Black Island Brewer
An Ode to Beer
OK, 20 lbs of grain will absorb 2-1/2 gallon of water (1/8 gallon per pound). Water weighs 8.3 lb per gallon so that's 20.75 lb for a total of 40.75 lb. To get back to 20 lb, you'll need to remove 20.75 lbs of sugar from 20 lbs of grain.
Something's not adding up.
No, you need to remove a combination of 20.75 pounds of sugar AND water.
The assertion was made that you'll never be able to squeeze all the water out of the grain, which is true. But you also don't have to squeeze it all out in order to end up with the same weight in your bag than you started with.
The dry-basis course grind is a measure of how much non-starch (husk & protein) remains behind from a congress mash. So if a grain has an 80% DBCG, then one pound of grain would yield .2 pounds of remaining material once it was dried back to it's original moisture level.
Of course, we don't get that ideal, but we get close: 60, 70, 80, 90% depending on our mash/lauter efficiency. So assuming 75%, that pound of grain mashed, lautered and dried would be reduced to 0.4 pounds (0.8 pounds x .75 efficiency = .6 pounds extracted, 1.0-0.6=0.4 pounds) So in a 20 pound mash, at the end of lautering, we'd have 20.75 pounds of water and 8 pounds of husk/protein/unconverted starch. Squeeze one gallon of water out of the bag and you're pretty much back to your initial grain weight.
Of course, maybe I'm completely wrong.