Why are air-to-water heat pumps more efficient at lower water temperatures?

[Garald]

Hi,

 

Everybody knows that air-to-water heat-pump systems are more efficient if they heat water up to 45 C (say) rather than 60 C. But *why* is this the case? There has to be a simple explanation.

 

(Note: 

- compare like with like (a radiator system with a radiator system, not with underfloor heating)

- no need to discuss the *other* ways in which a lower-temperature system is better (less heat loss through pipes, less corrosion).)

 

Assume basic university physics if needed. 

[DanDee]

Regardless of the type of heat pump, air/air, air/water, water/water, in order to make the refrigerant gas give out heat, it needs to be compressed to increase it's pressure.

Lower compression/pressure=low temperature heat

Higher compression/pressure=higher temperature heat

Lower compression uses less energy than higher compression

[SteamyTea]

Because it is based on the Carnot Cycle.

 

When a refrigerant gas is compressed it changes phase, from vapour to liquid (condensation) and when it is decompressed it changes back into a gas (vaporization). 

The heat capacity of the two phases is different.  Taking water as an example (because it is also a refrigerant gas and you can test things at home), when in the liquid state it take 4.18 kJ.kg^-1.K^-1 to change temperature, when a gas it takes 1.86 kJ.kg^-1.K^-1 to change temperature.

This may seem odd and you have to wonder where the extra energy goes when it is compressed, well it comes out as thermal energy.  It is that thermal energy that is used to heat the building.

There is a complication at the phase change temperature.  At the exact temperature that water changes from vapour to liquid, or liquid to solid, energy is released, and oddly, a lot of energy.  2500 kJ.kg^-1 in the case of water.  It is a huge amount of energy and drives the global weather system.

When a material is heated or cooled, it is not a linear process. but follow an exponential curve for both heating and cooling, with 'steps' at the phase change temperatures.  Isaac Newton formulised this in the 1701. 

The formula for this is T = (T₀-T₁)e(-kt)

There T is the final temperature after time, t.  T₀ and T₁ are the initial starting temperature and the final temperature. k is the system propertied and is closely related to Ideal Gas Law and the physical restraints of the heat pump i.e. shape and surface area of the pipework.

Knocking up a quick spreadsheet you get this.

 

 

It is easy to see that the temperature drops very rapidly at first then tails off, taking longer and longer to cool.

 

The reverse is the same when heating, except in reality energy is put into the compressor so I have adjusted the value of k to reflect this.

 

 

Now to answer your question "Why are air-to-water heat pumps more efficient at lower water temperatures"

It is down to where you are working on the chart.  Knowing that the heating, or cooling, is very rapid when there is a larger temperature difference, and the compressor and air fans have to work harder to pass the fluids though the relevant heat exchangers, which are physical restraints on the system, and knowing that energy is just power multiplied by time, you can see that the same time is taken to raise, or lower, the final temperature by ever decreasing amounts.

That is fairly simple, but in the real world, a heat pump is more complicated.

There would be 4 temperature/time charts.

Starting from outside and working inwards.

1: Outside air temperature and refrigerant temperature difference

2: Phase change within the refrigerant 

3: Refrigerant to transfer fluid temperature difference

4: Transfer fluid to internal air temperature difference

 

Throughout all those steps, there are different efficiencies, calculating them is our favourite pastime Solutions to Partial Differential Equations.

Personally I prefer to call them Partial Solution to Differential Equations.

It is for this very reason that collected data is used to create the CoP charts at varying temperatures and is often misleading when the underlying physics is not appreciated.

There is also the problem what the Celsius temperature scale is used which highlights the phase transition in the outside air temperature and humidity changes, giving an unexplained drop in CoP between 4°C and 0°C (the frosting window)  This usually shows up as a minima and is often misunderstood and thought of as a CoP of zero.

 

Note:

I used water as a refrigerant gas as you can play with this at home.  You can put a litre of water in a pan, turn on the ring and monitor the time it takes to get to boiling.  You can stop the experiment then and plot how long each K of temperature rise takes, giving you a heating curve.

But for extra marks, you can leave it on the ring and see how long it takes for all the water to evaporate.  The water will be at 373 K, but the time taken will be longer than you think.

You can also do the reverse, but with kilogram of ice melting in a bucket of water, initially it will loose mass quickly, but slow as time goes on.  It is because of this that the 'Ice Kings' could transfer Canadian ice to Barbados and store it in straw insulated building.  It was the tail end of the chart that allowed it to stay frozen for months.

 

In a heat pump, the cold side is between 250 and 260 K and as the ambient air gets closer to that temperature, the ability to heat the expanded gas diminishes, so the efficiency, with respect to time, is reduced.  Again this is often simplified in manufacture's data to a difference in CoP between outside air temperature and delivered temperature, with a smaller difference increasing the CoP.

Different refrigerant materials have different phase change temperatures, pressures and heat capacities, and what is used in Sweden for heating may not be ideal for cooling in Florida.

[Garald]

7 hours ago, DanDee said:

Regardless of the type of heat pump, air/air, air/water, water/water, in order to make the refrigerant gas give out heat, it needs to be compressed to increase it's pressure.

Lower compression/pressure=low temperature heat

Higher compression/pressure=higher temperature heat

Lower compression uses less energy than higher compression

 

Well, obviously heating a *given volume* of water to a higher temperature takes more energy than heating it to a lower temperature. But why is it more efficient, overall? (Operating at a lower yemperature, you presumably have to heat a larger volume of water.)

[SteamyTea]

17 minutes ago, Garald said:

Well, obviously heating a *given volume* of water to a higher temperature takes more energy than heating it to a lower temperature. But why is it more efficient, overall? (Operating at a lower yemperature, you presumably have to heat a larger volume of water.)

It takes a disproportionally higher amount of energy to heat it to a higher temperature.

This is down to the thermal losses being higher.

If you think of a sphere of water, it has the minimum surface area possible, that is how the molecules, within a water cylinder, can be modelled, but the ones that are touching the sides of the cylinder 'change shape' to a sphere cut in half, increasing the surface area to volume ratio, and therefore the thermal losses.

Keeping the temperature lower is, in effect, reducing the number of water molecules that are cut in half, this reduces the losses.

Edited June 18, 2023 by SteamyTea

[JohnMo]

Really it is similar to all mechanical things. A car for example can do 100mph, if you drove it a 100mph your fuel consumption would be worse than driving at 50mph. An easy experiment most have done, drive hard your fuel consumption suffers. Accelerate gently, moderate your speed fuel consumption decreases.

 

Why - it's having to do more work, it has to fight aerodynamic drag, move the mass of the vehicle during acceleration, the engine is producing more kW to maintain the speed.

 

So a heat pump is doing the same, the cooling fan is having to work harder, the same physical laws apply to that as the car moving through air. The compressor works harder, same as the car engine.

 

Just listen to the heat pump towards the end of a DHW heating cycle, it's noisy, on heating mode almost silent, if the flow temps are low. It's doing less work, requires less kW to maintain that work.

[SteamyTea]

2 minutes ago, JohnMo said:

Really it is similar to all mechanical things

That is a better explanation than the real physics that goes on, and probably did not take you two hours to type.

[Nic]

1 hour ago, JohnMo said:

Really it is similar to all mechanical things. A car for example can do 100mph, if you drove it a 100mph your fuel consumption would be worse than driving at 50mph. An easy experiment most have done, drive hard your fuel consumption suffers. Accelerate gently, moderate your speed fuel consumption decreases.

 

Why - it's having to do more work, it has to fight aerodynamic drag, move the mass of the vehicle during acceleration, the engine is producing more kW to maintain the speed.

 

So a heat pump is doing the same, the cooling fan is having to work harder, the same physical laws apply to that as the car moving through air. The compressor works harder, same as the car engine.

 

Just listen to the heat pump towards the end of a DHW heating cycle, it's noisy, on heating mode almost silent, if the flow temps are low. It's doing less work, requires less kW to maintain that work.

Brilliant explanation 👍🏼

[Garald]

14 minutes ago, SteamyTea said:

Because it is based on the Carnot Cycle.

 

Thanks - this is the key bit.

 

14 minutes ago, SteamyTea said:

 

When a material is heated or cooled, it is not a linear process. but follow an exponential curve for both heating and cooling, with 'steps' at the phase change temperatures.  Isaac Newton formulised this in the 1701. 

The formula for this is T = (T₀-T₁)e(-kt)

 

Well, sure, this is in an immediate corollary of Newton's law of cooling (which also explains why you need bigger radiators if you are operating at lower temperature).

 

Not sure that's enough to give a satisfactory explanation - it would seem you need thermodynamics for that, and that's later. At the very least you need to know that the change in entropy equals the exchanged heat divided by the temperature, no?

 

I'll read up quickly - I got off the bus at the end of vol 3 of the Berkeley Physics Course...

[Garald]

1 hour ago, JohnMo said:

Really it is similar to all mechanical things. A car for example can do 100mph, if you drove it a 100mph your fuel consumption would be worse than driving at 50mph. An easy experiment most have done, drive hard your fuel consumption suffers. Accelerate gently, moderate your speed fuel consumption decreases.

 

Why - it's having to do more work, it has to fight aerodynamic drag, move the mass of the vehicle during acceleration, the engine is producing more kW to maintain the speed.

 

So a heat pump is doing the same, the cooling fan is having to work harder, the same physical laws apply to that as the car moving through air. The compressor works harder, same as the car engine.

 

Just listen to the heat pump towards the end of a DHW heating cycle, it's noisy, on heating mode almost silent, if the flow temps are low. It's doing less work, requires less kW to maintain that work.

 

I think this is a good explanation of how the difference between the real world and a simple idealized model works further in favor of heating to lower temperatures. In the case of a heat pump, though, lower temperatures seem to be better even in the simple idealized model. 

[SteamyTea]

40 minutes ago, Garald said:

2 hours ago, SteamyTea said:

Because it is based on the Carnot Cycle.

 

Thanks - this is the key bit.

You cannot get better than the Carnot Cycle.  Everything else in Thermodynamics is a disappointment.

 

42 minutes ago, Garald said:

Well, sure, this is in an immediate corollary of Newton's law of cooling (which also explains why you need bigger radiators if you are operating at lower temperature).

Yes, it affects the value of k.

43 minutes ago, Garald said:

Not sure that's enough to give a satisfactory explanation - it would seem you need thermodynamics for that, and that's later. At the very least you need to know that the change in entropy equals the exchanged heat divided by the temperature, no?

Not sure.

J.K^-1.K^-1 just leaves J, or energy.

That seems to make sense, but not sure it is actually equal because if you break down the units for entropy, you get:

kg.m².s^-2.K^-1

So the amount of time you have a fixed temperature comes into play.

Time always (expletive deleted)s up physics.  Acceleration double so.

[Garald]

48 minutes ago, Garald said:

At the very least you need to know that the change in entropy equals the exchanged heat divided by the temperature, no?

 

Once you have that, you can deduce that the COP in the idealized model (Carnot heat pump) is 1/(1-(T_L/T_H)), where T_L is the outside temperature and T_H is the temperature at which you are heating the water of your system, both in Kelvins, obviously. So, clearly, you want T_H to be as low as possibly, and hope for T_L to be as high as possible, so that the COP is as large as possible.

 

The math was basically worked out by Sadi Carnot in 1824, at least in the simplest, most idealized case. Of course he was working with the reverse situation (how to extract as much work as possible from as little heat as possible) so the result is standing on its head. (If you are running a steam engine, it is better if the ratio of temperatures is high!)

 

 

PS. How do you typeset math here?

 

 

[Garald]

3 minutes ago, SteamyTea said:

Yes, it affects the value of k.

 

 

Aha, so you are saying you can deduce Carnot's cycle if you know k as a function of the pressure of your gas?

[SteamyTea]

2 minutes ago, Garald said:

PS. How do you typeset math here?

No idea.

There are a few Alt+xxx that help, but very limited.

The alternative is is type up in LATEX (or your favorite editor and screen capture it.

 

[SteamyTea]

2 minutes ago, Garald said:

 

Aha, so you are saying you can deduce Carnot's cycle if you know k as a function of the pressure of your gas?

Again I am not sure, I suspect that a correlation is possible, and that may point to the underlying physics.

When changing the value of k, you could change any of the physical properties of the system, it may be worth, for a giggle, knocking up a spreadsheet and basing k on the ideal gas laws PV/T = C, where P is pressure, V is volume and T is temperature.

An adjust factor may have to be introduced that is, in effect, describing the rest of the physical system.

 

I shall have to think about it more, but as I am out at the moment, it is a task to do when I get home, rather than a busy cafe on a cloudy Sunday.

[Garald]

9 minutes ago, SteamyTea said:

Again I am not sure, I suspect that a correlation is possible, and that may point to the underlying physics.

When changing the value of k, you could change any of the physical properties of the system, it may be worth, for a giggle, knocking up a spreadsheet and basing k on the ideal gas laws PV/T = C, where P is pressure, V is volume and T is temperature.

An adjust factor may have to be introduced that is, in effect, describing the rest of the physical system.

 

That's interesting (as are all alternative derivations). At any rate, 

 

increase in entropy = heat exchanged divided by temperature 

 

does the trick.

 

It turns out one can find everything (explained in too many words) in

https://en.wikipedia.org/wiki/Carnot_cycle

https://en.wikipedia.org/wiki/Heat_pump_and_refrigeration_cycle

(put together).

 

Time to look into vol. 5 of the Berkeley Physics Course...

[Garald]

Ah, the Carnot cycle is also in lecture 44 of vol I of Feynman's Lectures. Of course again this needs to stand on its head.

[SteamyTea]

1 hour ago, Garald said:

increase in entropy = heat exchanged divided by temperature 

 

does the trick

Yes.

Thinking about it, I seem to remember that it was an easy calculation.

Just the last two decades since I studied it have not been so easy.

 

You should find a friendly Thermodynamicist to work with. It is the ability to condense equations to the Occum's Razor point that we often find difficult.

[SteamyTea]

Time to stop thinking about entropy and move onto wave theory.

 

[Garald]

12 minutes ago, SteamyTea said:

 

 

You should find a friendly Thermodynamicist to work with. It is the ability to condense equations to the Occum's Razor point that we often find difficult.

 

Right. If I ever do a second stage of the renovation (raising the roof, insulating the north wall from the outside (the insulation we've put on the inside barely approaches R=4, counting the air gap, and one of the two layers is reflective material - who knows how that will behave long-term), etc.), then I might consider hiring a good thermal engineer (unless I've learned enough stuff on my own by then, perhaps).

And of course I should actually make either the remaining volumes of the Berkeley Physics course or Feynman's lecture into my bedside reading - or a more condensed version for more mathematically mature people, if it exists. 

[Garald]

1 minute ago, SteamyTea said:

Time to stop thinking about entropy and move onto wave theory.

 

 

Nice! I did take a course on that.

[Garald]

You might not believe it, but entropy does come into play in probabilistic number theory. It shows up uninvited in the exponents of probabilities of various things, once you go way off the center of the distribution and into the tails.

[SteamyTea]

2 minutes ago, Garald said:

You might not believe it, but entropy does come into play in probabilistic number theory. It shows up uninvited in the exponents of probabilities of various things, once you go way off the center of the distribution and into the tails.

What Claude Shannon found out I think.

In a way he is responsible for us chatting via miles of cables, servers, wireless transmitters.

Clever man, as was Feynman. I just reread 'And another thing'.

[Garald]

16 minutes ago, SteamyTea said:

What Claude Shannon found out I think.

 

 

Right, mathematically the two things are the same (- \sum_i p_i \log p_i), though establishing a clear link between entropy in thermodynamics and in information theory (without going through the math, in the sense of saying "well, we can deduce either from a few axioms, and the axioms are the same once you forget what the words mean", or something to that effect) is tricky and not necessarily that useful, from what I understand (though knowing that the math is the same *is* useful).

Edited June 18, 2023 by Garald

[Marvin]

Hi @Garald

 

Why are air-to-water heat pumps more efficient at lower water temperatures?

 

My simple explenation to your question is that air passing through the ASHP changes the temperature of the water(or air) output and the bigger the difference between the air temperature and the required water temp (hotter or colder) the harder the fan and compressor have to work.

 

After that, any further detail is unnecessary unless your an academic and couldn't be bothered to google it.

 

M