## A Simple Question. Mike? Tony? Anyone?

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• Further to that discussion about low-power continuous hobby stills without large boilers... I think I have the answer to this on paper, but it would be nice if
Message 1 of 3 , May 3, 2005
Further to that discussion about low-power continuous hobby stills
without large boilers...

I think I have the answer to this on paper, but it would be nice if
someone with the right training could say yea or nay. Anyway,
here's the question and guidelines...

Q) Calculate the amount of additional heat required to separate
this binary mixture into azeotrope and bottoms: 20 litres of 10:90
ethanol-water @ 70°C.

Working parameters (assumptions):

Assume an ideal lossless system (for simplicity).

The specific heat capacity of water is 4.18 kJ.kg-1.K-1.
The specific heat of vaporization of water is 2260 kJ.kg-1.

The specific heat capacity of ethanol is 2.44 kJ.kg-1.K-1.
The specific heat of vaporization of ethanol is 838 kJ.kg-1.

Given that a 10:90 ETOH:H2O mixture boils at 93°C, MY rough
calculations determine
The specific heat capacity of 10:90 ETOH:H2O is 3.63 kJ.kg-1.K-1.
The specific heat of vaporization of 10:90 ETOH:H2O is 1807 kJ.kg-1.

If any of the above is off, let me know.

Slainte!
regards Harry
• Harry wrote: Subject: [Distillers] A Simple Question. Mike? Tony? Anyone? Further to that discussion about low-power continuous hobby stills without large
Message 2 of 3 , May 3, 2005
Harry wrote:
Subject: [Distillers] A Simple Question. Mike? Tony? Anyone?

Further to that discussion about low-power continuous hobby stills
without large boilers...

I think I have the answer to this on paper, but it would be nice if
someone with the right training could say yea or nay. Anyway,
here's the question and guidelines...

Q) Calculate the amount of additional heat required to separate
this binary mixture into azeotrope and bottoms: 20 litres of 10:90
ethanol-water @ 70°C.

Working parameters (assumptions):

Assume an ideal lossless system (for simplicity).

The specific heat capacity of water is 4.18 kJ.kg-1.K-1.
The specific heat of vaporization of water is 2260 kJ.kg-1.

The specific heat capacity of ethanol is 2.44 kJ.kg-1.K-1.
The specific heat of vaporization of ethanol is 838 kJ.kg-1.

Given that a 10:90 ETOH:H2O mixture boils at 93°C, MY rough
calculations determine.
The specific heat capacity of 10:90 ETOH:H2O is 3.63 kJ.kg-1.K-1.
The specific heat of vaporization of 10:90 ETOH:H2O is 1807 kJ.kg-1.

If any of the above is off, let me know.
===================
Try working it with kJ/mol ... it's much easier!
Latent heat of vaporization of water is 40.639 kJ.mol and of ethanol is
39.33 kJ/mol, so they are both pretty well much the same for any mixture ...
take a mean of 40 kJ/mol.

In volume, 1 mol of water is 18.016 ml, and ethanol is 58.24 ml, so the same
amount of heat is needed to vaporize either volume. Just take it from there
and the answers drop out fairly painlessly.

What .... you want ME to do it????? :-))

All the best,
Mike N
• ... ethanol is ... mixture ... ... so the same ... from there ... Many thanks, Teech! :-)) Hmmm...guess I ll have to brush up on my self-taught Chem.Eng. Eh
Message 3 of 3 , May 3, 2005
--- In Distillers@yahoogroups.com, "Mike Nixon" <mike@s...> wrote:

> Try working it with kJ/mol ... it's much easier!
> Latent heat of vaporization of water is 40.639 kJ.mol and of
ethanol is
> 39.33 kJ/mol, so they are both pretty well much the same for any
mixture ...
> take a mean of 40 kJ/mol.
>
> In volume, 1 mol of water is 18.016 ml, and ethanol is 58.24 ml,
so the same
> amount of heat is needed to vaporize either volume. Just take it
from there
> and the answers drop out fairly painlessly.
>
> What .... you want ME to do it????? :-))
>
> All the best,
> Mike N