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clear;
clc;
//To find Approx Value
function[A]=approx(V,n)
A=round(V*10^n)/10^n;//V-Value n-To what place
funcprot(0)
endfunction
//Example 13.5
//Caption : Program to Calculate the Fraction of Heat Reacted for Various Cases
//CO(g) + H2O(g) --> CO2(g) + H2(g)
v_CO=-1;
v_H2O=-1;
v_CO2=1;
v_H2=1;
v=v_CO+v_H2O+v_CO2+v_H2;
//Calculate e(Fraction of Stream) in each case
//(a)
n_H2O_a=1;//mol
n_CO_a=1;//mol
T_a=1100;//[K]
P_a=1;//[bar]
x=10^4/T_a;
//at this x the value of ln K=0 form Graph
y=0;
nt=n_H2O_a+n_CO_a;
K=exp(y);
//y_H2O=(n_H2O+(v_H2O*e))/nt
//y_CO=(n_CO+(v_CO*e))/nt
//y_H2=(v_H2*e)/nt
//y_CO2=(v_CO2*e)/nt
y_H2O='(1-e)/2'
y_CO='(1-e)/2'
y_H2='e/2'
y_CO2='e/2'
//K=(y_H2*y_CO2)/(y_CO*y_H2O)
K='e^2/((1-e)^2)=1';
//Solving
e_a=0.5;
//(b) same as in (a) P_b=10bar
//Pressure has no effect on fraction of stream
e_b=e_a;
//(c) same as in (a) N2=2mols included
//Since N2 just act as diluent
//It only changes the total moles fraction remains the same
e_c=e_a;
//(d)
n_H2O_d=2;//mol
n_CO_d=1;//mol
T_d=1100;//[K]
P_d=1;//[bar]
x=10^4/T_d;
//at this x the value of ln K=0 form Graph
y=0;
nt=n_H2O_d+n_CO_d;
K=exp(y);
//y_H2O=(n_H2O+(v_H2O*e))/nt
//y_CO=(n_CO+(v_CO*e))/nt
//y_H2=(v_H2*e)/nt
//y_CO2=(v_CO2*e)/nt
y_H2O='(1-e)/3'
y_CO='(2-e)/3'
y_H2='e/3'
y_CO2='e/3'
//K=(y_H2*y_CO2)/(y_CO*y_H2O)
K='e^2/((1-e)(2-e))=1';
//Solving
e=approx(2/3,3);
e_d=approx(e/2,3);
//(e)
//Here the y_CO and y_H2O are interchanged
//No change in Fraction of stream
e_e=e_d;
//(f)
n_H2O_f=1;//mol
n_CO_f=1;//mol
n_CO2_f=1;;//[mol]
T_f=1100;//[K]
P_f=1;//[bar]
x=10^4/T_f;
//at this x the value of ln K=0 form Graph
y=0;
nt=n_H2O_f+n_CO_f+n_CO2_f;
K=exp(y);
//y_H2O=(n_H2O+(v_H2O*e))/nt
//y_CO=(n_CO+(v_CO*e))/nt
//y_CO2=(n_CO2+(v_CO2*e))/nt
//y_H2=(v_H2*e)/nt
y_H2O='(1-e)/3'
y_CO='(1-e)/3'
y_H2='(1+e)/3'
y_CO2='e/2'
//K=(y_H2*y_CO2)/(y_CO*y_H2O)
K='(e*(e+1))/((1-e)^2)=1';
//Solving
e_f=approx(1/3,3);
//(g)
n_H2O_g=1;//mol
n_CO_g=1;//mol
T_g=1650;//[K]
P_g=1;//[bar]
x=10^4/T_g;
//at this x the value of ln K=0 form Graph
y=-1.15;
nt=n_H2O_g+n_CO_g;
K=exp(y);
//y_H2O=(n_H2O+(v_H2O*e))/nt
//y_CO=(n_CO+(v_CO*e))/nt
//y_H2=(v_H2*e)/nt
//y_CO2=(v_CO2*e)/nt
y_H2O='(1-e)/2'
y_CO='(1-e)/2'
y_H2='e/2'
y_CO2='e/2'
//K=(y_H2*y_CO2)/(y_CO*y_H2O)
Exp='e^2/((1-e)^2)=0.316';
//Solving
p=poly([K -2*K K-1],'e','c');
root=roots(p);
e_g=approx(root(1),2);
//Other Root is negative and the Fraction of steam cannot be negative
disp('(a)1mol H20 and 1 mol CO T=1100K P=1bar')
disp('(b)1mol H20 and 1 mol CO T=1100K P=10bar')
disp('(c)1mol H20 and 1 mol CO 2mol N2 T=1100K P=1bar')
disp('(d)2mol H20 and 1 mol CO T=1100K P=1bar')
disp('(e)1mol H20 and 2 mol CO T=1100K P=1bar')
disp('(f)1mol H20 and 1 mol CO 1mol CO2 T=1100K P=1bar')
disp('(g)1mol H20 and 1 mol CO T=1650K P=1bar')
e=[e_a e_b e_c e_d e_e e_f e_g];
disp(e,'Fraction Of Steam Reacted in each case')
//End
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