how much volume does a lb of grain typically have?

[je52rm]

If I placed 1 lb. of grain in 1 gallon water how much will the new combined volume be? Just curious nothing more

Or will the grain absorb enough liquid that it will not affect the percieved volume?

 

[ajf]

1 lb grain occupies 0.32 quarts when saturated (according to Promash).

-a.

 

[je52rm]

1 lb grain occupies 0.32 quarts when saturated (according to Promash).

-a.

thank u!!!

 

[Tinga]

1 lb grain occupies 0.32 quarts when saturated (according to Promash).

-a.

does this account for water absorption?

 

[ajf]

Not quite sure what you mean by water absorption.
If the grains are fully saturated, then they will have absorbed a certain amount of water.
If you add 1 lb grain to 1 qt water, then the mixture will take up a volume of 1.32 qts.
If you add 1 lb grain to 2 qt water, it will take up 2.32 qts.
Again, according to Promash, grain absorbs 0.12 gallons per lb (very nearly 1 pint per lb).
So, if you drained the runnings in the above two examples you would get just over 1 pint in the first example or just over 3 pints in the second (assuming you don't lose anything to dead space). In both cases, the drained grains would occupy a volume of 0.32 qt.

-a.

 

[kpr121]

Ahhh I thought this was going to my chance to say tell the world my secret that a red solo cup filled to the top crease is almost exactly 8 ounces for most base and some specialty grains.

But thats dry, sorry I cant help.

 

[weirdboy]

Not quite sure what you mean by water absorption.
If the grains are fully saturated, then they will have absorbed a certain amount of water.
If you add 1 lb grain to 1 qt water, then the mixture will take up a volume of 1.032 qts.
-a.

Hang on a second there. You are not discounting the amount of water that is absorbed into the grain, which is something like .13 gallons per pound.

Or maybe you are accounting for that, and if that is the case, then your previous statement is wrong.

 

[je52rm]

Seems he would be right because were talking about volume not absorbtion. If the 1 lb. grain goes into 1 quart water it will be "absorbed" and then the resulting volume of the final product will be 1.32 quarts.

This is the way i understood it

think of it like this: after the grains are done absorbing water they are not taken out of the total volume. so even if .13 of the total begining water is absorbed into the grain the resulting volume of the saturated grain+total water(minus absorption) will be 1.32. is this correct AJF?

 

[weirdboy]

My take on the original post was that 1 lb of soaking wet grain occupies a volume of .32 quarts (0.08 gallons), without any other water there.

If that is true, then adding 1 lb of dry grain to 1 gallon of water will absorb about .13 gallons of water, and then occupy the remaining space, leaving:

1 gallon - .13 gallons absorbed + .08 gallons = 0.95 gallons.


Since that seems physically impossible, I have to assume the numbers are different.

 

[DarthMalt]

Hang on a second there. You are not discounting the amount of water that is absorbed into the grain, which is something like .13 gallons per pound.

Or maybe you are accounting for that, and if that is the case, then your previous statement is wrong.

I believe AJF is correct.

Maybe you're confusing me and I'm misunderstanding you, but why would the amount of water absorbed while the grains and water are in a mixture create a difference in final volume? Once in a mixture (that is the grains completely submerged in the water and the water fully absorbed) the volume will be same, no matter what. Regardless of amount of water absorbed, etc.

Also, I believe the question is about grain volume, not absorption.

The difference between the original water volume and the combined grain/water volume results in the volume of the grains.

If, as AJF states, 1lb. of dry grains is immersed in 1qt. water and the final volume is 1.32qts., the difference in volume (the volume of the grains) is 0.32qts.

 

[bob1852]

My take on the original post was that 1 lb of soaking wet grain occupies a volume of .32 quarts, without any other water there.

Right, but the volume of a quart of water is going to be a quart, regardless of whether or not the water is absorbed into the grain or not. Right? (assuming absorbed grain + absorbed water doesn't displace more than non-absorbed grain + total water)

EDIT: too slow...

 

[je52rm]

I believe AJF is correct.

Maybe you're confusing me and I'm misunderstanding you, but why would the amount of water absorbed while the grains and water are in a mixture make a difference in final volume? Once in a mixture (that is the grains completely submerged in the water and the water fully absorbed) the volume will be same, no matter what. Regardless of amount of water absorbed, etc.

exactly my thoughts

 

[ajf]

Hang on a second there. You are not discounting the amount of water that is absorbed into the grain, which is something like .13 gallons per pound.

Or maybe you are accounting for that, and if that is the case, then your previous statement is wrong.

I admit that I was wrong in my previous post. I inadvertently added a couple of 0's and specified volumes of 1.032 and 2.032. They should have been 1.32 and 2.32. I edited my post to correct these errors.

Seems he would be right because were talking about volume not absorbtion. If the 1 lb. grain goes into 1 quart water it will be "absorbed" and then the resulting volume of the final product will be 1.32 quarts.

This is the way i understood it

think of it like this: after the grains are done absorbing water they are not taken out of the total volume. so even if .13 of the total begining water is absorbed into the grain the resulting volume of the saturated grain+total water(minus absorption) will be 1.32. is this correct AJF?

Yes, that's the way I see it.

-a.

 

[william_shakes_beer]

I was just getting ready to post the OP's question, so please permit me to summarize the discussion here for the benefit of my feeble mind;

Given: I need to check that my kettle is big enough to soak X pounds of dry grains.

Therefore;

1. I put 1 gallon of clear water in the pot

2. I add 1 pound of dry grains

3. The grains ultimately soak up .13 gallons, leaving 0.87 gallons of theoretical" runnings, ignoring things such as equipment loss, etc.

4. Initially the dry grains float on the surface. As they soak up water, they sink into the water volume

5. Once fully saturated, the combination of grains and water occupy a total of 1 gallon plus .32 quarts, or .08 gallons. Therefore, to soak 1 pound of grains a pot of 1.08 gallons is required.

Therefore, If i have a grain bill of 12 pounds and want to end up with 7 gallons of runnings, I need a pot that will hold:

7 gallons + 1.56 gallons absorption + .96 gallons displacement=9.52 gallons (assuming I can boil "to the brim, which is silly, but this is beer math, what the hell)

AND I need to ultimately soak 7 gallons water plus 1.56 gallons absorption=8.56 gallons combined strike and sparge water.

I assume that mashing temps do not evaporate a volume of water sufficient to be significant for our purposes. (this is beer math in the basement, not in the lab)

 

[Cromacster]

I believe AJF is correct.

Maybe you're confusing me and I'm misunderstanding you, but why would the amount of water absorbed while the grains and water are in a mixture create a difference in final volume? Once in a mixture (that is the grains completely submerged in the water and the water fully absorbed) the volume will be same, no matter what. Regardless of amount of water absorbed, etc.

Wrong, when the grains absorb the water the density of the mixture changes.

D = M/V or V = M/D

The mass will stay constant, but the density changes when the water and grains are mixed. The volume changes depending on how much water is absorbed into the grain.

Edit:

1 lb grain occupies 0.32 quarts when saturated (according to Promash).

Missed the point that the grains are saturated. If saturated grains are added to water then it was just be addition of the two volumes. Assuming that the grains are densely packed. Even so there is probably some "empty space" in the saturated grains, thus the density of the mixture would still change slightly. Overthinking.....

 

[je52rm]

no offense to anybody but seems the initial question is getting twisted and turned. We are talking about VOLUME not absorbtion, density, ect.... As stated before by AZ IPA the volume of one gallon of water will always be the same (1 gallon = 1 gallon) no matter what u put into it (As long as the absorbed grain isnt removed). You cannot ADD ANYTHING to water that will lessen its VOLUME (Impossible) !!! Realize that the grain have mass. Even if the grain have airspace or whatever else inside (room to absorb water) they still have mass so when you add one mass to another mass the result will be higher than one or the other by themselves . we are trying to figure out:

With the addition of 1 lb. of grains to 1 quart water how much VOLUME will be gained?

.08 gallons volume added (1 gallon+.08= 1.08 total water+grain volume)

 

[DarthMalt]

.

 

[michaeltrego]

we are trying to figure out:

With the addition of 1 lb. of grains to 1 quart water how much VOLUME will be gained?

.08 gallons volume added (1 gallon+.08= 1.08 total water+grain volume)

I made a little spreadsheet with the following formula to ensure that my grain bill and strike water will fit in my mash tun:

Mash Volume (gal) = Weight of Grains (lbs) x (0.08 + Mash Thickness (qts/lb) /4 )

For example, 10 lbs. grain plus 3.75 gal. strike water would take up 4.55 gal. volume in your mash tun.

4.55 = 10 x (0.08 + 1.5 / 4)

 

[je52rm]

I made a little spreadsheet with the following formula to ensure that my grain bill and strike water will fit in my mash tun:

Mash Volume (gal) = Weight of Grains (lbs) x (0.08 + Mash Thickness (qts/lb) /4 )

For example, 10 lbs. grain plus 3.75 gal. strike water would take up 4.55 gal. volume in your mash tun.

4.55 = 10 x (0.08 + 1.5 / 4)

awesome except u should express the equation in this manner to eliminate bad calculations

.08 + (1.5 / 4) x 10 = 4.55

or .08 + (mash thickness (qts/lb) / 4) x total lbs grain = total mash volume

if someone used your equation and added the .08 and 1.5 first and then divided by 4 it doesnt calculate out correctly

 

[michaeltrego]

awesome except u should express the equation in this manner to eliminate bad calculations

.08 + (1.5 / 4) x 10 = 4.55

or .08 + (mash thickness (qts/lb) / 4) x total lbs grain = total mash volume

if someone used your equation and added the .08 and 1.5 first and then divided by 4 it doesnt calculate out correctly

Actually there shouldn't be any confusion at all if you follow the standard mathematical order of operations! And if you set yourself up with a calculator, it's even less confusing

 

[midfielder5]

If you are trying to see how many pounds you can mash in the MLT, try the Green Bay rackers calculator "Can I Mash It" (middle of page).
Green Bay Rackers--Mash Calculators

 

[michaeltrego]

Yep - I have used the Rackers Calculators as well, but I wanted to have a quick offline version.

 

[je52rm]

Actually there shouldn't be any confusion at all if you follow the standard mathematical order of operations! And if you set yourself up with a calculator, it's even less confusing

Order of operations

From Wikipedia, the free encyclopedia

In mathematics and computer programming, the order of operations (sometimes called operator precedence) is a rule used to clarify unambiguously which procedures should be performed first in a given mathematical expression.

For example, in mathematics and most computer languages multiplication is done before addition; in the expression 2 + 3 × 4, the answer is 14. Brackets, which have their own rules, may be used to avoid confusion, thus the preceding expression may also be rendered 2 + (3 × 4), but the brackets are not required as multiplication still has precedence without them.

as i said in my previous post it's a simplified version to "avoid confusion" for others that may not be "well versed" in all the aspects of math terminoligy. Your equation is CORRECT. I made it "less confusing" as it says can be done in the wiki article u posted.

 

[sssarahevt]

no offense to anybody but seems the initial question is getting twisted and turned. We are talking about VOLUME not absorbtion, density, ect.... As stated before by AZ IPA the volume of one gallon of water will always be the same (1 gallon = 1 gallon) no matter what u put into it (As long as the absorbed grain isnt removed). You cannot ADD ANYTHING to water that will lessen its VOLUME (Impossible) !!! Realize that the grain have mass. Even if the grain have airspace or whatever else inside (room to absorb water) they still have mass so when you add one mass to another mass the result will be higher than one or the other by themselves . we are trying to figure out:

With the addition of 1 lb. of grains to 1 quart water how much VOLUME will be gained?

.08 gallons volume added (1 gallon+.08= 1.08 total water+grain volume)

I think this answers my question as well, I just need to know if I start with x amount of water and have x pounds of grains bagged, how many gallons worth of room will this take in my mash kettle. I've got an 8 gallon kettle, so know I will need to either do the dip sparge in hot water then add it into the 8 gallon kettle, or top myself up to a full 5 gallons with cold water during wort chilling.
Sooo...if I'm mashing 15 pounds of grains, they will take up an additional 1.2 gallons of room, plus 5 gallons final beer, plus absorption lost and boil loss=my total mash volume (which without actually doing the math I would guess is approaching more than my 8 gallon kettle can contain). Am I reasoning this through correctly?

 

[dtwhite]

Are you talking about BIAB? If you want to do BIAB with the maximum amount of water possible, take your kettle volume and subtract the weight of your grain * 0.08. So if you have a 10 gallon kettle and theoretically filled it up to the brim and want to add 10 lbs of grain, you'd be able to put 10 gal - (10 lbs * 0.08) = 9.2 gallons of water in there.

Then, if you want to figure out how much volume you'd get after draining the bag, you take that volume and subtract what is absorbed by the grain, which is something like 0.13 gallons per lb of grain. So you'd end up with 9.2 gal - (10 lbs * 0.13) = 7.9 gallons.

Of course you'd probably want to put a safety factor of a half to a full gallon in there so you don't spill everywhere.


For your scenario of an 8 gallon kettle and 15 lbs of grain, I'd say that you could reasonably put 6.3 gallons in there and get 4.35 gallons out of it, leaving a half gallon of room in the kettle. If you want to get up to 7 gallons for your boil, I would probably squeeze the bag to get some extra liquid out, then pour some ~180 degree water through the grain bag into the wort until you reach the volume you want. If you don't squeeze the bag then you can do a simple subtraction to figure out the sparge volume (7.5 gal - 4.35 gal = 3.15 gallons to pour through), and your efficiency should be similar.

 

[Moto19]

I think a lot of people are confused by the volume of grain and when it's added to water. If you happened to have 1 quart of grain and added it to 4 quartes of water, you don't get 5 quarts volume of the new mixture. The grain will absorb the water as it will fill the empty spaces inside the grain and between the grains. You will end up with a total volume somewhere between 4 - 5 quarts. But that doesn't matter or any of the mathematics, because someone already did it for us in the previous threads. Just remember every lb of grain will totally add .32 quarts of volume when added to any amount of water.
I think people should think of the visual experiement, if you have a jar full of marbles and another half jar of sand. If you combined them, you wouldn't have a jar and a half of total solids. The sand would fall down and fill in the gaps of the marbles. That's the same idea if you think about a certain volume of grain being added to a certain volume of water. The volumes are not simply added, but somewhere between the 2 numbers is the answer.

I appreciate that people took the time to figure this out as it makes it a lot easier to find out truly how big a beer you can make, gravity wise, in a given kettle size.

 

[TomVA]

The 0.32 quart expansion per pound of grain is very useful information, but I don't know where the 0.13 gallon water absorption loss per pound of grain comes from. I take careful measurements and keep spreadsheets on every batch, and I average half of that water loss due to grain absorption. Perhaps it is because I do BIAB with full water volume and squeeze the grains?

TomVA

 

[pricelessbrewing]

1 lb of crushed grain increases volume by ~0.08 gallons.

Absorption rate doesn't factor into it, since it's apparent absorption, and the wort volume is greater than the water volume you add due to dissolved sugar, starches, protein etc.

 

[dtwhite]

The 0.32 quart expansion per pound of grain is very useful information, but I don't know where the 0.13 gallon water absorption loss per pound of grain comes from. I take careful measurements and keep spreadsheets on every batch, and I average half of that water loss due to grain absorption. Perhaps it is because I do BIAB with full water volume and squeeze the grains?

TomVA

Yes, squeezing the bag significantly lowers the .13 water loss figure. If you just suspended the bag over the kettle to let it drain without squeezing, you'd probably get closer to .13, but being able to squeeze the bag is a pretty clear advantage of BIAB as far as I'm concerned.

Just to reiterate:
Volume that grain adds to a mash while it's still in there: 0.08 gallons / lb
Volume that grain will take from the mash due to absorption: 0.13 (if simply draining / mash tun) to ~0.07 (if squeezing the BIAB bag)

 

[andy6026]

Sorry to revive a dead thread, but this is the most thorough analysis of mash volume requirements I've encountered. Unfortunately, the numbers arrived at here just don't correspond with my real life experience... ie., trying to mash and sparge 20 lbs of grain for a 10 gallon batch in a 48 qt cooler. Inevitably it just doesn't quite fit. But according to the calculations here, I should be able to fit this! Either these numbers are wrong, or my cooler manufacturer lied about the size of the cooler!

Let's assume, for simplicity sake, that I split my water bill in half. Let's also assume that I need to start with about 17.5 gallons of water to make a 10 gallon batch once grain absorption, dead space, trub, boil off, hop absorption, etc, is all taken into account.

Cooler size: 48 qts
Total water needed at beginning of process: 70 qts
Mash water bill: 35 qts
Sparge water bill: 35 qts.
Grain: 20 lbs
Volume of grain @ 0.32 qts per pound: 6.2 quarts

Total mash volume: 41.2 qts
Total sparge volume: 47.4 qts (assuming the grains are still saturated from the mash).

I'd have more play if I borrow more water from my sparge bill and fill the mash right to the brim, so theoretically, this cooler should handle this mash and have a around a gallon of space left for play for each the mash and sparge. I can't do it though... in real life it just won't fit. Maybe my calculation is wrong? If anything I think I've erred on the side of assuming more water than is actually used.

 

[mongoose33]

I skimmed this thread, and the first thought that came to mind is what kind of grain? I've got some rye that has tiny kernels compared to typical barley malt; I haven't weight out a pound of each to see their volumes, but I'll bet a pound of the Rye has less overall volume than a pound of the Barley.

If we're talking about crushed grain, then it may be different; it may also depend on the coarsness of the crush.

BTW, @andy6026 , I have a 48-quart igloo cooler I use for mashing. A typical grain bill has 12-13# of grain, and I'm underletting with about 8 to 8.25 gallons of water. Analogous to a BIAB with no sparge. That cooler is pretty full, but if I went to 20# of grain, I am not sure it would all fit, grain and water. If I were doing a mash water volume of 1.5 quarts per pound, that would be 30 quarts, 7.5 gallons. Might be close, but still tight.

 

[pricelessbrewing]

Aside from different grains absorbing different amounts, and having different kernel densities, there's thermal expansion which can realistically account for an additional 2-2.5% of your water volume at mash temps.

At this point we're still missing ~4 qts in your example @andy6026. I'd have to ask how you're measuring your water volumes, are you using a pitcher, weight, etc?

 

[bleme]

Measure 48 quarts of water into that cooler and see for yourself. If it doesn't fit, you know that either it isn't a 48 qt cooler or your quart measurement is off.

 

[andy6026]

As much as that would certainly answer my question, bleme, going through those motions is a little more effort than I want to expend to find out. I guess what I'm getting at is that regardless whether or not Coleman is lying, my 48 qt cooler just isn't cutting it for 10 gallon batches (so mongoose33 - we are agreed!), despite the measurements above saying that it should work. So I bought a 70 qt cooler today which I'm guesting should be perfect - although the point of threads like this is to answer such questions with more precision than that.

Bottom line so far peeps (at least for me): if you want to mash and sparge a 10 gallon batch in a 48 qt Coleman cooler without sparging twice, then you likely won't be able to do it and your floor's going to get wet... unless you like really low alcohol beer!

I've bought a 70 gallon cooler this afternoon and have converted it... will report back after my next brew day.

Ps: to answer pricelessbrewing's questions, I've long noticed that the factory volume markings on fermenter buckets are often waaaay off the mark (perhaps this is why you asked me how I measure), so I used a large kitchen measuring pitcher to re-mark a fermenter bucket which I've since used to calculate my water volumes on brew day.

Thanks everybody!

 

[pricelessbrewing]

Ps: to answer pricelessbrewing's questions, I've long noticed that the factory volume markings on fermenter buckets are often waaaay off the mark (perhaps this is why you asked me how I measure), so I used a large kitchen measuring pitcher to re-mark a fermenter bucket which I've since used to calculate my water volumes on brew day.

Thanks everybody!

See this may not be much better. In general the measurmental uncertainty is going to be half the smallest unit that's marked. So lets say you're filling your pitcher 9 times for the example of 37 qts for mash water, and it's only as accurate as a cup, this gives your re-marked fermenter bucket an uncertainty of about 37 qts +- 2.25. This is more than reasonable.

If however you use weight as the standard, you can be much more accurate.

Not saying it's the sole reason, but it's likely a contributing source of error.

 

[mongoose33]

I was thinking about this when I brewed today (yeah, I was able to take off a bit early). Below is a pic showing my 48-quart cube cooler where I have about 8+ gallons of water plus 12# 13 ounces of grain. It seems to me that I would have been able to add maybe 2 more gallons of water to this before I was pushing the very limits.

 

[davidabcd]

I was just getting ready to post the OP's question, so please permit me to summarize the discussion here for the benefit of my feeble mind;

Given: I need to check that my kettle is big enough to soak X pounds of dry grains.

Therefore;

1. I put 1 gallon of clear water in the pot

2. I add 1 pound of dry grains

3. The grains ultimately soak up .13 gallons, leaving 0.87 gallons of theoretical" runnings, ignoring things such as equipment loss, etc.

4. Initially the dry grains float on the surface. As they soak up water, they sink into the water volume

5. Once fully saturated, the combination of grains and water occupy a total of 1 gallon plus .32 quarts, or .08 gallons. Therefore, to soak 1 pound of grains a pot of 1.08 gallons is required.

Therefore, If i have a grain bill of 12 pounds and want to end up with 7 gallons of runnings, I need a pot that will hold:

7 gallons + 1.56 gallons absorption + .96 gallons displacement=9.52 gallons (assuming I can boil "to the brim, which is silly, but this is beer math, what the hell)

AND I need to ultimately soak 7 gallons water plus 1.56 gallons absorption=8.56 gallons combined strike and sparge water.

I assume that mashing temps do not evaporate a volume of water sufficient to be significant for our purposes. (this is beer math in the basement, not in the lab)

sorry for off topic. You get a plus one for your name.

 

[william_shakes_beer]

The numbers given here are estimates. In addition to absorption there is also displacement ( the grains take up space). To add a personal data point, I have always used these figures in my brew-in-a-bag setup to verify my container is sufficient. Of course, I need not be particularly precise because my mash tun/kettle is 15 gallons and I don't do batches over 10 gallons. There are going to be variances based on deadspace in the bottom of the tun, moisture content of the grains, etc. If the batches you usually do are in danger of overflowing the tun, you need a bigger tun. If your sensibilities require more precision, I'd suggest you get a graduated water measure, a red oak dowel and a sharpie. Pour 2 quarts of water into the empty tun with the sparge braid in place. Place the dowel in the center and draw a line. Repeat until adding another container of water overflows the tun. That's your real capacity. Now place 4 gallons of mash temp water from your graduated stick into the tun. Add 4 pounds of milled grains and let it sit 30 minutes or so. Poke in the stick and see what the apparent volume is. Thats your volume factor. Repeat for as many grains as required to satisfy your sensibilities.

 

[andy6026]

The numbers given here are estimates. In addition to absorption there is also displacement ( the grains take up space). To add a personal data point, I have always used these figures in my brew-in-a-bag setup to verify my container is sufficient. Of course, I need not be particularly precise because my mash tun/kettle is 15 gallons and I don't do batches over 10 gallons. There are going to be variances based on deadspace in the bottom of the tun, moisture content of the grains, etc. If the batches you usually do are in danger of overflowing the tun, you need a bigger tun. If your sensibilities require more precision, I'd suggest you get a graduated water measure, a red oak dowel and a sharpie. Pour 2 quarts of water into the empty tun with the sparge braid in place. Place the dowel in the center and draw a line. Repeat until adding another container of water overflows the tun. That's your real capacity. Now place 4 gallons of mash temp water from your graduated stick into the tun. Add 4 pounds of milled grains and let it sit 30 minutes or so. Poke in the stick and see what the apparent volume is. Thats your volume factor. Repeat for as many grains as required to satisfy your sensibilities.

This would be a very sensible way to determine the volumes required. Unfortunately it is more involved than I was hoping to do and I have already upgraded to a 70 qt cooler. I haven't used it yet, but I'm excited to test it out soon. What I was hoping for were some quick and easy calculations based on a cooler's manufacturer claims as to the volume of the tun and the general guides presented in this thread regarding grain volume and absorbency. From good estimates we should be able to determine with fairly decent precision how many pounds of grain can be mashed and sparged (and by that I mean single-sparged) in a 48 qt cooler, 70 qt cooler, 100 qt cooler, etc. I'm quite surprised that there isn't a handy chart like this available. There are 'can i mash it' charts that do exactly this, but only seem to take into account the mash stage. A proper chart would likely be very helpful for folks that are shopping for a new cooler and who are obviously not yet able to do the test above that you've suggested. Well, at least that was the predicament I was in...