# Carnot Cycle and Coefficient of Performance

bardia sepehrnia
Homework Statement:
A refrigeration system is used to cool down water from 15 C temperature and 150 kPa pressure to 5 C at a rate of 0.25 kg/s. Assume the refrigeration system works in a Carnot cycle between 0 C and 40C.
Determine:
a) The required rate of heat absorbed from the water
b) The maximum possible refrigeration system COP
c) The amount of power input that a refrigeration system required
d) The rate of heat rejected to the atmosphere
Relevant Equations:
Q=m*c*deltaT
Carnot efficiency = 1-(T_L/T_H)
COP=Q_L/W_net,in
I think I calculated part a correctly by extracting the cp (specific heat) of water from the table which is 4.188
Then calculated Q_dot by simply using the equation Q=m*c*deltaT=10.47kW

But I am stuck at part b, I know that the heat extracted from the water is the same as Q_L (rate of heat absorption from low temp reservoir). But how can I calculate the coefficient of performance (COP) if I don't have Q_H (rate of heat rejection to the high temp reservoir) or W_in (work input)

## Answers and Replies

Mentor
The maximum COP is obtained if the change in entropy of the reservoirs is zero. Based on this and the reservoir temperatures, what is the rate of heat rejection to the high temperature reservoir?

bardia sepehrnia
I'm not sure if I'm following you. I tried using another approach.
COP=Q_L/W_net and since w_net=Q_H-Q_L then we have: COP=Q_L/(Q_H-Q_L), and since this is a Carnot cycle, the Q_L and Q_H can be replaced with T_L and T_H. So we have

COP=T_L/(T_H-T_L)

Is this correct?

Mentor
I'm not sure if I'm following you. I tried using another approach.
COP=Q_L/W_net and since w_net=Q_H-Q_L then we have: COP=Q_L/(Q_H-Q_L), and since this is a Carnot cycle, the Q_L and Q_H can be replaced with T_L and T_H. So we have

COP=T_L/(T_H-T_L)

Is this correct?
It is correct, although it is definitely not how I would have done it. l would have written that the change in entropy of the pair of reservoirs is given by $$\Delta S=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}=0$$where ##Q_C## is the heat received by the cold reservoir at 0 C, and ##Q_H## is the heat rejected to the hot reservoir at 40 C. From this you can calculate ##Q_H##. The rest is easy.

bardia sepehrnia
bardia sepehrnia
Thank you. Got it now