About how much will 1bbl of spent grains weigh?

[akthor]

So I have my hoist system attached to my brew stand to lift my 1bbl basket out of the brew pot. I would like to test it before actually using it on a brew day. So does anyone know about how much 1bbl of spent grains will weigh?

 

[Qhrumphf]

A 1 bbl batch? A full 1bbl of grains? Which one?

I'd approximate double the grain weight in drained spent grain after absorption. Ie I'd expect a 100lb grain bill to weigh approximately 200 lbs as wet spent grain.

 

[madscientist451]

Lets say an average 1 bbl batch uses 60 lbs of grain and your absorption is .1 gallon per pound so you'd have to account for 6 gallons of water x 8.34 lbs or about 50 lbs so the whole thing would weigh about 110 lbs.
Some people use the 1/2 quart/lb for absorption, so 60 lbs x .5 quarts would be 30 quarts or 7.5 gallons x 8.34lbs/gallon or 62.5 lbs for a total of 120 lbs.

 

[akthor]

Yeah I meant a 1bbl batch

 

[doug293cz]

What everyone has forgotten so far is that after mash and lauter, you have removed a significant amount of the original grain weight from the spent grain.

Mash Efficiency = Wt of Extract in BK / Potential Extract Wt in Grain, so Wt of Extract in BK =
Mash Efficiency * Potential Extract Wt in Grain​

Typical extract potential is about 80%, so if mash efficiency is 75% for example, then the weight of the grain that has been transferred to the BK is:

75% * 80% * Grain Weight = 60% of Grain Wt​

So, the dry weight of the spent grain is only 40% of the original grain weight. But the amount of absorbed wort is more than the apparent grain absorption, since some of the volume of the collected wort is due to the extract in the wort, but for now, I'll leave that correction as an exercise for the reader.

Brew on

 

[madscientist451]

What everyone has forgotten so far is that after mash and lauter, you have removed a significant amount of the original grain weight from the spent grain.
So, the dry weight of the spent grain is only 40% of the original grain weight.

Doug is partially right. I never even considered the weight of the soluble sugars that are removed from the mash tun.
But I'm not so sure of Doug's calculation that 60% of the weight is removed, leaving only 40%.
So here are some calculations with the following assumptions: 10 lbs of grain,a 5 gallon batch, 6.5 gallons of pre- boil 1.040 wort.
6.5 gallons of water x 8.34 lbs/gallon= 54.21 lbs
54.21 lbs x 1.040=56.38 or a difference of 2.17 lbs.
So using the above assumptions, if the spent grain was dried out, it would weigh 7.83 lbs? (I'm including the question mark, because I've had another 16 hr day with 5 hrs of sleep, so I figured I messed this up somewhere)
So the spent grain would be 78% of its original weight? (using the above brewhouse assumptions) I'm throwing this out as a question, not making a boilerplate assertion.

 

[wilserbrewer]

So here are some calculations with the following assumptions: 10 lbs of grain,a 5 gallon batch, 6.5 gallons of pre- boil 1.040 wort.

6.5 gallons of water x 8.34 lbs/gallon= 54.21 lbs

54.21 lbs x 1.040=56.38 or a difference of 2.17 lbs.

So using the above assumptions, if the spent grain was dried out, it would weigh 7.83 lbs? .

I believe the fly in the ointment here is that 6.5 gallons of wort contains slightly less than 6,5 gallons of water, hence more sugar weight and less water weight than the assumption? 2.17 lbs of DME would not yield a gravity of 1.040 in 6.5 gallons???

Is it also that other compounds are extracted from the grain besides sugar? Proteins etc perhaps IDK?

 

[doug293cz]

Edit: Flaky analysis removed.

Brew on

 

[doug293cz]

To do the math a simpler way, you start with 10 lb of grain and 7.6 gal of water:

Mash Weight = 10 lb + 7.6 gal * 8.33 gal/lb = 73.31 lb​

Then you collect 6.5 gal of 1.040 wort:

Collected Wort Wt = 6.5 gal * 8.33 lb/gal * 1.040 =56.31 lb​

Then the residual mash weight must be:

73.31 lb - 56.31 lb = 17.0 lb​

Brew on

 

[_BullDog_]

Think you guys are getting overly exact... don't want the system to fall.... check it for 200lbs and you know you are good to go.

 

[Biscuits]

Think you guys are getting overly exact... don't want the system to fall.... check it for 200lbs and you know you are good to go.

Although, if OP has any way to weigh the spent grain after he brews...this could be a fun game to see who gets it right.

 

[doug293cz]

Think you guys are getting overly exact... don't want the system to fall.... check it for 200lbs and you know you are good to go.

I agree, the OP should proof test the hoist at 3 - 4X the expected max load. Always want to have a safety factor.

I went into detail because I thought is was an interesting question, and there is a common misconception that the answer is the weight of the original grain plus absorbed water (when it is actually less than that.)

Brew on

 

[augiedoggy]

Doug is partially right. I never even considered the weight of the soluble sugars that are removed from the mash tun.
But I'm not so sure of Doug's calculation that 60% of the weight is removed, leaving only 40%.
So here are some calculations with the following assumptions: 10 lbs of grain,a 5 gallon batch, 6.5 gallons of pre- boil 1.040 wort.
6.5 gallons of water x 8.34 lbs/gallon= 54.21 lbs
54.21 lbs x 1.040=56.38 or a difference of 2.17 lbs.
So using the above assumptions, if the spent grain was dried out, it would weigh 7.83 lbs? (I'm including the question mark, because I've had another 16 hr day with 5 hrs of sleep, so I figured I messed this up somewhere)
So the spent grain would be 78% of its original weight? (using the above brewhouse assumptions) I'm throwing this out as a question, not making a boilerplate assertion.

Not to ruin the discussion of dry spent grain but why does any of this matter here? Does the OP have a special way to dry the spent grain out before he uses the hoist to remove the grain from the kettle? I would think that the grain is going to weigh much more because it will be wet no?

 

[doug293cz]

Not to ruin the discussion of dry spent grain but why does any of this matter here? Does the OP have a special way to dry the spent grain out before he uses the hoist to remove the grain from the kettle? I would think that the grain is going to weigh much more because it will be wet no?

Post #9shows the calculation for the drained, but still wet, grain mass. Of course it will weigh more before draining is complete. The post you quoted is just trying to calculate the same number, and the dry spent grain weight is just an intermediate value in the chain of calculations.

Brew on

 

[doug293cz]

Doug is partially right. I never even considered the weight of the soluble sugars that are removed from the mash tun.
But I'm not so sure of Doug's calculation that 60% of the weight is removed, leaving only 40%.
So here are some calculations with the following assumptions: 10 lbs of grain,a 5 gallon batch, 6.5 gallons of pre- boil 1.040 wort.
6.5 gallons of water x 8.34 lbs/gallon= 54.21 lbs
54.21 lbs x 1.040=56.38 or a difference of 2.17 lbs.
So using the above assumptions, if the spent grain was dried out, it would weigh 7.83 lbs? (I'm including the question mark, because I've had another 16 hr day with 5 hrs of sleep, so I figured I messed this up somewhere)
So the spent grain would be 78% of its original weight? (using the above brewhouse assumptions) I'm throwing this out as a question, not making a boilerplate assertion.

This analysis makes the common error of assuming 6.5 gal of wort contains 6.5 gal of water, which it does not. °Plato is defined as the weight % of extract (mostly sugar) in the wort, thus the weight % of water in the wort is 100° - °Plato. So, to work the example in the quoted post:

6.5 gal of wort @ 1.040 SG @ 68°F/20°C = 6.5 gal * 8.33 gal/lb * 1.040 = 56.311 lb
1.040 SG = 9.994 °P, so
Extract Wt in BK = 56.311 lb * 0.09994 = 5.628 lb, and
Water Wt in BK = 56.311 lb * (1 - 0.09995) = 50.683 lb, so
Water Vol in BK = 50.683 / 8.33 = 6.0844 gal, and
Extract Vol in BK = 6.5 gal - 6.084 = 0.4156 gal​

Calculation of the dried spent grain weight is complicated by the fact that about 4% of the original grain weight is from absorbed water, and the residual dried material will contain unconverted grain material and dried extract (from the absorbed wort.) Plus, some of the dried extract weight is due to water that chemically combined with the starch during saccharification (i.e. didn't come from the original dried grain weight.) I have a spreadsheet that takes all of these factors into account (you can find a link to it here.) The calculated dried spent grain material weight is 2.817 lb, and the dried retained extract weight is 1.417 lb, so the total dried spent grain mass would weigh 4.234 lb. This is 42.3% of the original grain weight. In this specific example, the conversion efficiency was 91.6%. If instead we had 98% efficiency, the dried spent grain would weigh 2.34 lb + 1.54 lb = 3.88 lb, or 38.8% of the original grain weight. So, my previous ~40% of original grain weight estimate was right on.

I do all my calculations with volumes corrected to 68°F/20°C since that is the temp that was used to measure the °Plato to SG data. Water at 68°F/20°C has a density of 8.33 lb/gal. According to the Bettin & Spieweck polynomial, water has a density of 8.34 lb/gal at 55.7°F/13.2°C.

But, as noted by Augie, the dried spent grain weight directly relevant to the OP's question, as OP is interested in the wet spent grain weight.

Brew on