Answer to Question #279847 in Mechanical Engineering for sha

Question

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

Answer 1

Compressor :

$Q_{comp} = m (h_1-h_6) + W_{COMP} =0.05(377-284)-0.5 = -0.35kW$

Condenser:

$Q_{cond} = m(h3-h2) = 0.05(134-367) = -11.65kW$

Evoporator :

$Q_{evap}=m(h_5-h-4) = 0.05 (280-134) = 7.3kW$