CHE654
Modeling Chemical Processes
Using MATLAB
Homework Set #5 Solutions
Class-16
Prepared by
Dr. Hong-ming Ku
King Mongkut’s University of Technology Thonburi
Chemical Engineering Department
Chemical Engineering Practice School
53. Boiling and Draining of Ethanol in a Spherical Vessel
Consider the boiling of pure ethanol in a jacketed spherical vessel with a radius R = 1 meter, as shown in the diagram. The liquid is drained at the bottom of the vessel while some of it is boiled and escapes as vapor through the top of the vessel.
Steam TS 1 z CV
The following data are available:
MW = 46.07 λ(TB) = 3.858x107 J/kmol
ρ = 16.575 kmol/m3
CP = 1.4682x105 J/kmol-°C
log10 Pvap = 8.04494 − __1554.30__ T in °C and Pvap in mmHg T + 222.65
CV = 4.5 m3-atm-1/2/hr = valve constant
TS = 100 °C U = 2.0x106 J/hr-m2-°C P0 = 1 atm z0 = z(t=0) = 1 m
(a) Model this operation and use MATLAB to determine the time it takes for the vessel to completely empty, assuming that initially the liquid is at its boiling point. Note that the heat transfer area AT is not constant.
(b)Repeat the calculations in Part (a), assuming that there is no draining of the liquid at the bottom (i.e. the liquid leaves the vessel only through boiling). Determine the solution analytically (an exact solution is possible in this case).
Useful conversion factors and formulae: g = gravitational constant = 9.807 m/s2
1 atm = 1.01325x105 N/m2 1 N = 1 kg-m/s2
The volume of liquid V(z) in a spherical vessel as a function of its height z is given by V(z) = πRz2 − πz3
3
while the surface area S(z) is given by 1 atm v
S(z) = 4πRz − πz
Solution:
(a) First, calculate TB:
log10 (760) = 8.04494 – 1554.3/(TB + 222.65) TB = 78.33 °C
Overall mass balance:
ρ (dV/dt) = −v − ρCv√ P – P0
dV/dt = −0.06033 v − 4.5√ P – 1
But P = P0 + φgz = 1 atm + (763.61 kg/m3) ( 1 atm ) (9.807 m/s2)( 1 N ) 1.01325x105 N/m2 (1 kg-m/s2) P = 1 + 0.073908z dV/dt = −0.06033 v − 1.22337√z Energy balance: v = q/λ = (2.0x106)(100 – 78.33)AT = 1.12338(4πRz – πz2) 3.858x107 (2πz – πz2) (dz/dt) = –0.27109πz + 0.06777πz2 – 1.22337√z z(2 – z)(dz/dt) = –0.27109z + 0.06777z2 – 0.38941√z dz = 0.27109 – 0.06777z + 0.38941z-1/2 dt (z – 2)
[
]
∫
=
+
−
+
+
2)
ln(
b
bx
a
a
bx
a
dx
bx
a
x
s.t. z(t = 0) = 1 m
Determine t at z = 0 using MATLAB t = 1.6966 hours or 101.8 minutes to empty
The MATLAB Script File: %
% CHE654: Problem #53 %
% Boiling and Draining of Ethanol in a Spherical Vessel % clc clear format long drain_t = 1; initZ = 1; % Part (a) [t, z] = ode45('spherea',drain_t,initZ); while z(length(z),1) > 0 drain_t = drain_t + 0.01; [t, z] = ode45('spherea',drain_t,initZ); end [t, z] plot(t,z)
title('Boiling and Draining of Ethanol in a Spherical Tank') xlabel('Time in Hours')
ylabel('Ethanol Height in Meters') % Part (b) drain_t = 1; [t, z] = ode45('sphereb',drain_t,initZ); while z(length(z),1) > 0 drain_t = drain_t + 0.01; [t, z] = ode45('sphereb',drain_t,initZ); end [t, z]
The MATLAB M-File for the ODE: function dzdt = ode(t,z)
dzdt = zeros(1,1);
(b) Repeat the calculations in Part (a) analytically if there is no drainage. dV/dt = −0.06033 v (2πz – πz2) (dz/dt) = –0.27109πz + 0.06777πz2 (2 – z)(dz/dt) = –0.27109 + 0.06777z (z – 2)(dz/dt) = 0.06777(4 – z) 0 0 t
∫
z dz –∫
2 dz =∫
0.06777 dt 1 4 – z 1 4 – z 0 [4 – z – 4ln(4–z)] + 2ln(4–z) = 0.06777 t 0 4 – z – 2ln(4–z) = 0.06777 t 11.22741 – 0.802775 = 0.06777 t
t = 6.266 hours or 375.9 minutes MATLAB gives an answer of about 6.27 hours.
The MATLAB M-File for the ODE: function dzdt = ode(t,z)
dzdt = zeros(1,1);
54. Boiling of Water in a Closed Cylindrical Vessel
A closed cylindrical vessel (R = 0.5 m and H =5.0 m) contains 500 kg of water at its boiling temperature of 100 °C (373.15 K) as shown in the figures below. Let’s assume that initially at t = 0, the water (500 kg) is already at 100 °C and starts to boil. Initially, there is also 5 kg of water vapor (or steam) in the gas phase of the vessel. We assume the water vapor is an ideal gas and that the temperature in the gas and liquid phases are always in thermal equilibrium at 100 °C.
The following data are also known about water:
MW (molecular weight) = 18.015 kg/kmol
φ
(mass density) = 1000 kg/m3λ
(heat of vaporization) = 2256.84 kJ/kg CP (liquid heat capacity) = 4.1813 kJ/kg-KR (universal gas constant) = 0.08206 atm-m3/kmol-K
This problem is divided into two parts, namely Part 1 and Part 2.
Part 1: In this part, water is being boiled by a heating coil at TS = 200 °C with a constant heat transfer area A = 2.0 m2 and an overall heat transfer coefficient U = 100 kJ/min-m2-°C as shown in the figure for Part 1.
(a) Derive an analytical expression of the liquid height z as a function of time.
(b) Derive an analytical expression of the water gas pressure PG as a function of time. Plot your analytical solution in MATLAB and run the model until all the water has vaporized. Hint: You may check the correctness of your analytical solution by checking PG at t = 0 and at t when all water has been vaporized.
Answer the following questions:
How long does it take for water in Part 1 to boil off completely? ____________ minutes
Water VL v
z
PG VG
Figure for Part 1 Figure for Part 2
Steam
v PG VG
VL Water
What is the gas pressure in Part 1 at the following conditions? (i) PG = ____________ atm after 30 minute
(ii) PG = ____________ atm when water has boiled off completely
Part 2: Now, let’s study the same system in which the cylindrical vessel is being heated by steam in a jacketed chamber as shown in the figure for Part 2. In this case, the heat
transfer area decreases with the liquid height and is no longer constant. The heating now comes from the surrounding wall and the bottom of the vessel. We assume U and TS to be
the same as in Part 1.
(a) Derive an analytical expression of liquid volume VL as a function of time.
(b) Derive a single ODE for the gas pressure PG using the analytical expression in Part 2(a). Then use MATLAB (ode45) to solve the ODE for PG from t = 0 to t when all liquid is gone, and make a plot of the PG profile. Note that the final value of PG when the vessel contains only gas in Part 2(b) should be equal to that in Part 1(b).
Answer the following questions:
How long does it take for water in Part 2 to boil off completely? ____________ minutes
What is the gas pressure in Part 2 at the following conditions? (i) PG = ____________ atm after 30 minutes
(ii) PG = ____________ atm when water has boiled off completely
Solution:
Part 1(a): Constant heat transfer area Mass balance for the liquid phase:
Energy balance for the liquid phase:
v dt dVL =− φ
λ
φ
λ
φ
) ( ) ( ) ( ) ( B P B S L B P B P B S B P L T vC v T T UA dt dV T C T C v T T UA dt T C V d − − − = + − − =The mass balance equation then becomes:
When z = 0, t = 56.44 minutes
Part 1(b): Constant heat transfer area Mass balance for the vapor phase:
m m m kg kg t z z 0.6366 ) 5 . 0 )( )( / 1000 ( ) 500 ( ) 0 ( 3 2 0 = = = =
π
2 2 ( ) ( ) R T T UA dt dz T T UA dt dz R S B S Bφλπ
φλ
π
=− − → =− − t z t t R T T UA z z S B 01128 . 0 6366 . 0 ) 5 . 0 )( )( 84 . 2256 )( 1000 ( ) 100 200 )( 2 )( 100 ( 6366 . 0 ) ( 2 2 0 − = − − = − − =π
φλπ
B G G B G G RT V P n nRT V P = → = ) ( ) ( MW vRT dt V P d G G B =λ
) (TS TB UA v dt dn MW = = − ) (MW vRT dt dP V dt dV P G B G G G + = ) ( ) ( ) ( MW vRT dt dP V V dt V V d P G B L T L T G + − = − ) ( ) ( MW vRT dt dP V V dt dV P G B L T L G + − = −But
But
Let’s check PG at t = 0 and when water completely vaporized: At t = 0, atm PG 2.478 ) 5 . 0 927 . 3 ( ) 15 . 373 )( 08206 . 0 )( 015 . 18 / 0 . 5 ( 0 = − = ) ( ) ( MW vRT dt dP V V v P G B L T G + − = φ 927 . 3 ) 5 ( ) 5 . 0 ( 008862 . 0 5 . 0 ) ( ) ( 2 2 0 = = = − = − − = → − − = π π φλ φλ H R V and t T T UA V V T T UA dt dV T B S L L B S L − − = − = − φ λ φ G B B S G B G L T P MW RT T T UA P MW RT v dt dP V V ) ( ) ( ) ( ) ( − = + − 1000 015 . 18 ) 15 . 373 )( 08206 . 0 ( 8619 . 8 ) 008862 . 0 5 . 0 927 . 3 ( G PG dt dP t
[
G]
G P dt dP t = − +0.008862 ) 0.0088621699.733 427 . 3 ( t dt P dP t P P G G G G 3.427 0.008862 008862 . 0 733 . 1699 0 0 + = −∫
∫
+ = − 3.427 008862 . 0 427 . 3 ln 733 . 1699 255 . 1697 ln t PG t P t P G G 3.427 0.008862 427 . 3 255 . 1697 733 . 1699 427 . 3 008862 . 0 427 . 3 733 . 1699 255 . 1697 + = − → + = − t PG 008862 . 0 427 . 3 493 . 5816 733 . 1699 + − = atm PG 2.478 427 . 3 493 . 5816 733 . 1699 − = =At t = 56.44 min,
Check against the ideal gas law:
At t = 30.0 min, PG = 124.67 atm
MATLAB script to plot the derived equation:
clc clear time(1) = 0; PG(1) = 2.478; for t = 1:56 time(t+1) = t; PG(t+1) = 1699.733-5816.493/(3.427+0.008862*t); end time(58) = 56.44; PG(58) = 1699.733-5816.493/(3.427+0.008862*56.44); plot(time,PG)
title('Gas Pressure vs. Time') xlabel('Time in Minutes') ylabel('Gas Pressure in atm')
atm PG 218.64 ) 44 . 56 ( 008862 . 0 427 . 3 493 . 5816 733 . 1699 = + − = atm PG 218.58 ) 927 . 3 )( 015 . 18 ( ) 15 . 373 )( 08206 . 0 )( 5 500 ( = + =
Part 2(a): Varying heat transfer area
Time for water to completely vaporize = 71.44 min when VL = 0
Part 2(b): Varying heat transfer area
φλ π φλ ) ( 2 ) ( 2 B S L L B S L T T R V R U dt dV T T UA dt dV − + − = → − − = ) 4 7854 . 0 ( 004431 . 0 ) 84 . 2256 )( 1000 ( ) 100 200 ( 5 . 0 2 ) 5 . 0 ( ) 100 ( 2 L L L V V dt dV + − = − + − = π t V dt V dV L L L 01772 . 0 7854 . 2 4 7854 . 0 ln 004431 . 0 ) 4 7854 . 0 ( =− + → − = + 19635 . 0 ) 01772 . 0 exp( 69635 . 0 − − = t VL
− − = − φ λ G B B S G L T P MW RT T T UA dt dP V V ) ( ) ( ) ( − − + = −
φ
λ
π
G B B S L G L T P MW RT T T R V R U dt dP V V ) ( ) ( 2 ) ( 2 − − + = − 1000 ) 015 . 18 ( ) 15 . 373 )( 08206 . 0 ( 84 . 2256 ) 100 200 ( 5 . 0 2 ) 5 . 0 ( ) 100 ( ) 927 . 3 ( 2 G L G L P V dt dP Vπ
− + = − 1000 6997 . 1 ) 4 7854 . 0 ( 4310 . 4 ) 927 . 3 ( G L G L P V dt dP V 478 . 2 ) 0 ( . . 1000 6997 . 1 69635 . 0 12335 . 4 3421 . 12 01772 . 0 01772 . 0 = − − = − − G G t t G P t s P e e dt dPMATLAB Script File: clc clear format short simtim = [0:0.01:71.44]; initPG = [2.478];
[t, PG] = ode45('pg', simtim, initPG); [t, PG]
plot (t,PG)
title ('Gas Pressure vs Time When Heat Tranfer Area Is Not Constant') xlabel ('Time in Minutes')
ylabel ('Gas Pressure in Atm') function fPG = ode(t, PG) fPG = zeros(1,1); fPG(1) = 12.3421*exp(-0.01772*t)*(1.6997-PG(1)/1000)/(4.12335-0.69635* ... exp(-0.01772*t)); At t = 30 min, PG = 133.71 atm At t = 71.44 min, PG = 218.62 atm
58. Batch Reactor with a Series Reaction
Consider a batch reactor with a series reaction where component A reacts to form component B. Component B can also react reversibly to form component C. The reaction scheme can be characterized by:
k1 k2f
A B C k2r
Here k2f and k2r represent the kinetic rate constants for the forward and reverse reactions for the conversion of B to C, while k1 represents the rate constant for the conversion of A to B.
(a) Assuming that each of the reactions is first-order and constant volume, write down the 3 modeling equations for CA, CB, and CC, where CA, CB, and CC represent the concentrations (mol/volume) of components A, B, and C, respectively.
(b)Using the following definitions:
Dimensionless time, τ = k1 t
Conversion of A, x1 = (CA0 – CA) / CA0 Dimensionless concentration of B, x2 = CB / CA0
Ratio of rate constants, α = k2f / k1 Ratio of forward and reverse rate constants, β = k2r / k1
Derive a second-order ODE for the dimensionless concentration of B. Your ODE must contain only dimensionless quantities (x2, τ, α, and β).
(c) Solve the ODE in Part (b) analytically to find x2 as a function of τ, α, and β. (d) Using the following data:
k1 = 1.0 min-1 k2f = 1.5 min-1 k2r = 2.0 min-1 CA0 = CA (t = 0) = 3.0 mol/liter
CB0 = CB (t=0) = 0 mol/liter CC0 = CC (t=0) = 0 mol/liter
Solve for CB analytically as a function of time.
(e) Given the data in Part (d), find the maximum concentration of B and the reaction time at this concentration. If no such maximum exists, prove it mathematically. Repeat the above calculations for the case of k1 = 3 min-1, k2f = 1.5 min-1, k2r = 1 min-1 while the initial conditions remain the same.
(f) Validate your analytical solutions by solving the differential equations in Part (a) with MATLAB and plot the time profiles of components A, B, and C. Use both sets of rate constants (i.e. k1 = 1.0 min-1, k2f = 1.5 min-1, k2r = 2.0 min-1 and k1 = 3.0 min-1, k2f = 1.5 min-1, k2r = 1.0 min-1).
Solution:
k1 k2f A B C k2r
(a) There are 3 modeling equations which are : dCA/dt = –k1CA dCB/dt = k1CA – k2fCB + k2rCC dCC/dt = k2fCB – k2rCC (b) Define τ = k1t x1 = (CA0 – CA) / CA0 x2 = CB / CA0 α = k2f / k1 β = k2r / k1
Need a 2nd-order ODE for dimensionless concentration of B starting with: CA = CA0 e-k1t ⇒ 1 – x1 = e-τ or x1 = 1 – e-τ d2CB/dt2 = k1dCA/dt – k2fdCB/dt + k2rdCC/dt d2CB/dt2 = –k12CA0 e-k1t – k2f dCB/dt + k2r(k2f CB – k2rCC) = –k12CA0 e-k1t – k2f dCB/dt + k2r k2f CB – k2r[dCB/dt + k2f CB – k1 CA0 e-k1t] = ( k1k2r – k12) CA0 e-k1t – (k2r + k2f)dCB/dt
Introducing dimensionless variables:
d[dCB/dt]/dt = [1/(1/k1)dτ] [CA0dx2/(1/k1)dτ] = k12CA0d2x2/dτ2
–(k2r + k2f)dCB/dt = –k1(k2r + k2f)CA0dx2/dτ
(k1k2r – k12)CA0 e-k1t = (β-1)k12CA0e-τ
⇒ k12CA0d2x2/dτ2 = (β–1) k12CA0e-τ – (α+β)k12CA0dx2/dτ
⇒ d2
x2/dτ2 + (α+β)dx2/dτ + (1–β)e-τ = 0
d2x2/dττττ2 + (ααα+ββββ)dxα 2/dττττ = (ββββ–1)e-ττττ
(c) Solve for x2 analytically as a function of α, β, and τ
x2(τ) = C1e λ 1τ + C2eλ2τ + C3e-τ λ2 + (α+β)λ = 0 ⇒ λ1 = 0 , λ2 = –(α+β) ∴ x2(τ) = C1 + C2e-(α+β)τ + C3e-τ At t = 0 , CB = 0 ⇒ τ = 0 , x2 = 0 C1 + C2 + C3 = 0 dx2/dτ = –(α+β)C2e-(α+β)τ – C3e-τ d2x2/dτ2 = –(α+β)2C2e–(α+β)τ + C3e-τ d2x2/dτ2 + (α+β)dx2/dτ = C3(1–α–β)e-τ = (β–1)e-τ ⇒ C3 = (1–β)/(α+β–1) C1 + C2 + (1–β)/(α+β–1) = 0 At t = 0 or τ = 0 , dCB/dt = k1CA(t=0) – k2fCB(t=0) + k2rCC(t=0) = k1CA0
or dx2/dτ = 1 1 = –(α+β)C2 – C3 C2 = [–1/(α+β)][1+(1–β)/(α+β–1)] = –(1/α+β)[(α+β–1+1–β)/(α+β–1)] = α / [(α+β)(1–α–β)] C1 = –C2–C3 = α / [(α+β)(1–α–β)] + (β–1)/( α+β–1) = [α+(α+β)(β–1)]/[(α+β)(α+β –1)] ∴ ∴ ∴ ∴ x2(ττττ) = ββββ/(αααα+ββββ) + αααα/[(αααα+ββββ)(1–αα–ββββ)] eαα -(αααα+ββββ)ττττ + (1–ββββ)/(ααα+ββββ–1) eα -ττττ
(c) If k1 = 1 min-1 , k2f = 1.5 min-1 , k2r = 2 min-1 CA0 = 3 mol/liter , CB = CC = 0 when t = 0 ⇒ α = 1.5 and β = 2 x2(τ) = 0.57143 – 0.17143e-3.5τ – 0.4e-τ CB(t) = 1.71429 – 0.51429e-3.5t – 1.2e-t
(d) Find CB,max and tmax:
dCB/dt = 1.80e-3.5t + 1.2e-t = 0
⇒ –1.5e-3.5t = e-t
⇒ In (–1.5) – 3.5t = –t
⇒ no solution / root to the above equation!
Now k1 = 3 min-1 k2f = 1.5 min-1 k2r = 1 min-1 α = 0.5 β = 1/3 x2(τ) = 0.40 – 3.6e-0.8333τ – 4e-τ CB(t) = 1.2 + 10.8e-2.5t –12e-3t dCB/dt = –27e-2.5t + 36e-3t = 0 ⇒ –2.5t = In (1.3333) – 3t ⇒ tmax = 0.575 min ⇒ CB,max = 1.627 mol/liter (f)
The MATLAB Script File: %
% CHE654: Problem #58 %
% Batch Reactor with a Series Reaction % clc clear simtime=[0:0.1:5]; initC=[3,0,0]; [t,c] = ode45('batchr1',simtime,initC); [t,c] plot(t,c)
title('Batch Reactor with a Series Reaction') xlabel('Time in Minutes')
ylabel('Concentration of A, B, and C in Mole/Liter') legend('A','B','C')
%
[t,c] = ode45('batchr2',simtime,initC); [t,c]
plot(t,c)
title('Batch Reactor with a Series Reaction') xlabel('Time in Minutes')
ylabel('Concentration of A, B, and C in Mole/Liter') legend('A','B','C')
The MATLAB M-Files for ODE: function fv=batchr1(t,c) fv = zeros(3,1); fv(1) = -c(1); fv(2) = c(1)-1.5*c(2)+2*c(3); fv(3) = 1.5*c(2)-2*c(3); function fv=batchr2(t,c) fv = zeros(3,1); fv(1) = -3*c(1); fv(2) = 3*c(1)-1.5*c(2)+c(3); fv(3) = 1.5*c(2)-c(3);
The MATLAB Output: >> ans = 0 3.0000 0 0 0.1000 2.7145 0.2661 0.0194 0.2000 2.4562 0.4764 0.0674 0.3000 2.2225 0.6453 0.1322 0.4000 2.0110 0.7831 0.2059 0.5000 1.8196 0.8971 0.2833 0.6000 1.6464 0.9928 0.3608 0.7000 1.4898 1.0740 0.4363 0.8000 1.3480 1.1438 0.5082 0.9000 1.2197 1.2044 0.5759 1.0000 1.1036 1.2573 0.6391 1.1000 0.9986 1.3039 0.6975 1.2000 0.9036 1.3452 0.7512 1.3000 0.8176 1.3818 0.8006 1.4000 0.7398 1.4145 0.8457 1.5000 0.6694 1.4438 0.8868 1.6000 0.6057 1.4702 0.9242 1.7000 0.5480 1.4938 0.9582 1.8000 0.4959 1.5150 0.9891 1.9000 0.4487 1.5341 1.0172 2.0000 0.4060 1.5514 1.0426 2.1000 0.3674 1.5670 1.0656 2.2000 0.3324 1.5811 1.0865 2.3000 0.3008 1.5938 1.1054 2.4000 0.2722 1.6053 1.1226 2.5000 0.2463 1.6157 1.1381 2.6000 0.2228 1.6251 1.1521 2.7000 0.2016 1.6336 1.1648 2.8000 0.1824 1.6413 1.1763
2.9000 0.1651 1.6482 1.1867 3.0000 0.1494 1.6545 1.1961 3.1000 0.1351 1.6602 1.2046 3.2000 0.1223 1.6654 1.2124 3.3000 0.1107 1.6700 1.2193 3.4000 0.1001 1.6742 1.2256 3.5000 0.0906 1.6780 1.2314 3.6000 0.0820 1.6815 1.2365 3.7000 0.0742 1.6846 1.2412 3.8000 0.0671 1.6874 1.2454 3.9000 0.0607 1.6900 1.2493 4.0000 0.0549 1.6923 1.2527 4.1000 0.0497 1.6944 1.2559 4.2000 0.0450 1.6963 1.2587 4.3000 0.0407 1.6980 1.2613 4.4000 0.0368 1.6996 1.2636 4.5000 0.0333 1.7010 1.2657 4.6000 0.0302 1.7022 1.2676 4.7000 0.0273 1.7034 1.2693 4.8000 0.0247 1.7044 1.2709 4.9000 0.0223 1.7053 1.2723 5.0000 0.0202 1.7062 1.2736 ans = 0 3.0000 0 0 0.1000 2.2225 0.7212 0.0563 0.2000 1.6464 1.1648 0.1888 0.3000 1.2197 1.4227 0.3576 0.4000 0.9035 1.5589 0.5376 0.5000 0.6694 1.6167 0.7139 0.6000 0.4958 1.6264 0.8778 0.7000 0.3674 1.6073 1.0254 0.8000 0.2721 1.5730 1.1548 0.9000 0.2016 1.5319 1.2665 1.0000 0.1494 1.4890 1.3616 1.1000 0.1106 1.4478 1.4415 1.2000 0.0820 1.4098 1.5082 1.3000 0.0608 1.3758 1.5634 1.4000 0.0450 1.3462 1.6088 1.5000 0.0333 1.3206 1.6460 1.6000 0.0247 1.2990 1.6763 1.7000 0.0183 1.2809 1.7008 1.8000 0.0136 1.2658 1.7207 1.9000 0.0100 1.2533 1.7367 2.0000 0.0074 1.2430 1.7495 2.1000 0.0055 1.2346 1.7599 2.2000 0.0041 1.2278 1.7681 2.3000 0.0030 1.2223 1.7747
2.4000 0.0022 1.2178 1.7799 2.5000 0.0017 1.2142 1.7841 2.6000 0.0012 1.2113 1.7874 2.7000 0.0009 1.2090 1.7901 2.8000 0.0007 1.2071 1.7922 2.9000 0.0005 1.2057 1.7938 3.0000 0.0004 1.2045 1.7951 3.1000 0.0003 1.2036 1.7962 3.2000 0.0002 1.2028 1.7970 3.3000 0.0002 1.2022 1.7976 3.4000 0.0001 1.2018 1.7981 3.5000 0.0001 1.2014 1.7985 3.6000 0.0001 1.2011 1.7989 3.7000 0.0000 1.2009 1.7991 3.8000 0.0000 1.2007 1.7993 3.9000 0.0000 1.2005 1.7994 4.0000 0.0000 1.2004 1.7996 4.1000 0.0000 1.2003 1.7997 4.2000 0.0000 1.2003 1.7997 4.3000 0.0000 1.2002 1.7998 4.4000 0.0000 1.2002 1.7998 4.5000 0.0000 1.2001 1.7999 4.6000 0.0000 1.2001 1.7999 4.7000 0.0000 1.2001 1.7999 4.8000 0.0000 1.2001 1.7999 4.9000 0.0000 1.2000 1.8000 5.0000 0.0000 1.2000 1.8000
60. Isothermal Semi-Batch Reactor
(a) Consider an isothermal semi-batch reactor where a single reaction takes place in a solvent S, which is inert. In this reaction, 2 moles of component A react with one
k1 mole of component B to form one mole of component C: 2A + B ---> C. The reaction rate does not conform to the stoichiometry but is 1st-order with respect to each reactant as follows: rA = −k1CACB
FA = 15 liter/min FB = 10 liter/min
Initially (t = 0 min), the reactor contains 100 liters of solution and 300 moles of A. Assuming that all components have the same density of 60 mol/liter, derive 3 ODE equations needed to compute CA, CB, and CC, the concentrations (moles/liter) of A, B, and C, respectively. Use ode45 in MATLAB to solve for and plot (in a single graph) the concentrations of the 3 components. Run the model for 20 minutes with an increment of 0.5 minute.
The following experimental data have been obtained for this reaction when carried out in a batch reactor:
Time(minute) 0 5.0 7.5 12.0 15.5 25.0 32.0 40.0 CA(mol/liter) 2.0 1.65 1.52 1.40 1.28 1.17 1.10 1.06
CB(mol/liter) 0.5
(b) At the end of 20 minutes, the 2 feeds to the reactor are suddenly shut off, and 4,500 moles of a new component called D (same density as components A, B, and C) are charged to the reactor. That is, the reactor now operates in a batch mode. Component D reacts with component C to form A and B, and the reaction now looks as follows:
k1
2A + B C + D k2
with a reaction rate of rA = −k1CACB + k2CD (2nd -order forward and 1st-order reverse). The value of k2 has been measured to be 1.0 min-1. Derive analytically the concentration of A as a function of time, and compute CA at steady state based
on your derived equation. Your final expression should be simplified as much as possible and should not contain any parameters except t (time) and CA.
Useful Integrals: where B = a + bx + cx2 γ = 4ac – b2 ω = b + 2cx Solution:
(a) Must first determine the rate constant k using the graphical method. dCA = 2dCB CA – CA0 = 2(CB – CB0) CB = CB0 + ½ (CA – CA0) (dCA/dt) = –kCACB dCA = – kdt
CA[(CB0–0.5CA0)+0.5CA]
Integrate from CA0 to CA on the LHS and from 0 to t on the RHS. Integrate by partial fractions or use a table of integrals:
CA – [1/(CB0–0.5CA0)] ln[(CB0–0.5CA0)+0.5CA)/CA] = –kt CA0 ln(C /C ) – ln(C /C ) = (0.5C – C )kt
∫
+
−
=
+
x
bx
a
a
bx
a
x
dx
ln
1
)
(
0
tanh
2
0
2
0
tan
2
1 1<
−
−
−
=
−
=
>
+
− −∫
γ
γ
ω
γ
γ
ω
γ
γ
ω
γ
if
if
B
dx
if
So plot ln(CA/CB) vs. (0.5CA0– CB0)kt to obtain the value of k.
From MATLAB, get slope = 0.054203 k = 0.1084 liter/mole-min
dV = FA + FB = 15 + 10 = 25 V = 100 + 25t dt d(VCA)/dt = –kVCACB + ρAFA because 1 dNA = dCA = –kCACB V dt dt d(VCA)/dt = V(dCA/dt) + CA(dV/dt) = V(dCA/dt) + 25CA = –0.1084VCA[0 + 0.5(CA – 3.0)] + (60)(15) = –0.1084CA[0.5CA – 1.5]V + 900 dCA/dt = 0.1084CA(1.5 – 0.5CA) + (900 – 25CA)/(100 + 25t) s.t. CA(t = 0) = 3.0
d(VCB)/dt = –0.5kVCACB + ρBFB dCA/dt = 0.0542CA(1.5 – 0.5CA) + (600 – 25CB)/(100 + 25t) s.t. CB(t = 0) = 0 d(VCC)/dt = 0.5kVCACB dCC/dt = 0.0542CA(0.5CA – 1.5) – 25CC/(100 + 25t) s.t. CC(t = 0) = 0
The MATLAB Script File: %
% CHE654: Problem #60 %
% Isothermal Semi-Batch Reactor %
clc clear
format short e %
% First, determine the value of k t = [0 5.0 7.5 12.0 15.5 25.0 32.0 40.0]; CA = [2.0 1.65 1.52 1.40 1.28 1.17 1.10 1.06]; CB = 0.5+0.5*(CA-2.0); logy = log(CA./CB); [P,S] = polyfit(t,logy,1); P(1) k = P(1)/0.5 x = [0:2:40]; y = P(1)*x+P(2); % plot(t,logy,'o',x,y) %
% Now, model the semi-batch reactor for 20 minutes %
format short
simtime = [0:0.5:20]; initC = [3.0, 0, 0];
[t, z] = ode45('fexam', simtime, initC); [t, z]
plot(t,z)
xlabel('Time in Minutes')
ylabel('Concentration in Moles/Liter') legend('CA','CB','CC')
The MATLAB M-File for the ODE: function fC = ode(t, C) fC = zeros(3,1); fC(1) = 0.1084*C(1)*(1.5-0.5*C(1)) + (900-25*C(1))/(100+25*t); fC(2) = 0.0542*C(1)*(1.5-0.5*C(1)) + (600-25*C(2))/(100+25*t); fC(3) = 0.0542*C(1)*(0.5*C(1)-1.5) - 25*C(3)/(100+25*t); >> >> ans = 5.4203e-002 k = 1.0841e-001 ans = 0 3.0000 0 0 0.5000 6.4107 2.5387 0.1280 1.0000 8.4731 4.2365 0.5635 1.5000 9.5443 5.3176 1.2278 2.0000 10.0106 6.0053 1.9947 2.5000 10.1425 6.4559 2.7749 3.0000 10.1014 6.7650 3.5207 3.5000 9.9756 6.9878 4.2122 4.0000 9.8112 7.1556 4.8444 4.5000 9.6326 7.2869 5.4190 5.0000 9.4530 7.3932 5.9402 5.5000 9.2776 7.4809 6.4138 6.0000 9.1095 7.5548 6.8452 6.5000 8.9493 7.6175 7.2396 7.0000 8.7984 7.6719 7.6008 7.5000 8.6570 7.7198 7.9324 8.0000 8.5232 7.7616 8.2384 8.5000 8.3955 7.7977 8.5223 9.0000 8.2751 7.8299 8.7855 9.5000 8.1620 7.8588 9.0301 10.0000 8.0545 7.8844 9.2585 10.5000 7.9517 7.9069 9.4724 11.0000 7.8541 7.9270 9.6730 11.5000 7.7616 7.9453 9.8611
12.0000 7.6733 7.9616 10.0384 12.5000 7.5885 7.9761 10.2058 13.0000 7.5075 7.9891 10.3639 13.5000 7.4303 8.0009 10.5134 14.0000 7.3562 8.0114 10.6552 14.5000 7.2848 8.0208 10.7900 15.0000 7.2163 8.0292 10.9182 15.5000 7.1506 8.0368 11.0401 16.0000 7.0873 8.0436 11.1564 16.5000 7.0262 8.0497 11.2674 17.0000 6.9672 8.0551 11.3735 17.5000 6.9105 8.0599 11.4750 18.0000 6.8557 8.0642 11.5722 18.5000 6.8025 8.0679 11.6654 19.0000 6.7512 8.0713 11.7548 19.5000 6.7016 8.0742 11.8407 20.0000 6.6535 8.0768 11.9232 >>
(b) After 20 minutes, the initial conditions for the batch reactor are: The total liquid volume = 100+(25)(20) +4500/60 = 675 liters CA = (6.6535)(600)/675 = 5.9142
CD = 4500/675 = 4.6667 mol/liter dCA = 2dCB = –2dCD CA – CA0 = 2(CB – CB0) CB = CB0 + ½ (CA – CA0) and CA – CA0 = 2(CD0 – CD) CD = CD0 + ½ (CA0 – CA) (dCA/dt) = −k1CACB + k2CD = −k1CA[CB0 + ½ (CA – CA0)] + k2[CD0 + ½ (CA0 – CA)] = −0.1084CA[7.1794 + ½ (CA – 5.9142)] + 1.0[4.6667 + ½ (5.9142 – CA)] = CA[−0.77825 – 0.0542CA + 0.32055 – 0.5] + 7.6238 = 7.6238 – CA[0.0542CA + 0.9577] dCA = dt 7.6238–CA[0.0542CA+0.9577] dCA = dt 7.6238–0.9577CA–0.0542CA2 a = 7.6238, b = –0.9577, c = –0.0542 γ = 4ac – b2 = –2.5700 –(2/√–γ) tanh-1 (ω/√–γ) = –(2/√–γ) tanh-1 [(b+2cCA)/√–γ] –1.24756 tanh-1 [–0.62378(0.1084CA+0.9577)] = dt –1.24756 tanh-1 [–0.62378(0.1084CA+0.9577)] – 4.1206 = t
Check: when t = 0, CA does equal to 5.9142 mol/liter.
When ∝∝∝∝, CA = 5.9541 mol/liter from the above equation.
From the ODE: (dCA/dt) = 7.6238 – CA[0.0542CA + 0.9577] = 0 ---> CA = 5.9541 mol/liter same answer!
62. Isothermal Batch Reactor with a Series Reaction
Consider an isothermal batch reactor with a series reaction where 2 moles of component A react to form one mole of component B. Component B also reacts to form component C. The reaction scheme can be characterized as follows:
k1 k2
2A B C
Initially (t = 0 hr), CA = CA0, CB = 0, and CC = 0, where CA, CB, and CC represent the concentrations (mol/liter) of components A, B, and C, respectively.
(a) Assume constant volume and that the first reaction 2A → B is one-half order and the second reaction B → C is first-order, i.e.
dCA/dt = −k1CA1/2 dCB/dt = k1 CA1/2 − k2CB 2
Derive an analytical expression for CB as a function of time. (b) The following experimental data were obtained for component CA:
Time(hr) 0 0.05 0.15 0.35 0.60 0.85 1.0 CA(mol/liter) 1.0 0.93 0.79 0.54 0.30 0.13 0.06
CB was also measured to be 0.11 mol/liter at t = 1 hr. Determine the values of k1 and k2 and the time tmax at which CB is at its maximum. Hint: Use Polyfit function in MATLAB to help determine k1 and k2.
(c) Now, suppose the order of the above series reaction conforms to the stoichiometry, derive analytically a 1st-order ODE for CB, i.e.
dCB/dt + p(t)CB = q(t) but do not solve this ODE.
(d) When the reaction order conforms to the stoichiometry, the following experimental data were obtained for CA:
Time(hr) 0 0.03 0.06 0.10 0.15 0.20 0.30 CA(mol/liter) 1.0 0.76 0.63 0.51 0.39 0.33 0.25
It was also observed that CB reached a maximum of 0.14 mol/liter at t = 0.1 hr. Determine the values of k1 and k2.
Solution: k1 k2 2A B C (a) If rA = dCA/dt = -k1CA½ s.t. CA(0) = CA0 rB = dCB/dt = (k1/2)CA½ − k2CB s.t. CB(0) = 0 rC = dCC/dt = k2CB s.t. CC(0) = 0 Derive analytically CB(t): CA t
∫
dCA/CA½ = −k1∫
dt ⇒ 2(CA½ − CA0½) = −k1t CA0 0 CA(t) = (√CA0 − ½k1t)2 dCB/dt + k2CB = (k1/2)(√CA0 − ½k1t) CB(t) = C1e-k2t + C2t + C3 At t = 0 0 = C1 + C3 dCB/dt = −C1e-k2t + C2 dCB/dt + k2CB = −k2C1e-k2t + C2 + k2C1e-k2t + k2C2t + k2C3 k2C2t + (C2 + k2C3) = (k1/2)√CA0 − (k12/4)t ⇒ k2C2 = −(k12/4) or C2 = −(k12/4k2) C2 + k2C3 = k12√CA0C3 = (1/k2)[(k1/2)√CA0 + (k12/4k2)] = (k1/2k2)√CA0 + (k1/2k2)2 = (k1/2k2)(√CA0 + k1/2k2) C1 = −(k1/2k2)(√CA0 + k1/2k2) ∴ ∴ ∴ ∴ CB(t) = (k1/2k2)(√CA0 + (k1/2k2) [1 −−−− e-k2t] −−−− k12/4k2 t (b) Time (hr) 0 0.05 0.15 0.35 0.60 0.85 1.0 CA (mole/liter) 1.0 0.93 0.79 0.54 0.3 0.13 0.06
Has been observed: CB (t = 1 hr) = 0.11 mole/liter CA(t) = (1 − ½k1t)2 = 1 − k1t + (k12/4)t2
Use Polyfit in MATLAB to find the CA vs. t data to a 2nd-order degree polynomial (or plot CA1/2 vs. t)
From MATLAB : CA = 0.5729t2 – 1.5146t + 1.0023 Can obtain 2 values of k1 as :
k1 = 1.5146 and k12/4 = 0.5729 ⇒ k1 = 1.5138 So take the average value: k1 = 1.5142
CB(t) = (0.7571/k2)(1+0.7571/k2)(1 – e-k2t) – (0.5732/k2)t
CB(t = 1 hr) = 0.11 = (0.7571/k2)(1+0.7571/k2)(1 – e-k2) – 0.5732/k2
Use MATLAB to solve the above nonlinear equation for k2 k2 = 2.9818 hr-1
CB(t) = 0.3184(1 – e-2.9818t) – 0.1922t dCB/dt = 0.9494e-2.9818t – 0.1922 = 0
⇒ e-2.9818t = 0.2024 tmax = 0.536 hr CB,max = 0.151 mole/liter
(c) The reaction order conforms to stoichiometry
dCA/dt = –k1CA2 s.t. same initial conditions dCB/dt = (k1/2) CA2 – k2CB
Derive 1st-order ODE for CB, i.e. dCB/dt + p (t)CB = q(t)
CA t
∫
dCA/CA2 = –k1∫
dt ⇒ 1/CA – 1/CA0 = k1t CA0 0 ⇒ CA(t) = CA0/(CA0k1t + 1) ∴ dCB/dt + k2CB = (k1CA02/2)[CA0k1t + 1]–2 (d) Time (hr) 0 0.03 0.06 0.1 0.15 0.2 0.3 CA (mole/liter) 1.0 0.76 0.63 0.51 0.39 0.33 0.25CB,max has been observed at 0.14 mol/lit at t = 0.1 hr 1/CA = k1t + 1
Use Polyfit to regress 1/CA vs. t by fitting y = ax + b Slope of the straight line is k1 ⇒ y = 10.0676t + 1.0002 k1 = 10.0676
CB,max occurs at dCB/dt = 0
⇒ k2CB,max = (10.0676/2)(10.0676tmax + 1)–2 k2 = 8.9285 hr-1
67. Batch Distillation, II
A liquid mixture containing 60.0 mole% n-pentane and 40.0 mole% n-hexane is to be distilled in a batch still, and is initially charged with 100 gmoles of the mixture. The equilibrium relationship between the mole fraction x of n-pentane in the liquid and that in the vapor y has been correlated to the following equation:
y = 1.8804 x – 0.8804 x2
Note that this equation is only valid for n-pentane, and not necessarily for n-hexane. V (gmol/min)
yi
xi L (gmoles)
(a) Calculate the system temperature at which the above equilibrium relationship was established for the given liquid mixture (i.e. determine T in °C at which the above equation is valid). Also, compute the total system pressure P at the system
temperature.
Assume ideal gas and ideal liquid, and the vapor pressures of the two components are:
For n-pentane, log10 PVAP = 6.85221 − __1064.630__
T + 232.00
PVAP in mmHG
For n-hexane, log10 PVAP = 6.87776 − __1171.53_ T in °C T + 224.366
(b) Derive an analytical expression relating the amount of liquid left in the batch still L as a function of x (mole fraction of n-pentane in the still) and compute the value of
x after 90% of liquid has been vaporized.
Solution:
(a) Start with Raoult’s Law for ideal liquid:
But xc5 = 0.6 and P = xc5PC5VAP + xc6PC6VAP
0.8113[0.6*10[6.85221 – 1064.63/(T+232.0)] + 0.4*10[6.87776 – 1171.53/(T+224.366)]] = 0.6*10[6.85221 – 1064.63/(T+232.0)]
0.32452 (10[6.87776 – 1171.53/(T+224.366)]) – 0.11322 (10[6.85221 – 1064.63/(T+232.0)]) = 0
Nonlinear equation in T Use MATLAB for solve for the root or T Get T = 55.55 °°°°C
Also get, P = 845.86 + 196.69 = 1042.55 mmHg
(b) Start with mass balances: dL/dt = –V
d(xL)/dt = –yV or x(dL/dt) + L(dx/dt) = (1.8804x – 0.8804x2)(dL/dt)
xdL + Ldx = (1.8804x – 0.8804x2)dL Ldx = (0.8804x – 0.8804x2)dL
1.13585 dx = dL Integrate from x = 0.6 to x and L = 100
to L x( 1 – x) L 1.13585 1 + 1 = dL x 1 – x L ln (x) – ln(1 – x) = 0.8804 ln(L) x L ln x = 0.8804 ln(L) 1 – x 0.6 100
ln 0.6667 x = ln L 0.8804
1 – x 100
0.6667 x = L 0.8804
1 – x 100
