Relative Humidity ```Name: Melissa S. Status: student Age: 20s Location: N/A Country: N/A Date: 2000-2001 ``` Question: 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. Melissa, 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 Atmosphere. 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. Richard Barrans 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. Vince Calder Click here to return to the Weather Archives

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