Futzing with volume -- help me calculate ABV!

[bernerbrau]

My porter fermented from 1.073 to 1.026, so I have about 6.15% ABV.

My problem is I'm about to add 3 quarts of water, because my original volume was too low.

I'm thinking I can take another hydro reading after adding the water, then another at bottling time, and somehow figure out the final ABV from these four numbers.

Anyone with a good scientific knowledge of the processes behind the gravity/ABV numbers want to give me a hand?

 

[Yuri_Rage]

What is the current volume? That's all you need to know to figure the ABV after diluting it.

 

[bernerbrau]

I don't know the exact volume I suppose I could get a tape measure and guesstimate knowing that it's a 6.5 gallon carboy...

 

[Yuri_Rage]

If you don't know the exact volume, why are you adding 3 quarts?

 

[bernerbrau]

I'm guessing 4.28 gallons from the measurements I'm taking. It's not an exact science since the carboy isn't a perfect cylinder.

I was hoping the drop in gravity would be enough to deduce the original volume. Any chemists?

 

[Yuri_Rage]

FYI, it's a physics problem, not a chemistry problem.

 

[bernerbrau]

Using qbrew, I inferred the "real" volume to be something like 4.25 gallons by stepping back volume on my recipe until the target OG matched my measured OG. But since that's not totally precise, I wanted to see if there was a way to get a more mathematical precision.

 

[bernerbrau]

Eh, I know. Fluid dynamics. But it's usually chemists who work with test tubes and strange liquids

 

[Yuri_Rage]

This might help:

BT - Understanding Specific Gravity and Extract

I'm not sure that you're going to get the answer you need without an actual volume measurement, though.

 

[bernerbrau]

Well, if the SG of water is 1.0, and I am correct in assuming SG is density (mass per volume) compared to water...

Then, first, we have the gravity of water, and a given volume:

Gw = mw/vw = 1
and so the mass of the water mw = vw * 1

Then, we have the gravity of the beer and its original (unknown) volume:

G0 = m0/v0
and its mass m0 = v0 * G0

Finally, given the new gravity, and assuming that mass and volume are additive:

G1 = m1/v1 = (m0+mw)/(v0+vw)
G1 = (v0*G0+vw*1)/(v0+vw) [substitution]
(v0+vw)G1 = (v0*G0+vw) [multiply out]
v0*G1 + vw*G1 = v0*G0 + vw [reduce]
v0*G1 - v0*G0 = vw - vw*G1 [subtract out]
v0(G1-G0) = vw(1-G1) [simplify]
v0(G0-G1) = vw(G1-1) [multiply both sides by -1 and simplify]
v0 = vw(G1-1)/(G0-G1) [divide out]

So my original volume is vw(1-G1)/(G1-G0), where vw is the volume of water added, G0 is the gravity measured before adding water, and G1 is the gravity measured after.

Which means, if I measured a G1 of 1.022, adding 0.75 gallons, with my measured G0 of 1.026, that would mean my original volume was (0.75)(.022)/(.004) = 4.125 gallons

Someone want to check my math??

 

[Yuri_Rage]

Try this formula:

Make sure you use the same units for volume (i.e., 3 quarts is .75 gallons). The SG values are the measurements before and after dilution (SG1 is the SG after dilution - sorry if that's counterintuitive). If I did the algebra correctly, it should yield the volume prior to dilution.

EDIT: Using the numbers from your example above, (1.022 and 1.026), I get the same 4.125 result as your method above. Guess yours is a little simpler.

 

[Gordie]

It was my understanding there would be no math.

 

[bernerbrau]

Hmm, did I mess up my algebra then?

 

[Yuri_Rage]

I don't think you did. We get the same result.

I just used one of the formulas from the page I referenced and did some substitution and factoring. Mine can probably be further simplified (to look exactly like your answer above), but I'm done with algebra for today.

 

[bernerbrau]

Awesome, thanks for your help!!

 

[bernerbrau]

BTW, how did you get the formula graphic? Looks a lot more professional than mine.

 

[Yuri_Rage]

I hacked it together with GIMP and uploaded it to an image hosting site.

 

[bernerbrau]

Ah, a low tech solution to a high tech problem, I like that Thanks again!

 

[Rick500]

Since we're talking about volume (abv), wouldn't it go like this?:

6.15% * 4.25g = 0.261375g (so there's 0.261375 gallons of alcohol, which won't change)
4.25g + .75g = 5g (current volume + added water to get new total volume)
0.261375g / 5g = 0.052275 = ~5.23% abv (volume of alcohol per total volume)

 

[Yuri_Rage]

Yes, once we've determined the volume. Presently, his initial volume is an unknown, so that's what the formulae above are designed to solve.

What's funny is that the fidelity of the formula is probably only about as good as you can eyeball. Differences in SG of 1.0002 change the resulting volume by a few quarts. You'd need a much more accurate specific gravity reading to get better fidelity than +/- a quart or two.

 

[Rick500]

Ah, yeah, I see. Didn't follow the whole thread correctly.

 

[bernerbrau]

Got a second gravity reading of 1.022, putting me at 4.125, which is, sadly, accurate to +/- a whole gallon... So, knowing that I added 0.75 gal, I will be able to figure out volume and ABV measurements at bottling time, when I multiply the number of bottles by 22 ounces.