Any takers approximating evaporation rate?

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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
 

Gokul43201

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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 ?
 

Bystander

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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.
 
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part of answer maybe in 'psychrometry'

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.
 

Gokul43201

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Bystander said:
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.
This approximation is likely a good one at low concentrations, but the OP's sludge is more than 50% solid and getting more concentrated as drying happens. This is not a solution or suspension anymore, and the rate of diffusion of water through the solids may not be large enough to assume that water at the surface is "instantaneously" replenished. This process will be diffusion limited beyond some "concentration".
 

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Until the OP is expanded with a more detailed composition of "the sludge," there is not a whole lot to be said --- absolute statements about "50% solids" are particularly dangerous --- try it with concrete sometime.
 
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MOre information on this problem

I really appreciate the replies! I agree the it would be much easier to experimentally evaluate the rate and that is probably what I am going to do ultimately but I am trying to keep down the cost of unfruitful equipment.

I guess to simplify the question and get me into the ballpark it would be easier to assume a container full of pure water 2 meters deep having a surface area of 8 square meters. Use the vacuum above and the 180 cfm airflow and assume 50% humidity of the air before the pressure is reduced.

I understand that solute will raise the boiling point and make a big difference from the above assuptions as would any oils in the solution etc. I am trying to get into the ballpark. I am only trying to thicken this waste enough to make it landfill compatable. I also have a low enough volume of material that if it takes a week to decrease the moisture content to a satisfactory level it will be ok.

I can also get a different machine that has comparable airflow and can pull up to 24 inches of mercury vacuum but it will cost more and may not be necessary for this application.

I believe that the kind of processes are used in the food processing industry for concentrating tomato juice to ketchup etc.

Thank you everyone for your time and help.

Steven
 
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Re: Any takers approximating evaporation rate?!!!

can anyone help me how to calculate the evaporation rate of water at 95 degree celsius (kg/hr.m2 )
 

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