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+clear ;
+clc;
+// Example 22.2
+printf('Example 22.2\n\n');
+//page no. 652
+// Solution
+
+// Given
+T1 = 80 ;// Initial temperature -[degree F]
+T1 = 40 ;// final temperature -[degree F]
+
+// Additional data obtained from steam table at given temperatures and corresponding vapour pressures
+p1 = 0.5067 ;// Initial saturation pressure-[psia]
+p2 = 0.1217 ;// Final saturation pressure-[psia]
+V1 = 0.01607 ;// Initial specific volume - [cubic feet/lb]
+V2 = 0.01602 ;// Final specific volume - [cubic feet/lb]
+H1 = 48.02 ;//Initial specific enthalpy -[Btu/lb]
+H2 = 8.05 ;// Final specific  enthalpy -[Btu/lb]
+
+del_P = p2 - p1 ;// Change in pressure -[psia]
+del_V = V2 - V1 ;// Change in specific volume -[cubic feet/lb]
+del_H = H2 - H1 ;// Change in specific enthalpy -[Btu/lb]
+del_pV = p2*144*V2/778 - p1*144*V1/778 ;// Change in pv-[Btu]
+del_U = del_H - del_pV ;// Change in specific internal energy - [Btu/lb]
+del_E = del_U ;// Change in specific total energy(since KE=0,PE=0 and W=0) -[Btu/lb]
+
+printf('\nChange in pressure is %.3f psia .\n ',del_P);
+printf('\nChange in specific volume is %.5f cubic feet/lb (negligible value) .\n ',del_V);
+printf('\nChange in specific enthalpy is %.2f Btu/lb .\n ',del_H);
+printf('\nChange in specific internal energy is %.2f Btu/lb .\n ',del_U);
+printf('\nChange in specific total energy is %.2f Btu/lb .\n ',del_E);
+\ No newline at end of file