Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
LAB ASSESSMENT RUBRICS
Student Name: AmmarUllah Khan
Program
(20 pts)
(Excellent)
(20~18)
(Good)
(17-14)
(Average)
(13-10)
Presence in
Lab
20%
Programme
r was
present and
actively
participated
during Lab
Programmer
was few
minutes late
but actively
participated
during Lab
Programmer
was late by
half but
actively
participated
during Lab
Program is
stylistically
well
designed
Few
inappropriat
e design
choices (i.e.
poor
variable
names,
improper
indentation)
Several
inappropriat
e design
choices (i.e.
poor
variable
names,
improper
indentation)
Standards
20%
Program
execution
30%
Design and
correct output
20%
Documentatio
n
10%
Program
executes
correctly
with no
error’s
messages
Program
displays
correct
output with
no errors
Program is
well
documente
d
Program
Program
executes
executes
with a minor with a minor
(easily fixed (not easily
errors)
fixed error)
Output has
minor errors
Output has
multiple
errors
Missing one
required
comment
Missing two
or more
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comments
(Fair)
(09-05)
(Poor)
(04-00)
Programmer
Programmer
failed to show
was late by
up at all or
half and did
was in lab but
not actively
failed
participated
participate in
during Lab
activities
Program is
poorly written
Failed to
write
Program
Program does
Program does
not execute at
not execute
all
Output is
incorrect
Output is
completely
incorrect
Most
All
documentatio documentatio
n missing
n missing
Total Score =
Remarks if any:
Signature with date: __________________________
1
/20
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
TITLE:
Lab 7: Numerical differentiation and integration
DESCRIPTION:
Calculate the derivatives analytically and numerically for y= 4x^4 –3x^3 +5 at
0≤x≤3, interval = 0.1, and show them.
CODES:
1. m. file: lab71
close all; clc;
x=0:0.1:3;
m=length(x);
m1=m-1;% m-1.
p=[1 -3 0 0 5];
y= polyval(p,x);
dp=[1*4 -3*3 0 0 ];
dydx_a = polyval(dp,x);
dydx_n=diff(y)./diff(x);
fprintf('sol)\n x y dydx_anal dydx_nume \n');
for i=1:m1
fprintf(' %5.2f %12.3e %12.3e %12.3e \n',
x(i),y(i),dydx_a(i),dydx_n(i));
end
plot(x,y,'o-g', x,dydx_a,'or-',
x(1:m1),dydx_n,'^b-');
legend('given data','analytical differentiation',
'numerical differentiation.');
2
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
2. command window:
sol)
x
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.50
2.60
2.70
2.80
2.90
y
5.000e+00
4.997e+00
4.978e+00
4.927e+00
4.834e+00
4.688e+00
4.482e+00
4.211e+00
3.874e+00
3.469e+00
3.000e+00
2.471e+00
1.890e+00
1.265e+00
6.096e-01
-6.250e-02
-7.344e-01
-1.387e+00
-1.998e+00
-2.545e+00
-3.000e+00
-3.335e+00
-3.518e+00
-3.517e+00
-3.294e+00
-2.813e+00
-2.030e+00
-9.049e-01
6.096e-01
2.561e+00
dydx_anal
0.000e+00
-8.600e-02
-3.280e-01
-7.020e-01
-1.184e+00
-1.750e+00
-2.376e+00
-3.038e+00
-3.712e+00
-4.374e+00
-5.000e+00
-5.566e+00
-6.048e+00
-6.422e+00
-6.664e+00
-6.750e+00
-6.656e+00
-6.358e+00
-5.832e+00
-5.054e+00
-4.000e+00
-2.646e+00
-9.680e-01
1.058e+00
3.456e+00
6.250e+00
9.464e+00
1.312e+01
1.725e+01
2.187e+01
dydx_nume
-2.900e-02
-1.950e-01
-5.050e-01
-9.350e-01
-1.461e+00
-2.059e+00
-2.705e+00
-3.375e+00
-4.045e+00
-4.691e+00
-5.289e+00
-5.815e+00
-6.245e+00
-6.555e+00
-6.721e+00
-6.719e+00
-6.525e+00
-6.115e+00
-5.465e+00
-4.551e+00
-3.349e+00
-1.835e+00
1.500e-02
2.225e+00
4.819e+00
7.821e+00
1.126e+01
1.514e+01
1.951e+01
2.439e+01
3
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
RESULT & DISCUSSION:
We’ve calculated the value of derivative for every 0.1 step values of x. MATLAB
allows us to obtain a range of values for the given equation.
4
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
CASE STUDY:
CODES:
1. m. file: lab72
close all; clc;
t=[ 0 2.25 4.5 6.33 8.00 10.25 12.00 13.50 ...
15.60 17.85 19.60 27.00 30.00 38.00 41.00 45.00 47.00
57.00 63.00];
Cbr=[ 0.3335 0.2965 0.2660 0.2450 0.2255 0.2050 0.1910
0.1794 0.1632 0.1500 0.1429 0.1160 0.1053 0.0830 0.0767
0.0705 0.0678 0.0553 0.0482];
m=length(t);
m1=m-1;
t1=t(1:m1);
5
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
dCbrdt = diff(Cbr)./diff(t);
y = log(-dCbrdt);
x = log(Cbr(1:m1) );
p=polyfit(x,y,1);
n=p(1);
lnk=p(2); k=exp(lnk);
y_hat = polyval(p,x);
y_bar = mean(y);
r2= sum((y_hat - y_bar).^2)/sum((y - y_bar).^2);
subplot(2,1,1); plot(x,y,'o');
x_calc=-3:0.1:-0.7
y_calc= polyval(p,x_calc);
plot(x_calc,y_calc,'b-');
title('linear plot for ln(-dC_{Br2}/dt) vs.
ln(C_{Br2})');
xlabel('ln(C_{Br2})'); ylabel('ln(-dC_{Br2}/dt)');
s1=num2str(lnk,4);
s2=num2str(n,4);
s3=num2str(r2,4);
eq=['ln(-dCBr_2/dt)= ',s1,' + ',s2,' * ln(CBr_2)',' r^2 =
',s3];
text(-2.5,-7.1,eq);
fprintf('The calculated equation is %s.\n',eq);
dCbrdt_cal = -k*Cbr.^n;
subplot(2,1,2); plot(t1,dCbrdt,'bo',t,dCbrdt_cal,'g-');
legend('dC_{Br2}/dt (exp)','dC_{Br2}/dt (calc)')
xlabel('t (min)');
ylabel('dC_{Br2}/dt (g-mol/dm^3 .min)');
fprintf(' t(min) Cbr dCbr/dt ln(Cbr) ln(-dVbr/dt) \n');
for i=1:1:m1
fprintf(' %6.2f %12.3e %12.3e %12.3e %12.3e
\n',t(i),Cbr(i),dCbrdt(i),x(i),y(i) );
end
2. Command window:
The calculated equation is ln(-dCBr_2/dt)= -2.342 + 1.575 * ln(CBr_2)
r^2 = 0.9841.
t(min) Cbr dCbr/dt ln(Cbr) ln(-dVbr/dt)
0.00 3.335e-01 -1.644e-02 -1.098e+00 -4.108e+00
2.25 2.965e-01 -1.356e-02 -1.216e+00 -4.301e+00
4.50 2.660e-01 -1.148e-02 -1.324e+00 -4.468e+00
6.33 2.450e-01 -1.168e-02 -1.406e+00 -4.450e+00
8.00 2.255e-01 -9.111e-03 -1.489e+00 -4.698e+00
6
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
10.25 2.050e-01 -8.000e-03 -1.585e+00 -4.828e+00
12.00 1.910e-01 -7.733e-03 -1.655e+00 -4.862e+00
13.50 1.794e-01 -7.714e-03 -1.718e+00 -4.865e+00
15.60 1.632e-01 -5.867e-03 -1.813e+00 -5.138e+00
17.85 1.500e-01 -4.057e-03 -1.897e+00 -5.507e+00
19.60 1.429e-01 -3.635e-03 -1.946e+00 -5.617e+00
27.00 1.160e-01 -3.567e-03 -2.154e+00 -5.636e+00
30.00 1.053e-01 -2.788e-03 -2.251e+00 -5.883e+00
38.00 8.300e-02 -2.100e-03 -2.489e+00 -6.166e+00
41.00 7.670e-02 -1.550e-03 -2.568e+00 -6.470e+00
45.00 7.050e-02 -1.350e-03 -2.652e+00 -6.608e+00
47.00 6.780e-02 -1.250e-03 -2.691e+00 -6.685e+00
57.00 5.530e-02 -1.183e-03 -2.895e+00 -6.739e+00
RESULTS AND DISCUSSION:
The resulting graphs show the original non-linear graph and then the
linear regression version with natural logarithm applied.
7
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
CASE STUDY
1. m.file: lab73
2. close all; clc;
3. t=[ 0 2.25 4.5 6.33 8.00 10.25 12.00 13.50 ...
4. 15.60 17.85 19.60 27.00 30.00 38.00 41.00 45.00 47.00 57.00 63.00]';
5. CBr=[ 0.3335 0.2965 0.2660 0.2450 0.2255 0.2050 0.1910 0.1794 ...
6. 0.1632 0.1500 0.1429 0.1160 0.1053 0.0830 0.0767 0.0705 0.0678 0.0553
0.0482]'; % a)
7. x= t; y= CBr; y_bar = mean(y); % take x, y and ybar
8. p=polyfit(x,y,3); % make coefficients of the polynomial with
polyfit()
9. y_hat=polyval(p,x); % take calculate values with polyval()
10.
r2=sum((y_hat-y_bar).^2)/sum((y-y_bar).^2);
11.
fprintf('Sol)\nC_Br2 = %10.3e*t^3 +%10.3e*t^2 +%10.3e*t
+%10.3e, r^2=%6.4f\n\n',p,r2);
12.
figure(1); set(1,'position',[10 100 500 600],'color','w');
13.
subplot(311); plot(t,CBr,'r.',t,y_hat,'b-');
14.
xlabel('t [min]'); ylabel('C_{Br2}');
15.
legend('given-data','calculated-value')
16.
%b)
17.
m=length(t); m1=m-1; t1=t(1:m1); CBr1=CBr(1:m1); % to fit the
length, for later use.
18.
dp=[3*p(1) 2*p(2) p(3)]; % coefficients of derivate of 3rd
order polynomial.
19.
dCBrdt_a= polyval(dp, t1); % with polyval() % analytical
derivatives.
20.
dCBrdt_n=diff(CBr)./diff(t); % with diff(). array operation.
21.
subplot(312); plot(t1,dCBrdt_a,'r.', t1,dCBrdt_n,'bo');
22.
legend('analytical-diff', 'numerical-diff');
23.
xlabel('t [min]'); ylabel('dC_{Br2}/dt');
24.
% c) Applying equation (B) for analytical derivatives
25.
x_a = log(CBr1); % take x use for eq(B).
26.
y_a = log(-dCBrdt_a); % calculate y with analytical equation
for eq(B).Attention - sign.
8
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
27.
p_a=polyfit(x_a, y_a, 1); % calculate parameters with
polyfit().
28.
y_hat_a= polyval(p_a, x_a ); % take the calculated values with
polyval().
29.
y_bar_a = mean(y_a); % average with mean();
30.
r2_a= sum((y_hat_a - y_bar_a).^2)/sum((y_a - y_bar_a).^2);
31.
n_a=p_a(1); k_a=exp(p_a(2)); % calculate k and n using
coefficient of eq(B)
32.
% Applying equation (B) for numerical derivatives.
33.
% copy above(from x_a= ; to k_a= ;), and replace _n to _a.
34.
x_n = log(CBr1);
35.
y_n = log(-dCBrdt_n);
36.
p_n=polyfit(x_n, y_n, 1);
37.
y_hat_n= polyval(p_n, x_n );
38.
y_bar_n = mean(y_n);
39.
r2_n= sum((y_hat_n - y_bar_n).^2)/sum((y_n - y_bar_n).^2);
40.
n_n=p_n(1); k_n=exp(p_n(2));
41.
subplot(313); % lower figure is plotted with loglog(),
42.
% the values which is not taken by log are entered.
43.
% and take exp() to hat values for same x axis
44.
loglog(CBr1,-dCBrdt_a,'r.', CBr1,exp(y_hat_a),'r-', CBr1,dCBrdt_n,'bo',CBr1,exp(y_hat_n),'b-');
45.
46.
legend('analytical-diff-data','analytical-diffcalc','numerical-diff-data','numerical-diff-calc');
47.
48.
49.
50.
51.
xlabel('C_{Br2}'); ylabel('-dC_{Br2}/dt ');
fprintf('eq(B) : ln(-dCBr2/dt) = ln(k) + n ln(CBr2)\n')
fprintf(' calculated values k n r^2\n');
fprintf(' from analytical eq. %12.3e %12.3e %6.4f\n',k_a, n_a,
r2_a);
52.
fprintf(' from numerical eq. %12.3e %12.3e %6.4f\n',k_n, n_n,
r2_n);
53.Command window:
Sol)
C_Br2 = -2.043e-06*t^3 + 2.842e-04*t^2 +-1.428e-02*t + 3.266e-01,
r^2=0.9982
eq(B) : ln(-dCBr2/dt) = ln(k) + n ln(CBr2)
calculated values k n r^2
from analytical eq. 1.048e-01 1.603e+00 0.9404
9
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
from numerical eq. 9.615e-02 1.575e+00 0.9841
RESULTS & DISCUSSION:
After running the program we obtain the graphs with comparison between
analytical and numerical differentiation. The command window showed the
required values for k, n and r2 for both analytical and numerical differentiation
methods.
10
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
Task 2: Numerical differentiation
Programs:
Q4: Integrate from -1 to 1 with equal space of .
Function File: lab7ex4fun.m
function f=lab7ex4fun(x)
f= x.^3 +1 ; %integrand
1.mfile: lab7ex4
close all; clc;
h=0.1; % equal space a=x(1), ... ,x(n1)=b
x= (-1 : h : 1); % interval divide by n, the number of calculation point is
n+1.
n1=length(x); % find number of calculate point = 21
% given function
f= x.^3 +1 ;
% analytical solution
J_anal=x.^4/4-x(1).^4/4 + x - x(1); % F(x(i)) - F(a)
% for rectangular rule
n=n1-1;
for i=1:n % there are 20 central points.
sum_f = 0;
x_center(i)=(x(i+1)+x(i))/2; %% make central points.
for j=1 : i
sum_f = sum_f + x_center(j)^3 +1; % calculate function value of central
points and summing
end
J_rect(i) = h*sum_f;
end
% For trapezoidal rule
for i=1:n1 % loop for every calculation points.
if (i==1); sum_f=0; % no interval.
elseif (i==2); sum_f=( f(1)+f(i) )/2; % only two ends.
else % general cases i>=3.
sum_f=( f(1)+f(i) )/2; % calculate for two ends.
for k=2 : i-1; sum_f=sum_f+x(k)^3 +1; end % except of two ends.
end % end of if
J_trap(i) = h*sum_f;
end % end of for i =1:n1
%Simpson rule & Simpson 3/8 rule
warning off
for i=1:n1; J_simpson(i) =quad ('lab7ex4fun',x(1),x(i)); end
for i=1:n1; J_simpson8(i)=quad ('lab7ex4fun',x(1),x(i)); end
plot(x,f,'ro-',x,J_anal,'r*',x_center,J_rect,'bv-', ...
x,J_trap,'g^-',x,J_simpson,'mo-',x,J_simpson8,'k.-');
xlabel('x');ylabel('x^3');
legend('given data','analytical Integral','rectangular rule', ...
'trapezoidal rule','Simpson (quad)','Simpson (cubic)');
11
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
2. Command window:
Function File: lab7ex5fun.m
function J=lab7ex5fun(x,p) %integrand, receive p from main
J=polyval(p,x);
1.mfile: lab7ex5
close all; clc;
X =[0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85]; % make data
12
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
mrA =[0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125
0.00100];
FA0=0.867;
x=X; y=1./mrA; %% make variables for polyfit().
plot(x,y,'ro'); hold on; grid on;
p=polyfit(x,y,4); % calculate coef of degree 4 polynomial, % calculated p
is used in quad().
xn = 0: 0.01: 0.85; % make x-axis variable to draw in figure(interval 0.01)
yn = polyval(p,xn); % make y-axis variables;
plot(xn,yn,'b-');
Xp=X(1:2:9); mrAp=mrA(1:2:9); imrAp=y(1:2:9); % select variables for
printout.
fprintf('Plug Flow Reactor(PFR) Volume(V) for conversion(X)\n')
fprintf(' X -rA 1/(-rA) V \n');
for i=1: length(Xp)
J(i)=quad('lab7ex5fun',0,Xp(i),[],[],p);
V(i)=FA0*J(i);
fprintf('%7.2f %8.5f %8.2f %8.2f\n',Xp(i),mrAp(i),imrAp(i),V(i));
end
plot(Xp,V,'mo-'); xlabel('Conversion [-]'); ylabel('1/-r_A or V[dm^3]');
legend('1/(-r_A)_{exp}','1/(-r_A)_{cal}','V(dm^3)');
2.Command window:
Plug Flow Reactor(PFR) Volume(V) for conversion(X)
X -rA 1/(-rA) V
0.00 0.00530 188.68
0.00
0.20 0.00500 200.00 33.13
0.40 0.00400 250.00 71.93
0.60 0.00250 400.00 125.81
0.80 0.00125 800.00 224.52
13
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
Function File: lab7ex6fun.m
function J=lab7ex6fun(x,p) %integrand, receive p from main
J=polyval(p,x);
1.mfile: lab7ex6
close all; clc;
T =[ 50 100 150 200 273.15 300 400 500 ...
600 700 800 900 1000 1100 1200 1300 1400 1500];
Cp=[ 34.06 41.30 48.79 56.07 68.74 73.93 94.01 112.59 ...
128.70 142.67 154.77 163.35 174.60 182.67 189.74 195.85 201.21 205.89];
% make variable for polyfit(). In this, no transformation is required.
14
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
plot(T,Cp,'o'); hold on; grid on;
p=polyfit(T,Cp,5); % p is used in quad.
T_new = 40: 1: 1500; % for line drowing, make x-variable for plot(interval
1);
Cp_calc = polyval(p,T_new); % make y-variable with polyval();
plot(T_new,Cp_calc,'b-');
warning off
Tref=25+273.15; % make ref temp
fprintf('Temp[K] Cp mean_Cp \n');
for i=1: length(T)
J_Cp(i)=quad('lab7ex6fun',Tref,T(i),[],[],p);
Cp_mean(i)=J_Cp(i)/(T(i)-Tref);
fprintf('%7.2f %8.2f %8.2f \n',T(i),Cp(i),Cp_mean(i));
end
plot(T,Cp_mean,'ro-'); xlabel('Temp (K)'); ylabel('Cp (kJ/kg-mol K)');
legend('Cp','Cp-calc','Cp-mean');
new_x_tick=(0:200:1600); % declare variable for rearrange of tick mark as
like figure.
set(gca,'xtick',new_x_tick); % set tick mark rearrange (for x-axis values
change)
2.Command Window:
Temp[K] Cp mean_Cp
50.00 34.06 52.70
100.00 41.30 56.53
150.00 48.79 60.72
200.00 56.07 65.17
273.15 68.74 71.99
300.00 73.93 74.54
400.00 94.01 84.09
500.00 112.59 93.44
600.00 128.70 102.33
700.00 142.67 110.61
800.00 154.77 118.22
900.00 163.35 125.17
1000.00 174.60 131.51
1100.00 182.67 137.32
15
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
1200.00 189.74 142.69
1300.00 195.85 147.69
1400.00 201.21 152.34
1500.00 205.89 156.63
16
Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
Function File: lab7ex7fun.m
function g=lab7ex7fun(V)
global P R T a b
g=P-( R*T/(V-b) -a/(V*(V+b)*sqrt(T) ) );
lab7ex7fun2.m
function J=lab7ex7fun2(x,p) %integrand, receive p from main
J=polyval(p,x);
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Date: 11-1-2022
LAB MANUAL
CHEMICAL PROCESS OPTIMIZATION (CH-404)
Lab No: _7___
Roll No: CH-18062
1.mfile: lab7ex7
close all; clc;
global P R T a b
%given data
R=0.08206 ; % atm.L/mol.K
Tc=647.4 ; Pc=218.3 ; % critical values
a=0.42747*R^2*Tc^(5/2)/Pc;
b=0.08664*R*Tc/Pc;
% make T, P for enthalpy calculate
TC=100 ; % given temperature in Celsius
T0= 273.15; % reference temperature = 0 C
T=TC+T0; % convert to Kelvin
P=1; % given Pressure (atm)
% initial guess, calculate by ideal gas low
V0=R*T/P ; % the unit of molar volume is [L/mol]
V=fzero('lab7ex7fun',V0);
Z=P*V/R/T ; % compressibility factor
fprintf('\n Pressure Temp. Molar Vol Zfactor \n')
fprintf('% 7.2f %7.3f %7.3f %7.4f \n', P,T,V,Z);
% given data for enthalpy calculation
H_l_o=0; % kJ/kg
DHvap_o=2501.3; %kJ/kg
coef=[7.700, 0.04594e-2, 0.2521e-5, -0.8587e-9]; % given data
p=coef(length(coef):-1:1); % coeff of highest order for polyval is 1.
change coeff order
J_Cp = quad('lab7ex7fun2',T0,T,[],[],p); % integration [cal/gmol]
J_Cp = J_Cp*4.1868/18; % [kJ/kg] convert units.
H_v_o = H_l_o + DHvap_o + J_Cp; % enthalpy at ref temp [kJ/kg]
H_calib = 101.4/18*R*T*(Z-1- 1.5*a/(b*R*T^1.5)*log(1+b/V)); % kJ/kg
H_v = H_v_o + H_calib;
fprintf('\n DHvap_o J_Cp H_v^o H_calib H_v (kJ/kg_steam)\n')
fprintf('%8.2f %8.2f %8.2f %8.2f %8.2f \n', DHvap_o, J_Cp, H_v_o, H_calib,
H_v);
fprintf('The steam Table value is 2676.1 at same conditions.\n')
2.Command Window:
Pressure Temp. Molar Vol Zfactor
1.00 373.150 30.403 0.9929
DHvap_o J_Cp H_v^o H_calib H_v (kJ/kg_steam)
2501.30 188.04 2689.34
-3.25 2686.09
The steam Table value is 2676.1 at same conditions
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