I'm going to attempt my first all-grain with this recipe and I thought I would go for broke with a 11 gal. batch. Have John Palmer's book but not figuring a few things out, like his sparge calcs. I've read on the beginner forum to figure .5 gals per pound of grain for sparge, so with 22# pale, 2# Munich, and 1.5# Crystal, or 25.5# x .5 gals. equals 12.75 gals of sparge water. Some calculators assume I know how many gallons I want for initial boil, which I don't. Does this seem right? (this assumes you throw the crystal grain into the mash too - right?)
Also, EdWort mentions sparging 3 times. I assume that means dividing that 12.75 gallons into roughly 3, or a bit over 4 gallons per sparge? These are probably best answered elsewhere, but since the quantities are sort of recipe specific, hope you don't mind. cheers, JD
[edit] I fired up Brewtarget for the first time to see what I could do with that. (I'm on linux so not so easy to deal with some other softwares). I had it calculate my target boil size with target batch size at 11 gallons and it came up with 12.6 gallons for target boil size. For conversion, it says about 8 gallons and for sparge it says 8.2 gal. That sounds more realistic. Curious if I'm on the right track? thanks much, JD