Question #279847

An R-410a heat pump cycle shown in the figure below has an R-410a flow rate of 0.05 kg/s with 5 kW into the compressor. The following data are given:


State 1 2 3 4 5 6

P, kPa 3100 3050 3000 420 400 390

T, oC 120 110 45 -10 -5

h, kJ/kg 377 367 134 280 284


Calculate the heat transfer

a) from the compressor

b) from the R-410a in the condenser

c) to the R-410a in the evaporator


1
Expert's answer
2021-12-16T00:42:01-0500

Compressor :

Qcomp=m(h1h6)+WCOMP=0.05(377284)0.5=0.35kWQ_{comp} = m (h_1-h_6) + W_{COMP} =0.05(377-284)-0.5 = -0.35kW

Condenser:

Qcond=m(h3h2)=0.05(134367)=11.65kWQ_{cond} = m(h3-h2) = 0.05(134-367) = -11.65kW

Evoporator :

Qevap=m(h5h4)=0.05(280134)=7.3kWQ_{evap}=m(h_5-h-4) = 0.05 (280-134) = 7.3kW



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