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+// Example 7_6
+clc;funcprot(0);
+// Given data
+T_1=1000;// K
+P_1=0.5;// The inlet pressure in MPa
+P_2=0.1;// The exit pressure in MPa
+T_0=298;// K
+R=0.286;// kJ/kg.K
+
+// Calculation
+// From the air tables
+phi_1=2.968;// kJ/kg.K
+phi_2=phi_1-(R*log(P_1/P_2));// kJ/kg.K
+// Thus
+T_2=657.5;// K
+h_2=667.8;// kJ/kg
+h_1=1046.1;// kJ/kg
+h_0=298.2;// kJ/kg
+V_2=sqrt(2)*((h_1-h_2)*10^3)^(0.5);// m/s
+P_0=P_2;// MPa
+phi_0=1.695;// kJ/kg.K
+X_2=(h_2-h_0)+((V_2)^2/(2*1000))-(T_0*(phi_2-phi_0-(R*log(P_2/P_0))));// kJ/kg
+X_1=h_1-h_0-(T_0*(phi_1-phi_0-(R*log(P_1/P_0))));// The availability supplied in kJ/kg
+e_II=X_2/X_1;// The second law effectiveness for an ideal isentropic nozzle
+printf("\nThe second law effectiveness for an ideal isentropic nozzle,e_II=%1.2f",e_II);
+// The answer provided in the textbook is wrong