Answered

A refrigerator is used to cool water (Cp = 4.18 kJ/(kg.oC) & density  = 1 kg/L) from 23 oC to 5 oC in a continuous manner. The heat rejected in the condenser is 570 kJ/min and the power of the compressor is 2.65 kW. Determine: a) The rate at which water is cooled in L/min b) The refrigerator COP

Answer :

Answer:

Q=5.46 L/s

COP=2.58

Explanation:

Given that

Cp = 4.18 kJ/(kg.C

density  = 1 kg/L

Heat rejected Qr= 570 kJ/min

Power in put W= 2.65 KW

From first law of thermodynamics

Qr= W+ Qa

Qa=Heat absorbed

Qr= 570 kJ/min  = 9.5 KW

9.5 = 2.65 + Qa

Qa=6.85 KW

COP = Qa/W

COP = 6.58 / 2.65

COP=2.58

Lets take volume flow rate is Q

So mass flow rate of water m = ρ Q

Qa= m Cp ΔT

6.85 = 1 x Q x 4.18 ( 23-5)

Q=0.091 L/min

Q=5.46 L/s

Other Questions