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+clear ;
+clc;
+// Example 22.3
+printf('Example 22.3\n\n');
+//page no. 662
+// Solution fig.E22.3a
+
+//Lets take tank to be system
+// Given
+T = 600 ; // Temperature of steam  -[K]
+P = 1000 ;// Pressure of steam -[kPa]
+
+// Additional data for steam obtained from CD database at T and P
+U = 2837.73 ;// Specific internal energy-[kJ/kg]
+H = 3109.44 ;// Specific enthalpy -[kJ/kg]
+V = 0.271 ;// Specific volume -[cubic metre/kg]
+
+// Use eqn. 22.6 to get change in specific internal energy,by simplifing it with following assumption:
+//1. Change in KE and PE of system = 0, therefore change in total energy = change in internal energy
+//2. W = 0,work done by or on the system
+//3. Q = 0 , system is well insulated
+//4. Change in KE and PE of entering steam = 0
+//5. H_out = 0, no stream exits the system
+//6. Ut1 = 0, initially no mass exists in the system
+
+// By the reduced equation 
+Ut2 = H ;// Internal energy at final temperature-[kJ/kg]
+
+printf('\nThe specific internal energy at final temperature is %.2f kJ/kg. \nNow use two properties of the steam (P = %i kPa and Ut2 = %.2f kJ/kg) to find final temperature (T) from steam table. \nFrom steam table we get T = 764 K.',Ut2,P,Ut2);
+\ No newline at end of file