Matlab_code_for_Binary_Distillation

8/7/2019 Matlab_code_for_Binary_Distillation

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Assignment 01

Process Modeling and

Simulation

Name:K.S.M.S.Senarath

Index No: 050418H

Date of Sub: 02.02.2009


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01.Pendulum ProblemMomentum equation of a pendulum is given as below when initial faorce of

the pendulum is zero.

In Matlab code I took angle as y(1) and angular velocity as y(2).

Code in Pendulum M-File

function dydt=pendulum(t,y)g=9.81; % Gravitational Accelaration (ms-2)

M=3; % Mass of the rod (kg)

I=12; % Inertia (kgm2)

R=sqrt(2*I/M); % Radius (m)

f1=y(2);

f2=-M*g*R*sin(y(1))/I

dydt=[f1;f2];

Code in Command Window

>> y0=[pi/3,0];

>> tspan=[0,10];

>> [t,y]=ode45(@pendulum,tspan,y0);

>> plot(t,y)

>> xlabel('Time\it (S)')

>> grid

>>legend('\it Angular Acceleration','\it Angular

Velocity')


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Obtained Graph for Angular acceleration and angular Velocity

0 1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (S)

Angular Accelaration

Angular Velocity


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02.Temperature Variation in Plates of Binary Distillation TowerThis binary distillation tower distills Benzene and Toluene. Below equation

will be used for construct the MatLab model.

For the total mass balance of the system

F = W + D

Composition Mass Balance

F.XF= W.XW + D.XD

Equation of Top Operating Line

Equation of q Line

Where

F Feed Rate (kmol/hr)

X- Liquid composition of Benzene

Y- Gas Composition of Benzene

D Distilled product (kmol/hr)

D, XD

V, XD

F, XF

Lo

W, XW


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W - Bottom Product (kmol/hr)

In M-File

%-------------K.S.M.S.Senarath----------------%

%--------------------------050418H----------------------%

clear all

%--Liquid & Gas composition data for Benzene & Toluene--%

X= [0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800

0.900 0.950 1.000];

%Composition of liquid (Benzene)

Y= [0.000 0.208 0.372 0.507 0.612 0.713 0.791 0.857 0.912

0.959 0.980 1.000];

%Composition of gas (Benzene)

T= [110 105.3 101.5 98.0 95.1 92.3 89.7 87.3 85.0 82.7

81.4 80.1];

%Boiling points of mixtures

%--------Data of Distillation Tower----------%

F=120; %Feed rate (kmol/hr)

D=23.75; %Product Rate (kmol/hr)

Xf=0.45; %Feed composition

Xd=0.98; %Distillate composition

P=1; % Operating pressure (atm)

R=1.5; %Reflux Ratio

%---------Calculate W, Xw and coefficients of BOL--------%

W=F-D; %Bottom Flow Rate (kmol/hr)

Xw=(F*Xf-D*Xd)/W; %Bottom Composition


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Yf=R*Xf/(R+1)+Xd/(R+1); %Point of Intersection of TOL &

BOL

m=(Yf -Xw)/(Xf -Xw); %Gradient

c=Xw-m*Xw; %Intersection

fprintf ('Bottom Flow rate: %.2f kmol/hr \n',W)fprintf ('Bottom Composition: %.2f \n',Xw)

%----------Polynomials of Equilibrium Curves-----------%

A= polyfit(X,Y,5);

%Coefficients of the Polynomial (Y vs X)

B= polyfit(X,T,5);

%Coefficients of the Polynomial (Temperaturevs. X)

comp_temp=zeros(50,2);

%-----Composition Calculation of Partial Condenser)-----%

Z=zeros(1,6);

Z=A;

Z(6)=A(6)-Xd;

r=roots(Z);

for i=1:5

if (r(i,1)0)&(imag(r(i,1))==0)

x=r(i,1);

comp_temp (1,1)=x;

end

end

%--------------Plate Calculation-------------%

for i=1:49

x=comp_temp(i,1);

y=R*x/(R+1) + Xd/(R+1);

%Top Operating Line Equation

Z(6)=A(6)-y;


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a=roots(Z);

for j=1:5

if (a(j,1)0)&(imag(a(j,1))==0)

%Find the Correct between 0 & 1

x=a(j,1);if x>=Xf

%check the condition for feed composition

comp_temp(i+1,1)=x;

else

y=m*x + c;

%bottom operating line

Z(6)=A(6)-y;

a=roots (Z);

for k=1:5

if

(a(k,1)0)&(imag(a(k,1))==0)

%Correct Root

x=a(k,1);

if Xw 0

comp(k,1)=comp_temp(k,1);

else

break


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end

end

N=length(comp);

NN=1:1:N;T=polyval(B,comp);

subplot(2,1,1),plot(NN,comp,'r')

grid on

xlabel ('Plate No.(From Top)')

ylabel ('Liquid Composition-Benzene')

subplot (2,1,2),plot(NN,T,'b')

grid onxlabel ('Plate No:(From Top)')

ylabel ('Plate Temperature(0C)')

fprintf ('No of plates: %.0f \n',N-2)

Comp_Temp= [Composition T]

return

end

end

end

end


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Results

In Command Window

>> Bottom Flow rate: 96.25 kmol/hr

Bottom Composition: 0.32

No of plates: 12

Temperature Variation in plates of distillation tower

Matlab_code_for_Binary_Distillation

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