Heat loss from an outdoor fountain 1

[MarauderX]

Does anyone have recommendations on how to calculate the heat loss from a water fountain running in the winter? We are looking at the viability of running it with heaters, but it depends on the load.

Thanks in advance!

http://www.vanderweil.com

 

[reidh]

I am assuming you mean a decorative type water fountain and not a drinking water fountain.

For the convection heat loss:

Q=h*A*delta T

h=convective heat transfer coefficent
A=surface area of water exposed to air
delta T=ambient temperature-water temperature

For the conduction through the walls of the fountain/floor of the fountain:

Q=k*A*(delta T/delta X).

k=thermal conductivity of the wall material
A=surface area of water exposed to walls (or floor) of the fountain
delta T=water temperature-outside wall temperature (the outside wall temperature should be close to ambient temperature assuming a non-conductive wall material such as cement)
delta X=wall thickness


If the wall is made up of more than one material, you need to use this equation for each material since they will have a different K value. If you are just looking for an approximation, you can always guesstimate the total K value of the two materials.

Hope this helps.

 

[MarauderX]

Thanks, and I was referring to a decorative fountain. Any chance you might have approximate values for h?

Plus the water will be moving (churning)... which will make a difference to the heat loss. How can that be estimated?

http://www.vanderweil.com

 

[MintJulep]

Values of h and k are both functions of water and air velocity. Neither are constant with time.

Radiation cannot be ignored.

If the fountain is very active, latent heat loss through evaporation may be significant.

 

[Transient1]

How large is this fountain?

 

[MarauderX]

Not exactly sure how large the fountain is... it's going to be a flat fountain with three jets in the center. I'm sure radiant & evaporation will play a part.

http://www.vanderweil.com

 

[dcasto]

The heat loss from the water "churning" in air can be estimated using be applying an effiency to a cooling tower calculation. You'll need air temperature, air humiity, and all the other information listed riedh. The probblem will be in estimating the effiency to apply. Someone else with practicle experience might help here. For lack of anything better, I'd start with 30%. For a worst case to size your heater, I'd use 70%.

 

[reidh]

Marauderx,


I would argue that we could ignore radiation, I think you will find it is negligible compared to conduction and convection.

For Black Body Radiation, Q=Boltzman Constant*area*delta T^4

The boltzman constant for radiation is 5.67x10^-8 Watts*m^2*K^4

So if the area of the fountain is 50 m^2 and the temperature difference between the water and ambient is 25 C, you will have approximately 1 watt of heat transfer. This assumes an emissivity of 1.0 (a black body), so the actual number would be less for this situation. Your measurement error for conduction and convection will probably be greater than radiation, so I recommend ignoring it.

Good idea dcasto on the cooling tower calculation. Unfortunately I do not have any experience with these, but I will poke around some old books to see if I can find anything.

-Reidh

 

[sailoday28]

Don't forget to include the effect of loss of water due to evaporation.

 

[reidh]

I retract my earlier comment about radiaton. I apologize, it was late Friday night, my brain must have been away for the weekend.

The only radiaton that will likely be significant is that from a clear sky (the sun during the day or deep space on a clear night). Of course you would size your heater assuming a clear night.

Sorry for any confusion.

-Reidh

 

[rjw57]

How about freeze point depressants? Might not have to worry about heating it then. Check with your local water treatment business, like one that deals with spray coolers.
I can't think of anything worse than this type of service for freeze protection.

If you are really stuck on heaters, check the recommended sump heater size for a spray cooler with spray flow rates similar to your application. This, in my opinion, would be worst case.

 

[cmillburn]

A couple of sources of significant heat loss:
1. Convection due to area winds. The formula that is commonly used in pool heater sizing is 10.5 x surface area x delta T.

2. Evaporation of the water due to your decorative spray head. Look for cooling tower makeup water charts for your climate area. In short, x gallons of water evaporated x 8.33 lbs/gallon x 1000 BTU/lb = heat lost from the fountain.

Typically, the heat lost to the surrounding ground is not significant compared to the first two. However, if the fountain is over a parking garage, etc. you could experience the pool freezing from the bottom up!

Another point to consider is blowing drift from any spray heads - makes a great ice rink on adjacent walkways.

 

[eaboujaoudeh]

I may be a month late, but i hope that the info i will give u will help. I was asked to model a new method to heat a 100m3 indoor and 600m3 outdoor pools. I learned a lot about the convection and radiation coefficient of water. Usually for 3.5mph air the water coefficient of convection H=10.5 W/m2 K.As for the radiation it was very difficult to get, so i used approximate values calculated by someone else, usually for a 15 Degree Celsius difference in temperature the heat loss due to radiation would be near 157.6 W/m2.In the end my results matched a the engineering standards of pool heating in my country. But frankly in your case the fountain brings big troubles with it, and i think you should calculate its losses too, cause they are very significant.