Heat loss from an outdoor fountain
Heat loss from an outdoor fountain
(OP)
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!
Thanks in advance!





RE: Heat loss from an outdoor 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.
RE: Heat loss from an outdoor fountain
Plus the water will be moving (churning)... which will make a difference to the heat loss. How can that be estimated?
www.vanderweil.com
RE: Heat loss from an outdoor fountain
Radiation cannot be ignored.
If the fountain is very active, latent heat loss through evaporation may be significant.
RE: Heat loss from an outdoor fountain
RE: Heat loss from an outdoor fountain
www.vanderweil.com
RE: Heat loss from an outdoor fountain
RE: Heat loss from an outdoor fountain
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
RE: Heat loss from an outdoor fountain
RE: Heat loss from an outdoor fountain
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
RE: Heat loss from an outdoor fountain
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.
RE: Heat loss from an outdoor fountain
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.
RE: Heat loss from an outdoor fountain