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+/////////Chapter 10 Properties Of Steam
+///Example 10.23 Page No:203
+///Find Constant pressure process
+//Input data
+clc;
+clear;
+ms=1000; //Steam in Kg/h
+P=16; //Absolute pressure in bar
+x2=0.9; //Steam is dry
+t1=30+273; //temperature in degree celsius
+tsup=380; //tmperature rised in degree celsius
+
+//from steam table(pressure basis at 16 bar)
+h1=125.7; //in KJ/Kg
+ts=201.4; //In degree celsius
+hf=858.5; //in kJ/Kg
+hfg=1933.2; //in kJ/Kg
+hg=2791.7; //in kJ/Kg
+Cps=2.3;
+
+//Calculation
+h2=hf+x2*hfg; //Final enthalpy of wet steam in KJ/Kg
+Q1=(ms*(h2-h1))*(10^(-3)); //Constant pressure process in KJ/h
+h3=hg+Cps*(tsup-ts); //Final enthalpy of superheated steam in KJ/g
+Q2=(ms*(h3-h2))*(10^(-3)); //Suprheated steam in KJ/h
+
+//Output
+printf('Final enthalpy of wet steam= %f KJ/Kg \n ',h2);
+printf('Constant pressure process= %f KJ/h \n',Q1);
+printf('Final enthalpy of superheated steam= %f KJ/g \n',h3);
+printf('Suprheated steam= %f KJ/h \n',Q2);