Answer to Question #110485 in Thermal Power Engineering for Kashish Sharma

Question #110485
Refrigerant-134a enters the condenser of a heat pump at 180 psia and 140 ºF at a rate of 8.3 lbm/min and leaves as a saturated liquid. If the compressor consumes 103.25 Btu/min of power, determine the coefficient of performance of the heat pump unit.
1
Expert's answer
2020-04-28T12:26:25-0400

Given:-

"\\\\Inlet\\,pressure\\,of \\,the\\,condenser\\,(P_{1})=180\\,psi\\\\[10pt]\nInlet\\,Temperature\\,of \\,the\\,condenser\\,(T_{1})=140F\\,\\\\[10pt]\nFlow\\,rate\\,of\\,the\\,refrigerant\\,(m)=8.3\\,lbm\/min\\\\[10pt]\nExit\\,pressure\\,(P_{2})=180lbm\/min\\\\[10pt]\nQuality \\,of\\,refrigant\\,at\\,exit\\,of\\,condenser\\,(x_{2})=0\\\\[10pt]\nPower\\,input\\,to\\,the\\,compressure\\,(w_{net})=103.25\\,Btu\/min\\\\[10pt]"

Find coefficient of performance (C.O.P)

Now,

"\\\\From\\, Superheated\\, refrigerant\\,-134a\\,table\\\\[10pt]\nFrom\\,steam\\,table\\,at\\,(P_{1})=180\\,psi\\,and\\,T_{1}=35C\\\\[10pt]\nAt\\,,h_{1}=124.17\\,Btu\/lbm\\\\[10pt]\n\\\\From\\, Saturated\\, refrigerant\\,-134a\\,table\\\\[10pt]\nFrom\\,steam\\,table\\,at\\,(P_{2})=180\\,psi\\,and\\,x_{2}=0\\\\[10pt]\nAt\\,,h_{2}=51.5Btu\/lbm"

Now,

The heat rejected in the condenser is expressed as,


"\\\\Q=m(h_{1}-h_{2})\\\\[10pt]\nQ=8.3\\times (124.17-51.50)\\\\[10pt]\nQ=603.17\\,Btu\/min"

So,

coefficient of performance (C.O.P)

"\\\\(COP)_{Hp}=\\frac{Q_{H}}{W_{net}}\\\\[10pt]\n\n(COP)_{Hp}=\\frac{603.17}{103.25}=5.85\\\\[10pt]"

Therefore, COP of heat pump = 5.85


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