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Grain to gallon ratio

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CrashNebula

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So I have a newbie question. So when you are designing a recipe how do you figure out how many pounds of grain you want per gallon of beer? Is there and equation or a rule of thumb? Any help or point out the right direction would be greatly appreciated
 
Are you asking about your mash and sparge water volumes? For those I would recommend a water calculator. I typically go 1.5qts/lbs.

OR, are you asking... If I have a 5 gallon batch, how many pounds of grain should I use? Because the answer to that is... It varies greatly based on style of beer, types of grains used, your brew house efficiency...
 
On a related note, I read to take the number of lbs of malt and multiply it by .2 and it will give you the amount of water that is absorbed by the grain. That will help you calculate how much water to use to end up with six gallons (or whatever) for the boil. Some people use spreadsheets or brewing software, but I've found that formula to be a good general rule of thumb. So if you're using 12 lbs of grain x .2 is 2.4 gallons of water absorption. If you want 6 gallons of wort for your boil, you'll need to use about 8.5 gallons of water. As for how much grain to use, that depends on the style of beer, OG you're shooting for, etc.
 
I use 1.25 to 1.5 gallons per pound of grain, which is basic rule of thumb. It depends, for me, on how many pounds of grains in the bag are being crammed into my 5 gallon BK/MT.
 
When discussing the MASH ratio, I tend to go with 1.25 quarts per pound. But that is somewhat flexible and I've mashed with the full volume of water when doing BIAB.

When discussing how much grain goes into a 5 gallon recipe, then it's wide open. You can brew a recipe with very few grains (light lager, or Blonde, or Mild), and you can brew a recipe with more graint han can reasonably be converted by teh enzymes and lautered out of the mash tun (Barleywine, RIS, Belgian Strong, etc.)

I'd recommend finding a style you like and look up some recipes to see what they recommend for gravity. Calculate your necessary grain and water volumes to achieve that gravity.
 
I believe the question is about how much grain to use for a given recipe batch size, not about mash thickness.

The fun begins with a value called PPG - Points per Pound per Gallon. Each type of fermentable (grains, sugars) has a PPG value; this corresponds to its theoretical maximum sugar yield. It's the number of gravity points that a pound of the fermentable will yield to a gallon of water.

A typical base malt has 37 PPG. That means a pound of this malt, when mashed with one gallon of water, and we get every molecule of sugar possible, the resulting wort will have a gravity of 1.037.

Of course, we never get all the sugar. So we have to multiply the theoretical gravity by our predicted mash efficiency - let's say 75% - so the 1.037 becomes 1.028.

The equation for this is:
Pre-Boil Gravity Pts = (PPG x lb) / Pre-Boil Gal. x Mash Efficiency

Each fermentable has a different PPG and weight, so the equation must be done separately for each ingredient. The sum of all these gravity point values equals the predicted pre-boil SG.

You can then predict the OG by including the pre- and post-boil wort volumes:
Predicted OG = Pre-Boil SG x Pre-Boil Gal. / Ending Kettle Gal.

It's far easier to do this in software, which is what all the popular tools do - or build your own spreadsheet to handle it. Here's a list of common PPG values:
http://www.brewersfriend.com/fermentables/
 
On a related note, I read to take the number of lbs of malt and multiply it by .2 and it will give you the amount of water that is absorbed by the grain. That will help you calculate how much water to use to end up with six gallons (or whatever) for the boil. Some people use spreadsheets or brewing software, but I've found that formula to be a good general rule of thumb. So if you're using 12 lbs of grain x .2 is 2.4 gallons of water absorption. If you want 6 gallons of wort for your boil, you'll need to use about 8.5 gallons of water. As for how much grain to use, that depends on the style of beer, OG you're shooting for, etc.

0.2 gal/lb is much higher than the typical grain absorption rate. For traditional mash tuns and batch sparging, the typical grain absorption rate is about 0.125 gal/lb. With BIAB the grain absorption rate can range from 0.04 to 0.10 gal/lb depending on the bag draining and/or squeezing protocol. To measure your grain absorption rate, subtract your (first runnings wort volume plus MLT undrainable volume) from your strike water volume and divide by the total grain weight. Grain absorption rate is meaningless for a traditional fly sparging process, since you leave excess liquid in the MLT at the end of the sparge.

Brew on :mug:
 
It really depends on your mash efficiency. I usually start with about 2 lbs per gallon to get me an OG of 1.060. That is a pretty decent efficiency of about 80%. If you get closer to 70%, that will get you an OG of about 1.052 for 2 lbs.

There are recipe calculators out there that can help you. Just make sure you have the correct mash efficiency in the program.
 
0.2 gal/lb is much higher than the typical grain absorption rate. For traditional mash tuns and batch sparging, the typical grain absorption rate is about 0.125 gal/lb. With BIAB the grain absorption rate can range from 0.04 to 0.10 gal/lb depending on the bag draining and/or squeezing protocol. To measure your grain absorption rate, subtract your (first runnings wort volume plus MLT undrainable volume) from your strike water volume and divide by the total grain weight. Grain absorption rate is meaningless for a traditional fly sparging process, since you leave excess liquid in the MLT at the end of the sparge.

Brew on :mug:

I must have a really inefficient mash tun. I just brewed an imperial IPA using 16 lbs. of grain. I used nine gallons of water (mash and sparge combined) which yielded 6 gallons of wort. After the boil and removal of bittering hops, I'm down to 4.5 gallons. The x lbs. grain x .2 formula works perfectly for my set-up.
 
That's a loss of 0.1875 gal/lb to the mash. Is any portion of that from dead space (unrecovered but NOT absorbed liquid)?

If you were able to physically tip the mash tun to pour out any liquid, how much would flow out?
 
Here's a pic of my mash tun. Maybe I will gain some efficiency if I go to a cylindrical version. I like this one because of the size. I bet I couldn't have dumped out more than a 1/4 to 1/2 gallon if I had turned it completely upside down.

Mashtun_zpsawpxd4uf.jpg
 
Here's a pic of my mash tun. Maybe I will gain some efficiency if I go to a cylindrical version. I like this one because of the size. I bet I couldn't have dumped out more than a 1/4 to 1/2 gallon if I had turned it completely upside down.

Mashtun_zpsawpxd4uf.jpg

Fill it with water an inch or two above the bazooka screen, place it in the normal operating position, and then drain the water. Now dump out and measure the volume of all the water that didn't drain. This is your "undrainable volume." Given the position of your drain, I think you will find your undrainable volume is higher than you suspect.

Brew on :mug:
 
Fill it with water an inch or two above the bazooka screen, place it in the normal operating position, and then drain the water. Now dump out and measure the volume of all the water that didn't drain. This is your "undrainable volume." Given the position of your drain, I think you will find your undrainable volume is higher than you suspect.

Brew on :mug:
When I brew and the wort stops draining from the mash tun, I turn it up 90 degrees perpendicular to flat. I guess that would be vertical, lol. I'm sure some liquid remains behind, but I don't think it's significant.
 
When I brew and the wort stops draining from the mash tun, I turn it up 90 degrees perpendicular to flat. I guess that would be vertical, lol. I'm sure some liquid remains behind, but I don't think it's significant.

Wouldn't 45° be better to drain more wort from that corner?
 
I must have a really inefficient mash tun. I just brewed an imperial IPA using 16 lbs. of grain. I used nine gallons of water (mash and sparge combined) which yielded 6 gallons of wort. After the boil and removal of bittering hops, I'm down to 4.5 gallons. The x lbs. grain x .2 formula works perfectly for my set-up.

So 1.5 gallons of your calculated 4.5 gallons absorbed was actually boiled off, not absorbed. It shouldn't be included.
 
So 1.5 gallons of your calculated 4.5 gallons absorbed was actually boiled off, not absorbed. It shouldn't be included.

Right. It's not being counted. 16 lbs of grain x .2 = 3.2. Supposedly that represents the "absorption" of water by the grain, but it's really the "total loss" of water between absorption and unattainable wort. I started with nine gallons and ended up with six gallons, pre boil.
 
Question: Do you base your grain calculation on your Strike water volume, your target pre boil water volume, or your target post boil volume?
 
Question: Do you base your grain calculation on your Strike water volume, your target pre boil water volume, or your target post boil volume?
Grain bill weight is based on your desired OG (Original, or starting, specific Gravity, a measure of sugar concentration), your desired wort volume, and your process efficiency. Wort desired volume can be taken post-boil, as volume in fermenter, or final packaged volume. In the case of post-boil volume, you base your calculation on mash efficiency. In the case of fermenter volume, you base your calculation on brewhouse efficiency. In case of final packaged volume (whether in bottles or kegs), you base your calculation on packaged, or end-to-end efficiency. The difference between these different efficiencies is due to volume left behind going from BK to fermenter, and volume left behind going from fermenter to bottles/kegs. Mash efficiency is greater than brewhouse efficiency, and brewhouse efficiency is greater than packaged efficiency.

Brew on :mug:
 

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