Heat Loss from Pool: Info Reliability & Free Natural Convection

In summary, the website says that a pool is closer to the temperature of the air surrounding it, unless heated. If you are looking for calculations for heat loss on an indoor heated pool, they are dependant on the natatorium conditions as well as the pool conditions. The heat loss is 10.5 * Area * (PoolTemp - Inside DB)
  • #1
eaboujaoudeh
152
0
Hi all

I found values for heat transfer from a pool on the following website:
http://www.engineeringtoolbox.com/heat-loss-open-water-tanks-d_286.html
can someone please tell me how reliable is this information?
especially that they didn't mention free natural convection, or is it negligeable in this case w.r.t radiation and evaporation.
 
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  • #2
A pool is closer to the temperature of the air surrounding it, unless heated. If you are looking for calculations for heat loss on an indoor heated pool, they are dependant on the natatorium conditions as well as the pool conditions. The vapor pressure of the air, as a function of its temperature and RH.

Evap = .1 * Area * (PresWater - PresAir)

Evap = Evaporated Water in Lbs per hour
Area = Area of pool surface in square feet
PresWater = Pressure of Water Surface in inches HG (for 80 deg F water) = 1.0321
PresAir = Sat Pressure Room Air in inches HG (for 76 Deg F air dewpoint) = .88
 
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  • #3
Nevermind my previous post. That is for the quantity of water loss due to evaporation. The heat loss is:

Pool surface loss = 10.5 * Area * (PoolTemp - Inside DB)

PoolTemp = The heated pool temperature
Inside DB = The pool enclosure dry bulb temperature of the air
Area = Surface area of the pool
 
  • #4
thats a big difference between this value and the value given on the website ! and yes the pool is inside and heated (at least the first one)..my research is to design a heat exchanger for it.
i have another question in that matter, if i want to calculate the temperature of the pool water entering the heat exchanger(and leaving the pool to get reheated), what in ur opinion would be a good temperature to take?
 
  • #5
eaboujaoudeh said:
thats a big difference between this value and the value given on the website ! and yes the pool is inside and heated (at least the first one)..my research is to design a heat exchanger for it.
i have another question in that matter, if i want to calculate the temperature of the pool water entering the heat exchanger(and leaving the pool to get reheated), what in ur opinion would be a good temperature to take?

I know there is quite a discrepency in the calculation and the website data. I don't know why. I had placed the calculation in a Qbasic program I wrote ages ago, and I can't find my reference that gave me that 10.5 figure. I found a good source for what you are searching for

http://www.engineeringtoolbox.com/swimming-pool-heating-d_878.html" [Broken]

They only use 5 for their constant although they present it as a range usually of 5-7 where I got 10.5 I'll never know.:yuck: The rest seems the same as what I gave you.

Determining the temperature rise through the heater depends on how fast you want to be able to heat the water. This will depend more on how much you want to invest up front in the heater, more than how much it will cost to run it. Check out the calculation in the link. I think you will find what you are searching for.

Edit: I think I made that number so high because of activity of the water. It was a combination pool and whirlpool, spray features, which increase heat loss and evaporation.
 
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  • #6
yeah i found this website too:) thnx anyway.
what i meant in temp leaving the pool is that after steady state is reached and my pool is heated, i need to maintain that heat. so when i have to reheat the water I'm taking it that this water leaves the pool to be reheated at 23 Celsius, do u think its a fair approximation?
Thnx for everything, i appreciate it...and btw i was also wondering about that 10.5 value of yours. i found an old heat transfer book thati have, mayb i should use it, even though its more complicated then what i got from you or the website, but safetycomes first i guess:)
 
  • #7
maybe they are basing calculations involving Newton's law of cooling
 
  • #8
pakmingki said:
maybe they are basing calculations involving Newton's law of cooling

Well I'm not sure about that, i think they are kind of experimental cause they are measured relative to one external air temperature. Besides Newton's law of cooling doesn't allow us to measure the convection constants of air neither the radiation constants.
 
  • #9
eaboujaoudeh, I'm not sure I understand your concern. Once you size your heater unit to heat the pool, maintaining that temperature will not be a problem for the heater. It should cycle based on the deadband setting of the aquastat/thermostat on the heater.

this water leaves the pool to be reheated at 23 Celsius
The surface would be about room temperature, the bottom where the water would leave the pool would be a little colder. So, yes, this seems like a reasonable temperature.
 
  • #10
yeah, but u always have to see that ur input heat compensates for the loss of heat from the surface of the pool. like if evaporation causes 10kW loss, and ur heater only compensates for 8kW..so ur water is going to become cooler gradually
 
  • #11
eaboujaoudeh said:
yeah, but u always have to see that ur input heat compensates for the loss of heat from the surface of the pool. like if evaporation causes 10kW loss, and ur heater only compensates for 8kW..so ur water is going to become cooler gradually
That is correct. If your heater can't overcome the surface loss, you will never be able to bring it up to temperature from any starting point below the desired temperature.
 

What causes heat loss from a pool?

Heat loss from a pool is primarily caused by evaporation, conduction, and radiation. Evaporation occurs when the water in the pool turns into water vapor and escapes into the air. Conduction is the transfer of heat between the pool water and the surrounding surfaces, such as the pool walls and ground. Radiation is the transfer of heat in the form of electromagnetic waves from the pool's surface to the surrounding environment.

How can I prevent heat loss from my pool?

There are several ways to prevent heat loss from a pool. One way is to use a pool cover when the pool is not in use. This will reduce evaporation and keep the pool water warmer. Another way is to insulate the pool walls and bottom, which will reduce heat loss through conduction. Additionally, using a solar pool heater can help maintain the pool's temperature by using the sun's energy to heat the water.

Is information about heat loss from a pool reliable?

The reliability of information about heat loss from a pool can vary. It is important to make sure the information comes from a reputable source, such as a scientific study or a trusted pool company. Additionally, information from multiple sources can help confirm its reliability. It is also important to consider the specific conditions and factors of your pool, as they can affect the accuracy of the information.

What is free natural convection and how does it affect heat loss from a pool?

Free natural convection is the transfer of heat through the movement of fluids, such as air or water, due to differences in temperature. This can affect heat loss from a pool by causing warm air near the pool's surface to rise and be replaced by colder air, resulting in heat loss through evaporation. It can also cause warm water near the surface to sink and be replaced by colder water, resulting in heat loss through conduction.

How can I use natural convection to my advantage in reducing heat loss from my pool?

You can use natural convection to your advantage by ensuring good air circulation around the pool. This will help prevent the buildup of warm air near the pool's surface and reduce evaporation. Additionally, using windbreaks or installing a pool in a sheltered area can help reduce the effect of convection on heat loss. You can also consider using a pool cover or insulating the pool to minimize the impact of natural convection on heat loss.

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