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

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.

The following data are known about the system:

CV = 1×10-5 m3/Pa1/2-min (characteristic valve constant) P0= 1.01325×105 Pascal

Cross-sectional area of rectangular tank = 3 m2 Radius of sphere R = 2 m 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

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:

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

F1 = 0.1 m3/min F3 = 0.3 m3/min F2 P0 P2 P1 P z2 z1 Air 3 3 2 z z R V =

π

−

π

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:

Mass balance around the rectangular tank:

2 1 1 F F dt dV φ φ φ = − 2 1 3 2 1 ) 3 ( F F dt z z R d

φ

φ

π

π

φ

− = − 2 1 1 2 1 1 ) 2 ( F F dt dz z Rz −

π

= −

π

2 1 1 2 1 1 ) 0.1 4 ( C P P dt dz z z −

π

= − _V −

π

3 2 2 F F dt dV

φ

φ

φ

= − 3 . 0 2 1 2 2 =C P −P − dt dz A _V 2 1 5 1 2 1 1 ) 0.1 1 10 4 ( P P dt dz z z −

π

= − × − −

π

3 . 0 10 1 3 2 _{= ×} −5 _{1− 2 −} P P dt dz 1 0 1 P gz P = +

φ

P₂ =P+

φ

gz₂ nRT nRT P= =

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) 2 2 10 810600 3 30 ) 15 . 298 )( 8314 )( 981 . 0 ( z z P − = − = ∴ ) sec / / 1 )( sec)( / 80665 . 9 )( / 1000 ( 10 810600 2 2 2 3 2 2 kg m m z m N kg m z P + − − = ∴ 2 2 2 9806.65 10 810600 z z P + − = ∴ 1 1 101325 9806.65z P = + ∴ 1 . 0 10 810600 ) ( 65 . 9806 101325 10 333 . 3 2 2 1 6 2 ₋ − − − + × = − z z z dt dz

π

) 4 ( 10 810600 ) ( 65 . 9806 101325 10 1 1 . 0 1 1 2 2 1 5 1 z z z z z dt dz − − − − + × − = −

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;

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:

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.

(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. F2 F1 = 15 ft3/min z1 1 2 C A 2gz F = _d z2 F3 = 20 ft3/min

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:

Now, do variable transformation by defining:

u = 2.37990 – z11/2 z11/2 = 2.37990 – u ft z state steady at dt dz z dt dz z dt dz z gz A C F F dt dV d 664 . 5 0 ) 37990 . 2 ( 10 30279 . 6 30279 . 6 15 100 ) 3600 )( 2 . 32 ( 2 12 1 )( )( 6 . 0 ( 15 2 15 max 1 1 1 2 1 1 1 1 2 1 2 1 1 = → = − × = − =       − = − = − = −

π

∫

₌

∫

_×

−

₌

_×

−

−

1 2 2 0 2 1 1

10

30279

.

6

10

30279

.

6

37990

.

2

z t

t

dt

z

dz

du

u

du

z

dz

or

dz

z

du

2

2

(

2

.

37990

)

2

1

₂1 1 1 1 2 1 1

=

−

=

−

−

−

=

−

∫

=

−

∫

−

−

1 2 1 1

2

.

37990

2

37990

.

2

z

du

u

u

z

dz

1 1 2 2 1 1 2 2 1 1

2

2

.

37990

37990

.

2

ln

7598

.

4

z z

z

z

_











−

+













−

−

=





















−

−

−

=

965686

.

0

37990

.

2

ln

99016

.

75

)

414214

.

1

(

73198

.

31

2 1 1 2 1 1

z

z

t

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

steady-state 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:

So the two ODEs are:

        ₋ = − =             =       = − =               − = − = 2 2 1 2 1 2 2 2 2 2 3 2 2 2 1 3 2 2 1 3 2 2 20 30279 . 6 09296 . 5 20 30279 . 6 20 5 3 1 20 30279 . 6 3 1 20 2 z z dt dz z dt dz z r R H z from z H R V because z z H R dt d gz A C F F dt dV d

π

π

π

15 ) 0 ( 2 ) 0 ( . . 20 30279 . 6 09296 . 5 ) 37990 . 2 ( 10 30279 . 6 2 1 2 2 1 2 1 2 1 = =         − = − × = − z and z t s z z dt dz z dt dz

%

% 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 2.0000 15.0000 1.0000 2.0602 14.7463 2.0000 2.1191 14.4867 3.0000 2.1767 14.2207 4.0000 2.2331 13.9475 5.0000 2.2883 13.6664 6.0000 2.3424 13.3764 7.0000 2.3954 13.0766 8.0000 2.4474 12.7657 9.0000 2.4982 12.4423 10.0000 2.5481 12.1046 11.0000 2.5970 11.7506 12.0000 2.6450 11.3775 13.0000 2.6920 10.9821 14.0000 2.7382 10.5601 15.0000 2.7834 10.1058 16.0000 2.8279 9.6110 17.0000 2.8715 9.0645 18.0000 2.9143 8.4483 19.0000 2.9563 7.7329 20.0000 2.9975 6.8611

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.

(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

F2 M2 T2 F1 M1 T1 Tank 1 Tank 2 50 0 ₁ 1 2 1 _{= − = → =} M F F dt dM

Mass and energy balances around the second vessel

The two ODEs are:

(

)

1 1 2 2 1 1 , 1 1 ) (T T F C T FC T UA dt T C M d P P S P = − + − 1 1 2 2 1 1 1 1 1 UA(T T ) FC T FC T dt dM T dt dT M CP = S − + P − P    ₊ 1 2 1 1 2 2 ) 100 )( 5 . 0 )( 0 . 4 ( 50 T T T dt dT − + − = → − + = 2 1 1 08 . 0 04 . 0 4 T T dt dT 100 2 25 1 ₁ 2 = + T − dt dT T 100 0 1 2 1 2 _{= − = → =} M F F dt dM

(

)

2 2 1 1 2 2 , 2 2 ) (T T FC T F C T UA dt T C M d P P S P _{= − + −} 2 2 1 1 2 2 , 2 2 2 2 UA(T T ) FC T FC T dt dM T dt dT M C_P = _S − + _P − _P     ₊ 2 1 1 2 2 2 ) 120 )( 5 . 0 )( 0 . 4 ( 100 T T T dt dT − + − = 2 1 2 4 2 240 100 T T dt dT − + = 2 1 2 04 . 0 02 . 0 4 . 2 T T dt dT − + = 20 ) 0 ( . . 08 . 0 04 . 0 4 _{2 1 1} 1 _{= + − =} T t s T T dt dT dT

(b) Now combine the two ODEs to derive T1(t) and T2(t) First, take another derivative of the first-order ODE for T1:

Substituting the second ODE for T2 into the above equation:

Now eliminating T2 in the above equation:

This is a second-order ODE with constant coefficients which can be solved by the method of undetermined coefficients.

Now, we need to determine the values of c1, c2, and c3 using initial conditions:

dt dT dt dT dt T d 2 1 2 1 2 08 . 0 04 . 0 − =

[

]

dt dT T T dt dT T T dt T d 1 2 1 1 2 1 2 1 2 08 . 0 0016 . 0 0008 . 0 096 . 0 08 . 0 04 . 0 02 . 0 4 . 2 04 . 0 + − − = + − − = dt dT T dt dT T dt T d ₁ 1 1 1 2 1 2 08 . 0 100 2 25 0016 . 0 0008 . 0 096 . 0 −     _{+ −} − + = dt dT T dt dT T dt T d ₁ 1 1 1 2 1 2 08 . 0 16 . 0 0032 . 0 04 . 0 0008 . 0 096 . 0 + − − + − = dt dT T dt T d ₁ 1 2 1 2 12 . 0 0024 . 0 256 . 0 − − = 256 . 0 0024 . 0 12 . 0 1 1 2 1 2 = + + T dt dT dt T d 02535 . 0 09465 . 0 0 0024 . 0 12 . 0 2 _{+ + = → =− −} and

λ

λ

λ

3 2 1 1(t) c exp( 0.09465t) c exp( 0.02535t) c T = − + − + 20 ) 0 ( _{1 2 3} 1 =c +c +c = T ) 02535 . 0 exp( 02535 . 0 ) 09465 . 0 exp( 09465 . 0 1 2 1 t c t c dt dT − − − − =

Plugging the analytic expression into the ODE for T1:

Solving for c1 and c2:

Therefore,

[

]

70 63375 . 0 36625 . 2 10 100 ) 20 ( 2 02535 . 0 09465 . 0 25 ) 0 ( 100 2 25 2 1 2 1 2 1 1 2 = − − = − + − − = → − + = c c c c T T dt dT T ) 02535 . 0 exp( 10 4262 . 6 ) 09465 . 0 exp( 00896 . 0 2 4 1 2 1 2 t c t c dt T d − × + − = − 667 . 106 0024 . 0 256 . 0 0024 . 0 12 . 0 3 3 1 1 2 1 2 = = = + + c c T dt dT dt T d 667 . 86 2 1 + c =− c 9656 . 77 7014 . 8 ₂ 1 =− c =− c ) 02535 . 0 exp( 9656 . 77 ) 09465 . 0 exp( 7014 . 8 667 . 106 ) ( 1 t t t T = − − − −

[

]

[

106.667 8.7014exp( 0.09465 ) 77.9656exp( 0.02535)

]

100 2 ) 02535 . 0 exp( 9764 . 1 ) 09465 . 0 exp( 8236 . 0 25 100 2 25 2 1 1 2 − − − − − + − + − = − + = t t t t T T dt dT T ) 02535 . 0 exp( 5212 . 106 ) 09465 . 0 exp( 1872 . 3 334 . 113 ) ( 2 t t t T = + − − −

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

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:

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

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); ) 02535 . 0 exp( 9656 . 77 ) 09465 . 0 exp( 7014 . 8 667 . 106 ) ( 1 t t t T = − − − − ) 02535 . 0 exp( 5212 . 106 ) 09465 . 0 exp( 1872 . 3 334 . 113 ) ( 2 t t t T = + − − − 0 667 . 6 ) 02535 . 0 exp( 5556 . 28 ) 09465 . 0 exp( 8886 . 11 2 1 =T → − t − − t + = T 100 ) 02535 . 0 exp( 9656 . 77 ) 09465 . 0 exp( 7014 . 8 667 . 106 ) ( 1 t = − − t − − t = T 100 ) 02535 . 0 exp( 5212 . 106 ) 09465 . 0 exp( 1872 . 3 334 . 113 ) ( 2 t = + − t − − t = T

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 V T Tank Outlet F4 T4 Jacket Vj 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 = liquid density in tank V = liquid density in tank

F1 = tank inlet flow F2 = tank inlet flow

F3 = jacket flow (inlet and outlet) T1 = tank inlet temperature T2 = tank outlet temp T3 = jacket inlet temp T4 = jacket outlet temp

CP = liquid heat capacity in tank Pj = liquid density in jacket Cij = liquid heat capacity in jacket U = overall heat transfer coefficient A = heat transfer area

Vj = liquid volume in jacket Overall mass balance in the tank:

ρ ρ ρ 2 1 1 ) ( F F dt V d − =

Assume ρi = ρ and V is constant

⇒ ( ) =0

dt V

d ρ _⇒

F1 = F2

Energy balance in the tank:

) ( ) ( ) ( )) ( ( 2 4 2 2 1 2 2 T T UA T T C F T T C F dt T T C V d rep p ref p ref p − + − − − = −

ρ

ρ

ρ

Assume Cp is not a function of temperature 0 ) ( 2 2 2 2 2 1 2 2 T T UA T C F T C F T C F T C F dt dT C V dt dT C V p p ref p p ref j ref p P −

ρ

=

ρ

−

ρ

−

ρ

+

ρ

+ −

ρ

or ( 1 2) ( 4 2) 2 2 T T C V UA T T V F dt dT p − + − =

ρ

(1)

) ( ) ( 3 4 4 2 3 4 T T C V UA T T V F dt dT pi j j j − − − =

ρ

(2)

There are 2 modeling equations.

(b) min 1 3 1 ft F = 61.3 ₃ Fft Btu pC_P = _o 61.3 3 Fft Btu C pj pj = o F T₁ =50o T₂ =125oF V =10ft 3 F T o 200 3 = T F o 150 4 = 3 1ft V_j =

Find F3 and UA at steady state:

At steady–state: 2 ₌0 dt dT and 4 ₌0 dt dT Eq. (1) becomes 0 ) 125 150 ( ) 3 . 61 )( 10 ( ) 125 50 ( 10 1 = − + − UA

and Eq. (2) gives

0 ) 125 150 ( ) 3 . 61 )( 1 ( ) 150 200 ( 1 3 _{− −} UA _{− =} F ⇒ UA = 183.9 Btu/ oF min F3 = 1.50 ft3 / min

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

) ( 3 . 0 ) 50 ( 1 . 0 _{2 4 2} 2 T T T dt dT − + − = =5 +0.3T −₄ 0.4T₂ ) ( 3 ) 200 ( 5 . 1 _{4 4 2} 4 _{T T T} dt dT − − − = = 300 – 45 T4 + 3T2 dt dT dt dT dt T d _{4 2} 2 2 2 4 . 0 3 . 0 − =

[

]

dt dT T T 2 2 4 3 0.4 5 . 4 300 3 . 0 − + − = dt dT T T 2 2 4 0.9 0.4 35 . 1 90 − + − = But ( 0.4 5) 3 . 0 1 2 2 4 = + T − dt dT T So that dt dT T dt dT dt T d ₂ 2 2 2 2 2 4 . 0 5 . 22 9 . 0 5 . 4 90 − − + − = 5 . 112 9 . 0 9 . 4 2 ₂ 2 2 2 = + + T dt dT dt T d

Solve by method of undetermined coefficients:

t X z t x h z t c e c e T 1 2 1 , ( ) = + 2 ) 9 . 0 )( 1 ( 4 9 . 4 9 . 4 ; 0 9 . 0 9 . 4 2 2 _{+ + = =} − ± − X X X x1 = -0.1911 x2 = −4.7089 t t n t c e c e T_2, ( ) = ₁ −0.1911 + ₂ −4.7089 A t T_2,p( ) = (trial function) A e c e c T T t T₂( ) = _2,n + _2,p = ₁ −0.1911t + ₂ −4.7089t +

Solve for 3 constants C₁, C2 and A

At t ∞ : T (t ∞) = A = 125 So T₂ =c₁e −0.1911t +c₂e −4.7089t +125 At t = 0 , T2 = 50 oF 50 = C1 + C2 + 125 or C1 + C2 = -75 Recall T4 = _      − +0.4 5 3 . 0 1 2 2 T dt dT =

[

0.1911 4.7089 0.4 , 0.4 50 5

]

3 . 0 1 4.7089 2 1911 . 0 7089 . 4 1911 . 0 1 − 2 + + + − − − t − t − t − t e c e c e c e c T4 = 0.6963c₁e−0.1911t −14.363c₂e −4.7089t +150

At t = 0, T4 = 200 ⇒ 0.6963 C1 - 14.363 C2 = 50 0.6963 C1 − 0.6963C2 = −52.2225 C2 = −6.7880 C1 = −68.2120 ∴ T1 = 125 - 68.2120 t e−0.1911 - 6.7880 e−4.7089t T4 = 150 – 47.4960 e−0.1911t +97.4960 e−4.7089t

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

(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 CP (gas) = 125 kJ/kmol-K qf = 1.0 kmol/min

Tf = 25 °C q = 0.5 kmol/min

Q = 1.5x104 kJ/min T(t=0) = 298.15 K = 25 °C

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 P, V, T

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

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 1 125(298.15 – 0.5T) + 1.5x104 – 0.5T 100P 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')

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.) Pressure(atm) Temperature(K) 0.0000 1.0000 298.1500 0.2124 1.0499 305.1033 0.4248 1.0992 311.5360 0.6372 1.1479 317.4987 0.8497 1.1961 323.0360 1.0997 1.2521 329.0610 1.3497 1.3075 334.6070 1.5997 1.3624 339.7234 1.8497 1.4167 344.4532 2.0997 1.4705 348.8343 2.3497 1.5238 352.8998 2.5997 1.5767 356.6792 2.8497 1.6292 360.1986 3.0997 1.6813 363.4813 3.3497 1.7331 366.5477 3.5997 1.7846 369.4165 3.8497 1.8357 372.1042 4.0997 1.8866 374.6256 4.3497 1.9372 376.9941 4.5997 1.9875 379.2217 4.8497 2.0376 381.3194 5.0997 2.0875 383.2970 5.3497 2.1372 385.1634 5.5997 2.1866 386.9268 5.8497 2.2359 388.5946 6.0997 2.2850 390.1736 6.3497 2.3339 391.6698 6.5997 2.3827 393.0890 6.8497 2.4313 394.4363 7.0997 2.4798 395.7165

8.0997 2.6724 400.2482 8.3497 2.7203 401.2517 8.5997 2.7680 402.2096 8.8497 2.8157 403.1247 9.0997 2.8632 403.9994 9.3497 2.9107 404.8361 9.5997 2.9580 405.6369 9.8497 3.0053 406.4039 9.8872 3.0124 406.5164 9.9248 3.0195 406.6282 9.9624 3.0266 406.7392 10.0000 3.0337 406.8496 >>

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.

20 kg/min Tank 1 Tank 2 40 °C 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 UA = 40 kJ/min-°C TS = heating coil temperature = 200 °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

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 30 kg/min Mass balances: dM1 = –10 ---> M1 = 300 – 10t dt dM2 = 0 ---> M2 = 100 dt Energy balances: d[M1CPT1] = UA(TS – T1) – w1CPT1 dt -10 CP M1(dT1/dt) + CPT1(dM1/dt) = UA(TS – T1) – w1CPT1 300 – 10t 2 (300 – 10t) dT1 – 20T1 = 40(200 – T1) – 20T1 dt

0.05 dT1 = 0.1 dt ----> Integrate from T1 = 20 to T1 and from t = 0 to t. (200 – T1) (30 – t) ln 200 – T1 1/2 = ln 30 – t 180 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:

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 Correct

When Tank empties, M1 = 300 – 10t = 0 t = 30 minutes

Therefore, at t = 30 min: T2 = 91.85 °°°°C

(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 but v = q/λ dt = –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

1200 dV = –10.23562 – 5.5556V dt

1200 _______dV_______ = –dt 10.23562 + 5.5556V

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