A heat pump with refrigerant-134 as the working fluid is used to keep a space at

Question

A heat pump with refrigerant-134 as the working fluid is used to keep a space at ( 25 Celsius ) by absorbing heat from geothermal water that enters the evaporator at ( 50 C) at a rate of 0.065 kg/s and leaves at (40C). The refrigerant enters the evaporator at ( 20 C) with a quality of 23 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine (a) the degrees of subcooling of the refrigerant in the condenser, (b) the mass flow rate of the refrigerant, (c) the heating load and the COP of the heat pump, and (d) the theoretical minimum power input to the compressor for the same heating load.

Explanation / Answer

At T4 = 20 C and x4 = 0.23

We get P4 = 572.1 kPa

ALso look for hf and hfg values and calculate h = hf + x4.hfg = 121.24 kJ/kg

Now, Process 3-4 is isenthalpic.

so h3 = h4

P1 = P4 = 572.1 kPa and state is saturated vapour

h1 = hg = 261.59 kJ/kg

s1 = sg = 0.9223 kJ/kg

Process 1-2 is isentropic compression

P2 = 1400 kPa

s2 = s1 = 0.0223 kJ/kg

So by interpolation we get, h2 = 280 kJ/kg

Now, using the R 134 a graph or EES, in subcooled region

at P3 = 1400 kPa

h3 = 121.24 kJ

we get T3 = 48.6 C

Had it been saturated liquid at 3, the saturation temp at condenser pressure of 1400 kPa would be 52.4 C (From chart)

Degree of subcooling = Tsat - T3 = 52.4 - 48.6 = 3.8 C

b) In the evaporator we are using water to transfer heat to the refrigerant.

hw1 = hf @ 50 C = 209.34 kJ/kg

hw2 = hf @ 40 C = 167.53 kJ/kg

Using energy balance in the evaporator

m(dot)w (hw1 - hw2) = m(dot)R (h1 - h4)

m(dot)R = 0.065 (209.34 - 167.53) / (261.59 - 121.24) = 0.01936 kg/s

c) Power input to compressor W(dot)in = m(dot)R (h2 - h1) + Q(dot) out = 0.01936 (280 - 261.59) = 0.6564 kW

Heating Load, Q(dot)h = m(dot)R (h2 - h3) = 0.01936 (280 - 121.24) = 3.074 kW

COP = Q(dot)h / W(dot)in = 4.68

d) The reversible COP of the cycle = 1 / [1 - (Tl / Th)] = 1/ [1 - (25+273) / (50 + 273)] = 12.92

Hence the minimum power input = Q(dot)h / COP rev = 2.074 / 12.92 = 0.238 kW

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