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+//example 3
+//energy transport by mass
+clear
+clc
+vf=0.001053 //specific volume of saturated liquid water  in m3/kg
+vg=1.1594  //specific volume of water vapour in m3/kg
+ug=2519.2  //specific internal energy of water vapour kJ/kg
+hg=2693.1  //specific enthalpy of water vapour kJ/kg
+disp('Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established')
+disp(' Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure.')
+m=0.6/(vf*1000) //reduction in mass of liquid in pressure cooker in kg
+M=m/(40*60) //mass flow rate of steam in kg/s
+A=8*10^-6 //exit area in m^2
+V=M*vg/A //exit velocity in m/s
+e=hg-ug //flow energy of steam in kJ/kg
+TE=hg //total nergy of steam in kJ/kg
+E=M*hg //energy flow rate of steam leaving cooker in kW
+printf("\n Hence,The mass flow rate of the steam is = %.6f kg/s. \n",M); 
+printf("\n The exit velocity is = %.1f m/s. \n",V);
+printf("\n The total energy of the steam is = %.1f kJ/kf. \n",TE);
+printf("\n The flow energy of the steam is = %.1f kJ/kg. \n",e);
+printf("\n The rate at which energy leaves the cooker by steam is = %.3f kW. \n",E);
+\ No newline at end of file