎡ T∞ Δr ⎞ ⎤ ⎛ − g ′′′ (TN* ) π L ⎢ rN2 − ⎜ rN − ⎟ ⎥ − 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣   

b( N )

The matrices A and b are initialized and the resistances due to convection and conduction through the plastic are computed: A=spalloc(N,N,3*N); b=zeros(N,1); R_p=t_p/(k_p*2*pi*R_bale*L); R_conv=1/(h*2*pi*R_bale*L);

%resistance through plastic %resistance due to convection

The matrices A and b are filled in according to Eqs. (17), (19), and (21): %Node 1 A(1,1)=-k*2*pi*(r(1)+Deltar/2)*L/Deltar; A(1,2)=k*2*pi*(r(1)+Deltar/2)*L/Deltar; b(1)=-gen(T_g(i))*pi*L*(r(1)+Deltar/2)^2;

%Nodes 2 to (N-1) for i=2:(N-1) A(i,i)=-k*2*pi*(r(i)-Deltar/2)*L/Deltark*2*pi*(r(i)+Deltar/2)*L/Deltar; A(i,i-1)=k*2*pi*(r(i)-Deltar/2)*L/Deltar; A(i,i+1)=k*2*pi*(r(i)+Deltar/2)*L/Deltar; b(i)=-gen(T_g(i))*pi*L*((r(i)+Deltar/2)^2-(r(i)-Deltar/2)^2); end %Node N A(N,N)=-k*2*pi*(r(N)-Deltar/2)*L/Deltar-1/(R_p+R_conv); A(N,N-1)=k*2*pi*(r(N)-Deltar/2)*L/Deltar; b(N)=-gen(T_g(N))*pi*L*(r(N)^2-(r(N)-Deltar/2)^2)T_infinity/(R_p+R_conv);

The temperature distribution is obtained according to: T=A\b;

The successive substitution process occurs within a while loop that is terminated when some convergence error, err, goes below a tolerance, tol. The tolerance is set and the error is initialized to a value that will ensure that the loop executes at least once. Once the solution is obtained, it is compared with the guess value to determine an error. The guess values are reset and, if the error is not sufficiently small then the process is repeated. The code is shown below; the new lines are shown in bold: function[r,T]=P1p5_1(N) L = 1; R_bale= 1.524; t_p=0.00114; k_p=0.15; T_infinity=293.2; h=10; k=0.04;

%per unit length of bale (m) %bale radius (m) %plastic thickness (m) %plastic conductivity (W/m-K) %ambient temperature (K) %heat transfer coefficient (W/m^2-K) %hay conductivity (W/m-K)

Deltar=R_bale/(N-1); %distance between adjacent nodes (m) for i=1:N r(i,1)=Deltar*(i-1); %radial location of each node (m) end %initial guess for temperature distribution for i=1:N T_g(i,1)=T_infinity; end A=spalloc(N,N,3*N); b=zeros(N,1); R_p=t_p/(k_p*2*pi*R_bale*L); R_conv=1/(h*2*pi*R_bale*L); tol=0.1;

%resistance through plastic %resistance due to convection

%tolerance for convergence (K)

err=2*tol; %error initialization while(err>tol) %Node 1 A(1,1)=-k*2*pi*(r(1)+Deltar/2)*L/Deltar; A(1,2)=k*2*pi*(r(1)+Deltar/2)*L/Deltar; b(1)=-gen(T_g(i))*pi*L*(r(1)+Deltar/2)^2; %Nodes 2 to (N-1) for i=2:(N-1) A(i,i)=-k*2*pi*(r(i)-Deltar/2)*L/Deltark*2*pi*(r(i)+Deltar/2)*L/Deltar; A(i,i-1)=k*2*pi*(r(i)-Deltar/2)*L/Deltar; A(i,i+1)=k*2*pi*(r(i)+Deltar/2)*L/Deltar; b(i)=-gen(T_g(i))*pi*L*((r(i)+Deltar/2)^2-(r(i)-Deltar/2)^2); end %Node N A(N,N)=-k*2*pi*(r(N)-Deltar/2)*L/Deltar-1/(R_p+R_conv); A(N,N-1)=k*2*pi*(r(N)-Deltar/2)*L/Deltar; b(N)=-gen(T_g(N))*pi*L*(r(N)^2-(r(N)-Deltar/2)^2)T_infinity/(R_p+R_conv); T=A\b; %obtain temperature distribution err=sum(abs(T-T_g))/N %calculate error T_g=T; end end function[gv]=gen(T) %coefficients of function a=-1; %(W/m^3) b=0.01; %(W/m^3-K) gv=a+b*T; end

The temperature as a function of radius is shown in Figure 5.

Figure 5: Predicted temperature as a function of radial position

Problem 1.5-2 (1-12 in text): Mass Flow Meter (re-visited) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). Assume that the conductivity of the material that is used to make the test section is not actually constant as was assumed in Problem 1.3-9 (1-9 in text) but rather depends on temperature according to:

k = 10

W ⎡ W ⎤ + 0.035 ⎢ T − 300 [ K ]) 2 ( m-K ⎣ m-K ⎥⎦

a.) Develop a numerical model of the mass flow meter using MATLAB. Plot the temperature as a function of radial position for the conditions shown in Figure P1.3-9 (P1-9 in text) with the temperature-dependent conductivity. The inputs are entered in a MATLAB function that requires as an input the number of nodes (N): function[r,T_C]=P1p5_2(N) r_out=0.0254; %outer radius of test section (m) r_in=0.01905; %inner radius of test section (m) h_bar_out=10; %external convection coefficient (W/m^2-K) T_infinity=293.2; %air temperature (K) T_f=291.2; %fluid temperature (K) gv=1e7; %rate of generation (W/m^3) m_dot=0.75; %mass flow rate (kg/s) th_ins=0.00635; %thickness of the insulation (m) k_ins=1.5; %insulation conductivity (W/m-K) L=0.0762; %length of the test section (m) C=2500; %constant for convection relationship

The convection coefficient on the internal surface is computed: h_bar_in=C*m_dot^0.8;

%internal convection coefficient

A function is defined that returns the conductivity of the material: function[k]=k_t(T) %conductivity of the material % %Inputs: % T: temperature (K) % %Outputs: % k: conductivity (W/m-K) k=10+0.035*(T-300); end

A uniform distribution of nodes is used, the radial location of each node (ri) is:

ri = rin +

( i − 1) r − r ( ) ( N − 1) out in

for i = 1..N

(1)

where N is the number of nodes. The radial distance between adjacent nodes (Δr) is:

Δr =

( rout − rin ) ( N − 1)

(2)

DELTAr=(r_out-r_in)/(N-1); %distance between adjacent nodes (m) for i=1:N r(i)=r_in+(r_out-r_in)*(i-1)/(N-1); %position of each node (m) end

The system of equations is placed in matrix format. AX =b

(3)

The most logical technique for ordering the unknown temperatures in the vector X is: ⎡ X 1 = T1 ⎤ ⎢ X =T ⎥ 2 ⎥ X =⎢ 2 ⎢ ... ⎥ ⎢ ⎥ ⎣ X N = TN ⎦

(4)

Equation (4) shows that the unknown temperature at node i (i.e., Ti) corresponds to element i of vector X (i.e., Xi). The most logical technique for placing the equations into the A matrix is: ⎡ row 1 = control volume 1 equation ⎤ ⎢ row 2 = control volume 2 equation ⎥ ⎥ A=⎢ ⎢ ⎥ ... ⎢ ⎥ ⎣ row N = control volume N equation ⎦

(5)

In Eq. (5), the equation for control volume i is placed into row i. An energy balance is carried out on a control volume surrounding each node. For node 1, placed at the inner surface (Figure P1.5-2-1):

qconv ,in + qouter + g = 0

(6)

qouter g1 qconv ,in

T2 T1

Figure P1.5-2-1: Control volume around node 1.

The rate equation for convection is: qconv ,in = hin 2 π rin L (T f − T1 )

(7)

The rate equation for conduction is: Δr ⎞ (T2 − T1 ) ⎛ qouter = kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(8)

The rate equation for generation is: g = 2 π rin

Δr L g ′′′ 2

(9)

Substituting Eqs. (7) through (9) into Eq. (6) leads to: Δr ⎞ (T2 − T1 ) ⎛ + π rin Δr L g ′′′ = 0 hin 2 π rin L (T f − T1 ) + kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(10)

Equation (10) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ T1 ⎢ − hin 2 π rin L − kT =(T1 +T2 ) / 2 2 π ⎜ rin + = ⎟ ⎥ + T2 ⎢ kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣ −π rin Δr L g ′′′ − hin 2 π rin LT f

(11)

An energy balance on an internal node is shown in Figure P1.5-2-2: qinner + qouter + g = 0

qouter g qinner

Ti+1 Ti Ti-1

Figure P1.5-2-2: Control volume around internal node i.

(12)

The rate equations for conduction are: Δr ⎞ (Ti +1 − Ti ) ⎛ qouter = kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(13)

Δr ⎞ ( T − T ) ⎛ qinner = kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i 2 ⎠ Δr ⎝

(14)

The rate equation for generation is: g = 2 π ri Δr L g ′′′

(15)

Substituting Eqs. (13) through (15) into Eq. (12) for all of the internal nodes leads to: Δr ⎞ (T − T ) Δr ⎞ (Ti +1 − Ti ) ⎛ ⎛ kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i + kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L Δr Δr 2 ⎠ 2 ⎠ ⎝ ⎝ +2 π ri Δr L g ′′′ = 0 for i = 2.. ( N − 1)

(16)

Equation (16) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ Δr ⎞ L Δr ⎞ L ⎤ ⎛ ⎛ Ti ⎢ − kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ +Ti −1 ⎢ −kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ ⎥ + Ti +1 ⎢ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎦⎥ ⎝ ⎝ ⎣ ⎣ = −2 π ri Δr L g ′′′ for i = 2.. ( N − 1)

(17)

An energy balance on node N placed on the outer surface is shown in Figure P1.5-2-3: qinner + qair + g = 0

(18)

qair TN g qinner

TN-1

Figure P1.5-2-3: Control volume around internal node N.

The rate equation for the heat transfer with the air is: qair =

(R

(T∞ − TN )

ins

+ Rconv ,out )

(19)

where

⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎣ ⎦ Rins = 2 π L kins Rconv ,out =

1 2 π ( rout + thins ) L hout

(20)

(21)

R_ins=log((r_out+th_ins)/r_out)/(2*pi*L*k_ins); %resistance to conduction through insulation R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out); %resistance to convection from the outside surface of the insulation

The rate equation for conduction is: Δr ⎞ (T − T ) ⎛ qinner = kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N Δr 2 ⎠ ⎝

(22)

The rate equation for generation is: g = 2 π rout

Δr L g ′′′ 2

(23)

Substituting Eqs. (19), (22), and (23) into Eq. (18) leads to:

(T∞ − TN ) + 2 π r Δr L g ′′′ = 0 Δr ⎞ (T − T ) ⎛ kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N + out Δr 2 ⎠ 2 ⎝ ( Rins + Rconv,out )

(24)

Equation (24) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ ⎤ 1 Δr ⎞ L ⎛ TN ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ − ⎥ 2 ⎠ Δr ( Rins + Rconv ,out ) ⎥⎦ ⎝ ⎢⎣ ⎡ Δr ⎞ L ⎤ ⎛ +TN −1 ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ ⎥ 2 ⎠ Δr ⎦ ⎝ ⎣ T∞ = − π rout Δr L g ′′′ − ( Rins + Rconv,out )

(25)

Equations (11), (17), and (25) are N equations in the N unknown temperatures. Because they are non-linear, they must be linearized and a successive substitution method used. A guess temperature distribution ( Tˆi ) is assumed: %initial guess for temperature distribution for i=1:N Tg(i,1)=T_f; end

The matrix A is defined as a sparse matrix with at most 3N nonzero entries: %initialize A and b A=spalloc(N,N,3*N); b=zeros(N,1);

The solution is placed within a while loop that terminates when the error between the solution and the guess is less than some tolerance: err=999; tol=0.01; while(err>tol)

%initial value for error (K), must be larger than tol %tolerance for convergence (K)

The equation for node 1, Eq. (11), is linearized by using the guess temperature distribution to compute the conductivity: ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ T1 ⎢ − hin 2 π rin L − kT = Tˆ +Tˆ / 2 2 π ⎜ rin + = ⎟ ⎥ + T2 ⎢ kT =(Tˆ1 +Tˆ2 ) / 2 2 π ⎜ rin + ⎟ ( 1 2) 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣  

 

A1,1

A1,2

−π rin Δr L g ′′′ − hin 2 π rin LT f  

b1

A(1,1)=-h_bar_in*2*pi*r_in*L-... k_t((Tg(1)+Tg(2))/2)*2*pi*(r_in+DELTAr/2)*L/DELTAr; A(1,2)=k_t((Tg(1)+Tg(2))/2)*2*pi*(r_in+DELTAr/2)*L/DELTAr; b(1)=-pi*r_in*DELTAr*L*gv-h_bar_in*2*pi*r_in*L*T_f;

The equations for the internal nodes, Eq. (17), is also linearized:

(26)

⎡ Δr ⎞ L Δr ⎞ L ⎤ ⎛ ⎛ Ti ⎢ − kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr 2 ⎠ Δr ⎦⎥ ⎝ ⎝ ⎣ 

Ai ,i

⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ +Ti −1 ⎢ −kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ ⎥ + Ti +1 ⎢ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣ 

 

(27)

Ai ,i +1

Ai ,i −1

= −2 π ri Δr L g ′′′ for i = 2.. ( N − 1)  

bi

for i=2:(N-1) A(i,i)=-k_t((Tg(i)+Tg(i-1))/2)*2*pi*(r(i)-DELTAr/2)*L/DELTAr... -k_t((Tg(i)+Tg(i+1))/2)*2*pi*(r(i)+DELTAr/2)*L/DELTAr; A(i,i-1)=k_t((Tg(i)+Tg(i-1))/2)*2*pi*(r(i)-DELTAr/2)*L/DELTAr; A(i,i+1)=k_t((Tg(i)+Tg(i+1))/2)*2*pi*(r(i)+DELTAr/2)*L/DELTAr; b(i)=-2*pi*r(i)*DELTAr*L*gv; end

The equation for node N, Eq. (25), is linearized: ⎡ ⎤ 1 Δr ⎞ L ⎛ TN ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ − ⎥ 2 ⎠ Δr ( Rins + Rconv ,out ) ⎦⎥ ⎝ ⎣⎢ 

AN , N

⎡ Δr ⎞ L ⎤ ⎛ +TN −1 ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ ⎥ 2 ⎠ Δr ⎦ ⎝ ⎣  

(28)

AN , N −1

T∞ ( Rins + Rconv,out )  

= − π rout Δr L g ′′′ −

bN

A(N,N)=-k_t((Tg(N)+Tg(N-1))/2)*2*pi*(r_in-DELTAr/2)*L/DELTAr-... 1/(R_ins+R_conv_out); A(N,N-1)=k_t((Tg(N)+Tg(N-1))/2)*2*pi*(r_in-DELTAr/2)*L/DELTAr; b(N)=-pi*r_out*DELTAr*L*gv-T_infinity/(R_ins+R_conv_out);

The solution is obtained: X=A\b; T=X;

and used to compute the error between the assumed and calculated solutions is obtained: err =

1 N

∑ (T − Tˆ ) N

i =1

i

i

2

(29)

err=sqrt(sum((T-Tg).^2)/N) %compute rms error

The calculated solution becomes the guess value for the next iteration: Tg=T;

%reset guess values used to setup A and b

end

The solution is converted to degrees Celsius: T_C=T-273.2;

%convert to C

end

The solution is illustrated in Figure P1.5-2-4. 75 72.5

Temperature (°C)

70 67.5 65 62.5 60 57.5 55 52.5 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.5-2-4: Temperature as a function of radius.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in text) in the limit that the conductivity is constant. The conductivity function is modified temporarily so that it returns a constant value: function[k]=k_t(T) %conductivity of the material % %Inputs: % T: temperature (K) % %Outputs: % k: conductivity (W/m-K) k=10;%+0.035*(T-300); end

Figure P1.5-2-5 illustrates the temperature distribution predicted by the numerical and analytical solutions in the limit that k is constant.

80

Temperature (°C)

75 70 65 60

analytical model numerical model

55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.5-2-5: Temperature as a function of radius predicted by the analytical and numerical models in the limit that k is constant.

Problem 1.6-1 (1-13 in text): Temperature Sensor Error A resistance temperature detector (RTD) utilizes a material that has a resistivity that is a strong function of temperature. The temperature of the RTD is inferred by measuring its electrical resistance. Figure P1.6-1 shows an RTD that is mounted at the end of a metal rod and inserted into a pipe in order to measure the temperature of a flowing liquid. The RTD is monitored by passing a known current through it and measuring the voltage across it. This process results in a constant amount of ohmic heating that may tend to cause the RTD temperature to rise relative to the temperature of the surrounding liquid; this effect is referred to as a self-heating error. Also, conduction from the wall of the pipe to the temperature sensor through the metal rod can also result in a temperature difference between the RTD and the liquid; this effect is referred to as a mounting error.

Tw = 20°C

L = 5.0 cm h = 150 W/m -K 2

x

T∞ = 5.0°C

pipe

D = 0.5 mm k = 10 W/m-K RTD

qsh = 2.5 mW

Figure P1.6-1: Temperature sensor mounted in a flowing liquid.

The thermal energy generation associated with ohmic heating is q sh = 2.5 mW. All of this ohmic heating is assumed to be transferred from the RTD into the end of the rod at x = L. The rod has a thermal conductivity k = 10 W/m-K, diameter D = 0.5 mm, and length L = 5 cm. The end of the rod that is connected to the pipe wall (at x = 0) is maintained at a temperature of Tw = 20°C. The liquid is at a uniform temperature, T∞ = 5°C and the heat transfer coefficient between the liquid and the rod is h = 150 W/m2-K. a.) Is it appropriate to treat the rod as an extended surface (i.e., can we assume that the temperature in the rod is a function only of x)? Justify your answer. The input parameters are entered in EES. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q_dot_sh=0.0025 [W] k=10 [W/m-K] d=0.5 [mm]*convert(mm,m) L=5.0 [cm]*convert(cm,m) T_wall=convertTemp(C,K,20) T_f=convertTemp(C,K,5) h=150 [W/m^2-K]

"self-heating power" "conductivity of mounting rod" "diameter of mounting rod" "length of mounting rod" "temperature of wall" "temperature of liquid" “heat transfer coefficient”

The appropriate Biot number for this case is:

Bi =

hd 2k

(1)

"Extended surface approximation" Bi=h*d/(2*k)

The Biot number calculated by EES is 0.004 which is much less than 1.0 and therefore the extended surface approximation is justified. b.) Develop an analytical model of the rod that will predict the temperature distribution in the rod and therefore the error in the temperature measurement; this error is the difference between the temperature at the tip of the rod and the liquid. You may find it easiest to use Maple for this process. Figure 2 illustrates a differential control volume for the rod.

Figure 2: Differential control volume for the rod.

The energy balance suggested by Figure 2 is: q x = q x + dx + qconv

(2)

or, expanding the x+dx term:

q x = q x +

dq dx + qconv dx

(3)

The rate equations for conduction and convection are:

d 2 dT q x = − k π 4 dx

(4)

qconv = h π d dx (T − T f )

(5)

and

Substituting Eqs. (4) and (5) into Eq. (3) leads to: 0=

d ⎡ d 2 dT ⎤ k π − ⎢ ⎥ dx + h π d dx (T − T f ) dx ⎣ 4 dx ⎦

(6)

or

d 2T 4 h − (T − T f ) = 0 dx 2 k d

(7)

which is a non-homogeneous 2nd order differential equation. The general solution to Eq. (7) can be found in your text as Eq. (3.66) or in the handout on Extended Surfaces as Eq. (6-22). The easiest thing to do is enter the differential equation into Maple and let it solve it for you: > GDE:=diff(diff(T(x),x),x)-4*h*(T(x)-T_f)/(k*d)=0;

2 ⎛d ⎞ 4 h ( T( x ) − T_f ) GDE := ⎜⎜ 2 T( x ) ⎟⎟ − =0 kd ⎝ dx ⎠

> Ts:=dsolve(GDE);

Ts := T( x ) = e

⎛⎜ 2 h x ⎞⎟ ⎜ k d ⎟ ⎝ ⎠

_C2 + e

⎛⎜ 2 h x ⎞⎟ ⎜− k d ⎟ ⎝ ⎠

_C1 + T_f

The solution can be copied and pasted into EES (don’t forget that you may need to change your output display to Maple Notation to facilitate the copying process depending on your version of Maple): > Ts:=dsolve(GDE);

Ts := T(x) = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C2+exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C1+T_f

which can be copied to EES: Ts := T(x) = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C2+exp(-2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C1+T_f "solution copied from Maple"

The solution will need to be modified slightly so that it is compatible with EES (the _C1 must become C1, _C2 must be C2, Ts:= should be deleted and the T(x) must be just T: T = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*C2+exp(-2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*C1+T_f "solution copied from Maple and modified"

Trying to solve now should give the message that you have 12 variables but only 9 equations – you need to specify C1, C2, and x to have a completely specified problem. Let’s set x = 0: x=0

and concentrate on determining symbolic expressions for the boundary conditions. temperature at the pipe wall (x=0) is specified to be Twall. Using Maple:

The

> rhs(eval(Ts,x=0))=T_wall;

_C2+_C1+T_f = T_wall

which can be pasted into EES (and modified): C2+C1+T_f = T_wall

"wall boundary condition"

The boundary condition at the end of the rod with the sensor is associated with an energy balance on the interface:

kπ

d 2 dT 4 dx

= qsh

(8)

x= L

which can be evaluated symbolically in Maple: > k*pi*d^2*rhs(eval(diff(Ts,x),x=L))/4=q_dot_sh; 1/4*k*pi*d^2*(2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*_C2/(k^(1/2)*d^(1/ 2))-2*h^(1/2)*exp(-2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*_C1/(k^(1/2)*d^(1/2))) = q_dot_sh

Aren’t you glad you don’t have to do this by hand? The expression can be copied and pasted into EES to complete your solution: 1/4*k*pi*d^2*(2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*C2/(k^(1/2)*d^(1/2))-2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*C1/(k^(1/2)*d^(1/2))) = q_dot_sh "sensor boundary condition"

Check your units (Figure 3 shows the variable information window with the units set) to make sure that no errors were made.

Figure 3: Variable Information window.

c.) Prepare a plot of the temperature as a function of position and compute the temperature error. Comment out the specification that x=0 and prepare a parametric table that includes T and x. Alter x so that it varies from 0 to 0.05 and plot the result. You can convert the temperature to °C and position to cm for a better looking plot: x_cm=x*convert(m,cm) T_C=converttemp(K,C,T)

Figure 4 illustrates the temperature distribution; note that the temperature elevation at the tip with respect to the fluid is about 3.6 K and it represents the measurement error. For the conditions in the problem statement, it is clear that the measurement error is primarily due to the self-heating effect because the effect of the wall (the temperature elevation at the base) has died off after about 2.0 cm.

Figure 4: Temperature distribution in the mounting rod.

d.) Investigate the effect of thermal conductivity on the temperature measurement error. Identify the optimal thermal conductivity and explain why an optimal thermal conductivity exists. The temperature measurement error can be calculated from your solution by setting x = L: "Part d - temperature measurement error" x=L errT=T-T_f

Figure 5 illustrates the temperature measurement error as a function of the thermal conductivity of the rod material. Figure 5 shows that the optimal thermal conductivity, corresponding to the minimum measurement error, is around 100 W/m-K. Below the optimal value, the self-heating error dominates as the local temperature rise at the tip of the rod is large. Above the optimal value, the conduction from the wall dominates. The inset figures show the temperature distribution for high and low thermal conductivity in order to illustrate these different behaviors.

Figure 5: Temperature measurement error as a function of rod thermal conductivity. The inset figures show the temperature distribution at low conductivity and high conductivity.

Problem 1.6-2 (1-14 in text): Optimizing a Heat Sink Your company has developed a micro-end milling process that allows you to easily fabricate an array of very small fins in order to make heat sinks for various types of electrical equipment. The end milling process removes material in order to generate the array of fins. Your initial design is the array of pin fins shown in Figure P1.6-2. You have been asked to optimize the design of the fin array for a particular application where the base temperature is Tbase = 120°C and the air temperature is Tair = 20°C. The heat sink is square; the size of the heat sink is W = 10 cm. The conductivity of the material is k = 70 W/m-K. The distance between the edges of two adjacent fins is a, the diameter of a fin is D, and the length of each fin is L. array of fins k = 70 W/m-K

Tair = 20°C, h D a L

W = 10 cm

Tbase = 120°C

Figure P1.6-2: Pin fin array

Air is forced to flow through the heat sink by a fan. The heat transfer coefficient between the air and the surface of the fins as well as the unfinned region of the base, h , has been measured for the particular fan that you plan to use and can be calculated according to: ⎞ a ⎡ W ⎤⎛ h = 40 ⎢ 2 ⎥ ⎜⎜ ⎟ ⎣ m K ⎦ ⎝ 0.005 [ m ] ⎟⎠

0.4

⎛ ⎞ D ⎜⎜ ⎟⎟ ⎝ 0.01 [ m ] ⎠

−0.3

Mass is not a concern for this heat sink; you are only interested in maximizing the heat transfer rate from the heat sink to the air given the operating temperatures. Therefore, you will want to make the fins as long as possible. However, in order to use the micro-end milling process you cannot allow the fins to be longer than 10x the distance between two adjacent fins. That is, the length of the fins may be computed according to: L = 10 a . You must choose the most optimal value of a and D for this application. a.) Prepare a model using EES that can predict the heat transfer coefficient for a given value of a and D. Use this model to predict the heat transfer rate from the heat sink for a = 0.5 cm and D = 0.75 cm.

The input values are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs"

T_air=converttemp(C,K,20) T_base=converttemp(C,K,120) k=70 [W/m-K] W=10.0 [cm]*convert(cm,m)

"air temperature" "base temperature" "fin material conductivity" "base width"

The optimization parameters, a and D, are set to their initial values: "Optimization parameters" a=0.5 [cm]*convert(cm,m) D=0.75 [cm]*convert(cm,m)

"distance between adjacent fins" "diameter of fins"

The length of the fins is computed using the aspect ratio and the number of fins is determined according to:

⎛ W ⎞ N =⎜ ⎟ ⎝a+D⎠

2

(1)

The heat transfer coefficient is computed using the equation provided in the problem statement. L=10*a N=(W/(a+D))^2 h = 40 [W/m^2-K]*(a/0.005 [m])^(0.4)*(D/0.01 [m])^(-0.3)

"length of fins" "number of fins" "heat transfer coefficient"

The perimeter and cross-sectional area of each fin are computed according to:

p =π D

(2)

D2 4

(3)

Ac = π

The EES function for the fin efficiency of a constant cross-sectional area fin is used. The function is accessed using the Function Information selection from the Options menu and then selecting Fin Efficiency from the pull-down menu. Scroll to the Circular-Base Rectangular Fin (Figure 2(a)) and select Info to learn how to access this function (Figure 2(b)).

(a) (b) Figure 2: (a) Function Information window and (b) Help information for the Circular-Base Rectangular Fin.

The fin constant, mL, is computed according to: mL =

hp L k Ac

(4)

and used to call the function eta_fin_spine_rect which returns the fin efficiency, ηf. p=pi*D Ac=pi*D^2/4 mL=L*sqrt(h*p/(k*Ac)) eta=eta_fin_spine_rect(mL)

"perimeter of fin" "cross sectional area of fin" "fin constant" "fin efficiency"

The total area of the fins on the heat sink is:

Af = N p L

(5)

1 h Af η f

(6)

and so the total resistance of the fins are: Rf = A_f=p*L*N R_f=1/(h*A_f*eta)

"finned area" "resistance of fins"

The total area of the base of the heat sink that is not finned is: Auf = W 2 − N Ac

(7)

and the thermal resistance from the unfinned base is: Ruf =

1 h Auf

(8)

A_uf=W^2-N*Ac R_uf=1/(h*A_uf)

"unfinned area" "resistance of unfinned area"

The total resistance of the heat sink is the combination of Rf and Ruf in parallel:

Rtotal

⎛ 1 1 ⎞ =⎜ + ⎟⎟ ⎜R R f uf ⎝ ⎠

−1

(9)

and the total heat transfer rate is:

q = R_total=(1/R_f+1/R_uf)^(-1) "total thermal resistance of the heat sink" q_dot=(T_base-T_air)/R_total

(Tbase − Tair )

(10)

Rtotal

"heat transfer"

which leads to q = 291.7 W. b.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the distance between adjacent fins, a, for a fixed value of D = 0.75 cm. Be sure that the fin length is calculated using L = 10 a . Your plot should exhibit a maximum value, indicating that there is an optimal value of a. Figure 3 illustrates the heat transfer rate from the heat sink as a function of a for D = 0.75 cm.

Figure 3: Heat transfer rate as a function of the distance between adjacent fins for D = 0.75 cm.

c.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the diameter of the fins, D, for a fixed value of a = 0.5 cm. Be sure that the fin length is calculated using L = 10 a . Your plot should exhibit a maximum value, indicating that there is an optimal value of D. Figure 4 illustrates the heat transfer rate from the heat sink as a function of D for a = 0.5 cm.

Figure 4: Heat transfer rate as a function of the diameter of the fins for a = 0.5 cm.

d.) Determine the optimal value of a and D using EES' built-in optimization capability.

Comment out the optimization parameters (a and D) and access the optimization algorithms from the Calculate Menu by selecting Min/Max (Figure 5).

Figure 5: Find Minimum or Maximum Window

Select the variable to be minimized or maximized from the list on the left and the independent variables to be varied from the list on the right. You will need to provide a reasonable initial guess and bounds for the independent variables by selecting the Bounds button; note that it is not practical for a or D to be less than 1.0 mm. You can experiment with the different optimization methods and see which technique is more robust. I found the genetic optimization algorithm to work the best for this problem; with a sufficient number of individuals I identified an optimal design consisting of approximately 1500 very small fins of D = 1.1 mm separated by a = 1.4 mm. The associated rate of heat transfer is q = 352.2 W.

Problem 1.6-3: Finned Tube Water Heater A water heater consists of a copper tube that carries water through hot gas in a furnace, as shown in Fig. P1.6-3(a). The copper tube has an outer radius, ro,tube = 0.25 inch and a tube wall thickness of th = 0.033 inch. The conductivity of the copper is ktube = 300 W/m-K. Water flows through the pipe at a temperature of Tw = 30ºC. The heat transfer coefficient between the water and the internal surface of the pipe is hw = 500 W/m2-K. The external surface of the tube is exposed to hot gas at Tg = 500ºC. The heat transfer coefficient between the gas and the outer surface of the pipe is hg = 25 W/m2-K. Neglect radiation from the tube surface. Tg = 500°C 2 hg = 25 W/m -K ro,tube = 0.25 inch ktube = 300 W/m-K

Tw = 30°C 2 hw = 500 W/m -K Figure P1.6-3(a): Copper tube in a water heater. th = 0.033 inch

a.) At what rate is heat is added to the water for a unit length of tube, L = 1 m, for this configuration (W/m)? The known information is entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_o_tube=0.25 [inch]*convert(inch,m) th=0.033 [inch]*convert(inch,m) k_tube=300 [W/m-K] T_w=converttemp(C,K,30) h_w=500 [W/m^2-K] T_g=converttemp(C,K,500) h_g=25 [W/m^2-K] L=1 [m]

"outer tube radius" "tube thickness" "tube material conductivity" "water temperature" "water to tube heat transfer coefficient" "gas temperature" "gas to tube heat transfer coefficient" "unit length of tube"

The resistance network that represents this problem is shown in Figure 2.

Figure 2: Resistance network for unfinned tube.

The resistance network includes convection to the inner surface of the tube (Rconv,w),

1

Rconv , w =

hw 2 π ( ro ,tube − th ) L

,

(1)

conduction through the tube (Rcond),

Rcond

⎛ r ⎞ ln ⎜⎜ o ,tube ⎟⎟ ro ,tube − th ⎠ = ⎝ , 2 π ktube L

(2)

and convection from the external surface of the tube (Rconv,g), Rconv , g =

1 hg 2 π ro ,tube L

(3)

The heat transfer is provided by:

Tg − Tw

q = Rconv , w + Rcond

⎛ 1 1 +⎜ + ⎜R ⎝ conv , g Rrad

⎞ ⎟⎟ ⎠

−1

"Part a: unfinned tube" R_conv_w=1/(h_w*2*pi*(r_o_tube-th)*L) "internal convection resistance" R_cond=ln(r_o_tube/(r_o_tube-th))/(2*pi*k_tube*L) "conduction resistance" R_conv_g=1/(h_g*2*pi*r_o_tube*L) "external convection" q_dot=(T_g-T_w)/(R_conv_w+R_cond+R_conv_g) "heat transfer rate"

The heat transfer is q =443 W. b.) What is the dominant resistance to heat transfer in your water heater? The solution window is shown in Figure 3:

Figure 3:Solution window.

(4)

Notice that the values of Rconv,g is much higher than Rcond or Rconv,w and therefore the convection from the surface of the tube limits the heat transfer rate. In order to increase the capacity of the water heater, you decide to slide washers over the tube, as shown in Fig. P1.6-3(b). The washers are w = 0.06 inch thick with an outer radius of ro,washer = 0.625 inch and have a thermal conductivity of kwasher = 45 W/m-K. The contact resistance between the washer and the tube is Rc′′ = 5x10-4 m2-K/W. The distance between two adjacent washers is b = 0.25 inch. ro,washer = 0.625 inch b = 0.25 inch kwasher = 45 W/m-K w = 0.06 inch

Rc′′ = 5x10 m -K/W -4

2

Figure P1.6-3(b): Water heater with washers installed.

c.) Can the brass washers be treated as extended surfaces (i.e. can the temperature in the washers be considered to be only a function of radius)? Justify your answer with a calculation. The additional information is entered in EES: "Inputs for finned tube" w=0.06 [inch]*convert(inch,m) r_o_washer=0.625 [inch]*convert(inch,m) k_washer=45 [W/m-K] R``_c=5e-4 [m^2-K/W] b=0.25 [inch]*convert(inch,m)

"thickness of washers" "outer radius of washer" "conductivity of washer" "contact resistance" "distance between washers"

The Biot number associated with the washer must be the ratio of the resistance to conduction from the center of the washer to its edge (axially) to the resistance to heat transfer from its surface:

Bi =

Rcond , x Rconv , g , x

(5)

where the resistances in Eq. (5) are related to heat transfer axially and so:

Rcond , x =

w 2 kwasher Awasher

Rconv , g , x =

1 hg Awasher

(6)

(7)

and

Awasher = π ( ro2, washer − ro2,tube ) "Part c" A_washer=pi*(r_o_washer^2-r_o_tube^2) R_cond_x=w/(2*k_washer*A_washer) R_conv_g_x=1/(h_g*A_washer) Bi=R_cond_x/R_conv_g_x

(8)

"exposed area of one side of washer" "conduction resistance axially" "convection resistance axially" "Biot number"

The Biot number is 0.0004 which is much less than one and therefore the temperature gradient in the washer across its thickness is negligible relative to the temperature drop between its surface and the gas. Therefore, the extended surface model is valid and the washer can be treated as a fin. Assume that your answer to (c) showed that the washers can be treated as extended surfaces and therefore modeled as a fin with the appropriate fin resistance. d.) Draw a thermal resistance network that can be used to represent this situation. Be sure to draw and label resistances associated with convection through the water (Rconv,w), conduction through the copper tube (Rcond), heat transfer through contact resistance (Rcontact), heat transfer through the washers (Rwashers), and convection to gas from unfinned outer surface (Rconv,g,unfinned). The thermal resistance network is shown in Figure 5.

Figure 5: Resistance network representing finned tube.

e.) How much heat is added to the water with the washers installed on the tube for a 1 m length of tube? The value of Rconv,w and Rcond do not change from (a). The resistance associated with convection from the unfinned portion of the tube is:

Rconv , g ,unfinned =

1 ⎛ b ⎞ hg 2 π ro ,tube L ⎜ ⎟ ⎝b+w⎠

(9)

where the last term in the denominator is the fraction of the tube surface that is not occupied by the fins. The contact resistance is: Rcontact =

Rc′′ ⎛ w ⎞ 2 π ro ,tube L ⎜ ⎟ ⎝b+w⎠

(10)

where the last term in the denominator is the fraction of the tube surface that is occupied by the fins. "Part d" R_conv_g_unfinned=1/(h_g*2*pi*r_o_tube*L*(b/(b+w))) "convection from unfinned surface" R_contact=R``_c/(2*pi*r_o_tube*L*(w/(b+w))) "contact resistance"

The resistance of the fins is: Rwashers =

1 ⎛ L ⎞ η f hg 2 Awasher ⎜ ⎟ ⎝b+w⎠

(11)

where the last term in the denominator is the number of washers present on the tube. The fin efficiency, ηf, can be calculated using EES’ built-in functions for fin efficiency. Select Function Info from the Options menu and then select the button next to the list at the lower right of the top box. Select Fin Efficiency and then Dimensional Efficiency and scroll over until you find eta_fin_annular_rect (Figure 6).

Figure 6: Fin efficiency function information.

Select Paste and the function call will be inserted into your EES program. Note that Info provides more detailed help about the function. Modify the arguments of the function so that they match your variable names: eta_f=eta_fin_annular_rect(w, r_o_tube, r_o_washer, h_g, k_washer)

"fin efficiency"

Use the fin efficiency to calculate the total washer resistance: R_washers=1/(eta_f*h_g*2*A_washer*(L/(b+w))) "washer resistance"

The total heat transfer to the finned tube is:

(T

q finned =

g

Rconv , w + Rcond

− Tw )

⎛ 1 1 +⎜ + ⎜R ⎝ conv , g ,unfinned Rcontact + Rwashers

⎞ ⎟⎟ ⎠

−1

(12)

q_dot_finned=(T_g-T_w)/(R_conv_w+R_cond+(1/R_conv_g_unfinned+1/(R_contact+R_washers))^(-1)) "total heat transfer rate from finned tube"

The addition of the fins has increased the heat transfer rate to 1540 W.

Problem 1.6-4: Wire with Ohmic Heating A wire is subjected to ohmic heating (i.e., a current runs through it) while it is convectively cooled. The ends of the wire have fixed temperatures. Describe how you would determine whether the extended surface approximation (i.e., the approximation in which you treat the temperature of the wire as being one-dimensional along its length and uniform radially) is appropriate when solving this problem. The justification of the extended surface approximation is related to computing the Biot number, the ratio of the resistance that is being neglected (conduction radially within the wire) to the resistance(s) being considered (convection from the wire surface). For this problem, the Biot number would be:

rh k

(1)

where r is the radius of the wire, h is the heat transfer coefficient, and k is the wire conductivity. Note that the rate of ohmic heating is not important for this calculation because the ohmic heating must be both conducted radially and convected from the surface.

Problem 1.6-5 A cylindrical bracket that is L = 4 cm long with diameter D = 5 mm extends between a wall at TH = 100°C (at x= 0) and a wall at TC = 20°C (at x= L). The conductivity of the bracket is k = 25 W/m2-K. The cylinder is surrounded by gas at T∞ = 200°C and the heat transfer coefficient is h = 250 W/m2-K. a.) Is an extended surface approximation appropriate for this problem? Justify your answer. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" D=5 [mm]*convert(mm,m) L=4 [cm]*convert(cm,m) k=25 [W/m-K] h_bar=250 [W/m^2-K] T_H=converttemp(C,K,100 [C]) T_C=converttemp(C,K,20 [C]) T_infinity=converttemp(C,K,200 [C])

"diameter of strut" "length of strut" "conductivity of strut" "heat transfer coefficient" "hot end temperature" "cold end temperature" "ambient temperature"

The Biot number is:

Bi =

hD k

Bi=h_bar*D/k

(1) "Biot number"

which leads to Bi = 0.05, justifying the extended surface approximation. b.) Assume that your answer to (a) was yes. Develop an analytical model in EES. Plot the temperature as a function of position within the bracket. The development of the governing differential equation and the derivation of the general solution proceeds as discussed in Section 1.6.2 and leads to: T = C1 exp ( m x ) + C2 exp ( − m x ) + T∞

(2)

where C1 and C2 are undetermined constants and m is the fin constant:

m=

per h k Ac

(3)

where per is the perimeter of the bracket and Ac is the cross-sectional area of the bracket.

per = π D

(4)

Ac = π

D2 4

(5)

per=pi*D A_c=pi*D^2/4 m=sqrt(per*h_bar/(k*A_c))

"perimeter" "cross-sectional area" "fin constant"

The boundary conditions at x = 0 and x = L lead to:

TH = C1 + C2 + T∞

(6)

TC = C1 exp ( m L ) + C2 exp ( − m L ) + T∞

(7)

Equations (6) and (7) are entered in EES in order to determine C1 and C2: T_H=C_1+C_2+T_infinity T_C=C_1*exp(m*L)+C_2*exp(-m*L)+T_infinity

"boundary condition at x=0" "boundary condition at x=L"

and the solution is entered in EES: x=0 [m] T=C_1*exp(m*x)+C_2*exp(-m*x)+T_infinity T_degC=converttemp(K,C,T)

"axial position" "solution" "in C"

Figure 1 illustrates the temperature as a function of position in the bracket. 200 2

h = 2500 W/m -K 175

Temperature (°C)

150 2

h = 250 W/m -K

125 100

2

75

h = 25 W/m -K 2

h = 2.5 W/m -K

50 25 0 0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

Position (m) Figure 1: Temperature as a function of position for various values of the heat transfer coefficient.

c.) Overlay on your plot from (b) the temperature as a function of position with h = 2.5, 25 and 2500 W/m2-K. Explain the shape of your plots.

The requested plots are shown in Figure 1. As the heat transfer coefficient increases, the resistance between the bracket and the surrounding gas:

Rconv =

1 h per L

(8)

diminishes while the resistance to conduction along the bracket:

Rcond =

L k Ac

(9)

does not change. R_cond=L/(k*A_c) R_conv=1/(h_bar*per*L) R_cond\R_conv=R_cond/R_conv

"conduction resistance from T_H to T_C" "convection resistance from surface to ambient" "ratio of conduction to convection resistances"

At h = 2500 W/m2-K, Rcond/Rconv = 128 and therefore the bracket material very quickly equilibrates with the gas (see Figure 1). At h = 2.5 W/m2-K, Rcond/Rconv = 0.128 and therefore convection is not very important and the bracket material temperature distribution is nearly linear (i.e., the situation is close to conduction through a plane wall). d.) Plot the heat transfer from the wall at TH into the bracket (i.e., the heat transfer into the bracket at x = 0) as a function of h . Explain the shape of your plot. The heat transfer into the bracket at x = 0 is: ⎛ dT ⎞ q H = −k Ac ⎜ ⎟ ⎝ dx ⎠ x =0

(10)

Substituting Eq. (2) into Eq. (10) leads to: q H = − k Ac ( m C1 − m C2 ) q_dot_H=-k*(m*C_1-m*C_2)*A_c

(11)

"heat transfer rate at x=0"

Figure 2 illustrates the rate of heat transfer as a function of h and shows that as h approaches zero the rate of heat transfer approaches a constant value, consistent with conduction through a plane wall:

q H ,h →0 =

k Ac (TH − TC ) L

(12)

As h becomes large, the heat transfer is reduced and eventually changes sign as heat is transferred into the wall from the warmer gas.

Heat transfer from hot wall (W)

2.5

0

-2.5

-5

-7.5

-10 1

10

100

2

1000

Average heat transfer coefficient (W/m -K) Figure 2: Heat transfer from hot wall as a function of the heat transfer coefficient.

Problem 1.6-6 Your company has developed a technique for forming very small fins on a plastic substrate. The diameter of the fins at their base is D = 1 mm. The ratio of the length of the fin to the base diameter is the aspect ratio, AR = 10. The fins are arranged in a hexagonal close packed pattern. The ratio of the distance between fin centers and to the base diameter is the pitch ratio, PR = 2. The conductivity of the plastic material is k = 2.8 W/m-K. The heat transfer coefficient between the surface of the plastic and the surrounding gas is h = 35 W/m2-K. The base temperature is Tb = 60°C and the gas temperature is T∞ = 35°C. Do the analysis on a per unit of base area basis (A = 1 m2). a.) Determine the number of fins per unit area and the thermal resistance of the unfinned region of the base. The inputs are entered in EES. An aspect ratio is chosen to start the problem. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" D=1 [mm]*convert(mm,m) AR=10 [-] L=AR*D PR=2 [-] p=PR*D h_bar=35 [W/m^2-K] k=2.8 [W/m-K] T_b=converttemp(C,K,60 [C]) T_infinity=converttemp(C,K,35 [C]) A=1 [m^2]

"diameter" "aspect ratio" "length" "pitch ratio" "pitch (distance between fin centers)" "heat transfer coefficient" "thermal conductivity" "temperature of base" "temperature of ambient air" "per unit area of base"

The Biot number is:

Bi =

hD k

(1)

Bi=h_bar*D/k

"Biot number"

which leads to Bi = 0.0125, justifying the extended surface approximation. A unit cell of the hexagonal close packed pattern is examined, as shown in Figure 1. D p Acell

Figure 1: Hexagonal close pack array of fins.

The area of the triangular unit cell is:

⎛ p⎞ ⎛π ⎞1 Acell = ⎜ ⎟ p sin ⎜ ⎟ ⎝2⎠ ⎝ 3⎠2

(2)

The number of fins per unit area is:

N ′′ =

0.25 Acell

(3)

A_cell=(p/2)*p*sin(pi/3)/2 N``=0.25/A_cell

"area of unit cell" "number of fins per unit area"

which leads to N ′′ = 2.89x105 fins/m2. The number of fins on a 1 m2 base is: N fin = N ′′ A

(4)

The unfinned area of the base is: Aunfin = A − N fin π

D2 4

(5)

and the thermal resistance of the unfinned region is: Runfin = N_fin=N``*A A_unfin=A-N_fin*pi*D^2/4 R_unfin=1/(h_bar*A_unfin)

1 h Aunfin

(6)

"number of fins" "unfinned surface area" "thermal resistance of unfinned surface area"

which leads to Runfin = 0.0080 K/W. b.) You have been asked to evaluate whether triangular, parabolic concave, or parabolic convex pin fins will provide the best performance. Plot the heat transfer per unit area of base surface for each of these fin shapes as a function of aspect ratio. Note that the performance of these fins can be obtained from the EES functions eta_fin_spine_parabolic_ND, eta_fin_spine_parabolic2_ND, and eta_fin_spine_triangular_ND. Explain the shape of your plot. The area of the triangular fins is computed according to Eq. (7) (note that formulae for the fin area can be found in the Help Information for the fin efficiency functions).

D 2 ⎛D⎞ Afin = N fin π L +⎜ ⎟ 2 ⎝2⎠

2

(7)

The fin constant is computed according to: mL =

4h L kD

(8)

The fin efficiency (η) is computed with the eta_fin_spine_triangular_ND. "triangle spine fin" A_fin=N_fin*pi*D*sqrt(L^2+(D/2)^2)/2 mL=sqrt(4*h_bar/(k*D))*L eta=eta_fin_spine_triangular_ND(mL)

"finned area" "fin constant" "fin efficiency"

The fin resistance is computed according to: R fin =

1 h A fin η

(9)

The total resistance to heat transfer is: Rtotal

⎛ 1 1 =⎜ + ⎜R ⎝ fin Runfin

⎞ ⎟⎟ ⎠

−1

(10)

and the rate of heat transfer is: q =

R_fin=1/(h_bar*A_fin*eta) R_total=(1/R_fin+1/R_unfin)^(-1) q_dot=(T_b-T_infinity)/R_total

(Tb − T∞ ) Rtotal

(11)

"thermal resistance of finned surface area" "total thermal resistance" "total heat transfer"

which leads to q = 3128 W/m2. Figure 2 illustrates the rate of heat transfer for triangular pin fins as a function of aspect ratio. Note that increasing the aspect ratio increases the heat transfer because the surface area is larger for longer fins. However, as the aspect (and therefore the length) increases, the fins become less efficient.

2

Heat transfer per unit area (W/m )

5500 convex pin fin 5000 4500 4000 3500

triangular pin fin

3000

concave pin fin

2500 2000 1500 1000 2

4

6

8

10

12

14

16

18

20

Aspect ratio Figure 2: Rate of heat transfer per area as a function of the aspect ratio for various fin shapes.

In order to analyze the concave parabolic fins, the formulae for the fin area, fin constant, and efficiency are commented out and replaced with: {"triangle spine fin" A_fin=N_fin*pi*D*sqrt(L^2+(D/2)^2)/2 mL=sqrt(4*h_bar/(k*D))*L eta=eta_fin_spine_triangular_ND(mL) "concave parabolic fin" C_3=1+2*(D/L)^2 C_4=sqrt(1+(D/L)^2) A_fin=N_fin*pi*L^3*(C_3*C_4-L*ln(2*D*C_4/L+C_3)/(2*D))/(8*D) mL=sqrt(4*h_bar/(k*D))*L eta=eta_fin_spine_parabolic_ND(mL)

"finned area" "fin constant" "fin efficiency"}

"finned area" "fin constant" "fin efficiency"

The performance of the concave parabolic fins is overlaid onto Figure 2. In order to analyze the convex parabolic fins, these equations are replaced with: {"concave parabolic fin" C_3=1+2*(D/L)^2 C_4=sqrt(1+(D/L)^2) A_fin=N_fin*pi*L^3*(C_3*C_4-L*ln(2*D*C_4/L+C_3)/(2*D))/(8*D) mL=sqrt(4*h_bar/(k*D))*L eta=eta_fin_spine_parabolic_ND(mL) "convex parabolic" A_fin=N_fin*pi*D^4*((4*L^2/D^2+1)^1.5-1)/(6*L^2) mL=sqrt(4*h_bar/(k*D))*L eta=eta_fin_spine_parabolic2_ND(mL)

"finned area" "fin constant" "fin efficiency"} "finned area" "fin constant" "fin efficiency"

The performance of convex parabolic fins is also overlaid onto Figure 2. Figure 2 shows that the convex pin fin provides the best performance; this is because this type of fin has the most surface area. c.) Assume that part (b) indicated that parabolic convex fins are the best. You have been asked whether it is most useful to spend time working on techniques to improve (increase) the

aspect ratio or improve (reduce) the pitch ratio. Answer this question using a contour plot that shows contours of the heat transfer per unit area in the parameter space of AR and PR. The aspect ratio and pitch ratio are commented out and a parametric table is created with these variables and the total heat transfer. A contour plot is shown in Figure 3 and shows that it is much more beneficial to reduce the pitch ratio than the aspect ratio. 2 6451

1.9

7839

1.8

9227

1.7

Pitch ratio

10615 1.6

12003

1.5

13391

1.4

14779

1.3

16167

1.2

17555 18943

1.1 1 10

11

12

13

14

15

16

17

18

19

20

Aspect ratio Figure 3: Contour plot of refrigeration per unit area as a function of pitch ratio and aspect ratio.

Problem 1.7-1: Furnace Manipulator Arm You are designing a manipulator for use within a furnace. The arm must penetrate the side of the furnace, as shown in Figure P1.7-1. The arm has a diameter of D = 0.8 cm and protrudes Li = 0.5 m into the furnace, terminating in the actuator that can be assumed to be adiabatic. The portion of the arm in the manipulator is exposed to flame and hot gas; these effects can be represented by a heat flux of q ′′ = 1x104 W/m 2 and convection to gas at Tf = 500°C with heat transfer coefficient h f = 50 W/m2-K. The conductivity of the arm material is k = 150 W/m-K. The arm outside of the furnace has the same diameter and conductivity, but is exposed to air at Ta = 20°C with heat transfer coefficient ha = 30 W/m2-K. The length of the arm outside of the furnace is Lo = 0.75 m and this portion of the arm terminates in the motor system which can also be considered to be adiabatic. ′′ = 1x104 W/m 2 qrad x

k = 150 W/m-K D = 0.8 cm

Lo = 0.75 m Li = 0.5 m 2 ha = 30 W/m -K h f = 50 W/m -K = 20°C T = ° a T f 500 C Figure P1.7-1: Manipulator arm for a furnace. 2

a.)

Is an extended surface model appropriate for this problem? Justify your answer.

The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" qf_rad=1e4 [W/m^2] T_f=converttemp(C,K,500) h_f=50 [W/m^2-K] "heat transfer coefficient within furnace" L_i=0.5 [m] D=0.8 [cm]*convert(cm,m) k=150 [W/m-K] T_a=converttemp(C,K,20) h_a=30 [W/m^2-K] "heat transfer coefficient outside of furnace" L_o=0.75 [m] "manipulator arm length outside of furnace"

"radiant heat flux on arm" "air temperature within furnace"

"manipulator arm length within furnace" "diameter of arm" "arm conductivity" "air temperature outside of furnace"

The Biot number based on the heat transfer coefficient within the furnace will be largest because the highest heat transfer coefficient exists within the furnace:

Bii = Bi_i=(D/2)*h_f/k

D hf

(1)

2k

"Biot number based on internal heat transfer coefficient"

which leads to Bii = 0.013 which is sufficiently less than unity to allow an extended surface approximation model to be used. b.) Develop an analytical model of the manipulator arm; implement your model in EES. Plot the temperature as a function of axial position x (see Figure P1.7-1) for -Lo < x < Li. The differential equation that governs the temperature within the furnace (Ti) is derived using the differential energy balance shown in Figure 2.

Figure 2: Differential energy balance on manipulator arm within furnace.

The energy balance suggested by Figure 2 is:

′′ π D dx = q x + q x + qrad

dq dx + h f π D (Ti − T f ) dx dx

(2)

Substituting Fourier's law:

q x = −k

π D 2 dTi 4

dx

(3)

into Eq. (2) leads to: ′′ π D dx = qrad

or

d ⎡ π D 2 dTi ⎤ ⎢ −k ⎥ dx + h f π D (Ti − T f ) dx dx ⎣ 4 dx ⎦

(4)

4 hf d 2Ti 4 h f 4 q ′′ T T f − rad − = − i 2 dx kD kD kD

(5)

The solution to the ordinary differential equation is divided into its homogeneous (ui) and particular (vi) parts:

Ti = ui + vi

(6)

and substituted into Eq. (5) in order to obtain: 4h d 2ui 4 h f d 2 vi 4 h f 4 q ′′ − + − u vi = − f T f − rad i 2 2 dx kD dx kD kD kD   

  

homogeneous equation

(7)

particular equation

The solution to the homogeneous equation:

d 2ui 4 h f ui = 0 − dx 2 k D

(8)

ui = C1 exp ( mi x ) + C2 exp ( − mi x )

(9)

is

where

mi =

4 hf

kD

(10)

The solution to the particular equation: 4h d 2 vi 4 h f 4 q ′′ − vi = − f T f − rad 2 dx kD kD kD

(11)

is vi = T f +

So the general solution for Ti is:

′′ qrad hf

(12)

Ti = C1 exp ( mi x ) + C2 exp ( − mi x ) + T f +

′′ qrad hf

(13)

A similar set of steps leads to the general solution for the temperature outside of the furnace, To, which is valid for -Lo < x < 0: To = C3 exp ( mo x ) + C4 exp ( − mo x ) + Ta

(14)

where 4 ha kD

mo =

(15)

The fin constants, mi and mo, are computed: m_i=sqrt(4*h_f/(k*D)) m_o=sqrt(4*h_a/(k*D))

"fin constant inside furnace" "fin constant outside of furnace"

The boundary conditions for the solution must be obtained at the edges of each of the computational domains (i.e., at x= -Lo, x=0, and x=Li). The two ends of the arm are adiabatic; therefore:

dTi dx

dTo dx

=0

(16)

=0

(17)

x = Li

x =− Lo

The temperature at the interface between the two computational domains must be continuous:

Ti , x =0 = To , x =0

(18)

Also, the rate of energy transferred from the furnace to the ambient air must be the same regardless of which side of the interface between the computational domains you are on; that is, an interface balance at x = 0 between the two computational domains leads to: dTi dx

= x =0

dTo dx

(19) x =0

Substituting Eqs. (13) and (14) into Eqs. (16) through (19) leads to 4 equations for the constants of integration C1 and C4: C1 mi exp ( mi Li ) − C2 mi exp ( −mi Li ) = 0

(20)

C3 mo exp ( − mo Lo ) − C4 mo exp ( mo Lo )

(21)

C1 + C2 + T f +

′′ qrad = C3 + C4 + Ta hf

C1 mi − C2 mi = C3 mo − C4 mo

(22) (23)

These are entered in EES: C_1*m_i*exp(m_i*L_i)-C_2*m_i*exp(-m_i*L_i)=0 C_3*m_o*exp(-m_o*L_o)-C_4*m_o*exp(m_o*L_o)=0 C_1+C_2+T_f+qf_rad/h_f=C_3+C_4+T_a C_1*m_i-C_2*m_i=C_3*m_o-C_4*m_o

"adiabatic end at L_i" "adiabatic end at -L_o" "continuity of temperatures at x=0" "energy balance at x=0"

The solutions are entered; note that the variable x_bar is varied from 0 to 1 within a parametric table which corresponds to x_i going from 0 to L_i and x_o going from 0 to -L_o. x_i=x_bar*L_i x_o=-x_bar*L_o T_i=C_1*exp(m_i*x_i)+C_2*exp(-m_i*x_i)+T_f+qf_rad/h_f T_o=C_3*exp(m_o*x_o)+C_4*exp(-m_o*x_o)+T_a

The temperature distribution in the arm is shown in Figure 3.

Figure 3: Temperature distribution.

c.) Prepare a plot showing the maximum temperature at the end of the arm (within the furnace) as a function of the internal length of the arm (Li) for various values of the diameter (D).

Figure 4: Maximum temperature as a function of the arm length within the furnace for various values of the diameter.

Problem 1.7-2: Rotating Ring Figure P1.7-2 illustrates a metal ring of radius R = 5.0 cm that is rotating with an angular velocity ω = 0.1 rad/s. During each rotation, the ring material passes from compartment #1 containing hot fluid at Tf,1 = 200ºC to compartment #2 containing fluid at Tf,2 = 20ºC. The heat transfer coefficient between the ring surface and the fluid in compartments #1 and #2 are h1 = 10 W/m2-K and h2 = 20 W/m2-K, respectively. The ring has a circular cross-section with diameter d = 1.0 mm. For this problem you can assume that d/R restart; > ODE:=diff(diff(T(x),x),x)-h_rad*T(x)/(k*th)=-q_flux/(k*th)-h_rad*T_infinity/(k*th); 2 q_flux h_rad T_infinity ⎞ h_rad T( x ) ⎛d − ODE := ⎜ 2 T( x ) ⎟⎟ − =− ⎜ dx k th k th k th ⎝ ⎠

and solved: > Ts:=dsolve(ODE);

Ts := T( x ) = e

⎛ h_rad x ⎞ ⎜ ⎟ ⎜ k th ⎟ ⎝ ⎠

_C2 + e

⎛ ⎜− ⎜ ⎝

h_rad x ⎞ ⎟ k th ⎟⎠

_C1 +

q_flux + h_rad T_infinity h_rad

Symbolic expressions for the boundary conditions are obtained: > BC1:=(T_c-rhs(eval(Ts,x=0)))/Rc=-k*rhs(eval(diff(Ts,x),x=L));

T_c − _C2 − _C1 − BC1 := ⎛ ⎜ ⎛ ⎜ ⎜ ⎜ h_rad e ⎝ ⎜ −k ⎜ ⎜ k ⎝

> BC2:=rhs(eval(diff(Ts,x),x=L))=0;

h_rad L ⎞ ⎟ k th ⎟⎠

th

q_flux + h_rad T_infinity h_rad = Rc

_C2

−

h_rad e

⎛ ⎜− ⎜ ⎝

k

h_rad L ⎞ ⎟ k th ⎟⎠

th

⎞ ⎟ _C1 ⎟⎟ ⎟⎟ ⎠

⎛ h_rad L ⎞ ⎟ ⎜ ⎜ k th ⎟ ⎠ ⎝

BC2 :=

h_rad e k

_C2

th

h_rad e

−

⎛ ⎜− ⎜ ⎝

k

h_rad L ⎞ ⎟ k th ⎟⎠

_C1

th

=0

The original equations in EES are commented out and the expressions from Maple are copied to EES: {"boundary conditions" m=sqrt(h_rad/(k*th)) (T_c-C_1-C_2-q_flux/h_rad-T_infinity)/Rc=-k*(C_1*m-C_2*m) C_1*m*exp(m*L)-C_2*m*exp(-m*L)=0

"fin constant" "at x=0" "at x=L"

"solution" {x_bar=0 [-]} x=x_bar*L T=C_1*exp(m*x)+C_2*exp(-m*x)+q_flux/h_rad+T_infinity T_Celsius=converttemp(K,C,T)

"dimensionless position" "position" "temperature solution" "in C"}

"Maple solution" (T_c-C_2-C_1-1/h_rad*(q_flux+h_rad*T_infinity))/Rc = -k*(1/k^(1/2)/th^(1/2)*h_rad^(1/2)*& exp(1/k^(1/2)/th^(1/2)*h_rad^(1/2)*L)*C_2-1/k^(1/2)/th^(1/2)*h_rad^(1/2)*exp(-1/k^(1/2)/& th^(1/2)*h_rad^(1/2)*L)*C_1) "boundary condition at x=0" 1/k^(1/2)/th^(1/2)*h_rad^(1/2)*exp(1/k^(1/2)/th^(1/2)*h_rad^(1/2)*L)*C_2-1/k^(1/2)/th^(1/2)*& h_rad^(1/2)*exp(-1/k^(1/2)/th^(1/2)*h_rad^(1/2)*L)*C_1 = 0 "boundary condition at x=L"

The solution from Maple is copied into EES: T = exp(1/k^(1/2)/th^(1/2)*h_rad^(1/2)*x)*C_2+exp(-1/k^(1/2)/th^(1/2)*h_rad^(1/2)*x)*C_1& +1/h_rad*(q_flux+h_rad*T_infinity) "solution" x=x_bar*L "position" T_Celsius=converttemp(K,C,T) "in C"

The solution is overlaid onto the plot from (d) in Figure 4. 37.5 35

Temperature (°C)

32.5 30 27.5 25

solution from (d) Maple solution

22.5 20 17.5 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized position

Figure 4: Temperature as a function of position in the plate using Maple solution.

f.) Prepare a calibration curve for the flux meter - plot the heat flux as a function of the difference between the temperature at the center of the plate and the casing. Figure 5 illustrates the difference between the temperature at the center of the plate as a function of the applied heat flux. 1400

2

Heat flux (W/m )

1200 1000 800 600 400 200 0 0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

Temperature rise (K) Figure 5: Calibration curve for the flux meter.

g.) If the uncertainty in the measurement of the temperature difference is δΔT = 0.5 K then what is the uncertainty in the measurement of the heat flux? The uncertainty in the measurement of the heat flux is approximately:

δ q ′′ =

∂q ′′ δΔT δΔT

(20)

∂q ′′ = 55 W/m2-K according to Figure 5. Therefore, the uncertainty in the heat flux is δΔT approximately 22 W/m2.

where

Problem 1.8-1 (1-17 in text): Disk Brake Figure P1.8-1 illustrates a disk brake for a rotating machine. The temperature distribution within the brake can be assumed to be a function of radius only. The brake is divided into two regions. In the outer region, from Rp = 3.0 cm to Rd = 4.0 cm, the stationary brake pads create frictional heating and the disk is not exposed to convection. The clamping pressure applied to the pads is P = 1.0 MPa and the coefficient of friction between the pad and the disk is μ = 0.15. You may assume that the pads are not conductive and therefore all of the frictional heating is conducted into the disk. The disk rotates at N = 3600 rev/min and is b = 5.0 mm thick. The conductivity of the disk is k = 75 W/m-K and you may assume that the outer rim of the disk is adiabatic. coefficient of friction, μ = 0.15

stationary brake pads

clamping pressure P = 1 MPa b = 5 mm Rd = 4 cm

Ta = 30°C, h Rp = 3 cm center line k = 75 W/m-K

disk, rotates at N = 3600 rev/min Figure P1.8-1: Disk brake.

In the inner region of the disk, from 0 to Rp, is exposed to air at Ta = 30°C. The heat transfer coefficient between the air and disk surface depends on the angular velocity of the disk, ω, according to: 1.25

⎞ ω ⎡ W ⎤ ⎡ W ⎤⎛ h = 20 ⎢ 2 ⎥ + 1500 ⎢ 2 ⎥ ⎜⎜ ⎟ ⎣ m -K ⎦ ⎣ m -K ⎦ ⎝ 100 [ rad/s ] ⎟⎠

a.) Develop an analytical model of the temperature distribution in the disk brake; prepare a plot of the temperature as a function of radius for r = 0 to r = Rd. The inputs are entered in EES and the heat transfer coefficient is computed according to Eq. Error! Reference source not found.. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in b=5 [mm]*convert(mm,m) N=3600 [rev/min] omega=N*convert(rev/min,rad/s) mu=0.15 [-] P=1 [MPa]*convert(MPa,Pa) k=75 [W/m-K] Rd=4.0 [cm]*convert(cm,m) Rp=3.0 [cm]*convert(cm,m)

"thickness of disk" "rotational velocity of disk" "angular velocity of disk" "coefficient of friction" "clamping pressure" "conductivity" "outer radius of disk" "inner radius of pad"

Ta=converttemp(C,K,30) h=20[W/m^2-K]+1500 [W/m^2-K]*(omega/100 [rad/s])^1.25

"air temperature" "heat transfer coefficient"

In the outer region, region 1, the energy balance on a differential control volume is shown in Figure 2.

Figure 2: Differential energy balance in outer region, (region 1)

The energy balance suggested by Figure 2 is:

qr + q fh = qr + dr

(1)

where q fh is the rate of thermal energy generated by frictional heating. After expanding the r +

dr term, Eq. (1) becomes:

q fh =

dq dr dr

(2)

The rate equation for conduction is:

q = −b 2 π r k

dT1 dr

(3)

where T1 is the temperature in region 1. The force generated by the pad within the control volume is the product of the clamping pressure, the area of contact, and the coefficient of friction:

F = 4 π r dr P μ

(4)

Note that the factor of 4 in Eq. (4) is due to their being contact on both sides of the disk. The rate of frictional heating is the product of the force, the radius, and the angular velocity: q fh = 4 π r 2 dr P μ ω

Substituting Eqs. (3) and (5) into Eq. (2) leads to:

(5)

4 π r 2 dr P μ ω =

d ⎡ dT ⎤ −b 2 π r k 1 ⎥ dr ⎢ dr ⎣ dr ⎦

(6)

which can be rearranged: d ⎡ dT1 ⎤ 2Pμω 2 = − r r dr ⎢⎣ dr ⎥⎦ bk

(7)

d ⎡ dT1 ⎤ r = −β r 2 ⎢ ⎥ dr ⎣ dr ⎦

(8)

or

where

β=

2Pμω bk

(9)

Equation (8) can be directly integrated: ⎡ dT1 ⎤

∫ d ⎢⎣ r dr ⎥⎦ = − β ∫ r

2

dr

(10)

to achieve: r

dT1 r3 = − β + C1 dr 3

(11)

Equation (11) can be directly integrated again: ⎛ r 2 C1 ⎞ β dT = − ∫ 1 ∫ ⎜⎝ 3 + r ⎟⎠ dr

(12)

r3 T1 = − β + C1 ln ( r ) + C2 9

(13)

to achieve:

Equation (13) is the general solution for the temperature in region 1; the constants of integration will be selected in order to satisfy the boundary conditions. In the inner region, region 2, the energy balance on a differential control volume is shown in Figure 3.

Figure 3: Differential energy balance in inner region, (region 2)

The energy balance suggested by Figure 2 is: qr = qr + dr + qconv

(14)

After expanding the r + dr term, Eq. (14) becomes: 0=

dq dr + qconv dr

(15)

The rate equation for conduction remains the same: q = −b 2 π r k

dT2 dr

(16)

where T2 is the temperature in region 2. The rate equation for convection is: qconv = 4 π r dr h (T2 − Ta )

(17)

Substituting Eqs. (16) and (17) into Eq. (15) leads to: d ⎡ dT ⎤ −b 2 π r k 2 ⎥ dr + 4 π r dr h (T2 − Ta ) = 0 ⎢ dr ⎣ dr ⎦

(18)

d ⎡ dT2 ⎤ r − m 2 r T2 = −m 2 r Ta dr ⎢⎣ dr ⎥⎦

(19)

or

where

m=

2h bk

(20)

The solution to Eq. (19) can be divided into its homogeneous (u2) and particular (v2) parts: T2 = u2 + v2

(21)

d ⎡ dv2 ⎤ r − m 2 r v2 = − m 2 r Ta ⎢ ⎥ dr ⎣ dr ⎦

(22)

v2 = Ta

(23)

d ⎡ du2 ⎤ r − m 2 r u2 = 0 dr ⎢⎣ dr ⎥⎦

(24)

d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠

(25)

x=r

(26)

θ = u2

(27)

p =1

(28)

c=m

(29)

s =1

(30)

The solution to the particular equation:

is

The homogeneous equation:

is a form of Bessel's equation:

where

and the last term is negative. Following the flow chart provided in Section 1.8.4 of the book leads to n = 0, a = 1, and therefore the solution is: u2 = C3 BesselI ( 0, m r ) + C4 BesselK ( 0, m r )

(31)

The general solution for the temperature distribution in region 2 is therefore: T2 = C3 BesselI ( 0, m r ) + C4 BesselK ( 0, m r ) + Ta

(32)

Note that this could be obtained directly from Maple by entering Eq. (19): > restart; > ODE:=diff(r*diff(T2(r),r),r)-m^2*r*T2(r)=-m^2*r*Ta; 2 d ⎛d ⎞ ODE := ⎛⎜⎜ T2( r ) ⎞⎟⎟ + r ⎜⎜ 2 T2( r ) ⎟⎟ − m 2 r T2( r ) = −m 2 r Ta d r ⎝ ⎠ ⎝ dr ⎠

> T2s:=dsolve(ODE);

T2s := T2( r ) = BesselI( 0, m r ) _C2 + BesselK( 0, m r ) _C1 + Ta

The constants C1 through C4 in Eqs. (13) and (32) are obtained by applying the correct boundary conditions. At r = 0, the temperature must remain finite. The figures provided in Section 1.8.4 of the book or the limit capability in Maple show that BesselK(0,m r) will become infinite as r approaches zero: > limit(BesselI(0,m*r),r=0);

1 > limit(BesselK(0,m*r),r=0);

∞

therefore: C4 = 0

(33)

The temperature and temperature gradient at the interface between the regions must be continuous: T2, r = Rp = T1, r = Rp

(34)

and dT2 dr

= r = Rp

dT1 dr

(35) r = Rp

The temperature gradient at the outer rim must be zero: dT1 dr

=0 r = Rp

(36)

Substituting Eqs. (13) and (32) into Eqs. (33) through (36) leads to: C3 BesselI ( 0, m R p ) + Ta = − β

R 3p 9

C3 m BesselI (1, m R p ) = − β

+ C1 ln ( R p ) + C2 R p2 3

+

C1 Rp

Rd2 C1 −β + =0 3 Rd

(37)

(38)

(39)

Equations (37) through (39) are 3 equations for the unknown constants and can be solved in EES. beta=2*mu*P*omega/(k*b) m=sqrt(2*h/(k*b)) BesselI(0,m*Rp)*C_3+Ta=-1/9*beta*Rp^3+C_1*ln(Rp)+C_2 BesselI(1,m*Rp)*m*C_3=-1/3*beta*Rp^2+1/Rp*C_1 "equality of temperature gradient at r=Rp" -1/3*beta*Rd^2+1/Rd*C_1=0

"generation parameter" "fin parameter" "equality of temperature at r=Rp"

"zero temperature gradient at r=Rd"

The general solutions are entered in EES: T2 = BesselI(0,m*r2)*C_3+Ta T1 = -1/9*beta*r1^3+C_1*ln(r1)+C_2

"solution in region 2" "solution in region 1"

A dimensionless radius, the variable rbar, is defined in order to allow a Parametric Table to be generated where the variable r1 can be easily altered from Rp to Rd and the r2 can be easily altered from 0 to Rp: r1=Rp+(Rd-Rp)*rbar r2=rbar*Rp

Figure 4 illustrates the temperature distribution in the disk.

Figure 4: Temperature distribution in the disk

b.) If the disk material can withstand a maximum safe operating temperature of 750°C then what is the maximum allowable clamping pressure that can be applied? Plot the temperature distribution in the disk at this clamping pressure. What is the braking torque that results? The maximum operating temperature is obtained at r = Rd (see Figure 4). The clamping pressure that results in T1 at the outer rim reaching the maximum allowable temperature can be determined by commenting out the originally specified clamping pressure and specifying this temperature: {P=1 [MPa]*convert(MPa,Pa)} T_max_allowed=converttemp(C,K,750) rbar=1.0 T1=T_max_allowed

"clamping pressure" "maximum allowable temperature"

which leads to a clamping pressure of P = 0.57 MPa. The temperature distribution for this clamping pressure is shown in Figure 4. The torque applied by the pads (Tq) is obtained from the integral: Tq =

Rd

∫ 4 π r μ P dr 2

(40)

Rp

or 4 Tq = π μ P ⎡⎣ Rd3 − R 3p ⎤⎦ 3

(41)

which leads to Tq = 13.2 N-m. c.) Assume that you can control the clamping pressure so that as the machine slows down the maximum temperature is always kept at the maximum allowable temperature, 750°C. Plot the torque as a function of rotational speed for 100 rev/min to 3600 rev/min. A parametric table is created that includes the variables N and Tq,; N is varied from 100 rev/min to 3600 rev/min. The results are shown in Figure 5. Notice that it is possible to dramatically improve the performance of the brake if you can adjust the clamping pressure with speed.

Figure 5: Clamping pressure and torque as a function of rotational velocity.

Problem 1.8-2: Absorption in a Window Figure P1.8-2 illustrates a thin, disk-shaped window that is used to provide optical access to a combustion chamber. The thickness of the window is b and the outer radius of the window is Ro. The window is composed of material with conductivity k and absorption coefficient α. The combustion chamber side of the window is exposed to convection with hot gas at Tg and heat transfer coefficient h. Convection with the air outside of the chamber can be neglected. There is ′′ , that is incident on the combustion chamber side of the glass. The a radiation heat flux, qrad ′′ α b . The remainder amount of this radiation that is absorbed by the glass is, approximately, qrad ′′ (1 − α b ) , exits the opposite surface of the glass. The outer edge (at r = Ro) of this radiation, qrad

of the glass is held at temperature Tedge. combustion chamber

Tf , h

window, k and α

′′ radiant flux, qrad

b

x r Tedge

Ro

′′ unabsorbed radiant flux, (1 − α b )q rad

Tedge

outside of chamber Figure P1.8-2: Disk-shaped window.

You are to develop a 1-D, steady state analytical model that can predict the temperature distribution in the glass as a function of radial position, r. a.) How would you justify using a 1-D model of the glass? What number would you calculate in order to verify that the temperature does not vary substantially in the x direction? The Biot number should be computed in order to justify the extended surface approximation. The Biot number is the ratio of conduction resistance in the axial direction to convection resistance from the inner surface of the window: Bi =

Rcond , x Rconv

=

b h A bh = kA 1 k

(1)

Anything within a factor of 2 of Eq. (1) would be sufficient b.) Derive the ordinary differential equation in r that must be solved. Make sure that your differential equation includes the effect of conduction, convection with the gas within the chamber, and generation of thermal energy due to absorption. A differentially small control volume is shown in Figure 2.

Figure 2: Differentially small control volume.

The energy balance suggested by Figure 2 is: qr + qrad + qconv = qr + dr + qrad (1 − α b )

(2)

Expanding the r + dr term and simplifying leads to:

qrad + qconv =

dq dr + qrad (1 − α b ) dr

(3)

The rate equations are: ′′ 2 π r dr qrad = qrad

(4)

dT dr

(5)

q = − k 2 π r b

qconv = h 2 π r dr (T f − T )

(6)

Substituting Eqs. (4) through (6) into Eq. (3) leads to: ′′ 2 π r dr + h 2 π r dr (T f − T ) = qrad

d ⎡ dT ⎤ ′′ 2 π r dr (1 − α b ) k r b dr + qrad 2 π − dr ⎢⎣ dr ⎥⎦

(7)

Dividing through by (-k 2 π b dr) leads to: ′′ α h ⎛ qrad ⎞ d ⎡ dT ⎤ h r r T r T − = − + ⎜ ⎟ f dr ⎢⎣ dr ⎥⎦ k b kb ⎠ ⎝ k

(8)

c.) What are the boundary conditions for the ordinary differential equation that you derived in part (b)? At r = 0 the temperature must be finite. At the edge, the temperature is specified:

Tr = Ro = Tedge

(9)

d.) Solve the ordinary differential equation in order to obtain an expression for the temperature as a function of radius. The solution is split into its homogeneous (u) and particular (v) parts: T =u+v

(10)

The particular solution is a constant: −

⎛ q ′′ α h ⎞ h r v = − r ⎜ rad + Tf ⎟ kb kb ⎠ ⎝ k

(11)

or

v = Tf +

′′ α b qrad h

(12)

The homogeneous form of the differential equation is:

d ⎡ du ⎤ r ⎥ − m2 r u = 0 ⎢ dr ⎣ dr ⎦

(13)

where m=

h kb

(14)

Equation (13) is a form of Bessel's equation: d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ = 0 dx ⎝ dx ⎠

(15)

θ =u

(16)

x=r

(17)

p =1

(18)

c=m

(19)

s =1

(20)

where

The solution can be obtained by using the chart found in the notes:

n=

1− p 1−1 = =0 s − p + 2 1−1+ 2

(21)

2 =1 1−1+ 2

(22)

a=

n 1−1 = =0 a 2

(23)

so that u = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r )

(1-24)

and the solution is: T = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r ) + T f +

′′ α b qrad h

(1-25)

Because BesselK(0,0) becomes infinite, C2 = 0 and: T = C1 BesselI ( 0, m r ) + T f +

′′ α b qrad h

(1-26)

The boundary condition associated with Eq. (9) leads to: Tedge = C1 BesselI ( 0, m Ro ) + T f +

′′ α b qrad h

(1-27)

so that: ′′ α b qrad h C1 = BesselI ( 0, m Ro ) Tedge − T f −

(1-28)

and q ′′ α b ⎞ BesselI ( 0, m r ) q ′′ α b ⎛ + T f + rad T = ⎜ Tedge − T f − rad ⎟ h ⎠ BesselI ( 0, m Ro ) h ⎝

(1-29)

Problem 1.8-3: Cryogenic Thermal Standoff It is often necessary to provide a fluid outlet port that will allow very cold gas to escape from a cryogenic facility; for example, the boil-off of liquid nitrogen or liquid helium from within a vacuum vessel must be allowed to vent to the atmosphere. The seal between the base of the flange that contains the fluid passage and the surrounding vessel is usually made with an o-ring; these seals are convenient in that they are hermetic and easily demountable. However, most convenient o-ring materials do not retain their ductility at temperatures much below 0°C and therefore it is important that the o-ring be kept at or above this temperature so that it continues to provide a good seal; if the o-ring “freezes” then the cryogenic facility will lose its vacuum. The o-ring temperature is maintained at an appropriate level using a thermal standoff, as shown in Figure P1.8-3 The fluid passage is attached to the flange via a separate, slightly larger tube made of a stainless steel with thermal conductivity k = 15 W/m-K. This thermal standoff has an outer radius, rts = 1 cm, thickness tts = 1 mm, and length Lts = 5 cm. The cryogenic gas is at 77 K and you may assume that the point x = 0 in Figure P1.8-3 is at Tcold = 77 K. The inside of the tube is exposed to a vacuum and you may assume that it experiences negligible radiation heat transfer and no convective heat transfer. The outside of the tube is exposed to air at Tair = 20°C with heat transfer coefficient, h = 7 W/m2-K. The bottom of the tube (x = Lts) is welded to the flange. The flange has an outer radius rfl = 8 cm and a thickness tfl = 1.5 mm. The inside of the flange is exposed to vacuum and therefore, for the purposes of this problem, adiabatic. The outside of the flange is exposed to the same 20°C air with the same 7 W/m2-K heat transfer coefficient.

thermal stand off flange

vacuum o-ring seal

cryogenic gas

Tcold = 77 K

Lts = 5 cm Tair = 20°C 2 h = 7 W/m -K

x

k = 15 W/m-K tfl = 1.5 mm

tts = 1 mm rts = 1 cm

rfl = 8 cm Figure P1.8-3: Cryogenic thermal standoff.

a.) Is it appropriate to treat the thermal standoff and the flange as extended surfaces? That is, can the temperature within the stand-off be treated as being only a function of x and the temperature in the flange only a function of r? The input parameters are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" k=15 [W/m-K] r_ts=1.0 [cm]*convert(cm,m) t_ts=1.0 [mm]*convert(mm,m) L_ts=5.0 [cm]*convert(cm,m) T_cold=77 [K] T_air=converttemp(C,K,20 [C]) h=7 [W/m^2-K] r_fl=8.0 [cm]*convert(cm,m) t_fl=1.5 [mm]*convert(mm,m)

"thermal conductivity" "outer radius of thermal standoff" "thickness of thermal standoff" "length of thermal standoff" "cryogenic gas temperature" "air temperature" "heat transfer coefficient" "radius of flange" "thickness of flange"

The Biot numbers that must be calculated for the standoff and the flange (Bits and Bifl, respectively) are:

Bits =

h tts k

Bi fl =

h t fl

and

k

These are computed in EES: Bi_ts=t_ts*h/k Bi_fl=t_fl*h/k

"Biot number for thermal standoff" "Biot number for flange"

and found to be very small. b.) Develop an analytical model of the problem that can predict the temperature distribution as a function of x in the thermal stand-off and as a function of r in the flange. An energy balance on the differential control volume within the thermal standoff leads to:

q x = q x + dx + qconv or, after making the usual simplifications:

0=−

dTts ⎤ d ⎡ r t k dx + 2 π rts dx h (Tts − Tair ) 2 π ts ts dx ⎢⎣ dx ⎥⎦

which becomes the governing differential equation for a constant cross-sectional area fin that was solved in Section 1.6: d 2θts − mts2 θts = 0 dx 2 where mts =

h tts k

and

θts = Tts − Tair Equation (1) is solved by exponentials (or equivalently by sinh and cosh):

(1)

θts = C1 exp ( mts x ) + C2 exp ( − mts x )

(2)

where C1 and C2 are determined by the boundary coefficients. An energy balance on the differential control volume within the flange (see Figure 1) is: qr = qr + dr + qconv or, after making the usual simplifications: 0=−

dT fl ⎤ d ⎡ ⎢ 2 π r t fl k ⎥ dr + 2 π r dr (T fl − Tair ) dr ⎣ dr ⎦

which becomes Bessel's equation: d ⎡ dθ fl ⎤ 2 ⎢r ⎥ − m fl r θ fl = 0 dr ⎣ dr ⎦

(3)

where m fl =

h t fl k

and

θ fl = T fl − Tair Equation (3) is solved by 0th order modified Bessel functions:

θ fl = C3 BesselI ( 0, m fl r ) + C4 BesselK ( 0, m fl r )

(4)

where C3 and C4 are determined by the boundary coefficients. There must be four boundary conditions; two for each of the 2nd order differential equations. The temperature at the top of the thermal stand-off is specified:

Tts , x =0 = Tcold or, substituting into Eq. (2):

C1 + C2 = Tcold − Tair

The temperature at the bottom of the thermal standoff must be equal to the temperature at the inner edge of the flange:

Tts , x = Lts = T fl ,r = rts or

C1 exp ( mts Lts ) + C2 exp ( −mts Lts ) = C3 BesselI ( 0, m fl rts ) + C4 BesselK ( 0, m fl rts ) The rate of heat transfer out of the bottom of the thermal standoff must be equal to the rate that heat that is transferred into the inner edge of the flange: − k 2 π rts tts

dTts dx

= − k 2 π rts t fl Lts

dT fl dr

rts

or tts

dθts dx

= t fl

dθ fl

Lts

dr

rts

whic leads to: tts ⎡⎣C1 mts exp ( mts Lts ) − C2 mts exp ( −mts Lts ) ⎤⎦ = t fl ⎡⎣C3 m fl BesselI (1, m fl rts ) − C4 m fl BesselK (1, m fl rts ) ⎤⎦ Finally, the outer edge of the flange is assumed to be adiabatic (the small amount of convection from the edge can be neglected with little error): − k 2 π rfl t fl

dT fl dr

=0 r fl

or dθ fl dr

=0 r fl

which leads to: C3 m fl BesselI (1, m fl rfl ) − C4 m fl BesselK (1, m fl rfl ) = 0

The boundary condition equations are entered into EES:

"Solution parameters" m_ts=sqrt(h/(t_ts*k)) m_fl=sqrt(h/(t_fl*k)) "Boundary conditions" C_1+C_2=T_cold-T_air C_1*exp(m_ts*L_ts)+C_2*exp(-m_ts*L_ts)=C_3*BesselI(0,m_fl*r_ts)+C_4*BesselK(0,m_fl*r_ts) t_ts*(C_1*m_ts*exp(m_ts*L_ts)-C_2*m_ts*exp(-m_ts*L_ts))=t_fl*(C_3*m_fl*BesselI(1,m_fl*r_ts)C_4*m_fl*BesselK(1,m_fl*r_ts)) C_3*m_fl*BesselI(1,m_fl*r_fl)-C_4*m_fl*BesselK(1,m_fl*r_fl)=0

The solution is provided in EES using two parametric tables for the thermal standoff and flange temperature distributions. The first table is titled ‘Thermal Stand off’ and includes values of dimensionless position, x , ranging from 0 to 1; where: x=

x Lts

The coordinate, s, is also computed; s goes from 0 to Lts + (rfl – rts) as you move from the top of the thermal stand off to the edge of the flange. The use of s provides a convenient method for looking at the entire temperature distribution in the thermal standoff and flange. s=x

The ‘Thermal Stand off’ table includes columns that contain the values of θts and Tts. The solution for the temperature within the thermal stand off is only calculated if you are running the parametric table entitled ‘Thermal Stand off’. The selection of equations that are to be solved with each parametric table is facilitated by using the $IF PARAMETRICTABLE directive to check which table is being calculated. The equations located between the $IF and $ENDIF in the EES code below are executed only if the ‘Thermal Stand off’ parametric table is being calculated. $IF PARAMETRICTABLE='Thermal Stand off' x_bar=x/L_ts s=x theta_ts=C_1*exp(m_ts*x)+C_2*exp(-m_ts*x) Temp_ts=theta_ts+T_air Temp_ts_C=converttemp(K,C,Temp_ts) $ENDIF

A second parametric table entitled ‘Flange’ is includes values of dimensionless position in the flange, r : r=

( r − rts )

(r

fl

− rts )

and the coordinate, s, which is computed in the flange as:

s = r − rts + L

The table ‘Flange’ also includes the values θfl and Tfl. The required commands are included in a separate $IF statement: $IF PARAMETRICTABLE='Flange' r_bar=(r-r_ts)/(r_fl-r_ts) s=L_ts+r-r_ts theta_fl=C_3*BesselI(0,m_fl*r)+C_4*BesselK(0,m_fl*r) Temp_fl=theta_fl+T_air Temp_fl_C=converttemp(K,C,Temp_fl) $ENDIF

By sequentially running parametric table ‘Thermal Stand off’ and ‘Flange’ it is possible to determine the entire temperature distribution; the result is shown in Figure 2.

Figure 2: Temperature as a function of the coordinate s.

Notice that the solution satisfies each of the boundary conditions. The temperature at the top of the stand off is equal to 77 K and the temperature gradient at the edge of the flange is zero. The temperature at the intersection of the flange and the thermal stand off is continuous but there is a discontinuity in the temperature gradient related to the fact that the flange is slightly thicker (and therefore has a lower temperature gradient for the same conduction heat transfer) than the thermal stand off. The value of the thermal stand off is clear. If the o-ring seal is placed towards the outer radius of the flange then Figure 2 shows that the temperature will remain above freezing and therefore the o-ring will continue to function. Using the EES model it is possible to evaluate alternative, more effective designs (i.e., thermal stand off geometries that keep the temperature at the outer edge of the flange higher). Figure 3 illustrates the temperature at the edge of the flange (i.e., T fl ,r = rfl ) as a function of the thermal stand-off thickness for various values of its length.

Figure 3: Temperature at the edge of the flange as a function of the thermal stand off thickness for various values of the thermal stand off length.

Note that either increasing the thermal stand off length or decreasing its thickness will tend to make it a less efficient fin and therefore increase the temperature gradient due to conduction. It turns out that a good thermal stand off is a bad fin, isolating the tip of the fin (i.e., the flange) from the base of the fin (i.e., the cryogenic temperature).

Problem 1.8-4: Circular Fin Figure P1.8-4 shows a typical fin design that is fabricated by attaching a thin washer to the outer radius of a tube. The inner and outer radii of the fin are rin and rout, respectively. The thickness of the fin is th and the fin material has conductivity, k. The fin is surrounded by fluid at T∞ and the average heat transfer coefficient is h . The base of the fin is maintained at Tb and the tip is adiabatic. h , T∞ rout

th

rin

Tb

k

Figure P1.8-4: Circular fin.

Determine an analytical solution for the temperature distribution in the fin and the fin efficiency. The differential control volume shown in Figure 2 can be used to derive the governing equation.

Figure 2: Differential control volume.

An energy balance for the control volume is:

qr = qr + dr + qconv or

0= The conduction and convection terms are:

dq dr + qconv dr

q = − k 2 π r th

dT dr

qconv = 4 π r dr h (T − T∞ )

Combining these equations leads to: 0=

d ⎡ dT ⎤ dr + 4 π r dr h (T − T∞ ) = 0 −k 2 π r th ⎢ dr ⎣ dr ⎥⎦

which can be simplified to: d ⎡ dT ⎤ 2 r h 2r h r T=− T∞ − ⎢ ⎥ dr ⎣ dr ⎦ k th k th The solution is divided into a homogeneous and particular component:

T = Th + Tp which leads to: 2r h d ⎡ dTh ⎤ 2 r h d ⎡ dTp ⎤ 2 r h − − r Th + ⎢ r Tp = − T∞ ⎥ ⎢ ⎥ dr ⎣ dr ⎦ k th dr ⎣ dr ⎦ k th k th  

  

=0 for homogeneous differential equation

whatever is left is the particular differential equation

The solution to the particular differential equation: 2r h d ⎡ dTp ⎤ 2 r h Tp = − T∞ ⎢r ⎥− dr ⎣ dr ⎦ k th k th

is

Tp = T∞ The homogeneous differential equation is:

d ⎡ dTh ⎤ r − m 2 r Th = 0 ⎢ ⎥ dr ⎣ dr ⎦ where m is the fin parameter, defined as:

(1)

m=

2h kb

Equation (1) is a form of Bessel’s equation:

d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠

(2)

where (by comparing Eqs. (1) and (2)), p= 1, c = m, and s = 1. Referring to the flow chart presented in Section 1.8.4, the value of s-p+2 is equal to 2 and therefore the solution parameters n and a must be computed:

n=

1−1 =0 1−1+ 2

a=

2 =1 1−1+ 2

The last term in Eq. (1) is negative and therefore the solution to Eq. (1) is given by: n

(

Th = C1 x a BesselI n, c a x

1

a

)+C x

n

2

a

(

BesselK n, c a x

1

a

)

or Th = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r )

The solution to the governing differential equation is: T = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r ) + T∞

(3)

Note that Maple would provide this information as well: > restart; > ODE:=diff(r*diff(T(r),r),r)-m^2*r*T(r)=-m^2*r*T_infinity;

d ODE := ⎛⎜⎜ T( r ) ⎞⎟⎟ + d ⎝ r ⎠

2 ⎛d ⎞ ⎜ r ⎜ 2 T( r ) ⎟⎟ − m 2 r T( r ) = −m 2 r T_infinity ⎝ dr ⎠

> Ts:=dsolve(ODE);

Ts := T( r ) = BesselI( 0, m r ) _C2 + BesselK ( 0, m r ) _C1 + T_infinity

The boundary conditions must be used to obtain the constants C1 and C2. The base temperature is specified:

Tr = rin = Tb or: C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r ) + T∞ = Tb

(4)

The tip of the fin is adiabatic: − k 2 π rin

dT dr

=0 r = rout

or C1

d d ⎡⎣ BesselI ( 0, m r ) ⎤⎦ r = r + C2 ⎡⎣ BesselK ( 0, m r ) ⎤⎦ r = r = 0 out out dr dr

Using the rules for differentiating Bessel functions presented in Section 1.8.4 leads to: C1 m Bessel_I (1, m rout ) − C2 m Bessel_K (1, m rout ) = 0

(5)

The boundary condition equations, Eqs. (4) and (5), can be obtained using Maple: > BC1:=rhs(eval(Ts,r=r_in))=T_b;

BC1 := BesselI( 0, m r_in ) _C2 + BesselK ( 0, m r_in ) _C1 + T_infinity = T_b

> BC2:=rhs(eval(diff(Ts,r),r=r_out))=0;

BC2 := BesselI( 1, m r_out ) m _C2 − BesselK ( 1, m r_out ) m _C1 = 0

These equations can be copied into EES in order to obtain the solution for arbitrary conditions: "Boundary conditions" theta_b = C_1*BesselI(0, m*r_tube)+C_2*BesselK(0, m*r_tube) 0 = C_1*BesselI(1, m*r_fin)*m-C_2*BesselK(1, m*r_fin)*m "Temperature distribution" theta = C_1*BesselI(0, m*r)+C_2*BesselK(0, m*r)

Given arbitrary values of the variables T_b, T_infinity, m, r_in, and r_out, the EES code above will provide the temperature distribution. It is convenient to solve for the two constants explicitly and substitute them into the temperature distribution; we can let Maple accomplish this process and avoid the algebra. The first step is to solve the two boundary conditions equations simultaneously to obtain the unknown constants; this is done using the solve command in Maple where the first argument is the set of equations (BC1 and BC2) and the second are the arguments to be solved for (_C1 and _C2):

> constants:=solve({BC1,BC2},{_C1,_C2});

BesselK ( 1, m r_out ) ( −T_infinity + T_b ) BesselK ( 1, m r_out ) BesselI( 0, m r_in ) + BesselK ( 0, m r_in ) BesselI( BesselI( 1, m r_out ) ( −T_infinity + T_b ) , _C1 = BesselK ( 1, m r_out ) BesselI( 0, m r_in ) + BesselK ( 0, m r_in ) BesselI( 1, m r_o }

constants := { _C2 =

These equations for the constants can be substituted into the solution using the eval command, where the first argument is the base expression and the second contains the sub-expressions that must be substituted into the base expression: > Ts:=eval(Ts,constants);

BesselI( 0, m r ) BesselK ( 1, m r_out ) ( −T_infinity + T_b ) BesselK ( 1, m r_out ) BesselI( 0, m r_in ) + BesselK ( 0, m r_in ) BesselI( 1, m r_ BesselK ( 0, m r ) BesselI( 1, m r_out ) ( −T_infinity + T_b ) + BesselK ( 1, m r_out ) BesselI( 0, m r_in ) + BesselK ( 0, m r_in ) BesselI( 1, m r_out ) + T_infinity

Ts := T( r ) =

which can be copied into EES in place of the 3 original equations: "Explicit solution" T = BesselI(0,m*r)*BesselK(1,m*r_out)*(-T_infinity+T_b)/(BesselK(1,m*r_out)*BesselI(0,m*r_in)+& BesselK(0,m*r_in)*BesselI(1,m*r_out))+BesselK(0,m*r)*BesselI(1,m*r_out)*(-T_infinity+T_b)& /(BesselK(1,m*r_out)*BesselI(0,m*r_in)+BesselK(0,m*r_in)*BesselI(1,m*r_out))+T_infinity

So the temperature distribution through the circular fin is given by: ⎡ BesselK (1, m rout ) BesselI ( 0, m r ) + BesselI (1, m rout ) BesselK ( 0, m r ) ⎤⎦ T = T∞ + (Tb − T∞ ) ⎣ (6) ⎡⎣ BesselI (1, m rout ) BesselK ( 0, m rin ) + BesselI ( 0, m rin ) BesselK (1, m rout ) ⎤⎦ The heat transfer rate to the base of the fin, q fin , is obtained by applying Fourier’s law to evaluate the conduction heat transfer rate at the base of the fin: q fin = −k 2 π rin th Substituting Eq. (6) into Eq. (7) leads to:

dT dr

(7) r = rin

q fin

d ⎧ ⎫ BesselK (1, m rout ) ⎡⎣ BesselI ( 0, m r ) ⎤⎦ r = r ⎪ ⎪ in dr +⎪ ⎪ ⎪ BesselI (1, m rout ) BesselK ( 0, m rin ) + BesselI ( 0, m rin ) BesselK (1, m rout ) ⎪ = − k 2 π rin th (Tb − T∞ ) ⎨ ⎬ d ⎪ ⎪ BesselI (1, m rout ) ⎡⎣ BesselK ( 0, m r ) ⎤⎦ r = r in ⎪ ⎪ dr ⎪ BesselI (1, m r ) BesselK ( 0, m r ) + BesselI ( 0, m r ) BesselK (1, m r ) ⎪ out in in out ⎩ ⎭

Using the rules for differentiating Bessel functions, presented in Section 1.8.4, to evaluate the derivatives leads to:

⎡ BesselK (1, m rout ) BesselI (1, m rin ) -BesselI (1, m rout ) BesselK (1, m rin ) ⎤⎦ q fin = − k 2 π rin th (Tb − T∞ ) m ⎣ ⎡⎣ BesselI (1, m rout ) BesselK ( 0, m rin ) + BesselI ( 0, m rin ) BesselK (1, m rout ) ⎤⎦ (8) Maple achieves the same result: > q_dot_fin:=-k*2*pi*r_in*th*rhs(eval(diff(Ts,r),r=r_in));

q_dot_fin := − 2 k π r_in th ( −BesselI( 1, m r_in ) m BesselK ( 1, m r_out ) T_infinity + BesselI( 1, m r_in ) m BesselK ( 1, m r_out ) T_b + BesselK ( 1, m r_in ) m BesselI( 1, m r_out ) T_infinity − BesselK ( 1, m r_in ) m BesselI( 1, m r_out ) T_b )/( BesselK ( 1, m r_out ) BesselI( 0, m r_in ) + BesselK ( 0, m r_in ) BesselI ( 1, m r_out ) )

which can be cut and pasted directly into EES: "Fin heat transfer rate" q_dot_fin=-2*k*pi*r_in*th*(-BesselI(1,m*r_in)*m*BesselK(1,m*r_out)*T_infinity+& BesselI(1,m*r_in)*m*BesselK(1,m*r_out)*T_b+BesselK(1,m*r_in)*m*BesselI(1,m*r_out)& *T_infinity-BesselK(1,m*r_in)*m*BesselI(1,m*r_out)*T_b)/(BesselK(1,m*r_out)*& BesselI(0,m*r_in)+BesselK(0,m*r_in)*BesselI(1,m*r_out))

Finally, the fin efficiency (ηfin) is the ratio of the heat transfer rate to the heat transfer rate from an isothermal fin at the base temperature:

η fin =

q fin

2 π ( r − rin2 ) h (Tb − T∞ ) 2 out

Substituting Eq. (8) into the definition of the fin efficiency leads to:

η fin =

⎡⎣ BesselI (1, m rout ) BesselK (1, m rin ) -BesselK (1, m rout ) BesselI (1, m rin ) ⎤⎦ 2 rin 2 2 m ( rout − rin ) ⎡⎣ BesselI (1, m rout ) BesselK ( 0, m rin ) + BesselI ( 0, m rin ) BesselK (1, m rout ) ⎤⎦

which can be expressed as a function of the tube-to-fin radius ratio (rin/rout) and the product of the fin parameter and the fin radius (m rout).

η fin

⎡ ⎛ ⎛ rin ⎞ rin ⎞ ⎤ ⎛r ⎞ 2 ⎜ in ⎟ ⎢ BesselI (1, m rout ) BesselK ⎜ 1, m rout ⎟ -BesselK (1, m rout ) BesselI ⎜ 1, m rout ⎟⎥ rout ⎠ rout ⎠ ⎦ rout ⎠ ⎝ ⎝ ⎝ ⎣ = ⎤ ⎛ ⎛ r ⎞2 ⎞ ⎡ ⎛ ⎛ rin ⎞ rin ⎞ m rout ⎜ 1 − ⎜ in ⎟ ⎟ ⎢ BesselI (1, m rout ) BesselK ⎜ 0, m rout ⎟ BesselK (1, m rout ) ⎥ ⎟ + BesselI ⎜ 0, m rout rout ⎠ rout ⎠ ⎜ ⎝ rout ⎠ ⎟ ⎣ ⎝ ⎝ ⎦ ⎝ ⎠

The fin efficiency for a circular fin is shown in Figure 3 as a function of m rout for various values of rin/rout. Note that the fin radius can be corrected approximately to account for convection from the tip by adding the half-thickness of the fin; as previously discussed in Section 1.6.5, this correction is small and rarely worth considering.

Figure 3: Fin efficiency of a circular fin as a function of m rout for various values of rin/rout.

Problem 1.8-5 (1-18 in text): Optimizing a Fin Figure P1.8-5 illustrates a fin that is to be used in the evaporator of a space conditioning system for a space-craft. 2 h = 120 W/m -K T∞ = 20°C

x

th = 1 mm

L = 2 cm

ρb = 8000 kg/m3

k = 50 W/m-K ρ = 3000 kg/m3 Tb = 10°C thg = 2 mm

thb = 2 mm

Wb = 1 cm Figure P1.8-5: Fin on an evaporator.

The fin is a plate with a triangular shape. The thickness of the plate is th = 1 mm and the width of the fin at the base is Wb = 1 cm. The length of the fin is L = 2 cm. The fin material has conductivity k = 50 W/m-K. The average heat transfer coefficient between the fin surface and the air in the space-craft is h = 120 W/m2-K. The air is at T∞ = 20°C and the base of the fin is at Tb = 10°C. Assume that the temperature distribution in the fin is 1-D in x. Neglect convection from the edges of the fin. a.) Obtain an analytical solution for the temperature distribution in the fin. Plot the temperature as a function of position. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in h_bar=120 [W/m^2-K] k=50 [W/m-K] T_infinity=converttemp(C,K,20[C]) T_b=converttemp(C,K,10[C]) th_mm= 1 [mm] th=th_mm*convert(mm,m) L_cm=2 [cm] L=L_cm*convert(cm,m) W_b=1 [cm]*convert(cm,m)

"average heat transfer coefficient" "conductivity" "air material" "base temperature" "fin thickness in mm" "fin thickness" "fin length in cm" "fin length" "fin base width"

The differential control volume shown in Figure P1.8-5-2 is used to derive the governing differential equation: q x = q x + dx + qconv

(1)

x

qx qconv q x + dx

Figure P1.8-5-2: Differential control volume.

The rate of conduction and convection are:

dT dx

q x = − k Ac

qconv = h per (T − T∞ ) dx

(2) (3)

where Ac is the cross-sectional area for conduction and per is the perimeter. The width of the fin is a function of x:

W = Wb

x L

(4)

Therefore, Ac and per are:

Ac = Wb th

x L

(5)

per = 2Wb

x L

(6)

Substituting Eqs. (5) and (6) into Eq. (2) and (3) leads to: x dT L dx

(7)

x (T − T∞ ) dx L

(8)

q x = − k Wb th qconv = h 2Wb

Substituting Eqs. (7) and (8) into Eq. (1) leads to: 0= Simplifying:

d ⎡ x dT ⎤ x dx + h 2Wb (T − T∞ ) dx − k Wb th ⎢ ⎥ dx ⎣ L dx ⎦ L

(9)

d ⎛ dT ⎜x dx ⎝ dx

⎞ 2 2 ⎟ − m xT = −m xT∞ ⎠

(10)

where m2 =

2h k th

m=sqrt(2*h_bar/(k*th))

(11)

"solution parameter"

Maple is used to identify the solution to Eq. (10): > restart; > ODE:=diff(x*diff(T(x),x),x)-m^2*x*T(x)=-m^2*x*T_infinity;

2 d ⎞ ⎛d ODE := ⎛⎜⎜ T( x ) ⎞⎟⎟ + x ⎜⎜ 2 T( x ) ⎟⎟ − m 2 x T( x ) = −m 2 x T_infinity ⎠ ⎝ dx ⎝ dx ⎠

> Ts:=dsolve(ODE);

Ts := T( x ) = BesselI( 0, m x ) _C2 + BesselK( 0, m x ) _C1 + T_infinity

Therefore: T = C2 BesselI ( 0, m x ) + C1 BesselK ( 0, m x ) + T∞

(12)

The fin temperature at the tip must be bounded: Tx =0 = C2 BesselI ( 0, m 0 ) + C1 BesselK ( 0, m 0 ) + T∞ < ∞  

 

(13)

∞

1

The limit of the 0th order modified Bessel functions as x → 0 are evaluated using Maple: > limit(BesselI(0,m*x),x=0);

1

> limit(BesselK(0,m*x),x=0);

∞

Therefore, C1 must be zero: T = C2 BesselI ( 0, m x ) + T∞

The base temperature is specified; therefore:

(14)

Tb = C2 BesselI ( 0, m L ) + T∞

(15)

so:

(Tb − T∞ ) BesselI ( 0, m L )

C2 =

(16)

Substituting Eq. (16) into Eq. (14) leads to: T = (Tb − T∞ )

BesselI ( 0, m x ) + T∞ BesselI ( 0, m L )

x_bar=0.5 [-] x=x_bar*L T=(T_b-T_infinity)*BesselI(0,m*x)/BesselI(0,m*L)+T_infinity T_C=converttemp(K,C,T)

(17)

"dimensionless position" "position" "temperature" "in C"

Figure P1.8-5-3 illustrates the temperature as a function of position normalized by the fin length. 14 13.5

Temperature (°C)

13 12.5 12 11.5 11 10.5 10 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized position, x/L

Figure P1.8-5-3: Fin temperature as a function of dimensionless position.

b.) Calculate the rate of heat transfer to the fin. The rate of heat transfer to the fin is computed according to: q fin = k Wb th

dT dx

(18) x= L

Equation (18) is evaluated using Maple: > restart; > T:=(T_b-T_infinity)*BesselI(0,m*x)/BesselI(0,m*L)+T_infinity;

T :=

( T_b − T_infinity ) BesselI( 0, m x ) + T_infinity BesselI( 0, m L )

> q_dot_fin=k*W_b*th*eval(diff(T,x),x=L);

q_dot_fin =

k W_b th ( T_b − T_infinity ) BesselI( 1, m L ) m BesselI( 0, m L )

Therefore: q fin = k Wb th m (Tb − T∞ )

BesselI (1, m L ) BesselI ( 0, m L )

(19)

q_dot_fin=k*W_b*th*(T_b-T_infinity)*m*BesselI(1,m*L)/BesselI(0,m*L) "fin heat transfer rate"

which leads to q fin = -0.196 W (the heat transfer is negative because the base temperature is less than the ambient temperature). c.) Determine the fin efficiency. The fin efficiency is defined according to:

η fin =

q fin

h As (Tb − T∞ )

(20)

where As is the total surface area of the fin exposed to the fluid:

As = Wb L

(21)

Substituting Eqs. (19) and (21) into Eq. (20) leads to:

η fin =

k Wb th m (Tb − T∞ ) BesselI (1, m L ) h Wb L (Tb − T∞ ) BesselI ( 0, m L )

(22)

Substituting Eq. (11) into Eq. (22) and simplifying leads to:

η fin =

k th 2 h BesselI (1, m L ) 2 th k BesselI (1, m L ) = h L k th BesselI ( 0, m L ) L 2 h BesselI ( 0, m L ) N

(23)

2 BesselI (1, mL ) mL BesselI ( 0, mL )

(24)

1/ m

or

η fin =

eta_fin=2*BesselI(1,m*L)/(m*L*BesselI(0,m*L)) "fin efficiency"

which leads to ηfin = 0.8178. Figure P1.8-5-4 illustrates the fin efficiency as a function of the fin parameter mL. 1 0.9

Fin efficiency

0.8 0.7 0.6 0.5 0.4 0.3 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Fin parameter, mL

Problem P1.8-5-4: Fin efficiency as a function of the fin parameter, mL.

The fin has density ρ = 3000 kg/m3. The fin is installed on a base material with thickness thb = 2 mm and density ρb = 8000 kg/m3. The half-width between the gap between adjacent fins is thg = 2 mm. Therefore, the volume of the base material associated with each fin is thb Wb (th + 2 thg). d.) Determine the ratio of the absolute value of the rate of heat transfer to the fin to the total mass of material (fin and base material associated with the fin). The additional inputs are entered in EES: rho=3000 [kg/m^3] th_b=2 [mm]*convert(mm,m) th_g=2 [mm]*convert(mm,m) rho_b=8000 [kg/m^3]

"density of fin material" "thickness of base material" "half-width of gap between adjacent fins" "base material density"

The fin mass is given by: M fin =

Wb L th ρ 2

(25)

The mass of the associated base material is:

M b = Wb ( th + 2 thg ) thb ρb

(26)

The ratio of rate of the fin heat transfer to mass is: q fin M M_fin=W_b*L*th*rho/2 M_b=W_b*(th+2*th_g)*th_b*rho_b q\M=abs(q_dot_fin)/(M_fin+M_b)

=

(M

q fin fin

+ Mb ) "fin mass" "mass of base material" "ratio of heat transfer to mass"

(27)

which leads to q fin /M = 178.4 W/kg. e.) Prepare a contour plot that shows the ratio of the heat transfer to the fin to the total mass of material as a function of the length of the fin (L) and the fin thickness (th). A parametric table is generated that contains the variables L_cm, th_mm and q\M and has 400 rows. The value of the variable L_cm is varied from 1 cm to 10 cm every 20 rows and the value of th_mm is varied from 0.2 mm to 2 mm in increments of 20 rows. The table is run and used to generate the contour plot shown in Figure P1.8-5-5. 2

1.6

Fin thickness (mm)

115.9

139.3

1.8

127.6

162.7 151

1.4 1.2

174.4

186.1

1 197.8

0.8 209.5 W/kg

0.6 0.4 0.2 0 1

2

3

4

5

6

7

8

9

10

Fin length (cm)

Figure P1.8-5-5: Contours of heat transfer per mass in the parameter space of fin length and thickness.

f.) What is the optimal value of L and th that maximizes the absolute value of the fin heat transfer rate to the mass of material? According to Figure P1.8-5-5, the optimal design is approximately L = 3.3 cm and th = 0.58 mm. A more precise optimization can be carried out using EES' internal optimization feature. Maximizing q fin /M by varying L and th leads to q fin /M = 209.6 W/kg at L = 3.25 cm and th = 0.56 mm.

Problem 1.8-6: Cryosurgical Probe As an alternative to surgery, cancer tumors may be destroyed by placing cylindrically-shaped cryoprobes into the body, as shown in Figure P1.8-6. The probe surface is cooled causing the temperature of the surrounding tissue to drop to a lethal level, killing the tumor. tissue k = 0.6 W/m-K β = 40000 W/m3-K 2 q ′′p = 30000 W/m

Tb = 37°C rp = 5 mm

Figure 1.8-6: Cryosurgical probe.

The probe radius is rp = 5 mm and the heat flux at the surface of the probe (leaving the tissue) is q ′′p = 30000 W/m2. The tissue has conductivity k = 0.6 W/m-K. The blood flow through the tissue results in a volumetric heating effect ( g ′′′ ) that is proportional to the difference between the local temperature and the blood temperature, Tb = 37ºC: g ′′′ = β (Tb − T )

where β = 40000 W/m3-K. The temperature of the tissue far from the probe is Tb. Assume that the temperature distribution is 1-D and steady-state. a.) Develop an analytical model that can be used to predict the temperature distribution in the tissue. Implement your solution in EES and prepare a plot of the temperature distribution as a function of radius. The differential control volume shown in Figure 2 can be used to derive the governing equation. r

qr

g

qr + dr

dr

Figure 2: Differential control volume.

An energy balance for the control volume is:

g + qr = qr + dr

(1)

or

g =

dq dr dr

(2)

The conduction and blood perfusion terms are:

q = − k 2 π r L

dT dr

g = 2 π r dr L β (Tb − T )

(3) (4)

Combining these equations leads to: 2 π r dr L β (Tb − T ) =

d ⎡ dT ⎤ dr −k 2 π r L ⎢ dr ⎣ dr ⎥⎦

(5)

which can be simplified to: d ⎡ dT ⎤ β β r r T r Tb − = − dr ⎢⎣ dr ⎥⎦ k k

(6)

The solution is divided into a homogeneous and particular component: T = Th + Tp

(7)

β β d ⎡ d (Th + Tp ) ⎤ ⎢r ⎥ − r (Th + Tp ) = −r Tb dr ⎢ dr k k ⎥⎦ ⎣

(8)

β β β d ⎡ dTh ⎤ d ⎡ dT ⎤ − r Th + ⎢ r p ⎥ − r Tp = −r Tb r ⎢ ⎥ dr ⎣  dr  k dr ⎣ dr ⎦ k k ⎦   

(9)

which leads to:

or

= 0 for homogeneous ODE

whatever is left is particular ODE

The solution to the particular differential equation:

β β d ⎡ dTp ⎤ ⎢r ⎥ − r Tp = − r Tb dr ⎣ dr ⎦ k k

(10)

is Tp = Tb

(11)

β d ⎡ dTh ⎤ − r Th = 0 r ⎢ ⎥ dr ⎣ dr ⎦ k

(12)

The homogeneous differential equation is:

Equation (12) is a form of Bessel’s equation: d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠

(13)

where, by comparing Eqs. (12) and (13), p = 1, c = β / k , and s = 1. Referring to the flow chart presented in Section 1.8, the value of s-p+2 is equal to 2 and therefore the solution parameters n and a must be computed: n=

1−1 =0 1−1+ 2

(14)

a=

2 =1 1−1+ 2

(15)

The last term in Eq. (12) is negative; therefore the homogeneous solution is given by: n

(

Th = C1 x a BesselI n, c a x

1

a

)+C x 2

n

a

(

BesselK n, c a x

1

a

)

(16)

or, for this problem:

⎛ ⎛ β⎞ β⎞ Th = C1 BesselI ⎜⎜ 0, r ⎟⎟ + C2 BesselK ⎜⎜ 0, r ⎟ k ⎠ k ⎟⎠ ⎝ ⎝

(17)

Substituting Eqs. (11) and (17) into Eq. (7) leads to: ⎛ ⎛ β⎞ β⎞ T = C1 BesselI ⎜⎜ 0, r ⎟⎟ + C2 BesselK ⎜⎜ 0, r ⎟ + Tb k ⎠ k ⎟⎠ ⎝ ⎝

(18)

Note that Maple could be used to identify this solution as well; it is necessary to specify that the parameters β and k are positive so that Maple identifies the solution in terms of modified Bessel functions (as opposed to Bessel functions with complex arguments): > restart; > assume(beta>0); > assume(k>0); > ODE:=diff(r*diff(T(r),r),r)-r*beta*T(r)/k=-r*beta*T_b/k;

2 r β∼ T_b d ⎛d ⎞ r β∼ T( r ) ODE := ⎛⎜⎜ T( r ) ⎞⎟⎟ + r ⎜ 2 T( r ) ⎟ − =− ⎜ ⎟ k~ k~ ⎝ dr ⎠ ⎝ dr ⎠

> Ts:=dsolve(ODE);

⎛ Ts := T( r ) = BesselI⎜ 0, ⎜ ⎝

β∼ r ⎞ ⎛ ⎟ _C2 + BesselK⎜ 0, ⎟ ⎜ k~ ⎠ ⎝

β∼ r ⎞ ⎟ _C1 + T_b k~ ⎟⎠

The boundary conditions must be used to obtain the constants C1 and C2. As radius approaches infinity, the body temperature is recovered: Tr →∞ = Tb

(19)

⎛ ⎛ β⎞ β⎞ Tr →∞ = C1 BesselI ⎜⎜ 0, ∞ ⎟⎟ + C2 BesselK ⎜⎜ 0, ∞ ⎟ + Tb = Tb k ⎠ k ⎟⎠ ⎝ ⎝

(20)

⎛ ⎛ β⎞ β⎞ C1 BesselI ⎜⎜ 0, ∞ ⎟⎟ + C2 BesselK ⎜⎜ 0, ∞ ⎟=0 k ⎠ k ⎟⎠ ⎝ ⎝   

 

(21)

Substituting Eq. (18) into Eq. (19) leads to:

or

∞

0

The 0th order modified Bessel function of the 1st kind, BesselI(0,x), approaches ∞ as x → ∞ and 0th order modified Bessel function of the 2nd kind, BesselK(0,x), approaches 0 as x → ∞: > limit(BesselI(0,x),x=infinity); > limit(BesselK(0,x),x=infinity);

∞ 0

Therefore, C1 must be zero and Eq. (18) becomes: ⎛ β⎞ T = C2 BesselK ⎜⎜ 0, r ⎟ + Tb k ⎟⎠ ⎝

(22)

The heat flux into the probe (i.e., in the negative r-direction) at r = rp is specified, providing the additional boundary condition: k

dT dr

= q ′′p

(23)

r = rp

Substituting Eq. (22) into Eq. (23) leads to: k

⎤ ⎛ d ⎡ β⎞ = q ′′p ⎢C2 BesselK ⎜⎜ 0, r ⎟⎟ + Tb ⎥ dr ⎢⎣ k ⎠ ⎥⎦ r = rp ⎝

(24)

Using the rules for differentiating Bessel functions presented in Section 1.8.4 leads to: − k C2

⎛ β⎞ BesselK ⎜⎜1, rp ⎟⎟ = q ′′p k k ⎝ ⎠

β

(25)

which leads to: C2 = −

q ′′p ⎛ β⎞ β k BesselK ⎜1, rp ⎟ k ⎠ ⎝

(26)

Substituting Eq. (26) into Eq. (22) leads to: ⎛ BesselK ⎜ 0, r q ′′ ⎝ T = Tb − p ⎛ βk BesselK ⎜1, rp ⎝

β⎞

⎟ k ⎠ β⎞ ⎟ k ⎠

(27)

The solution can also be identified using Maple. Substitute C2 = 0 into the previously obtained solution: > Ts:=subs(_C2=0,Ts);

⎛ Ts := T( r ) = BesselK⎜⎜ 0, ⎝

β∼ r ⎞ ⎟ _C1 + T_b k~ ⎟⎠

Obtain an equation for the boundary condition associated with Eq. (23): > BC:=k*rhs(eval(diff(Ts,r),r=r_p))=qf_p;

β∼ r_p ⎞ ⎟ β∼ _C1 = qf_p k~ ⎟⎠

⎛ BC := − k~ BesselK⎜⎜ 1, ⎝

Substitute the solution to the boundary condition equation into the general solution: > subs(_C1=solve(BC,_C1),Ts);

T( r ) = −

β∼ r ⎞ ⎛ ⎟ qf_p BesselK⎜⎜ 0, k~ ⎟⎠ ⎝ + T_b β∼ r_p ⎞ ⎛ ⎟ β∼ k~ BesselK⎜⎜ 1, k~ ⎟⎠ ⎝

The solution is implemented in EES. The inputs are entered: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_p_mm=5 [mm] r_p=r_p_mm*convert(mm,m) q``_p=30000 [W/m^2] k=0.6 [W/m-K] beta=40000 [W/m^3-K] T_b=converttemp(C,K,37[C])

"probe radius, in mm" "probe radius" "probe heat flux" "tissue conductivity" "blood perfusion effect" "blood temperature"

Equation (27) is implemented in EES; the radius and temperature are converted to mm and ºC, respectively. T=T_b-q``_p*BesselK(0,r*sqrt(beta/k))/(sqrt(beta*k)*BesselK(1,r_p*sqrt(beta/k))) r_mm=r*convert(m,mm) T_C=converttemp(K,C,T)

Figure 3 illustrates the temperature as a function of radius. 50

Temperature (°C)

25 0 -25 -50 -75 -100 -125 5

10

15

20

25

30

Radius (mm)

Figure 3: Temperature in tissue as a function of radius.

"solution" "radius" "in C"

b.) The lethal temperature for cell death is Tlethal = -30ºC. Plot the radius of the cryolesion (i.e., the kill radius - all tissue inside of this radius is dead) as a function of the heat flux provided by the cryoprobe. The temperature is set to the lethal temperature. The variable r must be constrained to be positive in the Variable Information Window to avoid convergence errors. The kill radius as a function of heat flux is shown in Figure 4. 9

Kill radius (mm)

8 7 6 5 4 3 2 5.0x103

1.5x104

2.5x104

3.5x104

4.5x104

2

Probe flux (W/m )

Figure 4: Kill radius as a function of the cryoprobe heat flux.

Problem 1.8-7: Bracket A disk-shaped bracket connects a cylindrical heater to an outer shell, as shown in Figure P1.87(a). center line

shell rt rb

q

h , T∞

Tt

k th bracket r heater

h , T∞

Figure P1.8-7(a): Disk-shaped bracket.

The thickness of the bracket is th and it is made of material with conductivity k. The bracket extends radially from rb at the heater to rt at the outer shell. The temperature of the bracket location where it intersects the shell (at r = rt) is Tt. The heater provides a rate of heat transfer to the bracket, q , at r = rb. Both the upper and lower surfaces of the bracket are exposed to fluid at T∞ with average heat transfer coefficient h . a.) What calculation would you do in order to justify treating the bracket as an extended surface (i.e., justify the assumption that temperature is only a function of r); provide an expression in terms of the symbols in the problem statement. The appropriate Biot number is the ratio of the resistance to conduction across the thickness of the fin to the resistance to convection from the fin surface. The resistance to conduction across the thickness of the fin (i.e., in the x direction) is:

Rcond , x =

th 2 k π ( rt 2 − rb2 )

(1)

The resistance to convection from the fin surface is: Rconv =

1 h π ( rt 2 − rb2 )

(2)

The Biot number is therefore: Bi =

Rcond , x Rconv

h π ( rt − rb ) th h th = 1 2k 2 k π ( rt 2 − rb2 ) 2

=

2

(3)

For the remainder of the problem, assume that the bracket can be treated as an extended surface.

center line rt rb

q

h , T∞

Tt

h , T∞ r h →∞

T T∞

h →0 Tt rb

rt

r

Figure P1.8-7(b): Qualitative sketch of the temperature distribution expected if h → 0 and h → ∞.

b.) On the axes in Figure P1.8-7(b), sketch the temperature distribution that you would expect if the heat transfer coefficient h is very low ( h → 0). Note that the qualitative values of Tt and T∞ are indicated in the plot - your sketch should be consistent with these values. The sketch is shown in Figure P1.8-7(a) and has the following characteristics. 1. The slope at r = rb should be negative due to the heat transfer at the base. 2. The temperature at r = rt must be Tt. 3. The surfaces of the bracket are insulated if h → 0; therefore, the conduction heat transfer is constant in the r-direction. Because the area for conduction increases with r, the temperature gradient must decrease. c.) On the axes in Figure P1.8-7(c), sketch the temperature distribution that you would expect if the heat transfer coefficient h is very high ( h → ∞). Note that the qualitative values of Tt and T∞ are indicated in the plot - your sketch should be consistent with these values. The sketch is also shown in Figure P1.8-7(a) and has the following characteristics. 1. The slope at r = rb is negative due to the heat transfer at the base and is identical to the slope of the distribution from (b). 2. The temperature at r = rt must be Tt. 3. The temperature of the bracket otherwise tends to approach T∞ due to the strong coupling between the surface and the surrounding fluid. The temperature must approach T∞ from above at the base (because you are transferring heat from the bracket to the fluid) and from below at the tip (because you are transferring from the fluid to the bracket.

d.) What dimensionless number would you calculate in order to determine whether the actual temperature distribution is closer to your sketch from (b) or (c)? Provide an expression in terms of the symbols in the problem statement. The behavior of the bracket is governed by two resistances: the resistance to convection from the surface of the bracket, Eq. (2), and the resistance to conduction in the radial direction:

Rcond ,r

⎛r ⎞ ln ⎜ t ⎟ r = ⎝ b⎠ 2 π th k

(4)

The ratio of Rcond,r to Rconv governs the behavior:

Rcond ,r Rconv

⎛r ⎞ ln ⎜ t ⎟ 2 2 2 2 rb ⎠ h π ( rt − rb ) ⎛ rt ⎞ h ( rt − rb ) ⎝ = = ln ⎜ ⎟ 2 π th k 1 ⎝ rb ⎠ 2 th k

(5)

If Rcond,r/Rconv 1, then the distribution will approach your answer from (c). e.) Derive the governing ordinary differential equation for the bracket. An energy balance on a differential segment of the bracket is shown in Figure P1.8-7(c). center line rt q

rb

dr qr

qr + dr

Tt

qconv r Figure P1.8-7(c): Energy balance on a differential segment of the bracket.

The energy balance in Figure P1.8-7(c) leads to: qr = qr + dr + qconv

(6)

dq dr + qconv dr

(7)

or

0=

Substituting rate equations into Eq. (7) leads to: 0=

d ⎡ dT ⎤ dr + h 4 π r dr (T − T∞ ) −k 2 π r th ⎢ dr ⎣ dr ⎥⎦

(8)

or

d ⎡ dT ⎤ r − β 2 r T = − β 2 r T∞ ⎢ ⎥ dr ⎣ dr ⎦

(9)

where

β2 =

2h k th

(10)

f.) What are the boundary conditions for the ordinary differential equation from (e)? The boundary conditions are:

− k 2 π rb th

dT dr

= q

(11)

r = rb

and

Tr = rt = Tt

(12)

g.) Obtain a solution to your ordinary differential equation that includes two undetermined constants. The solution is split into a homogeneous and particular component:

T = Th + Tp

(13)

d ⎡ dTh ⎤ d ⎡ dTp ⎤ 2 β − + − β 2 r Tp = − β 2 r T∞ r r T r h ⎢ ⎥ ⎢ ⎥ dr ⎣ dr dr ⎣ dr ⎦ ⎦   

 

(14)

which leads to:

homogeneous ODE

particular ODE

The solution to the particular ordinary differential equation is:

Tp = T∞

(15)

Comparing the homogeneous ordinary differential equation to Bessel's equation and following the flow chart provided in Figure 1-54 of your notes leads to: Th = C1 BesselI ( 0,β r ) + C2 BesselK ( 0,β r )

(16)

Substituting Eqs. (15) and (16) into Eq. (13) leads to: T = C1 BesselI ( 0,β r ) + C2 BesselK ( 0,β r ) + T∞

(17)

h.) Write the two algebraic equation that could be solved to provide the undetermined constants (don't solve these equations). Substituting Eq. (17) into Eq. (12) leads to: C1 BesselI ( 0,β rt ) + C2 BesselK ( 0,β rt ) + T∞ = Tt

(18)

Substituting Eq. (17) into Eq. (11) leads to: − k 2 π rb th β ⎡⎣C1 BesselI (1,β rb ) − C2 BesselK (1,β rb ) ⎤⎦ = q

(19)

Problem 1.8-8 Figure P1.8-8 illustrates a triangular fin with a circular cross-section.

Figure P1.8-5: Fin on an evaporator.

The fin is surrounded by fluid at T∞ with heat transfer coefficient h . The base of the fin is at Tb and the fin conductivity is k. a.) Derive the governing differential equation and the boundary conditions for the problem. The x-coordinate is defined as starting from the tip of the fin and moving to the base. The crosssectional area is therefore: Ac = π

D2 x2 4 L2

(1)

x L

(2)

and the perimeter (assuming that L >> D) is:

per = π D A differential energy balance leads to:

q x = q x + dx + qconv

(3)

The rate of conduction and convection are:

q x = − k Ac

dT dx

qconv = h per (T − T∞ ) dx

(4) (5)

Substituting Eqs. (1), (2), (4), and (5) into Eq. (3) and expanding the x+dx term leads to: 0=

d ⎡ D 2 x 2 dT ⎤ x k π − ⎢ ⎥ + h π D (T − T∞ ) 2 dx ⎣ 4 L dx ⎦ L

(6)

Simplifying:

d ⎡ 2 dT ⎤ x − m 2 x (T − T∞ ) = 0 ⎢ ⎥ dx ⎣ dx ⎦

(7)

where m2 =

4h L kD

(8)

The boundary conditions are: Tx =0 must be bounded

(9)

Tx = L = Tb

(10)

b.) Normalize the governing differential equation and the boundary conditions. This process should lead to the identification of a dimensionless fin parameter that governs the solution. Identify the physical significance of this parameter. Dimensionless position and temperature difference are defined:

x L

(11)

T − T∞ Tb − T∞

(12)

x =

θ =

Substituting Eqs. (11) and (12) into Eq. (7) leads to:

(Tb − T∞ )

d ⎡ 2 dθ ⎤ x − m 2 L x (Tb − T∞ ) θ = 0 ⎢ ⎥ dx ⎣ dx ⎦

(13)

or d ⎡ 2 dθ ⎤ x − ( m 2 L ) x θ = 0 ⎢ ⎥ dx ⎣ dx ⎦

(14)

where

m2 L =

4 h L2 kD

(15)

is the dimensionless parameter identified by the process. The dimensionless parameter is nominally equal to the ratio of the resistance to conduction along the fin (Rcond) to the resistance to convection from the fin surface (Rconv):

Rcond ≈

8L π D2 k

(16)

Rconv ≈

2 π D Lh

(17)

Rcond 8 L π D L h 4 L2 h = = = m2 L 2 Rconv π D 2 k Dk

(18)

The normalized boundary conditions are:

θx =0 must be bounded

(19)

θx =1 = 1

(20)

c.) Solve the differential equation subject to the boundary conditions. The differential equation: d ⎡ 2 dθ ⎤ 2  ⎢ x  ⎥ − ( m L ) x θ = 0  dx ⎣ dx ⎦

(21)

is homogeneous and a form of the Bessel's equation: d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ = 0 dx ⎝ dx ⎠

(22)

where p = 2, c = m L , and s = 1. The parameter s - p + 2 is 1; therefore the solution parameters are: n=

1− 2 = −1 1− 2 + 2

(23)

2 =2 1− 2 + 2

(24)

a=

n 1− 2 1 = =− a 2 2

(25)

The solution is therefore:

θ = C1

(

BesselI −1, 2 m L x

1

2

) + C BesselK ( −1, 2 m 2

x

L x

1

2

)

(26)

x

The same solution can be identified in Maple: > restart; > assume(m2L,positive); > ODE:=diff(x^2*diff(q(x),x),x)-m2L*x*q(x)=0; 2 d ⎞ ⎛ ⎞ 2⎛ d ODE := 2 x ⎜⎜ q( x ) ⎟⎟ + x ⎜ 2 q( x ) ⎟ − m2L~ x q( x ) = 0 ⎜ ⎟ ⎝ dx ⎠ ⎝ dx ⎠

> qs:=dsolve(ODE);

qs := q( x ) =

_C1 BesselI( 1, 2 m2L~ x

x)

+

_C2 BesselK( 1, 2 m2L~ x

x)

Note that BesselI ( −1, x ) is equal to BesselI (1, x ) and so Maple has identified the same solution that we found manually. The boundary condition:

θx =0 must be bounded

(27)

requires that C2 in Eq. (26) must be zero. To see this, take the limit of the second term as x goes to zero. > limit(BesselK(-1,x)/x,x=0);

∞

Therefore:

θ = C1

(

BesselI −1, 2 m L x

x

1

2

)

(28)

The second boundary condition:

θx =1 = 1 leads to:

(29)

(

1 = C1 BesselI −1, 2 m L

)

(30)

Substituting Eq. (30) into Eq. (28) leads to:

θ =

(

BesselI −1, 2 m L x

(

BesselI −1, 2 m L

)

1

2

)

(31)

x

d.) Prepare a plot of dimensionless temperature as a function of dimensionless position for various values of the remaining dimensionless parameter, identified in (b). The solution is programmed in EES: "P1.8-8" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in msqrtL=10 [-] "fin parameter" theta_bar=BesselI(-1,2*msqrtL*sqrt(x_hat))/(BesselI(-1,2*msqrtL)*sqrt(x_hat)) "temperature solution"

and used to generate Figure 2. 1

Dimensionless temperature

0.9

0.1

0.2

0.5

0.8 1

0.7 0.6

2

0.5 0.4 0.3

5

0.2

1/2

mL

0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

= 10

0.8

0.9

1

Dimensionless position Figure 2: Dimensionless temperature as a function of dimensionless position for various values of the fin parameter.

e.) The fin efficiency is defined as the ratio of the heat transfer into the base of the fin to the heat transfer that would occur if the entire fin were isothermal and at the base temperature (i.e., if the fin material were infinitely conductive). Develop an equation for the fin efficiency and plot the fin efficiency as a function of the dimensionless fin parameter identified in (b). The fin heat transfer rate is:

q fin = k

π D 2 ⎛ dT ⎞

⎜ ⎟ 4 ⎝ dx ⎠ x = L

(32)

or q fin = k

π D 2 (Tb − T∞ ) ⎛ dθ ⎞ 4L

⎜ ⎟ ⎝ dx ⎠ x =1

(33)

The fin efficiency is therefore:

η fin =

2 q fin

π D L h (Tb − T∞ )

=k

2 π D 2 (Tb − T∞ ) ⎛ dθ ⎞ ⎜ ⎟ 4 π D L h (Tb − T∞ ) L ⎝ dx ⎠ x =1

(34)

which can be simplified to:

η fin =

 2 k D ⎛ dθ ⎞ 2 ⎛ dθ ⎞ = 2 m L ⎜ ⎟ ⎜ ⎟ 2 2 h L2 ⎝ dx ⎠ x =1 ⎝ dx ⎠ x =1

(35)

Substituting Eq. (31) into Eq. (35) leads to:

η fin

(

1 ⎛  2 d ⎜ BesselI −1, 2 m L x = ⎜ x BesselI −1, 2 m L m 2 L dx ⎜ ⎝

(

2

)

) ⎞⎟

⎟ ⎟ ⎠ x =1

(36)

The derivative in Eq. (36) is evaluated in Maple: > restart; > eval(diff(BesselI(-1,2*msqrtL*sqrt(x))/sqrt(x),x),x=1);

⎛ BesselI( 0, 2 msqrtL ) − 1 BesselI( 1, 2 msqrtL ) ⎞ msqrtL − 1 BesselI( 1, 2 msqrtL ) ⎜⎜ ⎟⎟ 2 msqrtL 2 ⎝ ⎠

> simplify(%);

BesselI( 0, 2 msqrtL ) msqrtL − BesselI( 1, 2 msqrtL )

Therefore, the fin efficiency is:

η fin

(

)

(

)

2 ⎡ BesselI 0, 2 m L m L − BesselI 1, 2 m L ⎤ ⎦ = ⎣ 2 BesselI −1, 2 m L m L

(

)

eta_fin=2*(BesselI(0,2*msqrtL)*msqrtL-BesselI(1,2*msqrtL))/(msqrtL^2*BesselI(-1,2*msqrtL)) "fin efficiency"

(37)

Figure 3 illustrates the fin efficiency as a function of the fin parameter m L . 1 0.9 0.8

Fin efficiency

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.01

0.1

1

10

100 1/2

Fin parameter, m L Fin efficiency as a function of the fin parameter.

1000

Problem 1.8-9 One side of a thin circular membrane is subjected to a flux of energy that varies according to: q ′′ = a r 2

(1)

One side of the membrane is exposed to fluid at T∞ with heat transfer coefficient h . The outer edge of the membrane is held at T∞. The radius of the membrane is ro and the thickness is th. The conductivity of the membrane material is k. Assume that the temperature distribution in the membrane is only a function of radius. a.) Derive the governing differential equation for the temperature in the membrane and the boundary conditions. A differential energy balance leads to: qr + q ′′ 2 π r dr = qr + dr + 2 π r dr h (T − T∞ )

(2)

Expanding the r+dr term leads to:

q ′′ 2 π r dr =

dq dr + 2 π r dr h (T − T∞ ) dr

(3)

dT dr

(4)

The rate of conduction heat transfer is:

q = − k 2 π r th Substituting Eqs. (1) and (4) into Eq. (3) leads to: a 2 π r 3 dr =

d ⎛ dT ⎜ −k 2 π r th dr ⎝ dr

⎞ ⎟ dr + 2 π r dr h (T − T∞ ) ⎠

(5)

⎞ ⎟ + r h (T − T∞ ) ⎠

(6)

a 3 ⎞ h r (T − T∞ ) = − r ⎟− k th ⎠ k th

(7)

which can be simplified to: a r3 =

d ⎛ dT ⎜ −k r th dr ⎝ dr

or d ⎛ dT ⎜r dr ⎝ dr The boundary conditions are:

Tr = ro = T∞

(8)

Tr =0 must be bounded

(9)

b.) Define a dimensionless temperature difference and radius. Use them to non-dimensionalize the governing differential equation and boundary conditions from (a). This process should lead to the identification of another dimensionless parameter. Explain the significance of this parameter. A dimensionless temperature is defined:

θ =

T − T∞ ΔT

(10)

where ΔT is a normalizing temperature difference. The normalizing temperature difference is defined based on the temperature difference that would be induced if the entire rate of energy transfer from the heat flux were transferred convectively from the membrane. The total rate of heat transfer from the flux is: ro

q = ∫ aN r 2 2 π r dr = 0

π a ro4

q ′′

2

(11)

The reference temperature difference is therefore: ΔT =

π a ro4 a ro2 q = = π ro2 h 2 π ro2 h 2 h

(12)

2 h (T − T∞ ) a ro2

(13)

r ro

(14)

which leads to:

θ = A dimensionless radius is defined:

r =

Substituting Eqs. (13) and (14) into Eq. (7) leads to: a ro d ⎛ dθ ⎞ h a ro3  a ro3 3   r r r θ − = − ⎜ ⎟ k th 2 h dr ⎝ dr ⎠ k th 2 h

(15)

or d ⎛ dθ ⎞ 2 2 3 ⎜ r ⎟ − m r θ = −2 m r dr ⎝ dr ⎠

(16)

where

m2 =

h ro2 k th

(17)

The parameter m2 is, approximately, the ratio of the resistance to conduction along the membrane in the radial direction to the resistance to convection from the membrane surface. The nondimensional boundary conditions are:

θr =1 = 0

(18)

θr =0 must be bounded

(19)

c.) Solve the normalized problem from (b). Prepare a plot of the dimensionless temperature as a function of the dimensionless radius for various values of the dimensionless parameter identified in (b). The solution is split into a homogeneous and nonhomogeneous component:

θ = θh + θp

(20)

Equation (20) is substituted into Eq. (16), leading to: d ⎛ dθh ⎞ d ⎛ dθp ⎞ 2 2 2 3    − + r m r θ ⎜ r ⎟ − m r θp = −2 m r ⎜ ⎟ h dr ⎝ dr ⎠ dr ⎜⎝ dr ⎟⎠  

 

= 0 for homogeneous differential equation

(21)

whatever is left is the particular differential equation

The solution to the particular differential equation is considered first. d ⎛ dθp ⎜ r dr ⎜⎝ dr

⎞ 2 2 3 ⎟⎟ − m r θp = −2 m r ⎠

(22)

Based on the form of the left side, a second order polynomial is assumed for the particular solution:

θp = a + b r + c r 2

(23)

Substituting Eq. (23) into Eq. (22) leads to: d ( r ( b + 2 c r ) ) − m2 r ( a + b r + c r 2 ) = −2 m2 r 3 dr

(24)

b + 4 c r − m 2 r ( a + b r + c r 2 ) = −2 m 2 r 3

(25)

or

Equating like coefficients of r leads to: b=0

(26)

4 c − m2 a = 0

(27)

−m2 b = 0

(28)

− m 2 c = −2 m 2

(29)

which leads to a= 8/m2, b = 0, and c = 2. Therefore, the particular solution is:

θp =

8 + 2 r 2 2 m

(30)

The solution to the homogeneous differential equation: d ⎛ dθh ⎜ r dr ⎝ dr

⎞ 2 ⎟ − m r θh = 0 ⎠

(31)

is obtained using the flow chart:

θh = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r )

(32)

The solution is:

θ = C1 BesselI ( 0, m r ) + C2 BesselK ( 0, m r ) + The same solution can be identified in Maple: > restart;

8 + 2 r 2 m2

(33)

> ODE:=diff(r*diff(q(r),r),r)-m^2*r*q(r)=-2*m^2*r^3; 2 d ⎛d ⎞ ODE := ⎛⎜⎜ q( r ) ⎞⎟⎟ + r ⎜⎜ 2 q( r ) ⎟⎟ − m 2 r q( r ) = −2 m 2 r3 ⎝ dr ⎠ ⎝ dr ⎠

> qs:=dsolve(ODE);

qs := q( r ) = BesselI( 0, m r ) _C2 + BesselK( 0, m r ) _C1 +

8 + 2 m 2 r2 m2

The boundary condition:

θr =0 must be bounded

(34)

is satisfied by evaluating the limits of the two Bessel functions: > limit(BesselI(0,r),r=0);

1 > limit(BesselK(0,r),r=0);

∞

which means that C2 in Eq. (33) must be zero:

θ = C1 BesselI ( 0, m r ) +

8 + 2 r 2 m2

(35)

The boundary condition:

θr =1 = 0

(36)

is enforced: C1 BesselI ( 0, m ) +

8 +2=0 m2

(37)

which leads to: ⎛ 8 ⎞ ⎜ 2 + 2⎟ m ⎠ C1 = − ⎝ BesselI ( 0, m ) Substituting Eq. (38) into Eq. (35) leads to:

(38)

⎛ 8 ⎞ ⎜ 2 + 2⎟ m ⎠ BesselI 0, m r + 8 + 2 r 2 θ = − ⎝ ( ) 2 BesselI ( 0, m ) m

(39)

Equation (39) is programmed in EES: "P1.8-9" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" m=1 [-] "dimensionless parameter - ratio of conduction to convection" theta_hat=-(8/m^2+2)*BesselI(0,m*r_hat)/BesselI(0,m)+8/m^2+2*r_hat^2 "solution"

and used to generate Figure 1, which shows the dimensionless temperature as a function of dimensionless radius.

Dimensionless temperature

1.5 m = 20

1.25 1

m = 10

0.75 m=5 0.5 m=2 0.25 0 0

m=1

0.1

0.2

0.3

0.4

0.5

m = 0.5 0.6

0.7

0.8

0.9

1

Dimensionless radius Figure 1: Dimensionless temperature as a function of dimensionless radius for various values of m.

Problem 1.9-1: A 4-Node Numerical Simulation of a Fuse A fuse is a long, thin piece of metal that will heat up when current is passed through it. If a large amount of current is passed through the fuse, then the material will melt and this protects the electrical components downstream of the fuse. Figure P1.9-1 illustrates a fuse that is composed of a piece of metal with a square cross-section (a x a) that has length L. The conductivity of the fuse material is k. The fuse surface experiences convection with air at temperature Ta with heat transfer coefficient h . Radiation from the surface can be neglected for this problem. The ohmic heating associated with the current passing through the fuse results in a uniform rate of volumetric thermal energy generation, g ′′′ . The two ends of the fuse (at x = 0 and x = L) are held at temperature Tb. You have been asked to generate a numerical model of the fuse. Figure P1.91 also shows a simple numerical model that includes only four nodes which are positioned uniformly along the length of the fuse. L x k , g ′′′

A node 1 node 2 A

Ta , h node 3 node 4

a

a section A-A Figure P1.9-1: Fuse and a four node numerical model of the fuse.

a.) How would you determine if the extended surface approximation was appropriate for this problem? The Biot number compares the resistance to conduction from the center to the edge of the fuse to the resistance to convection from the surface of the fuse; for this problem, the Biot number should be:

Bi =

ha 2k

(1)

Any formula that is within a factor of 2 of Eq. (1) is fine. For the remainder of this problem, assume that you can use the extended surface approximation. b.) Derive a system of algebraic equations that can be solved in order to predict the temperatures at each of the four nodes in Figure P1.9-1 (T[1], T[2], T[3], and T[4]). Your equations should include only those symbols defined in the problem statement. Do not solve these equations. An energy balance on a control volume around node 2 is shown in Figure 2.

Figure 2: Energy balance on a control volume around node 2

The energy balance suggested by Figure 2 is: g [ 2] + q LHS [ 2] + q RHS [ 2] + qconv [ 2] = 0

(2)

The rate equations for the terms in Eq. (2) are: g [ 2] =

L a2 g ′′′ 3

(3)

q LHS [ 2] =

3 k a2 (T [1] − T [ 2]) L

(4)

q RHS [ 2] =

3 k a2 (T [3] − T [ 2]) L

(5)

qconv [ 2] =

4a Lh (Ta − T [ 2]) 3

(6)

A similar process for node 3 leads to: g [3] + q LHS [3] + q RHS [3] + qconv [3] = 0

(7)

where: g [3] =

L a2 g ′′′ 3

(8)

q LHS [3] =

3 k a2 (T [ 2] − T [3]) L

(9)

q RHS [3] =

3 k a2 (T [ 4] − T [3]) L

(10)

qconv [3] =

4a Lh (Ta − T [3]) 3

(11)

The temperatures of the edge nodes are specified: T [1] = Tb

(12)

T [ 4] = Tb

(13)

Equations (2) through (13) can be solved (using, for example, EES) to provide the temperature at each node. c.) How would you determine the amount of heat transferred from the fuse to the wall at x = 0 using your solution from (b)? In order to determine the amount of heat transferred to the wall at x= 0, it is necessary to do an energy balance on the control volume that surrounds node 1. g [1] + q RHS [1] + qconv [1] = qwall

(14)

L a2 g [1] = g ′′′ 6

(15)

where

q RHS [1] =

3 k a2 (T [ 2] − T [1]) L

(16)

2a Lh (Ta − T [1]) 3

(17)

qconv [1] =

d.) Derive the differential equation and boundary conditions that you would need in order to solve this problem analytically. Show your steps clearly. A differential control volume is shown in Figure 3.

Figure 3: Differential control volume.

The energy balance suggested by Figure 3 is: q x + g = q x + dx + qconv

(18)

Expanding the x+dx term leads to:

q x + g = q x +

dq dx + qconv dx

(19)

or

g =

dq dx + qconv dx

(20)

The rate equations are:

g = a 2 g ′′′ dx

(21)

dT dx

(22)

q = − k a 2

qconv = 4 h a dx (T − Ta )

(23)

Substituting Eqs. (21) through (23) into Eq. (20) leads to:

a 2 g ′′′ dx =

d ⎡ dT ⎤ −k a 2 dx + 4 h a dx (T − Ta ) ⎢ dx ⎣ dx ⎥⎦

(24)

d 2T 4 h g ′′′ − − = − T T ( ) a dx 2 k a k

(25)

Tx =0 = Tb

(26)

Tx = L = Tb

(27)

or

The boundary conditions are:

Problem 1.9-2: Bracket (revisited) Reconsider the disk-shaped bracket that was discussed in Problem P1.8-7. You have decided to generate a numerical model of the bracket that has three nodes, positioned as shown in Figure P1.9-2. center line rt rb

node 3

node 2

node 1 r

Figure P1.9-2: A 3-node numerical model of the disk-shaped bracket.

a.) Derive a system of algebraic equations that can be solved in order to predict the temperatures at each of the three nodes in Figure P1.9-2 (T1, T2, T3). Your equations should include only those symbols defined in the problem statement as well as the radial locations of the three nodes (r1, r2, and r3). Do not solve these equations. The temperature at node 3 is specified:

T3 = Tt

(1)

An energy balance on node 2 is shown in Figure P1.9-2(b). center line qconv q LHS node 1

node 3 q RHS

node 2

r Figure P1.9-2(b): Energy balance on node 2.

and leads to:

q LHS + q RHS = qconv or

(2)

⎡⎛ r2 + r3 ⎞ 2 ⎛ r2 + r1 ⎞ 2 ⎤ T1 − T2 ) T3 − T2 ) ( ( + 2 π k th = 2π h − 2 π k th ⎛r ⎞ ln ⎜ 2 ⎟ ⎝ r1 ⎠

⎢⎜ ⎢⎣⎝

⎛r ⎞ ln ⎜ 3 ⎟ ⎝ r2 ⎠

2

⎜ ⎝

⎟ ⎠

2

⎟ ⎥ (T2 − T∞ ) ⎠ ⎥⎦

(3)

An energy balance on node 1 is shown in Figure P1.9-2(c). center line qconv q

node 3 qRHS

node 1

node 2

r Figure P1.9-2(c): Energy balance on node 1.

and leads to:

q + q RHS = qconv

(4)

or ⎡⎛ r1 + r2 ⎞ 2 2 ⎤ T2 − T1 ) ( q + 2 π k th = 2π h −r ⎛r ⎞ ln ⎜ 2 ⎟ ⎝ r1 ⎠

⎢⎜ ⎣⎢⎝

2

⎟ ⎠

1

⎥ (T1 − T∞ ) ⎦⎥

(5)

Problem 1.9-3 (1-19 in text): Fiber optic bundle A fiber optic bundle (FOB) is shown in Figure P1.9-3 and used to transmit the light for a building application. 2 h = 5 W/m -K T∞ = 20°C

rout = 2 cm

5 2 q ′′ = 1x10 W/m

x

fiber optic bundle

Figure P1.9-3: Fiber optic bundle used to transmit light.

The fiber optic bundle is composed of several, small diameter fibers that are each coated with a thin layer of polymer cladding and packed in approximately a hexagonal close-packed array. The porosity of the FOB is the ratio of the open area of the FOB face to its total area. The porosity of the FOB face is an important characteristic because any radiation that does not fall directly upon the fibers will not be transmitted and instead contributes to a thermal load on the FOB. The fibers are designed so that any radiation that strikes the face of a fiber is “trapped” by total internal reflection. However, radiation that strikes the interstitial areas between the fibers will instead be absorbed in the cladding very close to the FOB face. The volumetric generation of thermal energy associated with this radiation can be represented by: ⎛ x ⎞ φ q ′′ g ′′′ = exp ⎜ − ⎟ Lch ⎝ Lch ⎠ where q ′′ = 1x105 W/m2 is the energy flux incident on the face, φ = 0.05 is the porosity of the FOB, x is the distance from the face, and Lch = 0.025 m is the characteristic length for absorption of the energy. The outer radius of the FOB is rout = 2 cm. The face of the FOB as well as its outer surface are exposed to air at T∞ = 20°C with heat transfer coefficient h = 5 W/m2-K. The FOB is a composite structure and therefore conduction through the FOB is a complicated problem involving conduction through several different media. Section 2.9 discusses methods for computing the effective thermal conductivity for a composite. The effective thermal conductivity of the FOB in the radial direction is keff,r = 2.7 W/m-K. In order to control the temperature of the FOB near the face where the volumetric generation of thermal energy is largest, it has been suggested that high conductivity filler material be inserted in the interstitial regions between the fibers. The result of the filler material is that the effective conductivity of the FOB in the axial direction varies with position according to: ⎛ x ⎞ keff , x = keff , x ,∞ + Δkeff , x exp ⎜ − ⎟ ⎝ Lk ⎠ where keff,x,∞ = 2.0 W/m-K is the effective conductivity of the FOB in the x-direction without filler material, Δkeff,x = 28 W/m-K is the augmentation of the conductivity near the face, and Lk = 0.05 m is the characteristic length over which the effect of the filler material decays. The length of the FOB is effectively infinite. a.) Is it appropriate to use a 1-D model of the FOB?

The inputs are entered in EES and functions are defined to return the volumetric generation and effective conductivity in the x-direction: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function k_FOB(x) k_eff_x_infinity=2 [W/m-K] L_k=0.05 [m] Dk_eff_x=28 [W/m-K] k_FOB=k_eff_x_infinity+Dk_eff_x*exp(-x/L_k) end

"conductivity far from the face" "characteristic length of elevated conductivity" "conductivity elevation at the face due to filler material" "conductivity"

function gv_FOB(x) phi=0.05 [-] q``=1e5 [W/m^2] L_ch=0.025 [m] gv_FOB=phi*q``*exp(-x/L_ch)/L_ch end

"porosity" "incident heat flux" "characteristic length for absorption" "volumetric rate of thermal energy generation"

"Inputs" k_eff_r=2.7 [W/m-K] r_out=2 [cm]*convert(cm,m) h_bar=5 [W/m^2-K] T_infinity=converttemp(C,K,20[C])

"effective conductivity in the radial direction" "radius of FOB" "heat transfer coefficient" "ambient temperature"

A Biot number is defined according to:

Bi =

h rout keff , r

(1)

which leads to Bi = 0.037, justifying an extended surface model of the FOB. b.) Assume that your answer to (a) was yes. Develop a numerical model of the FOB. Nodes are positioned along the FOB. The FOB is infinitely long; however, the first L = 0.75 m of the bundle is simulated. Examination of the solution shows that this is sufficient to capture the end effects. L=0.75 [m] N=41 [-] Dx=L/(N-1) duplicate i=1,N x[i]=Dx*(i-1) end

"length of FOB to simulate" "number of nodes" "distance between adjacent nodes" "position of each node"

An energy balance on node 1 leads to: 2 π rout Δx 2 Δx h π r (T∞ − T1 ) + h 2 π rout keff , x , x =( x1 + x2 ) / 2 (T2 − T1 ) + g ′′′x = x1 π rout =0 (T∞ − T1 ) + 2 Δx 2 2 out

(2)

h_bar*pi*r_out^2*(T_infinity-T[1])+h_bar*2*pi*r_out*(Dx/2)*(T_infinity-T[1])+& pi*r_out^2*k_FOB((x[1]+x[2])/2)*(T[2]-T[1])/Dx+gv_FOB(x[1])*pi*r_out^2*Dx/2=0 "energy balance on node 1"

Energy balances on the internal nodes lead to: h 2 π rout Δx (T∞ − Ti ) + 2 π rout

Δx

2 π rout

Δx

keff , x , x =( xi + xi+1 ) / 2 (Ti +1 − Ti ) +

2 Δx = 0 keff , x , x =( xi + xi−1 ) / 2 (Ti −1 − Ti ) + g ′′′x = xi π rout

(3)

i = 2.. ( N − 1)

duplicate i=2,(N-1) h_bar*2*pi*r_out*Dx*(T_infinity-T[i])+pi*r_out^2*k_FOB((x[i]+x[i+1])/2)*(T[i+1]-T[i])/Dx+& pi*r_out^2*k_FOB((x[i]+x[i-1])/2)*(T[i-1]-T[i])/Dx+gv_FOB(x[i])*pi*r_out^2*Dx=0 "energy balance on internal nodes" end

The temperature of the last node is taken to be specified at the ambient temperature:

TN = T∞ T[N]=T_infinity

(4)

"node N temperature is specified"

The temperature is converted to Celsius: duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

"temperature in C"

Figure 2 illustrates the temperature distribution within the FOB.

140

Temperature (°C)

120

without filler material

100 80 with filler material

60 40 20 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Position (m) Figure 2: Temperature distribution within the FOB for the case where the filler material is filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0).

c.) Overlay on a single plot the temperature distribution within the FOB for the case where the filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0). Figure 2 shows the case where filler material is present and is removed. The reduction in the maximum temperature related to the addition of the filler material is evident in Figure 2.

Problem 1.9-4 (1-20 in text) An expensive power electronics module normally receives only a moderate current. However, under certain conditions it is possible that it might experience currents in excess of 100 amps. The module cannot survive such a high current and therefore, you have been asked to design a fuse that will protect the module by limiting the current that it can experience, as shown in Figure P1.9-4. L = 2.5 cm

ε = 0.9 Tend = 20°C

D = 0.9 mm

Tend = 20°C

T∞ = 5°C 2 h = 5 W/m -K

k = 150 W/m-K ρr = 1x10-7 ohm-m I = 100 amp Figure 1.9-4: A fuse that protects a power electronics module from high current.

The space available for the fuse allows a wire that is L = 2.5 cm long to be placed between the module and the surrounding structure. The surface of the fuse wire is exposed to air at T∞ = 20°C and the heat transfer coefficient between the surface of the fuse and the air is h = 5.0 W/m2-K. The fuse surface has an emissivity of ε = 0.90. The fuse is made of an aluminum alloy with conductivity k = 150 W/m-K. The electrical resistivity of the aluminum alloy is ρe = 1x10-7 ohm-m and the alloy melts at approximately 500°C. Assume that the properties of the alloy do not depend on temperature. The ends of the fuse (i.e., at x=0 and x=L) are maintained at Tend =20°C by contact with the surrounding structure and the module. The current passing through the fuse, I, results in a uniform volumetric generation within the fuse material. If the fuse operates properly, then it will melt (i.e., at some location within the fuse, the temperature will exceed 500°C) when the current reaches 100 amp. Your job will be to select the fuse diameter; to get your model started you may assume a diameter of D = 0.9 mm. Assume that the volumetric rate of thermal energy generation due to ohmic dissipation is uniform throughout the fuse volume. a.) Prepare a numerical model of the fuse that can predict the steady state temperature distribution within the fuse material. Plot the temperature as a function of position within the wire when the current is 100 amp and the diameter is 0.9 mm. The input parameters are entered in EES and the volumetric generation rate is computed: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" L=2.5 [cm]*convert(cm,m) d=0.9 [mm]*convert(mm,m) T_a=converttemp(C,K,20) T_end=converttemp(C,K,20) h=5.0 [W/m^2-K] e=0.90 [-]

"length" "diameter" "air temperature" "end temperature" "heat transfer coefficient" "emissivity"

k=150 [W/m-K] er=1e-7 [ohm-m] T_melt=converttemp(C,K,500) current=100 [amp]

"conductivity" "electrical resistivity" "melting temperature" "current"

"Volumetric generation" Ac=pi*d^2/4 Rst=er*L/Ac w_dot_ohmic=current^2*Rst g```_dot=w_dot_ohmic/(Ac*L)

"cross-sectional area" "resistance" "total dissipation" "volumetric rate of generation"

The appropriate Biot number for this case is:

Bi =

hd 2k

(1)

The Biot number is calculated according to: "Extended surface approximation" Bi=h*d/(2*k)

The Biot number calculated by EES is much less than 1.0 and therefore the extended surface approximation is justified. The development of the numerical model follows the same steps that were previously discussed in the context of numerical models of 1-D geometries. Nodes (i.e., locations where the temperature will be determined) are positioned uniformly along the length of the rod. The location of each node (xi) is: xi =

( i − 1) L ( N − 1)

i = 1..N

(2)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δx) is: Δx =

L ( N − 1)

This distribution is entered in EES: "Setup nodes" N=10 [-] duplicate i=1,N x[i]=(i-1)*L/(N-1) end Dx=L/(N-1)

"number of nodes" "position of nodes" "distance between nodes"

(3)

A control volume is defined around each node; the control surface bisects the distance between the nodes. The control volume shown in Fig. 2 is subject to conduction heat transfer at each edge ( qtop and qbottom ), convection ( qconv ), radiation ( qrad ), and generation ( g ). The energy balance is:

qtop + qbottom + qconv + qrad + g = 0

(4)

The conduction terms are approximated as: qtop =

kπ d2 (Ti −1 − Ti ) 4 Δx

qbottom =

kπ d2 (Ti +1 − Ti ) 4 Δx

(5)

(6)

The convection term is modeled according to: qconv = h π d Δx (Ta − Ti )

(7)

qrad = ε σ π d Δx (Ta4 − Ti 4 )

(8)

The radiation term is:

The generation term is: g = g ′′′ π

d2 Δx 4

(9)

Substituting Eqs. (5) through (9) into Eq. (4) leads to: kπ d2 kπ d2 (Ti −1 − Ti ) + (Ti +1 − Ti ) + ha π d Δx (Ta − Ti ) 4 Δx 4 Δx d2 +ε σ π d Δx (Ta4 − Ti 4 ) + g ′′′ π Δx = 0 for i = 2.. ( N − 1) 4

(10)

The nodes at the edges of the domain must be treated separately; the temperature at both edges of the fuse are specified: T1 = Tend

(11)

TN = Tend

(12)

Equations (10) through (12) are a system of N equations in an equal number of unknown temperatures which are entered in EES: "Numerical solution" T[1]=T_end T[N]=T_end duplicate i=2,(N-1) k*pi*d^2*(T[i-1]-T[i])/(4*dx)+k*pi*d^2*(T[i+1]-T[i])/(4*dx)+pi*d*dx*h*(T_aT[i])+pi*d*dx*e*sigma#*(T_a^4-T[i]^4)+g```_dot*pi*d^2*dx/4=0 end duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

Figure 2 illustrates the temperature distribution in the fuse for N = 100 nodes.

Figure 2: Temperature distribution in the fuse.

b.) Verify that your model has numerically converged by plotting the maximum temperature in the wire as a function of the number of nodes in your model. With any numerical simulation it is important to verify that a sufficient number of nodes have been used so that the numerical solution has converged. The key result of the solution is the maximum temperature in the wire, which can be obtained using the MAX command: T_max_C=max(T_C[1..N])

Figure 3 illustrates the maximum temperature as a function of the number of nodes and shows that the solution has converged for N greater than 100 nodes.

Figure 3: Maximum temperature as a function of the number of nodes.

c.) Prepare a plot of the maximum temperature in the wire as a function of the diameter of the wire for I=100 amp. Use your plot to select an appropriate fuse diameter. The number of nodes was set to 100 and the plot shown in Figure 4 was generated:

Figure 4: Maximum temperature as a function of diameter.

The maximum temperature reaches 500°C when the diameter is approximately 1.15 mm; this would provide a fuse that correctly limited the current.

Problem 1.9-5 An A triangular fin is shown in Figure P1.9-5.

Figure P1.9-5: Wedge fin

The fin infinitely long (in the z-direction) and can be treated as an extended surface. The thickness of the fin base is th = 1 cm and the length is L = 10 cm. The conductivity of the material is k = 24 W/m-K. The base temperature is Tb = 140°C and the ambient temperature is T∞ = 25°C. The heat transfer coefficient is h = 15 W/m2-K. The width of the fin, W, is much larger than its length. a.) Develop a numerical model of the fin. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" L=10 [cm]*convert(cm,m) th=1 [cm]*convert(cm,m) k=24 [W/m-K] T_infinity=converttemp(C,K,25 [C]) T_b=converttemp(C,K,140 [C]) h_bar=15 [W/m^2-K] W=1 [m]

"length of fin" "base thickness" "conductivity" "ambient temperature" "base temperature" "heat transfer coefficient" "per unit width of fin"

The nodes are positioned along the fin according to:

xi =

( i − 1) L ( N − 1)

for i = 1..N

(1)

where N is the number of nodes. The distance between adjacent nodes is: Δx =

L ( N − 1)

The cross-sectional area for conduction at each node is:

(2)

Ac ,i =

thW xi for i = 1..N L

"Let x=0 at the tip" N=11 [-] duplicate i=1,N x[i]=L*(i-1)/(N-1) Ac[i]=x[i]*th*W/L end Dx=L/(N-1)

(3)

"number of nodes" "position" "area" "distance between nodes"

The total surface area available for convection is: ⎛ th ⎞ As = 2 W L + ⎜ ⎟ ⎝2⎠

2

2

A_s=2*sqrt(L^2+(th/2)^2)*W

(4) "surface area"

The temperature at the base is fixed: TN = Tb

(5)

Energy balances on the internal nodes are: As

( Ac,i + Ac,i+1 ) T − T + k ( Ac,i + Ac,i−1 ) T − T = 0 for i = 2.. N − 1 (6) Δx h (T∞ − Ti ) + k ( i +1 i ) ( i −1 i ) ( ) L 2 Δx 2 Δx

An energy balance on the node at the tip is: As

( Ac,1 + Ac,2 ) T − T = 0 Δx h (T∞ − Ti ) + k ( 2 1) 2L 2 Δx

T[N]=T_b "base temperature" "internal node energy balances" duplicate i=2,(N-1) A_s*Dx/L*h_bar*(T_infinity-T[i])+k*(Ac[i]+Ac[i+1])*(T[i+1]-T[i])/(2*Dx)+k*(Ac[i]+Ac[i-1])*(T[i-1]T[i])/(2*Dx)=0 end A_s*Dx/L*h_bar*(T_infinity-T[1])/2+k*(Ac[1]+Ac[2])*(T[2]-T[1])/(2*Dx)=0

The solution is converted to Celsius. duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

(7)

b.) Plot the temperature distribution within the fin. Figure 2 illustrates the temperature as a function of position (recall that x is measured from the tip of the fin). 140 130

Temperature (°C)

120 110 100 90 80 70 60 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Position (m) Figure 2: Temperature distribution in the fin.

c.) Determine the fin efficiency. Compare your answer with the fin efficiency obtained from the EES function eta_fin_straight_triangular. The rate of heat transfer to the fin base is obtained by carrying out an energy balance on node N. q = As

( Ac, N + Ac, N −1 ) T − T Δx h (TN − T∞ ) + k ( N N −1 ) 2L 2 Δx

(8)

The maximum possible heat transfer is: qmax = As h (Tb − T∞ )

(9)

The fin efficiency is:

η=

q

(10)

qmax

q_dot=A_s*Dx*h_bar*(T[N]-T_infinity)/(2*L)+k*(Ac[N]+Ac[N-1])*(T[N]-T[N-1])/(2*Dx) "actual heat transfer rate" q_dot_max=A_s*h_bar*(T_b-T_infinity) "maximum possible heat transfer rate" eta=q_dot/q_dot_max "fin efficiency"

which leads to η = 0.6561. The EES function eta_fin_straight_triangular is also determined: eta_EES=eta_fin_straight_triangular(th,L,h_bar,k)

"fin efficiency from EES"

which leads to ηEES = 0.6556. d.) Plot the fin efficiency as a function of the number of nodes used in the solution. Figure 3 illustrates the predicted efficiency as a function of the number of nodes and suggests that you must use at least 10 nodes. 0.73 0.72 0.71

Efficiency

0.7 0.69 0.68 0.67 0.66 0.65 2

10

100

Number of nodes Figure 3: Predicted efficiency as a function of number of nodes.

500

Problem 1.9-6 Your company manufacturers heater wire. Heater wire is applied to surfaces that need to be heated and then current is passed through the wire in order to develop ohmic dissipation. A key issue with your product is failures that occur when a length of the wire becomes detached from the surface and therefore the wire is not well connected thermally to the surface. The wire in the detached region tends to get very hot and melt. The wire diameter is D = 0.4 mm and the current passing through the wire is current = 10 amp. the detached wire is exposed to surroundings at T∞ = 20ºC through convection and radiation. The average convection heat transfer coefficient is h = 30 W/m2-K and the emissivity of the wire surface is ε = 0.5. The length of the wire that is detached from the surface is L = 1 cm. The ends of the wire at x = 0 and x = L are held at Tend = 50ºC. The wire conductivity is k = 10 W/m-K and the electrical resistivity is ρe = 1x10-7 ohm-m. a.) Develop a numerical model of the wire and use the model to plot the temperature distribution within the wire. The inputs are entered in EES and, looking ahead to parts (d) and (e), functions are defined that return the conductivity and electrical resistivity: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function k(T) "Input T - temperature (K) Output k - conductivity (W/m-K)" k_o=10 [W/m-K] k=k_o end function rho_e(T) "Input T - temperature (K) Output rho_e - electrical resistivity (ohm-m)" rho_e_o=1e-7 [ohm-m] rho_e=rho_e_o end "Inputs" L_cm=1 [cm] L=L_cm*convert(cm,m) d=0.4 [mm]*convert(mm,m) current=10 [amp] h_bar=30 [W/m^2-K] e=0.5 [-] T_end=converttemp(C,K,50 [C]) T_infinity=converttemp(C,K,20[C])

The nodes are positioned along the wire according to:

"length of detached region, in cm" "length of detached region" "diameter of heater wire" "current" "heat transfer coefficient" "emissivity" "end temperatures" "surrounding temperatures"

xi =

( i − 1) L ( N − 1)

for i = 1..N

(1)

where N is the number of nodes. The distance between adjacent nodes is: Δx =

L ( N − 1)

(2)

The perimeter and cross-sectional area for conduction along the wire is: per = π D

(3)

D2 4

(4)

Ac = π N=21 [-] duplicate i=1,N x[i]=L*(i-1)/(N-1) end Dx=L/(N-1) per=pi*d A_c=pi*d^2/4

"number of nodes" "position of nodes" "distance between nodes" "perimeter of wire" "cross-sectional area of wire"

The temperatures of the nodes at the ends of the wire are set: T1 = Tend

(5)

TN = Tend

(6)

T[1]=T_end T[N]=T_end

Energy balances on the internal nodes lead to: k

T=

(Ti +Ti−1 ) 2

Ac

(Ti −1 − Ti ) + k Δx

T=

(Ti +Ti+1 ) 2 4 i

ε per Δx σ (T − T 4 ∞

Ac

(Ti +1 − Ti ) + ρ Δx

) + h per Δx (T

i = 2.. ( N − 1)

∞

e ,T =Ti

− Ti

Δx current 2 + Ac

)=0

(7)

duplicate i=2,(N-1) k((T[i]+T[i-1])/2)*A_c*(T[i-1]-T[i])/Dx+k((T[i]+T[i+1])/2)*A_c*(T[i+1]-T[i])/Dx& +rho_e(T[i])*Dx*current^2/A_c+e*per*Dx*sigma#*(T_infinity^4-T[i]^4)+h_bar*per*Dx*(T_infinity-T[i])=0 end duplicate i=1,N

T_C[i]=converttemp(K,C,T[i]) end

"temperature in C"

Figure 1 illustrates the temperature as a function of position in the wire. 600

Temperature (°C)

500 400 300 200 100 0 0

0.002

0.004

0.006

0.008

0.01

Position (m) Figure 1: Temperature as a function of position.

b.) Plot the maximum temperature in the wire as a function of the number of nodes in the numerical model. The maximum temperature (Tmax) is identified. T_max_C=max(T_C[1..N])

"maximum temperature in C"

Figure 2 illustrates the maximum temeprature as a function of number of nodes. 560

Maximum temperature (°C)

555 550 545 540 535 530 525 520 515 2

10

100

450

Number of nodes Figure 2: Maximum temperature as a function of the number of nodes.

c.) Plot the maximum temperature in the wire as a function of the length of the detached region. If the maximum temperature of the wire before failure is Tmax = 400ºC, then what is the maximum allowable length of detached wire? Figure 3 illustrates the maximum temperature in the wire as a function of the length of the detached region and shows that the detached region cannot exceed about 0.75 cm in length without resulting in failure. 900

-7

k = 10 W/m-K, ρe = 1x10 ohm-m

Maximum temperature (°C)

800 Material D

700 600

Material E

500 400 300 200 100 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Length of detached region (m) Figure 3: Maximum temperature as a function of the detached length for the nominal wire with k = 10 W/mK and ρe = 1x10-7 ohm-m as well as for materials D and E.

You are looking at methods to alleviate this problem and have identified an alternative material, material D, in which the electrical resistivity is ρe = 1x10-7 ohm-m but the conductivity increases with temperature according to: ⎡ W ⎤ ⎡ W ⎤ k = 10 ⎢ + 0.05 ⎢ T − 300 [ K ]) ⎥ 2 ⎥( ⎣mK ⎦ ⎣mK ⎦ d.) Modify your numerical model in order to simulate material D. Overlay on your plot from (c) the maximum temperature as a function of the length of the detached wire for material D. The function for the conductivity is modified: function k(T) "Input T - temperature (K) Output k - conductivity (W/m-K)" k_o=10 [W/m-K] alpha=0.05 [W/m-K^2] { k=k_o} k=k_o+alpha*(T-300 [K]) end

The maximum temperature as a function of detached length is shown in Figure 3. The maximum length of detached wire has increased to approximately 1.1 cm. You have identified another alternative material, material E, in which the thermal conductivity is k = 10 W/m-K but the electrical resistivity decreases with temperature according to: ⎡ ohm m ⎤ ρ = 1x10−7 [ ohm m ] − 1x10-10 ⎢ (T − 300 [ K ]) ⎣ K ⎥⎦ e.) Modify your numerical model in order to simulate material E. Overlay on your plot from (c) the maximum temperature as a function of the length of the detached wire for material E.

The function for the conductivity is restored and the function for electrical resistivity is modified: function k(T) "Input T - temperature (K) Output k - conductivity (W/m-K)" k_o=10 [W/m-K] alpha=0.05 [W/m-K^2] k=k_o { k=k_o+alpha*(T-300 [K])} end function rho_e(T) "Input T - temperature (K) Output rho_e - electrical resistivity (ohm-m)" rho_e_o=1e-7 [ohm-m] beta=1e-10 [ohm-m/K] {rho_e=rho_e_o} rho_e=rho_e_o-beta*(T-300 [K]) end

The maximum temperature as a function of detached length is shown in Figure 3. The maximum length of detached wire has increased to approximately 1.0 cm.

Problem 1.9-7: Flat Plate Solar Collector Figure P1.9-7 illustrates a flat plate solar collector. 2 q ′′s = 900 W/m

Tw = 50°C

ε=1

2 h = 5 W/m -K T∞ = 10°C

k = 75 W/m2-K th = 1.5 mm

L = 8 cm Figure 1.9-7: Flat plate solar collector.

The collector consists of a flat plate that is th = 1.5 mm thick with conductivity k = 75 W/m-K. The plate is insulated on its back side and experiences a solar flux of qs′′ = 900 W/m2 which is all absorbed. The surface is exposed to convection and radiation to the surroundings. The emissivity of the surface is ε = 1. The heat transfer coefficient is h = 5 W/m2-K and the surrounding temperature is T∞ = 10ºC. The temperature of the water is Tw = 50ºC. The centerto-center distance between adjacent tubes is 2 L where L = 8 cm. a.) Develop a numerical model that can predict the rate of energy transfer to the water per unit length of collector. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" th=1.5 [mm]*convert(mm,m) k=75 [W/m-K] e=1 [-] q``_s=900 [W/m^2] T_infinity=converttemp(C,K,10[C]) h_bar=5 [W/m^2-K] L=8 [cm]*convert(cm,m) T_w=converttemp(C,K,50 [C]) W=1 [m]

"thickness" "conductivity of collector plate" "emissivity of collector plate" "solar flux" "ambient temperature" "heat transfer coefficient" "half-width between tubes" "water temperature" "per unit length of collector"

Nodes are placed along the length of the collector. Only the region from 0 < x < L is considered due to the symmetry of the system. Therefore, the position of each node is:

xi = The distance between adjacent nodes is:

L ( i − 1) for i = 1..N ( N − 1)

(1)

Δx =

L ( N − 1)

N=21 [-] Dx=L/(N-1) duplicate i=1,N x[i]=L*(i-1)/(N-1) end

(2) "number of nodes"

"location of each node"

Energy balances on each of the internal nodes leads to: q LHS ,i + q RHS ,i + qs ,i + qconv ,i + qrad ,i = 0 for i = 2.. ( N − 1)

(3)

where

q LHS ,i =

k W th (Ti −1 − Ti ) Δx

(4)

q RHS ,i =

k W th (Ti +1 − Ti ) Δx

(5)

qs ,i = W Δx q s′′

(6)

qs ,i = W Δx h (T∞ − Ti )

(7)

qrad ,i = W Δx ε σ (T∞4 − Ti 4 )

(8)

"internal nodes" duplicate i=2,(N-1) q_dot_LHS[i]=k*W*th*(T[i-1]-T[i])/Dx "conduction from left hand side node" q_dot_RHS[i]=k*W*th*(T[i+1]-T[i])/Dx "conduction from right hand side node" q_dot_s[i]=W*Dx*q``_s "absorbed solar radiation" q_dot_conv[i]=W*Dx*h_bar*(T_infinity-T[i]) "convection" q_dot_rad[i]=W*Dx*e*sigma#*(T_infinity^4-T[i]^4) "radiation" q_dot_LHS[i]+q_dot_RHS[i]+q_dot_s[i]+q_dot_conv[i]+q_dot_rad[i]=0 "energy balance" end

The temperature of node 1 is assumed to be equal to the water temperature (neglecting any resistance to convection on the water-side):

T1 = Tw

(9)

q LHS , N + qs , N + qconv , N + qrad , N = 0

(10)

An energy balance on node N leads to:

where q LHS , N =

k W th (TN −1 − TN ) Δx W Δx q s′′ 2

(12)

W Δx h (T∞ − TN ) 2

(13)

W Δx ε σ (T∞4 − TN4 ) 2

(14)

qs , N = qs , N = qrad , N =

(11)

"node 1" T[1]=T_w "node N" q_dot_LHS[N]=k*W*th*(T[N-1]-T[N])/Dx q_dot_s[N]=W*Dx*q``_s/2 q_dot_conv[N]=W*Dx*h_bar*(T_infinity-T[N])/2 q_dot_rad[N]=W*Dx*e*sigma#*(T_infinity^4-T[N]^4)/2 q_dot_LHS[N]+q_dot_s[N]+q_dot_conv[N]+q_dot_rad[N]=0

"conduction from left hand side node" "absorbed solar radiation" "convection" "radiation" "energy balance"

The temperature distribution within the collector is shown in Figure 2: 333

Temperature (K)

331

329

327

325

323 0

0.02

0.04

0.06

0.08

0.1

Position (m)

Figure 2: Temperature as a function of position in the collector.

An energy balance on node 1 provides the rate of energy transfer to the water:

qwater = q RHS ,1 + qs ,1 + qconv ,1 + qrad ,1

(15)

where

q RHS ,1 =

k W th (T2 − T1 ) Δx W Δx q s′′ 2

(17)

W Δx h (T∞ − T1 ) 2

(18)

qs ,1 = qs ,1 = qrad ,1 =

(16)

W Δx ε σ (T∞4 − T14 ) 2

(19)

"node 1" q_dot_RHS[1]=k*W*th*(T[2]-T[1])/Dx "conduction from right hand side node" q_dot_s[1]=W*Dx*q``_s/2 "absorbed solar radiation" q_dot_conv[1]=W*Dx*h_bar*(T_infinity-T[1])/2 "convection" q_dot_rad[1]=W*Dx*e*sigma#*(T_infinity^4-T[1]^4)/2 "radiation" q_dot_water=q_dot_RHS[1]+q_dot_s[1]+q_dot_conv[1]+q_dot_rad[1] "energy balance"

which leads to qwater = 28.9 W. b.) Determine the efficiency of the collector; efficiency is defined as the ratio of the energy delivered to the water to the solar energy incident on the collector. The efficiency is calculated according to:

η= eta=q_dot_water/(L*W*q``_s)

qwater LW

(20)

"efficiency"

which leads to η = 0.402. c.) Plot the efficiency as a function of the number of nodes used in the solution. Figure 3 illustrates the efficiency as a function of the number of nodes and shows that at least 20-30 nodes are required for numerical convergence.

0.409 0.408 0.407

Efficiency (-)

0.406 0.405 0.404 0.403 0.402 0.401 2

10

100

225

Number of nodes

Figure 3: Efficiency as a function of the number of nodes.

d.) Plot the efficiency as a function of Tw - T∞. Explain your plot. Figure 4 illustrates the efficiency as a function of the water-to-ambient temperature difference. When the water temperature is low then the losses are low (but not zero, because the temperature of the copper plate is elevated by conduction). As the water temperature increases, the temperature of the plate increases and therefore the losses increase and efficiency drops. The drop in efficiency is dramatic for this type of unglazed collector and therefore the collector may be suitable for providing water heating for swimming pools (at low water temperature) but probably is not suitable for providing domestic hot water (at high water temperature). 1 0.9 0.8

Efficiency

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

10

20

30

40

50

60

70

80

Water to ambient temperature difference (K)

Figure 4: Solar collector efficiency as a function of the water-to-ambient temperature difference.

P1.1-2 (1-1 in text): Conductivity of a dilute gas Section 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at ideal gas conditions. Test the validity of this approximation by comparing the conductivity estimated using Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., low pressure argon) provided by the internal function in EES. Note that the molecular radius, σ, is provided in EES by the Lennard-Jones potential using the function sigma_LJ. a.) What is the value and units of the proportionality constant required to make Eq. (1-18) an equality? Equation (1-18) is repeated below: k∝

cv

σ

T MW

2

(1)

Equation (1) is written as an equality by including a constant of proportionality (Ck): k = Ck

cv

σ

2

T MW

(2)

MW T

(3)

Solving for Ck leads to:

Ck =

kσ 2 cv

which indicates that Ck has units m-kg1.5/s-kgmol05-K0.5. The inputs are entered in EES for Argon at relatively low pressure (0.1 MPa) and 300 K. "Problem 1.1-2" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in T=300 [K] F$='Argon' P_MPa=0.1 [MPa] P=P_MPa*convert(MPa, Pa)

"temperature" "fluid" "pressure, in MPa" "pressure"

The conductivity, specific heat capacity, Lennard-Jones potential, and molecular weight of Argon (k, cv, σ, and MW) are evaluated using EES' built-in funcions. Equation (3) is used to evaluate the proportionality constant. k=conductivity(F$,T=T,P=P) cv=cv(F$,T=T,P=P) MW=molarMass(F$) sigma=sigma_LJ(F$) C_k=k*sigma^2*sqrt(MW/T)/cv

"conductivity" "specific heat capacity at constant volume" "molecular weight" "Lennard-Jones potential" "constant of proportionality"

which leads to Ck = 2.619x10-24 m-kg1.5/s-kgmol0.5-K0.5. b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and 100 MPa on a semi-log plot with pressure on the log scale. At what pressure does the approximation given in Eq. (1-18) begin to fail at 300 K for argon? Figure 1 illustrates the constant of proportionality as a function of pressure for argon at 300 K. The approximation provided by Eq. (1-18) breaks down at approximately 1 MPa. 8x10-24

-K

5x10-24

Ck (m-kg

1.5

0.5

6x10-24

/s-kgmol

0.5

)

7x10-24

4x10-24 3x10-24 2x10-24 10-24 0x100 0.001

0.01

0.1

1

10

100

Pressure (MPa)

Figure 1: Constant of proportionality in Eq. (3) as a function of pressure for argon at 300 K.

Problem 1.2-3 (1-2 in text): Conduction through a Wall Figure P1.2-3 illustrates a plane wall made of a very thin (thw = 0.001 m) and conductive (k = 100 W/m-K) material that separates two fluids, A and fluid B. Fluid A is at TA = 100°C and the heat transfer coefficient between the fluid and the wall is hA = 10 W/m2-K while fluid B is at TB = 0°C with hB = 100 W/m2-K. thw = 0.001 m TA = 100°C

TB = 0°C 2 hB = 100 W/m -K

hA = 10 W/m -K 2

k = 100 W/m-K Figure P1.2-3: Plane wall separating two fluids

a.) Draw a resistance network that represents this situation and calculate the value of each resistor (assuming a unit area for the wall, A = 1 m2). Heat flowing from fluid A to fluid B must pass through a fluid A-to-wall convective resistance (Rconv,A), a resistance to conduction through the wall (Rcond), and a wall-to-fluid B convective resistance (Rconv,B). These resistors are in series. The network and values of the resistors are shown in Figure 2. 0.1

K W

0.0001

K W

0.01

K W

TA = 100°C

TB = 0°C Rconv , A =

1 hA A

Rcond =

tw kA

Rcond , B =

1 hB A

Figure 2: Thermal resistance network representing the wall.

b.) If you wanted to predict the heat transfer rate from fluid A to B very accurately, then which parameter (e.g., thw, k, etc.) would you try to understand/measure very carefully and which parameters are not very important? Justify your answer. The largest resistance in a series network will control the heat transfer. For the wall above, the largest resistance is Rconv,A. Therefore, I would focus on predicting this resistance accurately. This would suggest that hA is the most important parameter and the others do not matter much.

Problem 1.2-8 (1-3 in text): Frozen Gutters You have a problem with your house. Every spring at some point the snow immediately adjacent to your roof melts and runs along the roof line until it reaches the gutter. The water in the gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutter and causing water to run into your attic. The situation is shown in Figure P1.2-8. snow melts at this surface 2 Tout , hout = 15 W/m -K Ls = 2.5 inch

snow, ks = 0.08 W/m-K insulation, kins = 0.05 W/m-K 2 Tin = 22°C, hin = 10 W/m -K Lins = 3 inch plywood, L p = 0.5 inch, k p = 0.2 W/m-K

Figure P1.2-8: Roof of your house.

The air in the attic is at Tin = 22°C and the heat transfer coefficient between the inside air and the inner surface of the roof is hin = 10 W/m2-K. The roof is composed of a Lins = 3.0 inch thick piece of insulation with conductivity kins = 0.05 W/m-K that is sandwiched between two Lp = 0.5 inch thick pieces of plywood with conductivity kp = 0.2 W/m-K. There is an Ls = 2.5 inch thick layer of snow on the roof with conductivity ks = 0.08 W/m-K. The heat transfer coefficient between the outside air at temperature Tout and the surface of the snow is hout = 15 W/m2-K. Neglect radiation and contact resistances for part (a) of this problem. a.) What is the range of outdoor air temperatures where you should be concerned that your gutters will become blocked by ice? The input parameters are entered in EES and converted to base SI units (N, m, J, K) in order to eliminate any unit conversion errors; note that units should still be checked as you work the problem but that this is actually a check on the unit consistency of the equations. "P1.2-8: Frozen Gutters" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in T_in=converttemp(C,K,22) "temperature in your attic" L_ins=3 [inch]*convert(inch,m) "insulation thickness" L_p=0.5 [inch]*convert(inch,m) "plywood thickness" k_ins=0.05 [W/m-K] "insulation conductivity" k_p=0.2 [W/m-K] "plywood conductivity" k_s=0.08 [W/m-K] "snow conductivity" L_s=2.5 [inch]*convert(inch,m) "snow thickness" h_in=10 [W/m^2-K] "heat transfer coefficient between attic air and inner surface of roof" h_out=15 [W/m^2-K] "heat transfer coefficient between outside air and snow" A=1 [m^2] "per unit area"

The problem may be represented by the resistance network shown in Figure 2.

Figure 2: Resistance network representing the roof of your house.

The network includes resistances that correspond to convection with the inside and outside air:

Rconv ,out =

Rconv ,in =

1

(1)

hout A 1

(2)

hin A

where A is 1 m2 in order to accomplish the problem on a per unit area basis. There are also conduction resistances associated with the insulation, plywood and snow: Rins =

Lins kins A

Rp =

Lp

Rs = R_conv_out=1/(h_out*A) R_s=L_s/(k_s*A) R_p=L_p/(k_p*A) R_ins=L_ins/(k_ins*A) R_conv_in=1/(h_in*A)

(3)

(4)

kp A Ls ks A

(5) "outer convection resistance" "snow resistance" "plywood resistance" "insulation resistance" "inner convection resistance"

Which leads to Rconv,out = 0.07 K/W, Rs = 0.79 K/W, Rp = 0.06 K/W, Rins = 1.52 K/W and Rconv,in = 0.10 K/W. Therefore, the dominant effects for this problem are conduction through the insulation and the snow; the other effects (convection and the plywood conduction) are not terribly important since the largest resistances will dominate in a series network. If the snow at the surface of the room is melting then the temperature at the connection between Rs and Rp must be Ts = 0°C (see Figure 2). Therefore, the heat transferred through the roof ( q in Figure 2) must be:

q =

(Tin − Ts )

(6)

Rconv ,in + 2 R p + Rins

The temperature of the outside air must therefore be:

Tout = Ts − q ( Rs + Rconv ,out )

(7)

T_s=converttemp(C,K,0) "roof-to-snow interface temperature must be melting point of water" q_dot=(T_in-T_s)/(R_conv_in+2*R_p+R_ins) "heat transfer from the attic to the snow when melting point is reached" T_out=T_s-q_dot*(R_s+R_conv_out) "outside temperature required to reach melting point at roof surface" T_out_C=converttemp(K,C,T_out) "outside temperature in C"

which leads to Tout = -10.8°C. If the temperature is below this then the roof temperature will be below freezing and the snow will not melt. If the temperature is above 0°C then the water will not refreeze upon hitting the gutter. Therefore, the range of temperatures of concern are -10.8°C < Tout < 0°C. b.) Would your answer change much if you considered radiation from the outside surface of the snow to surroundings at Tout? Assume that the emissivity of snow is εs = 0.82. The modified resistance network that includes radiation is shown in Figure 3.

Figure 3: Resistance network representing the roof of your house and including radiation.

The additional resistance for radiation is in parallel with convection from the surface of the snow as heat is transferred from the surface by both mechanisms. The radiation resistance can be calculated approximately according to:

Rrad =

1

(8)

4T ε s σ A 3

where T is the average temperature of the surroundings and the snow surface. In order to get a quick idea of the magnitude of this resistance we can approximate T with its largest possible value (which will result in the largest possible amount of radiation); the maximum temperature of the snow is 0°C: e_s=0.82 [-]

"emissivity of snow"

R_rad=1/(4*T_s^3*e_s*sigma#*A)

"radiation resistance"

which leads to Rrad = 0.26 K/W. Notice that Rrad is much larger than Rconv,out; the smallest resistance in a parallel combination dominates and therefore the impact of radiation will be minimal. Furthermore, Rconv,out is not even a very important resistance in the original series circuit shown in Figure 2.

Problem P1.2-11 (1-4 in text) Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with kA = 1 W/m-K and B with kB = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A; the heating element provides q ′′ = 5000 W/m 2 of heat. The surface of the wall at x = 2L is exposed to fluid at Tf,in = 300 K with heat transfer coefficient hin = 100 W/m2-K. 2 q ′′ = 5000 W/m

insulated

material A kA = 1 W/m-K L = 1 cm

x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

Figure P1.2-11(a): Composite wall with a heater.

You may neglect radiation and contact resistance for parts (a) through (c) of this problem. a.) Draw a resistance network to represent this problem; clearly indicate what each resistance represents and calculate the value of each resistance. The input parameters are entered in EES: “P1.2-11: Heater" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q_flux=100 [W/m^2] L = 1.0 [cm]*convert(cm,m) k_A=1.0 [W/m-K] k_B=5.0 [W/m-K] T_f_in=300 [K] h_in=100 [W/m^2-K] A=1 [m^2]

"heat flux provided by the heater" "thickness of each layer" "conductivity of material A" "conductivity of material B" "fluid temperature at inside surface" "heat transfer on inside surface" "per unit area"

The resistance network that represents the problem shown in Figure 2 is:

Figure 2: Resistance network.

The resistances due to conduction through materials A and B are:

RA =

RB =

L

(1)

kA A

L

(2)

kB A

where A is the area of the wall, taken to be 1 m2 in order to carry out the analysis on a per unit area basis. The resistance due to convection is:

Rconv ,in =

1

(3)

hin A

"part (a)" R_A=L/(k_A*A) R_B=L/(k_B*A) R_conv_in=1/(h_in*A) "resistance to convection on inner surface"

"resistance to conduction through A" "resistance to conduction through B"

which leads to RA = 0.01 K/W, RB = 0.002 K/W, and Rconv,in = 0.01 K/W. b.) Use your resistance network from (a) to determine the temperature of the heating element. The resistance network for this problem is simple; the temperature drop across each resistor is equal to the product of the heat transferred through the resistor and its resistance. In this simple case, all of the heat provided by the heater must pass through materials A, B, and into the fluid by convection so these resistances are in series. The heater temperature (Thtr) is therefore:

Thtr = T f ,in + ( RA + RB + Rconv ,in ) q ′′ A

(4)

T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"

which leads to Thtr = 410 K. c.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (b). The temperatures at x = L and x = 2L can be computed according to:

Tx = L = T f ,in + ( RB + Rconv ,in ) q ′′ A

(5)

Tx = 2 L = T f ,in + Rconv ,in q ′′ A

(6)

T_L=T_f_in+(R_B+R_conv_in)*q_flux*A T_2L=T_f_in+R_conv_in*q_flux*A

"temperature at x=L" "temperature at x=2L"

which leads to Tx=L = 360 K and Tx=2L = 350 K. The temperature distribution is sketched on the axes in Figure 3.

Figure 3: Sketch of temperature distribution.

Notice that the temperature drop through the two larger resistances (RA and RB) are much larger than the temperature drop across the small resistance, RB. Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm. material C kC = 2 W/m-K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

insulated x

L = 1 cm

T f ,in = 300 K 2 hin = 100 W/m -K material B L = 1 cm k = 5 W/m-K B

Figure P1.2-11(b): Composite wall with Material C.

Neglect radiation and contact resistance for parts (d) through (f) of this problem. d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly indicate what each resistance represents and calculate the value of each resistance. There is an additional resistor corresponding to conduction through material C, RC, as shown below:

Notice that the boundary condition at the end of RC corresponds to the insulated wall; that is, no heat can be transferred through this resistance. The resistance to conduction through material C is:

RC = "part (b)" k_C=2.0 [W/m-K] R_C=L/(k_C*A)

L

(7)

kC A

"conductivity of material C" "resistance to conduction through C"

which leads to RC = 0.005 K/W. e.) Use your resistance network from (d) to determine the temperature of the heating element. Because there is no heat transferred through RC, all of the heat must still go through materials A and B and be convected from the inner surface of the wall. Therefore, the answer is not changed from part (b), Thtr = 410 K. f.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is consistent with your solution from (e). The answer is unchanged from part (c) except that there is material to the left of the heater. However, no heat is transferred through material C and therefore there is no temperature gradient in the material.

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a contact resistance between materials A and B, Rc′′ = 0.01 K-m 2 /W , and the surface of the wall at

x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient hout = 10 W/m2-K. material C kC = 2 W/m-K T f ,out = 400 K 2 hout = 10 W/m -K

2 q ′′ = 5000 W/m

material A kA = 1 W/m-K L = 1 cm

x

T f ,in = 300 K 2 hin = 100 W/m -K material B kB = 5 W/m-K

L = 1 cm L = 1 cm

Rc′′ = 0.01 K-m /W 2

Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.

Neglect radiation for parts (g) through (i) of this problem. g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly indicate what each resistance represents and calculate the value of each resistance. The additional resistances associated with contact resistance and convection to the fluid at the outer surface are indicated. Notice that the boundary condition has changed; heat provided by the heater has two paths ( qout and qin ) and so the problem is not as easy to solve.

The additional resistances are computed according to: Rconv ,out =

1

(8)

hout A

Rcontact =

Rc′′ A

"part (c)" R``_c=0.01 [K-m^2/W] h_out=10 [W/m^2-K] T_f_out=400 [K] R_contact=R``_c/A R_conv_out=1/(h_out*A) "convection resistance on outer surface"

(9)

"area specific contact resistance" "heat transfer coefficient" "fluid temperature on outside surface" "contact resistance"

which leads to Rcontact = 0.01 K/W and Rconv,out = 0.1 K/W. h.) Use your resistance network from (j) to determine the temperature of the heating element. It is necessary to carry out an energy balance on the heater: q ′′ A = qin + qout

(10)

The heat transfer rates can be related to Thtr according to: qin =

(T

htr

− T f ,in )

RA + Rcontact + RB + Rconv ,in

qout =

(T

htr

− T f ,out )

(11)

(12)

RC + Rconv ,out

These are 3 equations in 3 unknowns, Thtr, qout and qin , and therefore can be solved simultaneously in EES (note that the previous temperature calculations from part (b) must be commented out): {T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A T_L=T_f_in+(R_B+R_conv_in)*q_flux*A

"heater temperature" "temperature at x=L"

T_2L=T_f_in+R_conv_in*q_flux*A q_flux*A=q_dot_in+q_dot_out q_dot_in=(T_htr-T_f_in)/(R_A+R_contact+R_B+R_conv_in) q_dot_out=(T_htr-T_f_out)/(R_C+R_conv_out)

"temperature at x=2L"} "energy balance on the heater" "heat flow to inner fluid" "heat flow to outer fluid"

which leads to Thtr = 446 K. The other intermediate temperatures shown on the resistance diagram can be computed: Tx = L − = Thtr − RA qin

(13)

Tx = L + = Thtr − ( RA + Rcontact ) qin

(14)

Tx = 2 L = Thtr − ( RA + Rcontact + RB ) qin

(15)

Tx =− L = Thtr − RC qout

(16)

"intermediate temperatures" T_Lm=T_htr-R_A*q_dot_in T_Lp=T_htr-(R_A+R_contact)*q_dot_in T_2L=T_htr-(R_A+R_contact+R_B)*q_dot_in T_mL=T_htr-R_C*q_dot_out

which leads to Tx=L- = 400.4 K, Tx=L+ = 354.7 K, Tx=2L = 345.6 K, and Tx=-L = 443.8 K. i.) Sketch the temperature distribution on the axes provided below.

Problems 1.2-12 (1-5 in text): Floor Heater You have decided to install a strip heater under the linoleum in your bathroom in order to keep your feet warm on cold winter mornings. Figure P1.2-12 illustrates a cross-section of the bathroom floor. The bathroom is located on the first story of your house and is W = 2.5 m wide by L = 2.5 m long. The linoleum thickness is thL = 5 mm and has conductivity kL = 0.05 W/m-K. The strip heater under the linoleum is negligibly thin. Beneath the heater is a piece of plywood with thickness thp = 5 mm and conductivity kp = 0.4 W/m-K. The plywood is supported by ths = 6 cm thick studs that are Ws = 4 cm wide with thermal conductivity ks = 0.4 W/m-K. The centerto-center distance between studs is ps = 25.0 cm. Between each stud are pockets of air that can be considered to be stagnant with conductivity kair = 0.025 W/m-K. A sheet of drywall is nailed to the bottom of the studs. The thickness of the drywall is thd = 9.0 mm and the conductivity of drywall is kd = 0.1 W/m-K. The air above in the bathroom is at Tair,1 = 15°C while the air in the basement is at Tair,2 = 5°C. The heat transfer coefficient on both sides of the floor is h = 15 W/m2-K. You may neglect radiation and contact resistance for this problem. 2 Tair ,1 = 15°C, h = 15 W/m -K

strip heater thp = 5 mm

ps = 25 cm

linoleum, kL = 0.05 W/m-K plywood, kp = 0.4 W/m-K thL = 5 mm

ths = 6 cm Ws = 4 cm studs, ks = 0.4 W/m-K drywall, kd = 0.1 W/m-K air, ka = 0.025 W/m-K

thd = 9 mm 2 Tair ,2 = 5°C, h = 15 W/m -K

Figure P1.2-12: Bathroom floor with heater.

a.) Draw a thermal resistance network that can be used to represent this situation. Be sure to label the temperatures of the air above and below the floor (Tair,1 and Tair,2), the temperature at the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat input to the strip heater ( qh ) on your diagram. The resistance diagram corresponding to this problem is shown in Figure 2.

Figure 2: Resistance diagram representing the bathroom floor.

Starting at the left-hand side of Figure 2 (i.e., from the basement air), the resistances correspond to convection between the air in the basement and the drywall (Rconv), conduction through the drywall (Rd), conduction through the air (Rair) and studs (Rs) in parallel, conduction through the plywood (Rp), conduction through the linoleum (RL), and convection between the air in the bathroom and the linoleum (Rconv). b.) Compute the value of each of the resistances from part (a). The known values for the problem are entered in EES and converted to base SI units: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" W=2.5 [m] L=2.5 [m] T_air_1=converttemp(C,K,15) T_air_2=converttemp(C,K,5) h=15 [W/m^2-K] th_L=5.0 [mm]*convert(mm,m) k_L=0.05 [W/m-K] th_P=5.0 [mm]*convert(mm,m) k_P=0.4 [W/m-K] th_s=6.0 [cm]*convert(cm,m) W_s=4.0 [cm]*convert(cm,m) k_s=0.4 [W/m-K] p_s=25 [cm]*convert(cm,m) k_air=0.025 [W/m-K] th_d=9.0 [mm]*convert(mm,m) k_d=0.1 [W/m-K]

"width of bathroom" "length of bathroom" "air temperature in the bathroom" "air temperature in the basement" "heat transfer coefficient" "linoleum thickness" "linoleum thermal conductivity" "plywood thickness" "plywood thermal conductivity" "stud thickness" "stud width" "stud conductivity" "stud center-to-center distance" "air conductivity" "drywall thickness" "drywall conductivity"

The units for each of these variables is set in the Variable Information window (select Variable Information from the Options menu), Figure 3. The units of each of the additional variables that are added as you solve the problem should be set in the Variable Information window.

Figure 3: Variable Information window.

The area of the floor is:

A = LW

(1)

The convection resistance is computed according to:

Rconv =

1 hA

(2)

The conduction resistances of the linoleum, plywood, and drywall are computed: RL =

thL kL A

Rp =

thp

Rd = A=L*W R_conv=1/(h*A) R_L=th_L/(k_L*A) R_P=th_P/(k_P*A) R_d=th_d/(k_d*A)

(3)

(4)

kp A thd kd A

(5) "area for conduction through floor" "convection resistance" "linoleum resistance" "plywood resistance" "drywall resistance"

The conduction resistance of the studs is computed according to:

Rs =

ths ⎛W ⎞ ks A ⎜ s ⎟ ⎝ ps ⎠

(6)

Note that the area for conduction is the product of the area of the floor and the fraction of the floor occupied by the studs. The conduction resistance of the air is: ths

Rair = kair

(7)

⎛ p − Ws ⎞ A⎜ s ⎟ ⎝ ps ⎠

R_s=th_s/(k_s*A*W_s/p_s) R_air=th_s/(k_air*A*(p_s-W_s)/p_s)

"stud resistance" "air resistance"

These calculations lead to Rconv = 0.011 K/W, RL = 0.016 K/W, Rp = 0.002 K/W, Rd = 0.014 K/W, Rs = 0.15 K/W, and Rair = 0.46 K/W. c.) How much heat must be added by the heater to raise the temperature of the floor to a comfortable 20°C? If Ts in Figure 2 is 20°C then the heat transferred to the bathroom ( q1 ) is: q1 =

TL − Tair ,1

(8)

Rconv

T_L=converttemp(C,K,20) q_dot_1=(T_L-T_air_1)/R_conv

"linoleum temperature" "heat transferred to bathroom"

which leads to q1 = 469 W. The temperature of the heater is therefore: Th = TL + q1 RL

(9)

T_h=T_L+q_dot_1*R_L

"strip heater temperature"

which leads to a heater temperature, Th = 300.7 K. The heater must provide q2 (the heat transferred to the bathroom) as well as q1 (the heat transferred to the basement). q1 =

and the total heater power is:

Th − Tair ,2 −1

⎡1 1 ⎤ Rconv + Rd + ⎢ + ⎥ + Rp ⎣ Rs Rair ⎦

(10)

qh = q1 + q2

(11)

q_dot_2=(T_h-T_air_2)/(R_conv+R_d+(1/R_s+1/R_air)^(-1)+R_P) "heat lost to lower story" q_dot_h=q_dot_1+q_dot_2 "total heater power"

which leads to q2 = 161 W and qh = 630 W. d.) What physical quantities are most important to your analysis? What physical quantities are unimportant to your analysis? Figure 4 illustrates the values of the resistances on the resistance diagram.

Figure 4: Resistance values.

Examination of Figure 4 shows that Rp, Rd, and Rconv are unimportant relative to the amount of heat transferred to the basement; these resistances are small in a series combination. Therefore, you can expect that the conductivity of the drywall and plywood as well as their thickness are not very important. Furthermore, the resistance of the air is larger than the resistance of the studs; in a parallel combination, the larger resistance is not important. Therefore, the conductivity of air is likely not very important. The important quantities include the conductivity of the studs and their size as well as the thickness and conductivity of the linoleum and its thickness. The heat transfer coefficient is also important. e.) Discuss at least one technique that could be used to substantially reduce the amount of heater power required while still maintaining the floor at 20°C. Note that you have no control over Tair,1 or h. The heat transferred to the bathroom is given by Eq. (8); you cannot change h and therefore the value of Rconv is fixed. The only way for you can reduce the heater power is to reduce the amount of heat transferred to the basement. This can be done most effectively by increasing the resistance of the studs, perhaps by increasing their thickness or reducing their width.

Problem 1.2-15 (1-6 in text): The super ice-auger You are a fan of ice fishing but don't enjoy the process of augering out your fishing hole in the ice. Therefore, you want to build a device, the super ice-auger, that melts a hole in the ice. The device is shown in Figure P1.2-15.

2

h = 50 W/m -K T∞ = 5 ° C

ε = 0.9 insulation, kins = 2.2 W/m-K thp = 0.75 inch

heater, activated with V = 12 V and I = 150 A plate, kp = 10 W/m-K D = 10 inch thins = 0.5 inch thice = 5 inch

ρice = 920 kg/m3 Δifus = 333.6 kJ/kg

Figure P1.2-15: The super ice-auger.

A heater is attached to the back of a D = 10 inch plate and electrically activated by your truck battery, which is capable of providing V = 12 V and I = 150 A. The plate is thp = 0.75 inch thick and has conductivity kp = 10 W/m-K. The back of the heater is insulated; the thickness of the insulation is thins = 0.5 inch and the insulation has conductivity kins = 2.2 W/m-K. The surface of the insulation experiences convection with surrounding air at T∞ = 5°C and radiation with surroundings also at T∞ = 5°C. The emissivity of the surface of the insulation is ε = 0.9 and the heat transfer coefficient between the surface and the air is h = 50 W/m2-K. The super ice-auger is placed on the ice and activated, resulting in a heat transfer to the plate-ice interface that causes the ice to melt. Assume that the water under the ice is at Tice = 0°C so that no heat is conducted away from the plate-ice interface; all of the energy transferred to the plate-ice interface goes into melting the ice. The thickness of the ice is thice = 5 inch and the ice has density ρice = 920 kg/m3. The latent heat of fusion for the ice is Δifus = 333.6 kJ/kg. a.) Determine the heat transfer rate to the plate-ice interface. The inputs are entered in EES: "P1.2-15" $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 D=10 [inch]*convert(inch,m) th_ins=0.5 [inch]*convert(inch,m) k_ins=2.2 [W/m-K] th_p=0.75 [inch]*convert(inch,m) k_p=10 [W/m-K] e=0.9 [-] h_bar=50 [W/m^2-K] T_infinity=converttemp(C,K,5 [C]) V=12 [V] I=150 [A]

"diameter of ice fishing hole" "insulation thickness" "insulation conductivity" "plate thickness" "conductivity of plate" "emissivity" "air heat transfer coefficient" "ambient temperature" "battery voltage" "current"

T_ice=converttemp(C,K,0 [C]) th_ice=5 [inch]*convert(inch,m) rho_ice=920 [kg/m^3] DELTAi_fus=333.6 [kJ/kg]*convert(kJ/kg,J/kg)

"temperature of ice-water interface" "thickness of ice" "density of ice" "enthalpy of fusion of ice"

The power provided to the heater is the product of the voltage and current:

q = V I q_dot=V*I

(1) "power to melting plate"

A resistance network that can be used to represent this problem is shown in Figure P1.2-15-2. Rrad = 3.56 K/W q q2

q1

Tice = 0°C

T∞ = 5°C

Ttop

Rcond , p = 0.038 K/W Rcond ,ins = 0.114 K/W

Rconv = 0.395 K/W

The resistances include: Rcond,p = conduction through plate Rcond,ins = conduction through insulation Rrad = radiation resistance Rconv = convection resistance

Figure P1.2-15-2: The super ice-auger.

In order to compute the radiation resistance required to solve the problem, it is necessary to assume a value of Ttop, the temperature at the top of the insulation (this value will eventually be commented out in order to complete the problem): T_top=360 [K]

"guess for top surface temperature"

The cross-sectional area of the plate is computed:

Ac =

π D2

(2)

4

and the radiation resistance is computed according to:

Rrad =

1 ε Ac σ (T + T∞2 ) (Ttop + T∞ )

(3)

2 top

A_c=pi*D^2/4 R_rad=1/(e*A_c*sigma#*(T_top^2+T_infinity^2)*(T_top+T_infinity))

"area of hole" "radiation resistance"

Note that the equations should be entered, the units set, and the EES code solved line by line in order to debug the code in small segments. The convection resistance is computed according to:

Rconv =

1 Ac h

(4)

and the conduction resistances are computed according to:

Rcond ,ins =

Rcond , p =

thins Ac kins

(5)

thp

(6)

Ac k p

R_conv=1/(A_c*h_bar) R_cond_ins=th_ins/(k_ins*A_c) R_cond_p=th_p/(k_p*A_c)

"air convection resistance" "conduction resistance of insulation" "conduction resistance of plate"

The heat transfer from the heater to the ambient surroundings ( q1 in Figure P1.2-15-2) is:

q1 =

(Th − T∞ ) ⎛ 1 1 ⎞ Rcond ,ins + ⎜ + ⎟ ⎝ Rrad Rconv ⎠

−1

(7)

and the heat transfer to the ice is:

q2 =

(Th − Tice ) Rcond , p

(8)

where Th is the heater temperature. An energy balance on the heater leads to:

q = q1 + q2

(9)

Equations (7) through (9) are 3 equations in 3 unknowns ( q , q1 , and q2 ) and can be solved using EES: q_dot_1=(T_h-T_infinity)/(R_cond_ins+(1/R_rad+1/R_conv)^(-1)) "heat transfer to ambient" q_dot_2=(T_h-T_ice)/R_cond_p "heat transfer to ice" q_dot=q_dot_1+q_dot_2 "energy balance"

The temperature at the top of the plate can be computed based on the solution. Update the guess values for the problem (select Update Guess Values from the Calculate menu) and comment out the guessed value for Ttop: {T_top=360 [K]}

"guess for top surface temperature"

and calculate Ttop according to the resistance network:

Ttop = Th − q1 Rcond ,ins T_top=T_h-q_dot_1*R_cond_ins

(10) "recalculate top temperature"

The result is q2 = 1676 W. The values of the resistances are shown in Figure P1.2-15-2; notice that radiation does not play an important role in the problem because it is a large resistance in parallel with a much smaller one. The resistance to conduction through the plate is also unimportant since it is so small. The resistance to conduction through the insulation and convection are dominant. b.) How long will it take to melt a hole in the ice? An energy balance on the ice-to-plate interface leads to:

q2 = Ac Δi fus ρice

dthice dt

(11)

dthice is the rate at which the thickness of the ice is reduced. Because there is no energy dt lost to the water, the rate of ice melting is constant with ice thickness. Therefore the time required to melt the ice is estimated according to:

where

time

dthice = thice dt

q_dot_2=A_c*DELTAi_fus*dth_icedt*rho_ice dth_icedt*time=th_ice time_min=time*convert(s,min)

(12) "energy balance on ice interface" "time to melt ice" "in min"

which leads to time = 1178 s (19.6 min). c.) What is the efficiency of the melting process? The efficiency is defined as the ratio of the energy provided to the plate-to-ice interface to the energy provided to the heater:

η=

q2 q

eta=q_dot_2/q_dot

(13) "efficiency of process"

which leads to η = 0.93. d.) If your battery is rated at 100 amp-hr at 12 V then what fraction of the battery's charge is depleted by running the super ice-auger? The total amount of energy required to melt a hole in the ice is:

Q = q time

(14)

The energy stored in the battery (Ebattery) is the product of the voltage and the amp-hr rating. The fraction of the battery charge required is: f = Q=q_dot*time E_battery=100 [amp-hr]*V*convert(A-V-hr,J) f=Q/E_battery

which leads to f = 0.491.

Q Ebattery "total energy required" "car battery energy" "fraction of car battery energy used"

(15)

Problem 1.3-3 (1-7 in text): Critical Evaluation of a Solution One of the engineers that you supervise has been asked to simulate the heat transfer problem shown in Figure P1.3-3(a). This is a 1-D, plane wall problem (i.e., the temperature varies only in the x-direction and the area for conduction is constant with x). Material A (from 0 < x < L) has conductivity kA and experiences a uniform rate of volumetric thermal energy generation, g ′′′ . The left side of material A (at x = 0) is completely insulated. Material B (from L < x < 2L) has lower conductivity, kB < kA. The right side of material B (at x= 2L) experiences convection with fluid at room temperature (20°C). Based on the facts above, critically examine the solution that has been provided to you by the engineer and is shown in Figure P1.3-3(b). There should be a few characteristics of the solution that do not agree with your knowledge of heat transfer; list as many of these characteristics as you can identify and provide a clear reason why you think the engineer’s solution must be wrong. 250 200

L

L

material A

material B

kA

kB < kA

g ′′′A = g ′′′

g B′′′ = 0

x

h , T f = 20°C

Temperature (°C)

150 100 50 0 -50 Material A -100

0

Material B L

2L

Position (m)

(a) (b) Figure P1.3-3: (a) Heat transfer problem and (b) "solution" provided by the engineer.

1. The left side of material A is insulated; therefore, the temperature gradient should be zero. 2. Material A has a higher conductivity than material B; therefore, at x = L the temperature gradient should be larger in material B than in material A. 3. Heat is transferred to the fluid at 20°C; therefore the temperature at x = 2 L must be greater than 20°C.

PROBLEM 1.3-8 (1-8 in text): Hay Temperature Freshly cut hay is not really dead; chemical reactions continue in the plant cells and therefore a small amount of heat is released within the hay bale. This is an example of the conversion of chemical to thermal energy and can be thought of as thermal energy generation. The amount of thermal energy generation within a hay bale depends on the moisture content of the hay when it is baled. Baled hay can become a fire hazard if the rate of volumetric generation is sufficiently high and the hay bale sufficiently large so that the interior temperature of the bale reaches 170°F, the temperature at which self-ignition can occur. Here, we will model a round hay bale that is wrapped in plastic to protect it from the rain. You may assume that the bale is at steady state and is sufficiently long that it can be treated as a one-dimensional, radial conduction problem. The radius of the hay bale is Rbale = 5 ft and the bale is wrapped in plastic that is tp = 0.045 inch thick with conductivity kp = 0.15 W/m-K. The bale is surrounded by air at T∞ = 20°C with h = 10 W/m2-K. You may neglect radiation. The conductivity of the hay is k = 0.04 W/m-K. a.) If the volumetric rate of thermal energy generation is constant and equal to g ′′′ = 2 W/m3 then determine the maximum temperature in the hay bale. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in R_bale=5 [ft]*convert(ft,m) t_p=0.045 [inch]*convert(inch,m) k_p= 0.15 [W/m-K] h=10 [W/m^2-K] T_infinity=converttemp(C,K,20 [C]) L=1 [m] k = 0.04 [W/m-K] g_dot_v=2 [W/m^3]

"hay bale radius" "plastic thickness" "plastic conductivity" "heat transfer coefficient" "ambient temperature" "per unit length of bale" "conductivity of hay" "volumetric heat generation"

This is an example of a one-dimensional steady conduction problem with constant volumetric generation and therefore the formulae provided in Table 1-3 can used directly. The general solution is:

T =−

g ′′′ r 2 + C1 ln ( r ) + C2 4k

(1)

where C1 and C2 are constants selected to enforce the boundary conditions. The boundary condition at the center of the bale is either that the temperature remain bounded or that the temperature gradient be zero; either will lead to C1 = 0. An energy balance at the outer edge of the hay bale leads to: − k 2 π Rbale L

dT dr

= r = Rbale

Tr = Rbale − T∞ R p + Rconv

(2)

where Rp and Rconv are the thermal resistances associated with conduction through the plastic and convection from the outer surface of the bale, respectively:

Rp =

tp

(3)

k p 2 π Rbale L

Rconv =

1 h 2 π Rbale L

(4)

where L = 1 m for a problem that is done on a unit length basis. The temperature gradient and temperature at the outer radius of the bale are obtained using Eq. (1) with C1 = 0: dT dr

=− r = Rbale

Tr = Rbale = −

g ′′′ Rbale 2k

(5)

2 g ′′′ Rbale + C2 4k

(6)

Placing equations (2) through (6) into EES allows the constants of integration to be determined: R_p=t_p/(k_p*2*pi*R_bale*L) "thermal resistance associated with conduction through plastic" R_conv=1/(2*pi*R_bale*L*h) "thermal resistance associated with convection" dTdr_Rbale=-g_dot_v*R_bale/(2*k) "temperature gradient at outer edge" T_Rbale=-g_dot_v*R_bale^2/(4*k)+C_2 "temperature at outer edge" -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv) "interface energy balance"

The maximum temperature in the bale occurs at the center; according to Eq. (1) with C1 = 0, this temperature is given by: g ′′′ r 2 T =− + C1 ln ( r ) + C2 4k T_max=C_2 T_max_F=converttemp(K,F,T_max)

(7)

"maximum bale temperature" "maximum bale temperature in F"

The maximum temperature in the hay bale is 322.3 K or 120.6°F. b.) Prepare a plot showing the maximum temperature in the hay bale as a function of the hay bale radius. How large can the hay bale be before there is a problem with self-ignition? A parametric table is generated that contains the variables T_max_F and R_bale and used to generate Figure 1.

Figure 1: Maximum temperature as a function of the bale radius.

Note that a hay bale larger than approximately 2.1 m will result in a hay fire. Prepare a model that can consider temperature-dependent volumetric generation. Increasing temperature tends to increase the rate of chemical reaction and therefore increases the rate of generation of thermal energy according to: g ′′′ = a + bT where a = -1 W/m3 and b = 0.01 W/m3-K and T is in K. c.) Enter the governing equation into Maple and obtain the general solution (i.e., a solution that includes two constants). The governing differential equation is obtained as discussed in Section 1.3: g ′′′ r =

d ⎛ dT ⎞ ⎜ −k r ⎟ dr ⎝ dr ⎠

This ordinary differential equation is entered in Maple: > restart; > ODE:=(a+b*T(r))*r=diff(-k*r*diff(T(r),r),r);

2 d ⎞ ⎛d ⎞ ⎛ ⎜ T ( r ) k r ODE := ( a + b T( r ) ) r = −k ⎜⎜ T( r ) ⎟⎟ ⎟⎟ − 2 ⎜ d r ⎠ ⎝ ⎝ dr ⎠

and solved: > Ts:=dsolve(ODE);

(8)

⎛ Ts := T( r ) = BesselJ⎜⎜ 0, ⎝

b k

⎞ ⎛ r ⎟⎟ _C2 + BesselY⎜⎜ 0, ⎠ ⎝

b k

a ⎞ r ⎟⎟ _C1 − b ⎠

Note that the solution is given in the form of Bessel functions; ⎛ ⎛ b ⎞ b ⎞ a T = −C2 BesselJ ⎜⎜ 0, r ⎟⎟ + C1 BesselY ⎜⎜ 0, r ⎟⎟ − k k ⎝ ⎠ ⎝ ⎠ b

(9)

Even though we have not yet learned about Bessel functions, we can manipulate this solution within Maple. d.) Use the boundary conditions to obtain values for the two constants in your general solution (hint: one of the two constants must be zero in order to keep the temperature at the center of the hay bale finite). You should obtain a symbolic expression for the boundary condition in Maple that can be evaluated in EES. In part (a) we could not take the natural logarithm of 0 in Eq. (7) and therefore C1 was zero. A similar thing happens with the Bessel functions. We can evaluate the limits of the two Bessel functions as r → 0: > limit(BesselJ(0,r),r=0);

1

> limit(BesselY(0,r),r=0);

−∞

The BesselY function becomes infinite and therefore C1 in Eq. (9) must be 0. > Ts:=subs(_C1=0,Ts);

⎛ Ts := T( r ) = BesselJ⎜⎜ 0, ⎝

b k

a ⎞ r ⎟⎟ _C2 − b ⎠

The boundary condition at the outer surface of the hay does not change; the temperature and temperature gradient at Rbale can be evaluated symbolically using Maple: > dTdr_Rbale:=eval(diff(rhs(Ts),r),r=R_bale);

⎛ dTdr_Rbale := −BesselJ⎜⎜ 1, ⎝

b ⎞ R_bale ⎟⎟ k ⎠

b _C2 k

> T_Rbale:=eval(rhs(Ts),r=R_bale);

⎛ T_Rbale := BesselJ⎜⎜ 0, ⎝

b a ⎞ R_bale ⎟⎟ _C2 − k b ⎠

These symbolic expressions are cut and paste into EES and used to replace Eqs. (5) and (6) and provide a new constant C2: {g_dot_v=2 [W/m^3] a=-1 [W/m^3] "coefficients for volumetric generation function" b=0.01 [W/m^3-K]

"volumetric heat generation"}

R_p=t_p/(k_p*2*pi*R_bale*L) "thermal resistance associated with conduction through plastic" R_conv=1/(2*pi*R_bale*L*h) "thermal resistance associated with convection" {dTdr_Rbale=-g_dot_v*R_bale/(2*k) "temperature gradient at outer edge" T_Rbale=-g_dot_v*R_bale^2/(4*k)+C_2 "temperature at outer edge"} dTdr_Rbale = -BesselJ(1,(b/k)^(1/2)*R_bale)*(b/k)^(1/2)*C_2 "symbolic expressions cut and paste from Maple" T_Rbale = BesselJ(0,(b/k)^(1/2)*R_bale)*C_2-1/b*a -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv) "interface energy balance"

The maximum temperature is the temperature at the center of the bale; this is evaluated using Maple: > T_max=eval(rhs(Ts),r=0);

T_max = _C2 −

a b

and copied and pasted into EES: {T_max=C_2 T_max = C_2-1/b*a "symbolic expression cut and paste from Maple" T_max_F=converttemp(K,F,T_max)

"maximum bale temperature"}

"maximum bale temperature in F"

e.) Overlay on your plot from part (b) a plot of the maximum temperature in the hay bale as a function of bale radius when the volumetric generation is a function of temperature. The result is shown in Figure 1.

Problem 1.3-9 (1-9 in text): Mass Flow Meter Figure P1.3-9 illustrates a simple mass flow meter for use in an industrial refinery. T∞ = 20°C 2 hout = 20 W/m -K rout = 1 inch rin = 0.75 inch

insulation kins = 1.5 W/m-K test section 7 3 g ′′′ = 1x10 W/m k = 10 W/m-K m = 0.75kg/s T f = 18°C

L = 3 inch

thins = 0.25

Figure P1.3-9: A simple mass flow meter.

A flow of liquid passes through a test section consisting of an L = 3 inch section of pipe with inner and outer radii, rin = 0.75 inch and rout = 1.0 inch, respectively. The test section is uniformly heated by electrical dissipation at a rate g ′′′ = 1x107 W/m3 and has conductivity k = 10 W/m-K. The pipe is surrounded with insulation that is thins = 0.25 inch thick and has conductivity kins = 1.5 W/m-K. The external surface of the insulation experiences convection with air at T∞ = 20°C. The heat transfer coefficient on the external surface is hout = 20 W/m2-K. A thermocouple is embedded at the center of the pipe wall. By measuring the temperature of the thermocouple, it is possible to infer the mass flow rate of fluid because the heat transfer coefficient on the inner surface of the pipe ( hin ) is strongly related to mass flow rate ( m ). Testing has shown that the heat transfer coefficient and mass flow rate are related according to: ⎛ m ⎞ hin = C ⎜⎜ ⎟⎟ ⎝ 1[ kg/s ] ⎠

0.8

where C= 2500 W/m2-K. Under nominal conditions, the mass flow rate through the meter is m = 0.75 kg/s and the fluid temperature is Tf = 18°C. Assume that the ends of the test section are insulated so that the problem is 1-D. Neglect radiation and assume that the problem is steadystate. a.) Develop an analytical model in EES that can predict the temperature distribution in the test section. Plot the temperature as a function of radial position for the nominal conditions. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_out=1.0 [inch]*convert(inch,m) r_in=0.75 [inch]*convert(inch,m) h_bar_out=20 [W/m^2-K]

"outer radius of measurement section" "inner radius of measurement section" "external convection coefficient"

T_infinity=converttemp(C,K,20 [C]) T_f=converttemp(C,K, 18 [C]) k=10 [W/m-K] g```=1e7 [W/m^3] m_dot=0.75 [kg/s] th_ins=0.25 [inch]*convert(inch,m) k_ins=1.5 [W/m-K] L= 3 [inch]*convert(inch,m)

"ambient temperature" "fluid temperature" "conductivity" "volumetric rate of thermal energy generation" "mass flow rate" "thickness of insulation" "insulation conductivity" "length of test section"

The heat transfer coefficient on the internal surface is computed according to the specified mass flow rate: C=2500 [W/m^2-K] h_bar_in=C*(m_dot/1 [kg/s])^0.8

"constant for convection relationship" "internal convection coefficient"

The general solution to a 1-D problem in cylindrical coordinates with constant volumetric thermal energy generation was provided in Table 1-3, to within the unknown constants C1 and C2:

g ′′′ r 2 T =− + C1 ln ( r ) + C2 4k

(1)

dT g ′′′ r C1 =− + dr 2k r

(2)

The boundary condition at the outer edge of the test section is:

(

)

Tr = rout − T∞ ⎛ dT ⎞ − k 2 π rout L ⎜ = ⎟ ⎝ dr ⎠ r = rout ( Rins + Rconv ,out )

(3)

where Rins is the thermal resistance to conduction through the insulation (provided in Table 1-2): ⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎦ Rins = ⎣ 2 π L kins

(4)

and Rconv,out is the resistance to convection from the outer surface of the insulation: Rconv ,out =

1 2 π ( rout + thins ) L hout

R_ins=ln((r_out+th_ins)/r_out)/(2*pi*L*k_ins) R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out) T_r_out=-g```*r_out^2/(4*k)+C_1*ln(r_out)+C_2 dTdr_r_out=-g```*r_out/(2*k)+C_1/r_out

"resistance to conduction through insulation" "resistance to convection from outer surface" "temperature at outer surface of section" "temperature gradient at outer surface of section"

(5)

-k*2*pi*r_out*L*dTdr_r_out=(T_r_out-T_infinity)/(R_ins+R_conv_out) "boundary condition at r=r_out"

The boundary condition at the inner edge of the test section is:

⎛ dT ⎞ hin 2 π rin L T f − Tr = rin = −k 2 π rin L ⎜ ⎟ ⎝ dr ⎠ r = rin

(

)

(6)

T_r_in=-g```*r_in^2/(4*k)+C_1*ln(r_in)+C_2 "temperature at inner surface of section" dTdr_r_in=-g```*r_in/(2*k)+C_1/r_in "temperature gradient at inner surface of section" h_bar_in*2*pi*r_in*L*(T_f-T_r_in)=-k*2*pi*r_in*L*dTdr_r_in "boundary condition at r=r_in"

The EES code will provide the solution to the constants C1 and C2; note that it is not possible to eliminate the unit warnings that are associated with the argument of the natural logarithm in Eq. (1). In fact, if sufficient algebra was carried out, the equations could be placed in a form where the natural logarithm had a dimensionless argument. The location at which to evaluate the temperature (r) is specified in terms of a dimensionless radial position ( r ) that goes from 0 at the inner surface of the test section to 1 at the outer surface. The temperature is evaluated using Eq. (1): r_bar=0.5 [-] r=r_in+r_bar*(r_out-r_in) T=-g```*r^2/(4*k)+C_1*ln(r)+C_2 T_C=converttemp(K,C,T)

"dimensionless radial position" "radial position" "temperature" "in C"

Figure P1.3-9-2 illustrates the temperature as a function of radial position. 80

Temperature (°C)

75 70 65 60 55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure 1.3-9-2: Temperature as a function of radius.

b.) Using your model, develop a calibration curve for the meter; that is, prepare a plot of the mass flow rate as a function of the measured temperature at the mid-point of the pipe. The range of the instrument is 0.2 kg/s to 2.0 kg/s.

The dimensionless radial position is set to r =0.5, corresponding to the temperature of the center of the test section. Figure 1.3-9-3 illustrates the mass flow rate through the meter as a function of the measured temperature. 2 1.8

Mass flow rate (kg/s)

1.6 1.4 1.2

Tf = 28°C

1

Tf = 18°C

0.8

Tf = 8°C

0.6 0.4 0.2 0 40

60

80

100

120

140

160

Temperature (°C)

Figure 1.3-9-3: Mass flow rate as a function of the temperature at the center of the pipe wall for several values of the fluid temperature.

The meter must be robust to changes in the fluid temperature. That is, the calibration curve developed in (b) must continue to be valid even as the fluid temperature changes by as much as 10°C. c.) Overlay on your plot from (b) the mass flow rate as a function of the measured temperature for Tf = 8°C and Tf = 28°C. Is your meter robust to changes in Tf? The calibration curves generated at Tf = 8°C and Tf = 28°C are also shown in Figure 1.3-9-3. Notice that the fluid temperature has a large effect on the device. For example, if the measured temperature is 80°C then the mass flow rate could be anywhere from 0.45 kg/s to 0.75 kg/s depending on the fluid temperature. The meter is not robust to changes in Tf. In order to improve the meters ability to operate over a range of fluid temperature, a temperature sensor is installed in the fluid in order to measure Tf during operation. d.) Using your model, develop a calibration curve for the meter in terms of the mass flow rate as a function of ΔT, the difference between the measured temperatures at the mid-point of the pipe wall and the fluid. The temperature difference is calculated according to:

ΔT = Tr =0.5 − T f DT=T-T_f

"measured temperature difference"

Figure 1.3-9-4 illustrates the mass flow rate as a function of the temperature difference:

(7)

Mass flow rate (kg/s)

2 1.8

Tf = 28°C

1.6

Tf = 18°C

1.4

Tf = 8°C

1.2 1 0.8 0.6 0.4 0.2 0

30

40

50

60 70 80 90 100 Temperature difference (K)

110

120

Figure 1.3-9-4: Mass flow rate as a function of the temperature difference between the measured temperature at the center of the pipe wall and the fluid temperature for several values of the fluid temperature.

e.) Overlay on your plot from (d) the mass flow rate as a function of the difference between the measured temperatures at the mid-point of the pipe wall and the fluid if the fluid temperature is Tf = 8°C and Tf = 28°C. Is the meter robust to changes in Tf? The calibration curves for Tf = 8°C and Tf = 28°C are also shown in Figure 1.3-9-4; notice that the fluid temperature has almost no effect on the calibration curves and so the meter is robust to changes in the fluid temperature. f.) If you can measure the temperature difference to within δΔT = 1 K then what is the uncertainty in the mass flow rate measurement? (Use your plot from part (d) to answer this question.) The uncertainty in the measured mass flow rate that corresponds to an uncertainty in the temperature difference is evaluated according to:

⎛ ∂m ⎝ ∂ΔT

δ m = ⎜

⎞ ⎟ δΔT ⎠

(8)

From Figure 1.3-9-4 we see that the partial derivative of mass flow rate with respect to temperature difference decreases with flow rate. At high flow rates (around 2 kg/s), the partial derivative is approximately 0.08 kg/s-K which leads to an uncertainty of 0.08 kg/s. At low flow rates (around 0.2 kg/s), the partial derivative is approximately 0.04 kg/s-K which leads to an uncertainty of 0.04 kg/s. You can use the built-in uncertainty propagation feature in EES to assess uncertainty automatically. g.) Set the temperature difference to the value you calculated at the nominal conditions and allow EES to calculate the associated mass flow rate. Now, select Uncertainty Propagation from the Calculate menu and specify that the mass flow rate is the calculated variable while the temperature difference is the measured variable. Set the uncertainty in the temperature difference to 1 K and verify that EES obtains an answer that is approximately consistent with part (f).

The temperature difference is set to 50 K corresponding to approximately the middle of the range of the device. The mass flow rate is commented out and EES is used to calculate the mass flow rate from the temperature difference: DT=50 [K] {m_dot=0.75 [kg/s]}

"mass flow rate"

Select Uncertainty Propagation from the Calculate menu (Figure P1.3-9-5) and select the variable m_dot as the calculated variable and the variable DT as the measured variable.

Figure P1.3-9-5: Determine Propagation of Uncertainty dialog.

Select Set uncertainties and indicate that the uncertainty of the measured temperature difference is 1 K (Figure P1.3-9-6).

Figure P1.3-9-6: Uncertainties of Measured Variables dialog.

Select OK and then then OK again to carry out the calculation. The results are displayed in the Uncertainty Results tab of the Solution window (Figure P1.3-9-7).

Figure P1.3-9-7: Uncertainties Results tab of the Solution window.

The uncertainty calculated by EES is δ m = 0.031 kg/s, which falls between the bounds identified in part (e). h.) The nice thing about using EES to determine the uncertainty is that it becomes easy to assess the impact of multiple sources of uncertainty. In addition to the uncertainty δΔT, the constant C has an uncertainty of δC = 5% and the conductivity of the material is only known to within δk = 3%. Use EES' built-in uncertainty propagation to assess the resulting uncertainty in the mass flow rate measurement. Which source of uncertainty is the most important? Select Uncertainty Propagation from the Calculate menu and select the variable m_dot as the calculated variable and the variables DT, C, and k as the measured variables. Set the uncertainty of each of the measured variables according to the problem statement (Figure P1.3-9-8).

Figure P1.3-9-8: Uncertainties of Measured Variables dialog.

The results of the uncertainty calculation are shown in Figure P1.3-9-9.

Figure P1.3-9-9: Uncertainties Results tab of the Solution window.

Notice that the uncertainty has increased to δ m = 0.062 kg/s and that the dominant source of the uncertainty is related to C. The effect of the uncertainty in the conductivity is small (only 5.8% of the total). i.) The meter must be used in areas where the ambient temperature and heat transfer coefficient may vary substantially. Prepare a plot showing the mass flow rate predicted by your model for ΔT = 50 K as a function of T∞ for various values of hout . If the operating range of your

meter must include -5°C < T∞ < 35°C then use your plot to determine the range of hout that can be tolerated without substantial loss of accuracy. Figure P1.3-9-10 illustrates the mass flow rate as a function of T∞ for various values of hout . 0.9

2

h = 5 W/m -K

Mass flow rate (kg/s)

2

0.85

10 W/m -K 2 20 W/m -K

0.8

50 W/m -K

2

2

100 W/m -K

0.75 0.7 0.65 0.6 -10 -5

0

5

10 15 20 25 30 35 40 45 50 55

Air temperature (°C)

Figure P1.3-9-10: Mass flow rate predicted with ΔT = 50 K as a function of ambient temperature for various values of the air heat transfer coefficient.

The shaded region in Figure P1.3-9-10 indicates the operating temperature range (in the xdirection) and the region of acceptable accuracy (based approximately on the results of part (e)). Figure P1.3-9-10 shows that 5 W/m2-K < hout < 50 W/m2-K will keep you within the shaded region and therefore this is, approximately, the range of hout that can be tolerated without substantial loss of accuracy.

PROBLEM 1.4-2 (1-10 in text): Mass Flow Meter (revisited) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). The conductivity of the material that is used to make the test section is not actually constant as was assumed in Problem 1-9 but rather depends on temperature according to:

k = 10

W ⎡ W ⎤ + 0.035 ⎢ T − 300 [ K ]) 2 ( m-K ⎣ m-K ⎥⎦

a.) Develop a numerical model of the mass flow meter using EES. Plot the temperature as a function of radial position for the conditions shown in Figure P1.3-9 (Figure P1-9 in the text) with the temperature-dependent conductivity. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" r_out=1.0 [inch]*convert(inch,m) r_in=0.75 [inch]*convert(inch,m) h_bar_out=10 [W/m^2-K] T_infinity_C=20 [C] T_infinity=converttemp(C,K,T_infinity_C) T_f=converttemp(C,K, 18 [C]) g```=1e7 [W/m^3] m_dot=0.75 [kg/s] th_ins=0.25 [inch]*convert(inch,m) k_ins=1.5 [W/m-K] L= 3 [inch]*convert(inch,m) C=2500 [W/m^2-K] h_bar_in=C*(m_dot/1 [kg/s])^0.8

"outer radius of measurement section" "inner radius of measurement section" "external convection coefficient" "ambient temperature in C" "ambient temperature" "fluid temperature" "volumetric rate of thermal energy generation" "mass flow rate" "thickness of insulation" "insulation conductivity" "length of test section" "constant for convection relationship" "internal convection coefficient"

A function is defined that returns the conductivity of the material: Function k_t(T) "This function returns the conductivity of the test section material as a function of temperature" k_t=10 [W/m-K]+0.035 [W/m-K^2]*(T-300 [K]) end

A uniform distribution of nodes is used, the radial location of each node (ri) is:

ri = rin +

( i − 1) r − r ( ) ( N − 1) out in

for i = 1..N

(1)

where N is the number of nodes. The radial distance between adjacent nodes (Δr) is: Δr =

( rout − rin ) ( N − 1)

(2)

N=51 [-] DELTAr=(r_out-r_in)/(N-1) "Set up nodes" duplicate i=1,N r[i]=r_in+(r_out-r_in)*(i-1)/(N-1) end

"number of nodes" "distance between adjacent nodes (m)" "this loop assigns the radial location to each node"

An energy balance is carried out on a control volume surrounding each node. For node 1, placed at the inner surface (Figure P1.4-2-1): qconv ,in + qouter + g = 0

qouter g1 qconv ,in

(3)

T2 T1

Figure P1.4-2-1: Control volume around node 1.

The rate equation for convection is: qconv ,in = hin 2 π rin L (T f − T1 )

(4)

The rate equation for conduction is: Δr ⎞ (T2 − T1 ) ⎛ qouter = kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(5)

The rate equation for generation is: g = 2 π rin

Δr L g ′′′ 2

"Node 1" q_dot_conv_in=h_bar_in*2*pi*r_in*L*(T_f-T[1]) g_dot[1]=2*pi*r_in*L*DELTAr*g```/2 q_dot_outer[1]=k_t((T[1]+T[2])/2)*2*pi*(r[1]+DELTAr/2)*L*(T[2]-T[1])/DELTAr q_dot_conv_in+q_dot_outer[1]+g_dot[1]=0

(6)

"convection from fluid" "generation" "conduction from node 2" "energy balance on node 1"

An energy balance on an internal node is shown in Figure P1.4-2-2: qinner + qouter + g = 0

(7)

qouter

Ti+1

g

Ti

qinner

Ti-1

Figure P1.4-2-2: Control volume around internal node i.

The rate equations for conduction are: Δr ⎞ (Ti +1 − Ti ) ⎛ qouter = kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L 2 ⎠ Δr ⎝

(8)

Δr ⎞ ( T − T ) ⎛ qinner = kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i 2 ⎠ Δr ⎝

(9)

The rate equation for generation is: g = 2 π ri Δr L g ′′′

(10)

"Internal nodes" duplicate i=2,(N-1) q_dot_inner[i]=k_t((T[i]+T[i-1])/2)*2*pi*(r[i]-DELTAr/2)*L*(T[i-1]-T[i])/DELTAr "conduction from inner node" q_dot_outer[i]=k_t((T[i]+T[i+1])/2)*2*pi*(r[i]+DELTAr/2)*L*(T[i+1]-T[i])/DELTAr "conduction from outer node" g_dot[i]=2*pi*r[i]*L*DELTAr*g``` "generation" q_dot_inner[i]+q_dot_outer[i]+g_dot[i]=0 "energy balance on node i" end

An energy balance on node N placed on the outer surface is shown in Figure P1.4-2-3: qinner + qair + g = 0

(11)

qair TN g qinner

TN-1

Figure P1.4-2-3: Control volume around internal node N.

The rate equation for the heat transfer with the air is: qair =

(R

(T∞ − TN )

ins

+ Rconv ,out )

(12)

where ⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎦ Rins = ⎣ 2 π L kins Rconv ,out =

(13)

1 2 π ( rout + thins ) L hout

(14)

The rate equation for conduction is: Δr ⎞ (T − T ) ⎛ qinner = kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N Δr 2 ⎠ ⎝

(15)

The rate equation for generation is: g = 2 π rout

Δr L g ′′′ 2

(16)

"Node N" R_ins=ln((r_out+th_ins)/r_out)/(2*pi*L*k_ins) "resistance to conduction through insulation" R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out) "resistance to convection from outer surface" q_dot_air=(T_infinity-T[N])/(R_ins+R_conv_out) "heat transfer from air" q_dot_inner[N]=k_t((T[N]+T[N-1])/2)*2*pi*(r_out-DELTAr/2)*L*(T[N-1]-T[N])/DELTAr "conduction from node N-1" g_dot[N]=2*pi*r_out*L*DELTAr*g```/2 "generation" q_dot_air+q_dot_inner[N]+g_dot[N]=0 "energy balance on node N"

The solution is converted to degrees Celsius: duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

The solution is illustrated in Figure P1.4-2-4.

"convert solution to deg. C"

75 72.5

Temperature (°C)

70 67.5 65 62.5 60 57.5 55 52.5 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.4-2-4: Temperature as a function of radius.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in the text) in the limit that the conductivity is constant. The conductivity function is modified temporarily so that it returns a constant value: Function k_t(T) "This function returns the conductivity of the test section material as a function of temperature" k_t=10 [W/m-K]{+0.035 [W/m-K^2]*(T-300 [K])} end

The analytical solution from P1.3-9 is programmed and used to compute the analytical solution at each node: "Analytical solution from P1.3-9" k=k_t(300 [K]) "conductivity to use in the solution" T_r_out=-g```*r_out^2/(4*k)+C_1*ln(r_out)+C_2 "temperature at outer surface of section" dTdr_r_out=-g```*r_out/(2*k)+C_1/r_out "temperature gradient at outer surface of section" -k*2*pi*r_out*L*dTdr_r_out=(T_r_out-T_infinity)/(R_ins+R_conv_out) "boundary condition at r=r_out" T_r_in=-g```*r_in^2/(4*k)+C_1*ln(r_in)+C_2 "temperature at inner surface of section" dTdr_r_in=-g```*r_in/(2*k)+C_1/r_in "temperature gradient at inner surface of section" h_bar_in*2*pi*r_in*L*(T_f-T_r_in)=-k*2*pi*r_in*L*dTdr_r_in "boundary condition at r=r_in" duplicate i=1,N T_an[i]=-g```*r[i]^2/(4*k)+C_1*ln(r[i])+C_2 T_an_C[i]=converttemp(K,C,T_an[i]) end

"temperature" "in C"

Figure P1.4-2-5 illustrates the temperature distribution predicted by the numerical and analytical solutions in the limit that k is constant.

80

Temperature (°C)

75 70 65 60

analytical model numerical model

55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.4-2-5: Temperature as a function of radius predicted by the analytical and numerical models in the limit that k is constant.

c.) What effect does the temperature dependent conductivity have on the calibration curve that you generated in part (d) of Problem 1.3-9 (1-9)? The quantity measured by the meter is the difference between the temperature at the center of the pipe wall (T[26] when 51 nodes are used) and the fluid temperature: DT=T[26]-T_f

"temperature difference"

Figure P1.4-2-6 illustrates the calibration curve (i.e., the relationship between the temperature difference and the mass flow rate) with and without the temperature dependent conductivity included. 2

Mass flow rate (kg/s)

1.8 1.6 1.4 without temperature dependent conductivity

1.2 1

with temperature dependent conductivity

0.8 0.6 0.4 0.2 0 30

40

50

60

70

80

90

100

110

120

Temperature difference (K)

Figure P1.4-2-6: Calibration curve generated with and without the temperature dependent conductivity included.

PROBLEM 1.5-1 (1-11 in text): Hay Temperature (revisited) Reconsider Problem P1.3-8, but obtain a solution numerically using MATLAB. The description of the hay bale is provided in Problem P1.3-8. Prepare a model that can consider the effect of temperature on the volumetric generation. Increasing temperature tends to increase the rate of reaction and therefore increase the rate of generation of thermal energy; the volumetric rate of generation can be approximated by: g ′′′ = a + bT where a = -1 W/m3 and b = 0.01 W/m3-K. Note that at T = 300 K, the generation is 2 W/m3 but that the generation increases with temperature. a.) Prepare a numerical model of the hay bale using EES. Plot the temperature as a function of position within the hay bale. The input information is entered in EES and a function is used to define the volumetric generation: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function gen(T) "volumetric heat generation in wall" "Input - T, temperature [K]" "Output - gen, volumetric rate of heat generation [W/m^3]" a=-1 [W/m^3] b=0.01 [W/m^3-K] gen=a+b*T

"coefficients in generation function"

end "Inputs" L = 1 [m] R_bale= 5 [ft]*convert(ft,m) t_p=0.045 [inch]*convert(inch,m) k_p=0.15 [W/m-K] T_infinity=converttemp(C,K,20) h=10 [W/m^2-K] k=0.04 [W/m-K]

"per unit length of bale" "bale radius" "plastic thickness" "plastic conductivity" "ambient temperature" "heat transfer coefficient" "hay conductivity"

Nodes are distributed uniformly throughout the computational domain (which consists only of the hay, not the plastic), the location of each node (ri) is: ri =

(i − 1) R ( N − 1) bale

i = 1..N

(1)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δr) is: Δr =

Rbale ( N − 1)

(2)

"Setup grid" N=50 [-] duplicate i=1,N r[i]=(i-1)*R_bale/(N-1) end Deltar=R_bale/(N-1)

"number of nodes" "position of each node" "distance between adjacent nodes"

A control volume is defined around each node and an energy balance is written for each control volume. The control volume for an arbitrary, internal node (i.e., a node that is not placed on the edge or at the center of the hay) experiences conduction heat transfer passing through the internal surface ( q LHS ), conduction heat transfer passing through the external surface ( q RHS ), and heat generation within the control volume ( g ). A steady-state energy balance for the control volume is shown in Figure 1: q LHS + q RHS + g = 0

(3)

Figure 1: Internal node energy balance

Each of the terms in the energy balance in Eq. (3) must be modeled using a rate equation. Conduction through the inner surface is driven by the temperature difference between nodes i-1 and i through the material that lies between these nodes.

q LHS

Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ L 2 ⎠ ⎝ = (Ti −1 − Ti ) Δr

(4)

where L is the length of the bale (assumed to be 1 m, corresponding to doing the problem on a per unit length of bale basis). The conduction into the outer surface of the control volume is:

q RHS

Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟L 2 ⎠ ⎝ = (Ti +1 − Ti ) Δr

(5)

The generation is the product of the volume of the control volume and the volumetric generation rate, which is approximately: 2 2 ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ g = g ′′′ (Ti ) π L ⎢⎜ ri + r − − ⎟ ⎜ i ⎟ ⎥ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎢⎣⎝

(6)

where g ′′′ (Ti ) is the volumetric rate of generation evaluated at the nodal temperature Ti. Substituting Eqs. (4) through (6) into Eq. (3) leads to: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ L k 2 π ⎜ ri + 2 2 ⎟L ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ 2 ⎠ 2 ⎠ ⎝ ⎝ ′′′  π T T T T g T L r r − + − + + − − ( i −1 i ) ( i +1 i ) ( i ) ⎢⎜ i ⎟ ⎜ i ⎟ ⎥ = 0 (7) 2 ⎠ ⎝ 2 ⎠ ⎥⎦ Δr Δr ⎢⎣⎝ for i = 2...( N − 1) Figure 2 illustrates the control volume associated with the node that is placed on the outer surface of the hay (i.e., node N).

Figure 2: Control volume for node N located on hay outer surface

The energy balance for the control volume associated with node N is: q LHS + g = qout

(8)

where the conduction term is:

q LHS

Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎠ ⎝ = (TN −1 − TN ) , Δr

(9)

the generation term is: 2 ⎡ Δr ⎞ ⎤ ⎛ g = g ′′′ (TN ) π L ⎢ rN2 − ⎜ rN − ⎟ ⎥ , 2 ⎠ ⎦⎥ ⎝ ⎣⎢

(10)

Note that the volume in Eq. (10) is calculated differently from the volume in Eq. (6) because the control volume is half as wide radially. The heat transfer to the external air is: qout =

where

(TN − T∞ ) R p + Rconv

(11)

Rp =

thp

(12)

k p 2 π Rbale L

and

Rconv ,out =

1 h 2 π Rbale L

(13)

Substituting Eqs. (9) through (11) into Eq. (8) leads to: Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎡ 2 ⎛ Δr ⎞ ⎤ (TN − T∞ ) 2 ⎠ ⎝ ′′′  T T g T L r r π − + − − ( N −1 N ) ( N) ⎢ N ⎜ N ⎟ ⎥= Δr 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣

(14)

A similar procedure applied to the control volume associated with node 1 leads to: Δr ⎞ ⎛ k 2 π ⎜ r1 + ⎟L 2 ⎠ ⎝ (T2 − T1 ) + g ′′′ (T1 ) π Δr

Δr ⎞ ⎛ L ⎜ r1 + ⎟ =0 2 ⎠ ⎝ 2

(15)

Equations (7), (14), and (15) represent N equations in an equal number of unknowns; the solution of these equations provides the numerical solution. "Internal control volumes" duplicate i=2,(N-1) k*2*pi*(r[i]-Deltar/2)*L*(T[i-1]-T[i])/Deltar+k*2*pi*(r[i]+Deltar/2)*L*(T[i+1]T[i])/Deltar+gen(T[i])*pi*L*((r[i]+Deltar/2)^2-(r[i]-Deltar/2)^2)=0 end "node N" R_p=t_p/(k_p*2*pi*R_bale*L) "conduction resistance of plastic" R_conv=1/(h*2*pi*R_bale*L) "convection resistance" k*2*pi*(r[N]-Deltar/2)*L*(T[N-1]-T[N])/Deltar+gen(T[N])*pi*L*(r[N]^2-(r[N]-Deltar/2)^2)=(T[N]T_infinity)/(R_p+R_conv) "node 1" k*2*pi*(r[1]+Deltar/2)*L*(T[2]-T[1])/Deltar+gen(T[1])*pi*L*(r[1]+Deltar/2)^2=0

If the EES program is solved then the temperature distribution will be placed in the Arrays window. The temperature as a function of position is shown in Figure 3.

Figure 3: Temperature as a function of position within the bale

b.) Show that your model has numerically converged; that is, show some aspect of your solution as a function of the number of nodes in your solution and discuss an appropriate number of nodes to use. The maximum temperature (i.e., the temperature at the center of the bale) is shown in Figure 4 as a function of the number of nodes. The model is numerically converged after approximately N = 20.

Figure 4: Predicted maximum temperature as a function of the number of nodes

c.) Verify your numerical model by comparing your answer to an analytical solution in some, appropriate limit. The result of this step should be a plot which shows the temperature as a function of radius predicted by both your numerical solution and the analytical solution and demonstrates that they agree. The analytical solution derived in the problem 1.3-8 is used to compute the temperature at each nodal position: "Analytical solution from Problem 1.3-8" a=-1 [W/m^3] "coefficients for volumetric generation function" b=0.01 [W/m^3-K] dTdr_Rbale = -BesselJ(1,(b/k)^(1/2)*R_bale)*(b/k)^(1/2)*C_2 "symbolic expressions from Maple" T_Rbale = BesselJ(0,(b/k)^(1/2)*R_bale)*C_2-1/b*a -k*2*pi*R_bale*L*dTdr_Rbale=(T_Rbale-T_infinity)/(R_p+R_conv)"interface energy balance" duplicate i=1,N T_an[i]=BesselJ(0,sqrt(b/k)*r[i])*C_2-a/b end

Figure 3 illustrates the analytical solution overlaid on the numerical solution and demonstrates agreement. d.) Prepare a numerical model of the hay bale using MATLAB. Plot the temperature as a function of position within the hay bale. A new m-file is opened and formatted as a function with a single input (the number of nodes) and two outputs (vectors containing the radial position and temperature at each node). function[r,T]=P1p5_1(N) L = 1; R_bale= 1.524; t_p=0.00114; k_p=0.15; T_infinity=293.2; h=10; k=0.04;

%per unit length of bale (m) %bale radius (m) %plastic thickness (m) %plastic conductivity (W/m-K) %ambient temperature (K) %heat transfer coefficient (W/m^2-K) %hay conductivity (W/m-K)

end

A function is defined that returns the volumetric rate of generation as a function of temperature; the function is placed at the bottom of the same m-file so that it is accessible locally to P1p5_1. function[gv]=gen(T) %coefficients of function a=-1; %(W/m^3) b=0.01; %(W/m^3-K) gv=a+b*T; end

The radial position of each node is stored in the vector r. Deltar=R_bale/(N-1); %distance between adjacent nodes (m) for i=1:N r(i,1)=Deltar*(i-1); %radial location of each node (m) end

The problem is nonlinear because the generation rate depends on temperature; therefore, the method of successive substitution is used. An initial guess for the temperature distribution is stored in the vector T_g: %initial guess for temperature distribution for i=1:N T_g(i,1)=T_infinity; end

The guess values for temperature are used to setup the matrix A and vector b which contain the matrix formulation of the equations. The energy balance for node 1 is placed in row 1 of A. Δr ⎞ ⎛ k 2 π ⎜ r1 + ⎟L 2 ⎠ ⎝ (T2 − T1 ) + g ′′′ (T1* ) π Δr

Δr ⎞ ⎛ L ⎜ r1 + ⎟ =0 2 ⎠ ⎝ 2

(16)

where T1* is the guess value of the temperature or ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ 2 ⎢ k 2 π ⎜ r1 + 2 ⎟ L ⎥ ⎢ k 2 π ⎜ r1 + 2 ⎟ L ⎥ ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥ = − g ′′′ T * π L ⎛ r + Δr ⎞ T1 ⎢ − ( 1 ) ⎜⎝ 1 2 ⎟⎠ 2 Δr Δr ⎢ ⎥ ⎢ ⎥   

⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ b (1)  

 

A (1,1)

(17)

A (1,2)

The energy balances for the internal nodes are: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ L k 2 π ⎜ ri + 2 2 ⎟L ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ 2 ⎠ 2 ⎠ ⎝ ⎝ * (Ti −1 − Ti ) + (Ti +1 − Ti ) + g ′′′ (Ti ) π L ⎢⎜ ri + ⎟ − ⎜ ri − ⎟ ⎥ = 0 (18) 2 ⎠ ⎝ 2 ⎠ ⎥⎦ Δr Δr ⎢⎣⎝ for i = 2...( N − 1) or

⎡ Δr ⎞ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ L k 2 π ⎜ ri + 2 ⎟ L ⎥ ⎢ k 2 π ⎜ ri − 2 ⎟ L ⎥ ⎝ ⎠ − ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥+ Ti ⎢ − i −1 Δr Δr Δr ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥  

 

⎦ A ( i ,i −1)

A ( i ,i )

⎡ Δr ⎞ ⎤ ⎛ 2 2 ⎢ k 2 π ⎜ ri + 2 ⎟ L ⎥ ⎡⎛ Δr ⎞ ⎛ Δr ⎞ ⎤ ⎝ ⎠ * ′′′  ⎢ ⎥ = − + Ti +1 g (Ti ) π L ⎢⎜ ri ⎟ − ⎜ ri − ⎟ ⎥ Δr 2 ⎠ ⎦⎥ 2 ⎠ ⎝ ⎢ ⎥ ⎢⎝ ⎣  

⎢⎣ ⎥⎦  b (i )  

(19)

A ( i ,i +1)

for i = 2 .. ( N − 1)

The energy balance for node N is: Δr ⎞ ⎛ k 2 π ⎜ rN − ⎟ L 2 ⎡ 2 ⎛ Δr ⎞ ⎤ (TN − T∞ ) 2 ⎠ ⎝ * (TN −1 − TN ) + g ′′′ (TN ) π L ⎢ rN − ⎜ rN − ⎟ ⎥ = Δr 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣

(20)

or ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ k 2 π ⎜ rN − ⎟ L ⎥ ⎢ k 2 π ⎜ rN − 2 ⎟ L ⎥ ⎢ 1 2 ⎠ ⎝ ⎠ − ⎝ ⎥ + TN −1 ⎢ ⎥= TN ⎢ − Δr R p + Rconv ⎥ Δr ⎢ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥  

 

⎦ A ( N , N −1)

A( N , N )

(21)

2 ⎡ T∞ Δr ⎞ ⎤ ⎛ − g ′′′ (TN* ) π L ⎢ rN2 − ⎜ rN − ⎟ ⎥ − 2 ⎠ ⎥⎦ R p + Rconv ⎝ ⎢⎣   

b( N )

The matrices A and b are initialized and the resistances due to convection and conduction through the plastic are computed: A=spalloc(N,N,3*N); b=zeros(N,1); R_p=t_p/(k_p*2*pi*R_bale*L); R_conv=1/(h*2*pi*R_bale*L);

%resistance through plastic %resistance due to convection

The matrices A and b are filled in according to Eqs. (17), (19), and (21): %Node 1 A(1,1)=-k*2*pi*(r(1)+Deltar/2)*L/Deltar; A(1,2)=k*2*pi*(r(1)+Deltar/2)*L/Deltar; b(1)=-gen(T_g(i))*pi*L*(r(1)+Deltar/2)^2;

%Nodes 2 to (N-1) for i=2:(N-1) A(i,i)=-k*2*pi*(r(i)-Deltar/2)*L/Deltark*2*pi*(r(i)+Deltar/2)*L/Deltar; A(i,i-1)=k*2*pi*(r(i)-Deltar/2)*L/Deltar; A(i,i+1)=k*2*pi*(r(i)+Deltar/2)*L/Deltar; b(i)=-gen(T_g(i))*pi*L*((r(i)+Deltar/2)^2-(r(i)-Deltar/2)^2); end %Node N A(N,N)=-k*2*pi*(r(N)-Deltar/2)*L/Deltar-1/(R_p+R_conv); A(N,N-1)=k*2*pi*(r(N)-Deltar/2)*L/Deltar; b(N)=-gen(T_g(N))*pi*L*(r(N)^2-(r(N)-Deltar/2)^2)T_infinity/(R_p+R_conv);

The temperature distribution is obtained according to: T=A\b;

The successive substitution process occurs within a while loop that is terminated when some convergence error, err, goes below a tolerance, tol. The tolerance is set and the error is initialized to a value that will ensure that the loop executes at least once. Once the solution is obtained, it is compared with the guess value to determine an error. The guess values are reset and, if the error is not sufficiently small then the process is repeated. The code is shown below; the new lines are shown in bold: function[r,T]=P1p5_1(N) L = 1; R_bale= 1.524; t_p=0.00114; k_p=0.15; T_infinity=293.2; h=10; k=0.04;

%per unit length of bale (m) %bale radius (m) %plastic thickness (m) %plastic conductivity (W/m-K) %ambient temperature (K) %heat transfer coefficient (W/m^2-K) %hay conductivity (W/m-K)

Deltar=R_bale/(N-1); %distance between adjacent nodes (m) for i=1:N r(i,1)=Deltar*(i-1); %radial location of each node (m) end %initial guess for temperature distribution for i=1:N T_g(i,1)=T_infinity; end A=spalloc(N,N,3*N); b=zeros(N,1); R_p=t_p/(k_p*2*pi*R_bale*L); R_conv=1/(h*2*pi*R_bale*L); tol=0.1;

%resistance through plastic %resistance due to convection

%tolerance for convergence (K)

err=2*tol; %error initialization while(err>tol) %Node 1 A(1,1)=-k*2*pi*(r(1)+Deltar/2)*L/Deltar; A(1,2)=k*2*pi*(r(1)+Deltar/2)*L/Deltar; b(1)=-gen(T_g(i))*pi*L*(r(1)+Deltar/2)^2; %Nodes 2 to (N-1) for i=2:(N-1) A(i,i)=-k*2*pi*(r(i)-Deltar/2)*L/Deltark*2*pi*(r(i)+Deltar/2)*L/Deltar; A(i,i-1)=k*2*pi*(r(i)-Deltar/2)*L/Deltar; A(i,i+1)=k*2*pi*(r(i)+Deltar/2)*L/Deltar; b(i)=-gen(T_g(i))*pi*L*((r(i)+Deltar/2)^2-(r(i)-Deltar/2)^2); end %Node N A(N,N)=-k*2*pi*(r(N)-Deltar/2)*L/Deltar-1/(R_p+R_conv); A(N,N-1)=k*2*pi*(r(N)-Deltar/2)*L/Deltar; b(N)=-gen(T_g(N))*pi*L*(r(N)^2-(r(N)-Deltar/2)^2)T_infinity/(R_p+R_conv); T=A\b; %obtain temperature distribution err=sum(abs(T-T_g))/N %calculate error T_g=T; end end function[gv]=gen(T) %coefficients of function a=-1; %(W/m^3) b=0.01; %(W/m^3-K) gv=a+b*T; end

The temperature as a function of radius is shown in Figure 5.

Figure 5: Predicted temperature as a function of radial position

Problem 1.5-2 (1-12 in text): Mass Flow Meter (re-visited) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). Assume that the conductivity of the material that is used to make the test section is not actually constant as was assumed in Problem 1.3-9 (1-9 in text) but rather depends on temperature according to:

k = 10

W ⎡ W ⎤ + 0.035 ⎢ T − 300 [ K ]) 2 ( m-K ⎣ m-K ⎥⎦

a.) Develop a numerical model of the mass flow meter using MATLAB. Plot the temperature as a function of radial position for the conditions shown in Figure P1.3-9 (P1-9 in text) with the temperature-dependent conductivity. The inputs are entered in a MATLAB function that requires as an input the number of nodes (N): function[r,T_C]=P1p5_2(N) r_out=0.0254; %outer radius of test section (m) r_in=0.01905; %inner radius of test section (m) h_bar_out=10; %external convection coefficient (W/m^2-K) T_infinity=293.2; %air temperature (K) T_f=291.2; %fluid temperature (K) gv=1e7; %rate of generation (W/m^3) m_dot=0.75; %mass flow rate (kg/s) th_ins=0.00635; %thickness of the insulation (m) k_ins=1.5; %insulation conductivity (W/m-K) L=0.0762; %length of the test section (m) C=2500; %constant for convection relationship

The convection coefficient on the internal surface is computed: h_bar_in=C*m_dot^0.8;

%internal convection coefficient

A function is defined that returns the conductivity of the material: function[k]=k_t(T) %conductivity of the material % %Inputs: % T: temperature (K) % %Outputs: % k: conductivity (W/m-K) k=10+0.035*(T-300); end

A uniform distribution of nodes is used, the radial location of each node (ri) is:

ri = rin +

( i − 1) r − r ( ) ( N − 1) out in

for i = 1..N

(1)

where N is the number of nodes. The radial distance between adjacent nodes (Δr) is:

Δr =

( rout − rin ) ( N − 1)

(2)

DELTAr=(r_out-r_in)/(N-1); %distance between adjacent nodes (m) for i=1:N r(i)=r_in+(r_out-r_in)*(i-1)/(N-1); %position of each node (m) end

The system of equations is placed in matrix format. AX =b

(3)

The most logical technique for ordering the unknown temperatures in the vector X is: ⎡ X 1 = T1 ⎤ ⎢ X =T ⎥ 2 ⎥ X =⎢ 2 ⎢ ... ⎥ ⎢ ⎥ ⎣ X N = TN ⎦

(4)

Equation (4) shows that the unknown temperature at node i (i.e., Ti) corresponds to element i of vector X (i.e., Xi). The most logical technique for placing the equations into the A matrix is: ⎡ row 1 = control volume 1 equation ⎤ ⎢ row 2 = control volume 2 equation ⎥ ⎥ A=⎢ ⎢ ⎥ ... ⎢ ⎥ ⎣ row N = control volume N equation ⎦

(5)

In Eq. (5), the equation for control volume i is placed into row i. An energy balance is carried out on a control volume surrounding each node. For node 1, placed at the inner surface (Figure P1.5-2-1):

qconv ,in + qouter + g = 0

(6)

qouter g1 qconv ,in

T2 T1

Figure P1.5-2-1: Control volume around node 1.

The rate equation for convection is: qconv ,in = hin 2 π rin L (T f − T1 )

(7)

The rate equation for conduction is: Δr ⎞ (T2 − T1 ) ⎛ qouter = kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(8)

The rate equation for generation is: g = 2 π rin

Δr L g ′′′ 2

(9)

Substituting Eqs. (7) through (9) into Eq. (6) leads to: Δr ⎞ (T2 − T1 ) ⎛ + π rin Δr L g ′′′ = 0 hin 2 π rin L (T f − T1 ) + kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(10)

Equation (10) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ T1 ⎢ − hin 2 π rin L − kT =(T1 +T2 ) / 2 2 π ⎜ rin + = ⎟ ⎥ + T2 ⎢ kT =(T1 +T2 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣ −π rin Δr L g ′′′ − hin 2 π rin LT f

(11)

An energy balance on an internal node is shown in Figure P1.5-2-2: qinner + qouter + g = 0

qouter g qinner

Ti+1 Ti Ti-1

Figure P1.5-2-2: Control volume around internal node i.

(12)

The rate equations for conduction are: Δr ⎞ (Ti +1 − Ti ) ⎛ qouter = kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L Δr 2 ⎠ ⎝

(13)

Δr ⎞ ( T − T ) ⎛ qinner = kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i 2 ⎠ Δr ⎝

(14)

The rate equation for generation is: g = 2 π ri Δr L g ′′′

(15)

Substituting Eqs. (13) through (15) into Eq. (12) for all of the internal nodes leads to: Δr ⎞ (T − T ) Δr ⎞ (Ti +1 − Ti ) ⎛ ⎛ kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ L i −1 i + kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟L Δr Δr 2 ⎠ 2 ⎠ ⎝ ⎝ +2 π ri Δr L g ′′′ = 0 for i = 2.. ( N − 1)

(16)

Equation (16) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ Δr ⎞ L Δr ⎞ L ⎤ ⎛ ⎛ Ti ⎢ − kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ +Ti −1 ⎢ −kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ ⎥ + Ti +1 ⎢ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎦⎥ ⎝ ⎝ ⎣ ⎣ = −2 π ri Δr L g ′′′ for i = 2.. ( N − 1)

(17)

An energy balance on node N placed on the outer surface is shown in Figure P1.5-2-3: qinner + qair + g = 0

(18)

qair TN g qinner

TN-1

Figure P1.5-2-3: Control volume around internal node N.

The rate equation for the heat transfer with the air is: qair =

(R

(T∞ − TN )

ins

+ Rconv ,out )

(19)

where

⎡ ( r + thins ) ⎤ ln ⎢ out ⎥ rout ⎣ ⎦ Rins = 2 π L kins Rconv ,out =

1 2 π ( rout + thins ) L hout

(20)

(21)

R_ins=log((r_out+th_ins)/r_out)/(2*pi*L*k_ins); %resistance to conduction through insulation R_conv_out=1/(2*pi*(r_out+th_ins)*L*h_bar_out); %resistance to convection from the outside surface of the insulation

The rate equation for conduction is: Δr ⎞ (T − T ) ⎛ qinner = kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N Δr 2 ⎠ ⎝

(22)

The rate equation for generation is: g = 2 π rout

Δr L g ′′′ 2

(23)

Substituting Eqs. (19), (22), and (23) into Eq. (18) leads to:

(T∞ − TN ) + 2 π r Δr L g ′′′ = 0 Δr ⎞ (T − T ) ⎛ kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ L N −1 N + out Δr 2 ⎠ 2 ⎝ ( Rins + Rconv,out )

(24)

Equation (24) is rearranged to identify the coefficients that multiply each unknown temperature: ⎡ ⎤ 1 Δr ⎞ L ⎛ TN ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ − ⎥ 2 ⎠ Δr ( Rins + Rconv ,out ) ⎥⎦ ⎝ ⎢⎣ ⎡ Δr ⎞ L ⎤ ⎛ +TN −1 ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ ⎥ 2 ⎠ Δr ⎦ ⎝ ⎣ T∞ = − π rout Δr L g ′′′ − ( Rins + Rconv,out )

(25)

Equations (11), (17), and (25) are N equations in the N unknown temperatures. Because they are non-linear, they must be linearized and a successive substitution method used. A guess temperature distribution ( Tˆi ) is assumed: %initial guess for temperature distribution for i=1:N Tg(i,1)=T_f; end

The matrix A is defined as a sparse matrix with at most 3N nonzero entries: %initialize A and b A=spalloc(N,N,3*N); b=zeros(N,1);

The solution is placed within a while loop that terminates when the error between the solution and the guess is less than some tolerance: err=999; tol=0.01; while(err>tol)

%initial value for error (K), must be larger than tol %tolerance for convergence (K)

The equation for node 1, Eq. (11), is linearized by using the guess temperature distribution to compute the conductivity: ⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ T1 ⎢ − hin 2 π rin L − kT = Tˆ +Tˆ / 2 2 π ⎜ rin + = ⎟ ⎥ + T2 ⎢ kT =(Tˆ1 +Tˆ2 ) / 2 2 π ⎜ rin + ⎟ ( 1 2) 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣  

 

A1,1

A1,2

−π rin Δr L g ′′′ − hin 2 π rin LT f  

b1

A(1,1)=-h_bar_in*2*pi*r_in*L-... k_t((Tg(1)+Tg(2))/2)*2*pi*(r_in+DELTAr/2)*L/DELTAr; A(1,2)=k_t((Tg(1)+Tg(2))/2)*2*pi*(r_in+DELTAr/2)*L/DELTAr; b(1)=-pi*r_in*DELTAr*L*gv-h_bar_in*2*pi*r_in*L*T_f;

The equations for the internal nodes, Eq. (17), is also linearized:

(26)

⎡ Δr ⎞ L Δr ⎞ L ⎤ ⎛ ⎛ Ti ⎢ − kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr 2 ⎠ Δr ⎦⎥ ⎝ ⎝ ⎣ 

Ai ,i

⎡ Δr ⎞ L ⎤ ⎡ Δr ⎞ L ⎤ ⎛ ⎛ +Ti −1 ⎢ −kT =(Ti +Ti−1 ) / 2 2 π ⎜ rin − ⎟ ⎥ + Ti +1 ⎢ − kT =(Ti +Ti+1 ) / 2 2 π ⎜ rin + ⎟ 2 ⎠ Δr ⎦ 2 ⎠ Δr ⎥⎦ ⎝ ⎝ ⎣ ⎣ 

 

(27)

Ai ,i +1

Ai ,i −1

= −2 π ri Δr L g ′′′ for i = 2.. ( N − 1)  

bi

for i=2:(N-1) A(i,i)=-k_t((Tg(i)+Tg(i-1))/2)*2*pi*(r(i)-DELTAr/2)*L/DELTAr... -k_t((Tg(i)+Tg(i+1))/2)*2*pi*(r(i)+DELTAr/2)*L/DELTAr; A(i,i-1)=k_t((Tg(i)+Tg(i-1))/2)*2*pi*(r(i)-DELTAr/2)*L/DELTAr; A(i,i+1)=k_t((Tg(i)+Tg(i+1))/2)*2*pi*(r(i)+DELTAr/2)*L/DELTAr; b(i)=-2*pi*r(i)*DELTAr*L*gv; end

The equation for node N, Eq. (25), is linearized: ⎡ ⎤ 1 Δr ⎞ L ⎛ TN ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ − ⎥ 2 ⎠ Δr ( Rins + Rconv ,out ) ⎦⎥ ⎝ ⎣⎢ 

AN , N

⎡ Δr ⎞ L ⎤ ⎛ +TN −1 ⎢ − kT =(TN +TN −1 ) / 2 2 π ⎜ rout − ⎟ ⎥ 2 ⎠ Δr ⎦ ⎝ ⎣  

(28)

AN , N −1

T∞ ( Rins + Rconv,out )  

= − π rout Δr L g ′′′ −

bN

A(N,N)=-k_t((Tg(N)+Tg(N-1))/2)*2*pi*(r_in-DELTAr/2)*L/DELTAr-... 1/(R_ins+R_conv_out); A(N,N-1)=k_t((Tg(N)+Tg(N-1))/2)*2*pi*(r_in-DELTAr/2)*L/DELTAr; b(N)=-pi*r_out*DELTAr*L*gv-T_infinity/(R_ins+R_conv_out);

The solution is obtained: X=A\b; T=X;

and used to compute the error between the assumed and calculated solutions is obtained: err =

1 N

∑ (T − Tˆ ) N

i =1

i

i

2

(29)

err=sqrt(sum((T-Tg).^2)/N) %compute rms error

The calculated solution becomes the guess value for the next iteration: Tg=T;

%reset guess values used to setup A and b

end

The solution is converted to degrees Celsius: T_C=T-273.2;

%convert to C

end

The solution is illustrated in Figure P1.5-2-4. 75 72.5

Temperature (°C)

70 67.5 65 62.5 60 57.5 55 52.5 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.5-2-4: Temperature as a function of radius.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in text) in the limit that the conductivity is constant. The conductivity function is modified temporarily so that it returns a constant value: function[k]=k_t(T) %conductivity of the material % %Inputs: % T: temperature (K) % %Outputs: % k: conductivity (W/m-K) k=10;%+0.035*(T-300); end

Figure P1.5-2-5 illustrates the temperature distribution predicted by the numerical and analytical solutions in the limit that k is constant.

80

Temperature (°C)

75 70 65 60

analytical model numerical model

55 50 0.019

0.02

0.021

0.022

0.023

0.024

0.025

0.026

Radius (m)

Figure P1.5-2-5: Temperature as a function of radius predicted by the analytical and numerical models in the limit that k is constant.

Problem 1.6-1 (1-13 in text): Temperature Sensor Error A resistance temperature detector (RTD) utilizes a material that has a resistivity that is a strong function of temperature. The temperature of the RTD is inferred by measuring its electrical resistance. Figure P1.6-1 shows an RTD that is mounted at the end of a metal rod and inserted into a pipe in order to measure the temperature of a flowing liquid. The RTD is monitored by passing a known current through it and measuring the voltage across it. This process results in a constant amount of ohmic heating that may tend to cause the RTD temperature to rise relative to the temperature of the surrounding liquid; this effect is referred to as a self-heating error. Also, conduction from the wall of the pipe to the temperature sensor through the metal rod can also result in a temperature difference between the RTD and the liquid; this effect is referred to as a mounting error.

Tw = 20°C

L = 5.0 cm h = 150 W/m -K 2

x

T∞ = 5.0°C

pipe

D = 0.5 mm k = 10 W/m-K RTD

qsh = 2.5 mW

Figure P1.6-1: Temperature sensor mounted in a flowing liquid.

The thermal energy generation associated with ohmic heating is q sh = 2.5 mW. All of this ohmic heating is assumed to be transferred from the RTD into the end of the rod at x = L. The rod has a thermal conductivity k = 10 W/m-K, diameter D = 0.5 mm, and length L = 5 cm. The end of the rod that is connected to the pipe wall (at x = 0) is maintained at a temperature of Tw = 20°C. The liquid is at a uniform temperature, T∞ = 5°C and the heat transfer coefficient between the liquid and the rod is h = 150 W/m2-K. a.) Is it appropriate to treat the rod as an extended surface (i.e., can we assume that the temperature in the rod is a function only of x)? Justify your answer. The input parameters are entered in EES. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" q_dot_sh=0.0025 [W] k=10 [W/m-K] d=0.5 [mm]*convert(mm,m) L=5.0 [cm]*convert(cm,m) T_wall=convertTemp(C,K,20) T_f=convertTemp(C,K,5) h=150 [W/m^2-K]

"self-heating power" "conductivity of mounting rod" "diameter of mounting rod" "length of mounting rod" "temperature of wall" "temperature of liquid" “heat transfer coefficient”

The appropriate Biot number for this case is:

Bi =

hd 2k

(1)

"Extended surface approximation" Bi=h*d/(2*k)

The Biot number calculated by EES is 0.004 which is much less than 1.0 and therefore the extended surface approximation is justified. b.) Develop an analytical model of the rod that will predict the temperature distribution in the rod and therefore the error in the temperature measurement; this error is the difference between the temperature at the tip of the rod and the liquid. You may find it easiest to use Maple for this process. Figure 2 illustrates a differential control volume for the rod.

Figure 2: Differential control volume for the rod.

The energy balance suggested by Figure 2 is: q x = q x + dx + qconv

(2)

or, expanding the x+dx term:

q x = q x +

dq dx + qconv dx

(3)

The rate equations for conduction and convection are:

d 2 dT q x = − k π 4 dx

(4)

qconv = h π d dx (T − T f )

(5)

and

Substituting Eqs. (4) and (5) into Eq. (3) leads to: 0=

d ⎡ d 2 dT ⎤ k π − ⎢ ⎥ dx + h π d dx (T − T f ) dx ⎣ 4 dx ⎦

(6)

or

d 2T 4 h − (T − T f ) = 0 dx 2 k d

(7)

which is a non-homogeneous 2nd order differential equation. The general solution to Eq. (7) can be found in your text as Eq. (3.66) or in the handout on Extended Surfaces as Eq. (6-22). The easiest thing to do is enter the differential equation into Maple and let it solve it for you: > GDE:=diff(diff(T(x),x),x)-4*h*(T(x)-T_f)/(k*d)=0;

2 ⎛d ⎞ 4 h ( T( x ) − T_f ) GDE := ⎜⎜ 2 T( x ) ⎟⎟ − =0 kd ⎝ dx ⎠

> Ts:=dsolve(GDE);

Ts := T( x ) = e

⎛⎜ 2 h x ⎞⎟ ⎜ k d ⎟ ⎝ ⎠

_C2 + e

⎛⎜ 2 h x ⎞⎟ ⎜− k d ⎟ ⎝ ⎠

_C1 + T_f

The solution can be copied and pasted into EES (don’t forget that you may need to change your output display to Maple Notation to facilitate the copying process depending on your version of Maple): > Ts:=dsolve(GDE);

Ts := T(x) = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C2+exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C1+T_f

which can be copied to EES: Ts := T(x) = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C2+exp(-2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*_C1+T_f "solution copied from Maple"

The solution will need to be modified slightly so that it is compatible with EES (the _C1 must become C1, _C2 must be C2, Ts:= should be deleted and the T(x) must be just T: T = exp(2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*C2+exp(-2*h^(1/2)*x/(k^(1/2)*d^(1/2)))*C1+T_f "solution copied from Maple and modified"

Trying to solve now should give the message that you have 12 variables but only 9 equations – you need to specify C1, C2, and x to have a completely specified problem. Let’s set x = 0: x=0

and concentrate on determining symbolic expressions for the boundary conditions. temperature at the pipe wall (x=0) is specified to be Twall. Using Maple:

The

> rhs(eval(Ts,x=0))=T_wall;

_C2+_C1+T_f = T_wall

which can be pasted into EES (and modified): C2+C1+T_f = T_wall

"wall boundary condition"

The boundary condition at the end of the rod with the sensor is associated with an energy balance on the interface:

kπ

d 2 dT 4 dx

= qsh

(8)

x= L

which can be evaluated symbolically in Maple: > k*pi*d^2*rhs(eval(diff(Ts,x),x=L))/4=q_dot_sh; 1/4*k*pi*d^2*(2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*_C2/(k^(1/2)*d^(1/ 2))-2*h^(1/2)*exp(-2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*_C1/(k^(1/2)*d^(1/2))) = q_dot_sh

Aren’t you glad you don’t have to do this by hand? The expression can be copied and pasted into EES to complete your solution: 1/4*k*pi*d^2*(2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*C2/(k^(1/2)*d^(1/2))-2*h^(1/2)*exp(2*h^(1/2)*L/(k^(1/2)*d^(1/2)))*C1/(k^(1/2)*d^(1/2))) = q_dot_sh "sensor boundary condition"

Check your units (Figure 3 shows the variable information window with the units set) to make sure that no errors were made.

Figure 3: Variable Information window.

c.) Prepare a plot of the temperature as a function of position and compute the temperature error. Comment out the specification that x=0 and prepare a parametric table that includes T and x. Alter x so that it varies from 0 to 0.05 and plot the result. You can convert the temperature to °C and position to cm for a better looking plot: x_cm=x*convert(m,cm) T_C=converttemp(K,C,T)

Figure 4 illustrates the temperature distribution; note that the temperature elevation at the tip with respect to the fluid is about 3.6 K and it represents the measurement error. For the conditions in the problem statement, it is clear that the measurement error is primarily due to the self-heating effect because the effect of the wall (the temperature elevation at the base) has died off after about 2.0 cm.

Figure 4: Temperature distribution in the mounting rod.

d.) Investigate the effect of thermal conductivity on the temperature measurement error. Identify the optimal thermal conductivity and explain why an optimal thermal conductivity exists. The temperature measurement error can be calculated from your solution by setting x = L: "Part d - temperature measurement error" x=L errT=T-T_f

Figure 5 illustrates the temperature measurement error as a function of the thermal conductivity of the rod material. Figure 5 shows that the optimal thermal conductivity, corresponding to the minimum measurement error, is around 100 W/m-K. Below the optimal value, the self-heating error dominates as the local temperature rise at the tip of the rod is large. Above the optimal value, the conduction from the wall dominates. The inset figures show the temperature distribution for high and low thermal conductivity in order to illustrate these different behaviors.

Figure 5: Temperature measurement error as a function of rod thermal conductivity. The inset figures show the temperature distribution at low conductivity and high conductivity.

Problem 1.6-2 (1-14 in text): Optimizing a Heat Sink Your company has developed a micro-end milling process that allows you to easily fabricate an array of very small fins in order to make heat sinks for various types of electrical equipment. The end milling process removes material in order to generate the array of fins. Your initial design is the array of pin fins shown in Figure P1.6-2. You have been asked to optimize the design of the fin array for a particular application where the base temperature is Tbase = 120°C and the air temperature is Tair = 20°C. The heat sink is square; the size of the heat sink is W = 10 cm. The conductivity of the material is k = 70 W/m-K. The distance between the edges of two adjacent fins is a, the diameter of a fin is D, and the length of each fin is L. array of fins k = 70 W/m-K

Tair = 20°C, h D a L

W = 10 cm

Tbase = 120°C

Figure P1.6-2: Pin fin array

Air is forced to flow through the heat sink by a fan. The heat transfer coefficient between the air and the surface of the fins as well as the unfinned region of the base, h , has been measured for the particular fan that you plan to use and can be calculated according to: ⎞ a ⎡ W ⎤⎛ h = 40 ⎢ 2 ⎥ ⎜⎜ ⎟ ⎣ m K ⎦ ⎝ 0.005 [ m ] ⎟⎠

0.4

⎛ ⎞ D ⎜⎜ ⎟⎟ ⎝ 0.01 [ m ] ⎠

−0.3

Mass is not a concern for this heat sink; you are only interested in maximizing the heat transfer rate from the heat sink to the air given the operating temperatures. Therefore, you will want to make the fins as long as possible. However, in order to use the micro-end milling process you cannot allow the fins to be longer than 10x the distance between two adjacent fins. That is, the length of the fins may be computed according to: L = 10 a . You must choose the most optimal value of a and D for this application. a.) Prepare a model using EES that can predict the heat transfer coefficient for a given value of a and D. Use this model to predict the heat transfer rate from the heat sink for a = 0.5 cm and D = 0.75 cm.

The input values are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs"

T_air=converttemp(C,K,20) T_base=converttemp(C,K,120) k=70 [W/m-K] W=10.0 [cm]*convert(cm,m)

"air temperature" "base temperature" "fin material conductivity" "base width"

The optimization parameters, a and D, are set to their initial values: "Optimization parameters" a=0.5 [cm]*convert(cm,m) D=0.75 [cm]*convert(cm,m)

"distance between adjacent fins" "diameter of fins"

The length of the fins is computed using the aspect ratio and the number of fins is determined according to:

⎛ W ⎞ N =⎜ ⎟ ⎝a+D⎠

2

(1)

The heat transfer coefficient is computed using the equation provided in the problem statement. L=10*a N=(W/(a+D))^2 h = 40 [W/m^2-K]*(a/0.005 [m])^(0.4)*(D/0.01 [m])^(-0.3)

"length of fins" "number of fins" "heat transfer coefficient"

The perimeter and cross-sectional area of each fin are computed according to:

p =π D

(2)

D2 4

(3)

Ac = π

The EES function for the fin efficiency of a constant cross-sectional area fin is used. The function is accessed using the Function Information selection from the Options menu and then selecting Fin Efficiency from the pull-down menu. Scroll to the Circular-Base Rectangular Fin (Figure 2(a)) and select Info to learn how to access this function (Figure 2(b)).

(a) (b) Figure 2: (a) Function Information window and (b) Help information for the Circular-Base Rectangular Fin.

The fin constant, mL, is computed according to: mL =

hp L k Ac

(4)

and used to call the function eta_fin_spine_rect which returns the fin efficiency, ηf. p=pi*D Ac=pi*D^2/4 mL=L*sqrt(h*p/(k*Ac)) eta=eta_fin_spine_rect(mL)

"perimeter of fin" "cross sectional area of fin" "fin constant" "fin efficiency"

The total area of the fins on the heat sink is:

Af = N p L

(5)

1 h Af η f

(6)

and so the total resistance of the fins are: Rf = A_f=p*L*N R_f=1/(h*A_f*eta)

"finned area" "resistance of fins"

The total area of the base of the heat sink that is not finned is: Auf = W 2 − N Ac

(7)

and the thermal resistance from the unfinned base is: Ruf =

1 h Auf

(8)

A_uf=W^2-N*Ac R_uf=1/(h*A_uf)

"unfinned area" "resistance of unfinned area"

The total resistance of the heat sink is the combination of Rf and Ruf in parallel:

Rtotal

⎛ 1 1 ⎞ =⎜ + ⎟⎟ ⎜R R f uf ⎝ ⎠

−1

(9)

and the total heat transfer rate is:

q = R_total=(1/R_f+1/R_uf)^(-1) "total thermal resistance of the heat sink" q_dot=(T_base-T_air)/R_total

(Tbase − Tair )

(10)

Rtotal

"heat transfer"

which leads to q = 291.7 W. b.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the distance between adjacent fins, a, for a fixed value of D = 0.75 cm. Be sure that the fin length is calculated using L = 10 a . Your plot should exhibit a maximum value, indicating that there is an optimal value of a. Figure 3 illustrates the heat transfer rate from the heat sink as a function of a for D = 0.75 cm.

Figure 3: Heat transfer rate as a function of the distance between adjacent fins for D = 0.75 cm.

c.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the diameter of the fins, D, for a fixed value of a = 0.5 cm. Be sure that the fin length is calculated using L = 10 a . Your plot should exhibit a maximum value, indicating that there is an optimal value of D. Figure 4 illustrates the heat transfer rate from the heat sink as a function of D for a = 0.5 cm.

Figure 4: Heat transfer rate as a function of the diameter of the fins for a = 0.5 cm.

d.) Determine the optimal value of a and D using EES' built-in optimization capability.

Comment out the optimization parameters (a and D) and access the optimization algorithms from the Calculate Menu by selecting Min/Max (Figure 5).

Figure 5: Find Minimum or Maximum Window

Select the variable to be minimized or maximized from the list on the left and the independent variables to be varied from the list on the right. You will need to provide a reasonable initial guess and bounds for the independent variables by selecting the Bounds button; note that it is not practical for a or D to be less than 1.0 mm. You can experiment with the different optimization methods and see which technique is more robust. I found the genetic optimization algorithm to work the best for this problem; with a sufficient number of individuals I identified an optimal design consisting of approximately 1500 very small fins of D = 1.1 mm separated by a = 1.4 mm. The associated rate of heat transfer is q = 352.2 W.

Problem 1.7-3 (1-15 in text): Material Processing Figure P1.7-3 illustrates a material processing system. oven wall temperature varies with x gap filled with gas th = 0.6 mm kg = 0.03 W/m-K

u = 0.75 m/s Tin = 300 K

D = 5 cm x extruded material k = 40 W/m-K α = 0.001 m2/s

Figure P1.7-3: Material processing system.

Material is extruded and enters the oven at Tin = 300 K with velocity u = 0.75 m/s. The material has velocity u = 0.75 m/s and diameter D = 5 cm. The conductivity of the material is k = 40 W/m-K and the thermal diffusivity is α = 0.001 m2/s. In order to precisely control the temperature of the material, the oven wall is placed very close to the outer diameter of the extruded material and the oven wall temperature distribution is carefully controlled. The gap between the oven wall and the material is th = 0.6 mm and the oven-to-material gap is filled with gas that has conductivity kg = 0.03 W/m-K. Radiation can be neglected in favor of convection through the gas from the oven wall to the material. For this situation, the heat flux experienced by the material surface can be approximately modeled according to: ′′ ≈ qconv

kg th

(Tw − T )

where Tw and T are the oven wall and material temperatures at that position. The oven wall temperature varies with position x according to: ⎛ x ⎞ Tw = T f − (T f − Tw,0 ) exp ⎜ − ⎟ ⎝ Lc ⎠

where Tw,0 is the temperature of the wall at the inlet (at x = 0), Tf = 1000 K is the temperature of the wall far from the inlet, and Lc is a characteristic length that dictates how quickly the oven wall temperature approaches Tf. Initially, assume that Tw,0 = 500 K, Tf = 1000 K, and Lc = 1 m. Assume that the oven can be approximated as being infinitely long. a.) Is an extended surface model appropriate for this problem? The inputs are entered in EES: $UnitSystem SI MASS DEG PA C J

$Tabstops 0.2 0.4 0.6 0.8 3.5 k=40 [W/m-K] u=0.75 [m/s] T_f=1000 [K] T_w_0=500 [K] L_c=1 [m] T_in=300 [K] alpha=0.001 [m^2/s] k_g=0.03 [W/m-K] th=0.6 [mm]*convert(mm,m) D=5 [cm]*convert(cm,m)

"conductivity" "velocity" "wall temperature far from the inlet" "wall temperature at the inlet" "characteristic length which oven wall approaches T_f" "inlet temperature" "thermal diffusivity" "gas conductivity" "oven-to-material gap thickness" "diameter"

The Biot number is the ratio of the resistance that is neglected (internal conduction) to the resistance that is considered (conduction across the gap): Bi = Bi=(k_g/th)*D/(2*k)

kg D th 2 k

(1)

"Biot number"

which leads to Bi = 0.031. This is sufficiently less than 1 to justify an extended surface model. b.) Assume that your answer to (a) was yes. Develop an analytical solution that can be used to predict the temperature of the material as a function of x. An energy balance on a control volume differential for a differential (in x) segment of the material is shown in Figure P1.7-3-2. kg th

per dx (T − Tw )

(ρ u Ac cT )x dT ⎞ ⎛ ⎜ −k Ac ⎟ dx ⎠ x ⎝

(ρ u Ac cT )x+dx

dx

dT ⎞ ⎛ ⎜ −k Ac ⎟ dx ⎠ x+ dx ⎝

Figure P1.7-3-2: Energy balance on a differential control volume.

The energy balance suggested by Figure P1.7-3-2 is:

( ρ u Ac cT ) x + ⎛⎜ −k Ac ⎝

k dT ⎞ dT ⎞ ⎛ + g per dx (T − Tw ) ⎟ = ( ρ u Ac cT ) x + dx + ⎜ − k Ac ⎟ dx ⎠ x dx ⎠ x + dx th ⎝

(2)

where c is the specific heat capacity, Ac is the cross-sectional area and per is the perimeter of the material:

D2 4

(3)

per = π D

(4)

Ac = π

A_c=pi*D^2/4 per=pi*D

"cross-sectional area" "perimeter"

Expanding the terms in Eq. (2) and simplifying: 0 = ρ u Ac c

dT d 2T k − k Ac 2 + g per (T − Tw ) dx dx th

(5)

Rearranging Eq. (5) and dividing through by k Ac leads to:

k per d 2T u dT k g per T =− g Tw − − 2 dx th k Ac α dx th k Ac

(6)

Substituting the wall temperature variation into Eq. (6) leads to:

⎡ ⎛ x d 2T u dT − − m 2 T = − m 2 ⎢T f − (T f − Tw,0 ) exp ⎜ − 2 dx α dx ⎝ Lc ⎣

⎞⎤ ⎟⎥ ⎠⎦

(7)

where m= m=sqrt(4*k_g/(th*k*D))

k g per th k Ac

(8)

"fin parameter"

The boundary conditions are the inlet temperature: Tx =0 = Tin

(9)

and the temperature must approach Tf as x approaches infinity:

Tx →∞ = T f

(10)

The solution is broken into a homogeneous and particular component:

T = Th + Tp and substituted into Eq. (7):

(11)

⎡ d 2Tp u dTp ⎛ x ⎞⎤ d 2Th u dTh 2 2 2 m T m T m T T T exp − − + − − = − − − ( ) ⎢ ⎜ − ⎟⎥ ,0 h p f f w dx 2 α  dx dx 2 α dx ⎝ Lc

⎠⎦ 

 ⎣   homogeneous ordinary

(12)

particular ordinary differential equation

differential equation

The solution to the homogeneous differential equation is: ⎡⎛ u + u 2 + 4 α 2 m 2 Th = C1 exp ⎢⎜ 2α ⎢⎣⎜⎝

⎡⎛ u − u 2 + 4 α 2 m 2 ⎞ ⎤ ⎟ x ⎥ + C2 exp ⎢⎜ ⎟ ⎥ 2α ⎢⎣⎜⎝ ⎠ ⎦

⎞ ⎤ ⎟ x⎥ ⎟ ⎥ ⎠ ⎦

(13)

The particular solution is obtained by the method of undetermined coefficients; the assumed form of the particular solution is: ⎛ x Tp = C3 exp ⎜ ⎝ Lc

⎞ ⎟ + C4 ⎠

(14)

and substituted into the particular differential equation: ⎛ x C3 exp ⎜ 2 Lc ⎝ Lc

⎡ ⎞ ⎛ x ⎞ ⎛ x⎞ ⎛ x ⎞⎤ u C3 exp ⎜ - ⎟ − m 2C3 exp ⎜ - ⎟ − m 2 C4 = −m 2 ⎢T f − (T f − Tw,0 ) exp ⎜ − ⎟ ⎥ ⎟+ ⎠ α Lc ⎝ Lc ⎠ ⎝ Lc ⎠ ⎝ Lc ⎠ ⎦ ⎣ (15)

Equation (15) provides one equation for C3 that is obtained by considering the exponential terms: C3 =

m 2 (T f − Tw,0 )

(16)

⎛ 1 ⎞ u − m2 ⎟ ⎜ 2+ ⎝ Lc α Lc ⎠

and another equation for C4 that is obtained by considering the constant terms:

C4 = T f

(17)

Substituting Eqs. (13), (14), (16), and (17) leads to: ⎡⎛ u + u 2 + 4 α 2 m 2 ⎞ ⎤ ⎡⎛ u − u 2 + 4 α 2 m 2 ⎢ ⎥ ⎢ ⎟ x + C2 exp ⎜ T = C1 exp ⎜ ⎟ ⎥ 2α 2α ⎢⎣⎜⎝ ⎢⎣⎜⎝ ⎠ ⎦ m 2 (T f − Tw,0 ) ⎛ x ⎞ exp ⎜ - ⎟ + T f + ⎛ 1 ⎞ u ⎝ Lc ⎠ − m2 ⎟ ⎜ 2+ ⎝ Lc α Lc ⎠

⎞ ⎤ ⎟ x⎥ ⎟ ⎥ ⎠ ⎦

(18)

The constants C1 and C2 are obtained by considering the boundary conditions. Substituting Eq. (18) into Eq. (10) leads to:

⎡⎛ u + u 2 + 4 α 2 m 2 C1 exp ⎢⎜ 2α ⎢⎣⎜⎝

⎞ ⎤ ⎟ ∞⎥ + Tf = Tf ⎟ ⎥ ⎠ ⎦

(19)

which can only be true if C1 = 0. Therefore:

⎡⎛ u − u 2 + 4 α 2 m 2 T = C2 exp ⎢⎜ 2α ⎢⎣⎜⎝

⎞ ⎤ m2 (T f − Tw,0 ) ⎛ x ⎞ ⎥ ⎟x + exp ⎜ - ⎟ + T f ⎟ ⎥ ⎛ 1 ⎝ Lc ⎠ ⎠ ⎦ ⎜ + u − m 2 ⎞⎟ 2 ⎝ Lc α Lc ⎠

(20)

Substituting Eq. (20) into Eq. (9) leads to:

C2 +

m 2 (T f − Tw,0 ) ⎛ 1 ⎞ u − m2 ⎟ ⎜ 2+ ⎝ Lc α Lc ⎠

+ T f = Tin

(21)

or C2 = Tin − T f −

m 2 (T f − Tw,0 ) ⎛ 1 ⎞ u − m2 ⎟ ⎜ 2+ ⎝ Lc α Lc ⎠

C_2=T_in-T_f-m^2*(T_f-T_w_0)/(1/L_c^2+u/(alpha*L_c)-m^2)

(22)

"boundary condition at x=0"

The solution for the material temperature and the wall temperature are entered in EES: x=0.5 [m] "position" T=C_2*exp(((u-sqrt(u^2+4*alpha^2*m^2))/(2*alpha))*x)+m^2*(T_f-T_w_0)*& exp(-x/L_c)/(1/L_c^2+u/(alpha*L_c)-m^2)+T_f "temperature of the material" T_w=T_f-(T_f-T_w_0)*exp(-x/L_c) "wall temperature"

c.) Plot the temperature of the material and the temperature of the wall as a function of position for 0 < x < 20 m. Plot the temperature gradient experienced by the material as a function of position for 0 < x < 20 m. Figure P1.7-3-3 illustrates the temperature of the material and the wall as a function of position.

1000

Temperature (K)

900

wall

800 700 material

600 500 400 300 0

2.5

5

7.5

10

12.5

15

17.5

20

Position (m)

Figure P1.7-3-3: Temperature of the material and the wall as a function of position.

The temperature gradient is evaluated by differentiating Eq. (20): ⎛ u − u 2 + 4 α 2 m2 dT = C2 ⎜ ⎜ dx 2α ⎝

⎡⎛ u − u 2 + 4 α 2 m 2 ⎞ ⎟ exp ⎢⎜ ⎟ ⎜ 2α ⎠ ⎣⎢⎝

⎞ ⎤ m 2 (T f − Tw,0 ) ⎛ x ⎟ x⎥ − exp ⎜ ⎟ ⎝ Lc ⎠ ⎦⎥ Lc ⎛⎜ 1 + u − m 2 ⎞⎟ 2 ⎝ Lc α Lc ⎠

⎞ ⎟ (23) ⎠

dTdx=C_2*((u-sqrt(u^2+4*alpha^2*m^2))/(2*alpha))*exp(((u-sqrt(u^2+4*alpha^2*m^2))/(2*alpha))*x)& -m^2*(T_f-T_w_0)*exp(-x/L_c)/(1/L_c^2+u/(alpha*L_c)-m^2)/L_c "temperature gradient"

Figure P1.7-3-4 illustrates the temperature gradient as a function of position. 70

Temperature gradient (K/m)

60 50 40 30 20 10 0 0

2.5

5

7.5

10

12.5

15

17.5

20

Position (m)

Figure P1.7-3-4: Temperature gradient in the material as a function of position.

The parameter Lc can be controlled in order to control the maximum temperature gradient experienced by the material as it moves through the oven. d.) Prepare a plot showing the maximum temperature gradient as a function of Lc. Overlay on your plot the distance required to heat the material to Tp = 800 K (Lp). If the maximum temperature gradient that is allowed is 60 K/m then what is the appropriate value of Lc and the corresponding value of Lp. The value Lp is obtained:

T_p=800 [K] T_p=C_2*exp(((u-sqrt(u^2+4*alpha^2*m^2))/(2*alpha))*L_p)+& m^2*(T_f-T_w_0)*exp(-L_p/L_c)/(1/L_c^2+u/(alpha*L_c)-m^2)+T_f

which leads to Lp = 10.18 m.

14

90

80

maximum temperature gradient

13

70

12

60

11

50

10 distance at which T = 800 K

40 0

1

2

3

4

Position at which T = 800 K

Maximum temperature gradient (K/m)

The maximum temperature gradient can be obtained by using EES' optimization routines. Setup a parametric table that includes the variables L_c, x, dTdx, L_p, and L_c. The value of L_c that is set in the Equations window is commented out and the values of L_c in the table are varied from 0.1 to 5 m. Min/Max Table is selected from the Calculate menu. The value of dTdx is maximized by varying x with bounds from 0 to some large value. The maximum temperature gradient and value of Lp are shown Figure P1.7-3-5 as a function of Lc. Figure P1.7-3-5 indicates that Lc should be equal to 1.8 m in order to control the temperature gradient, which leads to Lp = 11 m.

9 5

Lc (m)

Figure P1.7-3-5: Maximum temperature gradient and Lp as a function of Lc.

Problem 1.7-4 (1-16 in text): Solar Collector Tube The receiver tube of a concentrating solar collector is shown in Figure P1.7-4. qs′′

Ta = 25°C 2 ha = 25 W/m -K r = 5 cm th = 2.5 mm k = 10 W/m-K

φ

Tw = 80°C 2 hw = 100 W/m -K Figure P1.7-4: A solar collector

The receiver tube is exposed to solar radiation that has been reflected from a concentrating mirror. The heat flux received by the tube is related to the position of the sun and the geometry and efficiency of the concentrating mirrors. For this problem, you may assume that all of the radiation heat flux is absorbed by the collector and neglect the radiation emitted by the collector to its surroundings. (Chapter 10 will provide information on the radiation characteristics of surfaces that will allow a more complete evaluation of solar collectors.) The flux received at the collector surface ( qs′′ ) is not circumferentially uniform but rather varies with angular position; the flux is uniform along the top of the collector, π < φ < 2π rad, and varies sinusoidally along the bottom, 0 < φ < π rad, with a peak at φ = π/2 rad. ⎪⎧qt′′+ ( q ′′p − qt′′) sin (φ ) for 0 < φ < π qs′′ (φ ) = ⎨ ⎪⎩qt′′ for π < φ < 2 π where qt′′ = 1000 W/m2 is the uniform heat flux along the top of the collector tube and q ′′p = 5000 W/m2 is the peak heat flux along the bottom. The receiver tube has an inner radius of r = 5.0 cm and thickness of th = 2.5 mm (because th/r restart; > ODE_b:=diff(diff(T_b(phi),phi),phi)-r^2*h_bar_a*T_b(phi)/(k*th)-r^2*h_bar_w*T_b(phi)/(k*th)=r^2*h_bar_a*T_a/(k*th)-r^2*h_bar_w*T_w/(k*th)-(qf_t+(qf_p-qf_t)*sin(phi))*r^2/(k*th);

2 2 2 ⎛d ⎞ r h_bar_a T_b( φ ) r h_bar_w T_b( φ ) = ODE_b := ⎜ 2 T_b( φ ) ⎟⎟ − − ⎜ dφ k th k th ⎝ ⎠ r2 h_bar_a T_a r2 h_bar_w T_w ( qf_t + ( qf_p − qf_t ) sin( φ ) ) r2 − − − k th k th k th

> T_b_s:=dsolve(ODE_b);

T_b_s := T_b( φ ) = e

⎛ r h_bar_a + h_bar_w φ ⎞ ⎜ ⎟ ⎜ ⎟ k th ⎝ ⎠

_C2 + e

⎛ r h_bar_a + h_bar_w φ ⎞ ⎜− ⎟ ⎜ ⎟ k th ⎝ ⎠

_C1 + (

−r ( −qf_p + qf_t ) ( h_bar_a + h_bar_w ) sin( φ ) 2

+ ( r2 ( h_bar_a + h_bar_w ) + k th ) ( h_bar_a T_a + h_bar_w T_w + qf_t ) ) ( r2 ( h_bar_a + h_bar_w ) + k th ) ( h_bar_a + h_bar_w ) )

(

> ODE_t:=diff(diff(T_t(phi),phi),phi)-r^2*h_bar_a*T_t(phi)/(k*th)-r^2*h_bar_w*T_t(phi)/(k*th)=r^2*h_bar_a*T_a/(k*th)-r^2*h_bar_w*T_w/(k*th)-qf_t*r^2/(k*th); 2 2 2 ⎛d ⎞ r h_bar_a T_t ( φ ) r h_bar_w T_t( φ ) = ODE_t := ⎜ 2 T_t( φ ) ⎟⎟ − − ⎜ dφ k th k th ⎝ ⎠ 2 r h_bar_a T_a r2 h_bar_w T_w qf_t r2 − − − k th k th k th

> T_t_s:=dsolve(ODE_t);

⎛ r h_bar_a + h_bar_w φ ⎞ ⎜ ⎟ ⎜ ⎟ k th ⎝ ⎠

T_t_s := T_t( φ ) = e _C2 + e h_bar_a T_a + h_bar_w T_w + qf_t + h_bar_a + h_bar_w

⎛ r h_bar_a + h_bar_w φ ⎞ ⎜− ⎟ ⎜ ⎟ k th ⎝ ⎠

_C1

Note that although both solutions are given with constants of integration C1 and C2 it is clear that these constants cannot be the same. Here, the constants for the general solution for Tt will be C3 and C4. The solutions are copied into EES and manipulated slightly to obtain: "Solutions" T_b = exp(r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*phi)*C_2+& exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*phi)*C_1+& (-r^2*(h_bar_a+h_bar_w)*(-qf_p+qf_t)*sin(phi)+(h_bar_a*T_a+& T_w*h_bar_w+qf_t)*(r^2*(h_bar_a+h_bar_w)+k*th))/(h_bar_a+& h_bar_w)/(r^2*(h_bar_a+h_bar_w)+k*th) T_t = exp(r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*phi)*C_4+& exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*phi)*C_3+& (h_bar_a*T_a+h_bar_w*T_w+qf_t)/(h_bar_a+h_bar_w)

There are four unknown constants of integration (C1 through C4). Therefore, 4 boundary conditions are required to obtain the four constants of integration (two for each solution). The temperature must be continuous at both of the interfaces between the top and bottom domains:

Tb ,φ =0 = Tt ,φ = 2π

(3)

Tb ,φ =π = Tt ,φ =π

(4)

Also, the energy flowing between the regions must be conserved. An interface energy balance at φ = π rad provides: −k

L th dTb r dφ

= −k φ =π

L th dTt r dφ

φ =π

which implies that the temperature gradient at φ = π rad is continuous in both domains: dTb dφ

= φ =π

dTt dφ

(5) φ =π

A similar equation results for the interface at φ = 0 rad. dTb dφ

= φ =0

dTt dφ

(6) φ = 2π

Maple can carry out the symbolic manipulation of the solution while EES does the algebra to determine the constants. To obtain the left and right hand sides of Eq. (3): > T_b_0:=rhs(eval(T_b_s,phi=0));

T_b_0 := _C2 + _C1 + > T_t_2pi:=rhs(eval(T_t_s,phi=2*pi));

h_bar_a T_a + h_bar_w T_w + qf_t h_bar_a + h_bar_w

⎛ 2 r h_bar_a + h_bar_w π ⎞ ⎜ ⎟ ⎜ ⎟ k th ⎝ ⎠

⎛ 2 r h_bar_a + h_bar_w π ⎞ ⎜− ⎟ ⎜ ⎟ k th ⎝ ⎠

T_t_2pi := e _C2 + e h_bar_a T_a + h_bar_w T_w + qf_t + h_bar_a + h_bar_w

_C1

These two expressions can be cut and pasted into EES and, with minimal modification, used to set the boundary condition associated with Eq. (3). The necessary modifications include changing _C1 and _C2 to C_1 and C_2 in the equation for T_b_0 and changing _C1 and _C2 to C_3 and C_4 in the equation for T_t_2pi. "Boundary conditions" "Temperature equality at phi=0" T_b_0 = C_2+ C_1+(h_bar_a*T_a+h_bar_w*T_w+qf_t)/(h_bar_a+h_bar_w) "temperature in bottom domain at phi=0" T_t_2pi = exp(2*r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_4+& exp(-2*r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_3+& (h_bar_a*T_a+h_bar_w*T_w+qf_t)/(h_bar_a+h_bar_w) "temperature in top domain at phi=2 pi"

T_b_0=T_t_2pi

The EES and Maple text listed above seems long and complicated however very little of it needed to be entered manually; the process of solving a relatively complex heat transfer problem is reduced to a relatively straightforward integration of two powerful pieces of software. The process is repeated for Eq. (4), in Maple: > T_b_pi:=rhs(eval(T_b_s,phi=pi));

T_b_pi := e

⎛ r h_bar_a + h_bar_w π ⎞ ⎜ ⎟ ⎜ ⎟ k th ⎝ ⎠

_C2 + e

⎛ r h_bar_a + h_bar_w π ⎞ ⎜− ⎟ ⎜ ⎟ k th ⎝ ⎠

_C1 + (

−r2 ( −qf_p + qf_t ) ( h_bar_a + h_bar_w ) sin( π ) + ( r2 ( h_bar_a + h_bar_w ) + k th ) ( h_bar_a T_a + h_bar_w T_w + qf_t ) ) ( r2 ( h_bar_a + h_bar_w ) + k th ) ( h_bar_a + h_bar_w ) ) > T_t_pi:=rhs(eval(T_t_s,phi=pi));

⎛ r h_bar_a + h_bar_w π ⎞ ⎜ ⎟ ⎜ ⎟ k th ⎝ ⎠

⎛ r h_bar_a + h_bar_w π ⎞ ⎜− ⎟ ⎜ ⎟ k th ⎝ ⎠

T_t_pi := e _C2 + e h_bar_a T_a + h_bar_w T_w + qf_t + h_bar_a + h_bar_w

(

_C1

The symbolic equations determined by Maple are then entered in EES with the same modification for _C1 and _C2 noted above: "Temperature equality at pi" T_b_pi = exp(r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_2+& exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_1+& (-r^2*(-qf_p+qf_t)*(h_bar_a+h_bar_w)*sin(pi)+(r^2*(h_bar_a+h_bar_w)+k*th)& *(h_bar_a*T_a+h_bar_w*T_w+qf_t))/(r^2*(h_bar_a+h_bar_w)+k*th)/(h_bar_a+h_bar_w) "temperature in bottom domain at phi = pi" T_t_pi = exp(r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_4+& exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_3+& (h_bar_a*T_a+h_bar_w*T_w+qf_t)/(h_bar_a+h_bar_w) "temperature in top domain at phi = pi" T_b_pi=T_t_pi

Equations (5) and (6) are dealt with in the same way. Maple is used to evaluate the symbolic expressions for the required derivatives: > dTbdphi_0:=rhs(eval(diff(T_b_s,phi),phi=0));

dTbdphi_0 := −

r h_bar_a + h_bar_w _C2 r h_bar_a + h_bar_w _C1 − k th k th

r2 ( −qf_p + qf_t ) r2 ( h_bar_a + h_bar_w ) + k th

> dTtdphi_2pi:=rhs(eval(diff(T_t_s,phi),phi=2*pi));

⎛ 2 r h_bar_a + h_bar_w π ⎞ ⎟ ⎜ ⎟ ⎜ k th ⎠ ⎝

dTtdphi_2pi :=

r h_bar_a + h_bar_w e k th

⎛ 2 r h_bar_a + h_bar_w π ⎞ ⎟ ⎜− ⎟ ⎜ k th ⎠ ⎝

−

r h_bar_a + h_bar_w e k th

> dTbdphi_pi:=rhs(eval(diff(T_b_s,phi),phi=pi));

dTbdphi_pi :=

−

r h_bar_a + h_bar_w e k th

r h_bar_a + h_bar_w e k th 2 r ( −qf_p + qf_t ) cos( π ) − 2 r ( h_bar_a + h_bar_w ) + k th

> dTtdphi_pi:=rhs(eval(diff(T_t_s,phi),phi=pi));

dTtdphi_pi :=

r h_bar_a + h_bar_w e k th

r h_bar_a + h_bar_w e k th

_C2

_C1

⎛ r h_bar_a + h_bar_w π ⎞ ⎟ ⎜ ⎟ ⎜ k th ⎠ ⎝

⎛ r h_bar_a + h_bar_w π ⎞ ⎟ ⎜− ⎟ ⎜ k th ⎠ ⎝

−

_C1

⎛ r h_bar_a + h_bar_w π ⎞ ⎟ ⎜ ⎟ ⎜ k th ⎠ ⎝

⎛ r h_bar_a + h_bar_w π ⎞ ⎟ ⎜− ⎟ ⎜ k th ⎠ ⎝

_C2

_C2

_C1

These expressions are entered in EES (with changes to _C1 and _C2) in order to provide the final two boundary conditions. "Temperature gradient equality at 0" dTbdphi_0 = r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*C_2-& r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*C_1-& r^2*(-qf_p+qf_t)/(r^2*(h_bar_a+h_bar_w)+k*th) "gradient in bottom domain at phi =0" dTtdphi_2pi = r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*exp(2*r/k^(1/2)/th^(1/2)*& (h_bar_a+h_bar_w)^(1/2)*pi)*C_4-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*& exp(-2*r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_3 "gradient in top at phi = 2 pi" dTbdphi_0=dTtdphi_2pi "Temperature gradient equality at pi" dTbdphi_pi = r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*exp(r/k^(1/2)/th^(1/2)*& (h_bar_a+h_bar_w)^(1/2)*pi)*C_2-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*& exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_1-r^2*(-qf_p+qf_t)*cos(pi)& /(r^2*(h_bar_a+h_bar_w)+k*th) "gradient in bottom at phi = pi" dTtdphi_pi = r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*exp(r/k^(1/2)/th^(1/2)*&

(h_bar_a+h_bar_w)^(1/2)*pi)*C_4-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)& *exp(-r/k^(1/2)/th^(1/2)*(h_bar_a+h_bar_w)^(1/2)*pi)*C_3 "gradient in top at phi = pi" dTbdphi_pi=dTtdphi_pi

The analytical solution is converted to Celsius: T_b_C=converttemp(K,C,T_b) T_t_C=converttemp(K,C,T_t)

"bottom temperature in C" "top temperature in C"

and plotted by setting up two parametric tables. The first table contains the variables phi and T_b_C (where phi is varied from 0 to π) while the second table contains the variables phi and T_t_C (where phi is varied from π to 2π). The temperature distribution is shown in Figure 2.

Figure 2: Temperature distribution around the circumference of the collector tube for various values of the tube conductivity.

It is possible to adjust any of the input parameters within EES the solution remains valid because the constants are evaluated symbolically. For example, Figure 2 also illustrates how the solution varies as the conductivity of the receiver tube changes.

Problem 1.8-1 (1-17 in text): Disk Brake Figure P1.8-1 illustrates a disk brake for a rotating machine. The temperature distribution within the brake can be assumed to be a function of radius only. The brake is divided into two regions. In the outer region, from Rp = 3.0 cm to Rd = 4.0 cm, the stationary brake pads create frictional heating and the disk is not exposed to convection. The clamping pressure applied to the pads is P = 1.0 MPa and the coefficient of friction between the pad and the disk is μ = 0.15. You may assume that the pads are not conductive and therefore all of the frictional heating is conducted into the disk. The disk rotates at N = 3600 rev/min and is b = 5.0 mm thick. The conductivity of the disk is k = 75 W/m-K and you may assume that the outer rim of the disk is adiabatic. coefficient of friction, μ = 0.15

stationary brake pads

clamping pressure P = 1 MPa b = 5 mm Rd = 4 cm

Ta = 30°C, h Rp = 3 cm center line k = 75 W/m-K

disk, rotates at N = 3600 rev/min Figure P1.8-1: Disk brake.

In the inner region of the disk, from 0 to Rp, is exposed to air at Ta = 30°C. The heat transfer coefficient between the air and disk surface depends on the angular velocity of the disk, ω, according to: 1.25

⎞ ω ⎡ W ⎤ ⎡ W ⎤⎛ h = 20 ⎢ 2 ⎥ + 1500 ⎢ 2 ⎥ ⎜⎜ ⎟ ⎣ m -K ⎦ ⎣ m -K ⎦ ⎝ 100 [ rad/s ] ⎟⎠

a.) Develop an analytical model of the temperature distribution in the disk brake; prepare a plot of the temperature as a function of radius for r = 0 to r = Rd. The inputs are entered in EES and the heat transfer coefficient is computed according to Eq. Error! Reference source not found.. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in b=5 [mm]*convert(mm,m) N=3600 [rev/min] omega=N*convert(rev/min,rad/s) mu=0.15 [-] P=1 [MPa]*convert(MPa,Pa) k=75 [W/m-K] Rd=4.0 [cm]*convert(cm,m) Rp=3.0 [cm]*convert(cm,m)

"thickness of disk" "rotational velocity of disk" "angular velocity of disk" "coefficient of friction" "clamping pressure" "conductivity" "outer radius of disk" "inner radius of pad"

Ta=converttemp(C,K,30) h=20[W/m^2-K]+1500 [W/m^2-K]*(omega/100 [rad/s])^1.25

"air temperature" "heat transfer coefficient"

In the outer region, region 1, the energy balance on a differential control volume is shown in Figure 2.

Figure 2: Differential energy balance in outer region, (region 1)

The energy balance suggested by Figure 2 is:

qr + q fh = qr + dr

(1)

where q fh is the rate of thermal energy generated by frictional heating. After expanding the r +

dr term, Eq. (1) becomes:

q fh =

dq dr dr

(2)

The rate equation for conduction is:

q = −b 2 π r k

dT1 dr

(3)

where T1 is the temperature in region 1. The force generated by the pad within the control volume is the product of the clamping pressure, the area of contact, and the coefficient of friction:

F = 4 π r dr P μ

(4)

Note that the factor of 4 in Eq. (4) is due to their being contact on both sides of the disk. The rate of frictional heating is the product of the force, the radius, and the angular velocity: q fh = 4 π r 2 dr P μ ω

Substituting Eqs. (3) and (5) into Eq. (2) leads to:

(5)

4 π r 2 dr P μ ω =

d ⎡ dT ⎤ −b 2 π r k 1 ⎥ dr ⎢ dr ⎣ dr ⎦

(6)

which can be rearranged: d ⎡ dT1 ⎤ 2Pμω 2 = − r r dr ⎢⎣ dr ⎥⎦ bk

(7)

d ⎡ dT1 ⎤ r = −β r 2 ⎢ ⎥ dr ⎣ dr ⎦

(8)

or

where

β=

2Pμω bk

(9)

Equation (8) can be directly integrated: ⎡ dT1 ⎤

∫ d ⎢⎣ r dr ⎥⎦ = − β ∫ r

2

dr

(10)

to achieve: r

dT1 r3 = − β + C1 dr 3

(11)

Equation (11) can be directly integrated again: ⎛ r 2 C1 ⎞ β dT = − ∫ 1 ∫ ⎜⎝ 3 + r ⎟⎠ dr

(12)

r3 T1 = − β + C1 ln ( r ) + C2 9

(13)

to achieve:

Equation (13) is the general solution for the temperature in region 1; the constants of integration will be selected in order to satisfy the boundary conditions. In the inner region, region 2, the energy balance on a differential control volume is shown in Figure 3.

Figure 3: Differential energy balance in inner region, (region 2)

The energy balance suggested by Figure 2 is: qr = qr + dr + qconv

(14)

After expanding the r + dr term, Eq. (14) becomes: 0=

dq dr + qconv dr

(15)

The rate equation for conduction remains the same: q = −b 2 π r k

dT2 dr

(16)

where T2 is the temperature in region 2. The rate equation for convection is: qconv = 4 π r dr h (T2 − Ta )

(17)

Substituting Eqs. (16) and (17) into Eq. (15) leads to: d ⎡ dT ⎤ −b 2 π r k 2 ⎥ dr + 4 π r dr h (T2 − Ta ) = 0 ⎢ dr ⎣ dr ⎦

(18)

d ⎡ dT2 ⎤ r − m 2 r T2 = −m 2 r Ta dr ⎢⎣ dr ⎥⎦

(19)

or

where

m=

2h bk

(20)

The solution to Eq. (19) can be divided into its homogeneous (u2) and particular (v2) parts: T2 = u2 + v2

(21)

d ⎡ dv2 ⎤ r − m 2 r v2 = − m 2 r Ta ⎢ ⎥ dr ⎣ dr ⎦

(22)

v2 = Ta

(23)

d ⎡ du2 ⎤ r − m 2 r u2 = 0 dr ⎢⎣ dr ⎥⎦

(24)

d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠

(25)

x=r

(26)

θ = u2

(27)

p =1

(28)

c=m

(29)

s =1

(30)

The solution to the particular equation:

is

The homogeneous equation:

is a form of Bessel's equation:

where

and the last term is negative. Following the flow chart provided in Section 1.8.4 of the book leads to n = 0, a = 1, and therefore the solution is: u2 = C3 BesselI ( 0, m r ) + C4 BesselK ( 0, m r )

(31)

The general solution for the temperature distribution in region 2 is therefore: T2 = C3 BesselI ( 0, m r ) + C4 BesselK ( 0, m r ) + Ta

(32)

Note that this could be obtained directly from Maple by entering Eq. (19): > restart; > ODE:=diff(r*diff(T2(r),r),r)-m^2*r*T2(r)=-m^2*r*Ta; 2 d ⎛d ⎞ ODE := ⎛⎜⎜ T2( r ) ⎞⎟⎟ + r ⎜⎜ 2 T2( r ) ⎟⎟ − m 2 r T2( r ) = −m 2 r Ta d r ⎝ ⎠ ⎝ dr ⎠

> T2s:=dsolve(ODE);

T2s := T2( r ) = BesselI( 0, m r ) _C2 + BesselK( 0, m r ) _C1 + Ta

The constants C1 through C4 in Eqs. (13) and (32) are obtained by applying the correct boundary conditions. At r = 0, the temperature must remain finite. The figures provided in Section 1.8.4 of the book or the limit capability in Maple show that BesselK(0,m r) will become infinite as r approaches zero: > limit(BesselI(0,m*r),r=0);

1 > limit(BesselK(0,m*r),r=0);

∞

therefore: C4 = 0

(33)

The temperature and temperature gradient at the interface between the regions must be continuous: T2, r = Rp = T1, r = Rp

(34)

and dT2 dr

= r = Rp

dT1 dr

(35) r = Rp

The temperature gradient at the outer rim must be zero: dT1 dr

=0 r = Rp

(36)

Substituting Eqs. (13) and (32) into Eqs. (33) through (36) leads to: C3 BesselI ( 0, m R p ) + Ta = − β

R 3p 9

C3 m BesselI (1, m R p ) = − β

+ C1 ln ( R p ) + C2 R p2 3

+

C1 Rp

Rd2 C1 −β + =0 3 Rd

(37)

(38)

(39)

Equations (37) through (39) are 3 equations for the unknown constants and can be solved in EES. beta=2*mu*P*omega/(k*b) m=sqrt(2*h/(k*b)) BesselI(0,m*Rp)*C_3+Ta=-1/9*beta*Rp^3+C_1*ln(Rp)+C_2 BesselI(1,m*Rp)*m*C_3=-1/3*beta*Rp^2+1/Rp*C_1 "equality of temperature gradient at r=Rp" -1/3*beta*Rd^2+1/Rd*C_1=0

"generation parameter" "fin parameter" "equality of temperature at r=Rp"

"zero temperature gradient at r=Rd"

The general solutions are entered in EES: T2 = BesselI(0,m*r2)*C_3+Ta T1 = -1/9*beta*r1^3+C_1*ln(r1)+C_2

"solution in region 2" "solution in region 1"

A dimensionless radius, the variable rbar, is defined in order to allow a Parametric Table to be generated where the variable r1 can be easily altered from Rp to Rd and the r2 can be easily altered from 0 to Rp: r1=Rp+(Rd-Rp)*rbar r2=rbar*Rp

Figure 4 illustrates the temperature distribution in the disk.

Figure 4: Temperature distribution in the disk

b.) If the disk material can withstand a maximum safe operating temperature of 750°C then what is the maximum allowable clamping pressure that can be applied? Plot the temperature distribution in the disk at this clamping pressure. What is the braking torque that results? The maximum operating temperature is obtained at r = Rd (see Figure 4). The clamping pressure that results in T1 at the outer rim reaching the maximum allowable temperature can be determined by commenting out the originally specified clamping pressure and specifying this temperature: {P=1 [MPa]*convert(MPa,Pa)} T_max_allowed=converttemp(C,K,750) rbar=1.0 T1=T_max_allowed

"clamping pressure" "maximum allowable temperature"

which leads to a clamping pressure of P = 0.57 MPa. The temperature distribution for this clamping pressure is shown in Figure 4. The torque applied by the pads (Tq) is obtained from the integral: Tq =

Rd

∫ 4 π r μ P dr 2

(40)

Rp

or 4 Tq = π μ P ⎡⎣ Rd3 − R 3p ⎤⎦ 3

(41)

which leads to Tq = 13.2 N-m. c.) Assume that you can control the clamping pressure so that as the machine slows down the maximum temperature is always kept at the maximum allowable temperature, 750°C. Plot the torque as a function of rotational speed for 100 rev/min to 3600 rev/min. A parametric table is created that includes the variables N and Tq,; N is varied from 100 rev/min to 3600 rev/min. The results are shown in Figure 5. Notice that it is possible to dramatically improve the performance of the brake if you can adjust the clamping pressure with speed.

Figure 5: Clamping pressure and torque as a function of rotational velocity.

Problem 1.8-5 (1-18 in text): Optimizing a Fin Figure P1.8-5 illustrates a fin that is to be used in the evaporator of a space conditioning system for a space-craft. 2 h = 120 W/m -K T∞ = 20°C

x

th = 1 mm

L = 2 cm

ρb = 8000 kg/m3

k = 50 W/m-K ρ = 3000 kg/m3 Tb = 10°C thg = 2 mm

thb = 2 mm

Wb = 1 cm Figure P1.8-5: Fin on an evaporator.

The fin is a plate with a triangular shape. The thickness of the plate is th = 1 mm and the width of the fin at the base is Wb = 1 cm. The length of the fin is L = 2 cm. The fin material has conductivity k = 50 W/m-K. The average heat transfer coefficient between the fin surface and the air in the space-craft is h = 120 W/m2-K. The air is at T∞ = 20°C and the base of the fin is at Tb = 10°C. Assume that the temperature distribution in the fin is 1-D in x. Neglect convection from the edges of the fin. a.) Obtain an analytical solution for the temperature distribution in the fin. Plot the temperature as a function of position. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in h_bar=120 [W/m^2-K] k=50 [W/m-K] T_infinity=converttemp(C,K,20[C]) T_b=converttemp(C,K,10[C]) th_mm= 1 [mm] th=th_mm*convert(mm,m) L_cm=2 [cm] L=L_cm*convert(cm,m) W_b=1 [cm]*convert(cm,m)

"average heat transfer coefficient" "conductivity" "air material" "base temperature" "fin thickness in mm" "fin thickness" "fin length in cm" "fin length" "fin base width"

The differential control volume shown in Figure P1.8-5-2 is used to derive the governing differential equation: q x = q x + dx + qconv

(1)

x

qx qconv q x + dx

Figure P1.8-5-2: Differential control volume.

The rate of conduction and convection are:

dT dx

q x = − k Ac

qconv = h per (T − T∞ ) dx

(2) (3)

where Ac is the cross-sectional area for conduction and per is the perimeter. The width of the fin is a function of x:

W = Wb

x L

(4)

Therefore, Ac and per are:

Ac = Wb th

x L

(5)

per = 2Wb

x L

(6)

Substituting Eqs. (5) and (6) into Eq. (2) and (3) leads to: x dT L dx

(7)

x (T − T∞ ) dx L

(8)

q x = − k Wb th qconv = h 2Wb

Substituting Eqs. (7) and (8) into Eq. (1) leads to: 0= Simplifying:

d ⎡ x dT ⎤ x dx + h 2Wb (T − T∞ ) dx − k Wb th ⎢ ⎥ dx ⎣ L dx ⎦ L

(9)

d ⎛ dT ⎜x dx ⎝ dx

⎞ 2 2 ⎟ − m xT = −m xT∞ ⎠

(10)

where m2 =

2h k th

m=sqrt(2*h_bar/(k*th))

(11)

"solution parameter"

Maple is used to identify the solution to Eq. (10): > restart; > ODE:=diff(x*diff(T(x),x),x)-m^2*x*T(x)=-m^2*x*T_infinity;

2 d ⎞ ⎛d ODE := ⎛⎜⎜ T( x ) ⎞⎟⎟ + x ⎜⎜ 2 T( x ) ⎟⎟ − m 2 x T( x ) = −m 2 x T_infinity ⎠ ⎝ dx ⎝ dx ⎠

> Ts:=dsolve(ODE);

Ts := T( x ) = BesselI( 0, m x ) _C2 + BesselK( 0, m x ) _C1 + T_infinity

Therefore: T = C2 BesselI ( 0, m x ) + C1 BesselK ( 0, m x ) + T∞

(12)

The fin temperature at the tip must be bounded: Tx =0 = C2 BesselI ( 0, m 0 ) + C1 BesselK ( 0, m 0 ) + T∞ < ∞  

 

(13)

∞

1

The limit of the 0th order modified Bessel functions as x → 0 are evaluated using Maple: > limit(BesselI(0,m*x),x=0);

1

> limit(BesselK(0,m*x),x=0);

∞

Therefore, C1 must be zero: T = C2 BesselI ( 0, m x ) + T∞

The base temperature is specified; therefore:

(14)

Tb = C2 BesselI ( 0, m L ) + T∞

(15)

so:

(Tb − T∞ ) BesselI ( 0, m L )

C2 =

(16)

Substituting Eq. (16) into Eq. (14) leads to: T = (Tb − T∞ )

BesselI ( 0, m x ) + T∞ BesselI ( 0, m L )

x_bar=0.5 [-] x=x_bar*L T=(T_b-T_infinity)*BesselI(0,m*x)/BesselI(0,m*L)+T_infinity T_C=converttemp(K,C,T)

(17)

"dimensionless position" "position" "temperature" "in C"

Figure P1.8-5-3 illustrates the temperature as a function of position normalized by the fin length. 14 13.5

Temperature (°C)

13 12.5 12 11.5 11 10.5 10 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized position, x/L

Figure P1.8-5-3: Fin temperature as a function of dimensionless position.

b.) Calculate the rate of heat transfer to the fin. The rate of heat transfer to the fin is computed according to: q fin = k Wb th

dT dx

(18) x= L

Equation (18) is evaluated using Maple: > restart; > T:=(T_b-T_infinity)*BesselI(0,m*x)/BesselI(0,m*L)+T_infinity;

T :=

( T_b − T_infinity ) BesselI( 0, m x ) + T_infinity BesselI( 0, m L )

> q_dot_fin=k*W_b*th*eval(diff(T,x),x=L);

q_dot_fin =

k W_b th ( T_b − T_infinity ) BesselI( 1, m L ) m BesselI( 0, m L )

Therefore: q fin = k Wb th m (Tb − T∞ )

BesselI (1, m L ) BesselI ( 0, m L )

(19)

q_dot_fin=k*W_b*th*(T_b-T_infinity)*m*BesselI(1,m*L)/BesselI(0,m*L) "fin heat transfer rate"

which leads to q fin = -0.196 W (the heat transfer is negative because the base temperature is less than the ambient temperature). c.) Determine the fin efficiency. The fin efficiency is defined according to:

η fin =

q fin

h As (Tb − T∞ )

(20)

where As is the total surface area of the fin exposed to the fluid:

As = Wb L

(21)

Substituting Eqs. (19) and (21) into Eq. (20) leads to:

η fin =

k Wb th m (Tb − T∞ ) BesselI (1, m L ) h Wb L (Tb − T∞ ) BesselI ( 0, m L )

(22)

Substituting Eq. (11) into Eq. (22) and simplifying leads to:

η fin =

k th 2 h BesselI (1, m L ) 2 th k BesselI (1, m L ) = h L k th BesselI ( 0, m L ) L 2 h BesselI ( 0, m L ) N

(23)

2 BesselI (1, mL ) mL BesselI ( 0, mL )

(24)

1/ m

or

η fin =

eta_fin=2*BesselI(1,m*L)/(m*L*BesselI(0,m*L)) "fin efficiency"

which leads to ηfin = 0.8178. Figure P1.8-5-4 illustrates the fin efficiency as a function of the fin parameter mL. 1 0.9

Fin efficiency

0.8 0.7 0.6 0.5 0.4 0.3 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Fin parameter, mL

Problem P1.8-5-4: Fin efficiency as a function of the fin parameter, mL.

The fin has density ρ = 3000 kg/m3. The fin is installed on a base material with thickness thb = 2 mm and density ρb = 8000 kg/m3. The half-width between the gap between adjacent fins is thg = 2 mm. Therefore, the volume of the base material associated with each fin is thb Wb (th + 2 thg). d.) Determine the ratio of the absolute value of the rate of heat transfer to the fin to the total mass of material (fin and base material associated with the fin). The additional inputs are entered in EES: rho=3000 [kg/m^3] th_b=2 [mm]*convert(mm,m) th_g=2 [mm]*convert(mm,m) rho_b=8000 [kg/m^3]

"density of fin material" "thickness of base material" "half-width of gap between adjacent fins" "base material density"

The fin mass is given by: M fin =

Wb L th ρ 2

(25)

The mass of the associated base material is:

M b = Wb ( th + 2 thg ) thb ρb

(26)

The ratio of rate of the fin heat transfer to mass is: q fin M M_fin=W_b*L*th*rho/2 M_b=W_b*(th+2*th_g)*th_b*rho_b q\M=abs(q_dot_fin)/(M_fin+M_b)

=

(M

q fin fin

+ Mb ) "fin mass" "mass of base material" "ratio of heat transfer to mass"

(27)

which leads to q fin /M = 178.4 W/kg. e.) Prepare a contour plot that shows the ratio of the heat transfer to the fin to the total mass of material as a function of the length of the fin (L) and the fin thickness (th). A parametric table is generated that contains the variables L_cm, th_mm and q\M and has 400 rows. The value of the variable L_cm is varied from 1 cm to 10 cm every 20 rows and the value of th_mm is varied from 0.2 mm to 2 mm in increments of 20 rows. The table is run and used to generate the contour plot shown in Figure P1.8-5-5. 2

1.6

Fin thickness (mm)

115.9

139.3

1.8

127.6

162.7 151

1.4 1.2

174.4

186.1

1 197.8

0.8 209.5 W/kg

0.6 0.4 0.2 0 1

2

3

4

5

6

7

8

9

10

Fin length (cm)

Figure P1.8-5-5: Contours of heat transfer per mass in the parameter space of fin length and thickness.

f.) What is the optimal value of L and th that maximizes the absolute value of the fin heat transfer rate to the mass of material? According to Figure P1.8-5-5, the optimal design is approximately L = 3.3 cm and th = 0.58 mm. A more precise optimization can be carried out using EES' internal optimization feature. Maximizing q fin /M by varying L and th leads to q fin /M = 209.6 W/kg at L = 3.25 cm and th = 0.56 mm.

Problem 1.9-3 (1-19 in text): Fiber optic bundle A fiber optic bundle (FOB) is shown in Figure P1.9-3 and used to transmit the light for a building application. 2 h = 5 W/m -K T∞ = 20°C

rout = 2 cm

5 2 q ′′ = 1x10 W/m

x

fiber optic bundle

Figure P1.9-3: Fiber optic bundle used to transmit light.

The fiber optic bundle is composed of several, small diameter fibers that are each coated with a thin layer of polymer cladding and packed in approximately a hexagonal close-packed array. The porosity of the FOB is the ratio of the open area of the FOB face to its total area. The porosity of the FOB face is an important characteristic because any radiation that does not fall directly upon the fibers will not be transmitted and instead contributes to a thermal load on the FOB. The fibers are designed so that any radiation that strikes the face of a fiber is “trapped” by total internal reflection. However, radiation that strikes the interstitial areas between the fibers will instead be absorbed in the cladding very close to the FOB face. The volumetric generation of thermal energy associated with this radiation can be represented by: ⎛ x ⎞ φ q ′′ g ′′′ = exp ⎜ − ⎟ Lch ⎝ Lch ⎠ where q ′′ = 1x105 W/m2 is the energy flux incident on the face, φ = 0.05 is the porosity of the FOB, x is the distance from the face, and Lch = 0.025 m is the characteristic length for absorption of the energy. The outer radius of the FOB is rout = 2 cm. The face of the FOB as well as its outer surface are exposed to air at T∞ = 20°C with heat transfer coefficient h = 5 W/m2-K. The FOB is a composite structure and therefore conduction through the FOB is a complicated problem involving conduction through several different media. Section 2.9 discusses methods for computing the effective thermal conductivity for a composite. The effective thermal conductivity of the FOB in the radial direction is keff,r = 2.7 W/m-K. In order to control the temperature of the FOB near the face where the volumetric generation of thermal energy is largest, it has been suggested that high conductivity filler material be inserted in the interstitial regions between the fibers. The result of the filler material is that the effective conductivity of the FOB in the axial direction varies with position according to: ⎛ x ⎞ keff , x = keff , x ,∞ + Δkeff , x exp ⎜ − ⎟ ⎝ Lk ⎠ where keff,x,∞ = 2.0 W/m-K is the effective conductivity of the FOB in the x-direction without filler material, Δkeff,x = 28 W/m-K is the augmentation of the conductivity near the face, and Lk = 0.05 m is the characteristic length over which the effect of the filler material decays. The length of the FOB is effectively infinite. a.) Is it appropriate to use a 1-D model of the FOB?

The inputs are entered in EES and functions are defined to return the volumetric generation and effective conductivity in the x-direction: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function k_FOB(x) k_eff_x_infinity=2 [W/m-K] L_k=0.05 [m] Dk_eff_x=28 [W/m-K] k_FOB=k_eff_x_infinity+Dk_eff_x*exp(-x/L_k) end

"conductivity far from the face" "characteristic length of elevated conductivity" "conductivity elevation at the face due to filler material" "conductivity"

function gv_FOB(x) phi=0.05 [-] q``=1e5 [W/m^2] L_ch=0.025 [m] gv_FOB=phi*q``*exp(-x/L_ch)/L_ch end

"porosity" "incident heat flux" "characteristic length for absorption" "volumetric rate of thermal energy generation"

"Inputs" k_eff_r=2.7 [W/m-K] r_out=2 [cm]*convert(cm,m) h_bar=5 [W/m^2-K] T_infinity=converttemp(C,K,20[C])

"effective conductivity in the radial direction" "radius of FOB" "heat transfer coefficient" "ambient temperature"

A Biot number is defined according to:

Bi =

h rout keff , r

(1)

which leads to Bi = 0.037, justifying an extended surface model of the FOB. b.) Assume that your answer to (a) was yes. Develop a numerical model of the FOB. Nodes are positioned along the FOB. The FOB is infinitely long; however, the first L = 0.75 m of the bundle is simulated. Examination of the solution shows that this is sufficient to capture the end effects. L=0.75 [m] N=41 [-] Dx=L/(N-1) duplicate i=1,N x[i]=Dx*(i-1) end

"length of FOB to simulate" "number of nodes" "distance between adjacent nodes" "position of each node"

An energy balance on node 1 leads to: 2 π rout Δx 2 Δx h π r (T∞ − T1 ) + h 2 π rout keff , x , x =( x1 + x2 ) / 2 (T2 − T1 ) + g ′′′x = x1 π rout =0 (T∞ − T1 ) + 2 Δx 2 2 out

(2)

h_bar*pi*r_out^2*(T_infinity-T[1])+h_bar*2*pi*r_out*(Dx/2)*(T_infinity-T[1])+& pi*r_out^2*k_FOB((x[1]+x[2])/2)*(T[2]-T[1])/Dx+gv_FOB(x[1])*pi*r_out^2*Dx/2=0 "energy balance on node 1"

Energy balances on the internal nodes lead to: h 2 π rout Δx (T∞ − Ti ) + 2 π rout

Δx

2 π rout

Δx

keff , x , x =( xi + xi+1 ) / 2 (Ti +1 − Ti ) +

2 Δx = 0 keff , x , x =( xi + xi−1 ) / 2 (Ti −1 − Ti ) + g ′′′x = xi π rout

(3)

i = 2.. ( N − 1)

duplicate i=2,(N-1) h_bar*2*pi*r_out*Dx*(T_infinity-T[i])+pi*r_out^2*k_FOB((x[i]+x[i+1])/2)*(T[i+1]-T[i])/Dx+& pi*r_out^2*k_FOB((x[i]+x[i-1])/2)*(T[i-1]-T[i])/Dx+gv_FOB(x[i])*pi*r_out^2*Dx=0 "energy balance on internal nodes" end

The temperature of the last node is taken to be specified at the ambient temperature:

TN = T∞ T[N]=T_infinity

(4)

"node N temperature is specified"

The temperature is converted to Celsius: duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

"temperature in C"

Figure 2 illustrates the temperature distribution within the FOB.

140

Temperature (°C)

120

without filler material

100 80 with filler material

60 40 20 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Position (m) Figure 2: Temperature distribution within the FOB for the case where the filler material is filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0).

c.) Overlay on a single plot the temperature distribution within the FOB for the case where the filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0). Figure 2 shows the case where filler material is present and is removed. The reduction in the maximum temperature related to the addition of the filler material is evident in Figure 2.

Problem 1.9-4 (1-20 in text) An expensive power electronics module normally receives only a moderate current. However, under certain conditions it is possible that it might experience currents in excess of 100 amps. The module cannot survive such a high current and therefore, you have been asked to design a fuse that will protect the module by limiting the current that it can experience, as shown in Figure P1.9-4. L = 2.5 cm

ε = 0.9 Tend = 20°C

D = 0.9 mm

Tend = 20°C

T∞ = 5°C 2 h = 5 W/m -K

k = 150 W/m-K ρr = 1x10-7 ohm-m I = 100 amp Figure 1.9-4: A fuse that protects a power electronics module from high current.

The space available for the fuse allows a wire that is L = 2.5 cm long to be placed between the module and the surrounding structure. The surface of the fuse wire is exposed to air at T∞ = 20°C and the heat transfer coefficient between the surface of the fuse and the air is h = 5.0 W/m2-K. The fuse surface has an emissivity of ε = 0.90. The fuse is made of an aluminum alloy with conductivity k = 150 W/m-K. The electrical resistivity of the aluminum alloy is ρe = 1x10-7 ohm-m and the alloy melts at approximately 500°C. Assume that the properties of the alloy do not depend on temperature. The ends of the fuse (i.e., at x=0 and x=L) are maintained at Tend =20°C by contact with the surrounding structure and the module. The current passing through the fuse, I, results in a uniform volumetric generation within the fuse material. If the fuse operates properly, then it will melt (i.e., at some location within the fuse, the temperature will exceed 500°C) when the current reaches 100 amp. Your job will be to select the fuse diameter; to get your model started you may assume a diameter of D = 0.9 mm. Assume that the volumetric rate of thermal energy generation due to ohmic dissipation is uniform throughout the fuse volume. a.) Prepare a numerical model of the fuse that can predict the steady state temperature distribution within the fuse material. Plot the temperature as a function of position within the wire when the current is 100 amp and the diameter is 0.9 mm. The input parameters are entered in EES and the volumetric generation rate is computed: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" L=2.5 [cm]*convert(cm,m) d=0.9 [mm]*convert(mm,m) T_a=converttemp(C,K,20) T_end=converttemp(C,K,20) h=5.0 [W/m^2-K] e=0.90 [-]

"length" "diameter" "air temperature" "end temperature" "heat transfer coefficient" "emissivity"

k=150 [W/m-K] er=1e-7 [ohm-m] T_melt=converttemp(C,K,500) current=100 [amp]

"conductivity" "electrical resistivity" "melting temperature" "current"

"Volumetric generation" Ac=pi*d^2/4 Rst=er*L/Ac w_dot_ohmic=current^2*Rst g```_dot=w_dot_ohmic/(Ac*L)

"cross-sectional area" "resistance" "total dissipation" "volumetric rate of generation"

The appropriate Biot number for this case is:

Bi =

hd 2k

(1)

The Biot number is calculated according to: "Extended surface approximation" Bi=h*d/(2*k)

The Biot number calculated by EES is much less than 1.0 and therefore the extended surface approximation is justified. The development of the numerical model follows the same steps that were previously discussed in the context of numerical models of 1-D geometries. Nodes (i.e., locations where the temperature will be determined) are positioned uniformly along the length of the rod. The location of each node (xi) is: xi =

( i − 1) L ( N − 1)

i = 1..N

(2)

where N is the number of nodes used for the simulation. The distance between adjacent nodes (Δx) is: Δx =

L ( N − 1)

This distribution is entered in EES: "Setup nodes" N=10 [-] duplicate i=1,N x[i]=(i-1)*L/(N-1) end Dx=L/(N-1)

"number of nodes" "position of nodes" "distance between nodes"

(3)

A control volume is defined around each node; the control surface bisects the distance between the nodes. The control volume shown in Fig. 2 is subject to conduction heat transfer at each edge ( qtop and qbottom ), convection ( qconv ), radiation ( qrad ), and generation ( g ). The energy balance is:

qtop + qbottom + qconv + qrad + g = 0

(4)

The conduction terms are approximated as: qtop =

kπ d2 (Ti −1 − Ti ) 4 Δx

qbottom =

kπ d2 (Ti +1 − Ti ) 4 Δx

(5)

(6)

The convection term is modeled according to: qconv = h π d Δx (Ta − Ti )

(7)

qrad = ε σ π d Δx (Ta4 − Ti 4 )

(8)

The radiation term is:

The generation term is: g = g ′′′ π

d2 Δx 4

(9)

Substituting Eqs. (5) through (9) into Eq. (4) leads to: kπ d2 kπ d2 (Ti −1 − Ti ) + (Ti +1 − Ti ) + ha π d Δx (Ta − Ti ) 4 Δx 4 Δx d2 +ε σ π d Δx (Ta4 − Ti 4 ) + g ′′′ π Δx = 0 for i = 2.. ( N − 1) 4

(10)

The nodes at the edges of the domain must be treated separately; the temperature at both edges of the fuse are specified: T1 = Tend

(11)

TN = Tend

(12)

Equations (10) through (12) are a system of N equations in an equal number of unknown temperatures which are entered in EES: "Numerical solution" T[1]=T_end T[N]=T_end duplicate i=2,(N-1) k*pi*d^2*(T[i-1]-T[i])/(4*dx)+k*pi*d^2*(T[i+1]-T[i])/(4*dx)+pi*d*dx*h*(T_aT[i])+pi*d*dx*e*sigma#*(T_a^4-T[i]^4)+g```_dot*pi*d^2*dx/4=0 end duplicate i=1,N T_C[i]=converttemp(K,C,T[i]) end

Figure 2 illustrates the temperature distribution in the fuse for N = 100 nodes.

Figure 2: Temperature distribution in the fuse.

b.) Verify that your model has numerically converged by plotting the maximum temperature in the wire as a function of the number of nodes in your model. With any numerical simulation it is important to verify that a sufficient number of nodes have been used so that the numerical solution has converged. The key result of the solution is the maximum temperature in the wire, which can be obtained using the MAX command: T_max_C=max(T_C[1..N])

Figure 3 illustrates the maximum temperature as a function of the number of nodes and shows that the solution has converged for N greater than 100 nodes.

Figure 3: Maximum temperature as a function of the number of nodes.

c.) Prepare a plot of the maximum temperature in the wire as a function of the diameter of the wire for I=100 amp. Use your plot to select an appropriate fuse diameter. The number of nodes was set to 100 and the plot shown in Figure 4 was generated:

Figure 4: Maximum temperature as a function of diameter.

The maximum temperature reaches 500°C when the diameter is approximately 1.15 mm; this would provide a fuse that correctly limited the current.