//Chapter-6, Illustration 11, Page 317 //Title: Refrigeration cycles //============================================================================= clc clear //INPUT DATA T1=-18;//Temperature at point 1 in oC T3=27;//Temperature at point 3 in oC rp=4;//Pressure ratio m=0.045;//mass flow rate in kg/s y=1.4;//Ratio of specific heats Cp=1.005;//Specific heat at constant pressure in kJ/kg-K //CALCULATIONS x=(y-1)/y;//Ratio T2=(rp^x)*(273+T1);//Temperature at point 2 in K Tmax=T2-273;//Maximum temperature in oC T4=((1/rp)^x)*(273+T3);//Temperature at point 4 in K Tmin=T4-273;//Minimum temperature in oC qL=Cp*(T1-Tmin);//Heat rejected Wcin=Cp*(Tmax-T1);//Compressor work Wtout=Cp*(T3-Tmin);//Turbine work Wnet=Wcin-Wtout;//Net work done COP=qL/Wnet;//Co-efficient of performance Qref=m*qL;//Rate of refrigeration in kW //OUTPUT mprintf('Maximum temperature in the cycle is %3.0foC \n Minimum temperature in the cycle is %3.0foC \n COP is %3.2f \n Rate of refrigeration is %3.2f kW',Tmax,Tmin,COP,Qref) //==============================END OF PROGRAM=================================