## Re: Rate of collection equation

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• ... so ... Borrowed from Mike Nixon s excellent paper in the Distillers/files section...
Message 1 of 6 , Nov 1, 2006
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--- 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
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
• I knew that!!!!! :) blanikdog ... From: Harry To: new_distillers@yahoogroups.com Subject: [new_distillers] Re: Rate of collection equation Date: Wed, 01
Message 2 of 6 , Nov 1, 2006
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I knew that!!!!! :) blanikdog

----- Original Message -----
From: Harry
To: new_distillers@yahoogroups.com
Subject: [new_distillers] Re: Rate of collection equation
Date: Wed, 01 Nov 2006 09:13:12 -0000

--- 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
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

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My SOARING, SAILING AND SKYDIVING web page
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• Damn you Harry! I had to break out the headache pills on that one! lol. Ok, well i think you helped explain to me about why the collection rate slows down on
Message 3 of 6 , Nov 1, 2006
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Damn you Harry! I had to break out the headache pills on that one!
lol. Ok, well i think you helped explain to me about why the
collection rate slows down on my pot still when the tail start to come
through. If 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram
water every minute, that is quite a noticeable difference, and a good
way to tell when your tails are coming. However, i'm still trying to
wrap my head around those equations. You spit all that out just to see
how fast my head would spin, didn't you? As far as calculations go,
1.5 x 46 = 69grams = 2.43oz, now this is by weight. Does 16 fluid oz
weigh 1 pound? Depends on what the fluid is. Anyway, it would come
out to 71.97 ml. Does that sound about right? If so, does that mean
that 2kw would be 3 x 46 = 138?

-Tyler

--- In new_distillers@yahoogroups.com, "Harry" <gnikomson2000@...> wrote:
>
> --- 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
>
• ... come ... good ... to ... see ... oz ... Sore head or no, methinks you re beginning to understand it. :-) Slainte! regards Harry
Message 4 of 6 , Nov 2, 2006
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--- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@...>
wrote:
>
> Damn you Harry! I had to break out the headache pills on that one!
> lol. Ok, well i think you helped explain to me about why the
> collection rate slows down on my pot still when the tail start to
come
> through. If 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram
> water every minute, that is quite a noticeable difference, and a
good
> way to tell when your tails are coming. However, i'm still trying
to
> wrap my head around those equations. You spit all that out just to
see
> how fast my head would spin, didn't you? As far as calculations go,
> 1.5 x 46 = 69grams = 2.43oz, now this is by weight. Does 16 fluid
oz
> weigh 1 pound? Depends on what the fluid is. Anyway, it would come
> out to 71.97 ml. Does that sound about right? If so, does that mean
> that 2kw would be 3 x 46 = 138?
>
> -Tyler

Sore head or no, methinks you're beginning to understand it. :-)

Slainte!
regards Harry
• Using the equation 1.5 x 46 = 69grams = 2.43oz = 71.97ml to estemate the amount of distillate produced per hour in a still that theoretically has no heat
Message 5 of 6 , Nov 10, 2006
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Using the equation "1.5 x 46 = 69grams = 2.43oz = 71.97ml" to estemate
the amount of distillate produced per hour in a still that
theoretically has no heat loss, I am going to attemp to do some
thinking out loud.

If i made a 50 gallon (189.26ltr) wash at 14% ethenol, and ran it
through a stripping run to collect 62.9ltr of 40% ethenol, i need to
find out how long it would take.

If i am using a 3kw electric heated boiler with a non-packed stripping
column, suing the equation above, (1kw = 71.97ml per min.) x 3 = (3kw
= 215.91ml per min.), i assume that i sould use the equation: "62.9 /
.21591", which would give me 291.3 minutes = 4.855 hours.

So if my messed up calculations are correct, i can strip a 50 gallon
wash from 14% to 40% in 4.855 hours at 3kw heat input.

Did i do that right?

If that is right, then a 4.8kw heater should distill it at 345.456ml
per min, and get it done in 3 hours.

I guess at this point i need to run my still at that heat input, and
measure the collection rate. When i compare that to my calculations,
it sould let me know how much heat loss my still has.

I was thinking, If i am using a non packed, non refluxing column, i
shouldn't really have to worry about the column "choking", right? It
seems to me that i just need to make sure that i can effectively slow
down and cool the vapors coming into the condenser so i don't have
steam coming out.

Well now that i gave myself a headache, could someone tell me if i am
right on any of this, or do i need to go back to middle school?

-Tyler
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