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+clear;
+clc;
+disp('Example 7.5');
+
+//  aim : To determine
+//  the  final condition of steam...
+//  the change in specific entropy during hyperbolic process
+
+//  Given values
+P1 = 2;//  pressure, [MN/m^2]
+t = 250;//  temperature, [C]
+P2 = .36;//  pressure, [MN/m^2]
+P3 = .06;//  pressure, [MN/m^2]
+
+//  solution
+
+//  (a)
+//  from steam table
+s1 = 6.545;//  [kJ/kg K]
+//  at .36 MN/m^2
+sg = 6.930;//  [kJ/kg*K]
+
+sf2 = 1.738;//  [kJ/kg K]
+sfg2 = 5.192;//  [kJ/kg K]
+vg2 = .510;//  [m^3]
+
+//  so after isentropic expansion, steam is wet
+//  hence, s2=sf2+x2*sfg2, where x2 is dryness fraction
+//  also
+s2 = s1;
+//  so
+x2 = (s2-sf2)/sfg2;
+//  and
+v2 = x2*vg2;//  [m^3]
+
+//  for  hyperbolic process
+//  P2*v2=P3*v3
+//  hence
+v3 = P2*v2/P3;//  [m^3]
+	
+mprintf('\n (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is  =  %f  m^3/kg\n',v3);
+
+//  (b)
+//  at this condition
+s3 = 7.609;//  [kJ/kg*K]
+//  hence
+change_s23 = s3-sg;//  change in specific entropy during the hyperblic process[kJ/kg*K]
+mprintf('\n (b) The change in specific entropy during the hyperbolic process is  =  %f kJ/kg K\n',change_s23);
+
+// In the book they have taken sg instead of s2 for part (b), so answer is not matching
+
+//  End
+