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+clc
+clear
+printf("Example 5.1 | Page number 119 \n\n");
+//Find the work output in KJ/kg
+//Given data
+Q = -24; //KJ/kg
+
+p1 = 5e5; //N/m^2
+t1 = 227; //°C
+V1 = 50; //m/s
+v1 = 0.78; //m^3/kg
+
+p2 = 1e5; //N/m^2
+t2 = 57; //°C
+V2 = 100; //m/s
+v2 = 0.97; //m^3/kg
+
+g = 9.81; //m/s^2 //acceleration due to gravity
+
+delta_z = -5; //m
+Cv = 0.7; //KJ/kg
+delta_u = Cv*(t2 - t1); //KJ/kg //change in internal energy //u2-u1
+
+//Solution
+delta_h = delta_u + (p2*v2 - p1*v1)*.001; //KJ/kg //change in enthalpy //h2-h1
+
+W_x = Q - (delta_h + (V2^2 - V1^2)/2*.001 + g*delta_z*.001); //kJ/kg //Work Output
+
+printf("Work output = %.2f KJ/kg",W_x);