CHE654_2012_Homework4_Solutions.pdf

CHE654 Modeling Chemical Processes Using MATLAB

Homework Set #4 Solutions Class-16

Prepared by

Dr. Hong-ming Ku

King Mongkut’s University of Technology Thonburi Chemical Engineering Department Chemical Engineering Practice School  July 2004-2012 – Use with Permission of the Author Only

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42. Modeling Liquid Heights in a Two-Vessel System An open spherical vessel is connected to a closed rectangular tank through an open valve as shown in the figure below. Water flows continuously into the spherical vessel at a volumetric flow rate of F1 with some of the water in the sphere also flowing into the rectangular tank at a rate of F2. Water then flows out of the rectangular tank at a flow rate of F3, and the temperature inside the tank is always maintained at 298.15 K. We wish to study the dynamics of the liquid heights in the two vessels using MATLAB. F1 = 0.1 m3/min

P Air P0 F3 = 0.3 m3/min

z2

z1

P2

P1

F2

The following data are known about the system: CV = 1×10-5 m3/Pa1/2-min (characteristic valve constant) Pascal Cross-sectional area of rectangular tank = 3 m2 Height of rectangular tank = 10 m Initially at t = 0, z1 = 3 m, z2 = 2 m, P = 1.01325×105 Pascal Gravitational acceleration g = 9.80665 m/s2

P0 = 1.01325×105 Radius of sphere R = 2 m

φ (water) = 1000 kg/m3

Universal gas constant R = 8314 m3-Pa/kmol-K = 8314 J/kmol-K Liquid volume inside a sphere as a function of liquid height z is given by: V =π R z − 2

π z3 3

Use MATLAB’s ode45 to determine whether the liquid heights in the two vessels will ever be equal, i.e. z1 = z2. If so, report the time (accurate to one decimal place) at which this happens and the height.. If z1 is never equal to z2, determine when the two liquid heights are at the closest. Run your model for 20 minutes, which should be sufficient to answer the questions. Answer the following questions: Are the two liquid heights ever equal?

Yes

No

If yes, z1 = z2 = ____________ meters Time at which the two liquid heights are equal or at the closest = _____________ minutes 2

Solution: Answer the following questions: Are the two liquid heights ever equal?

Yes

No

If yes, z1 = z2 = ________ meters Time at which the two liquid heights are equal or at the closest = ____0_____ minutes . Solution: Mass balance around the spherical vessel: dV φ 1 = φF1 − φF2 dt

φ

d (π R z − 2 1

π z3 3

)

dt

(2πRz1 − π z12 )

dz1 = F1 − F2 dt

(4π z1 − π z12 )

dz1 = 0.1 − CV P1 − P2 dt

(4π z1 − π z12 )

dz1 = 0.1 − 1 × 10 −5 P1 − P2 dt

= φF1 − φF2

Mass balance around the rectangular tank:

φ

dV2 = φF2 − φF3 dt

A2

3

dz 2 = 1 × 10 −5 P1 − P2 − 0.3 dt

dz 2 = CV P1 − P2 − 0.3 dt

P1 = P0 + φgz1

P2 = P + φgz 2

P=

nRT nRT = VG VT − V2

n=

PVG (101325)(24) (t = 0) = = 0.981 kmol RT (8314)(298.15) 3

∴P =

(0.981)(8314)(298.15) 810600 = 30 − 3 z 2 10 − z 2

∴ P2 =

810600 + (1000 kg / m 3 )(9.80665 m 2 / sec)( z 2 m)(1 N / kg − m / sec 2 ) 10 − z 2

∴ P2 =

810600 + 9806.65 z 2 10 − z 2

∴ P1 = 101325 + 9806.65 z1

dz1 = dt

0.1 − 1 × 10 −5 101325 + 9806.65( z1 − z 2 ) −

810600 10 − z 2

z1 (4 − z1 )π

dz 2 810600 = 3.333 × 10 −6 101325 + 9806.65( z1 − z 2 ) − − 0.1 dt 10 − z 2 subject to

z1(t = 0) = 3 m

z2(t = 0) = 2 m

% % Solve for Liquid Heights: Problem #42 % clc clear simtime = [0:0.1:20]; % Length of simulation initz = [3, 2]; % Initial values at t=0 [t, z] = ode45('fz', simtime, initz); % Solve the ODEs [t, z] plot(t,z); xlabel('Time in Minutes') ylabel('Values of z1 and z2') legend('z1', 'z2')

function fz = ode(t, z)

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fz = zeros(2,1); % Initialize fz as 2x1 matrix to zeros fz(1) = (0.1-1e-5*sqrt(101325+9806.65*(z(1)-z(2))-(81060/(10-z(2)))))/(z(1)*(4-z(1))*pi); fz(2) = 3.333e-6*sqrt(101325+9806.65*(z(1)-z(2))-(810600/(10-z(2))))-0.1;

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44. Modeling Liquid Heights in a Two-Vessel System Consider two tanks in series as shown where the water flow out of the first tank enters the second tank. The first tank is a cubic vessel with a width of 10 ft, a length of 10 ft, and a height of 10 ft, whereas the second tank is a cone-shaped vessel with a radius of 5 ft at the top and a height of 20 ft. The first tank is filled with water at a volumetric flow rate F1 and is drained through a hole (radius = 1 inch) at the bottom. Studies in fluid mechanics have identified the relation between the volume flow through the bottom hole and the liquid height as:

F2 = C d A 2gz1 where A is the area of the hole, g is the acceleration due to gravity (32.2 ft/sec2), and Cd is an experimentally determined value that depends partly on the type of fluid (for water, Cd = 0.6). Initially at t = 0, the cubic tank is filled with 2 ft of water and the cone-shaped tank is filled with 15 ft of water. F1 = 15 ft3/min

z1

F2

z2 F3 = 20 ft3/min

(a) Derive an analytical expression of liquid height z1 of the cubic tank as a function of time. Does z1 ever reach the steady-state, and if so, what is this value? Based on your analytical answer, also comment on whether z1 reaches a maximum, reaches a minimum, overflows, or goes to zero and the time for that to happen. In your derivation, you are not allowed to use tables of integrals to perform the integration. Instead, use substitution and be careful with your unit conversions. (b) Derive an ODE that describes the liquid height z2 in the second tank. Together with the ODE for z1 in Part (a), use MATLAB (ode45) to solve for z2 as a function of time and plot both z1 and z2 as a function of time. Run the simulation for 20 minutes.

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Answer the following questions: z1 (t = 10 minutes) = __2.5481__ ft

z2 (t = 10 minutes) = __12.1046__ ft

Solution: (a)

Mass balance around Tank 1: dV1 1 = F1 − F2 = 15 − C d A 2 gz1 = 15 − (0.6)(π )(  dt  12 

100

2

2(32.2)(3600) z1

dz1 = 15 − 6.30279 z1 dt

dz1 = 6.30279 × 10 − 2 (2.37990 − z1 ) dt dz1 = 0 at steady state → z1max = 5.664 ft dt

∫

dz1

z1

2

t

2.37990 − z1

= ∫ 6.30279 × 10 −2 dt = 6.30279 × 10 −2 t 0

Now, do variable transformation by defining: u = 2.37990 – z11/2

z11/2 = 2.37990 – u

1

1

1 − du = − z1 2 dz1 or dz1 = −2 z12 du = −2(2.37990 − u )du 2

∫

z1

2

dz1 2.37990 − z1

= −2∫

2.37990 − u du u z1

z1

1 1     2 = −4.7598 ln 2.37990 − z1  + 22.37990 − z12   2  2

1   2.37990 − z12   t = 31.73198(1.414214 − z ) − 75.99016 ln  0.965686    1 2 1

7

Check: At z1 = 2 ft, the analytical expression does give t = 0. As t

∞, 2.37990 –z11/2 = 0 so that ln(0)

∞, or z1 = 5.664 ft.

Comments: Notice that the liquid height in Tank 1 will reach a maximum steadystate height at 5.664 ft. However, this value will not change after it has been reached because this is when F1 is equal to F2. That is, liquid flow in is exactly equal to liquid flow out. By plugging in the maximum liquid height into the analytical solution, one can see that it will take t ∞, to reach steady state. (b)

Mass balance around Tank 2 now:

dV2 = F2 − F3 = C d A 2 gz1 − 20 dt 2 d 1  R  3  π    z 2  = 6.30279 z1 − 20 dt  3  H   2

1 R H because V2 = π   z 23 from z 2 =  r 3 H R 2

dz  5  π   z 22 2 = 6.30279 z1 − 20 dt  20   6.30279 z1 − 20  dz 2  = 5.09296 2   dt z 2  

So the two ODEs are:

dz1 = 6.30279 × 10 −2 (2.37990 − z1 ) dt  6.30279 z1 − 20  dz 2  = 5.09296 2   dt z 2   s.t. z1 (0) = 2 and z 2 (0) = 15

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% % CHE654: Problem #44 % Liquid Heights in a Two-Tank System % clc clear format short simtime = [0:1:20]; initz = [2 15]; [t, z] = ode45('fz', simtime, initz); [t, z] plot(t, z) function fz = ode(t, z) fz = zeros(2,1); fz(1) = (15-6.30279*sqrt(z(1)))/100; fz(2) = 5.09296*(6.30279*sqrt(z(1))-20)/z(2)^2;

ans = 0 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 19.0000 20.0000

2.0000 2.0602 2.1191 2.1767 2.2331 2.2883 2.3424 2.3954 2.4474 2.4982 2.5481 2.5970 2.6450 2.6920 2.7382 2.7834 2.8279 2.8715 2.9143 2.9563 2.9975

15.0000 14.7463 14.4867 14.2207 13.9475 13.6664 13.3764 13.0766 12.7657 12.4423 12.1046 11.7506 11.3775 10.9821 10.5601 10.1058 9.6110 9.0645 8.4483 7.7329 6.8611

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48. Energy Balance in a Two-Tank System Consider the following two-tank system in which water in the first tank flows into the second one with a flow rate of F1 = 2 kg/min and vice versa with a flow rate F2 = 2 kg/min. Water in both vessels is being heated with a heating coil with the same constant heat transfer area of A = 0.5 m2 and the same heat transfer coefficient of U = 4 kcal/m2-min-°C. However, the temperature of the heating coil in Tank 1 (100 °C) is lower than that of the heating coil in Tank 2 (120 °C). We define M1, T1, M2, and T2 as the mass and the temperature of water in the first vessel and in the second vessel, respectively. Tank 1

Tank 2

F1 M1

M2

T2

T1 F2

(a) Using the following data about the system, derive two ODEs that describe T1 and T2 as a function of time. CP (water) = 1.0 kcal/kg-°C Initially at t = 0: M1 = 50 kg of water, T1 = 20 °C, M2 = 100 kg of water, T2 = 10°C (b) Solve the two ODEs in Part (a) and derive an analytical expression for T1 and T2 as a function of time. (c) What is the domain of this system, i.e. the maximum time the derived model is valid for? Also, determine the time at which the temperatures in Tank 1 and Tank 2 are equal.

Answer the following questions: Domain of this system = ____________ minutes T1 = T2 = ____________ °C when t = ____________ minutes

Solution: (a) Mass and energy balances around the first vessel dM 1 = F2 − F1 = 0 dt

→

M 1 = 50

10

d (M 1C P T1 ) = UA(TS ,1 −T 1) + F2 C P T2 − F1C P T1 dt dT dM 1   C P  M 1 1 + T1 = UA(TS −T 1) + F2 C P T2 − F1C P T1 dt dt  

50

dT1 = (4.0)(0.5)(100 −T 1) + 2T2 − 2T1 dt

dT1 = 4 + 0.04T2 − 0.08T1 dt

→

T2 = 25

dT1 + 2T1 − 100 dt

Mass and energy balances around the second vessel dM 2 = F1 − F2 = 0 dt

→

M 1 = 100

d (M 2 C P T2 ) = UA(TS , 2 −T 2) + F1C P T1 − F2 C P T2 dt dT dM 2   C P  M 2 2 + T2 = UA(TS , 2 −T 2) + F1C P T1 − F2 C P T2 dt dt   100

dT2 = (4.0)(0.5)(120 −T 1) + 2T1 − 2T2 dt

100

dT2 = 240 + 2T1 − 4T2 dt

dT2 = 2.4 + 0.02T1 − 0.04T2 dt The two ODEs are: dT1 = 4 + 0.04T2 − 0.08T1 dt dT2 = 2.4 + 0.02T1 − 0.04T2 dt

s.t.

s.t.

11

T1 (0) = 20

T2 (0) = 10

(b) Now combine the two ODEs to derive T1(t) and T2(t) First, take another derivative of the first-order ODE for T1: d 2T1 dT dT = 0.04 2 − 0.08 1 2 dt dt dt Substituting the second ODE for T2 into the above equation: d 2T1 dT dT = 0.04[2.4 + 0.02T1 − 0.04T2 ] − 0.08 1 = 0.096 + 0.0008T1 − 0.0016T2 − 0.08 1 2 dt dt dt Now eliminating T2 in the above equation: d 2T1 dT  dT  = 0.096 + 0.0008T1 − 0.0016 25 1 + 2T1 − 100 − 0.08 1 2 dt dt  dt  d 2T1 dT dT = 0.096 + 0.0008T1 − 0.04 1 − 0.0032T1 + 0.16 − 0.08 1 2 dt dt dt d 2T1 dT = 0.256 − 0.0024T1 − 0.12 1 2 dt dt d 2T1 dT + 0.12 1 + 0.0024T1 = 0.256 2 dt dt This is a second-order ODE with constant coefficients which can be solved by the method of undetermined coefficients.

λ2 + 0.12λ + 0.0024 = 0

→

λ = −0.09465 and − 0.02535

T1 (t ) = c1 exp(−0.09465t ) + c 2 exp(−0.02535t ) + c3 Now, we need to determine the values of c1, c2, and c3 using initial conditions: T1 (0) = c1 + c 2 + c3 = 20 dT1 = −0.09465c1 exp(−0.09465t ) − 0.02535c 2 exp(−0.02535t ) dt

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d 2T1 = 0.00896c1 exp(−0.09465t ) + 6.4262 × 10 −4 c 2 exp(−0.02535t ) 2 dt

Plugging the analytic expression into the ODE for T1: d 2T1 dT + 0.12 1 + 0.0024T1 = 0.256 = 0.0024c3 2 dt dt c3 = 106.667

c1 + c 2 = −86.667 T2 = 25

dT1 + 2T1 − 100 dt

T2 (0) = 25[− 0.09465c1 − 0.02535c 2 ] + 2(20) − 100 = 10

→

− 2.36625c1 − 0.63375c 2 = 70

Solving for c1 and c2: c1 = −8.7014

c 2 = −77.9656

Therefore,

T1 (t ) = 106.667 − 8.7014 exp(−0.09465t ) − 77.9656 exp(−0.02535t )

T2 = 25

dT1 + 2T1 − 100 dt

T2 = 25[0.8236 exp(−0.09465t ) + 1.9764 exp(−0.02535t )] + 2[106.667 − 8.7014 exp(−0.09465t ) − 77.9656 exp(−0.02535t )] − 100

T2 (t ) = 113.334 + 3.1872 exp(−0.09465t ) − 106.5212 exp(−0.02535t )

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(c) Now compute the time at which T1 = T2:

T1 (t ) = 106.667 − 8.7014 exp(−0.09465t ) − 77.9656 exp(−0.02535t ) T2 (t ) = 113.334 + 3.1872 exp(−0.09465t ) − 106.5212 exp(−0.02535t ) T1 = T2 → 11.8886 exp(−0.09465t ) − 28.5556 exp(−0.02535t ) + 6.667 = 0 Using the solve function in MATLAB gives: t = 57.07 minutes with T = 88.28 °C The domain of this system is when either T1 or T2 reaches the boiling-point temperature of water at 100 °C, whichever comes first. For Tank 1:

T1 (t ) = 106.667 − 8.7014 exp(−0.09465t ) − 77.9656 exp(−0.02535t ) = 100

Again, using the solve function in MATLAB gives: t = 97.01 minutes For Tank 2:

T2 (t ) = 113.334 + 3.1872 exp(−0.09465t ) − 106.5212 exp(−0.02535t ) = 100

Again, using the solve function in MATLAB gives: t = 81.97 minutes

Therefore, the domain is t = 81.97 minutes. The following are results from solving the two ODEs numerically using MATLAB to confirm the analytical answers: clc clear format short simtim = [0:1:80]; initT = [20; 10]; [t, T] = ode45('fT', simtim, initT); [t, T] plot (t,T) title ('Temperatures in a Two-Tank System') xlabel ('Time in Minutes') ylabel ('Temperature in Degrees Celcius') legend ('T1', 'T2')

function fT = ode(t, T) fT = zeros(2,1); fT(1) = 4 + 0.04*T(2)-0.08*T(1); fT(2) = 2.4 + 0.02*T(1) - 0.04*T(2);

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49. Mass and Energy Balance in a Stirred Tank Heater Consider the following stirred tank heater shown below, where the tank inlet stream is received from another process unit. A heat transfer fluid is circulated through a jacket to heat the fluid in the tank. Assume that no change of phase occurs in either the tank liquid or the jacket liquid. The following symbols are used: Fi = volumetric flowrate of stream i, and Ti = temperature of stream i. Tank Inlet

F1

T1 Jacket Inlet F3 T3

Jacket Outlet F4 T4

V

T

Jacket

Vj

Tank Outlet F2 T2

Additional assumptions are: 1. The liquid levels in both the tank and the jacket are constant. 2. There is perfect mixing in both the tank and the jacket. 3. The rate of heat transfer from the jacket to the tank is governed by the equation Q = UA(T4 – T2), where U is the overall heat transfer coefficient and A is the area of heat exchange. (a) Write the dynamic modeling equations (ODEs) to find the tank and jacket temperatures. Do not use any numerical values – leave these equations in terms of the process parameters and variables. Be sure to define any new symbols you introduce into the equations. (b) Assume that both the tank fluid and the jacket fluid are water. The steady-state values of this system variables and some parameters are: F1 = 1.0 ft3/min ρCP (in tank) = ρCP (in jacket) = 61.3 Btu/°F-ft3 T1 = 50 °F T2 = 125 °F V = 10 ft3 T3 = 200 °F T4 = 150 °F Vj = 1 ft3 Solve for F3 and UA (show units) at steady-state. (c) If initially (t = 0), T2 = 50 °F and T4 = 200 °F, solve for T2 and T4 from the ODEs in Part (a) analytically as a function of time.

Solution: (a)

Define variables and process parameters: ρ V

= =

liquid density in tank liquid density in tank

16

F1 F2 F3 T1 T2 T3 T4 CP Pj Cij U A Vj

= = = = = = = = = = = = =

tank inlet flow tank inlet flow jacket flow (inlet and outlet) tank inlet temperature tank outlet temp jacket inlet temp jacket outlet temp liquid heat capacity in tank liquid density in jacket liquid heat capacity in jacket overall heat transfer coefficient heat transfer area liquid volume in jacket

Overall mass balance in the tank: d (Vρ ) = F1 ρ1 − F2 ρ dt

Assume

⇒

ρi d (Vρ ) =0 dt

=

ρ and V is constant

⇒

F1 = F2

Energy balance in the tank: d (VρC p (T2 − Tref )) dt

= F2 ρC p (T1 − Tref ) − F2 ρC p (T2 − Trep ) + UA(T4 − T2 )

Assume Cp is not a function of temperature 0 VρC P

or

dTref dT2 − V ρC p = F2 ρC p T1 − F2 ρC pTref − F2 ρC p T2 + F2 ρC pTref + UA(T j − T2 ) dt dt

dT2 F2 UA = (T1 − T2 ) + (T4 − T2 ) dt V VρC p

Similarly for the jacket:

17

(1)

dT4 F3 UA = (T3 − T4 ) − (T4 − T2 ) dt Vj V j ρ j C pi

(2)

There are 2 modeling equations.

(b)

F1 = 1 ft

3

pC P = 61 . 3 Btu

min

o

Fft 3

p j C pj = 61.3 Btu

T 1 = 50 o F

T 2 = 125 o F

V = 10ft

T 3 = 200 o F

T 4 = 150 o F

V j = 1ft

3

3

Find F3 and UA at steady state:

At steady–state:

dT 2 =0 dt

and

dT 4 =0 dt

Eq. (1) becomes

1 UA (50 − 125) + (150 − 125) = 0 10 (10)(61.3)

and Eq. (2) gives

F3 UA ( 200 − 150) − (150 − 125) = 0 1 (1)(61.3)

⇒

(c)

Btu/ oF min

UA =

183.9

F3

1.50 ft3 / min

=

Dynamic profile of both T’s analytically ! T2 (t = 0) = 50 oF T4 (t = 0) = 400 oF

18

o

Fft 3

dT 2 = 0 . 1(50 − T 2 ) + 0 . 3(T 4 − T 2 ) dt

= 5 + 0 . 3T 4 − 0 . 4T 2

dT 4 = 1 . 5( 200 − T 4 ) − 3(T 4 − T 2 ) dt

=

300 – 45 T4 + 3T2

d 2T 2 dT dT 2 = 0.3 4 − 0.4 2 dt dt dt

= 0 . 3[300 − 4 . 5T 4 + 3T 2 ] − 0 . 4

= 90 − 1 . 35T 4 + 0 . 9T 2 − 0 . 4

But

T4 =

dT 2 dt

dT 2 dt

1 dT 2 ( + 0 . 4T 2 − 5) 0 . 3 dt

So that

d 2T 2 dT dT 2 = 90 − 4 . 5 2 − 0 . 9T 2 + 22 . 5 − 0 . 4 2 dt dt dt 2 d T2 dT + 4 . 9 2 + 0 . 9T 2 = 112 . 5 2 dt dt

Solve by method of undetermined coefficients:

Homogeneous solution is

19

T z ,h (t ) = c 1e

X

2

x 1t

+ c ze

X 2t

− 4 . 9 ± 4 . 9 2 − 4 (1)( 0 . 9 ) + 4. 9 X + 0 . 9 = 0; X = 2

x1

=

-0.1911

x2

−4.7089

=

T 2 ,n (t ) = c 1e −0.1911 t + c 2 e −4. 7089 t

T 2 , p (t ) = A

(trial function)

T 2 (t ) = T 2 ,n + T 2 , p = c 1e −0.1911 t + c 2e −4.7089 t + A

Solve for 3 constants C 1 , C2 and A ∞

At t

T 2 = c 1e

So

:

−0 . 1911 t

At t = 0 ,

50 Recall

=

∞)

T (t

+ c 2e

T2

−4 . 7089 t

=

A

=

125

+ 125

50 oF

=

=

C1 + C2 + 125

T4

=

1 0.3

or

C1 + C2

=

-75

dT 2   dt + 0 . 4T 2 − 5   

[

1 − 0 . 1911 c 1e −0.1911 t − 4 . 7089 c 2 e −4.7089 t + 0 . 4c , e −0.1911 t + 0 . 4c 2e − 4.7089 t + 50 − 5 0.3

T4 =

0 . 6963 c 1e

−0 . 1911 t

− 14 . 363 c 2 e

20

−4. 7089 t

+ 150

]

At t = 0, T4

=

200

⇒

0.6963 C1 - 14.363 C2 = 50

0.6963 C1 − 0.6963C2 = −52.2225

C2 = −6.7880 ∴

−0 . 1911 t

C1 = −68.2120

- 6.7880 e −4.7089 t

T1

= 125 - 68.2120 e

T4

= 150 – 47.4960 e −0.1911 t + 97.4960 e −4.7089 t

21

50. Mass and Energy Balance in a Gas Surge Drum Consider a gas surge drum with variable inlet and outlet molar flowrates, qf and q, respectively. Assume that heat is being added to the tank at a rate of Q and ideal gas behavior in the drum.

Tf , qf, Pf

q, P, T

P, V, T

(a) Write the modeling equations (ODEs) that describe how the temperature, T, and pressure, P, inside the drum vary with time. Note that for a gas, the accumulation term on the left-hand side of the energy equation is dH – d(PV) = energy in – energy out dt dt where dH/dt = d(ρCPVT)/dt and CP is assumed constant. For liquids, the d(PV)/dt term is considered negligible (incompressible fluid and constant volume). So do not forget the PV term in your energy balance. (b) Solve the two ODEs in the Part (a) using MATLAB and find the values of P and T after 10 minutes. The data are: V = 100 m3

R = 8.205x10-2 m3-atm/kmol-K = 8.315 kJ/kmol-K qf = 1.0 kmol/min q = 0.5 kmol/min T(t=0) = 298.15 K = 25 °C

CP (gas) = 125 kJ/kmol-K Tf = 25 °C Q = 1.5x104 kJ/min P(t=0) = 1 atm

Solution: a) What are 2 unknowns that change with time? P and T Overall mass balance: d(PV/RT)/dt = qf – q Constant volume drum

(V/R)[d(P/T)/dt] = qf – q

(1/T)(dP/dt) – (P/T2)(dT/dt) = (R/V)( qf – q ) For gas, the accumulation term on the left hand side of the energy equation is:

dH/dt – d(PV)/dt

where dH/dt = d(ρCPVT)/dt

22

The energy equation is: dH/dt – d(PV)/dt = qf Hf – qH + Q

But

d(PV)/dt = P dV/dt + V dP/dt

CPVd(ρT)/dt = CPVρ(dT/dt) + CPVT(dρ/dt) But ρ = P/RT

dρ/dt = (1/RT)(dP/dt) – (P/RT2)(dT/dt)

CPVP/RT (dT/dt) + CPVT)

=

(1/RT) (dP/dt) – (P/RT2) (dT/dt)

(CPVP/RT) – (CPVP/RT)

(dT/dt) + CPV/R (dP/dt)

CPV/R (dP/dt) − V (dP/dt) = qf CpTf – qCpT + Q

dP/dt = (R/V){1/(Cp – R)}{ Cp (qf Tf – qT) + Q} (Equation 1)

dT/dt = (T2/P) {(1/T)(dP/dt) – (R/V) (qf – q)} = (RT/PV){1/(Cp – R)}{ Cp (qf Tf – qT) + Q} – (RT2/PV)(qf – q)

dT/dt = (RT/PV)[{1/(Cp – R)}{Cp(qf Tf – qT) + Q} – T(qf – q)] (Equation 2)

b) Solve Part (a) using MATLAB At t = 0, T = 2981.5 K and P = 1 atm

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dP/dt =

(8.205x10-2)

1

125(298.15 – 0.5T) + 1.5x104

100

125 – 8.315

dP/dt = 7.03175x10-6 (52268.75 – 62.5T) = 0.36754 – 4.39484x10-4T (Equation 1)

dT/dt =

(8.205x10-2)T 100P

1

125(298.15 – 0.5T) + 1.5x104 – 0.5T

125 – 8.315

dT/dt = (T/P) [0.36754 – 8.49735x10-4T] (Equation 2)

P is in atm and T is in Kelvin.

The MATLAB Script File: % % CHE 654: Problem #50 % % Mass and Energy Balance in a Gas Surge Drum % P and T are the variables in atm and Kelvin respectively % clc clear simtime = 10; initpt = [1 298.15]; [t,pt] = ode45('fpt',simtime,initpt); pres = pt(:,1); temp = pt(:,2); size = length(pres); fprintf ('Time(min.) Pressure(atm) Temperature(K)\n'); for i = 1:size fprintf ('%8.4f ',t(i)) fprintf ('%8.4f ',pres(i)) fprintf ('%8.4f \n',temp(i)); end plot(t,pres) title('Pressure in the Gas Surge Tank') xlabel('Time in Minutes')

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ylabel('Pressure in atm') plot(t,temp) title('Temperature in the Gas Surge Tank') xlabel('Time in Minutes') ylabel('Temperature in Kelvin')

The MABLAB M-File for the ODE: function fv = ode(t,pt) fv = zeros(2,1); fv(1) = 0.36754-4.39484e-4*pt(2); fv(2) = pt(2)/pt(1)*(0.36754-8.49735e-4*pt(2));

The MABLAB Output: Time(min.) 0.0000 0.2124 0.4248 0.6372 0.8497 1.0997 1.3497 1.5997 1.8497 2.0997 2.3497 2.5997 2.8497 3.0997 3.3497 3.5997 3.8497 4.0997 4.3497 4.5997 4.8497 5.0997 5.3497 5.5997 5.8497 6.0997 6.3497 6.5997 6.8497 7.0997 7.3497 7.5997 7.8497

Pressure(atm) Temperature(K) 1.0000 298.1500 1.0499 305.1033 1.0992 311.5360 1.1479 317.4987 1.1961 323.0360 1.2521 329.0610 1.3075 334.6070 1.3624 339.7234 1.4167 344.4532 1.4705 348.8343 1.5238 352.8998 1.5767 356.6792 1.6292 360.1986 1.6813 363.4813 1.7331 366.5477 1.7846 369.4165 1.8357 372.1042 1.8866 374.6256 1.9372 376.9941 1.9875 379.2217 2.0376 381.3194 2.0875 383.2970 2.1372 385.1634 2.1866 386.9268 2.2359 388.5946 2.2850 390.1736 2.3339 391.6698 2.3827 393.0890 2.4313 394.4363 2.4798 395.7165 2.5282 396.9339 2.5764 398.0926 2.6245 399.1962

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8.0997 8.3497 8.5997 8.8497 9.0997 9.3497 9.5997 9.8497 9.8872 9.9248 9.9624 10.0000 >>

2.6724 2.7203 2.7680 2.8157 2.8632 2.9107 2.9580 3.0053 3.0124 3.0195 3.0266 3.0337

400.2482 401.2517 402.2096 403.1247 403.9994 404.8361 405.6369 406.4039 406.5164 406.6282 406.7392 406.8496

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51. Mass and Energy Balance in a 2-Tank System Consider the following 2 tanks (both cylindrical vessels) in series used to store a liquid solution. Tank 1 is heated while Tank 2 is not. Liquid is drawn from Tank 1 into Tank 2 at the rate of 10 kg/min. At the same time, a feed (40 °C) enters Tank 2 at the rate of 20 kg/min, and an outflow of 30 kg/min leaves the vessel. Initially (at t = 0), Tank 1 is charged with 300 kg of the solution at a temperature of 20 °C, while Tank 2 is charged with 100 kg at a temperature of 30 °C.

Tank 1

20 kg/min 40 °C

Tank 2

T1 T2 10 kg/min

30 kg/min

(a) Assuming that Tank 1 is being heated with a heating coil that remains submerged at all times (hence, the heat transfer area remains constant), derive an analytical expression for T2, the temperature inside Tank 2 as a function of time. Also, compute T2 at the time when Tank 1 is completely emptied. Use the following data: CP = liquid heat capacity = 2.0 kJ/kg-°C TS = heating coil temperature = 200 °C

UA = 40 kJ/min-°C

Assume there is no phase change in either tank (i.e. no boiling occurs). (b) Consider again a similar 2-tank system in which Tank 1 is jacketed and is heated with steam at 200 °C as shown in the figure below. In this case, the heat transfer area is no longer constant and will vary with the liquid volume inside Tank 1. 20 kg/min 40 °C

T1 Steam

T2 10 kg/min

30 kg/min

This liquid solution is known to boil at 150 °C. Determine the time it takes for the liquid in Tank 1 to boil off completely (including the time to heat the liquid to the boiling temperature). Use the following data:

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CP = liquid heat capacity = 2.0 kJ/kg-°C U = overall heat transfer coefficient = 25 kJ/m2-min-°C TS = steam temperature = 200 °C φ = liquid mass density = 1200 kg/m3 R = radius of both cylindrical vessels = 0.3 m λ = heat of vaporization at 150 °C = 1500 kJ/kg Use the initial conditions given in Part (a)

Solution: (a) Derive an analytical expression for T2 as a function of time. 20 kg/min 40 °C

T1 T2 10 kg/min

Mass balances: dM1 = –10 dt

-----> M1 = 300 – 10t

dM2 = 0 dt

-----> M2 = 100

Energy balances: d[M1CPT1] = UA(TS – T1) – w1CPT1 dt

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CP M1(dT1/dt) + CPT1(dM1/dt) = UA(TS – T1) – w1CPT1 300 – 10t

2 (300 – 10t) dT1 – 20T1 = dt

40(200 – T1) – 20T1

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30 kg/min

0.05

dT1

=

(200 – T1)

ln

200 – T1 180

1/2

0.1 dt

----> Integrate from T1 = 20 to T1 and from t = 0 to t.

(30 – t)

= ln

30 – t 30

T1 = 200 – 0.2(30 – t)2

d[M2CPT2] = w1CPT1 + w2CPTF – w3CPT2 dt 0 CP M2(dT2/dt) + CPT2(dM2/dt) = w1CPT1 + wFCPTF – w2CPT2

200 dT2 = 20T1 + 1600 – 60T2 dt 10 dT2 = [200 – 0.2(30 – t)2] + 80 – 3T2 dt dT2 + 0.3T2 = 10 + 1.2t – 0.02t2 dt 1st-order constant-coefficient heterogeneous ODE! Solve using the method of undetermined coefficient. T2 = c1exp[-0.3t] + c2t2 + c3t + c4 At t = 0, T2 = 30

so that c1 + c4 = 30

(dT2/dt) = -0.3c1exp[-0.3t] + 2c2t + c3 = 10 + 1.2t – 0.02t2 – 0.3c1exp[-0.3t] – 0.3c2t2 – 0.3c3t – 0.3c4 Comparing the LHS and RHS: (0.02 + 0.3c2)t2 = 0

c2 = –0.06667

1.2t = (2c2 + 0.3c3)t

2c2 + 0.3c3 = 1.2

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c3 = 4.4445

10 = c3 + 0.3c4

c4 = 18.5183

c1 = 11.4817

T2 = 11.4817e-0.3t – 0.06667t2 + 4.4445t + 18.5183

Check: At t = 0, T2 = 11.4817 + 18.5183 = 30 °C

When Tank empties, M1 = 300 – 10t = 0

Correct

t = 30 minutes

T2 = 91.85 °C

Therefore, at t = 30 min:

(b) First calculate the time for liquid in Tank 1 to reach 150 °C: T1 = 200 – 0.2(30 – t)2 = 150

t = 14.19 minutes

When calculating this time, I treated the heat transfer area as being constant. In reality, it is not because the liquid moves up and down although there is no boiling yet. If you take into account the non-constant area, the answer is 14.94 minutes, but the math is somewhat complex. This gives z1 = 0.4442 m.

Energy balance after boiling occurs in Tank 1:

dM1 = –10 – v dt

but v = q/λ

= –10 – UAT(200 – T1)/λ = –10 – UAT(200 – T1)/λ = –10 – (25)(4V/D + πD2/4 )(200 – 150)/1500 = –10 – (25)(6.6667V+0.28274)(50)/1500 = –10.23562 – 5.5556V But φ = M1/V or M1 = φV V(t = 14.94 min) = 0.4442π(0.3)2 = 0.1256 m3

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1200 dV = –10.23562 – 5.5556V dt 1200 _______dV_______ 10.23562 + 5.5556V

=

–dt

Integrate from V = 0.25 to 0 (when the vessel is empty) and time from 0 to t: (1200/5.5556) ln[10.23562/(10.23562+5.5556*0.1256)] = -t

t = 14.24 minutes So the total time it takes to boil off Tank 1 = 14.94 + 14.24 = 29.18 minutes

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