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I'm puzzled by the fact that the CoP (Coefficient of Perfomance) of residential heat pumps is usually stated to be between 4 and 5, meaning that for every kWh of electricity used by the pump one gets between 4 to 5 kWh of heat transferred into the house. How's that even possible?

Here's my reasoning: the gaseous fluid in the power pump gets compressed inside the house, releasing heat into the residence. Fine. The compressed liquid fluid is moved to the outside where it is allowed to expand, and in the process will cool down even more. The compressed gas then absorbs energy from the environment outside the house to become a gas again. The expanded gas is then moved inside the house and compressed (thus heated up again), closing the circle.

  1. I can see that (discarding losses) one can put in 1 kWh to compress gas inside the house to release 1 kWh of heat. So far so good, so how can manufacturers state that the CoP (Coefficient of Performance) is 3 to 5?
  2. Also, the expanded gas coming from outside the house will be at outside ambient temperature, so we are actually introducing a cold substance into the house which would mean that the CoP would be even less.

Can somebody care to elaborate? It seems that I'm missing something vital in my reasoning. Thanks!

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  • $\begingroup$ Did you try to perform a calculation to see what COP you could get? $\endgroup$
    – Themis
    Apr 27, 2023 at 13:43
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    $\begingroup$ It’s like it gets free energy from thin air! $\endgroup$ Apr 27, 2023 at 23:10
  • $\begingroup$ If you moved the evaporator into the residence, all energy used by the pump would contribute to heating the home: CoP = 1. Doesn't it make more sense that a heat pump should only ever have a CoP > 1? $\endgroup$
    – Sam
    Apr 28, 2023 at 15:22

6 Answers 6

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A heat pump takes heat from outside and puts it into your house. The heat doesn't (only) come from the power that operates the pump. So it's acting like a refrigerator, but the inside of the refrigerator is ambient air, and the outside of the refrigerator is your house.

The maximum possible efficiency for this process can be inferred from the Carnot efficiency of a refrigerator. If you do work $W$ on a carnot cycle based refrigerator, a heat $Q_C$ comes out of the cold system, and a heat $Q_H=W+Q_C$ goes into the hot system. The carnot efficiency equation is: $$ \frac{Q_C}{W}=\frac{T_C}{T_H-T_C} $$ In this case, heat going into the hot system is actually what we want to achieve, so I can say: $$ \frac{Q_H}{W}=1+\frac{T_C}{T_H-T_C}=\frac{T_H}{T_H-T_C} $$ Notice that if $T_H$ is close to $T_C$ this efficiency can become infinite. That is, if you only want your house to be slightly warmer than outside, a heat pump can be infinitely efficient. And it becomes less efficient as the temperature difference gets bigger. For example if you want your house to be 300K but its 270K (freezing) outside, this maximum possible efficiency is 10. But obviously heat pumps aren't perfect Carnot cycles, so this maximum efficiency isn't achieved.

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  • $\begingroup$ If efficiency is close to infinite for very small temperature differences, it seems like in theory we could have a bunch of daisy chained pumps each raising the temperature a tiny amount to achieve a higher efficiency combined than simply raising the temperature in a single step. I'm guessing what prevents this from being practical is the imperfections of each stage. $\endgroup$
    – Michael
    Apr 30, 2023 at 15:59
  • $\begingroup$ @Michael you haven't yet understood how fundamentally unbeatable the Carnot efficiency is. This scheme does not work, and not just because of imperfections. If you actually apply the equations you see that the way it goes to infinity doesn't support this idea $\endgroup$
    – AXensen
    May 1, 2023 at 22:35
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Other answers have given the basic equations so I won't repeat them here.

The essential point is that a heat pump uses its input energy not simply to produce heat, but to TRANSFER heat out of one body and into another.

Let's compare this to pumping water. For a given amount of input energy to drive the pump, you get a given amount of water transferred from one place to another (e.g. up to the top of a hill or pushed through a constriction).

With a heat pump the overall aim is to take heat out of a body at a lower temperature and deliver that energy into a body at a higher temperature (possibly plus some further energy coming from the electrical supply). When put like this you can see that the total energy arriving at the hotter body could easily be more than the amount consumed from the electrical supply.

In fact, if you were not aware of the rules around entropy, you might think that you could get really huge efficiencies, where you spend a joule of electrical power to move a megajoule of heat. In fact there are fundamental limits on heat movements which is why the ratios in practice are more modest (4 or 5 as you say in the question).

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  • $\begingroup$ This is why they're called heat pumps: they pump heat, much like regular pumps pump water or air. Via controlled compression and expansion, no less. $\endgroup$
    – No Name
    Apr 29, 2023 at 5:38
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I can see that (discarding losses) one can put in 1 kWh to compress gas inside the house to release 1 kWh of heat. is not correct.

As the vapour is compressed to form a liquid with the expenditure of $1\,\rm kWh$ of energy say $5\,\rm kWh$ of latent heat is released in the process.

The liquid is then transported outside/underground and allowed to vaporise which requires $5\,\rm kWh$ of heat to be abstracted from the atmosphere/ground.

The vapour is then transported back to the compressor where the cycle begins again.

In this example you will note that for the expenditure of $1\,\rm kWh$ to run the compressor $5\,\rm kWh$ of heat has been released into the building resulting in a CoP of five.

So you are "transporting" heat from outside a building to inside the building.

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How can a residential heat pump have a CoP bigger than 1?

Without getting into the details of the cycle as has been done in the other answers, the short answer is conservation of energy.

A diagram for heat pump/refrigeration is shown below. Note that conservation of energy requires that the heat delivered to the high temperature environment equals the heat taken from the low temperature environment plus the work input of the compressor.

The Coefficient of Performance (COP) is defined as the desired heat transfer divided by the work input to cause the transfer. For a heat pump, the desired heat transfer is heat out to the high temperature environment. So for the heat pump

$$COP=\frac{Q_H}{W_{IN}}=\frac{Q_{C}+W_{IN}}{W_{IN}}=\frac{Q_{C}}{W_{IN}}+1\gt 1$$

The only reason why more heat is supplied to the high temperature environment than work input is the evaporation and compression processes transfer latent heat. If instead we simply had a standard electrical heater, no heat would be transferred form the outside ($Q_{C}=0$) and we would have COP=1, all the heat transfer being Joule heating.

Hope this helps.

enter image description here

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Kelvin proposed this idea of heating houses in cold countries more efficiently by using the thermodynamic method than burning the fuel directly inside the house. The method promises 4 to 6 fold increase in the heat delivered into the house using the thermodynamic method for every kilo of fuel used.

If you believe the second law of thermodynamics to be true, which many a great scientist, including the likes of Einstein, vouch for, there is no escape than accepting the impossible.

It surprises me that many get agitated when they find COP of ideal heat pump is greater than one, while they don't wouldn't at all worry when the efficiency of ideal heat engine, the Carnot heat, is less than one!

If we accept one we have no option but to accept the other.

We need to throw out both and start construction of thermodynamics anew with a modified concept of heat. The sooner the better.

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You may mix up heat and temperature. By pressurizing the gas, we do not add heat to it, we just increase the temperature. This increed temperature then allows the heat to flow from the fluid to the house; when depressurizing the fluid, it now expands with less heat then before, so it is colder then the outside and can be heated there.

Also, the heat capacity (how mucht heat (Joules) per Temperature difference (Celsius)) is highly depending on pressure and volume; we may change this by pressurizing the fluid and get even more heat out of it, as the temperature increases a lot if we yust decrease the heat capacity of a substance with a given heat.

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