Mechanical Engineering - Heat Transfer, Refrigeration and Air Conditioning - Discussion

@Patel Harshad.

The answer is 13.

Wkt, Ideal COP= Low Temp/(High Temp-Low Temp)= 271/42=6.45;
Actual COP= Refrigerating Effect/Work (or Power) Input =(1/4)*6.45=1.6125
So, 1.6125= 30/Work (or Power) Input (or) Work (or Power) Input=18.6 kW.

Here Actual COP of 1.6125 indicates that 18.6 kW of power input is needed for a evaporator to extract 30 kW of heat from the refrigerated space.

(Notes: Refrigerating Effect means " the quantity of heat that each kg of refrigerant absorbs from the refrigerated space to produce useful cooling". So, Refrigerating Effect=Heat Extracted)

A cold storage is to be maintained at -5*c while the surroundings are at 35*c. that heat leakage from the surroundings into the cold storage is estimated to be 29 kw.

The actual c.o.p of the refrigeration plant is one-third of an ideal plant working between the same temperatures. Find the power required to run the plant.

Here T1 = 273 - 2 = 271 and T2 = 273 + 40 = 313
So, COP = T1/T2 - T1 = 6.45
NOW actual cop is 1/4 of ideal cop
So,
ACTUAL COP = 1/4 * 6.45 = 1.6
basically COP = heat extracted/w.d.
So, POWER = 30/1.6 = 18.6 answer.

@Mohamme Thank you so much.

Thanks, @Mohamme Riyas.

@Mohit : We know COP = TEMP OF COLD BODY/TEMP OF HOT BODY - TEMP OF COLD BODY.

So here COP = 271/(313-271) = 6.45.

Actual COP = 1/4th of Ideal COP = 1/4(6.45) = 1.6125.

We also know actual COP = Heat extracted/Work input.

1.6125 = 30/WI.

Therefore Power = 30/1.6125 = 18.6 Kw.

What is the solution?