CHE656
Modeling Chemical Processes
Using MATLAB
Homework Set #1 Solutions
Class-17
Prepared by
Dr. Hong-ming Ku
King Mongkut’s University of Technology Thonburi
Chemical Engineering Department
Chemical Engineering Practice School
 July 2004 - 2013 – Use with Permission of the Author Only
1
2. System of Linear Algebraic Equations, II
Use MATLAB to solve the following system of linear algebraic equations, correct to 4
decimal places:
x1
2x1
–3x1
–x2
–2x1
6x1
+
–
–
+
+
–
3x2
2x2
2x2
2x4
3x4
2x2
+
+
+
+
+
–
2x3 – x5 + 4x6
4x3 + x4 + x5
4x3 + 5x4 + 2x5 – x6
6x5 + 4x6
4x5 + 5x6
x3 – 3x4 + 4x5 + 2x6
= –2
= 1
= 0
= 3
= –6
=5
Solution:
MATLAB Script File:
%
% Problem #2: System of Linear Algebraic Equations
%
clc
clear
A = [A = [1 3 2 0 -1 4; 2 -2 4 1 1 0; -3 -2 4 5 2 -1; 0 -1 0 2 6 4; -2 0 0 3 4 5; 6 -2 -1 -3 4 2];
b = [-2; 1; 0; 3; -6; 5];
x = inv(A)*b
MATLAB Output (from x1
x6):
x=
-1.2770
3.4089
2.4459
-3.6267
4.2148
-2.9067
2
4. Pressures in a Pipeline Network
Consider the following interconnected liquid pipeline network which contains 4 valves
with the following valve coefficients. Calculate the pressure at each location/node
indicated in the network.
Cv,1 = 0.4 ft3-(psia)1/2/hr
Cv,3 = 0.2 ft3-(psia)1/2/hr
Cv,2 = 0.3 ft3-(psia)1/2/hr
Cv,4 = 0.1 ft3-(psia)1/2/hr
1
P = 40 psia
Cv,1
Pump
Cv,2
Cv,3
2
Cv,4
P = 14.7 psia
Reservoir
Solution:
MATLAB Script File:
%
% CHE656 Problem #4: Pressures in a Pipeline Network
%
clc
clear
format short
%
% These equations can be set up for the pressure at each node using mass balance.
%
% At node 1: Q1 = 0.4*sqrt(40-p1) = 0.3*sqrt(p1-14.7) + 0.2*sqrt(p1-p2)
% At node 2: Q2 = 0.2*sqrt(p1-p2) = 0.1*sqrt(p2-14.7)
%
p = solve('0.1*sqrt(40-p1)-0.2*sqrt(p1-14.7)-0.3*sqrt(p1-p2)=0', ...
'0.3*sqrt(p1-p2)-0.4*sqrt(p2-14.7)=0')
double(p.p1)
double(p.p2)
The MATLAB output:
ans =
27.6883
P1
ans =
25.0906
P2
3
6. Simple Calculations and Array Manipulations in MATLAB, II
Consider the following arrays:
A =
1
3
10 5
6 10
3
π
5
30
4
8
B = log10(A)
C =
2 4 4
8 12 24
0 -1 5
8
7
6
Write MATLAB expressions in an M-file (script file) to do the following (use “format
short” in all calculations unless otherwise told):
(a) Determine the sum of the second column of B.
(b) Select the first 3 columns and the first 3 rows of A and B and add their determinants
.
(c) Determine the square root of the product (a scalar) between the last row of C and
the second column of A.
(d) Determine the sum of the second row of A divided element-by-element by the third
column of C.
(e) Evaluate the hyperbolic secant of (0.01 * A * C).
(f) Evaluate the following function: tan-1 (0.1 * B * C) in 6 decimal places.
Solution:
MATLAB Script File:
%
% ChE 656 Problem #6: Simple Calculations and Array Manipulations in MATLAB
%
clc
clear
format short
A = [1 3 5; 10 5 30; 6 10 4; 3 pi 8]
B = log10(A)
C = [2 4 4 8; 8 12 24 7; 0 -1 5 6]
%
% Determine the sum of the second row of B
%
column2_B = B(:,2);
sum_column2_B = sum(column2_B)
%
% Select the first 3 columns and first 3 rows of A and B and add their determinants
%
A33 = A(1:3,1:3);
B33 = B(1:3,1:3);
sum_det = det(A33) + det(B33)
%
% Determine the square root of the product between the 2nd column of A and the
% last row of C
4
%
row3_C = C(3,:);
column2_A = A(:,2);
product = row3_C*column2_A;
square_root = sqrt(product)
%
% Determine the sum of the second row of A divided element-by-element by the
% the third column of C.
%
row2_A = A(2,:);
col3_C = C(:,3)';
row2A_col3C = row2_A./col3_C;
Sum = sum(row2A_col3C)
%
% Evaluate the hyperbolic secant of 0.01*A*C
%
A_C = 0.01*A*C;
hyperbolic_secant_AC = sech(A_C)
%
% Evalute the inverse tangent of 0.1*B*C
%
B_C = 0.1*B*C;
inverse_tangent_BC = atan(B_C);
fprintf ('inverse_tangent_BC\n')
for i = 1:4
fprintf ('\n')
eachrow = inverse_tangent_BC(i,:);
fprintf ('%10.6f', eachrow)
================= Solutions ===================
A=
1.0000 3.0000 5.0000
10.0000 5.0000 30.0000
6.0000 10.0000 4.0000
3.0000 3.1416 8.0000
B=
0 0.4771 0.6990
1.0000 0.6990 1.4771
0.7782 1.0000 0.6021
0.4771 0.4971 0.9031
C=
2
4
4
8
5
8
0
12
-1
24 7
5 6
sum_column2_B =
2.6732
sum_det =
490.5800
square_root =
7.9906
Sum =
8.7083
hyperbolic_secant_AC =
0.9671
0.8436
0.6878
0.9534
0.9417
0.7967
0.4649
0.9189
0.6431
0.0899
0.1165
0.5188
0.8481
0.1044
0.4567
0.6779
inverse_tangent_BC
0.364629
0.649348
0.762714
0.458148
0.465764
0.828914
0.967387
0.608790
0.981121
1.229583
1.250270
1.071958
0.645652
1.139944
1.034867
0.904380
6
9. Using MATLAB to Solve the Peng-Robinson Equation of State
The Peng-Robinson equation of state contains 2 empirical parameters a and b, and its
conventional form is given by:
P=
where
RT
a
− 2
Vm − b
Vm + 2bVm − b 2
 αR 2TC2 

a = 0.45724
P
C


 RT
b = 0.07780 C
 PC



α = [1 + (0.37464 + 1.54226ω − 0.26992ω 2 )(1 − Tr0.5 )]2
Tr =
T
TC
 P VAP (Tr = 0.7) 

PC


ω = −1.0 − log10 
The variables are defined by:
P
Vm
T
R
TC
PC
PVAP
=
=
=
=
=
=
=
pressure in MPa
molar volume in cm3/gmole
temperature in K
gas constant (8.314 cm3-MPa/gmole-K)
the critical temperature (150.86 K for argon)
the critical pressure (4.898 MPa for argon)
vapor pressure (0.4946 MPa at Tr = 0.7 for argon)
Alternatively, this equation of state can be expressed in polynomial forms as given below:
A=
aP
R 2T 2
B=
bP
RT
Z 3 + ( B − 1) Z 2 + ( A − 2 B − 3B 2 ) Z + (− AB + B 2 + B 3 ) = 0
where Z =
PVm
, the compressibility factor
RT
7
Use MATLAB to compute the molar volume of argon using both the conventional form
and the polynomial form at T = 105.6 K and P = 0.498 MPa. The two solutions should
be the same. Because the two equation forms are both cubic in molar volume, you
should obtain 3 different answers in volume. Briefly describe the physical meaning or
significance of each answer.
Answers:
There are 3 roots or molar volumes for argon from the Peng-Robinson equation of
state. The roots are 1580.9562, 134.3314, and 27.7585 cm3/gmol. The largest number
represents the molar volume of argon in the gaseous state, while the smallest number
represents the molar volume of argon in the liquid state at the given temperature and
pressure. The middle number has no physical meaning or significance.
Solution:
MATLAB Script File:
%
% CHE656: Problem #9
%
clc
clear
format short
% Solve the traditional form of the equation
R = 8.314;
Tc = 150.86;
Pc = 4.898;
Pvap = 0.4946;
w = -1.0 - log10(Pvap/Pc);
T = 105.6;
P = 0.498;
Tr = T/Tc;
alpha = (1+(0.37464+1.54226*w-0.26992*w^2)*(1-Tr^0.5))^2;
a = 0.45724*(alpha*R^2*Tc^2/Pc);
b = 0.07780*R*Tc/Pc;
syms V
F = P - R*T/(V-b) + a/(V^2+2*b*V-b^2);
Vm = solve(F);
Vm = double(Vm);
fprintf ('The three molar volumes from tradtional form are:\n')
fprintf ('%10.4f\n',Vm)
% Solve the polynomial form of the equation
A = a*P/(R^2*T^2);
B = b*P/(R*T);
syms Z
F = Z^3+(B-1)*Z^2+(A-2*B-3*B^2)*Z+(-A*B+B^2+B^3);
%a2 = B-1
%a1 = A-2*B-3*B^2
%a0 = -A*B+B^2+B^3
8
Z = solve(F);
Z = double(Z);
Vm = (R*T/P).*Z;
fprintf ('\n')
fprintf ('The three molar volumes from polynomial form are:\n')
fprintf ('%10.4f\n',Vm)
MATLAB Output :
The three molar volumes from tradtional form are:
1580.9562
27.7585
134.3314
The three molar volumes from polynomial form are:
1580.9562
27.7585
134.3314
9
12. Statistical Analyses Using Matrices in MATLAB
You are to use MATLAB to carry out some statistical analyses on a set of data given in
the table below.
Undergr. GPA
ChEPS GPA
3.58
3.33
3.28
3.27
3.15
3.33
2.97
3.09
3.22
2.89
3.05
2.90
3.10
3.01
3.31
3.45
3.22
3.19
3.14
3.16
3.45
3.97
3.71
3.57
3.54
3.61
3.75
3.72
3.42
4.00
3.66
3.89
3.28
3.71
3.64
3.64
3.64
3.64
3.61
3.78
3.64
3.85
The data consist of undergraduate GPAs of 21 students from
ChEPS Class-11 and their ChEPS GPAs upon graduation. We
like to find out if there is a strong correlation between these two
set of data. You are asked to carry out a number of statistical
analyses using MATLAB’s matrix functionality. You must first
create a 21×2 matrix that contain the data given in the table.
From that matrix, do the following:
1. Compute the maximum, minimum, mean, and median of
the students’ undergraduate GPAs and ChEPS GPAs.
2. Compute the standard deviations of undergraduate
GPAs and ChEPS GPAs using the following formula:
1 N
∑ ( xi − µ ) 2
N − 1 i =1
where N = total data size and µ = mean of the data. Do
not use the built-in function in MATLAB to compute
this number.
s=
3. Compute the Pearson’s correlation coefficient r, which
indicates the correlation between undergraduate GPA
and ChEPS GPA, as given by the formula below:
r=
N (∑ xy ) − (∑ x ∑ y )
[N (∑ x ) − (∑ x) ] [N (∑ y ) − (∑ y ) ]
2
2
2
where x = undergraduate GPA and y = ChEPS GPA.
Please note that you are not allowed to use for loop or any control statements in your
calculations. Use only vectors, matrices, and their manipulations. Finally, use fprintf to
output your results that look like the table below. The spacing is not important as long
as your table looks neat. Note that I want 2 decimal places for the maximum and the
minimum and 3 decimal places for the mean, the median, and the standard deviation.
For the correlation coefficient, I want an answer in 4 decimal places.
Maximum Minimum Mean Median STDEV
------------- ------------- --------- ----------- --------Underg GPA
x.xx
x.xx
x.xxx
x.xxx
x.xxx
ChEPS GPA
x.xx
x.xx
x.xxx
x.xxx
x.xxx
Pearson correlation coefficient r = 0.xxxx
10
2
Maximum
Minimum
Mean
Median
STDEV
Underg GPA
________
________
________
________
________
ChEPS GPA
________
________
________
________
________
Pearson’s correlation coefficient r = ____________
Solution:
MATLAB Script File:
clc
clear
format short
% CHE656: Problem #12
GPA = [3.58 3.97; 3.33 3.71; 3.28 3.57; 3.27 3.54; 3.15 3.61; ...
3.33 3.75; 2.97 3.72; 3.09 3.42; 3.22 4.00; 2.89 3.66; ...
3.05 3.89; 2.90 3.28; 3.10 3.71; 3.01 3.64; 3.31 3.64; ...
3.45 3.64; 3.22 3.64; 3.19 3.61; 3.14 3.78; 3.16 3.64; ...
3.45 3.85];
U_GPA = GPA(:,1);
C_GPA = GPA(:,2);
n = length(U_GPA);
max_U_GPA = max(U_GPA);
min_U_GPA = min(U_GPA);
ave_U_GPA = mean(U_GPA);
med_U_GPA = median(U_GPA);
max_C_GPA = max(C_GPA);
min_C_GPA = min(C_GPA);
ave_C_GPA = mean(C_GPA);
med_C_GPA = median(C_GPA);
%
% mu_U_GPA = std(U_GPA)
% mu_C_GPA = std(C_GPA)
%
square = (U_GPA - ave_U_GPA).^2;
sum_square = sum(square);
std_U_GPA = sqrt(sum_square/(n-1));
square = (C_GPA - ave_C_GPA).^2;
sum_square = sum(square);
std_C_GPA = sqrt(sum_square/(n-1));
%
term1 = n*sum(U_GPA.*C_GPA);
term2 = sum(U_GPA)*sum(C_GPA);
term3 = n*sum(U_GPA.^2)-(sum(U_GPA)).^2;
term4 = n*sum(C_GPA.^2)-(sum(C_GPA)).^2;
r = (term1-term2)/sqrt((term3*term4));
fprintf ('
Maximum Minimum Mean Median STDEV\n')
fprintf ('
------- ------- ------- ------- -------\n')
fprintf ('Underg GPA %6.2f %9.2f %10.3f %9.3f %9.3f\n', max_U_GPA, ...
11
min_U_GPA, ave_U_GPA, med_U_GPA, std_U_GPA)
fprintf ('ChEPS GPA %7.2f %9.2f %10.3f %9.3f %9.3f\n', max_C_GPA, ...
min_C_GPA, ave_C_GPA, med_C_GPA, std_C_GPA)
fprintf ('\n')
fprintf ('Pearson correlation coefficient r = %6.4f\n', r)
Output from MATLAB:
12