Yes, the destillasjon setup should be OK. Actually it is very simple
since this method does not require any analysis of the distillate.
Maybe I should try the full distillation method just to be sure.
I have also done a Rayleigh (single pot) distillation calculation to
verify that the remaining alcohol is negligible. And this was
confirmed (less than 0.05 vol% before volume adjustment). I agree that
I should use the calculation method which comes out of the analysis
(yours and mine)until someone points to any error in them. But it is
strange that this other correlation is around on several web pages. I
have tried to send these people a mail asking them how it is derived
or from where their source is. But no replies so far.

(Ben Rotter) wrote in message . com>...
> > Does someone know which is the right calculation method and why there
> > are two different methods around for the same procedures? And does
> > somebody know how the numbers in the tables are calculated?
>
> The first calculation (from Monash Scientific) looks right to me.
>
> I derive it as follows (and get the same answer as yourself and the
> Monash page):
>
> The SG of the unboiled wine, SG1 = (Vw/Vt*Pw + Va/Vt*Pa)/Pw
> whe
> Vw = volume of water in sample
> Va = volume of alcohol in sample
> Vt = total volume of sample = Vw + Va
> Pw = density of water
> Pa = density of alcohol
>
> When the sample is distilled, the alcohol will be driven off. This
> will leave the original water content but the Vt of this distilled
> sample will now be equal to Vw. (This assumes no reconstitution with
> water of course, which is different to the experiment, but the concept
> for calculation purposes is correct). Thus,
>
> The SG of the distilled sample, SG2 = (Pw*Vw/(Vt-Va))/Pw
>
> Skipping all the algebra in between leads to:
>
> SG1 - SG2 = Va * (Pa/(Vt*Pw) - 1/Vt)
>
> So Va = (SG1-SG2) / (Pa/(Vt*Pw) - 1/Vt)
>
> In your case (units are litres & kilograms), Va = -0.017 / ((0.789 /
> 1*0.25) - 1/0.25)) = 0.020014 L
>
> The alcohol content = Va/Vt * 100 = 0.02014/0.25*100 = 8.05 %
>
> I don't know why the second reference (http://www.winemakeri.com) is
> different, but I'm not sure I'd trust it.
>
> > The alcohol % derived from the tables looks much closer to the
> > expected value going by must initial gravity.
>
> Are you sure your distillation setup and procedure are correct?
>
> Ben
>
> Improved Winemaking
> http://members.tripod.com/~BRotter/