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calculating Water evaporation rate

Could someone link me up with some information, either text or web based, concerning evaporation rate of water, in a slightly reduced pressure environment. I need to design a settling tank for a vacuum system to separate a fine oxide powder from water. We produce the slurry at a rate of about 20ml/min. I would like calculate an area and depth for my tank such that the water never accumulates above some predetermined level. The system runs continuously. It may be possible to slightly heat the container.
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 Recognitions: Gold Member Homework Help Science Advisor "Langmuir evaporation rate." Reduced pressure? You're sweeping the space over the tank at what rate? Temperature? Temperature of surroundings? Stirred? Unstirred?
 Mentor Blog Entries: 9 The mixture will be unstirred at about -3Kpa. I will have 2 input lines bringing slurry to the settling container and 1 line drawing a vacumn. The only thing moving will be air and water.

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calculating Water evaporation rate

Do I have to "rendition" a mod to get information? Right now, all I can do is make the wild guess that you've been tasked with recovering cerium oxide or some other high-priced abrasive from a polishing process, for recycle, disposal, or other purpose; the system has been hooked to a shop vac to aspirate waste from the "lap;" there is no air movement; the "settling" reservoir could be made of glass, metal, or be the shop vac; the waste collection is to run continuously without any attention or maintenance; the evaporation rate is then a function of heat leak from ambient through the walls of the the settling reservoir; heat leak equals delta T times thermal conductivity times area and is equal to evaporation rate times enthalpy of evaporation.

3kPa is the stall pressure of the shop vac? Is it some other vacuum system with a known volume pumping rate? Is the slurry stream aspirating air? What are the system inputs and outputs? Air and slurry being turned to saturated air and solid waste for disposal? Slurry being turned into water vapor and recovered abrasive? You wanta know how often to tell the "tech" to empty the shop vac?
 Mentor Blog Entries: 9 You seem to want this to be complicated. I thought it was a pretty simple question. It seems to me that this is a pretty straightforward Physical Chemistry problem. I do not have a text which covers it so I need a reference that will address what the rate of evaporation of water at about room temp will be. It would be nice if the pressure above the water could be factored in. Consider this system: A cylinder x cm in diam and Y cm high. It is maintained at z kPa (z can be negative) and held at Temperature T. There is a flow of F cm3 / min flowing through holes in the lid of a system.

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 Quote by Integral You seem to want this to be complicated. I thought it was a pretty simple question.
"Simple?" Trivially. Answer the questions, and you're done.

 It seems to me that this is a pretty straightforward Physical Chemistry problem.
Technically, it's mass transport; again, answering the questions saves you a lot of grief.

 I do not have a text which covers it so I need a reference that will address what the rate of evaporation of water at about room temp will be.
"Literal" answer to your question, CRC plus Langmuir, 20-25 mm Hg (3 - 3.5 kPa) is 200 mol/(m2 . s); vapor condenses at the same rate.

 It would be nice if the pressure above the water could be factored in.
The Poynting correction to vapor pressures under an atmosphere or less of inert gas or other media is less than 0.1%, and is a likewise trivial correction to evaporation rate.

 Consider this system: A cylinder x cm in diam and Y cm high. It is maintained at z kPa (z can be negative) and held at Temperature T. There is a flow of F cm3 / min flowing through holes in the lid of a system.
You want the transport rate for the system you've described? 20 cm3/min, room T, 97-8 kPa inert blanket? Milligram of water vapor or less per minute to the "vacuum" system; raise T to 98-99 C, and boost that to 15-20 milligrams per minute; these are maxima for optimized "vacuum" system geometry; service time to liquid overflow to "vacuum" is (Vsediment bowl/20cm3)min. at room T, 0.1-0.2% less at near boiling T.

Since it's a "pretty simple question," now that you've been given the answers, you're on your own.
 Recognitions: Gold Member Science Advisor Staff Emeritus Bystander, I did the calculation on a piece of paper and get a number that seems way high. After plugging in and evaluating constants in Langmuir, you have: $$\mu = 2.65 \cdot 10^{25} p / \sqrt{MT}~,~~(p~in ~atm.)$$ At, T ~ 300K, p (vapor pressure of water) is, going by your numbers, about 0.03 atm. M is 18g/mol That gives me $\mu \approx 10^{22}~ molecules/cm^2 \cdot sec \approx 0.3 g/cm^2\cdot sec$ This is obviously too high. What did I screw up on? (no volatile solutes, 3kPa is a tiny correction, corrections to temperature are small and container is assumed to be thermaized to ambient, volume of liquid is assumed to be large compared to surface so evaporative cooling is negligible and I can ignore the effect of heat leaks,... what else?)

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 Quote by Gokul43201 (snip)That gives me $\mu \approx 10^{22}~ molecules/cm^2 \cdot sec \approx 0.3 g/cm^2\cdot sec$ This is obviously too high. What did I screw up on?(snip)
Once you've double-checked your calculations, when faced with a conflict between "intuition" and science, go with the science. Compare your result with the pressure derivation from kinetic theory of gases if you need another stake in "intuition's" heart.

You're "right on the money."
 if stirred and maintained temperatured at 80 C...
 ^___^
 Hi..I got problem to look for Vapor pressure at the dew point according to the temperature of the ambient air of the room, in kPa when I calculate evaporation rate of water at 95 deg C . Can someone advice me how to get it.

 Quote by Integral Could someone link me up with some information, either text or web based, concerning evaporation rate of water, in a slightly reduced pressure environment. I need to design a settling tank for a vacuum system to separate a fine oxide powder from water. We produce the slurry at a rate of about 20ml/min. I would like calculate an area and depth for my tank such that the water never accumulates above some predetermined level. The system runs continuously. It may be possible to slightly heat the container.

Hi..I got the evaporation rate formuar at google & calculated as follow.

Y = Latent heat of evaporation - 2,270 kJ/kg

Kpa mmHG
Pw Kpa 84.64 634.8
Pa Kpa 4.2 31.5
V m/s 0.1
Hv kj/kg 2270
*

ASHRAE Formula for Evaporation rate of water

M/A = (42.6+37.6Vw)(Pw-Pa)
Hv
M/A= 12.32114009 kg/m2.hr
Water evaporation rate, M/A = 12.32 kg/m2.hr at 95o C

But I think, Pa may be not correct & I am not sure 12.32 kg/m2.hr is correct..

Can someone help me to correct my calculation?
 Im no rocket scientist just an ordinary individual with minimal mathematics in physics, Need assistance greatly appreciated. I need to figure out what the evaporation rate is for a pool that is constantly circulating water with a surface area of 38,640'.....Any takers?

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 Quote by flores.maryc Im no rocket scientist just an ordinary individual with minimal mathematics in physics, Need assistance greatly appreciated. I need to figure out what the evaporation rate is for a pool that is constantly circulating water with a surface area of 38,640'.....Any takers?
Maybe you should have opened a dedicated new thread in the homework section. Obviously there may not be a no simple answer to this question.

The closest relationship to evaporation rate could be Clausius Clappeyron, but that may not help much.

Obviously for practical purposes, the evaporation rate is mainly a function of energy available (temp of water, absorption of solar and IR energy flux), relative humidity and windspeed.

For the energy part, mind that latent heat of the vaporisation requires about 2500 joules per gram water.