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ss7ja [257]
3 years ago
11

A refrigerator operating on the Carnot cycle is used to make ice. Water freezing at 32oF is the cold reservoir. Heat is rejected

to a river at 72oF. Determine the work required to freeze 2000 lbm of ice if the energy required to freeze ice is 144 Btu/lbm. What is the power input required in kW and hp is this operation is carried out in 1 hour? [20] Hint: use the max. possible COP equations.
Engineering
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

Explanation:

From the given information:

Water freezing temp. T_L = 32 ^0 \ F

Heat rejected temp T_H = 72 ^0 \ F

Recall that:

The coefficient of performance is:

COP_{ref} = \dfrac{T_L}{T_H - T_L} \\ \\  = \dfrac{32+460}{(72 +460) -(32+460)} \\ \\ =\dfrac{492}{532 -492} \\ \\ = \dfrac{492}{40} \\ \\ COP_{ref} = 12.3

Again:

The efficiency given by COP can be represented by:

COP = \dfrac{Q_L}{W} \\ \\ W = \dfrac{Q_L}{COP} \\ \\ W = \dfrac{2000 \ lbm \times 144 \ Btu/lbm}{12.3} \\ \\ W = 23414.63 \ Btu

Finally; the power input in an hour can be determined by using the formula:

Power= \dfrac{W}{t} \\ \\  Power = \dfrac{23414.63 \ Btu}{ 1 \ hr} \\ \\  Power = \dfrac{23414.634 \times 1055.056 \ J}{1 \times 3600} \\ \\  Power  = 6.86 \ kW

In hp; since 1 kW = 1.34102 hp

6.86kW will be = (6.86 × 1.34102) hp

= 9.199 hp

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