Well, you can’t; that isn’t enough information. But the rest of the variables don’t have a huge impact. I assumed a 5.25 gal post-boil volume, and a weighted average potential extract of 36 point-gal/lb. So the grist weight would be: 5.25*70/(36*.72) = 14.2 lb. That sets the grain absorption at 1.7 gal, at which point calculating lauter efficiency for a double batch sparge is simple. For a 6.0 gal or 7.0 gal pre-boil volume, with equal runnings:

0.9*((3.0/4.7)+(3.0/4.7)(1.7/4.7)) = 0.78

0.9*((3.5/5.2)+(3.5/5.2)(1.7/5.2)) = 0.80

You can see that pre-boil volume doesn’t make much of a difference. The 0.9 is a fudge factor to account for the volume contributed by dissolved sugars. So the conversion efficiency is the overall mash efficiency divided by the lauter efficiency: 0.72/0.79 ~ 91%. Throwing in a little wiggle room for all the assumed values, you can still be almost certain the conversion efficiency was in the 85-95% range.

Kai’s spreadsheet would let you determine that, BTW. You just have to plug in all your numbers and then adjust the conversion efficiency until the OG lines up with what you got.[/quote]

Thanks for this, I was missing how to calculate lautering efficiency. I also figured out how to maniupulate Kai’s batch sparge simulator by adjusting CE to figure out brewhouse efficiency.

Should I add deadspace to the above formula? Or, doesn’t it make a big difference in lautering efficiency? Also, it appears the simulator assumes you drain before adding any sparge water. What I do is add water to ensure the first drain is 1/2 my preboil amount. To simulate this should I add the top-off water to the mash water field?

Thanks again!

Mark