# Evaporation/Distilling/Flow rate calculations

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I was unable to cut and paste my topics questions (probably my virus scanner), so I have attached my queries in a word doc...

They are technical in nature, so if I have taken liberties - I apologize.

Would appreciate any technical guidance or reference.

Regards,

David S

EvaporativeFlowRate&TimetoSpec.doc

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The liquid you are distilling contains no alcohol?

You seem to mention evaporation, and are focused on the geometry of the distillation vessel, however useful vaporization will be as a result of boiling and not evaporation. So, geometry is of less importance, and the critical factor with regards to timing will be how much power you have available for your heating element (which you do not mention).

You may run into issues with the final sugar content of your wash, with internal elements I suspect you may caramelize and burn on the elements.

Are trying to reduce the content of your liquid to 1/3rd of the original volume through the process, using only evaporation? I suspect this process would take significant time as the vapor would likely condense on the vessel walls above the liquid, because you'll have a difficult time generating sufficient vapor volume to push through to the condenser.

The other interesting scenario is that your volatile flavor components might find their way distilled out in the very early stages of the process, probably within the first gallon or two processed, if I were to wager a guess.

Sounds interesting, do you need to be vague to protect your commercial process or can you share more details?

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But to your original question, some very rough napkin math - assuming 10kw boiler power, 225 liters volume, no efficiency losses, and a boiler content of nothing but pure water, you can expect to vaporize approximately 15.86 liters of water per hour. Again, I have no idea the contents of your boiler, so this is based on water alone. (36,000kJ/hr / 2,270kJ/kg = 15.86kg/hr)

Assuming starting temp is 15c and 100c is the ending temp, 225 liters, the same 10kw boiler power, you can expect approximately 2.25 hours for heat up.

Putting it together, you'll need to reduce 225 to 75 liters for the liquid in the boiler to increase from 3 to 9 brix. This means you need to boil off 150 liters, which will take 9.5 hours (again with the 10kw power, no losses). So a total of 11.75 hours, but more realistically with losses probably nearer to 13 or 14.

This also assumes that your condenser can handle 100% of the input power and you don't need to reduce the input heat. Some more rough napkin math puts the condenser sizing somewhere around .5 meter of 10 sections of .5 inch pipe in shell (this will largely depend on your input water temps and potential flow rates.

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I would say the "true" calculations are more than likely kept secrets or those developed by paying for them.

You are left with empirical equations which also use "constants" that depend on the design of the hardware overall heat transfer coefficient- a ring around the rosie.

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Cheers James,

MG Thermal Consulting - there not big on revealing how - just selling the product - but I need to get close and rely on real time operational measurements.

James - doc attached

trouble cutting and pasting or logs out if to long typing.

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• 2 months later...

But to your original question, some very rough napkin math - assuming 10kw boiler power, 225 liters volume, no efficiency losses, and a boiler content of nothing but pure water, you can expect to vaporize approximately 15.86 liters of water per hour. Again, I have no idea the contents of your boiler, so this is based on water alone. (36,000kJ/hr / 2,270kJ/kg = 15.86kg/hr)

Assuming starting temp is 15c and 100c is the ending temp, 225 liters, the same 10kw boiler power, you can expect approximately 2.25 hours for heat up.

Putting it together, you'll need to reduce 225 to 75 liters for the liquid in the boiler to increase from 3 to 9 brix. This means you need to boil off 150 liters, which will take 9.5 hours (again with the 10kw power, no losses). So a total of 11.75 hours, but more realistically with losses probably nearer to 13 or 14.

This also assumes that your condenser can handle 100% of the input power and you don't need to reduce the input heat. Some more rough napkin math puts the condenser sizing somewhere around .5 meter of 10 sections of .5 inch pipe in shell (this will largely depend on your input water temps and potential flow rates.

James,

Not questioning your math abilities, but how did you come up with these numbers? My napkin just has a ring from condensation on my beer

Thanks,

NAB

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Jesus I have no idea, I don't even remember typing this. Clearly I was on my fourth or fifth napkin at the time.

KW = (Liters * Delta T) / (790 * Heat Up Time in Hours)

10kw = (225 * 85) / (790 * 2.4)

It would take 2.4hrs to heat 225 liters of water from 15c to 100c. (I'm guessing I was using 80 and not 85 for delta T above)

10kwh is 36,000kJ

Latent heat of evaporation of water is 2270kJ/kg

At 10kw, you can boil off 15.86kg in an hour (or 15.86 liters in an hour).

150l / 15.86l/hr = 9.5hr

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ok I'm following you. Now, why do you know this information? I suddenly feel like I'm behind the times a bit.. lol actually I could see that being really useful in still designs

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