Attn. all math geeks

[mr.jay]

Can somebody please check my math, here. I'm working with a new kind of product, and my figures don't seem 100% accurate. Solve for x:

A train leaving Michigan is travelling south at 72 MPH....Oops, sorry, wrong problem. Here it goes:

The fruit = 11.234 # per gallon @ 70 Brix

10 gallons will be added to 7.5 Bbls of beer.

Q) 10 gallons of fruit will increase 7.5 Bbls of beer (x) degrees plato.

A) Assuming that 1# of sugar raises 1 gallon of water 1.046, I came up with
2.3 degrees plato.

Am I close?

[pennbrew2]

11.234 # per gal X 10 gal X .7 = 78.6 # of extract you're adding

78.6 lbs / 7.5 bbl = 10.5 lbs. extract per bbl

Using my handy Siebel Brew Computer (everyone should own one of these, they're only a few bucks),

12 deg. P wort has 32.4 lbs. extract per bbl.

if you add 10.5 lbs. extract per bbl,

42.9 lbs/bbl equals about 15.65 deg. P.

So in this example, it added 3.65 deg. P.

I didn't account for the volume increase (almost a half bbl) of the 10 gallons, but you get the idea.

---Guy

[pennbrew2]

Actually, you should account for the volume increase because it is significant.

So, given 7.5 bbl wort at 12 deg P,

you have 7.5 x 32.4 lbs/bbl = 243 lbs. total extract before fruit addition;

add 78.6 lbs. extract from the fruit gives you 321.6 lbs. total extract.

7.5 bbls plus 10 gallons gives you 7.823 bbls.

321.6 / 7.823 is 41.1 lbs. extract per bbl.

Back to the Siebel Brew Computer, 41.1 lbs/bbl equals a shade over 15 deg. P.

So in this more precise example, it raised the gravity about 3.0 deg. P.

---Guy

[campbell.brian]

Stumbled on this old post and figured I would add my $0.02.

If you want the contribution in gravity to 7.5 BBLs, work the math as if you were adding 7.5 BBLs of water:

(7.5 + 10/31) * D°P = 10/31 * 70°P

°P = (10/31)(70) / (7.5 + 10/31)

°P = +2.89