Contributors to weight of finished wort

[McKnuckle]

Consider if you will, a kettle full of cooled wort. There are no hops in it, only liquid, sugars, and break material.

The apparent volume of the wort, measured with a ruler in the kettle to the nearest 1/16" and visually confirmed in the fermenter, is 2.080 gallons.
The wort weighs 8.049 Kg. Specific gravity measured with a hydrometer is 1.056.

8.049 Kg of pure water would be 8.049 Liters. I’ve been accounting for sugar by dividing this by OG, or 8.049 / 1.056 = 7.622 L of water in the solution. This converts to 2.014 gallons. If I consider my measured volume (2.080 gal), that leaves me with 0.066 gal to account for, a little over one cup. I’m guessing this is the break material.

Does this analysis of the wort components sound correct?

 

[kh54s10]

It seems reasonable, but my question is why you even care?

 

[McKnuckle]

I care because accurate volume measurement helps me hone in on elements of my process, and on my efficiency. I'm not too interested in defending that, but I'd like to be sure I'm understanding the physical science behind the approach, hence this post.

I'm still not sure that my use of SG to determine the proportion of water in the wort is entirely correct.

 

[thekraken]

I break my wort down into masses of water and "extract" using a couple functions derived from this conversation: https://www.homebrewtalk.com/showthread.php?t=531216&page=2

It might interest you.

 

[ajdelange]

Consider if you will, a kettle full of cooled wort. There are no hops in it, only liquid, sugars, and break material.

The apparent volume of the wort, measured with a ruler in the kettle to the nearest 1/16" and visually confirmed in the fermenter, is 2.080 gallons.
The wort weighs 8.049 Kg. Specific gravity measured with a hydrometer is 1.056.

That tells you that you have 8.049/(1.046*.998203) = 7.709 L equal to 7.79/3.785 = 2.058 gal of liquid at room temperature assuming you measured SG at room temperature. More to the point 1.056 is 13.805°P meaning that you have 8.049*.130847 1.05319 kg extract and 8.049*(1 - .130847) = 6.99581 kg water. At 20 ° that's 6.99581/0.998203 = 7.0084 L or 1.856 gal.

 

[McKnuckle]

Thank you AJ - you used 1.046 in your calculations but my post cited 1.056. I'll redo the math and try to parse through this. Appreciate your insight.

 

[ajdelange]

Easily enough fixed and I've done so.