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Calculation for adding sugar?
(Guy Therrien) wrote in message . com>...
> How about the following formula:
> S = W (B - A) / (100-B)
>
> Where S = weight of must to be added to increase must to a desired Brix
> W = weight of must
> B = desired brix
> A = original brix of must
>
> HTH.
> Guy
A follow up on my previous post, trying to get to the bottom of this:
We have:
Vs - starting volume of must
Bs - starting Brix (not BS 
and sugar by weight in a given volume is V*B/100 (V in litres, weight
in kg)
Then the basic formula is:
starting sugar + sugar addition = resulting sugar:
Vs*Bs/100 + dW = Ve*Be/100
(dW is added sugar by weight, e means "ending")
This gives:
Vs*Bs + 100*dW = (Vs+dV)*Be (dV is volume increase, so Ve =
Vs+dV)
100*dW = Vs*Be - Vs*Bs + Be*dV
= Vs*dB + Be*dV
Here is the funny part - the dV is affected by Be, and therefore by
Bs. If we assume that a given weight of sugar increases volume by a
fixed amount, which seems to make sense and is a pretty standard
assumption, then the formula says that you'd need more sugar the
higher your Bs is to get the same dB. If this is the case, you can
figure out how much volume increase you get from 1kg sugar and you'll
have the final formula. But the result seems counterintuitive to me, I
still can't figure out why the starting Brix would matter?
The other possibility is that the Bs doesn't matter. For this to work,
dV would have to proportionally decrease with the rising B, i.e., the
Be*dV should be constant for given dW, Vs, and dB. If this is the
case, you'd have to figure out the formula between dW, Be and dV,
which should be posible by doing some experiments. Again, I can't
really explain why this would be the case, but it makes intuitively
more sense to me. However, it goes against the assumption that volume
increase depends directly on weight, which seems pretty much
universal, so I don't know...
One way or another, some experimentation should tell you which is
correct - hopefully one of them is!
Pp
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