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A liter of vapor is equivalent to....

These usually have the means for returning the spirits to the column and usually allow metering of the take off to help in the reflux.

A liter of vapor is equivalent to....

Postby Destructo Mutt » Fri Jul 17, 2009 4:07 pm

can someone check my math?

1 kilowatt vaporizes 1.5 g/mol of either water or ethanol (or a combination of the two) every minute.

molar mass of ethanol is 46.07 g/mol, density is 0.789 g/cc at STP (yes, it drops as temperature climbs)

molar mass of water is 18.0 g/mol, density is 1.0 g/cc at STP (yes, it drops as temperature climbs)

for ethanol then: 1.5 g/mol x 46.07 g/mol = 69.1 g/mol / 45 liters => 1.5 g = 1 liter of vapor, or 1.95cc = 1 liter of vapor.

for water: 18.0 g/mol x 1.5 g/mol = 27 g/mol; 27 g/mol / 45 l/min => 0.6g = 1 liter of vapor, or 1 liter of vapor equals 0.6 cc of water.

obviously it's going to be less at temperatures greater than STP, but this somewhat close (water drops to 0.96 at 93C).
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Re: A liter of vapor is equivalent to....

Postby Harry » Fri Jul 17, 2009 6:24 pm

1 gram mole ethanol = 46gram
1 gram mole water = 18 gram

At around 90 degC 1 gram mole of any vapor occupies 30 liters.

So divide the weight by 30 to get how much is in 1 liter

1 liter ethanol vapor = 1.53g a/w when condensed.
1 liter water vapor = 0,6g w/w when condensed


So your figures are about right. But you coulda saved yerself some headache.


Compare it with this article extract from Mike Nixon that's been in the Distillers files since Mar/2005 (told ya you was missin' lots by not bein' a member :D :D ).

<extract>
1 Watt = 1 Joule/sec
So 1 kW = 1 kJ/sec
So a 1 kW heater gives 60 kJ/minute


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
</extract>
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Various experimental gear.

Re: A liter of vapor is equivalent to....

Postby Zymurgy Bob » Fri Jul 17, 2009 10:12 pm

DestructoMutt wrote:can someone check my math?

1 kilowatt vaporizes 1.5 g/mol of either water or ethanol (or a combination of the two) every minute.

molar mass of ethanol is 46.07 g/mol, density is 0.789 g/cc at STP (yes, it drops as temperature climbs)

molar mass of water is 18.0 g/mol, density is 1.0 g/cc at STP (yes, it drops as temperature climbs)

for ethanol then: 1.5 g/mol x 46.07 g/mol = 69.1 g/mol / 45 liters => 1.5 g = 1 liter of vapor, or 1.95cc = 1 liter of vapor.

for water: 18.0 g/mol x 1.5 g/mol = 27 g/mol; 27 g/mol / 45 l/min => 0.6g = 1 liter of vapor, or 1 liter of vapor equals 0.6 cc of water.

obviously it's going to be less at temperatures greater than STP, but this somewhat close (water drops to 0.96 at 93C).


The math's ok, once I figured out some units confusion (it's 18.0g/mol x 1.5 mol = 69.1g) and that 45 liters is a rough approximations of the volume of the 1.556mol EtOH=47.62L and 1.476mol water = 45.17L.

So 1.91ml of EtOH (measured at STP) boils to 1L of steam at 100C, and .588ml of water does the same to 1L of steam at 100C.

What do we do with those numbers?
Zymurgy Bob, a simple potstiller http://www.kelleybarts.com/zymurgy-bob- ... e-spirits/

You can make whisky in a reflux still, you can make vodka in a potstill,
and you can eat chicken noodle soup with a crescent wrench. But...
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Re: A liter of vapor is equivalent to....

Postby Harry » Fri Jul 17, 2009 10:42 pm

Zymurgy Bob wrote:What do we do with those numbers?



I'm not sure what Dmutt has in mind but these are possibilities...

You decide how much top product you want to produce, then figure how much energy is needed to do it.

You can also interpolate how much bottoms will arrive at the boiler during the distilling period. But both of these will only be close approximations as neither product (tops or bottoms) is a pure product. They both contain percentages of the other substance. And distilling always has 'loss' issues, small but they do matter.

You can also figure what volume of column space is needed to hold said traffic (both vapours & liquids) during operations. IOW it's the beginnings of reflux still design, before you start cutting copper. Just like Mike Nixon intended.
Slainte!
regards Harry
http://distillers.tastylime.net/library/

Winning the hearts & minds; one post at a time.
(you will be assimilated. resistance is futile.)
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Location: Paradise aka Cairns Qld Australia
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Various experimental gear.

Re: A liter of vapor is equivalent to....

Postby Harry » Fri Jul 17, 2009 11:03 pm

While we're on the subject of column design...how many of you make allowance for the physical volume of the mesh packing? If it takes up 20% of the column empty volume, you need to add ~25% more height to arrive at you theoretical numbers, otherwise your power in/ vapour speed balance suffers. Because there's less physical volume than the theoretical numbers, the vapour will be travelling too fast for thorough separation, so purity is reduced.

Either 25% more filled height, or a power reduction and consequently less output per timeslice will keep the purity up to spec.

To determine your packing volume, plug the empty column bottom and fill it with water. Pour the water into a measuring jug and note it. Now pack your column and repeat the water fill from the jug. What's left in the jug is the volume of column space the packing takes up. It's easy with a millilitre graduated jug. Then figure the ratio of packing to empty column volume as a percentage and add sufficient column length (again thinking 'filled') to compensate, thus bringing your theoretical vapour speeds, volumes and power back into design specs.

[Edit] Failure to observe and allow for the above volume reduction is probably the reason why some people cannot achieve the azeotrope. The 95.6% mark is somewhat undershot by 2 or 3 percent due to entrained impurities through inadequate separation caused by off-spec design.
Slainte!
regards Harry
http://distillers.tastylime.net/library/

Winning the hearts & minds; one post at a time.
(you will be assimilated. resistance is futile.)
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Location: Paradise aka Cairns Qld Australia
Equipment type: 25 Lt 1500W elec. boiler, 2" x 40" packed column VM, x-flow O/head HX, crimp-path Liebig HX, VM proportioning valve, all in copper.
Various experimental gear.

Re: A liter of vapor is equivalent to....

Postby Destructo Mutt » Sat Jul 18, 2009 1:41 am

DestructoMutt wrote:1.95cc (of liquid ethanol) = 1 liter of (ethanol) vapor.
DestructoMutt wrote:1 liter of (water) vapor equals 0.6 cc of (liquid) water.


these are the ones i'm looking at. and mr. nixon hadn't spelled out the one for ethanol, so i did have to do the math, at least to make sure i understood what i was doing. i was contemplating the apparent "auto shut down" feature of VM and CM heads. but i didn't know how much liquid was the equivalent to a liter of vapor.

45 liters per minute => 750ml per second; if it were 100%ABV in the boiler, then you would be getting ~1.5ml liquid per second, at no reflux; with a 1:9 reflux ratio you would get ~0.15 ml/sec of liquid. however it's not 100%ABV in the boiler. and as the wash is depleted of alcohol, less and less of it is present for refluxing....

water would be ~0.05ml/sec of liquid at 9:1 RR.....which is about what you get when the "auto shut-down" feature of CM and VM rears it's head. so it's not so much a "shut-down" as it is that the system is doing what it's designed to do --- letting 1/10th (or whatever RR you have decided upon) of the vapor "escape" as product. and for water, that's very little.....

Thank you for coming back Harry.
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