Question

A carnot refrigeration cycle absorbs heat at 270k and rejects heat at 300k. (44) (a) calculate cop of this cycle. (b) if the cycle is absorbing 1130 kj/min at 270k, how much kj of work is required per second? (c) if the carnot heat pump operates between same temperature limits then what is the cop? (d) how much kj/min heat will be delivered by heat pump at 300 k if it absorbs 1130kj/min at 270k? 2.

Answer 1

Answer:

Given that,

TemperatureT

1

=270K

TemperatureT

2

=300K

Absorbed heat Q=1260KJ/min=21KJ/s

We know that,

Coefficient of performance is

C.O.P =

T

2

−T

1

T

2

C.O.P=

300−270

270

C.O.P=9

Using formula of work done

C.O.P=

W

Q

W=

9

21

W=2.33KJ/s

The work required is2.33KJ/s.

Hence, This is the required solution