13 Problem Solving with MATLAB.pdf

8/11/2019 13 Problem Solving with MATLAB.pdf

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Cutlip and Shacham: Problem Solving in Chemical and Biochemical Engineering

Chapter 5

Problem Solving with MATLAB

Cheng-Liang Chen

PSELABORATORYDepartment of Chemical Engineering

National TAIWAN University


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where P = pressure in atm

V = molar volume in liters/g-mol

T = temperature in K

R = gas constant (R= 0.08206 (atm-liter/g-mol-K))T e = critical temperature in K

P e = critical pressure in atm

The compressibility factor is given by

z = P VRT

Equation (1) can be written, after considerable algebra, in terms of thecompressibility factor as a cubic equation (see Seader and Henley)

f(z) =z

3

z2

qz r= 0 (5)where r = A2B, q = B2 + B A2

A2 = 0.42747

Pr

T5/2r

B = 0.0866

PrTr

in whichPr is the reduced pressure (P/Pc) andTr is the reduced temperature

(T /Tc). Equation (5) can be solved analytically for three roots. Some of these


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roots are complex. Considering only the real roots, the sequence of calculationsinvolves the steps

C =f

33

+g

22

f = 3q 1

3 g =

27r 9q 227

IfC >0, there is one real solution for z given by

z = D+ E+ 1/3

D = (g/2 +

C)1/3 E = (g/2

C)1/3

IfC


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the maximal zk is selected as the true compressibility factor. (Note: let z = 0 forC


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%filename P5_01A_CCL

R = 0.08206;%Gas constant (L-atm/g-mol-K)

Tc = 647.4; %Critical temperature (K)

Pc = 218.3; %Critical pressure (atm)

a = 0.42747 * R ^ 2 * Tc ^ (5 / 2) / Pc;%Eq. (4-2), RK equation constb = 0.08664 * R * Tc / Pc; %Eq. (4-3),RK equation consta

Pr = 10 %1.2;%0.1; %Reduced pressure (dimensionless)

Tr = 3 %1; %Reduced temperature (dimensionless)

Asqr = 0.42747 * Pr / (Tr ^ 2.5);%Eq. (4-8)

B = 0.08664 * Pr / Tr; %Eq. (4-9)

r = Asqr * B; %Eq. (4-6)q = B ^ 2 + B - Asqr; %Eq. (4-7)

g = (-27 * r - (9 * q) - 2) / 27;%Eq. (4-12)

f = (-3 * q - 1) / 3; %Eq. (4-11)

C = (f / 3) ^ 3 + (g / 2) ^ 2; %Eq. (4-10)

if (C > 0)E1 = 0 - (g / 2) - sqrt(C); %Eq. (4-15)

D = (0 - (g / 2) + sqrt(C)) ^ (1 / 3);%Eq. (4-14)

E = sign(E1) * abs(E1) ^ (1 / 3);%Eq. (4-15)

z = D + E + 1 / 3;%Eq. (4-13), Compressibility factor (dimensionl

else

z = 0 ;

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end

P = Pr * Pc; %Pressure (atm)

T = Tr * Tc; %Temperature (K)

V = z * R * T / P %Molar volume (L/g-mol)

Pr =

10

Tr =

3

V =

0.0837

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%filename P5_01B_CCL

clear, clc, format compact, format short g

Tc = 647.4; %Critical temperature (K)

Pc = 218.3; %Critical pressure (atm)

Tr_set=[1 1.2 1.5 2 3];Pr_set(1) = 0.1;

Pr_set(2) = 0.2;

i = 2 ;

while Pr_set(i)


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end

end

end

disp( Compressibility Factor Versus Tr and Pr);

disp( Tabular Results);disp( );

disp(Pr\Tr 1.0 1.2 1.5 2 3 );

Res=[Pr_set z];

disp(Res);

subplot(1,2,1)

plot(Pr_set,z(:,1),-,Pr_set,z(:,2),+,Pr_set,z(:,3),*,...Pr_set,z(:,4),x,Pr_set,z(:,5),o,LineWidth,2);

set(gca,FontSize,14,Linewidth,2)

legend(Tr=1,Tr=1.2,Tr=1.5,Tr=2,Tr=3);

title(\bf Compressibility Factor Versus Tr and Pr,FontSize,12)

xlabel(\bf Reduced Pressure (Pr),FontSize,14);

ylabel(\bf Compressibility Factor (z),FontSize,14);

disp( Pause; Please press any key to continue ... )

pause

P_set=Pr_set.*Pc;

disp( Molar Volume Versus Tr and P);

disp( Tabular Results);

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disp( );

disp(Pr\Tr 1.0 1.2 1.5 2 3 );

Res=[P_set V];

disp(Res);

subplot(1,2,2)loglog(P_set,V(:,1),-,P_set,V(:,2),+,P_set,V(:,3),*,...

P_set,V(:,4),x,P_set,V(:,5),o,LineWidth,2);

legend(Tr=1,Tr=1.2,Tr=1.5,Tr=2,Tr=3);


title(\bf Molar Volume Versus Tr and P,FontSize,12)

xlabel(\bf Pressure (atm),FontSize,14);ylabel(\bf Molar Volume (L/g-mol),FontSize,14);

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function [z, V] = RKfun(Tr,Pr,Tc,Pc)

R = 0.08206; %Gas constant (L-atm/g-m

a = 0.42747 * R ^ 2 * Tc ^ (5 / 2) / Pc;%Eq. (4-2), RK equation const

b = 0.08664 * R * Tc / Pc; %Eq. (4-3),RK equation constant

Asqr = 0.42747 * Pr / (Tr ^ 2.5); %Eq. (4-8)B = 0.08664 * Pr / Tr; %Eq. (4-9)

r = Asqr * B; %Eq. (4-6)

q = B ^ 2 + B - Asqr; %Eq. (4-7)

g = (-27 * r - (9 * q) - 2) / 27; %Eq. (4-12)

f = (-3 * q - 1) / 3; %Eq. (4-11)

C = (f / 3) ^ 3 + (g / 2) ^ 2; %Eq. (4-10)if (C > 0)

E1 = 0 - (g / 2) - sqrt(C); %Eq. (4-15)

D = (0 - (g / 2) + sqrt(C)) ^ (1 / 3); %Eq. (4-14)

E = sign(E1) * abs(E1) ^ (1 / 3); %Eq. (4-15)

z = D + E + 1 / 3; %Eq. (4-13), Compressibility factor (dimenselse

z = 0 ;

end

P = Pr * Pc; %Pressure (atm)

T = Tr * Tc; %Temperature (K)

V = z * R * T / P; %Molar volume (L/g-mol)

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Calculation of Flow Rate In A Pipeline

Concepts Utilized: Application of the general mechanical energy balance for

incompressible fluids, and calculation of flow rate in a pipeline for various pipediameters and lengths.

Numerical Methods: Solution of a single nonlinear algebraic equation andalternative solution using the successive substitution method.

Problem Statement:The following figure shows a pipeline that delivers water at a constanttemperature T = 60oF from point 1 where the pressure is P1= 150 psig and theelevation is z1= 0 ft to point 2 where the pressure is atmospheric and theelevation is z2= 300 ft.

The density and viscosity of the water can be calculated from the following

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equations.

= 62.122 + 0.0122T 1.54 104T2 + 2.65 107T3 2.24 1010T4

ln = 11.0318 + 1057.51

T+ 214.624

where T is in oF, is in lbm/ft3 , and is in lbm/fts.

(a) Calculate the flow rate q (in gal/min) for a pipeline with effective length of

L= 1000 ft and made of nominal 8-inch diameter schedule 40commercial steelpipe. (Solution: v= 11.61 ft/s, gpm =1811 gal/min)

(b) Calculate the flow velocities in ft/s and flow rates in gal/min for pipelines at60oF with effective lengths ofL= 500, 1000, . . . 10, 000 ft and made of nominal

4-, 5-, 6- and 8-inch schedule 40 commercial steel pipe. Use the successivesubstitution method for solving the equations for the various cases and presentthe results in tabular form. Prepare plots of flow velocity v versusD andL, andflow rate qversus D andL.

(c) Repeat part (a) at temperatures T = 40, 60, and 100oF and display the results

in a table showing temperature, density, viscosity, and flow rate.

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Solution:

The general mechanical energy balance on an incompressible liquid applied to thiscase yields

1

2v2 + gz+

gcP

+ 2

fFLv2

D = 0 (4

20)

where v is the flow velocity in ft/s, g is the acceleration of gravity given byg= 32.174 ft/s2, z =z2 z1 is the difference in elevation (ft), gc is aconversion factor (in English units gc= 32.174ft-lbm/lbfs2), P =P2P1 is thedifference in pressure lbm/ft

2), fF is the Fanning friction factor, L is the length of

the pipe (ft) andD is the inside diameter of the pipe (ft). The use of thesuccessive substitution method requires Equation (4-20) to be solved for v as

v=

gz+

gcP

12 2fFLv

2

D

The equation for calculation of the Fanning friction factor depends on theReynolds number, Re= vD/, where is the viscosity in lbm/ft-s. For laminarflow (Re


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function P5_2C_CCL

clear, clc, format short g, format compact

D_list=[4.026/12 5.047/12 6.065/12 7.981/12]; % Inside diameter of pi

T = 60; %Temperature (deg. F)

for i = 1:4D = D_list(i);

j=0;

for L=500:500:10000

j = j+1;

L_list(j)=L; % Effective length of pipe (ft)

[v(j,i),fval]=fzero(@NLEfun,[1 20],[],D,L,T);if abs(fval)>1e-10

disp([No Conv. for L = num2str(L) and D = num2str(D)]);

end

q(j,i) = v(j,i) * pi * D ^ 2 / 4* 7.481 * 60; %Flow rate (gpm)

end

end

disp( Flow Velocity (ft/s) versus Pipe Length and Diameter);


disp();

disp( L\D D=4" D=5" D=6" D=8");

Res=[L_list v];

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disp(Res);

subplot(1,2,1)

plot(L_list,v(:,1),-,L_list,v(:,2),+,L_list,v(:,3),*,...

L_list,v(:,4),x,LineWidth,2);

set(gca,FontSize,14,Linewidth,2)legend( D=4", D=5", D=6", D=8");

title(\bf Flow Velocity,FontSize,12)

xlabel(\bf Pipe Length (ft),FontSize,14);

ylabel(\bf Velocity (ft/s),FontSize,14);


pausedisp( Flow Rate (gpm) versus Pipe Length and Diameter);

disp( Tabular Results); disp();

disp( L\D D=4" D=5" D=6" D=8");

Res=[L_list q(:,1) q(:,2) q(:,3) q(:,4)];

disp(Res);

subplot(1,2,2)

plot(L_list,q(:,1),-,L_list,q(:,2),+,L_list,q(:,3),*,...

L_list,q(:,4),x,Linewidth,2);


legend( D=4", D=5", D=6", D=8");

title(\bf Flow rate,FontSize,14)

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xlabel(\bf Pipe Length (ft),FontSize,14);

ylabel(\bf Flow rate (gpm),FontSize,14);

function fv = NLEfun(v,D,L,T)epsilon = 0.00015;%Surface rougness of the pipe (ft)

rho = 62.122+T*(0.0122+T*(-0.000154+T*(0.000000265-...

(T*0.000000000224)))); %Fluid density (lb/cu. ft

deltaz = 300; %Elevation difference (ft)

deltaP = -150; %Pressure difference (psi)

vis = exp(-11.0318 + 1057.51 / (T + 214.624)); %Fluid viscosity (lbm/pi = 3.1416; %The constant pi

eoD = epsilon / D; %Pipe roughness to diameter ratio (dimensionless)

Re = D * v * rho / vis; %Reynolds number (dimesionless)

if (Re < 2100) %Fanning friction factor (dimensionless)

fF = 16 / Re;

else

fF=1/(16*log10(eoD/3.7-(5.02*log10(eoD/3.7+14.5/Re)/Re))^2);

end

fv=v-sqrt((32.174*deltaz+deltaP*144*32.174/rho)...

/(0.5-(2*fF*L/D))); %velocity (ft/s)

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Flow Velocity (ft/s) versus Pipe Length and Diameter

Tabular Results

L\D D=4" D=5" D=6" D=8"

500 10.773 12.516 14.15 17.035

1000 7.4207 8.6048 9.7032 11.6131500 5.9721 6.9243 7.8051 9.3295

2000 5.1188 5.9361 6.6912 7.9953

2500 4.5409 5.2674 5.9382 7.0953

3000 4.1168 4.7769 5.3861 6.4362

3500 3.7888 4.3975 4.9592 5.927

4000 3.5255 4.093 4.6166 5.51854500 3.3082 3.8416 4.3338 5.1815

5000 3.1249 3.6297 4.0953 4.8973

5500 2.9677 3.4478 3.8907 4.6535

6000 2.8309 3.2896 3.7128 4.4415

6500 2.7106 3.1504 3.5561 4.2548

7000 2.6036 3.0266 3.4169 4.0889

7500 2.5077 2.9156 3.292 3.9402

8000 2.4211 2.8154 3.1793 3.8059

8500 2.3424 2.7244 3.0769 3.6838

9000 2.2706 2.6412 2.9832 3.5723

9500 2.2046 2.5648 2.8972 3.4698


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9000 90.099 164.71 268.65 557.05

9500 87.48 159.94 260.91 541.07

10000 85.064 155.54 253.76 526.33

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Adiabatic Operation of A Tabular Reactor for

Cracking of Acetone

Concepts Utilized: Calculation of the conversion and temperature profile in anadiabatic tubular reactor. Demonstration of the effect of pressure and heatcapacity change on the conversion in the reactor.

Numerical Methods:

Solution of simultaneous ordinary differential equations.

Problem Statement:

The irreversible, vapor-phase cracking of acetone (A) to ketene (B) and methane(C) that is given by the reaction

CH3COCH3 CH2CO+ CH4is carried out adiabatically in a tubular reactor. The reaction is first order withrespect to acetone and the specific reaction rate can be expressed by

ln(k) = 34.3434222

T (4-26)

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wherek is in s1 andT is in K. The acetone feed flow rate to the reactor is 8000kg/hr, the inlet temperature is T= 1150 K and the reactor operates at theconstant pressure ofP= 162 kPa (1.6 atm). The volume of the reactor is 4 m3.

The material balance equations for the plug-flow reactor are given by

dFAdV

= rA (4-27)

dFBdV

= rA (4-28)dFC

dV =

rA

(4-29)

where FA, FB, andFCare flow rates of acetone, ketene, and methane in g-mol/s,respectively and rA is the reaction rate ofA in g-mol/m

3s. The reaction is firstorder with respect to acetone, thus

rA= kCAwhere CA is the concentration of acetone in g-mol/m

3. For a gas phase reactor,using the appropriate units of the gas constant, the concentration of the acetonein g-mol/m3 is obtained by

CA=

1000yAP

8.31T


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reaction and the molar heat capacities.

H = 80770 + 6.8(T 298) 0.00575(T2 2982) 1.27 106(T3 2983)CpA = 26.60 + 0.183T

45.86

106T2

CpB = 20.04 + .0945T 30.95 106T2CpC = 13.39 + 0.077T 18.71 106T2

Galculate the final conversion and the final temperature ofP = 1.6, 1.8, . . . , 5.0atm, for acetone feed rates ofFA0= 10, 20, 30, 35, and38.3 mol/s where nitrogenis fed to maintain the total feed rate38.3 mol/s in all cases. Present the results intabular form and prepare plots of final conversion versus P andFA0 and finaltemperature versus P andFA0.

function P5_03C_CCL

clear, clc, format short g, format compactFA0set = [10 20 30 35 38.3]; %Feed rate of acetone in kg-mol/s

P_set(1)=1.6;

i=1;

while P_set(i)


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end

n_P = size(P_set);

for i=1:5

FA0=FA0set(i);

for j=1:n_PP=P_set(j)*101.325; % Pressure in kPa

y0=[FA0; 0; 0; 1035; 0];

[V,y] = ode45(@ODEfun,[0 4],y0,[],FA0,P);

Xfin(j,i)=y(end,5);

Tfin(j,i)=y(end,4);

endend

% -------------------------------------------------------------------

disp( Final Conversion versus FA0 and Pressure);


disp();

disp( Pressure FA0=10 FA0=20 FA0=30 FA0=35 FA0=38.3 );

Res=[P_set Xfin(:,1) Xfin(:,2) Xfin(:,3) Xfin(:,4) Xfin(:,5)];

disp(Res);

subplot(1,2,1)

plot(P_set,Xfin(:,1),-,P_set,Xfin(:,2),+,P_set,Xfin(:,3),*,...

P_set,Xfin(:,4),x,P_set,Xfin(:,5),o,LineWidth,2);

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legend(FA0=10,FA0=20,FA0=30,FA0=35,FA0=38.3);

title(\bf Final Conversion versus FA0 and Pressure,FontSize,12)

xlabel(\bf Pressure (atm),FontSize,14);

ylabel(\bf Final Conversion,FontSize,14);disp( Pause; Please press any key to continue ... )

pause

disp( Final Temperature versus FA0 and Pressure);


disp();

disp( Pressure FA0=10 FA0=20 FA0=30 FA0=35 FA0=38.3 );Res=[P_set Tfin(:,1) Tfin(:,2) Tfin(:,3) Tfin(:,4) Tfin(:,5)];

disp(Res);

subplot(1,2,2)

plot(P_set,Tfin(:,1),-,P_set,Tfin(:,2),+,P_set,Tfin(:,3),*,...

P_set,Tfin(:,4),x,P_set,Tfin(:,5),o,Linewidth,2);


legend(FA0=10,FA0=20,FA0=30,FA0=35,FA0=38.3);

title(\bf Final Temperature versus FA0 and Pressure,FontSize,12)

xlabel(\bf Pressure (atm),FontSize,14);

ylabel(\bf Temperature (K),FontSize,14);

% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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function dYfuncvecdV = ODEfun(V,Yfuncvec,FA0,P);

FA = Yfuncvec(1);

FB = Yfuncvec(2);

FC = Yfuncvec(3);

T = Yfuncvec(4);XA = Yfuncvec(5);

k = 8.2E14 * exp(-34222 / T); %Reaction rate constant in s-1

FN = 38.3 - FA0; %Feed rate of nitrogene in kg-mol/s

yA = FA / (FA + FB + FC + FN); %Mole fraction of acetone

CA = yA * P * 1000 / (8.31 * T); %Concentration of acetone in k-mol/m

yB = FB / (FA + FB + FC + FN); %Mole fraction of keteneyC = FC / (FA + FB + FC + FN); %Mole fraction of methane

rA = -k * CA; %Reaction rate in kg-mole/m3-s

deltaH = 80770 + 6.8 * (T - 298) - .00575 * (T ^ 2 - 298 ^ 2)...

- 1.27e-6 * (T ^ 3 - 298 ^ 3);

CpA = 26.6 + .183 * T - 45.86e-6 * T ^ 2; %Heat capacity of acetone

CpB = 20.04 + 0.0945 * T - 30.95e-6 * T ^ 2; %Heat capacity of ketene

CpC = 13.39 + 0.077 * T - 18.71e-6 * T ^ 2; %Heat capacity of methane

CpN = 6.25 + 8.78e-3 * T - 2.1e-8 * T ^ 2; %Heat capacity of nitrogen

dFAdV = rA; %Differential mass balance on acetone

dFBdV = -rA; %Differential mass balance on ketene

dFCdV = -rA; %Differential mass balance on methane

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dTdV = (-deltaH)*(-rA)/(FA*CpA+FB*CpB+FC*CpC+FN*CpN);

%Differential enthalpy balance

dXAdV = -rA / FA0; %Conversion of acetone

dYfuncvecdV = [dFAdV; dFBdV; dFCdV; dTdV; dXAdV];

Final Conversion versus FA0 and Pressure

Tabular Results

Pressure FA0=10 FA0=20 FA0=30 FA0=35

1.6 0.31358 0.27759 0.26394 0.25964

1.8 0.32043 0.28346 0.26946 0.26505

2 0.32651 0.28867 0.27436 0.269862.2 0.33197 0.29335 0.27877 0.27419

2.4 0.33693 0.2976 0.28277 0.27811

2.6 0.34147 0.30149 0.28643 0.2817

2.8 0.34565 0.30508 0.2898 0.28501

3 0.34952 0.3084 0.29293 0.288083.2 0.35313 0.31149 0.29584 0.29094

3.4 0.3565 0.31438 0.29856 0.29361

3.6 0.35967 0.3171 0.30112 0.29612

3.8 0.36266 0.31966 0.30353 0.29849

4 0.36548 0.32208 0.30581 0.30073

4.2 0.36815 0.32438 0.30797 0.30285


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0.30752

Pause; Please press any key to continue ...

Final Temperature versus FA0 and Pressure

Tabular Results

Pressure FA0=10 FA0=20 FA0=30 FA0=35Columns 1 through 5

1.6 911.84 908.14 907.47 907.47

1.8 909.02 905.32 904.66 904.67

2 906.5 902.81 902.16 902.18

2.2 904.24 900.55 899.91 899.93

2.4 902.18 898.49 897.87 897.892.6 900.3 896.61 895.99 896.02

2.8 898.56 894.87 894.26 894.29

3 896.94 893.26 892.66 892.69

3.2 895.44 891.76 891.16 891.2

3.4 894.03 890.35 889.76 889.8

3.6 892.7 889.03 888.44 888.48

3.8 891.45 887.78 887.19 887.24

4 890.27 886.59 886.02 886.07

4.2 889.15 885.47 884.9 884.95

4.4 888.08 884.4 883.84 883.89

4.6 887.07 883.39 882.82 882.88


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Column 6

907.53

904.74

902.25900.01

897.97

896.1

894.38

892.78

891.29889.89

888.58

887.34

886.17

885.05883.99

882.98

882.02

881.1

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Correlation of The Physical Properties of Ethane

Concepts Utilized:

Correlations for heat capacity, vapor pressure, and liquid viscosity for an ideal gas.

Numerical Methods:

Polynomial, multiple linear, and nonlinear regression of data with linearization andtransformation functions.

Problem Statement:

Tables F-l through F-4 of Appendix F present values for different propertiesof ethane (ideal gas heat capacity, vapor pressure, and liquid viscosity) as functionof temperature. Various regression models will be fitted to the properties ofAppendix using MATLAB.

(a) Construct a MATLAB function which solves the linear regression problemXb = y, where X is the matrix of the independent variable values, y is thevector of dependent variable values, and b is the vector of the linear regressionmodel parameters. The input parameters of the function are X, y, and alogical variable which indicates whether there is a free parameter. The returnedparameters are: and the respective confidence intervals, the calculated values

of the dependent variable ycalc, the linear correlation coefficient R2

, and the

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variance. Test the function by fitting the Wagner equation to vapor pressuredata of ethane from TableF-3 of Appendix F.

(b) Fit 3rd- and 5th-degree polynomials to the heat capacity of ethane for for the

data given in Tables F-l and F-2 of Appendix F by using the multiple linearregression function developed in (a). Compare the quality of the representationof the various data sets with the polynomials of different degrees.

(c) Fit the Antoine equation to liquid viscosity of ethane given in Table F-4 ofAppendix F.

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% filename P5_04A_CCL


prob_title = ([Vapor Pressure Correlation for Ethane]);

ind_var_name=[\bf Functions of Reduced Temp.];

dep_var_name=[\bf Logarithm of Reduced Pressure];fname=input(Please enter the data file name > ); % VPfile.txt

xyData=load(fname);

X=xyData(:,2:end);

y=xyData(:,1);

[m,n]=size(X);

freeparm=input(Input 1 if there is a free par., otherwise input 0>)[Beta,ConfInt, ycal,Var, R2]=MlinReg(X,y,freeparm);

disp([ Results, prob_title]);

Res=[];

if freeparm==0, nparm = n-1; else nparm = n; end

for i=0:nparm

if freeparm==0; ii=i+1; else ii=i; end

Res=[Res; ii Beta(i+1) ConfInt(i+1)];

end

disp( Parameter No. Beta Conf_int);

disp(Res);

disp([ Variance , num2str(Var)]);

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disp([ Correlation Coefficient , num2str(R2)]);


pause

subplot(1,2,1)

plot(y,y-ycal,*,Linewidth,2) % residual plotset(gca,FontSize,14,Linewidth,2)

title([\bf Residual, prob_title],FontSize,12)

xlabel([dep_var_name \bf (Measured)],FontSize,14)

ylabel(\bf Residual,FontSize,14)


pausesubplot(1,2,2)

plot(X,ycal, r-,X,y,bo,Linewidth,2)

title([\bf Cal/Exp Data prob_title],FontSize,12)


xlabel([ind_var_name],FontSize,14)

ylabel([dep_var_name],FontSize,14)

% VPfile.txt : data file provided elsewhere

Please enter the data file name > VPfile.txt

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Input 1 if there is a free parameter, otherwise input 0> 0

Results,Vapor Pressure Correlation for Ethane

Parameter No. Beta Conf_int

1 -6.4585 0.09508

2 1.2895 0.21514

3 -1.6712 0.26773

4 -1.2599 0.29417

Variance 9.3486e-005

Correlation Coefficient 1



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%


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% filename P5_04B_CCL


prob_title = ([Heat Capacity of Ethane]);

ind_var_name=[\bf Normalized Temp];

dep_var_name=[\bf Heat Capacity J/kmol*K ];fname=input(Please enter the data file name > ); % CPfile.txt

xyData=load(fname);

X=xyData(:,2:end);

y=xyData(:,1);

[m,n]=size(X);

freeparm=input(Input 1 if there is a free par., otherwise input 0>)[Beta, ConfInt,ycal, Var, R2]=MlinReg(X,y,freeparm);

disp([ Results, prob_title]);

Res=[];

if freeparm==0, nparm = n-1; else nparm = n; end

for i=0:nparm

if freeparm, ii=i+1; else ii=i; end

Res=[Res; ii Beta(i+1) ConfInt(i+1)];

end

disp( Parameter No. Beta Conf_int);

disp(Res);

disp([ Variance , num2str(Var)]);


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Chen CL 41

C l Ch i l E ilib i


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Complex Chemical Equilibrium

by Gibbs Energy Minimization

Concepts Utilized:Formulation of the chemical equilibrium problem as a Gibbs energy minimizationproblem with atomic balance constraints. The use of Lagrange multipliers tointroduce the constraints into the objective function. Conversion of theminimization problem into a system of nonlinear algebraic equations.

Numerical Methods:

Solution of a system of nonlinear algebraic equations.

Problem Statement:

Ethane reacts with steam to form hydrogen over a cracking catalyst at a

temperature ofT= 1000 K and pressure ofP = 1 atm. The feed contains 4moles ofH2O per mole ofCH4. Balzisher et al. suggest that only the compoundsshown in Table 4-10 are present in the equilibrium mixture (assuming that nocarbon is deposited). The Gibbs energies of formation of the various compoundsat the temperature of the reaction (1000K) are also given in Table 4-10. Theequilibrium composition of the effluent mixture is to be calculated using these

data.

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Table 4-10: Compounds Present in Effluentof Steam Cracking Reactor

Gibbs Energy Feed Effluent

No. Comp. kcal/gm-mol gm-mol Ini. Est.1 CH4 4.61 0.001

2 C2H4 28.249 0.001

3 C2H2 40.604 0.001

4 CO2

94.61 0.993

5 CO 47.942 1.6 O2 0. 0.0001

7 H2 0. 5.992

8 H2O 46.03 4 1.

9 C2H6 26.13 1 0.001

Formulate the problem as a constrained minimization problem. Introduce theconstraints into the objective function using Lagrange multipliers and differentiatethis function to obtain a system of nonlinear algebraic equations.

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Solution:

The objective function to be minimized is the total Gibbs energy given by

min

ni

G

RT

= i=1

cni Goi

RT

+ ln ni

ni (4 49)

where ni is the number of moles of component i, c is the total number ofcompounds, R is the gas constant, and Go is the Gibbs energy of pure componenti at temperature T. The minimization of Equation (4-49) must be carried outsubject to atomic balance constraints

Oxygen Balance 0 =g1 = 2n4+ n5+ 2n6+ n7 4 (4 50)Hydrogen Balance 0 =g2 = 4n1+ 4n2+ 2n3+ 2n7+ 2n8+ 6n9 14 (4 51)

Carbon Balance 0 =g3 = n1+ 2n2+ 2n3+ n4+ n5+ 2n9 2 (4 52)

The identification of the various components is given in Table 4-10. These threeconstraints can be introduced into the objective functions using Lagrangemultipliers: 1, 2,3. The extended objective function is

minni,j

F = i=1

cni GoiRT

+ ln ni

ni+

3

j=1

jgj (4 53)

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Th diti f i i f this f ti t ti l i t is th t ll th


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The condition for minimum of this function at a particular point is that all thepartial derivatives ofF with respect to ni andj vanish at this point. The partialderivative ofF with respect to n1 for example, is

F

n1 = Go

1RT + ln

n1ni + 4

2+ 3 (4 54)

The other partial derivatives with respect to ni can be obtained similarly. If it isexpected that the amount of a particular compound at equilibrium is very close tozero, it is preferable to rewrite the equation in a form that does not require

calculation of the logarithm of a very small number. Rearranging Equation (4-54),for example, yields

n1

niexp

Go1RT

+ 42+ 3

= 0

The partial derivatives ofFwith respect to 1, 2, and 3 areg1, g2, andg3,respectively.

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function P5 05A1 CCL


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function P5_05A1_CCL


xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001 ...

0.001 0.001 0.001 0.0001]; % initial guess vector

disp(Variable values at the initial estimate);

fguess = MNLEfun(xguess);

disp( Variable Value Function Value)

for i=1:size(xguess,2);

disp([xint2str(i) num2str(xguess(i)) num2str(fguess(i))]);

end

options = optimset(Diagnostics,[off],TolFun,[1e-9],...TolX,[1e-9]);

xsolv = fsolve(@MNLEfun,xguess,options);

disp(Variable values at the solution);

fsolv = MNLEfun(xsolv);


for i=1:size(xguess,2);disp([xint2str(i) num2str(xsolv(i)) num2str(fsolv(i))])

end

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function fx = MNLEfun(x)


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lamda1 = x(1);

lamda2 = x(2);

lamda3 = x(3);

H2 = x(4);

H2O = x(5);

CO = x(6);

CO2 = x(7);

CH4 = x(8);

C2H6 = x(9);

C2H4 = x(10);C2H2 = x(11);

O2 = x(12);

R = 1.9872;

sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;

fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4; %Oxygen balance

fx(2,1) = 4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo

fx(4,1) = log(H2 / sum) + 2 * lamda2;

fx(5,1) = -46.03 / R + log(H2O / sum) + lamda1 + 2 * lamda2;

fx(6,1) = -47.942 / R + log(CO / sum) + lamda1 + lamda3;

fx(7,1) = -94.61 / R + log(CO2 / sum) + 2 * lamda1 + lamda3;

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fx(8 1) = 4 61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;


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fx(8,1) = 4.61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;

fx(9,1) = 26.13 / R + log(C2H6 / sum) + 6 * lamda2 + 2 * lamda3;


fx(11,1) = C2H2-exp(-(40.604/R+2*lamda2+2*lamda3))*sum;

fx(12,1) = O2 - exp(-2 * lamda1) * sum;

Variable values at the initial estimate

Variable Value Function Value

x1 10 -0.0138

x2 10 0

x3 10 0x4 5.992 19.5944

x5 1 4.6407

x6 1 -6.3214

x7 0.993 -19.8127

x8 0.001 43.2161

x9 0.001 84.0454

x10 0.001 65.1117

x11 0.001 0.001

x12 0.0001 9.9981e-005

Optimization terminated: no further progress can be made.

Trust-region radius less than 2*eps.

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Problem may be ill-conditioned or Jacobian may be inaccurate


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Problem may be ill conditioned or Jacobian may be inaccurate.

Try using exact Jacobian or check Jacobian for errors.

Variable values at the solution


x1 10+7.44066e-015i -0.013798+4.9091e-

x2 10-3.7436e-008i -0.015082-0.0002223

x3 10-1.4705e-008i -0.0054569-7.4103e-

x4 5.992+8.1803e-010i 19.5947+4.04813e

x5 1+8.1803e-010i 4.6411+4.0488e-006i

x6 1+8.182e-010i -6.3211+4.109e-006i

x7 0.993+8.182e-010i -19.8124+4.10898ex8 0.00029165+6.9845e-010i 41.9842+6.3

x9 -0.00037496-3.7053e-005i 83.0696-3.

x10 -5.5221e-009-1.0635e-009i 53.0235-

x11 0.0010003+8.1796e-010i 0.0010003+8

x12 0.00010027+8.1823e-010i 0.00010025

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function P5 05A2 CCL


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function P5_05A2_CCL


xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...






disp([xint2str(i) num2str(xguess(i)) num2str(fguess(i))]);

end

options = optimset(Diagnostics,off,TolFun,1e-9,...TolX,1e-16,NonlEqnAlgorithm,gn);

xsolv = fsolve(@MNLEfun,xguess,options);


fsolv=MNLEfun(real(xsolv));


for i=1:size(xguess,2);disp([x int2str(i) num2str(real(xsolv(i))) num2str(fsolv(i))

end

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lamda1 = x(1);

lamda2 = x(2);

lamda3 = x(3);

H2 = x(4);

H2O = x(5);

CO = x(6);

CO2 = x(7);

CH4 = x(8);

C2H6 = x(9);

C2H4 = x(10);C2H2 = x(11);

O2 = x(12);

R = 1.9872;


fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4;

fx(2,1) =4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo

fx(4,1) = log(H2 / sum) + 2 * lamda2;





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Chen CL 52



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x1 24.4197 0

x2 0.25306 1.7764e-015

x3 1.5598 0

x4 5.3452 -1.1102e-016

x5 1.5216 2.1094e-015

x6 1.3885 2.2204e-016

x7 0.54492 -3.3307e-015

x8 0.066564 0

x9 1.6707e-007 1.3323e-015

x10 9.5412e-008 1.3323e-015x11 3.157e-010 1.4387e-020

x12 7.0058e-021 1.5466e-021

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function P5 05B CCL


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function P5_05B_CCL


xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...






disp([x int2str(i) num2str(xguess(i)) num2str(fguess(i))])

end

pote=[0 0 0 2 2 2 2 2 2 2 2 2];tol=1e-9;

maxit=100;

derfun=0;

print=0;

[xsolv,y,dy,info]=conles(@MNLEfun,xguess,pote,[],...

tol,maxit,derfun,print);disp(Variable values at the solution);

fsolv = MNLEfun(xsolv);



disp([x int2str(i) num2str(xsolv(i)) num2str(fsolv(i))])


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Chen CL 55



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function fx MNLEfun(x)

lamda1 = x(1);

lamda2 = x(2);

lamda3 = x(3);

H2 = x(4);

H2O = x(5);

CO = x(6);

CO2 = x(7);

CH4 = x(8);

C2H6 = x(9);

C2H4 = x(10);C2H2 = x(11);

O2 = x(12);

R = 1.9872;


fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4;


fx(4,1) = log(H2 / sum) + 2 * lamda2;




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fx(8,1) = 4.61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;


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fx(8,1) 4.61 / R log(CH4 / sum) 4 lamda2 lamda3;



fx(11,1) = C2H2-exp(-(40.604/R+2*lamda2+2*lamda3))*sum;

fx(12,1) = O2 - exp(-2 * lamda1) * sum;

Variable values at the initial estimate


x1 10 -0.0138

x2 10 0

x3 10 0x4 5.992 19.5944

x5 1 4.6407

x6 1 -6.3214

x7 0.993 -19.8127

x8 0.001 43.2161

x9 0.001 84.0454

x10 0.001 65.1117

x11 0.001 0.001

x12 0.0001 9.9981e-005

Variable values at the solution


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x1 24.4197 0


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x2 0.25306 -1.7764e-015

x3 1.5598 0

x4 5.3452 -1.1102e-016

x5 1.5216 -1.6653e-015

x6 1.3885 -1.5543e-015

x7 0.54492 5.5511e-015

x8 0.066564 -4.4409e-016

x9 1.6707e-007 -1.6964e-013

x10 9.5412e-008 -2.589e-013

x11 3.157e-010 9.3058e-025x12 5.4592e-021 -3.9873e-035

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function P5_05C_CCL


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xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...






disp([x int2str(i) num2str(xguess(i)) num2str(fguess(i))]);

end

pote=[0 0 0 2 2 2 2 2 2 2 2 2];tol=1e-9; maxit=100;

derfun=0; print=0;

[xsolv,y,dy,info] = conles(@MNLEfun,xguess,pote,[],...

tol,maxit,derfun,print);


fsolv = MNLEfun(xsolv);disp( Variable Value Function Value)


disp([x int2str(i) num2str(xsolv(i)) num2str(fsolv(i))]);

end

H2 =xsolv(4);

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H2O =xsolv(5);


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;

CO =xsolv(6);

CO2 =xsolv(7);

CH4 =xsolv(8);

C2H6=xsolv(9);

C2H4=xsolv(10);

C2H2=xsolv(11);

O2 =xsolv(12);

R = 1.9872;


G_O2 = O2 * log(abs(O2 / sum));G_H2 = H2 * log(H2 / sum);

G_H2O = H2O * (-46.03 / R + log(H2O / sum));

G_CO = CO * (-47.942 / R + log(CO / sum));

G_CO2 = CO2 * (-94.61 / R + log(CO2 / sum));

G_CH4 = CH4 * (4.61 / R + log(abs(CH4 / sum)));

G_C2H6 = C2H6 * (26.13 / R + log(abs(C2H6 / sum)));G_C2H4 = C2H4 * (28.249 / R + log(abs(C2H4 / sum)));

G_C2H2 = C2H2 * (40.604 / R + log(abs(C2H2 / sum)));

ObjFun=G_H2+G_H2O+G_CO+G_O2+G_CO2...

+G_CH4+G_C2H6+G_C2H4+G_C2H2

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lamda1 = x(1);

lamda2 = x(2);

lamda3 = x(3);

H2 = x(4);

H2O = x(5);

CO = x(6);

CO2 = x(7);

CH4 = x(8);

C2H6 = x(9);

C2H4 = x(10);C2H2 = x(11);

O2 = x(12);

R = 1.9872;


fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4; %Oxygen balance


fx(4,1) = log(H2 / sum) + 2 * lamda2;





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x1 24.4197 0


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x2 0.25306 -1.7764e-015

x3 1.5598 0

x4 5.3452 -1.1102e-016

x5 1.5216 -1.6653e-015

x6 1.3885 -1.5543e-015

x7 0.54492 5.5511e-015

x8 0.066564 -4.4409e-016

x9 1.6707e-007 -1.6964e-013

x10 9.5412e-008 -2.589e-013

x11 3.157e-010 9.3058e-025x12 5.4592e-021 -3.9873e-035

ObjFun =

-104.34

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Thank You for Your Attention

Questions Are Welcome

13 Problem Solving with MATLAB.pdf

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