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clear ;
clc;
//Page No. 419
// Example 13.8
printf('Example 13.8\n\n');
// Solution fig E13.8
T1c = 15 ;// Temperature of F & P -[degree C]
T1 = 273 + T1c ;// Temperature of F & P -[K]
P1 = 105 ;// Pressure of F & P -[kPa]
// F analysis
F_CO2 = 1.2/100 ;// Volume fraction
F_odr = 98.8/100 ;// Volume fraction
// P analysis
P_CO2 = 3.4/100 ;// Volume fraction
P_odr = 96.6/100 ;// Volume fraction
Tc_CO2 = 7 ;//Temperature CO2 -[degree C]
T_CO2 = 273 + Tc_CO2 ;// Temperature CO2 -[K]
P_CO2 = 131 ;// Pressure of CO2 -[kPa]
CO2 = 0.0917 ;// Volume flow rate of CO2-[cubic metre/min]
// Convert given volume flow rate of CO2 at temperature of F & P
nw_CO2 = (CO2 * T1 * P_CO2)/(T_CO2 * P1) ;// volume flow rate of CO2 at temperature of F & P-[cubic metre]
// Solve P & F by following eqns. obtained by component balance of CO2 and total balance
// F(F_odr) = P(P_odr) - others balance
// F + nw_CO2 = P - Total balance
// Solving by matrix method
a = [F_odr -P_odr;1 -1];// Matrix formed by coefficients of unknown
b = [0;-nw_CO2] ;// Matrix formed by constants
x = a\b ;// matrix of solution, x(1) = F;x(2) = P
F = x(1) ;//Volume flow rate of entering gas-[cubic metre/min]
P = x(2) ;//Volume flow rate of product [cubic metre/min]
printf('Volume flow rate of entering gas is %.2f cubic metre/min',F);
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