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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 5,Page 166
+//Title: Minimum power and maximum efficiency
+//================================================================================================================
+clear 
+clc
+
+//INPUT
+T_ice=0;//temperature of the ice to be produced in degree celsius
+m=5000;//rate at which ice is to be produced in kg/hour
+T_water=0;//temperature of water used to produce ice in degree celsius
+T_amb=40;//ambient temperature in degree celsius
+T_source=100;//temperature of the source for operating heat engine in degree celsius
+lambda_fusion=6.002;//latent heat of fusion of water in kJ/mol at 0 degree celsius
+molar_mass=18*10^-3;//molar mass of water in kg/mol
+
+//CALCULATION
+T_L=T_water+273.15;//sink temperature of the refrigerating unit in K
+T_H=T_amb+273.15;//source temperature of the refrigerating unit in K
+COP=(T_L)/(T_H-T_L);//calculation of COP of the refrigerating unit using Eq.(5.20) (no unit)
+Q_L=((m/3600)/molar_mass)*(lambda_fusion);//calculation of the energy from the sink taken up by the refrigerator in kW
+W=(Q_L)/(COP);//calculation of the minimum power required to operate the refrigerator using Eq.(5.20)in kW
+T1=T_source+273.15;//temperature of the source of the heat engine in K
+T2=T_amb+273.15;//temperature of the sink of the heat engine in K
+n_heatengine=(T1-T2)/T1;//calculation of the efficiency of heat engine using Eq.(5.18) (no unit)
+Q1=W/n_heatengine;//calculation of the energy absorbed by the heat engine using Eq.(5.1) in kW
+//calculation of the ratio of energy rejected by both the devices to ambient atmosphere to the energy absorbed by the refrigerator (no unit)
+energy_ratio=(Q1+Q_L)/Q_L;
+
+//OUTPUT
+mprintf("\n The minimum power required to operate the refrigerator=%0.2f kW\n",W);
+mprintf("\n The maximum possible efficiency of the heat engine=%0.4f \n",n_heatengine);
+mprintf("\n Ratio of the energy rejected to the ambient atmosphere to the energy absorbed from the water=%0.4f \n",energy_ratio);
+
+//===============================================END OF PROGRAM===================================================