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Evaporation and Humidity
Name: Chris
Status: other
Age: 50s
Location: N/A
Country: N/A
Date: 2000-2001
Question:
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?
ie
0% humidity = a rate of 1
10% humidity = a rate of .9
90% humidity = a rate of .1
or is there another formula?
Replies:
Chris,
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
Meteorologist
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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