Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片

Solution of Nonlinear
Equations
Chapter 2

02Solution of Nonlinear Equations
 Finding the solution to a nonlinear equation or a
system of nonlinear equations is a common
problem arising from the analysis and design of
chemical processes in steady state.
 the relevant MATLAB commands used to deal
with this kind of problem are introduced.
 the solution-finding procedure through using the
Simulink simulation interface will also be
demonstrated.
2

2.1 Relevant MATLAB commands and the
Simulink solution interface
 2.1.1 Nonlinear equation of a single variable
3
 For example, one intends to obtain the volume of 1 g-
mol propane gas at P = 10 atm and T = 225.46 K
from the following Van der Waals P-V-T equation:

2.1.1 Nonlinear equation of a single variable
• where a = 8.664 atm (L/g-mol), b = 0.08446 L/g-mol, and the
ideal gas constant R is 0.08206 atmL/g-molK.
x = fzero(‘filename’, x0)
4

2.1.1 Nonlinear equation of a single variable
──────────────── PVT.m ────────────────
function f=PVT(V)
%
% Finding the solution to the Van der Waals equation (2.1-2)
%
% given data
%
a=8.664; % constant, atm (L/g-mol)^2
b=0.08446; % constant, L/g-mol
R=0.08206; % ideal gas constant, atm-L/g-mol‧K
P=10; % pressure, atm
T=225.46; % temperature, K
%
% P-V-T equation
%
f= (P+a/V^2)*(V-b)-R*T;
────────────────────────────────────
5

>> V=fzero('PVT', 1) % the initial guess value is set to 1
V =
1.3204
>> V=fzero ('PVT', 0.1) % use 0.1 as the initial guess value
V =
0.1099
6

>> P=10;
>> T=225.46;
>> R=0.08206;
>> V0=R*T/P; % determine a proper initial guess value by the
ideal gas law
>> V=fzero('PVT', V0) % solving with V0 as the initial guess
value
V=
1.3204
7

>> options= optimset('disp', 'iter'); % display the iteration process
>> V=fzero('PVT', V0, options)
Search for an interval around 1.8501 containing a sign change:
Func-count a f(a) b f(b) Procedure
1 1.85012 3.62455 1.85012 3.62455 initial interval
3 1.7978 3.22494 1.90245 4.03063 search
5 1.77612 3.06143 1.92413 4.20061 search
7 1.74547 2.83234 1.95478 4.44269 search
9 1.70211 2.51286 1.99813 4.78826 search
11 1.64081 2.07075 2.05944 5.28301 search
13 1.5541 1.46714 2.14614 5.99373 search
15 1.43149 0.66438 2.26876 7.01842 search
16 1.25808 -0.340669 2.26876 7.01842 search
8

Search for a zero in the interval [1.2581, 2.2688]:
Func-count X f(x) Procedure
16 1.25808 0.340669 initial
17 1.30487 0.0871675 interpolation
18 1.32076 0.00213117 interpolation
19 1.32038 1.80819e-005 interpolation
20 1.32039 3.64813e-009 interpolation
21 1.32039 0 interpolation
Zero found in the interval [1.25808, 2.26876]
V=
1.3204
9

Solution by Simulink:
Step 1: start Simulink
10

Step 2:
11

Step 3: Edit Fcn
(10+8.664/u (1)^2)*(u (1)-0.08446)-0.08206*225.46
12

Step 4: solution display
13

Step 5: Key in the initial guess value
14

Step 6: Click
building a dialogue box
15

Step 1-2:
Step 3: Edit Fcn using variable names, P, R, T, a, and b
(P + a/ u (1) ^2)*(u (1)b) R*T
16

Step 4-5:
17

Step 6:
18

Step 6:
19

Step 7: Mask Subsystem
20

Step 8: key in the parameter values
21

Step 9: Execute the program
22

2.1.2 Solution of a system of nonlinear equations
23

• CSTR
24
• The kinetic parameters are as follows: B = 8, Da = 0.072,  =
20, and  = 0.3.
x=fsolve(‘filename’, x0)

Step 1: Provide the function
──────────────── cstr.m ────────────────
function f=fun(x)
%
% steady-state equation system (2.1-6) of the CSTR
%
% kinetic parameters
%
B=8;
Da=0.072;
phi=20;
beta=0.3;
%
% nonlinear equations in vector form
%
f= [-x(1)+Da*(1-x(1))*exp(x(2)/(1+x(2)/phi))
-(1+beta)*x(2)+B*Da*(1-x(1))*exp(x(2)/(1+x(2)/phi))];
─────────────────────────────────────────────────
25

Step 2:
>> x=fsolve('cstr', [0.1 1]) % solve the equations with the initial guess [0.1 1]
x=
0.1440 0.8860
>> ezplot('-x1+0.072*(1-x1)*exp(x2/(1+x2/20))', [0, 1, 0, 8]) % plotting (2.1-6a)
>> hold on
>> ezplot('-(1+0.3)*x2+8*0.072*(1-x1)*exp(x2/(1+x2/20))', [0, 1, 0, 8]) % plotting
(2.1-6b)
>> hold off
26

27

• (0.15, 1), (0.45, 3), and (0.75, 5)
>> x_1=fsolve('cstr', [0.15 1]) % the first solution
x_1=
0.1440 0.8860
>> x_2=fsolve('cstr', [0.45 3]) % the second solution
x_2=
0.4472 2.7517
>> x_3=fsolve('cstr', [0.75 5]) % the third solution
x_3=
0.7646 4.7050
28

Step 1:
29

Step 2:
In Fcn: -u(1)+Da*(1-u(1))*exp(u(2)/(1+u(2)/phi))
In Fcn 1: - (1+beta)*u(2)+B*Da*(1-u(1))*exp(u(2)/(1+u(2)/phi))
30

Step 2:
31

Step 3:
32

Step 4: Create a Subsystem for
33

Step 5: Create a dialogue box
34

Step 6: Input system parameters and initial guess values
35

Step 7:
36

2.2 Chemical engineering examples
Example 2-2-1
Boiling point of an ideal solution
Consider a three-component ideal mixture of which the vapor
pressure of each component can be expressed by the following
Antoine equation:
where T represents the temperature of the solution (C) and Pi
0
denotes the saturated vapor pressure (mmHg). The Antoine
coefficients of each component are shown in the following table
(Leu, 1985):
37

Example 2-2-1
Component Ai Bi Ci
1 6.70565 1,211.033 220.790
2 6.95464 1,344.800 219.482
3 7.89750 1,474.080 229.130
38
Determine the boiling point at one atmospheric pressure for
each of the following composition conditions:
Condition x1 x2 x3
1 0.26 0.38 0.36
2 0.36 0.19 0.45

Problem formulation and analysis:
Raoult’s law:
39

MATLAB program design:
─────────────── ex2_2_1.m ───────────────
%
% Example 2-2-1 boiling point of an ideal solution
%
% main program: ex2_2_1.m
% subroutine (function file): ex2_2_1f.m
%
clear
clc
%
global A B C X_M
%
% given data
%
A= [6.70565 6.95464 7.89750]; % coefficient A
B= [-1211.033 -1344.800 -1474.080]; % coefficient B
C= [220.790 219.482 229.130]; % coefficient C
%
40

─────────────── ex2_2_1.m ───────────────
X= [0.25 0.40 0.35
0.35 0.20 0.45]; % component concentrations for each case
%
[m, n]=size(X); % n: the number of components
% m: the number of conditions
%
for i=1: m
X_M=X(i, :);
T(i) =fzero('ex2_2_1f', 0);
end
%
% results printing
%
for i=1: m
fprintf('condition %d, X(l)=%.3f, X(2)=%.3f, X(3)=%.3f, boiling point=…
%.3f ℃n',i,X(i,1),X(i,2),X(i,3),T(i))
end
─────────────────────────────────────────────────
41

──────────────── ex2_2_1f.m ────────────────
%
% example 2-2-1 boiling point of an ideal solution
% subroutine (function file): ex2_2_1f.m
%
function f= ex2_2_1f(T)
global A B C X_M
Pi=10.^(A+B./(C+T)); % saturated vapor pressure of each component
p=sum(X_M.*Pi); % total pressure
f=p-760;
end
─────────────────────────────────────────────────
42

Execution results:
>> ex2_2_1
condition 1, X(l)=0.250, X(2)=0.400, X(3)=0.350, boiling
point=82.103C
condition 2, X(l)=0.350, X(2)=0.200, X(3)=0.450, boiling
point=77.425C
43

Example 2-2-2
Equilibrium concentrations in a reaction system
3H2 + CO ⇌ CH4 + H2O, 𝐾1 = 69.18
CO + H2O ⇌ CO2 + H2, 𝐾2 = 4.68
CO2 + C(𝑠) ⇌ 2CO, 𝐾3 = 5.60 × 10−3
5H2 + 2CO ⇌ C2H6 + 2H2O, 𝐾4 = 0.14
where Ki is the equilibrium constant under the operating
condition of 500C and 1 atmospheric pressure. Assume that 1
mole of CO and 3 moles of H2 exist initially in the system,
determine the equilibrium concentration of each component
(mole fraction) when the reaction system reaches an equilibrium.
44

Analysis of the question:
Let y1, y2, …, and y6 stand for the mole fractions of H2, CO, CH4,
H2O, CO2, and C2H6, respectively. Besides, let x1, x2, x3, and x4
be the reaction rates of individual equations in (2.2-5a-d).
45

Analysis of the question:
46
D = 4  2x1 + x3  4x4

─────────────── ex2_2_2.m ───────────────
function ex2_2_2 % notice the format of the first line
%
% Example 2-2-2 Equilibrium concentrations in a reaction system
%
clear
clc
%
% initial guess values
%
x0= [0.7 0 -0.1 0];
%
% solving by calling the embedded function file ex2_2_2f
%
options= optimset('disp', 'iter');
x=fsolve(@ex2_2_2f, x0, options); % notice the use of the function handle @
%
% calculate the mole fraction
%
47

─────────────── ex2_2_2.m ───────────────
y=ex2_2_2y(x);
%
% results printing
%
disp(' ')
disp('reaction rate of each equation')
%
for i=1:4
fprintf('x(%d) =%.3e n', i, x (i))
end
%
disp(' ')
disp('mole fraction of each component')
%
for i=1:6
fprintf('y(%i) =%.3e n', i, y (i))
end
48

─────────────── ex2_2_2.m ───────────────
%
% chemical equilibrium equations
%
function f=ex2_2_2f(x)
%
y= ex2_2_2y(x);
%
% the set of nonlinear equations
%
f= [y(3)*y(4)/(y (1)^3*y(2))-69.18
y(5)*y(1)/(y(2)*y(4))-4.68
y(2)^2/y(5)-5.60e-3
y(6)*y(4)^2/(y(1)^5*y(2)^2)-0.14];
%
% the subroutine for calculating the mole fraction
%
function y=ex2_2_2y(x)
49

─────────────── ex2_2_2.m ───────────────
% chemical equilibrium equations
%
function f=ex2_2_2f(x)
%
y= ex2_2_2y(x);
%
% the set of nonlinear equations
%
f= [y(3)*y(4)/(y (1)^3*y(2))-69.18
y(5)*y(1)/(y(2)*y(4))-4.68
y(2)^2/y(5)-5.60e-3
y(6)*y(4)^2/(y(1)^5*y(2)^2)-0.14];
%
% the subroutine for calculating the mole fraction
%
function y=ex2_2_2y(x)
D=4-2*x(1) +x(3)-4*x(4);
50

─────────────── ex2_2_2.m ───────────────
if D == 0
disp('D is zero')
return
end
y(1)=(3-3*x(1) +x(2)-5*x(4))/D;
y(2)=(1-x(1)-x(2) +2*x(3)-2*x(4))/D;
y(3)=x(1)/D;
y(4)=(x(1)-x(2) +2*x(4))/D;
y(5)=(x(2)-x(3))/D;
y(6)=x(4)/D;
─────────────────────────────────────────────────
51

Execution results:
>> ex2_2_2
reaction rate of each equation
x (1) =6.816e-001
x (2) =1.590e-002
x (3) =-1.287e-001
x (4) =1.400e-005
52

Execution results:
>> ex2_2_2
mole fraction of each component
y(1) =3.872e-001
y(2) =1.797e-002
y(3) =2.718e-001
y(4) =2.654e-001
y(5) =5.765e-002
y(6) =5.580e-006
53

Example 2-2-3
Analysis of a pipeline network
Figure 2.1 systematically illustrates a pipeline network installed
on a horizontal ground surface. In the network, raw materials are
fed to eight locations denoted as P1, P2, … , P7, and P8, and the
flow rate (m3/min) required at each of the eight locations is listed
as follows (Leu, 1985):
P1Q = 3.0, P2Q = 2.0, P3Q = 5.0, P4Q = 3.0
P5Q = 2.0, P6Q = 2.0, P7Q = 5.0, P8Q = 2.0
54

55
Due to flow frictions, the head loss h(m) for each 1/2 side of
the pipeline (such as P1 to P5) is given by
Note that the head loss is a function of the flow rate Q
(m3/min) in the pipeline. Based on the required flow rates
PiQ determine the corresponding flow rates, Q1 to Q9, in the
pipeline and the head at each of the eight locations.

56
Figure 2.1 Schematic diagram of the pipeline network.

mass balance
Mass balance at P2: Q3 = Q5 + P2Q
Mass balance at P3: Q7 + Q9 = P3Q
Mass balance at P5: Q2 = Q3 + Q4 + P5Q
Mass balance at P7: Q6 + Q8 = Q9 + P7Q
57

Head loss balance by considering hP1P5
+hP5P8
+hP8P7
+hP1P4
+hP4P7
:
58
Head loss balance by considering
hP5P2
+hP2P6
+hP6P6
=hP5P8
+hP8P7
+hP7P3
:

─────────────── ex2_2_3.m ───────────────
function ex2_2_3
%
% Example 2-2-3 pipeline network analysis
%
clear
clc
%
global PQ % declared as a global parameter
%
PQ=[3 2 5 3 2 2 5 2]; % flow rate required at each location
%
Q0=2*ones(1, 9); % initial guess value
%
% solving
%
Q=fsolve(@ex2_2_3f, Q0,1.e-6); % setting the solution accuracy as 1.e-6
%
59

─────────────── ex2_2_3.m ───────────────
h(3)=10;
h(6)=h(3)+0.5*abs(Q(7))^0.85*Q(7);
h(2)=h(6)+0.5*abs(Q(5))^0.85*Q(5);
h(7)=h(3)+0.5*abs(Q(9))^0.85*Q(9);
h(8)=h(7)+0.5*abs(Q(6))^0.85*Q(6);
h(5)=h(2)+0.5*abs(Q(3))^0.85*Q(3);
h(4)=h(7)+0.5*abs(Q(8))^0.85*Q(8);
h(1)=h(5)+0.5*abs(Q(2))^0.85*Q(2);
%
% results printing
%
disp(' ')
disp('flow at each location (m^3/min)')
for i=1:9
fprintf('Q(%i)=%7.3f n',i,Q(i))
end
disp(' ')
disp('head at each location (m)')
60

─────────────── ex2_2_3.m ───────────────
for i=1:8
fprintf('h(%i)=%7.3f n',i,h(i))
end
%
% nonlinear simultaneous equations
%
function f=ex2_2_3f(Q)
%
global PQ
%
f=[Q(3)-Q(5)-PQ(2)
Q(7)+Q(9)-PQ(3)
Q(1 )-Q(8)-PQ(4)
Q(2)-Q(3)-Q(4)-PQ(5)
Q(5)-Q(7)-PQ(6)
Q(6)+Q(8)-Q(9)-PQ(7)
Q(4)-Q(6)-PQ(8)
61

─────────────── ex2_2_3.m ───────────────
0.5*abs(Q(2))^0.85*Q(2)+0.5*abs(Q(4))^0.85*Q(4)+0.5*abs(Q(6))^0.85*Q(6)-...
abs(Q(1))^0.85*Q(1)-0.5*abs(Q(8))^0.85*Q(8)
0.5*abs(Q(3))^0.85*Q(3)+0.5*abs(Q(5))^0.85*Q(5)+0.5*abs(Q(7))^0.85*Q(7)-...
0.5*abs(Q(4))^0.85*Q(4)-0.5*abs(Q(6))^0.85*Q(6)-0.5*abs(Q(9))^0.85*Q(9)];
─────────────────────────────────────────────────
62

Execution results:
>> ex2_2_3
flow at each location (m^3/min)
Q(1)= 8.599
Q(2)= 12.401
Q(3)= 5.517
Q(4)= 4.884
Q(5)= 3.517
Q(6)= 2.884
Q(7)= 1.517
Q(8)= 5.599
Q(9)= 3.483
63
head at each location (m)
h(1)= 80.685
h(2)= 16.202
h(3)= 10.000
h(4)= 27.137
h(5)= 27.980
h(6)= 11.081
h(7)= 15.031
h(8)= 18.578

Example 2-2-4
A material drying process through heat conduction and
forced convection
Figure 2.2 schematically depicts a drying process (Leu, 1985),
where a wet material with thickness of 6 cm is placed on a hot
plate maintained at 100C.
64
Figure 2.2 A material drying process through heat conduction and forced convection.

To dry the material, the hot wind with a temperature of 85C and
partial vapor pressure of 40 mmHg is blowing over the wet
material at a flow rate of 3.5 m/s. The relevant operation
conditions and assumptions are stated as follows:
(1) During the drying process, the heat conduction coefficient of
the wet material is kept constant, and its value is measured to
be 1.0 kcal/m·hr·C.
(2) The convective heat transfer coefficient h (kcal/m·hr·C)
between the hot wind and the material surface is a function of
mass flow rate of wet air G (kg/hr), which is given by
h = 0.015G 0.8
(3) The saturated vapor pressure Ps (mmHg), which is a function
of temperature ( C), can be expressed by the Antoine
equations as follows:
65

(4) The latent heat of water vaporization rm (kcal/kg) is a
function of temperature T (C), which is given by
rm = 597.65 – 0.575 T
(5) The mass flow rate of wet air G and its mass transfer
coefficient M are measured to be the values of 1.225×104
kg/hr and 95.25 kg/m2·hr, respectively.
Based on the operating conditions mentioned above, determine
the surface temperature of the material Tm and the constant drying
rate Rc, which is defined as
Rc = M(Hm – H)
66

67
h = 0.015G 0.8
rm = 597.65 – 0.575T
Tp = 100 C
L = 0.06 m
k = 1 kcal/m·hr· C
Ta = 85 C
G = 1.225×104 kg/hr
M = 95.25 kg/m2· hr

68

─────────────── ex2_2_3.m ───────────────
function ex2_2_4
%
% Example 2-2-4 A material drying process via heat conduction and forced convection
%
clear
clc
%
global h Ta M k L Tp H
%
G=1.225e4; % mass flow rate of wet air (kg/hr)
M=95.25; % mass transfer coefficient (kg/m^2.hr)
k=1; % heat transfer coefficient of the wet material (kcal/m.hr.℃)
Tp=100; % hot plate temperature (℃)
L=0.06; % material thickness (m)
Ta=85; % hot wind temperature (℃)
P=40; % partial pressure of water vapor (mmHg)
H=18.02*P/(28.97*(760-P)); % humidity of hot air (%)
h=0.015*G^0.8; % heat transfer coefficient (kcal/m^2.hr. ℃)
69

─────────────── ex2_2_3.m ───────────────
%
% solving for Tm
%
Tm=fzero(@ex2_2_4f, 50);
%
Ps=ex2_2_4Ps(Tm); % eq. (2.2-12)
%
% calculating constant drying rate
%
Hm=18.02*Ps/(28.97* (760-Ps)); % eq. (2.2-17)
Rc=M*(Hm-H); % eq. (2.2-14)
%
% results printing
%
fprintf('the surface temperature of the material =%.3f ℃n', Tm)
fprintf('the constant drying rate = %.3f kg/m^2.hrn', Rc)
%
70

─────────────── ex2_2_3.m ───────────────
% nonlinear function
%
function f = ex2_2_4f(Tm)
%
global h Ta M k L Tp H
%
Ps= ex2_2_4Ps(Tm); % eq.(2.2-12)
%
Hm=18.02*Ps/(28.97*(760-Ps)); % eq.(2.2-17)
rm=597.65-0.575*Tm; % eq.(2.2-13)
%
f=h*(Ta-Tm)-M*(Hm-H)*rm-k/L*(Tm-Tp); % eq.(2.2-15)
%
% subroutine for calculating the saturated vapor pressure
%
function Ps=ex2_2_4Ps(Tm)
if Tm < 55
71

─────────────── ex2_2_3.m ───────────────
Ps=10^(7.7423-1554.16/(219+Tm));
elseif Tm >= 55
Ps=10^(7.8097-1572.53/(219+Tm));
end
─────────────────────────────────────────────────
72

Execution results:
>> ex2_2_4
the surface temperature of the material = 46.552C
the constant drying rate = 3.444 kg/m^2.hr
73

Example 2-2-5
A mixture of 50 mol% propane and 50 mol% n-butane is to be
distilled into 90 mol% propane distillate and 90 mol% n-butane
bottoms by a multistage distillation column. The q value of raw
materials is 0.5, and the feeding rate is 100 mol/hr. Suppose the
binary distillation column is operated under the following
operating conditions (McCabe and Smith, 1967):
1) The Murphree vapor efficiency EMV is 0.7;
2) The entrainment effect coefficient ε is 0.2;
3) The reflux ratio is 2.0;
4) The operating pressure in the tower is 220 Psia;
74

5) The equilibrium coefficient (K = y/x), which depends on the
absolute temperature T (R), is given by
ln K = A/T+B+CT
where the coefficients A, B, and C under 220 Psia for propane
and n-butane are listed in the table below.
75
Component A B C
Propane 3,987.933 8.63131 0.00290
n-Butane 4,760.751 8.50187 0.00206

Answers to the following questions:
1) The theoretical number of plates and the position of the
feeding plate.
2) The component compositions in the gas phase and the liquid
phase at each plate.
3) The temperature of the re-boiler and the bubble point at each
plate.
76

(1) Overall mass balance:
77
W = F  D
Name of
variable Physical meaning Name of
variable Physical meaning
D Flow rate of the distillate xD Composition of the distillate
F Flow rate of the feed xw Composition of the bottoms
W Flow rate of the bottoms zF Composition of the feed

(1) Overall mass balance:
78

79

(2) Material balance on the feed plate:
80
q value Condition of the feed
q > 1 The temperature of the feed is lower than the boiling point
q = 1 Saturated liquid
0 < q < 1 Mixture of liquid and vapor
q = 0 Saturated vapor
q < 0 Overheated vapor

(3) Material balance made for the re-boiler
(4) Stripping section
81

(5) Enriching section
𝑥 𝑛+1 =
𝐿𝑦𝑛 + 𝜀𝑉𝑥 𝑛 − 𝐷𝑥 𝐷
𝐿 + 𝜀𝑉
(6) Feed plate
𝑥𝑓+1 =
𝑉′𝑦𝑓 + 𝜀𝑉′𝑥𝑓 + 1 − 𝑞 𝐹𝑦 𝐹 − 𝐷𝑥 𝐷
𝐿 + 𝜀𝑉′
(A) Calculation of the flash point at the feed plate:
82
, i =1, 2

where
Ki = exp (Ai / T + Bi + CiT)
(B) Calculation of the flash point at the feed plate:
83

84

─────────────── ex2_2_5.m ───────────────
function ex2_2_5
%
% Example 2-2-5 analysis of a continuous multi-stage binary distillation
%
clear; close all
clc
%
% declare the global variables
%
global A B C zf q
global A B C xc
%
% operation data
%
zf=[0.5 0.5]; % feed composition
xd=[0.9 0.1]; % distillate composition
xw=[0.1 0.9]; % composition of the bottom product
F=100; % feed flow rate (mol/h)
85

─────────────── ex2_2_5.m ───────────────
q=0.5; % q value of the feed
Emv=0.7; % Murphree vapor efficiency
Ee=0.2; % entrainment coefficient
RD=2; % reflux ratio
%
A=[-3987.933 -4760.751]; % parameter A
B=[8.63131 8.50187]; % parameter B
C=[-0.00290 -0.00206]; % parameter C
%
% calculating all flow rates (D,W,L,V, LS, and VS)
%
D=F*(zf(1)-xw(1))/(xd(1)-xw(1)) ; % distillate flow rate
W=F-D; % flow rate of the bottom product
L=RD*D; % liquid flow rate in the enriching section
V=L+D; % gas flow rate in the enriching section
LS=L+q*F; % liquid flow rate in the stripping section
VS=V-(1-q)*F; % gas flow rate in the stripping section
%
86

─────────────── ex2_2_5.m ───────────────
% calculating the flash point at the feed plate
%
Tf=fzero(@ex2_2_5f1,620); % flash point
K=exp(A/Tf+B+C*Tf);
yf=zf./(1 +q*(1./K-1)); % gas phase composition at the feed plate
xf=yf./K; % liquid phase composition at the feed plate
Tf=Tf-459.6; % converted to ℉
%
% equilibrium in the re-boiler
%
xc=xw;
Tw=fzero(@ex2_2_5f2,620);
Kw=exp(A/Tw+B+C*Tw);
yw=Kw.*xw; % gas phase composition in the re-boiler
Tw=Tw-459.6; % bubble point (℉) of the re-boiler
%
% calculating the liquid phase composition in the re-boiler
87

─────────────── ex2_2_5.m ───────────────
m=1;
x(m,:)=xw;
y(m,:)=yw;
x(m+1,:)=(VS*y(m,:)+W*xw)/LS;
%
% calculating for the stripping section
%
T(m)=Tw;
while T(m)>Tf
m=m+1;
xc=x(m,:);
T (m)=fzero(@ex2_2_5f2,620);
K(m,:)=exp(A/T(m)+B+C*T(m));
ys(m,:)=K(m,:).*x(m,:);
y(m,:)=Emv*(ys(m,:)-y(m-1,:))+y(m-1,:);
x(m+1,:)=(VS*y(m,:)+Ee*VS*x(m,:)+W*x(m-1,:))/(LS+Ee*VS);
T(m)=T(m)-459.6;
end
88

─────────────── ex2_2_5.m ───────────────
%
% composition calculation
%
n=m+1;
x(n,:)=(VS*yf+Ee*VS*xf+(1-q)*F*yf-D*xd)/(L+Ee*VS);
xc=x(n,:);
T(n)=fzero(@ex2_2_5f2,620);
K(n,:)=exp(A/T(n)+B+C*T(n));
ys(n,:)=K(n,:).*x(n,:);
y(n,:)=Emv*(ys(n,:)-y(n-1,:))+y(n-1,:);
T(n)=T(n)-459.6;
x(n+ 1,:)=(V*y(n,:)+Ee*V*x(n,:)-D*xd)/(L+Ee*V);
%
% calculation for the enriching section
%
while y(n,1)<xd(1)
n=n+1;
89

─────────────── ex2_2_5.m ───────────────
xc=x(n,:);
T(n)=fzero(@ex2_2_5f2,620);
K(n,: )=exp(A/T(n)+B+C*T(n));
ys(n,:)=K(n,:).*x(n,:);
y(n,:)=Emv*(ys(n,:)-y(n-1,:))+y(n-1,:);
x(n+ 1,:)=(V*y(n,:)+Ee*V*x(n,:)-D*xd)/(L+Ee*V);
T(n)=T(n)-459.6;
end
%
N=n+1; % total number of plates
%
% distillate concentration
%
x(N,:)=y(n,:);
%
% results printing
%
90

─────────────── ex2_2_5.m ───────────────
disp(' ')
disp(' ')
disp('The bubble point and compositions at each plate.')
disp(' ')
disp('plate liquid gas temperature')
disp('no. phase phase')
disp('---- ------------------- -------------------- -----------')
for i=1:n
fprintf('%2i %10.5f %10.5f % 10.5f %10.5f %10.2fn',i,x(i,1),x(i,2),…
y(i,1), y(i,2),T(i))
end
fprintf('%2i %10.5f %10.5fn',N,x(N,1),x(N,2))
fprintf('n total no. of plates= %d; the feed plate no.= %d',N,m)
fprintf('n distillate concentration = [%7.5f %10.5f]n', x(N,1),x(N,2))
%
% composition chart plotting
%
91

─────────────── ex2_2_5.m ───────────────
plot(1 :N,x(:, 1),1 :N,x(:,2))
hold on
plot([m m],[0 1],'--')
hold off
xlabel('plate number')
ylabel('liquid phase composition')
title('liquid phase composition chart')
text(3,x(3,1),'leftarrow it{x_1} ','FontSize', 10)
text(3,x(3,2),' leftarrowit{x_2} ','FontSize', 10)
text(m,0.5,' leftarrow position of the feeding plate ','FontSize',10)
axis([1 N 0 1])
%
% flash point determination function
%
function y=ex2_2_5f1(T)
global A B C zf q
K=exp(A./T+B+C*T);
yf=zf./(1 +q*(1./K-1));
92

─────────────── ex2_2_5.m ───────────────
xf=yf./K;
y=sum(yf)-1 ;
%
% bubble point determination function
%
function y=ex2_2_5f2(T)
global A B C xc
K=exp(A./T+B+C.*T);
yc=K.*xc;
y=sum(yc)-1;
─────────────────────────────────────────────────
93

Execution results:
>> ex2_2_5
The bubble point and compositions at each plate.
plate liquid gas temperature
no. phase phase
---- ----------------------------- -------------------------------- ---------------
1 0.10000 0.90000 0.19145 0.80855 196.35
2 0.16097 0.83903 0.26333 0.73667 189.16
3 0.20325 0.79675 0.33070 0.66930 184.28
4 0.26578 0.73422 0.41292 0.58708 177.25
5 0.33394 0.66606 0.49800 0.50200 169.85
6 0.41040 0.58960 0.58324 0.41676 161.88
7 0.44630 0.55370 0.63418 0.36582 158.27
8 0.48858 0.51142 0.67733 0.32267 154.12
9 0.54813 0.45187 0.72610 0.27390 148.47
10 0.61815 0.38185 0.77825 0.22175 142.10
11 0.69448 0.30552 0.82978 0.17022 135.50
12 0.77156 0.22844 0.87685 0.12315 129.18
13 0.84365 0.15635 0.91688 0.08312 123.57
14 0.91688 0.08312
94

Execution results:
>> ex2_2_5
total no. of plates= 14; the feed plate no.= 6
distillate concentration = [0.91688 0.08312]
95

Example 2-2-6
Analysis of a set of three CSTRs in series.
96
In each of the reactors, the following liquid phase reaction
occurs

The reactor volumes and the reaction rate constants are listed in
the following table:
If the concentration of component A fed into the first reactor is C0
= 1 g-mol/L and the flow rate v is kept constant at the value of
0.5 L/min, determine the steady-state concentration of component
A and the conversion rate of each reactor.
97
Reactor
number
Volume, Vi
(L)
Reaction rate constant, Ki
(L/g – mol·min)
1 5 0.03
2 2 0.05
3 3 0.10

steady state, mass
98
, i = 1, 2, 3

─────────────── ex2_2_6.m ───────────────
function ex2_2_6
%
% Example 2-2-6 Analysis of a set of three CSTRs in series
%
Clear; clc;
%
global v VV kk cc
%
% given data
%
C0=1; % inlet concentration (g-mol/L)
v=0.5; % flow in the reactor (L/min)
V=[5 2 3]; % volume of reactor (L)
k=[0.03 0.05 0.1]; % reaction rate constant (L/g.mol.min)
%
% start calculating
%
99

─────────────── ex2_2_6.m ───────────────
C=zeros(n+1,1); % initialize concentrations
x=zeros(n+1,1); % initialize conversion rates
C(1)=C0; % inlet concentration
x(1)=0; % initial conversion rate
for i=2:n+1
VV=V(i-1);
kk=k(i-1);
cc=C(i-1);
C(i)=fzero(@ex2_2_6f,cc); % use inlet conc. as the initial guess value
x(i)=(C0-C(i))/C0; % calculating the conversion rate
end
%
% results printing
%
for i=2:n+1
fprintf('The exit conc. of reactor %d is %.3f with a conversion rate of… %.3f.n',i-
1,C(i),x(i))
100

─────────────── ex2_2_6.m ───────────────
end
%
% reaction equation
%
function f=ex2_2_6f(C)
global v VV kk cc
f=v*cc-v*C-kk*VV*C^1.5;
─────────────────────────────────────────────────
101

Execution results:
>> ex2_2_6
The conc. at the outlet of reactor 1 is 0.790 with a conversion rate of 0.210.
102

2.4 Summary of MATLAB commands related to
this chapter
a. Solution of a single-variable nonlinear equation:
The command format:
x=fzero(fun, x0)
x=fzero(fun, x0, options)
x=fzero(fun, x0, options, Pl , P2, ... )
[x, fval]=fzero(... )
[x, fval, exitflag]=fzero(... )
[x, fval, exitflag, output]=fzero(... )
103

b. Solution of a set of nonlinear equations
The command format:
x=fsolve(fun, x0)
x=fsolve(fun, x0, options)
x=fsolve(fun, x0, options, Pl , P2, ... )
[x, fval]=fsolve(... )
[x, fval, exitflag]=fsolve(... )
[x, fval, exitflag, output] =fsolve(... )
[x, fval, exitflag, output, Jacobian] =fsolve(... )
104

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