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evaporative heat loss

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brendanschulte

Civil/Environmental
Feb 12, 2003
1
In order to better implement an environmental management system for a hotel, I am trying to calculate the quantity of energy(btu's or kilowatthours) lost through evaporation from 4 hot tubs in Jackson Wyoming. It gets quite cold and dry here and I am considering some ideas to save some cash by either purchasing custom made covers or adjusting the amount of thermal energy used to heat the water at night when the hot tubs are not in service. They are closed from 11pm until 6am so there is 7 hours during which they are not in use.
The data i can get:
-ambient temp
-tub temp
-input Q
-output Q
-total storage
-daily average humidity
-total surface area
-evapotranspiration rate

let me know what other factor that i am forgetting and which ones in my list are irrelevant

The only way I can convince my GM that any action is worth the time is produce a proposal with dollar figures.
thanks
Brendan
 
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There have been at least one or two threads on this very topic (heat loss from the surface of a pool) in the past. I am not certain that those threads were in the "H.T. & Thermodynamics engineering" forum.

Try a "keyword search" - at the top of this page - to see if you can find some of those discussions.

 
The equation for heat loss because of evaporation is
Q = HxAx(VPs-VPa)

H = Latent Heat transfer rate in Btu/Hr/Sq.Ft/in.Hg
If airflow is parallel H at 0cfm is 100 and at 400fpm 275. As the relationship is linear you can intrapolate for the middle range.

If the flow is transverse then at 50fpm 250 and at 400fpm 550. Below 50fpm the value is nonlinear.

A = Total surface area in sq.ft
VPs = vapor pressure of saturated air at water temperature in inches Hg
VPa = Vapor pressure of in air above surface in inches Hg.

Apart from this you have to consider convective heat transfer. (For a rough estimate you can consider natural convection over a flat plate)
 
Talk to some companies who make electric heat tracing. I remember 'one' of them (and can't remember who it was and don't seem to have the information at work) had a really good technical section including the heat losses from open tanks containing water. Sounds just what you needed.

Sounds like you have a few leads above, this might be another check you want to follow up with.
 
Brendan,

A good reference to have on hand is available from a company called "Spirax Sarco" (they make steam & condensate system equipment). The book is called "Design of Fluid systems -Hook-Ups"

Despite the odd title it is an excellent text for the newbie

Table #25 in that book is titled "Heat Loss From Open Tanks" and losses are tabulated; both radiation and, of course, the biggie... evaporative.

For a tank at 100F, the radiation losses are 70 BTU/hr-sq. ft and 160 BTU/hr-sq. ft Total is 230... Losses thru the hot tub wall can be neglected. Ambient is 60F

Lets say you have 100 sq ft and that electricity costs $.05/kWhr

On a per hour basis: (230)(100)(1/3413)(0.05) = $.34/hr
On a per night basis ~$2.38

Not a lot, but you must also add in the cost of water evaporated and chemicals.....

Does anyone have a different approach ?

MJC

 
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