PseudoChef
Well-Known Member
So I'm looking to make the PM/AG plunge using the touted 5 gallon converted cooler (AG for the "normal beers" and more of a PM (but more mash than extract) for the big beers.) However, in reading Palmer, I can't get a good grasp on beginning boil volumes (forgive me if I'm missing something).
"Sparging is the rinsing of the grain bed to extract as much of the sugars from the grain as possible without extracting mouth-puckering tannins from the grain husks. Typically, 1.5 times as much water is used for sparging as for mashing (e.g., 8 lbs. malt at 2 qt./lb. = 4 gallon mash, so 6 gallons of sparge water). The temperature of the sparge water is important. The water should be no more than 170°F, as husk tannins...etc."
So you start with that 10 gallons wort and boil down to 5? So if the total grain bill is 10 lbs or so (touching the upper limit of space in the 5 gallon converted cooler) that means 2 qt/lb = 5 gallon mash and 7.5 gallon sparge? So now you're boiling 12.5 gallons down to 5?
Sorry if this simple seeming question was too long winded and thanks in advance.
"Sparging is the rinsing of the grain bed to extract as much of the sugars from the grain as possible without extracting mouth-puckering tannins from the grain husks. Typically, 1.5 times as much water is used for sparging as for mashing (e.g., 8 lbs. malt at 2 qt./lb. = 4 gallon mash, so 6 gallons of sparge water). The temperature of the sparge water is important. The water should be no more than 170°F, as husk tannins...etc."
So you start with that 10 gallons wort and boil down to 5? So if the total grain bill is 10 lbs or so (touching the upper limit of space in the 5 gallon converted cooler) that means 2 qt/lb = 5 gallon mash and 7.5 gallon sparge? So now you're boiling 12.5 gallons down to 5?
Sorry if this simple seeming question was too long winded and thanks in advance.