--- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@...>

wrote:

>

> I need the equation to find out the rate that i can theoretically

> collect distillate from a pot still running no reflux at a certain

> power input. I understand that no still will run without heat loss,

so

> the numbers might be off a little. I was just hoping to be able to

> calculat the collection rate for a particular heat input, without

> having to actually make it. Might also help when trying to determine

> the rate of collection for a given reflux ratio.

>

> -Tyler

>

Borrowed from Mike Nixon's excellent paper in the Distillers/files

section...

http://groups.yahoo.com/group/Distillers/files/Vapor%20Speed%20101%20-%

20Mike%20Nixon.htm

...or TinyURL http://tinyurl.com/y797ow

<quote>

The latent heat of vaporisation of both ethanol and water are about

the same... 40 kJ/gram mole

(this is just a happy coincidence ... other substances have different

LHVs, so please don't generalize)

So 1 kW vaporises 60/40 = 1.5 gram moles of either ethanol or water,

or any mix of them, every minute.

1 gram mole ethanol = 46gram

1 gram mole water = 18 gram

So 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram water every

minute.

1 gram mole of any vapor occupies 22.4 liters at STP (standard temp

and pressure)

Vapor expands when heated, and at around 90 deg C 1 gram mole of any

vapor occupies 30 liters.

So 1 kW, which vaporises 1.5 gram mole of either ethanol or water each

minute gives 1.5 x 30 = 45 liters vapor/minute ie. 45/60 liters/second

But 1 liter = 61.024 cubic inches, so 1 kW will give 61.024 x 45/60

cubic inches of ethanol or water vapor each second, = 45.77 cubic

inches/second

</quote>

Does the brain still hurt, Tyler? :-)

Slainte!

regards Harry