## Any takers approximating evaporation rate?!!!

I am trying to estimate the rate of water loss from a 13'L x 6'W x 6'D mixing container at room temperature under a 12 inches of mercury vacuum. There is constant mixing of the fluid (dirty water). There is 180 cfm air movement through the top of the container across the surface of the mixing solution. The mixture is about 50% water. The container is 4.5' full. I am concentrating sludge.

Any insight is appreciated,

Steven

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 Recognitions: Gold Member Science Advisor Staff Emeritus A diagram of the set-up will be required. I don't see how you can maintain a 12" (or about 300 torr) vacuum and have 180 cfms of air flow. Let me ask you to not be very hopeful on this one. The binding energy of water to the solid component of the slurry is a strong function of the solid material itself (and is not usually small compared to the latent heat). Without a knowledge of the solids in the slurry, this will be impossible. Even with a knowledge of the solids, an ab initio calculation is out of the question. If all the physical, chemical and geometric conditions are specified, an order of magnitude estimate may be doable... But realistically speaking, I would be extremely skeptical about any calculation done here. In fact, I doubt that any numbers available for sludge dewatering are generated computationally. Wouldn't it be much easier to determine the evaporation rate experimentally ?
 Recognitions: Gold Member Homework Help Science Advisor Approximation? Things to consider: evaporation rate is going to be a function of surface area, temperature, and composition of the air you're passing over the tank, and to a lesser extent, the dissolved solids content of the water in the slurry, and to a yet lesser extent, the nature of the suspended solids. Vapor pressure of water at "room temp" is 15 - 20 mm Hg, 2 -2.5 kPa. Evaporation rate of pure water into a perfectly dry gas stream is going to be 10-13 g/m2s (from a Langmuir argument). This will be reduced by the mole fraction of material in solution as a first approximation. It will be further reduced by the water content of the air being passed over the surface being dried. It is further reduced by the area of the surface obscured by dry "scum" forming (stirring keeps this effect small). Eight square meters at room T? You can get 100 g/s maximum. How well the air is dried, and the amount of crap dissolved in the water are going to cut that down.

## Any takers approximating evaporation rate?!!!

If I remember well, the measurement of moisture in a flow of moist air is called psychrometry.
I also remember that the mixture air/water vapor is somewhat particular in the sense that the "Lewis" number is nearly 1. (not completely sure about "Lewis")
I think this "Lewis" number compares heat flow and the energy cost of mass flow (evaporation).
The trick should be now to use the known scaling laws for heat flow and get from that the mass flow.

Still, I am not sure about the impact of the sludge concentration. However I guess that as long as the moisture content is high, the evaporation rate is not influenced. It begin to decreased when the available water is somewhat 'bond' to the solids. Then, this can only be determined experimentally, as a correcino factor. Indeed, it is clear that some solids are hydrophiles and others hydrophobes. It is also clear that water can participate is not kind of gel. Therefore, the detailled chemistry could enter on stage then.

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