After having done a couple extract batches I'm hoping to do my first stovetop BIAB tomorrow afternoon. I bought a NB 5gallon Cream Ale kit and plan to split it in half and make two separate batches BIAB-style. I'm trying to calculate the best way to manage a 2.5gal batch on my stovetop with only a 4-gallon pot available to me. The various online calculators show that to mash with the full boil volume I'd need 4 gallons of strike water, which my pot wouldn't be able to hold in addition to 4lbs of grain.
My plan is to mash with less water and then top off either pre-or post boil? I realize full boil volumes are ideal but I don't have the equipment for it.
My tentative plan is to mash the 4lbs grain in 1.5 gallons of water(1.5 quart/lb ratio, which I gather is pretty normal), allowing for 0.5 gallons of water absorption by the grain. I'll then top off the remaining ~1 gallon of wort with another 1.5 gallons to reach a pre-boil volume of 3.5 gallons, and then boil for 1hr, allowing for 1 gal/hr boil-off. This should leave me with 2.5 gallons going into the fermentor.
Does this make sense or is there something I'm not taking into account?
My plan is to mash with less water and then top off either pre-or post boil? I realize full boil volumes are ideal but I don't have the equipment for it.
My tentative plan is to mash the 4lbs grain in 1.5 gallons of water(1.5 quart/lb ratio, which I gather is pretty normal), allowing for 0.5 gallons of water absorption by the grain. I'll then top off the remaining ~1 gallon of wort with another 1.5 gallons to reach a pre-boil volume of 3.5 gallons, and then boil for 1hr, allowing for 1 gal/hr boil-off. This should leave me with 2.5 gallons going into the fermentor.
Does this make sense or is there something I'm not taking into account?