Name: Melissa S.
I realize I am technically out of the age range, but am
hoping you can help a fellow student. I am trying to calculate the water
losss between air of 10% rH and air that is 85% rH. They are both at 24C
and are at 5000 ft. Is there a good way to calculate a the volume of
water lost when 1L of 10% rH air replaces 1L of 85% rH
air? Thank you.
A few formulas are needed to figure this out.
First, we need to get an estimate of the
atmospheric pressure at 5000 feet. It is
approximately 24.89 inches of mercury,
or 842.9 millibars (mb), from the U.S. Standard
Next, we need to determine the saturation
vapor pressure e(w) (the pressure exerted by
water vapor when the air is saturated).
e(w) = [1.0007 + (.00000346 * P)] * [6.1121 exp[(17.502*T)/(T+240.97)]]
with T being temperature, P being pressure in mb,
* being multiplication, and exp being the natural
logarithm "e" value 2.7183, which is taken to the
exponent of the expression in the brackets. Calculators
usually have a key with this "e" to the x exponent on
them ("e" here is not to be confused with our symbol of
e as vapor pressure).
This gives a saturation vapor pressure e(w) of 29.94 mb.
We can now figure out the vapor pressure e (the pressure exerted by
water vapor in the air) from the relative humidity.
For RH = 85% (we use the decimal equivalent 0.85),
e = RH * e(w) = 0.85 * 29.94 = 25.45 mb
For RH = 10%,
e = RH * e(w) = 0.10 * 29.94 = 2.99 mb
Now we can determine the amount of water per liter,
grams per liter (g/l) = (0.2169 * e)/(T + 273.2)
for 85% RH, g/l = (0.2169 * 25.45)/(24 + 273.2) = 0.0186
for 10% RH, g/l = (0.2169 * 2.99)/(24 + 273.2) = 0.0022
So, per liter, we have gone from 0.0186 to 0.0022 grams,
a difference of 0.0164 grams.
The density of liquid water at the Earth's surface is 0.9973
grams per cubic centimeter at 24 degrees C. If we adjust
this linearly for our pressure divided by surface pressure
(842.9/1010.3) = 0.832, the density would be 0.83 grams
per cubic centimeter at 5000 feet. If our difference in
water vapor (0.0164 grams) was compressed to liquid form,
it would occupy only .0198 cubic centimeters, or 0.0000198
liters. The most important thing is the dramatic reduction
in grams per liter of water vapor.
You can see how complicated such calculations can become.
David R. Cook
Atmospheric Research Section
Environmental Research Division
Argonne National Laboratory
You have to know the (equilibrium) vapor pressure of air at the temperature.
(You also have to know the temperature, which is 24 C.) That tells you the
amount of water in air at 100 % rh. (Yes, you need to do some conversions
here, from pressure to moles/liter to grams/liter or whatever units you
need.) Then the amount of water "lost" will be (85% - 10%) = 75% of the
amount at saturation.
The rate of evaporation is a complicated function of air flow, surface area,
stirring of the water in addition to the temperature and relative humidity,
so it is not possible to express rate of evaporation in terms of absolute
loss per minute. However, you can say that the evaporation rate at 0 % R.H.
will be a maximum, and the evaporation rate at 100% R.H. will be zero
whatever the value of the other uncontrolled variables. Hope this helps a
little. Evaporation rate is a quantity that is often thought of as being an
"absolute" quantity when it, in fact, depends on several variables.
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Update: June 2012