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authorpriyanka2015-06-24 15:03:17 +0530
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+// Given :-
+T = 373.15 // initial temperature of saturated liquid in kelvin
+T0 = 293.15 // in kelvin
+P0 = 1.014 // in bar
+
+// Part(a)
+// From table A-2
+ug = 2506.5 // in kj/kg
+uf = 418.94 // in kj/kg
+vg = 1.673 // in m^3/kg
+vf = 1.0435*(10**(-3)) // in m^3/kg
+sg = 7.3549 // in kj/kg.k
+sf = 1.3069 // in kj/kg.k
+
+
+// Calculations
+// Energy transfer accompanying work
+etaw = 0 // since p = p0
+// Exergy transfer accompanying heat
+Q = 2257 // in kj/kg,obtained from example 6.1
+etah = (1-(T0/T))*Q
+
+// Exergy destruction
+ed = 0 // since the process is accomplished without any irreversibilities
+deltae = ug-uf + P0*(10**5)*(vg-vf)/(10**3)-T0*(sg-sf)
+
+// Results
+printf( ' Part(a)the change in exergy is %.2f kJ/kg.',deltae)
+printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',etaw)
+printf( ' The exergy transfer accompanying heat is %.2f kJ/kg',etah)
+printf( ' The exergy destruction is %.2f kJ/kg.',ed)
+
+
+// Part(b)
+Deltae = deltae // since the end states are same
+Etah = 0 // since process is adiabatic
+// Exergy transfer along work
+W = -2087.56 // in kj/kg from example 6.2
+Etaw = W- P0*(10**5)*(vg-vf)/(10**3)
+// Exergy destruction
+Ed = -(Deltae+Etaw)
+
+// Results
+printf( ' Part(b)the change in exergy is %.2f kJ/kg.',Deltae)
+printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',Etaw)
+printf( ' The exergy transfer accompanying heat is %.2f kJ/kg.',Etah)
+printf( ' The exergy destruction is %.2f kJ/kg.',Ed)