# Evaporation of Water: Error in Formula?

1. Jun 29, 2008

### t3sseract

I am trying to calculate the rate (g/m2) of water evaporation given Temperature, Humidity, Air Speed, elevation.

The best source of an equation for this I have found is from engineeringtoolbox.com, but I have found a possible error in their calculation. (I am in a far off land with no access to real textbooks)

The equation:
g = Θ A (xs - x) (1)
where
g = amount of evaporated water (kg/h)
Θ = (25 + 19 v) = evaporation coefficient (kg/m2h)

However, later on the example switches air speed units:
g = ( 25 + 19 (0.5 (m/s)) ) ( 25 (m) 20 (m) ) ( 0,019826 (kg/kg) - 0.0115 (kg/kg)) / 3600
= 0.04 kg/s

When I use this equation with m/s and kg/s, I get an inflated evaporation rate (their example alone provides and incredible evaporation rate for a pool).

Can anyone give me a confirmation of this equation?

Is there a more recommended equation for calculating evaporation(and a source)?

Thank you
:3

2. Jul 1, 2008

### RTW69

This should work.

M-dot= (A)(42.6+37.6V)(Pw-Pa)/Hv

where

m-dot= evaporation rate, kg/hr
A= surface area, m^2
V= air velocity over water surface, m/s
Pw= saturation vapor pressure at water temperature, mm Hg
Ps= saturation vapor pressure of air dew point, mm Hg
Hv= latent heat of vaporization of water at pool temperature, about 2270 KJ/Kg

3. Jul 1, 2008

### t3sseract

This equation is also giving me inflated evaporation rates:

Using:
A = 9m2
V = 0.5m/s
Pw (at 25 deg. C) = 11.76 mmHg
Ps (at 50% RH) = 23.53 mmHg

Gives:
m-dot = 2.84 kg/hr

At this rate my pool will empty almost as fast as I can fill it.

I am using the following equations for computing the Pressures (in Pa, and in visualbasic)

Public Function Vap_Pres_Sat(TempK As Double) As Double

Vap_Pres_Sat = 22105649.25 * Exp((-27405.526 + 97.5413 * TempK + -0.146244 * TempK ^ 2 + 0.00012558 * TempK ^ 3 + -0.000000048502 * TempK ^ 4) / (4.34903 * TempK - 0.0039381 * TempK ^ 2))

End Function

Public Function Vap_Pres(TempK As Double, RH As Double) As Double

Vap_Pres = Vap_Pres_Sat(TempK) * RH

End Function

4. Jul 2, 2008

### RTW69

Pw@25 C = 23.71 mmHG

Ps@ 18 C = 15.46 mmHG ( 25 C air temp, 50% RH)

m-dot= 2 KG/hr, that is about .0022 cm/hr drop in water level for a 9 m^2 surface area

For a vapor pressure calculator goto:

www.csgnetwork.com/vaporpressurecalc.html

To review evaporation rate information goto:

"Measurement and analysis of Evaporation from an inactive outdoor swimming pool"

www.p2pays.org/ref/19/18985.pdf

5. May 3, 2010

### poolguy

I don't have a formula for you, however, I can tell you from experience that you can expect no more than a 1/2 inch of evaporation, for a full day, in the hottest of weather. You can expect increased evaporation if you are in a very windy area. If you think your evaporating too much water, turn the pool off overnight, if you've lost water overnite chances are you have a leak.

6. May 7, 2010

### Bob S

See

http://www.rlmartin.com/rspec/whatis/equations.htm [Broken]

Bob S

Last edited by a moderator: May 4, 2017