94% mash efficiency? is that possible ?

[odie]

Just did a Kolsch this weekend. I used a couple online calculators and both gave me 94% mash efficiency. I thought 80% was about the best that can be normally be done.

BIAB. 7 gallons total mash water and 6.5 gallons left after pulling the bag and a few good squeezes. Total mash time was about 3 hours with constant recirc sine I have a large dead space below the basket holding my bag. Mash temp was 148-150' with a digital controller maintaining constant temps during the recirc and then stepped up to 170 for mash out.

grain bill: 7.5# total
6# German Kolsch malt
1# American pilsner malt
0.5# American Vienna malt

SG 1.040
OG 1.044

 

[Oginme]

Yes, it can be done. Not sure where you heard the 80% maximum value from but it may have been a different metric, such as extractable sugars by total weight of grain. Mash efficiency is based upon the amount of sugars extracted versus the total potential extractable sugars in the grains. I have hit the low 90's before on low gravity brews but found it unsustainable with a high degree of variability.

 

[odie]

the 80% is just kinda a typical average of what I see a on all these posts about BIAB efficiency. It seems (not a scientific statistic by any means) that many brag about hitting 80-85% with their BIAB methods. So me hitting 10-15% higher seems out of line. As though I screwed up a measurement. Not sure how much or in what way the size of the grain bill comes into play until you get to very large bills.

 

[RM-MN]

With a 3 hour time for mash there isn't any reason not to hit 100% conversion. From there on the limit is how much sugars you leave in the bag of grains. Did you do any kind of sparge?

 

[odie]

I'm sure pretty much all the starch converted. No sparge. full volume mash. but I did squeeze the bag pretty good. I'm just surprised that 94% of potential sugar actually made in into the kettle.

 

[wilserbrewer]

To what % accuracy did you measure all weights and volumes?

Efficiency #’s are only as accurate as the data gathered

Not trying to be accusatory, but it is very easy to miss measurements by a few percent...do that twice and actual efficiency could be 5-6 % off or more.

With 100% conversion and leaving .045 gal / lb wort absorbed by the grain you left behind about 6% of the wort and 6% of the sugar....

Seems believable. Hence why efficiency typically increases for smaller beers and decreases for larger beers, as more wort is left behind or absorbed by the grain as grain bills increase.

 

[doug293cz]

Count me a skeptic. 0.5 gal absorption with 7.5 lb of grain is an absorption rate of 0.067 gal/lb, which is reasonable. The maximum theoretical lauter efficiency for your volumes and grain weight is 87.4%, and mash efficiency must be less than or equal to lauter efficiency. Got to be some measurement errors somewhere. To get your mash efficiency and pre-boil SG, the weighted grain potential needs to be 1.038, and the conversion efficiency needs to be 108%.

Brew on

 

[dmtaylor]

Yup, been there a few times. But of course I keep on fiddling with mash pH and mill gaps and whatnot so I haven't gotten efficiency that high in a while. Didn't know how to quit while I was ahead!? But yes high efficiency like that is totally totally possible, regardless of naysayers. And like I always say, it ain't bragging if it's truth. Note: It is certainly much easier to get high efficiency on a low gravity beer like 1.044 vs. a higher gravity around 1.055 or higher.

 

[Qhrumphf]

I routinely hit 93-95% efficiency, but that's a dialed in old school false bottom lauter and continuous sparge (and oversparging by every accepted homebrewer metric). It can certainly be done. With a no-sparge BIAB I'm skeptical though. Unless you had virtually no liquid left in your grain after squeezing...

 

[eric19312]

That’s a high efficiency for sure. Three hour recirculation in your mash if open system could of resulted in substantial evaporation, meaning you got more than 92.85% (6.5/7gal) out of your grain...

That depends on measurement accuracy. Probably of the grain. Did you weigh the grains or get them pre weighed?

 

[odie]

mashed 3 hrs with continuous recirc and false bottom. covered and insulated kettle/mash tun...

I squeezed the crap outta the bag. It felt very "dry" when I dumped it. With a 7.5# grain bill it was pretty damn easy to squeeze...unlike my monster 23# ish grain bill a couple weeks ago. No matter how hard I squeezed...the spent grains where very, very sticky and wet when I tossed them out.

I only weighed the grains when building the grain bill. I didn't weigh the drained bag. perhaps a new thing to start doing?

 

[Oginme]

I only weighed the grains when building the grain bill. I didn't weigh the drained bag. perhaps a new thing to start doing?

Volume of water in less the volume of the wort out will give you grain retention. You do need to make sure you correct the volume for temperature if they are not measured around the same temperature.

Weighing the grains becomes a more complicated equation. Dry weight of the spent grains is ???

 

[McKnuckle]

It's not practical to weigh grains before and after the mash and perform simple subtraction to determine how much water remains absorbed. Sure; the post-mash grain retains water that wasn't there before, but it has also relinquished considerable weight in both original starch and water content to the collected wort. So that exercise is not going to be conclusive.

It's more accurate to weigh the collected wort and divide it by the pre-boil gravity to get the weight of the collected water in kg, swap that for liters, which can then be compared to the original volume of water. But that's a pain with hot liquid, so not something I'd do more than once.

 

[doug293cz]

Let's do a little math: A pre-boil SG of 1.040 is 9.99°Plato. That means that Extract_Weight / (Strike_Water_Weight + Extract_Weight) = 0.0999. Water at 68°F has a density of 8.3304 lb/gal, so your strike water weight was 58.31 lb. Solving the first equation for extract weight, we get Extract_Weight = 6.475 lb.
So the total weight of the wort created in the mash is 58.31 lb + 6.475 lb = 64.79 lb, and the wort volume in the mash is 68.79 / (8.3304 * 1.040) = 7.48 gal. The temp corrected (from 150°F to 68°F) pre-boil volume is 6.38 gal. So the percentage of the total wort that you collected in the BK is 100% * 6.38 / 7.48 = 85.4%. This is your lauter efficiency. (This calculation does not account for the small amount of additional water due to the moisture content of the grain - about 4% of the grain weight. The calculation in my previous post accounted for the grain moisture. That's why it's slightly different.)

So, you created 7.48 gal (corrected to 68°F) of 1.040 wort in the mash. That's 7.48 * 40 = 299.2 gravity points. With a 7.5 lb grain bill, your pts/lb would have to be (assuming 100% conversion) 299.2 / 7.5 = 39.9 pts/lb, or a grain potential of 1.0399. Never seen grain with that high a potential. Even if you did have a grain bill with that potential, your mash efficiency would still only have been ~85.4% Edit: The previous does not account for the moisture content of the malt, which is usually about 4%. If we account for this, the dry grain weight is only 7.2 lbs, and the dry basis potential would have to be 299.2 / 7.2 = 41.6 pts/lb, or 1.0416.

Brew on

 

[Qhrumphf]

Brainstorming here (after a few pints mind you).

If you have malt COAs available, you should be able to rough out a weight calculation more precise than below.

Figure base malt is usually circa 80% extract and 4ish% moisture. Let's call that 79% and 4% to factor in specialty grains (assuming some specialties are higher moisture and some lower so assuming it balances out).

If 100% of your starches are gelatinized, converted, and extracted, and 100% of the moisture (both added from mash water and pre-existing in the grain itself) is removed, your spent grain should weigh approximately 17% of it's original weight (with a small margin of error for protein).

Obviously that's not practical to achieve especially outside of a laboratory setting. Anything beyond that is gonna be wort left behind. Since you're no sparge BIABing, use your preboil gravity, convert it to Plato, multiply that by the difference in actual spent weight vs that max number, and you have theoretical weight of sugar remaining on your grain. Then divide that by the total potential extract of your grain (we're assuming 79%, so your grain weight*0.79, but if you have those COAs for each malt you could be more accurate, weighting averages for both extract and moisture). That'd be your efficiency. And using mass only such that temp and volume shouldn't factor.

 

[Qhrumphf]

Now to sit here and drunkenly wait for @doug293cz to tell me that what I came up with is drunken hogwash

 

[doug293cz]

Now to sit here and drunkenly wait for @doug293cz to tell me that what I came up with is drunken hogwash

Not hogwash. All that stuff gets taken care of in my spreadsheet available here, except the weight of the spent grains. Could be added, but why? Not really much to be learned from weighing the bag full of spent grains.

Brew on

 

[Qhrumphf]

Not hogwash. All that stuff gets taken care of in my spreadsheet available here, except the weight of the spent grains. Could be added, but why? Not really much to be learned from weighing the bag full of spent grains.

Brew on

If you're no sparge BIAB, may be more precise/accurate/simple than relying on volume and temp corrections to determine efficiency.

But as soon as sparging and/or dead space get involved, loses the simplicity.

 

[doug293cz]

If you're no sparge BIAB, may be more precise/accurate/simple than relying on volume and temp corrections to determine efficiency.

But as soon as sparging and/or dead space get involved, loses the simplicity.

Don't think it would be simpler. There's a lot of math required to get to the answer. If anyone really wants a method for weighing a wet sticky bag, I might look into a procedure, and see how much the math can be simplified.

Brew on

 

[Qhrumphf]

Don't think it would be simpler. There's a lot of math required to get to the answer. If anyone really wants a method for weighing a wet sticky bag, I might look into a procedure, and see how much the math can be simplified.

Brew on

Dunno. Seems pretty easy.

7.5*0.16= 1.2 lb minimum sugar-free moisture-free spent grain (assuming OPs grain bill is closer to 80% extract). Say he weighs it at 5 lbs. At his 1.040 or 10P, thats 3.8*0.1= 0.38 lb residual sugar, then 0.38/(0.80*7.5)=0.063 loss, or 6.3%, meaning 93.7% efficiency.

Seems like less calculation than your math above. But maybe that's just me.

In more realistic fashion, even after squeezing the hell out of it he probably had a more realistic 8-9 lbs of spent grain weight. But we'll never know.

 

[odie]

Only the chickens know...they got all the grain...

I'm skeptical of what I came up with as well...that's why I posted it here...

either way, I'm gonna drink it when it's done

 

[doug293cz]

Dunno. Seems pretty easy.

7.5*0.16= 1.2 lb minimum sugar-free moisture-free spent grain (assuming OPs grain bill is closer to 80% extract). Say he weighs it at 5 lbs. At his 1.040 or 10P, thats 3.8*0.1= 0.38 lb residual sugar, then 0.38/(0.80*7.5)=0.063 loss, or 6.3%, meaning 93.7% efficiency.

Seems like less calculation than your math above. But maybe that's just me.

In more realistic fashion, even after squeezing the hell out of it he probably had a more realistic 8-9 lbs of spent grain weight. But we'll never know.

Where did the 0.16 factor come from? I get the min dry weight of the spent grain as:

Grain dry weight = 7.5 lb * 0.96 = 7.2 lb (4% moisture assumed)
Spent dry weight = 7.2 lb * (1 - 0.8) = 1.44 lb (80% dry basis extract assumed)​

So the spent dry weight multiplier is 0.96 * 0.2 = 0.192, not 0.16.

Then your 0.38/(0.80*7.5) should actually be 0.38/(0.96*0.80*7.5) to account for the moisture of the starting malt, so loss is 0.066 or 6.6%.

If we go back and calculate what OP's actual wet spent grain mass weighed (using the linked spreadsheet, and assuming the required 41.6 pts/lb grain potential (90% dry basis extract potential) for OP's numbers to work), then we get 9.5 lb. In this case the spent dry weight multiplier is 0.96 * 0.1 = 0.096, and the min dry weight of the spent grain is 7.5 * 0.096 = 0.72 lb. So the retained wort weight is 9.5 lb - 0.72 lb = 8.78 lb, which is 0.878 lb of extract. So the fraction of retained extract is 0.878 / (0.96 * 0.9 * 7.5) = 0.135 => 13.5%, or 86.5% mash efficiency.

And, there is a subtle problem with doing the math as above: The dry spent grain weight does not actually equal the dry fresh grain weight minus the created extract weight. This is because the extract contains water (picked up during starch hydrolysis) that was not in the dry grain. Thus the spent dry grain weight is actually higher than the above calculation determines - 0.96 lb vs. 0.72 lbs, because the extract contain 0.24 lb of hydrolysis water.

The other problem is that doing the calculation this way only works for 100% conversion efficiency, and zero undrainable wort in the MLT. The method I laid out in a previous post for calculating lauter efficiency does not suffer from any of the drawbacks mentioned in this post.

Brew on

 

[Qhrumphf]

Where did the 0.16 factor come from? I get the min dry weight of the spent grain as:

Grain dry weight = 7.5 lb * 0.96 = 7.2 lb (4% moisture assumed)
Spent dry weight = 7.2 lb * (1 - 0.8) = 1.44 lb (80% dry basis extract assumed)​

So the spent dry weight multiplier is 0.96 * 0.2 = 0.192, not 0.16.

Then your 0.38/(0.80*7.5) should actually be 0.38/(0.96*0.80*7.5) to account for the moisture of the starting malt, so loss is 0.066 or 6.6%.

If we go back and calculate what OP's actual wet spent grain mass weighed (using the linked spreadsheet, and assuming the required 41.6 pts/lb grain potential (90% dry basis extract potential) for OP's numbers to work), then we get 9.5 lb. In this case the spent dry weight multiplier is 0.96 * 0.1 = 0.096, and the min dry weight of the spent grain is 7.5 * 0.096 = 0.72 lb. So the retained wort weight is 9.5 lb - 0.72 lb = 8.78 lb, which is 0.878 lb of extract. So the fraction of retained extract is 0.878 / (0.96 * 0.9 * 7.5) = 0.135 => 13.5%, or 86.5% mash efficiency.

And, there is a subtle problem with doing the math as above: The dry spent grain weight does not actually equal the dry fresh grain weight minus the created extract weight. This is because the extract contains water (picked up during starch hydrolysis) that was not in the dry grain. Thus the spent dry grain weight is actually higher than the above calculation determines - 0.96 lb vs. 0.72 lbs, because the extract contain 0.24 lb of hydrolysis water.

The other problem is that doing the calculation this way only works for 100% conversion efficiency, and zero undrainable wort in the MLT. The method I laid out in a previous post for calculating lauter efficiency does not suffer from any of the drawbacks mentioned in this post.

Brew on

Came from my drunkenly neglecting that extract potential is dry basis, and taking 80% from full weight including moisture, rather than without. You're entirely correct.

I would also assume that at least a portion of malt protein would be soluble as well.

I did say initially it's a rough calc. But potentially no more so than flawed volumes or temp corrections from thermal expansion (the only realistic way the OP overestimated efficiency so badly).

 

[doug293cz]

Came from my drunkenly neglecting that extract potential is dry basis, and taking 80% from full weight including moisture, rather than without. You're entirely correct.

I would also assume that at least a portion of malt protein would be soluble as well.

I did say initially it's a rough calc. But potentially no more so than flawed volumes or temp corrections from thermal expansion (the only realistic way the OP overestimated efficiency so badly).

Yep, some malt proteins, and other "stuff" is soluble, and ends up as part of the extract. That's why the "correct" term is extract and not sugar. Sugar (including dextrins) makes up about 90% of extract from what I have read.

Agreed, no efficiency calc can be any better than the measurements that went into it. GIGO applies in spades.

Also, calculation complexity for the user goes down substantially if the math is incorporated into a spreadsheet that accepts the needed inputs and spits out the results. Here's a screenshot of a little spreadsheet I whipped up the other night to cover the math in this post. User only has to input the Blue values.

Brew on

 

[Qhrumphf]

Yep, some malt proteins, and other "stuff" is soluble, and ends up as part of the extract. That's why the "correct" term is extract and not sugar. Sugar (including dextrins) makes up about 90% of extract from what I have read.

Agreed, no efficiency calc can be any better than the measurements that went into it. GIGO applies in spades.

Also, calculation complexity for the user goes down substantially if the math is incorporated into a spreadsheet that accepts the needed inputs and spits out the results. Here's a screenshot of a little spreadsheet I whipped up the other night to cover the math in this post. User only has to input the Blue values.

View attachment 662199

Brew on

The only issue with saying "extract" from malt in homebrew circles is people often think you mean....malt extract. "Sugar", while not technically accurate as there's other stuff in there, conveys the point more clearly.

 

[NobleNewt]

This thread caught my eye last night. I'm not an efficiency seeker. I've got a BIAB process that works quite well for me and routinely get 80% mash efficiency and 70% BH efficiency. Just for kicks, I crushed my grain slightly finer than I typically do.

Lo and behold, I got close to 90% mash efficiency on today's brew! I say close, because in the end, volume measurements are at best close approximations. Took gravity measurements with both my hydro and refractometer just to be safe and both were the same across pre-boil and post-boil measurements.

My BH efficiency rounded out around 75%. All to say, I got a little more wort at the same OG as predicted which means a little more bang for my buck!

 

[doug293cz]

This thread caught my eye last night. I'm not an efficiency seeker. I've got a BIAB process that works quite well for me and routinely get 80% mash efficiency and 70% BH efficiency. Just for kicks, I crushed my grain slightly finer than I typically do.

Lo and behold, I got close to 90% mash efficiency on today's brew! I say close, because in the end, volume measurements are at best close approximations. Took gravity measurements with both my hydro and refractometer just to be safe and both were the same across pre-boil and post-boil measurements.

My BH efficiency rounded out around 75%. All to say, I got a little more wort at the same OG as predicted which means a little more bang for my buck!

Yeah, efficiency should not be a primary goal, but brewers who are way under performing often want to make some improvement. What I write is an attempt to help people know how to diagnose and understand their issues, and let people know what is actually possible (so they don't have unrealistic expectations.)

Brew on

 

[NobleNewt]

Yeah, efficiency should not be a primary goal, but brewers who are way under performing often want to make some improvement. What I write is an attempt to help people know how to diagnose and understand their issues, and let people know what is actually possible (so they don't have unrealistic expectations.)

Brew on

It's super helpful for sure. In the end, I only really gained about 3 SG points with my added efficiency. The biggest help a brewer will have is increasing that efficiency from, say, the low 60s up to 70 or 75 percent. When you're already at 80% mash efficiency, there's not a whole lot of room to improve.

In my case, even though I had a great mash efficiency, I was still inefficient in the brewhouse which led to some losses. All-in-all, I just got more beer for the same price!

 

[doug293cz]

It's super helpful for sure. In the end, I only really gained about 3 SG points with my added efficiency. The biggest help a brewer will have is increasing that efficiency from, say, the low 60s up to 70 or 75 percent. When you're already at 80% mash efficiency, there's not a whole lot of room to improve.

In my case, even though I had a great mash efficiency, I was still inefficient in the brewhouse which led to some losses. All-in-all, I just got more beer for the same price!

Thanks.

I don't write much about brewhouse eff being much lower than the mash eff, since it is all due to wort getting 'lost' between the BK and fermenter, and not very interesting from a diagnosis standpoint.

Brew on