Question

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.

Answer 1

Given:-

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

Find coefficient of performance (C.O.P)

Now,

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

Now,

The heat rejected in the condenser is expressed as,

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

So,

coefficient of performance (C.O.P)

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

Therefore, COP of heat pump = 5.85