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+// Given:-
+p1 = 3.0 // entry pressure in Mpa
+p2 = 0.5 // exit pressure in Mpa
+T1 = 320.0 // entry temperature in degree celcius
+T0 = 25.0 // in degree celcius
+p0 = 1.0 // in atm
+
+// From table A-4
+h1 = 3043.4 // in kj/kg
+s1 = 6.6245 // in kj/kg.k
+h2 = h1 // from reduction of the steady-state mass and energy rate balances
+s2 = 7.4223 // Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k
+
+// From table A-2
+h0 = 104.89 // in kj/kg
+s0 = 0.3674 // in kj/kg.k
+
+// Calculations
+ef1 = h1-h0-(T0+273)*(s1-s0) // flow exergy at the inlet
+ef2 = h2-h0-(T0+273)*(s2-s0) // flow exergy at the exit
+// From the steady-state form of the exergy rate balance
+Ed = ef1-ef2 // the exergy destruction per unit of mass flowing is
+
+// Results
+printf( ' The specific flow exergy at the inlet is %.2f kJ/kg.',ef1)
+printf( ' The specific flow exergy at the exit is %.2f kJ/kg.',ef2)
+printf( ' The exergy destruction per unit of mass flowing is %.2f kJ/kg.',Ed)