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2 years ago

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A refrigerator is used to cool water from 23°C to 5°C in a continuous manner. The heat rejected in the condenser is 570 kJ/min a

nd the power is 2.65 kW. Determine the rate at which water is cooled in L/min and the COP of the refrigerator. The specific heat of water is 4.18 kJ/kg·°C and its density is 1 kg/L.

1 answer:

GenaCL600 [577]2 years ago

7 0

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

U = W+ q

q = Heat absorbed

U = internal energy

W = workdone

U = 570 kJ/min = 9.5 KW

9.5 = 2.65 + q

q = 6.85 KW

COP = q/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

q = m Cp ΔT

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

Q=0.091 L/min

Q=5.46 L/s

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