# Help understanding Coefficient of Performance and the working of heat pumps **URL:** \ **Category:** Factual Questions\ **Created:** [January 14, 2010, 12:06pm UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922 "2010-01-14T12:06:15Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![raindog](https://avatars.discourse-cdn.com/v4/letter/r/2bfe46/32.png) [@raindog](https://boards.straightdope.com/u/raindog)\ **Post date:** [January 15, 2010, 4:35am UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922/21 "2010-01-15T04:35:21Z")
> [@beowulff](#): > > I don’t know about geothermal systems, but my heatpump periodically changes modes in the winter so that the outside unit doesn’t ice up. Also, I can see the temperature drop on my outside thermometer when it’s in heating mode. It’s actually going into air conditioning to defrost the system. To hasten the melting process, it disables the fan outside. But since A/C in the middle of winter isn’t cool (hehe), the system energizes your electric back up heat to mitigate the A/C. It wants to get that done as rapidly as possible. ---
**Author:** ![raindog](https://avatars.discourse-cdn.com/v4/letter/r/2bfe46/32.png) [@raindog](https://boards.straightdope.com/u/raindog)\ **Post date:** [January 15, 2010, 4:54am UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922/22 "2010-01-15T04:54:24Z")
> [@Tilty](#): > > Hi everyone, > > I was wondering if someone could help me understand a coefficient of performance and more specifically how heat pumps work. > > A friend of mine mentioned he was getting an air-source heat pump installed as it was very efficient and could heat his water to 49C using the air outside, even if the air was at freezing. We got to discussing how they worked. > > I initially got hung up with the fact that I thought it was generating heat, and since it had a CoP of \>1, it effectively must be generating more energy than it was using. This didn’t sound right. Something to do with perpetual motion machines, free energy and the violation of the first law of thermodynamics. I certainly didn’t remember reading about the end of our reliance on oil, nor do I remember reading about any of the laws of thermodynamics being repealed 😃 > > I understand now that the CoP is not a measure of efficiency and that these pumps (in fact any heat pump, including ones such as air conditioners) achieve these \> 1 results by using less work to move heat/energy from one place to another, than the amount of energy they move. Since the measurements are made on an absolute basis (i.e the temperature is measured in Kelvin as opposed to Celsius) the movement of energy from air at 273K to heat water to 322K requires moving only a percentage of the available thermal energy in the air, into the water. > > Well, at least, that is how I currently understand it. > > Where I am getting confused, is how you ‘move’ energy from one source to another. I can grasp the concept that these pumps can use 1kW to move 3kW of energy from one source to another, I’m just having a hard time visualising how they ‘move’ this energy. > > Anyone have the Cliff notes for ‘A Dummies Guide to Heat Pumps’ or more accurately ‘How to explain heat pumps, very slowly indeed’? > > Thanking you in advance! > > Tilty Air conditioning systems, and heat pumps are simply moving heat (btus) from one place to another. If you place a cold Coke in your 70° kitchen a heat transfer takes place; Btus from the ambient air in the kitchen transfer into the Coke and in time the Coke will be the same temperature of the kitchen: 70°. So…hot always “runs” to cold, right? Nature always want to equalize. It’s a little harder than our Coke analogy however. How do we essentially “harvest” btus from the outside ambient conditions when it’s cold out side? (Of course, “cold” is just the absence of heat) Answer: by creating a surface----a Coke can-----that’s colder them the surrounding ambient, from which heat (btus) will transfer— an arrangement where hot will run to cold. So if it’s 20° outside------even though we perceive that to be cold—or, rather, lacking any heat----there is some heat in that air. If I put a surface----in this case a refrigeration coil----that I’ve engineered to be 5°, the btus in that 20° air will transfer to the 5° coil. Heat (what seems to be cold) has run to cold[er]. In other words, cold has run to colder. Now I can take those btus and essentially concentrate them----kind of round them up—and heat your home. The perfect medium to do that is refrigerant–largely because if it’s low boiling and condensing temperatures/pressure. The refrigerant is pumped through the system where it “accepts” heat/btus through "boiling (and changes to vapor (think:steam) and them is pumped somewhere where the heat is “rejected” through condensing. (think:liquid) And the process repeats again and again and again. ---
**Author:** ![raindog](https://avatars.discourse-cdn.com/v4/letter/r/2bfe46/32.png) [@raindog](https://boards.straightdope.com/u/raindog)\ **Post date:** [January 15, 2010, 5:09am UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922/23 "2010-01-15T05:09:00Z")
> [@](#): > > Originally Posted by Tilty View Post > The, quite stunning, clear answers in this thread for which I am incredibly grateful would lead me to believe that the input air temperature will always be lower when it leaves the system. Is that correct? > [@YamatoTwinkie](#): > > Yes, that’s right. A heat pump is basically air-conditioning the world while heating your house. There might be some controls on it to prevent the temperature from getting too low and icing up, but those come at reduced efficiency (while its running backwards to prevent ice buildup) I agree with YamatoTwinkie. Of course, in the summer the heat pump is doing the reverse—it’s heating the world, while air conditioning your home. In reality, of course, it’s simply moving heat from one place to another. (taking the heat of compression out of the equation for now) So in the summer (and I assume you were talking about the air temperatures outside as opposed to inside) the air temperature will be higher when it leaves the system. (while in both summer and winter the opposite is happening inside) ---
**Author:** ![Bill\_Door](https://avatars.discourse-cdn.com/v4/letter/b/50afbb/32.png) [@Bill\_Door](https://boards.straightdope.com/u/Bill_Door)\ **Post date:** [January 15, 2010, 11:34am UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922/24 "2010-01-15T11:34:10Z")
The heat pump COP complicates the often discussed theory about whether setting back the thermostat at night during heating season saves energy. If you use strictly heat generating central heat, cutting back the thermostat will always save energy. Heat loss, which is what you have to make up, is always proportional to the delta T between the house and the environment. Actually, since as your house cools off it heats the outside, it’s the log mean square of the delta T, but the outside doesn’t heat much, so forget about it. My heat pump (COP maybe 3.5, depending on the outside temperature), if the temperature set on the thermostat is 5 degrees higher than the temperature in the house, will start the emergency heat strips (COP 1, or a little bit less). So if I cut back the thermostat 10 degrees at night and crank it back up in the morning until my house heats up that first 5.1 degrees the efficiency of my heating takes a hit, because the folks that designed it figured that a five degree difference meant the heat pump wasn’t keeping up. I have to bump the temperature in a couple of stages to get the heat pump to do the work without relying on the resistance heating. ---
**Author:** ![raindog](https://avatars.discourse-cdn.com/v4/letter/r/2bfe46/32.png) [@raindog](https://boards.straightdope.com/u/raindog)\ **Post date:** [January 15, 2010, 1:14pm UTC](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922/25 "2010-01-15T13:14:03Z")
> [@Bill\_Door](#): > > The heat pump COP complicates the often discussed theory about whether setting back the thermostat at night during heating season saves energy. If you use strictly heat generating central heat, cutting back the thermostat will always save energy. Heat loss, which is what you have to make up, is always proportional to the delta T between the house and the environment. Actually, since as your house cools off it heats the outside, it’s the log mean square of the delta T, but the outside doesn’t heat much, so forget about it. > > My heat pump (COP maybe 3.5, depending on the outside temperature), if the temperature set on the thermostat is 5 degrees higher than the temperature in the house, will start the emergency heat strips (COP 1, or a little bit less). So if I cut back the thermostat 10 degrees at night and crank it back up in the morning until my house heats up that first 5.1 degrees the efficiency of my heating takes a hit, because the folks that designed it figured that a five degree difference meant the heat pump wasn’t keeping up. > > I have to bump the temperature in a couple of stages to get the heat pump to do the work without relying on the resistance heating. Every stat I’m aware of will do this unfortunately; if you’re more than a couple (3 give or take 1) degrees off setpoin it will bring in the back up heat. One potential fix is to install an outdoor thermostat that will disable the electric heat if the outside temperature is above the balance point. It will force the heat pump to do the work (because above the balance point it can) although it will obviously take longer to get there. [Previous page](https://boards.straightdope.com/t/help-understanding-coefficient-of-performance-and-the-working-of-heat-pumps/524922.md?page=1)