Evaporation under vacuum; surface area

[Jay167]

I run a vacuum distillation unit that is used to distill ethylene glycol. The old glycol is subjected to 25Hg of vacuum at a temp of 275 degrees F. There are burner tubes submerged in the glycol to heat it up. The vessel has a diameter of 36 inches and is 124 inches long. We fill the vessel half full, 18-20 inches high, heat it under vacuum and collect the evaporation. The new set up has us filling the vessel to 31 inches high. I stated that the lower surface area would make it so we are collecting evaporation at a slower rate. Am I right? Is there a formula I can use to prove my assumption. Thanks

 

[mfb]

If the pump is strong enough to keep pressure significant below the pressure where the liquid boils, evaporation will be proportional to the surface area, the proportionality constant will depend on pressure, temperature and chemical composition.
R=k*A if you absolutely need a formula (with the evaporation rate R, the area A and k as constant).

If the pump is weak, then the surface area won't matter, and just the pumping rate matters.

 

[Jay167]

Thank you for taking the time to send a reply. here are the answers to your questions if it helps to get a better answer. I explained that in the old set up, we filled the machine 20 inches high so we had a surface area of 34 inches wide by 124 inches long. We would collect 25 gallons of evaporate (93% glycol 6.2% water .8% light oils) per hour. Now that we fill to 31 inches we only get 16 in cross section by 124 long. I say we will collect slightly under half of what we used to.
Our vacuum pump is a constant 24-27Hg of Vacuum. The temp fluctuates between 270F (burner turn on point) and 275F (burner high limit setting). The usual composition of the glycol mixture is 40% glycol 52% water 8% light oils, other automotive fluids some suspened solids etc. We have a level sensor hooked to a pump that feeds the vessel to keep the fluid level a constant as well.