Today I wanted to run through a calculation that determines how quickly water evaporates from the surface of a lake. I am a member of a private lake in my hometown, and I often have to sit through member meetings about the minutia of water law that are so boring that I’d rather stuff a mentholated eucalyptus cough drop up my nose immediately after ramming a carrot peeler up there and rooting it around. (I hate it when that happens.) Yes, they’re that bad. Here in the parched West everybody is downstream from us so we can’t just use up all of the water as it runs out of the mountains. Water laws hope to prevent all of us evil Coloradoans from sucking up all of the H2O and leaving the rest of the U.S. 
the man said, “In the West whiskey is for drinking, and water is for fighting.” I am definitely not going to get into water laws here. That’s too complicated and boring for even me to tackle. However, since water is a strictly regulated resource out here I thought that I’d try to see how much of that resource we loose to evaporation. Seems reasonably simple, right? Well…not so much.
the man said, “In the West whiskey is for drinking, and water is for fighting.” I am definitely not going to get into water laws here. That’s too complicated and boring for even me to tackle. However, since water is a strictly regulated resource out here I thought that I’d try to see how much of that resource we loose to evaporation. Seems reasonably simple, right? Well…not so much.
Just finding a single equation that takes into account all of the factors that effect evaporation rate was a chore. Eventually I stumbled across a series of empirical formulas that were developed by the U.S. Air Force to monitor the evaporation rates of pools of jet fuel. Here they are:
E = (4.161 x 10-5)(u0.75)(TF)(M)(PS/PH)
PH = 760e(65.3319-(7245.2/TA)-(8.22lnTA)+(0.0061557TA))
If TP = 0 °C or less, then TF = 1.0
If TP > 0 °C, then TF = 1.0 + 0.0043 TP2
Where:
E = evaporation flux, (kg/min)/m² of pool surface
u = wind speed just above the liquid surface, m/s
TA = absolute ambient temperature, K
TF = pool liquid temperature correction factor, dimensionless
TP = pool liquid temperature, °C
M = pool liquid molecular weight, dimensionless
PS = pool liquid vapor pressure at ambient temperature, mmHg
PH = hydrazine vapor pressure at ambient temperature, mmHg
O.K., let’s begin by sorting out some of the information that we need to know about the lake. As far as lakes go, this one isn’t exactly huge. It is about 660,000 m2. As a reference, the largest lake in Colorado (Grand Lake Lake Michigan which has a surface area of a whopping 58 billion m2. The average air temp in town is 72°F in July which is the hottest month (yes, I know; you’re all jealous), and the average wind speed is 6.9 mph. This should be enough information to get us where we need to go. After consulting several charts and plugging and chugging the numbers through the machinery listed above I churned out a surprisingly large result. We lose around 1098 gallons of water every minute to evaporation from the relatively small lake. That’s about 66,000 gal per hour, 1.6 million gallons per day, or 578 billion gallons per year! If we extrapolate this to a lake the size of Lake Michigan the quantity of water lost to the atmosphere due to evaporation becomes astronomical. 5.0 x 1016 gallons evaporate off of that lake every year which drastically affects the local climate. Checking these numbers against both the EPA and Stiver & Mackay methods gives the same result. After mulling over these results for a bit it becomes easy to see how open ditches and ponds for agricultural irrigation are so inefficient.
What about applying our new equations to something else that’s a little more obscure? Want to figure out how much water evaporates from toilet bowls each year in the U.S. U.S.
Please, use the knowledge you’ve gained here responsibly…if you’re still awake.
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