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+clc
+clear
+//INPUT DATA
+cpic=1.94;//specific pressure inkJ/kgK
+cpv2=2.805;//specific pressure in kJ/kgK
+t21=303;//condenser temperature in K
+t1=258;//evaporator temperature in K
+t31=293;//subcooled temperature in K
+p1=2.36;//pressure in MPa
+p2=11.67;//pressure in MPa
+hf1=112.3;//enthalpy in kJ/kg
+hfg1=1313.7;//enthalpy in kJ/kg
+hg1=1426;//enthalpy in kJ/kg
+hf2=323.1;//enthalpy in kJ/kg
+hfg2=1145.9;//enthalpy in kJ/kg
+hg2=1469;//enthalpy in kJ/kg
+sf1=0.457;//entropy in kJ/kgK
+sf2=1.204;//entropy in kJ/kgK
+sg1=5.549;//entropy in kJ/kgK
+sg2=4.984;//entropy in kJ/kgK
+t2=369.7;//from t-s diagram temperature in K
+nac=0.8;//adiabatic efficiency in percentage
+vsa=2.96;//volume in kg/min
+N=1200;//speed in rpm
+
+//CALCULATIONS
+h2=hg2+cpv2*(t2-t21);//enthalpy in kJ/kg
+Rc=10*1000*(4.1868*30+335+1.94*5)/(24*60);//Refrigeration capacity in kJ/min
+Re=hg1-hf2;//Refrigeration effect in kJ/kg
+m=Rc/Re;//mass flow rate of refrigerant in kg/min
+h2a=((h2-hg1)/nac)+hg1;//enthalpy in kJ/kg
+t2a=((t2-t1)/nac)+t1;//Temperature in k
+d=(vsa*0.509*4/(3.14*1.2*N))^(1/3);//piston displacement of compressor in m
+l=1.2*d;//length of piston displacement in m
+w=(h2a-hg1)/0.95;//workdone in kJ/kg
+wac=m*w/60;//Power of the compressor motor in kW
+copa=(Re/wac)*(m/60);//COP of air 
+
+//OUTPUT
+printf('(a)Refrigeration capacity is %3.1f kJ/min \n (b)Mass flow rate of refrigerant is %3.2f kg/min \n (c)The discharge temperature is %3.2f K \n (d)Piston displacement of the compressor is d %3.4f m \n   l is %3.4f m \n(e)Power of the compressor motor is %3.2f kW \n (f)COPa is %3.3f',Rc,m,t2a,d,l,wac,copa)
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