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+// Example 12_9
+clc;clear;funcprot(0);
+//From example 12_7
+//Given values
+P_0=1000;// kPa;
+T_0=800;// K
+Ma_1=2;// Exit Mach number
+a=20;// Throat area in cm^2
+//Properties
+R=0.287;// kJ/kg.k
+C_p=1.005;// kJ/kg.k
+k=1.4;//The specific heat ratio of air 
+
+// Calculation
+//(a)
+//From example 12_7
+P_01=1.0;// MPa
+P_1=0.1278; // MPa
+T_1=444.5;// K
+rho_1=1.002;// kg/m^3
+// From table A-14,For Ma_1=2,we read
+Ma_2=0.5774
+P_02=0.7209*P_01;// MPa
+printf('(a)The stagnation pressure,P_02=%0.3f MPa\n',P_02);
+P_2=4.5000*P_1;// MPa
+printf('The static pressure,P_2=%0.3f MPa\n',P_2);
+T_2=1.6875*T_1;// K
+printf('The static temperature,T_2=%0.0f K\n',T_2);
+rho_2=2.6667*rho_1;// kg/m^3
+printf('The static density,rho_2=%0.2f kg/m^3\n',rho_2);
+
+//(b)
+//gradS=s2-s1
+gradS=(C_p*(log(T_2/T_1)))-(R*log((P_2/P_1)));
+printf('(b)The entropy change across the shock,s2-s1=%0.4f kJ/kg.K\n',gradS);
+
+//(c)
+c_2=sqrt(k*R*T_2*1000);// The speed of sound at the exit conditions in m/s
+V_2=Ma_2*c_2;
+printf('(c)The exit velocity,V_2=%0.0f m/s\n',V_2);
+
+//(d)
+//The mass flow rate in this case is the same as that determined in Example 12_7:
+V_1=517.5;// m/s
+rho_c=2.761;// kg/m^3
+m=rho_c*(a*10^-4)*V_1;// kg/s
+printf('(d)The mass flow rate,m=%0.2f kg/s\n',m);