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

Step-by-step 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) = (T_L)/(T_H - T_L) \\ \\ = (32+460)/((72 +460) -(32+460)) \\ \\ =(492)/(532 -492) \\ \\ = (492)/(40) \\ \\ COP_(ref) = 12.3`

Again:

The efficiency given by COP can be represented by:

`COP = (Q_L)/(W) \\ \\ W = (Q_L)/(COP) \\ \\ W = (2000 \ lbm * 144 \ Btu/lbm)/(12.3) \\ \\ W = 23414.63 \ Btu`

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

`Power= (W)/(t) \\ \\ Power = (23414.63 \ Btu)/( 1 \ hr) \\ \\ Power = (23414.634 * 1055.056 \ J)/(1 * 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