//Engineering and Chemical Thermodynamics //Example 2.20 //Page no :89 clear ; clc ; //solution (a) //Given data Cv = 3/2 * 8.314 ; Cp = 5/2 * 8.314 ; n = 1; R = 8.314 ; T1 = 1000 ; //[K] P1 = 10 ; //[bar] T2 = 1000 ; //[K] P2 = 0.1 ; //[bar] T3 = 300 ; //[K] T4 = 300 ; //[K] k = Cp / Cv ; P3 = P2 * (T3 / T2)^(k/(k-1)); //[bar] P4 = P1 * (T4 / T1)^(k/(k-1)) ; //[bar] // (1) del_U_12 = 0 ; // As process 1-2 is isothermal W_12 = n * R * T1 * log(P2 / P1); Q_h_12 = W_12 ; disp(" Example: 2.20 Page no : 89") ; printf('(a)\n (1)\n del_U = %d J',del_U_12) ; printf('\n Work = %d J',W_12) ; printf('\n Heat = %d J',Q_h_12) ; //(2) Q_23 = 0 ; // As adiabatic process del_U_23 = n * Cv *(T3 - T2) ; W_23 = del_U_23 ; printf('\n (2)\n del_U = %g J',del_U_23) ; printf('\n Work (J) = %d J',W_23) ; printf('\n Heat (J) = %d J',Q_23) ; //(3) del_U_34 = 0 ; // As isothermal process W_34 = n * R * T3 * log(P4 / P3) ; // Eqn E2.20.A Q_c_34 = del_U_34 - W_34 ; printf('\n (3)\n del_U = %g J',del_U_34) ; printf('\n Work = %d J',W_34) ; printf('\n Heat = %d J',Q_c_34) ; //(4) Q_41 = 0 ; // As adiabatic process del_U_41 = n * Cv * (T1 - T4) ; W_41 = del_U_41 ; printf('\n (4)\n del_U = %g J',del_U_41) ; printf('\n Work = %d J',W_41) ; printf('\n Heat = %d J',Q_41) ; //Solution (b) //Users can refer figure E2.20 //Solution (c) W_total = W_12 + W_23 + W_34 + W_41 ; Q_absor = Q_h_12 ; effi = W_total / Q_absor ; printf('\n\n(c) efficiency = %g',effi) //Solution (d) x = 1 - T3 / T1 ; printf('\n\n(d) 1 - Tc/Th = %g',x); disp(" i.e Efficiency = 1 - Tc/Th"); //Solution (e) disp("(e) The process can be made more efficient by raising Th or by lowering Tc ."); disp("Table E2.20B") ; disp(" T(K) P(bar) v(m^3/mol)") ; P = [P1 , P2 , P3 , P4 ] ; T = [T1 , T2 , T3 , T4 ] ; for i = 1:4 v(i) = R * T(i) * 10^-5/ P(i) ; printf("\n %d %.4f %f \n",T(i) ,P(i) ,v(i)) ; end