Research info on heat wasted evaporating inside water?
user-1091831
| Posted in Mechanicals on
I’m wondering if any has come across research documents or done their own research on how much heat goes into evaporating water that’s inside, and ultimately whether or not it is a large or small amount of losses. I know of a building with a wash bay that is heated with modines, and am wondering how much of the heat used is attributed to evaporating un-drained water. Maybe there is a way to eliminate the excess water in a more cost effective manner, prior to its next use. I’m not looking to get into specific numbers with this, just a broad sense, and any research that might be available would be great. Thanks.
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I came across something on this topic just yesterday...:
In "H2NO - Mechanical Systems and Moisture Control" Andy Ask says "The 5,000 or so pounds of lumber that frames an average home could accumulate 150 gallons of water before dry-in. About 100 gallons of that water will have to be removed by the building's mechanical systems sooner or later." There's no source referenced for those numbers, though - but I'm betting it's based on the typical variations in moisture content of lumber, 7-15% or so.
So, from that: 100 gallons x 8.35 pounds/gallon x 1,000 Btu/lb to vaporize water and we get 835 kBtu, or 245 kWh. So, depending on where you live, that'd cost you somewhere between $30 and $60 if you did all the drying with an electric heater. (But the book approaches this topic more from the side of dehumidification).
Not exactly the same situation as yours, but it does give a feel for the magnitude of it - around 30-60 cents per gallon evaporated. Electrically, anyway. Gas would be cheaper. Oddly enough, natural gas has about 1,000 Btu/cuft - the same as the energy needed to evaporate a pound (American pint) of water. So 8 cuft of gas to evaporate a gallon, and at around $10 per 1000 cuft, that's 8 cents per gallon of water evaporated. Efficiency of the boiler not included - so call it 10 cents/gallon.
But your mileage may vary. If condition does not improve after one week or a rash develops, stop use and see a doctor. At participating locations only. This is not a toy. Etc.
Cramer,
Nice find, and thanks for your response. Those numbers are useful and I've come across many of them myself. The other main variable I'd need to factor in is the air leakage component. Generally, I don't think it'd be a big deal. But the wash bay I referred to has two huge bay doors that are extremely leaky. Nonetheless, I've been more just curious than anything. Thanks
This how would do it (and have on past projects). Use simplified formulas for evaporation rates (similar for pool evaporation rates) in LB/hr, then multiple by the latent heat of evaporation of water (avg=1050 btuh/lb), that will give you Qlat.(btu/hr)
Determine your evaporation rate (Wp). You need to know the approximate wetted area (A), Saturation vapor pressure at the water surface (Pw), Saturation Vapor Pressure of the air (Pa) and than estimate your activity factor (Fa). The formula to use is
Wp = .1A (Pw-Pa) Fa.
the heat required would be
Qlat = Wp*1050 btu/hr
For your Vapor pressures (Pw, Pa) you will need to estimate the water tempertures and the air temperature+humidity. You should be able to find steam tables online that will give you the vapor pressures at that temperature and humidity. For the activity factor (Fa), assume .25 for a standing pool. The more activity, the higher the activity factor. For example a pool with a water slide would have an activity factor of 1.5. Use your best judgement to determine your activity factor.