Question

An ideal vapor-compression refrigeration cycle that uses refrigerant R-134a as its working fluid maintains a condenser at 800 kPa and the evaporator at -12C. (a) Determine this system's COP and the amount of power required to service a 150kW cooling load. (b) Determine the P,T, h, S and exergy of R-134a at all four states of the entire cycle. Assume the ambient temperature to be 25C.

Answer 1

Sol: Given T1=-12degreec p2 =p3 =800 kpa from R-134a properties table at T1 =-12degreec h1hg=243.3 KJ/kg: p1=psat=185.37 kra s1=sg=0.93911kj/kg k from super heated tables at p2=800kra & s1=s2=0.93911 kJ/kg k h2=267.29+276.45-267.29/0.9480-0.9183(0.93911-0.9183) h2=273.708 kJ/kg & T2=31.31+(40.31.31)(0.93911-0.918)/0.948-0.9183 T2=37.4degreec at 800 kra, h3=h4=95.47 kJ/Kg k h3=95.47 kJ/kg: s3=sF=0.35404 kJ/kg k Therefore at state 4: h3=h4= 95.47 kJ/Kg Therefore h4=hf+x4hfg material implication at T=-12degreec, hf=35.92 kJ/kg, hfg=207.38 kJ/kg Therefore x4=95.47-35.92/207.38=0.287 at T=-12degreec, Sf=0.14504 kJ/Kg k, Stg=0.79406 kJ/kg k Therefore S4=sf+x4 Sfg=0.14504+0.287times0.79406 S4=0.373 KJ/Kg k Therefore P4=Psat=185.37 kPa given cooling, Q=150 KW Qc=m(h1-h4)=150 kw m=150/243.3-95.47 m=1.0146 kg/s Therefore compressor power, wc=m(h2-h1) =1.0146(273.708-243.3) wc=30.854 kw Therefore cop of the cycle is cop=Qc/wnet cop=150/30.854=4.86 cop= energy (E)=h.T0 S at state(1): E1=h1-T0S1=243.3-298times0.93911=-36.554 kJ/Kg state(2): E2=h2-T0S2=273.708-293(0.93911)=-6.1467 KJ/kg E3=h3-T0S3=95.47-298times0.35404=-10.034 kJ/kg E4=h4-T0S4=95.47-298times-373=-15.684 kJ/kg