Water Volumes

[sasky7777]

I am tinkering around with some recipes, as I make the eventual transition to All Grain. My question is with a big beer. Let's say I need something like 20 lb of grain to hit my OG, this calculates to over 7 gallon of water in the mash at 1.5 qt/lb. After I sparge, I am way over the volume I need for a 6.5 gallon boil and eventually a 5 gallon batch. This may sound be a stupid question, but what do I need to change, or what am I not seeing?

 

[BigB]

You are forgetting that the grain will absorb quite a bit of mash water. Further, with that much grain, you would have a much smaller sparge. Finally, you would probably want to do a 7 gallon boil for 90 minutes anyways to drive off all DMS. EDIT: One other thing to consider is reducing the water:grain ratio to 1.25 qt/pound

 

[Hammy71]

To echo what BigB said. For many 'big' beers, like an Imperial Stout, you will probably end up doing a 90 minute or longer boil to get you down to your target 5 gallons.

 

[Misplaced_Canuck]

I use 1.1qt per pound, never had a stuck mash... You can go down to 1.0qt/lb easily. Under that it becomes a bit of a stiff mash.

M_C

 

[sasky7777]

Just to be sure I got this right. I guess I should fill a measured amount in my pot (7 gallons?) to see what my losses are over 90 minutes, and then when I mash out, just collected the first running plus whatever sparge I need to make it to the correct volume that will boil down to 5 gallons over a 90 minute boil. How will I know that I hit my OG, or is it just experimentation until I know my efficiency of my set-up?

 

[jeffmeh]

For evaporation rates in a given kettle, there will be some variability based upon boil vigor and relative humidity, but a good starting estimate is 0.00729 gallons, per square inch of the kettle mouth, per hour.

Works for me.