Q:

Can you calculate the rate at which water will flash to steam given the temperature and psia? For example, assuming approximately .8 PSIA and 100 degrees F, at what rate would the water evaporate? Would lowering the pressure or increasing the temperature change the evaporation rate significantly?
Thanks.

- Rick Gresham (age 51)

Portland, OR, USA

- Rick Gresham (age 51)

Portland, OR, USA

A:

The answer is yes, the rate that water evaporates can indeed be calculated, but it depends on a few more things than you mention. The evaporation rate is influenced by 1) The temperature of the water at the air-water surface 2) The humidity of the air 3) The area of the air-water surface 4) The temperature of the air (more on this below) In a real-world situation of evaporating water, none of these four quantities above remains constant because the process of evaporation itself changes them. Water evaporating takes quite a lot of heat away -- 540 calories per gram -- when it evaporates. That’s enough to cool down 540 grams of water by a degree, or 50 grams of water a little more than ten degrees. If you are not very careful to replace the lost heat energy during the evaporation, the temperature will go down. And even then the temperature right at the surface will be lower than elsewhere in the water and it will depend on 5) water currents convecting heat and the ability to keep the temperature constant at 100 degrees F. For a similar reason, the air near the surface of the water will become more saturated with water as the water evaporates. The evaporation rate will depend on: 6) airflow past the water/air surface. These factors explain why: People sweat to stay cool. People say "It’s not the heat, it’s the humidity." People use fans to keep themselves cool on hot days. Spraying a fine mist of water in the air will make it evaporate faster (increased total surface area). I am guessing your 0.8 PSIA is an indication of the partial pressure of water in the air, and that covers the humidity dependence mentioned above. Looking in a table of vapor-pressures (this tells me what the pressure of water vapor liquid water is in equilibrium with at different temperatures), I see that the vapor pressure of water at 100 degrees F is 0.96 pounds/square inch. If the vapor pressure is less than 0.8, the water would not evaporate at all, and in fact in that atmosphere, water would condense. Irving Langmuir developed a way to measure vapor pressure by measuring the evaporation rate, which we will now flip around backwards. His reasoning is that the rate at which molecues are lost due to evaporation to a gas with no partial pressure of the evaporating substance is the same as the rate at which molecules of the substance would hit the surface if it were in equilibrium with the vapor (because in equilibrium the evaporation rate and re-condensation rate cancel each other out). His expression is: (mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) ) (from Zemansky and Dittman, Heat and Thermodynamics, McGraw Hill, copyright dates from 1937 to 1981 in my copy). The ambient partial pressure of water is the 0.8 PSI you gave me, and the vapor pressure is the 0.96 PSI (we’re going to have to convert this to different units to make it work out). T is the temperature in degrees Kelvin -- 310 degrees above absolute zero. R is the gas constant which is 8.314 Joules/(mol degree K). Naturally the units of measurement in formulas like these are a disaster. I would suggest using standard Metric (SI) units for everything. Multiply the pressures in PSI by 6895 to get pressurs in Pascals (Newtons/sq. M), and the molecular weight should be in kg/mol (to cancel out the moles in the gas constant R). For water this is 0.018 kg/mole. Doing the arithmetic gives a rather largeish number -- 1.1 kg per square meter per second. Keep in mind that this assumes that the temperature is maintained at 100 degrees at the interface and the the humidity is maintained right up to the interface, meaning that you have to blow air with a big fan at a high rate to carry away the evaporating water, and even this might not be enough (or you are in a near vacuum with only a tiny pressure of water vapor over the water), and also that there are no impurities on the water (a layer of oil will spoil the whole thing, and sweat contains lots of oils). A real situation involves the fact that the humidity near the interface is much higher than even a short distance away, and that the water vapor must diffuse away. This effect will slow the evaporation down quite a a lot because the evaporation rate is proportional to the difference between the vapor pressure and the partial pressure of the substance, and diffusion can only take water away so fast. As the water evaporates, the partial pressure of water in the gas right over the water will be nearly equal to the vapor pressure, and then it will drop as you go away from the surface, and how steeply this drops (which depends on the airflow rate and how long the water has been there evaporating) determines the rate at which water will diffuse away. Even with a fan blowing air past the surface, the process is limited by diffusion very close to the surface because a thin layer of air (called the "boundary layer") right next to the surface does not move relative to the surface. I won’t do the work on the diffusion because it depends too much on the details of the setup. The diffusion constant for Nitrogen is 0.185 cm**2/sec at room temperature and 1 atm. Lowering the partial pressure of water will raise the evaporation rate as mentioned above. Lowering the air pressure will increase the diffusion rate. The partial pressure of water at 100 degrees F is so high in fact, that bubbles will form spontaneously in the fluid and cause it to boil rapidly if the water is placed in a vacuum (upsetting the surface area because of all of the bubbles). Increasing the temperature will increase the evaporation rate. It appears in the denominator (in the square root) above, but a much more important dependence comes in with the vapor pressure. vapor pressure is proportional to exp(-latent heat/RT) (F. Reif, Fundamentals of Statistical and Thermal Physics, McGraw Hill, 1965). The latent heat is the 540 cal/gram (watch the units again!). T again is in the absolute scale. Furthermore, as the temperature increases the diffusion rate increases too, but this is the temperature of the air, not the water -- hence the dependence on the air temperature I mentioned earlier. Tom |

*(republished on 07/24/06)*

Q:

related question - what about condensation rate?
Let"s say that instead of 0.8 PSIA we had 1.1 PSIA. The same calculation applies but with the direction reversed?

- John (age 21)

New York, NY, USA

- John (age 21)

New York, NY, USA

A:

Yes, you're right. Unfortunately the complications still apply too. As water condenses, that leaves a depleted layer behind with less water vapor. How big an effect that is depends on how rapidly the atmosphere is stirred by winds, etc. Likewise as the vapor condenses it deposits energy in the top of the liquid, heating it up. How hot it gets (and hence how much evaporation results) depends on how well-stirred the liquid is.

Mike W.

Mike W.

*(published on 01/21/08)*

Q:

Another related question - evaporation rate in a closed container:
Evaporation in a closed container will proceed only until there are as many molecules returning to the liquid as there are escaping (=saturated vapor pressure).
Is there a appropriate equation to calculate the conditions for such a equilibrium state? For example, I want to find out wheter a small amount of a fluid in a relatively large container at a constant temperature of the fluid can be prevented from net evaporation by increasing the pressure of the ambient air/gas in the container or by reducing the size of the container or increasing the amount of the fluid, respectively.

- Marco W. (age 35)

Zürich, Switzerland

- Marco W. (age 35)

Zürich, Switzerland

A:

Yes indeed there's an equation. It's called the Clapeyron equation, and for most conditions a simpler approximate version called the Clausius-Clapeyron equation works fine. The vapor pressure that's in equilibrium with the liquid depends strongly on the temperature, and the parameters in the equation are different for every substance.

Here's a link to a discussion of the equations: http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation,

These equations can also be found in any standard thermodynamics text.

The pressure obtained in the equation is the partial pressure due to the vapor from the liquid. Increasing the pressure of ambient air generally has very little effect, unless the liquid is made of the same molecules as the air, nitrogen or oxygen. So what one does is calculate the partial pressure at the given temperature, convert (using the ideal gas law pV=nRT, where p is pressure, V container volume, n moles of vapor, R the gas constant, T the absolute temperature) to figure out how many moles of vapor that would be. If the answer comes out smaller than the moles of liquid put in, the remaining liquid will be in equilibrium with the vapor. If it comes out bigger, what will happen instead is that all the liquid will vaporize.

Mike W.

Here's a link to a discussion of the equations: http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation,

These equations can also be found in any standard thermodynamics text.

The pressure obtained in the equation is the partial pressure due to the vapor from the liquid. Increasing the pressure of ambient air generally has very little effect, unless the liquid is made of the same molecules as the air, nitrogen or oxygen. So what one does is calculate the partial pressure at the given temperature, convert (using the ideal gas law pV=nRT, where p is pressure, V container volume, n moles of vapor, R the gas constant, T the absolute temperature) to figure out how many moles of vapor that would be. If the answer comes out smaller than the moles of liquid put in, the remaining liquid will be in equilibrium with the vapor. If it comes out bigger, what will happen instead is that all the liquid will vaporize.

Mike W.

*(published on 03/11/08)*

Q:

Your answer to calculating the evaporation rate of water is of great interest to us, but you cite Zemansky and Dittman, Heat and Thermodynamics as your source. But I have that book (7th edition) and cannot find the equation or derivation. Can you tell me where to find the equation ? And can you clarify how to find the ambient partial pressure ? Is this equation valid for water below the boiling point ?
Thanks.

- Paul (age 65)

Boonville, CA, USA

- Paul (age 65)

Boonville, CA, USA

A:

Paul- I'll give a partial answer, without taking the time to track down the source of the equation.

If you look at this equation, what it's saying is that the rate at which mass is leaving the liquid when there's no vapor above the liquid is equal to the equilibrium mass density of the vapor times a typical thermal speed of one of the molecules, with a numerical factor thrown in that comes from averaging over all the directions the molecules can be moving. When there is some vapor already above the liquid, the net rate is reduced by whatever fraction of the equilibrium pressure that ambient partial pressure is. If the ambient partial pressure is already equal to the equilibrium pressure, the net rate is zero- as many molecules go from vapor to liquid as go the other way.

So the equation makes sense. In equilibrium, the rate at which molecules leave the surface is the same as the rate that they enter the surface. But you can calculate that entrance rate by assuming that the molecules in the equilibrium gas are moving with random thermal speeds. There are two simple assumptions made here- that the vapor forms a classical ideal gas (usually true) and that when a molecule from the vapor hits the liquid the chance it sticks is 100%, which I guess is pretty close to true.

The equation should work fine for water below the boiling point. However, in using this equation there's one thing you have to be very careful about. If there's no wind, the ambient partial pressure near the surface builds up to nearly the equilibrium value, so the net evaporation rate drops. How far it stays below equilibrium depends on the wind, which has no fundamental equation. You're also assuming that the evaporation rate isn't so large as to cool the surface much. That depends largely on whether the liquid is well-stirred, again not described by some fundamental equation.

As for what the ambient partial pressure is far from the surface, that's just the equilibrium vapor pressure times the relative humidity (for water), by definition of relative humidity. There's no basic equation to tell you what the relative humidity should be. It depends on what's happening with the weather. You can get instruments that will measure it, or check a weather report.

Mike W.

If you look at this equation, what it's saying is that the rate at which mass is leaving the liquid when there's no vapor above the liquid is equal to the equilibrium mass density of the vapor times a typical thermal speed of one of the molecules, with a numerical factor thrown in that comes from averaging over all the directions the molecules can be moving. When there is some vapor already above the liquid, the net rate is reduced by whatever fraction of the equilibrium pressure that ambient partial pressure is. If the ambient partial pressure is already equal to the equilibrium pressure, the net rate is zero- as many molecules go from vapor to liquid as go the other way.

So the equation makes sense. In equilibrium, the rate at which molecules leave the surface is the same as the rate that they enter the surface. But you can calculate that entrance rate by assuming that the molecules in the equilibrium gas are moving with random thermal speeds. There are two simple assumptions made here- that the vapor forms a classical ideal gas (usually true) and that when a molecule from the vapor hits the liquid the chance it sticks is 100%, which I guess is pretty close to true.

The equation should work fine for water below the boiling point. However, in using this equation there's one thing you have to be very careful about. If there's no wind, the ambient partial pressure near the surface builds up to nearly the equilibrium value, so the net evaporation rate drops. How far it stays below equilibrium depends on the wind, which has no fundamental equation. You're also assuming that the evaporation rate isn't so large as to cool the surface much. That depends largely on whether the liquid is well-stirred, again not described by some fundamental equation.

As for what the ambient partial pressure is far from the surface, that's just the equilibrium vapor pressure times the relative humidity (for water), by definition of relative humidity. There's no basic equation to tell you what the relative humidity should be. It depends on what's happening with the weather. You can get instruments that will measure it, or check a weather report.

Mike W.

*(published on 05/07/09)*

Q:

Our town is considering building a new municipal swimming pool. Part of the argument (at a recent public forum on the topic) for needing a new pool is the city officials telling us that our 240,000 gal. pool is losing 70,000 gal. every day. A woman behind me asked the question, "How much is lost from evaporation?" Is it possible for a layman to calculate a correct order-of-magnitude value for our pool's evaporation? What measurements and equations would a person need?

- Jim Rasmussen (age 52)

Hampton, IA US

- Jim Rasmussen (age 52)

Hampton, IA US

A:

Hello Jim,

We get questions like this frequently. It's a toughie. The answer depends on a number of parameters: surface area, temperature, wind speed, humidity, etc. There are so many variables that there is no simple (or even complicated) formula. The best way to determine the answer is to perform a simple experiment. Get a bucketful of water from the pool, let it sit for 24 hours and then measure the drop in the level of water. You can make a pretty good estimate of your local evaporation rate by multiplying by the ratio of surface areas of the pool and bucket, times the volume of water that has evaporated from the bucket.

It does seem, however, that 70,000 gallons a day is a bit much for evaporation. I'd really look for some other mechanism, leakage, splash over, or...

Here is a web site that explains the experiment I suggested in greater detail. Good luck.

http://wiki.answers.com/Q/How_much_water_evaporates_from_a_pool

LeeH

We get questions like this frequently. It's a toughie. The answer depends on a number of parameters: surface area, temperature, wind speed, humidity, etc. There are so many variables that there is no simple (or even complicated) formula. The best way to determine the answer is to perform a simple experiment. Get a bucketful of water from the pool, let it sit for 24 hours and then measure the drop in the level of water. You can make a pretty good estimate of your local evaporation rate by multiplying by the ratio of surface areas of the pool and bucket, times the volume of water that has evaporated from the bucket.

It does seem, however, that 70,000 gallons a day is a bit much for evaporation. I'd really look for some other mechanism, leakage, splash over, or...

Here is a web site that explains the experiment I suggested in greater detail. Good luck.

http://wiki.answers.com/Q/How_much_water_evaporates_from_a_pool

LeeH

*(published on 06/22/09)*

Q:

In regards to the source of the equation, I believe the original work can be traced back to this reference:
"Evaporation and surface structure of liquids". By G. Wyllie. Published in Proc. R. Soc. Lond. A. 1949, 197, pp. 383-395.
The equation they report in this paper is:
(2pi*mkT)^(-1/2) * (saturation vapour pressure - actual pressure)
where m is the molecule's mass, k is botzmann's constant, and T is the temp in Kelvin. The units work out to (1 / sq meters * seconds)

- Scott (age 29)

Tucson, az

- Scott (age 29)

Tucson, az

A:

Thanks!

Mike W.

Mike W.

*(published on 09/03/10)*

Q:

The equation listed on this page for the mass loss rate:
(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )
appears to be incorrect.
This equation implies that the mass loss rate is proportional to sqrt(1/T). If this is the case, then as the temperature increases the mass loss rate decreases. This is not true.

- Joe Atherton

UK

- Joe Atherton

UK

A:

Whoops. We posted that comment from a reader without checking. On further consideration, we didn't even check what the expression was intended to equal.

However, if you'll note, the equation also includes the saturation vapor pressure. Even if the saturation vapor density N/V weren't increasing with T (it is), the ideal gas law p=NkT/V would say that the expression would come out an increasing function of T.

In fact, if you include that (and assume the initial vapor pressure is zero) one gets about sqrt(kT/m) N/V, where N/V is the saturation molecular concentration in the vapor. (I'm not worrying here about the numerical factor). So that is essentially the typical thermal speed times the concentration of molecules. That should be the evaporation rate in units of molecules/(area*time).

p.s. In the various notes sent in, there were two forms of the equation. One, described here, gives the number lost per area*time. The other includes an extra factor of m and gives the mass loss per area*time.

Mike W.

However, if you'll note, the equation also includes the saturation vapor pressure. Even if the saturation vapor density N/V weren't increasing with T (it is), the ideal gas law p=NkT/V would say that the expression would come out an increasing function of T.

In fact, if you include that (and assume the initial vapor pressure is zero) one gets about sqrt(kT/m) N/V, where N/V is the saturation molecular concentration in the vapor. (I'm not worrying here about the numerical factor). So that is essentially the typical thermal speed times the concentration of molecules. That should be the evaporation rate in units of molecules/(area*time).

p.s. In the various notes sent in, there were two forms of the equation. One, described here, gives the number lost per area*time. The other includes an extra factor of m and gives the mass loss per area*time.

Mike W.

*(published on 09/05/10)*

Q:

There is a debate in the water damage restoration world. Does a dehumidifier, by taking water vapor out of the air, create evaporation in structural materials that have been water damaged?
Is evaporation only created by heat?
I have seen equations for calculating the evaporation rate, is humidity the part of the equation that slows down that rate or does lowering the humdity actually create evaporation?

- Ben Justesen (age 32)

Moses Lake, WA, USA

- Ben Justesen (age 32)

Moses Lake, WA, USA

A:

That's an interesting question. Evaporation, in the sense of water molecules leaving the surface, will happen all the time. What you guys care about, however, is net evaporation, the difference between the bare rate of water molecules leaving and the rate of others coming in from the air to the surface. Drying out the air increases the net evaporation rate by reducing the rate at which molecules come in from the air.

All the evaporation requires heat, because it take energy to break a water molecule away from the stuff it's stuck to. Even at room temperature, that heat can flow in from the surroundings. The rate at which the water molecules break loose is very sensitive to temperature, so the people who say you need to heat things up have a very good point. If you want the drying to occur fast enough to avoid much mold growth, etc., then heating things up is a big help. Running dehumidifiers also helps, as I mentioned, but even if the air is perfectly dry, that can't get the net rate any higher than the bare evaporation rate, which is highly temperature sensitive.

Mike W.

All the evaporation requires heat, because it take energy to break a water molecule away from the stuff it's stuck to. Even at room temperature, that heat can flow in from the surroundings. The rate at which the water molecules break loose is very sensitive to temperature, so the people who say you need to heat things up have a very good point. If you want the drying to occur fast enough to avoid much mold growth, etc., then heating things up is a big help. Running dehumidifiers also helps, as I mentioned, but even if the air is perfectly dry, that can't get the net rate any higher than the bare evaporation rate, which is highly temperature sensitive.

Mike W.

*(published on 10/17/11)*

Q:

I'm an engineer (did research in Urbana). I also studied Meteorology. But I wanted to get some additional insight into some “claims” over water evaporating from a man-made canal essentially due to the effects of direct sunlight. (From what I understand, water evaporation is subject to air temp vs water temp, air humidity, air flow rate as well as the surface area of the water and the turbulent flow rate of the water.) So, there has been significant politically motivated drama surrounding a Canal Top Solar Field in India (the land of hot and dry air) where, loosely covering a man-made water canal (maybe 10 meters wide) can save about 9 million liters of water (2.5 mil gals) per year. This is claimed to be … only by covering less than 1 km of a 458 km long canal. (If you perform a web search, you will see that the solar panel field is raised above the surface where air can generously flow around the water.) My question is whether the claim of loosely covering a small section of flowing water (1.7 m/s) can save up to 2.5 million gallons of water per year or whether this is huge political drama? (I guess the bigger question is asking if sunlight directly heats up water or only water particulate and flow bed surfaces?) Please let me know your thoughts.

- Sunny (age 30)

Sacramento, CA

- Sunny (age 30)

Sacramento, CA

A:

This estimate actually looks very realistic to me, based on a crude calculation. Although all the factors you mention do show up in the net water evaporation rate, the key point is that for every kg of water that evaporates there must be a net heat flow of about 2MJ into the water to supply the latent heat. The sunlight (say 1000 W/m^{2}) hitting that 10^{4}m^{2} will supply a lot of energy. Some, mostly infrared, will be directly absorbed by the water. A lot of the visible light will be absorbed by those rocks at the canal bottom, and most of that heat should then be transferred to the flowing water, although some will diffuse out into the ground underneath. Within the rather large error bars of my ignorance of the details of the sunlight pattern in the area, the color of the rocks, etc., that 2.5 M gal/year sounds just about right.

This is purely a matter of keeping that extra energy from getting into the water, and makes no assumption that the flow of air over the water is reduced at all.

Mike W.

*(published on 09/23/13)*

Q:

I am doing my research for the science fair, and my question was does water evaporate the fastest in dirt, sand, or by itself. Can you think of a way that I can guess an accurate answer without doing the experiment?

- jaden (age 11)

Irvine, CA, USA

- jaden (age 11)

Irvine, CA, USA

A:

No, I can't think of a good way to guess. Here's a few of the complications that make it hard. A plain water surface allows the wind to blow on it directly. That might help speed up evaporation. Sand has lots of little crevices that might partly fill up, increasing the surface area. That might speed evaporation. Dirt might have a rough wet surface, also increasing surface area and increasing the evaporation rate. But maybe the dirt is mostly some clay with a smooth surface, and that might reduce evaporation.

Experiments are great things!

Mike W.

posted without checking by Lee until he returns from Paris

*(published on 10/15/13)*