Question

In a residential heat pump, Refrigerant-134a enters the condenser at 800 kPa and 50oC at a rate of 0.022 kg/s and leaves at 750 kPa subcooled by 3oC. The refrigerant enters the compressor at 200 kPa superheated by 4oC. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, (d) determine the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 kPa and 800 kPa.

Answer 1

a) 0.813

b) 4.38 KW

c) COP = 5.13

d) 3.91 KW , COP = 6.17

Step-by-step explanation:

Data obtained A-13 tables for R-134a:

`h_(1) = 247.88 (KJ)/(kg) \\s_(1) = 0.9575 (KJ)/(kgK)\\h_(2s) = 279.45 (KJ)/(kg)\\h_(2) = 286.71 (KJ)/(kg)\\h_(3) = 87.83 (KJ)/(kg)`

The isentropic efficiency of the compressor is determined by :

`n_(C) = (h_(2s) - h_(1) )/(h_(2) - h_(1) )\\= (279.45 - 247.88 )/(286.71 - 247.88)\\= 0.813`

The rate of heat supplied to the room is determined by heat balance:

`Q = m(flow) * (h_(2) -h_(3))\\= (0.022)*(286.71 - 87.83)\\\\= 4.38KW`

Calculating COP

`COP = (Q_(H) )/(W) \\COP = (Q_(H) )/(m(flow) * (h_(2) - h_(1)) )\\\\COP = (4.38)/((0.022)*(286.71-246.88))\\\\COP = 5.13`

Part D

Data Obtained:

`h_(1) = 244.5 (KJ)/(kg) \\s_(1) = 0.93788 (KJ)/(kgK)\\h_(2) = 273.31 (KJ)/(kg)\\h_(3) = 95.48 (KJ)/(kg)`

The rate of heat supplied to the room is determined by heat balance:

`Q = m(flow) * (h_(2) -h_(3))\\= (0.022)*(273.31 - 95.48)\\\\= 3.91KW`

Calculating COP

`COP = (Q_(H) )/(W) \\COP = (Q_(H) )/(m(flow) * (h_(2) - h_(1)) )\\\\COP = (4.38)/((0.022)*(273.31-244.5))\\\\COP = 6.17`