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I did my own research, and according to The Engineering Toolbox,the formula for the evaporation rate of water is as follows:

$\mathrm{gh = Θ A (x_s - x)}$

$\mathrm{gh}$ = amount of evaporated water per hour ($\mathrm{kg/h)}$

$\mathrm{Θ = (25 + 19 v)}$ = evaporation coefficient ($\mathrm{kg/m^2 > h}$)

$\mathrm{v}$ = velocity of air above the water surface ($\mathrm{m/s}$)

$\mathrm{A}$ = water surface area ($\mathrm{m^2}$)

$\mathrm{x_s}$ = humidity ratio in saturated air at the same temperature as the water surface ($\mathrm{kg/kg}$) (kg H2O in kg Dry Air)

$\mathrm{x}$ = humidity ratio in the air ($\mathrm{kg/kg}$) (kg H2O in kg Dry Air)

However, later I realized that something is wrong with the formula. There is no variable for water temperature.

The formula for humidity ratio in the air is:

$\mathrm{x = 0.62198 p_w / (p_a - p_w)}$ (2)

$\mathrm{p_w}$ = partial pressure of water vapor in moist air (Pa, psi)

$\mathrm{p_a}$ = atmospheric pressure of moist air (Pa, psi)

The formula for humidity ratio in saturated air is the same, except partial pressure of water vapor in moist air is replaced with saturation pressure of water vapor in moist air.

And the formula for water vapor saturation/partial pressure is:

$\mathrm{p_{ws} = e(77.3450 + 0.0057 T - 7235 / T) / T*8.2}$
(1)

$\mathrm{p_w = p_{ws} * HU}$

$\mathrm{p_{ws}}$ = water vapor saturation pressure (Pa)

e = the constant 2.718.......

T = dry bulb temperature of the moist air (K)

HU = humidity ratio (%)

So, according to this website, temperature-wise the formula is determined solely by the air temperature without considering the water temperature. Is it because the formula assumes the equivalence of the two temperatures for simplification or because I'm missing a variable? I searched for other methods, but so far no website is as comprehensive as this one, and I don't want to re-program everything again...

This is a different question from the existing ones. I've read many of them and the answers are not specific enough for my situation. Here I need to know how the temperature of the water plays a role in the formula for the rate of evaporation of water.

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  • $\begingroup$ chemistry.stackexchange.com/questions/55244/… $\endgroup$ – Mithoron Aug 7 '16 at 16:47
  • $\begingroup$ Thank you for your attention. But I saw that. This is a very different question, and the answers there don't satisfy my question. $\endgroup$ – Bingkongmaster Aug 7 '16 at 16:49
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    $\begingroup$ there was another similar question, but generally there's no way to have exact formula. $\endgroup$ – Mithoron Aug 7 '16 at 16:51
  • $\begingroup$ I'm no expert in this field, but for all I know the formula I've found is empirical formula. I thought if chemistry experts saw the formula, even if they do not know such made-up formula, they might recognize either 1. the simplification of the formula to remove the water temperature variable or 2. my stupidity for missing the water temperature variable. $\endgroup$ – Bingkongmaster Aug 7 '16 at 17:01
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    $\begingroup$ I have a feeling it's probably going to be assuming the equivalence of the ambient temperature and the water surface temperature. $\endgroup$ – Eashaan Godbole Aug 21 '17 at 20:18
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Air temperature is a factor in the vapor pressure value. Lower temps will have lower vapor coefficients. enter image description here

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