What Is the Coefficient of Performance of a Carnot Heat Pump?

In summary: In this equation, Q_h is the heat pump's efficiency, and W is the work done by the heat pump. So the COP would be negative if the heat pump were to waste energy.
  • #1
~angel~
150
0
This question is killing me because I just can't seem to get it.

A heat pump is used to heat a house in winter; the inside radiators are at T_h and the outside heat exchanger is at T_c. If it is a perfect (e.g., Carnot cycle) heat pump, what is K_pump, its coefficient of performance?

Give your answer in terms of T_h and T_c.

According to the hints, you're meant to work out the efficiency of the pump in terms of Q_c and Q_h. I thought it was 1 + (Q_c/Q_h)...

The textbook states that Q_c/Q_h = - T_c/T_h, but in the hints, it states Q_h/Q_c = T_h/T_c.

I'm totally confused.

Please help.
 
Physics news on Phys.org
  • #2
I don't know how to solve the problem, but I wanted to point out that Q_h/Q_c = T_h/T_c and Q_c/Q_h = T_c/T_h are the same equation written differently. I don't know what the negative symbolizes, but if it was left out, then it must be somewhat arbitrary in meaning.

Hope this helps, good luck.
 
  • #3
It might be confusing with redundant data. The expression should be Qh/(Qh-Qc) or Th/(Th-Tc)
 
  • #4
~angel~ said:
The textbook states that Q_c/Q_h = - T_c/T_h, but in the hints, it states Q_h/Q_c = T_h/T_c.

I'm totally confused.

Please help.

I would say this equation represents such an ideal behavior of a Carnot heat pump. Entropy variation of the system must be 0 for being a cyclic machine, and entropy variation of universe (system+surroundings must be also 0 for being a reversible machine). So that the variation of entropy of the surroundings must counterbalance each other (in each focus).

On the other hand you should be careful with the sign convention. Maybe the book refers to different sign convention in each sentence. I always take the absolute value of the heats and put externally the convenient sign.

Also be careful because the COP of a heat pump is defined as [tex]COP=Q_h/W[/tex].
 

1. How does a heat pump work?

A heat pump works by transferring heat from one location to another. It uses a refrigerant to absorb heat from the air or ground outside and then compresses and releases it into the air inside. This process can be reversed to provide cooling in the summer.

2. What factors affect the performance of a heat pump?

The performance of a heat pump can be affected by several factors, including outdoor temperature, humidity levels, size and efficiency of the unit, and proper installation and maintenance.

3. How do I know if my heat pump is operating efficiently?

To determine the efficiency of your heat pump, you can check the unit's coefficient of performance (COP). This is a ratio of the heat output to the energy input and should be greater than 1. The higher the COP, the more efficient the heat pump is operating.

4. What is the ideal temperature setting for a heat pump?

The ideal temperature setting for a heat pump depends on the outdoor temperature. Generally, a heat pump is most efficient when the outdoor temperature is above freezing. It is recommended to set the thermostat to a temperature that is comfortable for you and your household.

5. How can I improve the performance of my heat pump?

To improve the performance of your heat pump, you can ensure proper installation and regular maintenance, including changing air filters and cleaning the outdoor unit. You can also consider upgrading to a more efficient heat pump model or adding insulation to your home to reduce the workload on the unit.

Similar threads

Replies
3
Views
552
  • Introductory Physics Homework Help
Replies
1
Views
765
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
3K
Back
Top