ChE 441 Problem Set 8

Problem Set #8 Problem 10A.2: Heat loss from a rectangular fin

Calculate the heat loss from a rectangular fin (see Figure 10.7-1) for the following conditions: o

Air temperature 350 F o Wall temperature 500 F Thermal conductivity of fin 60 BTU/hr-ft-F Thermal conductivity of air 0.0022 BTU/hr-ft-F Heat transfer coefficient 120 BTU/hr-ft-F Length of fin 0.2 ft Width of fin 1.0 ft Thickness of fin 0.16 in Solution

Begin with Equation 10.7-14 we can obtain the following heat loss expression: actual rate of heat loss from the fin η = rate of heat loss from an isothermal fin at Tw

 − ∙        ∙ ∙∙ ∙  −  ∙     ∙ ∙   −   Q = 2WLh Tw

Ta

η

in which η is given by Equation 10.7-16: η

=

tanh tanh N N

hL hL2

 wit  with hN=

kB

For the conditions given in this problem,

N=

and

hL hL2 kB

η

=

=

(120 BTU/hr ft 2 F)(0.2 ft)2 = 3.46 3.4641 41 BTU 0.08 60 ( ft) 12 hr ft 2 F

tanh tanh N

=

N As a result, the heat loss expression gives: Q = 2WLh T0

Ta

 η  η

tanh (3.4641) 3.4641

= 2 1.0 1.0 ft  0.2 0.2 ft  120

= 2074 2074 BTU/h BTU/hrr

= 0.28 0.2881 81

BTU

hr ft 2 F

500 500

350 F 0.2881 0.2881

Problem 10A.6: Insulating power of a wall

The "insulating power" of a wall can ca n be measured by means of the arrangement shown in the figure. One places a plastic panel against the wall. In the panel two thermocouples are mounted flush with the panel surfaces. The thermal conduc tivity and thickness of the plastic panel are known. From the measured mea sured steady-state temperatures shown in the figure (Figure 10A.6), calculate: (a) The steady-state heat flux through the wall (and panel): Solution

−−  ∙ ∙  −

Using Equation 10.6-9 and the data given for the plastic panel: BTU 0.075 (69 61 F) k 12 (T1 T2 ) hr ft 2 F q0 = = = 14.3 14.3 Btu/hr Btu/hr ft 2 0.502 (x2 x1) ( ft) 12

∙

(b) The "thermal resistance" (wall thickness divided by thermal cond uctivity). Solution

− −

∙∙

The thermal resistance of the wall is: T2 T3 (61 0 F) ft 2 hr F R 23 = = = 4.2 BTU q0 BTU 14.3 2 hr ft 

∙

10B.2: Viscous heating in slit flow

Find the temperature profile for the viscous heating problem shown in Figure 10.4-2, when given the following boundary conditions: at x=0, T=T0; at x=b, qx=0. Solution

−  − − ∂∂ −     ∂∂    −   −  −    −    −    −  − 

Equation of Change for Energy (Equation 10.4-5): vb q x  μ x b B.C.#1: x=b at qx=0, so

C1 =

k 

T

T x

T=

B.C. #2: x=0 at T=T0

=

vb

k 

b

k 

b

b

vb

+ μx

μb

vb

2

vb

2

b

μb

= C1

μx

vb

 μ x

x

2

x

1

x

=0

b

2

2

2

b

x2

2b

+ C2

T0 = 0 + C2 so C2 = T0

T Rearrange:

T0 =

2 μb

vb

k 

b

T

2

x

b

T0

2 μvb

k 

1

=

1 x

2 b

=

2 μvb

x

1 x

k 

b

2 b

2 μvb

x

1 x

k 

b

2 b

2

2

10B.6: Insulation thickness for a furnace wall.

A furnace wall consists of three layers: (i) a layer of heat -resistant or refractory brick, (ii) a layer of insulating brick, and (iii) a steel plate, 0.25 in. thick, for mechanical protection. Calculate the thickness of each layer of brick to give minimum total wall thickness if the heat loss through the 2 wall is to be 5000 BTU/ft -hr, assuming that the layers are in excellent the rmal contact. The following information is available: Material

Maximum allowable temperature

Thermal conductivity (BTU/hr-ft-F) at o 100 F 1.8 0.9 26.1

o

Refractory Brick Insulating Brick Steel

2600 F o 2000 F ----

Thermal conductivity (BTU/hr-ft-F) at o 2000 F 3.6 1.8 ----

Solution

Let the regions be labeled as a s follows: Refractory brick: "01" Insulating brick: "02" Steel: "03" We can use the formulas given in Equations 10.6-8, 9 and 10. o The minimum wall thickness will occur when T1=2000 F, so we should increase the temperature tempe rature  o to be on the safe side, so let To=2500 F. So for region "01" the thickness thick ness must be

x1

−  −    −  −   x0 =

k 01 T0

T1

q0

1 4.1 + 3.6 2500 2500 =2 5000

2000

Thus, for Refractory brick,

x1

x0 = 0.3 0.39

Here we have taken the thermal conductivity of the refractory brick to be the arithmetic average o o of the values of the thermal conductivity at 2000 F and 2500 F, the latter of which was estimated  by linear extrapolation from the given data. For the remaining two regions, we add Equations 10.6-9 and 10.6-10 to get:

T1

− − − T3 = q 0

Taking the steel temperature to be 100,

x2

x1

k 12

+

x3

x2

k 23

2000 So for insulating brick,

−

100 = 5000

x2

−

 −  x2

x1

1 0.9 + 1.8 2

x1 = 0.51 0.51 ft 

+



1 12 26.1

(0.25)