Settling the Debate: Air Compressors & Koi Ponds

[AlanWF]

I wonder if someone can help settle a debate I am having over using air compressors to aerate garden koi ponds? A typical pond will have a small electric air pump (44W rated, delivers say 50 litres a minute at 1.5 metres deep in a 10000L pond) blowing air through a diffuser to add oxygen and create movement in the water. The air leaving the pump is of course hot once it is compressed, my friend is convinced this heat is transferred to the pond, I say no it will all be lost when the air returns to the atmosphere at normal pressure. Also where does the 44W input end up?
Since the debate started the subject of evaporation has occurred and what effect this may have on the water temperature particularly in winter when air temperatures will be lower than the water in the pond so any thoughts on the significance of this would be great.

 

[Bystander]

air returns to the atmosphere at normal pressure

And at pond temperature. The temperature rise in the air stream results in loss of heat to the pond.

where does the 44W input end up

Assuming a submerged air/circulating pump, it winds up in the pond water.

 

[AlanWF]

Thanks for your prompt response. To be clear, am I correct in saying the heat of compression is lost when the air expands into the water? If the air is say, 10C and the water is at 10C also will the water just end up gaining the 44W (less losses as they are externally fitted and not usually insulated)? Ignoring evaporation or other factors for the sake of this example.

 

[Bystander]

externally fitted

You're saying the pump is not submerged in the pond?

 

[AlanWF]

Correct. As far as I know all air pumps are fitted outside of the ponds. Many will be in a sort of pump shed but lots are outdoors with basic shelter roof over.
The water circulation pumps are about 50% submersible and 50% external. This amuses me as the trend is for low energy pumps but fitted externally- where waste heat is lost to the air! There may be some downside to lifespan and maintenance of submersible though, I am a newcomer to the hobby so can't speak from long experience.

 

[Bystander]

50 litres a minute at 1.5 metres deep

Fifty liters/min at 15 kPa is about 10 W mechanical work on the air, and that goes into the pond. The rest of it is waste heat from the pump itself and goes off into the air moving over the pump. If the entire 10 W went into evaporation losses, you'd be losing 16 cm³/hr. You know how much water you have to add and how often --- compare that loss from being open to the air to a pint or two a day from heating with very slightly compressed air.

 

[sophiecentaur]

If the pond is in full sun, it will be receiving at least a kW of solar power for quite a long period. On a cloudy day it may be getting only a few 100W. It copes well enough with that sort of variation in heat supply. Even if your 44W pump were submerged (and there are plenty of such pumps), the difference in 'heat supplied' to the pond would be negligible. Stop worrying and enjoy the fish.

 

[AlanWF]

If the pond is in full sun, it will be receiving at least a kW of solar power for quite a long period. On a cloudy day it may be getting only a few 100W. It copes well enough with that sort of variation in heat supply. Even if your 44W pump were submerged (and there are plenty of such pumps), the difference in 'heat supplied' to the pond would be negligible. Stop worrying and enjoy the fish.

Thank you for replying.
The question is somewhat a pedantic one I know but I am trying to win a debate over does the fact that the air leaving the pump is "hot" this means it heats the pond water. I know there are lots of other variables, air/water temp, humidity etc. which will have a much larger influence but my question is -

To be clear, am I correct in saying the heat of compression is lost when the air expands into the water? If the air is say, 10C and the water is at 10C also will the water just end up gaining the 44W (less losses as they are externally fitted and not usually insulated)? Ignoring evaporation or other factors for the sake of this example.

My apology if my question is not couched properly, and if I am asking in the wrong place I am sorry and could you recommend a better forum for this type of inquiry?

 

[sophiecentaur]

Sorry. My reply was more suited to a Fish Keeper's Forum.
It's correct that compressing a gas can increase its temperature and doing work on expansion can lower its temperature. So yes, it will happen, just as it does with diving compressors, which need heat exchangers to deal with all that heat. However,. . . . .
The pressure changes involved would be very small (0.1Bar ish) and I wouldn't expect the air in the pump would have time to lose much of its energy to the pump body, so it would involve an adiabatic compression and expansion, with almost no net internal energy change over the cycle.
As an Engineer, basically, I cannot ignore the significance of all the other factors, though. I would imagine that the heat generated in the motor itself (resistance etc) could also affect the air temperature. Also, the bubbles of dry air are likely to cause a measurable amount of evaporative cooling and it would be hard to isolate those and other factors from each other. As I stated in my rather fatuous reply, the actual values of the variables involved are what counts. and the Sun beats all the others. You would really need to do some experimentation under controlled conditions - for instance, you could make sure that the injected air was already saturated, to eliminate the evaporative effect. You could run the pump into a small volume of water and measure temperature change.
I think my point about Adiabatic changes is the most relevant for your argument with your friend.

 

[AlanWF]

Sorry. My reply was more suited to a Fish Keeper's Forum.
It's correct that compressing a gas can increase its temperature and doing work on expansion can lower its temperature. So yes, it will happen, just as it does with diving compressors, which need heat exchangers to deal with all that heat. However,. . . . .
The pressure changes involved would be very small (0.1Bar ish) and I wouldn't expect the air in the pump would have time to lose much of its energy to the pump body, so it would involve an adiabatic compression and expansion, with almost no net internal energy change over the cycle.
As an Engineer, basically, I cannot ignore the significance of all the other factors, though. I would imagine that the heat generated in the motor itself (resistance etc) could also affect the air temperature. Also, the bubbles of dry air are likely to cause a measurable amount of evaporative cooling and it would be hard to isolate those and other factors from each other. As I stated in my rather fatuous reply, the actual values of the variables involved are what counts. and the Sun beats all the others. You would really need to do some experimentation under controlled conditions - for instance, you could make sure that the injected air was already saturated, to eliminate the evaporative effect. You could run the pump into a small volume of water and measure temperature change.
I think my point about Adiabatic changes is the most relevant for your argument with your friend.

Thank you again for your time. It does seem that I will need to try some home experimentation to get further with this one. Your point about the Adiabatic nature of the process is probably the one I was looking for regarding the "hot" pipes.

 

[sophiecentaur]

Yeah. I think the Adiabatic thing is the most relevant to your argument - although not to your fish pond. ;)