Help understanding Coefficient of Performance and the working of heat pumps

Factual Questions
1

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reading about the end of our reliance on oil, nor do I remember reading about any of the laws of thermodynamics being repealed :smiley:

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

2

Just about every house has at least one heat pump, and most have two. Your refrigerator is a heat pump: it pumps heat out of the fridge/freezer and into your kitchen air. Your house A/C is another heat pump: it pumps heat out of your house and into the great outdoors.

There are several ways for heat pumps to work, but the most common involves an evaporative refrigerant. This used to be Freon, but now R134 is in common use. The basic cycle looks like this:

  1. Refrigerant gas is supplied to a compressor. The compressor drives up the pressure of the refrigerant. The mechanical work of compression heats it up, so now you’ve got a hot, high-pressure gas (~200 psi).

  2. heat is extracted from the hot, high-pressure gas in a condenser coil. For central A/C on a house, the condenser coil is the big noisy thing in your backyard (the compressor is typically out there with it). A fan blows air over a heat exchanger, cooling the hot high-pressure gas into a cool, high-pressure liquid.

  3. This liquid passes through an orifice, causing a drop in pressure, and then enters an evaporator coil. For central A/C on a house, the evaporator coil is positioned in the ventilation system, somewhere downstream of the air filter (the big flat thing you replace a couple times a year) and upstream of the furnace burner. At this new low pressure (~25 psi), the liquid can evaporate, but it takes thermal energy to make that happen (for any given substance, a gas at temperature X contains much more energy than liquid at the same temperature). Just like sweat cools your skin, the evaporation of this liquid refrigerant cools the coils of the evaporator, taking thermal energy from them, giving you a cool, low-pressure gas.

  4. The cool, low-pressure gas - which contains the thermal energy extracted from the air in your house - heads back to the compressor to start the cycle all over again.

Now that we understand how a real heat pump works for cooling things, let’s take your house A/C system apart, and turn it around. Move the condenser coil indoors, and the evaporator coil outdoors. You are now cooling/refrigerating the great outdoors, and dumping that heat inside your house. You’ve used X amount of electrical energy to do it, and dumped X+Y heat energy into your house. The coefficient of performance is (X+Y)/X.

Whereas the efficiency of a heat engine must necessarily be less than one, the coefficient of performance of a heat pump must necessarily be greater than one. Higher numbers mean better performance, i.e. you’re using less input power to move more heat around. Example:

A COP of 2 means you get two joules of thermal energy from the condenser for every 1 joule of mechanical work you input at the compressor. One of those output joules is the work you put in, and the other is heat you picked up from the evaporator.

A COP of 1.1 means you only get 1.1 joules of thermal energy from the condenser for every 1 joule of mechanical work you input at the compressor. One of those output joules is the work you put in, and the remaining 0.1 joule is heat you picked up from the evaporator.

3

I’ve got a geothermal system, and still can’t get my brain around how they work all the time, even with excellent explanations such as the one posted by Joe.

In such systems, a lot of air needs to move and move often. My blower runs almost non-stop on the coldest days (normal), and the air never feels warm, even though it is about 90 degrees. I still have trouble grasping how the goethermal system runs through earth that is about 55 degrees and extracts enough energy to pump in 90 degree air… then I just have to review the explanations to assure myslef it works, then I feel the 90 degree air and it feels cold… but I know it’s because it’s below my body temp and that other hot air systems are pumping in less air, but hotter-than-body-temp air.

4

One thing left out of the explnation is that a gas at a higher pressure will condence at a higher temperture. (I am lazzy so I am not going to dig out my PT chart). At the condencer pressure in the 200 psi range and 180 degrees. If the condencer is out side it heats out side, if it is in the furnace it heats the air flowing across it.

And liquid at a lower pressure will evaporate at a lower temperature. 50 PSI is around 35 degrees. if the evaporator is out side cools out side, if inside cools air inside.

Also when a liquid changes state it absorbs energy. Example at 14.7 PSIA to change the temp of 1 pound water 1 degree takes approx 1 BTU. To change one poound of water at 212 degrees to one pound of steam at 212 degrees takes 970 BTUs.

Heats pumps work well because in cooling mode the heat from the space plus the waste heat of compression are both dumped outside, but in heating mode the heat from outside and the wasste heat of compression are both dumped in the living space.

5

What trips people up is the concept of “hot” and “cold.”
32°F is “cold” to most people, and they can’t understand how there can be any heat to extract from air of that temperature, but it’s much hotter than the temperature of the evaporating refrigerant which boils at -30°F.

6

Thought experiment:

Say you had an enormous warehouse where the interior temperature is 50 degrees, and the outside temperature is 70 degrees, and you want to heat up the warehouse. Obviously, you’d just open a window and let nature take care of it.

Since opening up a window (say with a small electric motor) requires almost no energy to do, and you’re capable of getting almost limitless heat from the setup (the warehouse is really big), your COP is:
(huge number)/(tiny number) = close to infinite. In fact, you could even put up a fan attached to a generator in the window and it would generate electricity, so your power efficiency is now much greater than 100% as well.

So say the warehouse starts heating up and is now at an interior temperature of 69 and an outside temperature at 70. Well, you’re still getting some heat coming in, so your COP drops to:

(ok number)/(tiny number)= still pretty close to infinite. Your fan generator stops spinning very fast so you’re really only somewhat larger than 100% efficiency power wise now.

So now the warehouse and the exterior finally both stabilize at 70 degrees, and you still want to heat it. But wait, Just one degree previously and we were looking at COPs of ~infinity and efficiencies of >100%, so there must be some way to still get large COPs and decent efficiencies by “stealing” the heat from the ouside and putting it to use, right?

And that’s exactly what a heatpump does. Even when temperatures outside are lower than inside, you can still “steal” heat from the outside and bring it inside, but your COP starts taking a hit by doing so. Eventually if the temperature differences are too severe, it starts costing more energy to transport heat from the outside than you get out of it, at which point COP drops below 100% and you’re better off just using an electric resistance heater.

7

The ubiquitous R134 refrigerant cycle is not the only option out there. In fact, a circulating refrigerant is not particularly required. Witness thermoelectric cooling: electrical current is made to flow around a circuit composed of two dissimilar metals, with the result that heat is transported from one junction to the other. The physics behind this mechanism are somewhat advanced, but it really does work. This method is much less efficient than the R134 cycle, but it has no moving parts, which means it’s cheaper to manufacture, can be scaled down to very small size (e.g. CPU coolers in PCs), is maintenance-free, and has excellent long-term reliability.

The low cost of thermoelectric heat pumps has facilitated their use in camp coolers. I’ve got one. It’s powered by 12 volts DC (plug it into your car lighter, or a wall wart), and it’s reversible: switch the polarity on the plug, and instead of pumping heat out of the cooler, it pumps heat * into* the cooler, turning it into a food warmer. :cool:

8

Alright… in the simplest terms tell me how my geothermal system running through 55 degree F earth heats my house to 70?

I swear, I will eventually get this!

9

It takes energy from that 55-degree earth (leaving it somewhat colder) and pumps it to your house (making it warmer).

Everyone knows that left to itself, heat wants to flow from hotter to cooler - “downhill” as it were. You need a pump to make it go the other way. If the “hill” (read: temperature difference) isn’t too steep, the pump can move a fair amount of heat without consuming a great deal of energy doing the pumping.

10

In coils that are in contact with the ground, refrigerant is allowed to expand through an orifice. This turns the liquid refrigerant into gas, which requires energy. This energy is removed from the surrounding ground. The refrigerant now contains more heat energy than it did before it was allowed to expand. This gas is then compressed (which adds additional heat to it). This heat is rejected to your house - the hot, dense gas is condensed to liquid in coils which are cooled by your house’s air. The cycle then repeats.

The key to all of this is the evaporation of the refrigerant. If the ground was so cold that the refrigerant couldn’t evaporate, this system wouldn’t work. As the outside temperature get’s colder, there is less energy available to be pumped into your house.

11

Well, let’s run through that R134 cycle I described earlier:

  1. The compressor increases the pressure on gaseous R134. Now you have hot, high-pressure gaseous R134, probably around 150F.

  2. That hot, high-pressure gas circulates through a condenser somewhere in your house’s ductwork; a blower pushes house air past the condenser, cooling the R134 and causing it to condense into a liquid - and heating hour house air.

  3. The cooled (but not yet cold), high-pressure liquid R134 leaves the condenser and travels through an orifice into the evaporator, which is in contact with your geothermal heat source (what is it, groundwater?). The orifice causes a pressure drop. At this new low pressure in the evaporator (~25 psi), the R134 boils at a suitably low temperature, about 7F. As the gaseous cold R134 moves through the evaporator, it sucks in heat from your 55-degree geothermal source.

  4. The now-warm (but not yet hot), low-pressure gas heads back to the compressor; see step one.

There are thermodynamic tables (now computer programs) that tell you how much energy a substance contains as a function of its temperature, pressure, and phase (liquid/gas/solid); when you measure the temperatures and pressures of the R134 at various points in the cycle - after the compressor, after the condenser, after the orifice, after the evaporator - you can precisely account for both the mechanical work supplied to the compressor, and the thermal energy extracted from your geothermal source (and dumped into your house).

12

Wow - I think I’m starting to understand this.

Could I ask for clarification on one point please, though I know it may sound silly?

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?

I was told that the system being put in at my friends would always pump air out at 5C. Hence he has mounted the pipes to point towards a flat roof of his so that it never ices up or snow gathers on it causing a weight/stress issue.

This would appear, to me at least, contradictory to the claims that it will work down to -20C.

Thank you for all the effort all of the respondents in this thread have put in. It takes a lot of skill to explain a complicated subject matter so ‘easily’.

Tilty

13

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.

14

Just a note: I have a geothermal system. Entire unit, aside from piping running through earth, is inside the house.

15

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)

16

Ah, the equivalent of a frost-free refrigerator: valves change the routing to (temporarily) run hot refrigerant through the evaporator, melting any frost that has accumulated on the coils.

17

134a is not used in residential A/C systems.

The most common is R22, or R410a.

18

Technically your refrigerator is a heat pump, and so is an A/C system.

Technically

As a practical matter, they are not, and the typical use of the term “heat pump” is an appliance that has the ability to both heat and cool.

19

You know I’ve seen much redneck engineering in my life, some of it pretty clever. But I’ve never seen anyone put an A/C window unit in backwards when its cold. I’ve often figured it would work better than nothing (unless it got too cold) but I’ve never seen it. I guess most rednecks like me don’t have quite so much edumacation.

20

It might help to distinguish between temperature and heat content. Heres an example, k?

And we’ll need to start with a definition of sorts: A Btu is the amount of energy required to heat 1 pound of water 1 degree. (a pound of water is about a pint.)In more practical terms, a btu is about the energy (heat) in a single match. So…if I raise a pint of water from 69° to 70° I’ve put 1 btu in the water.

So…let’s take 2 examples and see the difference between temperature and heat content.

First I take 50,000 btus (matches) and I apply them to a shot glass of water. For discussion sake, we’ll assume the water won’t convert to steam or boil off. The shot glass is 2000°. (making the numbers up)

Then I take 50,000 btus (matches) and I apply them to an **Olympic size swimming pool. ** Note that I’ve put the same amount of btus into both bodies of water. Yet the swimming pool is 65°. Suppose I double the btus I put into the pool to 100,000 btus (matches). The temperature rises to 75°.

It’s obvious why the temperature in the pool is lower then the shot glass: The btus are “spread out”; they’re disbursed into a wider “area” of sorts. We might also say the btus in the shot glass are “concentrated” into a smaller area.

But what a heat pump does is “harvest” btus, not temperature. In fact, if you were harvesting btus from the shot glass it would be easier, perhaps, but you would run out of btus much faster than the swimming pool that has 100,000 btus in it. Simply put, the swimming pool has a higher heat content.

In a geothermal heat pump there are millions and millions of btus in the earth right around your home. They are sufficiently “spread out” that the earth stays about 55° below the frost line. That makes it harder, but surely not impossible, to “harvest” those spread out btus.

So…If I put hundreds of feet of pipe in the earth and harvest those btus, and bring them into your home and run them through a heat exchanger where I essentially concentrate them I can heat your home.

If you look at the examples above, I’m less concerned with temperature than I am total btus. If I manipulate how concentrated they are----in other words, if I take them out of a swimming pool and put them in a shot glass -----I will raise the temperature even though the heat content----the amount of btus—stayed the same.