Question

A carnot engine operates between temperature levels of 600k and 300k, It drives a carnot refrigerator which provides cooling at 250k and discards heat at 300k. Determine a numerical value for the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load).

Answer 1

the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load) is 2.5

Step-by-step explanation:

Efficiency is defined as the ratio of desired output / input

Efficiency of carnot engine =

`n_C = (W)/(Q_H) \\= 1 - (300)/(600) \\= 0.5`

so,

`W = 0.5Q_H\\= (250)/(300 - 250) \\= 5`

in this case

`W = 0.2Q_L`

work is the same because we use work from carnot engine to power the refrigerator

Hence,

`0.2Q_L = 0.5Q_H\\(Q_L)/(Q_H) = (0.5)/(0.2) \\= 2.5`

the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load) is 2.5