//Engineering and Chemical Thermodynamics //Example 8.17 //Page no :406 clear ; clc; //Given T = 300 ; //[K] A = 6235 ; //[J/mol] P_a_sat = 100 * 10^3 ; //[Pa] P_b_sat = 50 * 10^3 ; //{Pa} R = 8.314 ; w = 1/(R * T) ; function Z817 = f817(R) x_a_a = R(1) ; x_a_b = R(2) ; Z817(1) = x_a_b * exp(A * (1 - x_a_b) ^ 2 * w) - x_a_a * exp(A * (1 - x_a_a) ^ 2 * w) ; // E8.17A Z817(2) = (1 - x_a_b) * exp(A * ( x_a_b) ^ 2 * w) - (1 - x_a_a) * exp(A * (x_a_a) ^ 2 * w ) ; // E8.17B endfunction x0 = [0.75 ; 0.1] ; [z,fxs,m] = fsolve(x0,f817) ; disp(" Example: 8.17 Page no : 406") ; printf("\n The compositions are : x_a_a = %.3f and x_a_b = %.3f",z(1,1), z(2,1)) ; P = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat + z(2,1) * exp(A * z(1,1) ^ 2 * w) * P_b_sat ; printf("\n Total pressure = %d kPa",P * 10^-3) ; y_a = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat / P ; printf("\n y_a = %.3f" , y_a ) ;