Question

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.

Answer 1

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