Carnot Cycle and Coefficient of Performance

• Engineering
• bardia sepehrnia
In summary, the heat extracted from the water is the same as the heat rejected to the high temperature reservoir.
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)

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?

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?

bardia sepehrnia said:
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
Thank you. Got it now

What is a Carnot Cycle?

A Carnot Cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

What is the Coefficient of Performance (COP) in relation to the Carnot Cycle?

The Coefficient of Performance (COP) is a measure of the efficiency of a heat pump or refrigeration system, which is based on the Carnot Cycle. It is defined as the ratio of the desired output (heating or cooling) to the required input (energy or work).

What is the maximum efficiency of a Carnot Cycle?

The maximum efficiency of a Carnot Cycle is determined by the temperature of the heat source and heat sink. It is given by the equation: efficiency = (T1 - T2) / T1, where T1 is the temperature of the heat source and T2 is the temperature of the heat sink. This means that the efficiency increases as the temperature difference between the two increases.

What factors affect the Coefficient of Performance in a real-world heat pump or refrigeration system?

In a real-world system, the Coefficient of Performance is affected by various factors such as the efficiency of the compressor, heat exchangers, and expansion valve, as well as the temperature and pressure of the refrigerant. Other factors like external conditions and maintenance of the system can also impact the COP.

What are the practical applications of the Carnot Cycle and Coefficient of Performance?

The Carnot Cycle and Coefficient of Performance have practical applications in the design and analysis of heat pumps, refrigeration systems, and other thermodynamic processes. They also play a role in understanding the limitations of energy conversion and the potential for improving efficiency in these systems.

Similar threads

• Engineering and Comp Sci Homework Help
Replies
1
Views
4K
• Engineering and Comp Sci Homework Help
Replies
4
Views
3K
• Thermodynamics
Replies
3
Views
675
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
795
• Thermodynamics
Replies
12
Views
2K
• Thermodynamics
Replies
20
Views
1K
• Thermodynamics
Replies
7
Views
1K
• Thermodynamics
Replies
5
Views
1K
• Engineering and Comp Sci Homework Help
Replies
4
Views
2K