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Name: Chris
Status: other
Age: 50s
Location: N/A
Country: N/A
Date: 2000-2001

Say there was a container of water exposed to moving air. Assume the evaporation rate is measured at a constant temperature from 0% humidity to 100% humidity. Would the evaporation rate be inversely proportional to the rise in humidity?

0% humidity = a rate of 1
10% humidity = a rate of .9
90% humidity = a rate of .1
or is there another formula?


In your example, I assume that the speed and direction of the airflow and the temperature of the water are also held constant. If so, the evaporation rate is inversely proportional to relative humidity (RH), but more importantly, it is inversely proportional to vapor pressure. Relative humidity is temperature dependent, so in your example where the temperature doesn't change, it is okay to think in terms of RH. However, it is the vapor pressure of the air that is really the controlling factor. The proportionality is not quite a linear inverse; the evaporation rate is also affected by the air density, which decreases slightly as more water vapor is added to the air from the water surface (which also tends to decrease the air temperature very slightly). Therefore the evaporation rate will be somewhat less near 100% RH than at 10% RH.

The very controlled conditions that you describe in your example are good to illustrate what affects evaporation, but in the real world, there are many other factors to consider. Some very complicated models have been developed to estimate evaporation over, for instance, a field or crops. I used one of these in my graduate work. Despite it's attempting to consider all factors, it did not work well. That was 25 years ago and some people are still trying to modify the model to work better; they are not having much success. Soil moisture at different levels in the soil and the effects of plant canopies make evaporation measurements and modeling a very challenging field, one that I have been involved in for 25 years.

David Cook
Argonne National Laboratory

I think the answer is yes assuming the air flow rate is sufficiently slow that you evaporate just enough water at the constant temperature to bring the water content of the air from whatever it is to 100% humidity and it's water vapor saturated air that exits the system.

Also assumes no impurities and you can in fact keep the temperature constant.

FYI the heat of vaporization of water at 25 C is about 10 Kcal / mol That's a lot of heat to evaporate just 18 cm^3 of a liquid.

Vince Calder

You are right. Here's why: evaporation is a process that can be described by standard equations of chemical kinetics. It is a dynamic competition between water molecules in the liquid moving to the vapor phase (evaporation) and molecules of water vapor moving to the liquid phase (condensation). At any given temperature, the rate of evaporation is directly proportional to the concentration of water in the liquid phase, which is constant as long as there is some liquid water present. (If something is dissolved in the water, such as salt, this condition changes.) The rate of condensation at any given temperature is proportional to the concentration of water molecules in the vapor phase. 100% humidity is the concentration at which the condensation rate equals the evaporation rate. (The concentration of water vapor that makes 100% humidity depends on the temperature - because the evaporation and condensation rates depend on the temperature.)

The overall evaporation rate is the true evaporation rate minus the condensation rate. For simplicity, let's call the true evaporation rate 1. The condensation rate is some multiple of the water vapor concentration. It will be convenient to express the water vapor concentration as relative humidity - at a constant temperature, this is just a multiple of the vapor concentration. So,

condensation rate = kH, where

k = some constant proportionality factor (the "rate constant"), and
H = the relative humudity.

We know that the condensation rate = the evaporation rate at H = 100%. Thus,

kH = 1
k(100%) = 1
k = 1

So in our simplified units, the condensation rate = the relative humidity.

The overall evaporation rate then is

rate = 1 - kH
= 1 - H

which gives exactly the values you guessed.

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois

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