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Sir Col
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Heated Pond
« on: Apr 1st, 2004, 11:39am » |
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My brother-in-law has a query, which I am unable to answer, and I would be most grateful for any help on this matter. It relates to the approximate costs of heating an outdoor pond in which he keeps Koi carp.
Throughout the winter period the water is kept at a temperature of 10[pm]0.6oC by a 2kW heater.
The pond is a relatively small cuboid in desgin, measuring 7'x6'x3.5', and containing about 915 gallons (UK) of water.
He has a plastic cover which, the manufacturers claim, will save around 50% in heating bills.
I appreciate that information is very limited (for example, I don't know how much heat would be lost through the walls of the pond), but is it possible to build a model to make estimates of costs (based on running time of heater) with and without the cover? Maybe a function in terms of the external temperature?
Thanks in anticipation of any help. 
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SWF
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Re: Heated Pond
« Reply #1 on: Apr 5th, 2004, 9:33pm » |
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I will make a quick estimate, even though the heat loss depends on many factors. Since the plastic cover supposedly cuts the heat loss by 50%, and the cover presumably only covers the surface of the water, the amount conducted through the ground must be less than half of that lost from the surface of the water.
First estimate rate of heat loss from the uncovered water surface. You need to find a heat transfer coefficient, h, and that varies with geometry and other factors, but correlations can be found in handbooks. At the temperature given with static air, I estimate h=4.2 W/(m2-deg C). With a slight breeze this number may triple or more.
The rate heat is lost from the surface is found by multiplying h by the area of the surface (about 2 square meters) and the temperature difference between the water and the air (maybe 10C). For still air (h=4.2) that gives 84 Watts of heat lost from the pond's surface, and that is the power need for a heater to maintain a steady temperature. Next, the effect of the plastic cover and heat conducted through the ground can be estimated- I'd guess 42 Watts conducted through the ground, and the cover saves 42 watts, still leaving a total of 84 Watts.
From that you can make your own estimates based on expected average outdoor temperature, and whether there is a breeze.
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Sir Col
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Re: Heated Pond
« Reply #2 on: Apr 6th, 2004, 2:44am » |
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Thanks, SWF.
If you don't mind, I have a question...
Sticking with your example, and supposing the average enviromental temperature is 20oC (10 degrees above the minimum), your method would estimate a total of 84 watts of heat loss (42W from the surface and 42W through pond walls). Is this a measure of the power required to maintain the minimum temperature?
If it represents the average power to maintain temperature, and the heater is 2kW, is it reasonable to say that, as 2000/84~=24, running the heater for one hour would be equilvalent to 24 hours of providing the required 84 watts? In other words, we would expect the heater to "kick-in" for a period of about one hour throughout the day?
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SWF
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Re: Heated Pond
« Reply #3 on: Apr 6th, 2004, 7:06pm » |
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If the air is 20C, no power is needed to maintain the water above 10C. I was assuming the air was 0C and water was 10C. The value of 42W through the pond walls was a conservative guess, and I expect it to be less.
The power I gave is what would be needed to maintain the water at a steady temperature of 10C when air is 0C and still. 84 Watts of heat leave the water, and 84 Watts of electricity dissipated as heat replaces it. Your method of calculation looks right. Heater would need to on at full power for a total of about 1 hour per day. But maybe the heater can operate constantly at a reduced power.
For your climate, you can probably look up how many degree-days a typical winter is for you. Then you can figure out how many W-hr of energy would be used per season, and how much that would cost.
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: Heated Pond
« Reply #4 on: Apr 7th, 2004, 2:31am » |
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"So, Sir Col, a heater is required to keep the pond at 10 degrees BELOW the environmental temperature of 20 degrees, eh? Duh!" *slaps head* 
Thank you for all your help, SWF, it is much appreciated.
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