# What is the formula for evaporation rate of water?

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

$$g_\mathrm h = \Theta A (x_\mathrm s - x)$$

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

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

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

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

$$x_\mathrm s$$ = humidity ratio in saturated air at the same temperature as the water surface ($$\mathrm{kg/kg}$$) ($$\mathrm{kg}$$ $$\ce{H2O}$$ in $$\mathrm{kg}$$ dry air)

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

$$x = 0.62198 p_\mathrm w/(p_\mathrm a - p_\mathrm w)$$

$$p_\mathrm w$$ = partial pressure of water vapor in moist air (Pa, psi)

$$p_\mathrm a$$ = 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:

$$p_\mathrm{ws} = \mathrm e^{\frac{77.3450 + 0.0057 T - 7235 / T}{T\cdot8.2}}$$

$$p_\mathrm w = p_\mathrm {ws} \cdot HU$$

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

$$T$$ = 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.

• chemistry.stackexchange.com/questions/55244/… – Mithoron Aug 7 '16 at 16:47
• Thank you for your attention. But I saw that. This is a very different question, and the answers there don't satisfy my question. – Bingkongmaster Aug 7 '16 at 16:49
• there was another similar question, but generally there's no way to have exact formula. – Mithoron Aug 7 '16 at 16:51
• 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. – Bingkongmaster Aug 7 '16 at 17:01
• I have a feeling it's probably going to be assuming the equivalence of the ambient temperature and the water surface temperature. – Eashaan Godbole Aug 21 '17 at 20:18

The temperature of water is implicitly covered by actual and saturated vapour content, or equivalent vapour pressures.

There is deterministic relation between relative humidity, air temperature and temperature of water, supposed to be in dynamic equilibrium.

In such a state, the cooling effect of water evaporation is balanced by the thermal transfer from warmer air.

This relation is used in meteorology/climatology as the standard method to determine humidity via the psychrometric tables of temperature of "dry" and "wet" precise thermometers.

Effectively, the wet bulb temperature is somewhere between air temperature and the air dew point.

They are standard items in the white meteorological wooden box 2m above the lawn.

See psychrometrics particularly the sections dry and wet bulb temperature.

The main article Wet bulb temperature.

• @EdV Wind speed, $v$, has been factored into the mass transfer coefficient, $\Theta$. – Eashaan Godbole Sep 4 '19 at 17:35

Air temperature is a factor in the vapor pressure value. Lower temps will have lower vapor coefficients.