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Q & A: More on calculating evaporation rates

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Most recent answer: 10/22/2007
Q:
As for the evaporation rate calcalation; what would the rate be if the pool were 1 meter deep, 10 meters on a side, square; the ground being arbitrarily deep with a density of 4g./cm^3, temperature of 70 degrees F, with a heat conductivity of 0.25 cal/sec/cm/(difference in degrees C) and 200 watts/meter^2 of light at a wavelength of 500nanometers over an arbitrarily large area that includes the pool and some ground?
- Jonathan
waltham
A:
Hi Jonathan,

John from Bermuda asked a  but all we did was talk about the variables which affect the evaporation rate calculation. There are still some variables missing, but the main ingredients are there to estimate an evaporation rate (I'll let you do the calculation however).

When water evaporates, it requires 540 calories per gram of energy to break the loose bonds between the water molecules and let them move freely. In this case, since you have supplied information relating to the rate at which energy is delivered to the water, then balancing the energy input with that required for evaporation is the best way to handle the problem. The actual temperature of the water will change so that the evaporation rate's energy requirement matches the energy input. One conversion constant you need to go from calories to Joules (one Watt is one Joule per second) is: one calorie is 4.18 Joules.

A couple of things that are missing from your list:
1) Air humidity
2) Wind speed
3) Air pressure (we can assume it's one atmosphere, but water will boil at room temperature if placed in a vacuum); air pressure also affects the diffusion rate which is important in the evaporation rate.

Some things are tough to calculate, like the effect of convection in the water and the air.

As a first approximation, you can just take the sunlight energy and balance that against the energy needed by the evaporation, and the answer will probably be pretty close to what you want. 200 Watts/m^2 seems to be by far the biggest contribution to the energy flow into the pool; all of the other factors will be smaller. Just figure out how much water per m^2 has to evaporate to take the 200 Watts.

To do a better job, however, you have figure out how much thermal energy is lost to the air by conduction and convection and how much to the ground by conduction. To get these, you need the temperature of the water and the air and the air flow. The water temperature, as mentioned before, will slowly change until it settles down at some equilibrium temperature for which the energy flow adds up, so it does not need to be specified in the problem. The energy flow into the ground requires a solution of the heat equation, which is how to use the thermal conductivity number you gave (I assume you gave the ground temperature and not the water temperature -- the ground temperature far away from the pool is needed anyway). The water vapor diffuses away from the surface, and that rate depends on knowing the diffusion constant in air which depends on the air pressure.

All of these ingredients need to be calculated because their relative importances depend on the parameters you gave and the ones I mentioned, to make the energy flow all add up. You then solve for the temperature, and then find the fraction of the energy that is used for evaporation. But the main factors are just the sunlight and evaporation, and ignoring the conduction might get you a good approximate answer.

Tom

(published on 10/22/2007)

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