I'm trying to come up with a method to calculate the quantity of water needed to brew a desired volume in primary. Here are my thoughts:
The recipe is formulated for a 5 gallon batch. So I would assume that the desired volume in primary is 5 gallons even though some of the volume would be lost during raking to secondary or bottling.
The recipe contains 7 lbs of liquid malt extract and 1 lb of grains. The recipe calls for steeping the grains in 2 gallons of water for 25 minutes at 155 degrees. At the end of the 30 minutes there will be loss due absorption into the grains. From what I've seen this can be estimated at 0.15 gal/lb of grains. So ideally at the end of the steep I would be left with 1.85 gallons.
The next step is adding water to the steeped grains, bringing the mixture to a boil, and then adding in the liquid malt extract. I measured the volume of the liquid malt extract to be 0.6 gallons. So, with the 1.85 gallons from the steeped grains and 0.6 gallons from the liquid malt extract I would have 2.45 gallons. The unknown is how much water to add to the pot to obtain the desired volume in primary taking into account boil time, boiling loss rate, cooling loss, trub loss, and volume added from yeast starter. Let's call this unknown water volume "wort boil water". From what I've seen the boiling loss rate can be estimate at 20% per hour, the cooling loss is 4%, and the trub loss is 0.25 gal. According to Mr Malty Pitching Rate Calculator my starter will be 0.61 gal.
1. 2 gal steeping water
2. Lose 0.5 gal due to absorption of grains
3. Add X.XX gal of wort boil water [So, equation would be (step 1-step 2+step 4-step 5-step 6-step 7+step 8)-desired volume in primary]
4. Add 0.6 gal of liquid malt extract
5. Lose X.XX gal due to boiling loss [So, equation would be (step 1-step 2+step 3+step 4)*0.10]
6. Lose X.XX gal due to cooling loss [So, equation would be (step 5*0.04)]
7. Lose 0.25 gal due to trub loss
8. Add 0.61 gal of yest starter
Can anyone help fill in the X's? Thanks!
Hopefully this makes sense as I'm sure you can tell I was thinking this up while typing.
The recipe is formulated for a 5 gallon batch. So I would assume that the desired volume in primary is 5 gallons even though some of the volume would be lost during raking to secondary or bottling.
The recipe contains 7 lbs of liquid malt extract and 1 lb of grains. The recipe calls for steeping the grains in 2 gallons of water for 25 minutes at 155 degrees. At the end of the 30 minutes there will be loss due absorption into the grains. From what I've seen this can be estimated at 0.15 gal/lb of grains. So ideally at the end of the steep I would be left with 1.85 gallons.
The next step is adding water to the steeped grains, bringing the mixture to a boil, and then adding in the liquid malt extract. I measured the volume of the liquid malt extract to be 0.6 gallons. So, with the 1.85 gallons from the steeped grains and 0.6 gallons from the liquid malt extract I would have 2.45 gallons. The unknown is how much water to add to the pot to obtain the desired volume in primary taking into account boil time, boiling loss rate, cooling loss, trub loss, and volume added from yeast starter. Let's call this unknown water volume "wort boil water". From what I've seen the boiling loss rate can be estimate at 20% per hour, the cooling loss is 4%, and the trub loss is 0.25 gal. According to Mr Malty Pitching Rate Calculator my starter will be 0.61 gal.
1. 2 gal steeping water
2. Lose 0.5 gal due to absorption of grains
3. Add X.XX gal of wort boil water [So, equation would be (step 1-step 2+step 4-step 5-step 6-step 7+step 8)-desired volume in primary]
4. Add 0.6 gal of liquid malt extract
5. Lose X.XX gal due to boiling loss [So, equation would be (step 1-step 2+step 3+step 4)*0.10]
6. Lose X.XX gal due to cooling loss [So, equation would be (step 5*0.04)]
7. Lose 0.25 gal due to trub loss
8. Add 0.61 gal of yest starter
Can anyone help fill in the X's? Thanks!
Hopefully this makes sense as I'm sure you can tell I was thinking this up while typing.
