Originally Posted by janzik
Just as a reference, I generally get 80% eff on my batches and thats how I write it up in Beersmith.
I normally only make 5.5 gallon batches, so what I did was take one of my recipes and scaled it to 11 gallons.
Based on Mosher's tables, he provides a 66.6% to 33.3% split and a a 50/50. I'd like to do the 50/50, so 2 5.5 gallon batches into 2 brew kettles.
I was under the impression that half of my total volume would make a post boil 1.084 5.5g and a post boil 1.042 5.5g. If I'm doing 60 minute boils, I would need 6.5 gallons for each pre boil and if I did a 90 minute boil I would need 7 gallons each (I only say 90 because my original IPA recipe was a 90 minute mash/boil, but I don't care one way or the other what I do here).
Since I've never done this and it's been a while since I've measured 1st running gravity, I'm not sure what to expect to actually happen, but was basing it on Mosher's table.
the way that seems to work is he gives you a formula for figuing out how high a gravity wort you need by adding your two batches (volume x gravity in P) + (volume x gravity in P)= then divide by the total volume. so in you case:
5.5 gals x 20 p wort (.083)= 110* gal
5.5 gals x 10.5p wort(.042) = 57.75*gal
110 + 57.75 = 167.75 total * gals
167.75 / 11gals. batch = 15.25 p gravity wort (.062-3) so 11 gals of 1.062-3 wort is needed.
you may think that you could just drop half the volume first runnings and pull the 20 Plato wort out. But according to his other formula..
Batches split into two equal volumes have a different ratio: ~58% of the extract will be in the first half of the runoff and ~42% in the second half. For example, a 11-gal master batch designed to produce 15.25 °P wort contains 167.75 °gals (11 X 15.25 = 167.75). The first half will be 17.64 °P (167.75 X 0.58/5.5 =17.637P).073; The second half will be 12.81°P (167.75 X 0.42/5.5 = 12.81 P).051.
So according to that, you wont be able to pull even 5 gallons (4.82 gals) of first runnings and stay at or near 20*Plato (1.083). You still need to account for grist absorbtion, tun loss, evaporaton on top of that. You would directly Top Off the boiler for first runnings evaporation water loss with this method with acified water.
My thought is you'll pull around 4 and some gals. at 1.083. with a 062 063 wort then 6++ lesser wort.
So you'd need to collect more then half and boil it down then add extra sparge for the second beer to come out where you are heading.
Maybe this is where his 66% 33% formula comes in. Your first runnings would be 66% of the total batch size, your second runnings would be the 33/34% then add sparge to the second runnings.
The first runnings extra volume would have to be figured in on part of your evaporation
I would think it would be more like 6.25 gals boiled to 5.5 and 4.75 gallon sparge water added to hit 5.5 finished.
What do you guys think?