# How can heat exchangers have COP>1

1. Oct 17, 2011

### Low-Q

Hi,

I am building a new house. This house will be heated by a heat exchanger. I have been told that heat exchangers have a COP of 3-4 in practice (depending on inside and outside temperature, and the cooling medium).

What I do not understand, is the COP>1 part. In order to "charge" the house with heat, by extracting heat from outside air, it will require energy input to the pump. The house is leaking heat back to the outside air, so the heat exchanger must run in order to keep the same temperature in the house.

The house is insolated well, but that would also mean that a normal electric oven must add small portions of energy to keep the air inside at the same temperature....

I simply cannot understand how input energy is greater than output energy, and that heat exchangers does not require the same energy to sustain the temperature insde as a conventional electric oven does.

Can someone explain this in detail?

br.

Vidar

2. Oct 17, 2011

### Staff: Mentor

You mean a heat pump, not a heat exchangers.

Since a heat pump doesn't make heat directly, it just moves it from one place to another, it can have a COP (which is different than the efficiency) of greater than 1. Consider this: if you are pumping water, the energy input and efficiency are determined by flow rate and pressure. But what if you were more interested in the fact that the water was very warm compared to what you need. Then the energy you get out of the water and the pumping energy aren't directly related to each other anymore and you can get more heat energy out than mechanical energy put in. Same idea with a heat pump or air conditioner (which also has a COP of at least 3).

3. Oct 17, 2011

### Low-Q

OK, so the heat pump is just moving heat from the outside air to the inside at "no" cost except the energy required to maintain the pressure difference between hot and cold zone?
Is it correct to say that a heat pump is "focusing" the heat in the enormous volume of heat containing air at a small point (The heat pump itself)?

But wouldn't the potential difference in temperature be the same as the energy required to run the pump? If so, I am a little confused - still...

Vidar

Last edited: Oct 17, 2011
4. Oct 17, 2011

### Andrew Mason

It is taking some of the thermal energy of a larger mass of cooler matter (the ground) and transferring it to a smaller mass of warmer matter (the air in your house). If that is what you mean by "focusing", yes.
The heat pump does the work needed to transfer the thermal energy. The second law of thermodynamics says that heat only flows on its own from hotter bodies to cooler bodies. To make it flow the other way, you must do work.
The greater the temperature difference, the greater the work (input energy) that must be done in order to move a given quantity of heat from the outside to the inside.

Incidentally, if your heat pump has a COP of only 3 or 4, you are not going to be saving much energy or reducing CO2 emissions overall if you use electricity that is generated by burning fossil fuels. A coal fired power plant is about 1/3 efficient: one unit of electricity output for three units of heat from burning fossil fuel. And there are losses from transmission that reduce that electrical energy a little further in getting it to your house. So if your COP is 3 you will be getting heat in your house that is about the same as if you burned fossil fuel in a furnace. If you are getting your electricity from a hydro, wind or solar source it is a different matter.

AM

5. Oct 17, 2011

### Low-Q

Does it exist "power plants" using heat pump technology? I mean, if it is possible to harness the heat from thin air and then "focus" that heat energy to make it more efficient for a heat-to-mechanical work engines (turbines or similar), wouldn't it be possible to generate electric power that is greater than the energy required to run the pump? In my mind, that should not be possible, but if the energy is present already in the air, it might be possible after all?

Vidar

6. Oct 17, 2011

### Staff: Mentor

No. Efficiency of a heat engine and heat pump are inverses of each other. So if a heat pump has a cop of 4, then the best a heat engine could do for converting back to mechanical power is 25%, for no net gain.

7. Oct 17, 2011

### Low-Q

Ok. Thanks. Now I understand much better.

8. Sep 16, 2012

### Low-Q

This thread is old, but I have one question that I came up with here the other day. If a heat pump have a COP of 4, and I also have a heat pump with COP 2, would that last heat pump become a heat engine with 50% efficiency that is being able to be "self powered" by the COP 4 heat pump? Or is the calculations wrong regarding the relationship between COP and efficiency.

Br.

Vidar

9. Sep 17, 2012

### Low-Q

My guess is that the big area in the ground cools down as the "self running" system runs. The ground will be supplied with heat from surrounding groundwater or soil.

Is it possible to increase the efficiency of a heat engine this way - as questioned in post 8?

Anyone?

Vidar

10. Sep 18, 2012

### Staff: Mentor

The performance of an ideal heat pump and heat engine is dictated by the temperature difference produced/utilized. So the DT produced by a 4:1 COP heat pump will be much, much too small to run a 50% efficient heat engine. If you try to increase the DT out of the heat pump, the COP will go down. If you try to use a lower DT to run the heat engine, the efficiency will go down.

You can google for the efficiency equations and play with them -- they are not hard to use.