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+clear;
+clc;
+
+// Illustration 7.12
+// Page: 252
+
+printf('Illustration 7.12 - Page: 252\n\n')
+// solution
+
+//****Data****//
+Tempg1 = 32;// [OC]
+Tempw1 = 28;// [OC]
+//******//
+
+H1 = 90;// [kJ/kg]
+H1_prime = 72;// [kJ/kg dry air]
+H2_prime = 163.6;// [kJ/kg dry air]
+deff('y = f17(H2)','y = (H2-H1)-(H2_prime-H1_prime)');
+H2 = fsolve(2,f17);// [kJ/kg dry air]
+// Slope of Operating Line same as Operating Line as Illustration 7.11
+slopeOperat = (163.5-72)/(45-29);
+deff("[y] = f18(Temp)","y = slopeOperat*(Temp-Tempg1)+H1");
+Temp = 30:0.01:45;
+// Equilibrium Data:
+// Data  = [Temp.(OC),H_star(kJ/kg)]
+Data_star = [29 100;32.5 114;35 129.8;37.5 147;40 166.8;42.5 191;45 216];
+scf(5);
+plot(Data_star(:,1),Data_star(:,2),Temp,f18);
+xgrid();
+legend("Equilibrium Line","operating Line");
+xlabel("Liquid Temperature, C");
+ylabel("Enthalphy Of Air Water vapour, kJ/kg dry air");
+// The Value for NtoG & HtoG will be same as in Illustration 7.11
+NtoG = 3.25;
+HtoG = 2.22;// [m]
+// By hit & trial method:
+Temp = 37.1;// [OC]
+printf("The Temperature to which water is to be cooled is %f OC\n",Temp);
+\ No newline at end of file