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+//example 3
+//Calculating reversible work and irreversibility
+clear
+clc
+//Form the Steam Tables at state 1
+u1=1243.5 //initial specific internal energy in kJ/kg
+s1=4.4819 //initial specific entropy in kJ/kg-K
+v1=28.895 //initial specific volume in m^3/kg
+v2=2*v1 //final specific volume in kg/m^3
+u2=u1 //initial specific internal energy in kJ/kg
+//These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg
+T2=9.1+273.2 //final temp. in K
+x2=0.513 //quality in final state
+s2=4.644 //final specific entropy in kJ/kg
+V1=1 //volume of part of A in m^3
+m=V1/v1 //mass flow rate in kg/s
+To=20+273.2 //Room temperature in K
+Wrev=To*m*(s2-s1) //reversible work in kJ
+I=Wrev //irreversibility of the process
+printf("\n The irreversibility is I=%.3f kJ/kg.\n",I) \ No newline at end of file