Huge Stout: Split Mash or Sparge?

[deadwolfbones]

I'm a full-volume/no-sparge BIAB brewer. I use a Robobrew V3 (no pump).

I recently picked up a 5 gallon whiskey barrel, and I'm going to be making a big-ass pastry stout to fill it with—6 gallons total, estimated OG 1.120.

Since I'll never fit 24 lbs of grain in the Robobrew for a full-volume mash, my initial thought was to do two mashes in sequence and combine in the kettle before boiling. But that would take a long time.

Then I thought, I can ask someone at the brew club to borrow their kettle/bag and do two mashes concurrently and combine in the boil kettle (the Robobrew, in this case). I got someone to loan me their Blichmann 10g kettle and matching bag. Cool!

Then I realized that I could do the entire batch in the 10g kettle if I do a little less than full volume and do a dunk or pourover sparge. Say, 6 gal mash and 3 gal sparge.

I'm not sure which would be the best course of action. The first would take more time but keep me with equipment/processes I know and trust. The second would take less time, but involve using unfamiliar equipment. The third would take less time, but involve using unfamiliar processes.

Anyone have tips on how to proceed, considerations I might have missed, etc?

 

[RM-MN]

Option 3. Using the big kettle and the bag with a sparge at the end should get you the highest efficiency which is what you are going to need for such a big beer. The process with that system is to fill to the volume you can make fit with the grain and heat it to strike temp. Drop in the bag, stir in the grains, wait. Pull bag when the mash is done, dunk it in another vessel (fermenter buckets work well) and pull it out to drain. Add the wort from the dunk sparge to the original kettle to boil. If you can hang the bag over the boil kettle or the other vessel to drain as the water comes to a boil you can add that to the boil.

 

[brewman !]

Option 4: reiteritative or polygyle mashing. Google is your friend.

 

[deadwolfbones]

Option 4: reiteritative or polygyle mashing. Google is your friend.

Hooo boy, I think I'm gonna pass for now, but thanks for the suggestion.

EDIT: Actually, this looks somewhat interesting. Might give it a go.

 

[brewman !]

Jamil Zainasheff at Heretic Brewing now uses reiterative mashing in their process for high gravity brews.

 

[deadwolfbones]

Yeah, I'm just trying to figure out how doing it using BIAB would differ from their method. It also wouldn't save me time like option 3. It wouldn't be any more efficient than option 2, as far as I can tell?

 

[deadwolfbones]

Now I'm thinking I'm definitely going to make a smaller partigyle stout with the post-mash grains, though.

 

[brewman !]

Reiterative mashing works great with BIAB. Just sparge the bag, hoist it out, change the grains and mash again. Couldn't be simpler. Way, way better efficiency.

 

[deadwolfbones]

Reiterative mashing works great with BIAB. Just sparge the bag, hoist it out, change the grains and mash again. Couldn't be simpler. Way, way better efficiency.

I have always done full-volume BIAB (no sparge, but squeezed pretty hard). Is it possible to do reiterative mashing that way? Whether full-volume or sparged, is reiterative mashing more efficient than doing two concurrent full-volume mashes with 1/2 the water in each mash?

 

[doug293cz]

Work the problem backwards. You want an OG of 1.120, so first figure out what your pre-boil SG needs to be by determining how much boil-off you are willing to do. The required pre-boil SG is given by:

Pre-Boil SG = 1 + (Target OG - 1) * Post-Boil Volume / Pre-Boil Volume
​

So, if you want 6 gal @ 1.120 and only want to boil off 1.5 gal then the pre-boil SG needs to be: 1 + 0.120 * 6.0 / (6.0 + 1.5) = 1.096. If you are willing to boil off 3.0 gal then the pre-boil SG only needs to be 1.067.

You are going to need 120 [pts/gal] * 6 [gal] = 720 total gravity points in your BK, so your grain bill (assuming a weighted 36 pts/lb) will have to be: 720 [pts] / (36 [pts/lb] * Mash Efficiency) = 20 [lb] / Mash Efficiency. Thus at 75% mash efficiency you would need 20 / 0.75 = 26.7 lbs of grain to start. But, lauter efficiency is a function of sparge process, grain absorption rate, and grain bill weight / pre-boil volume (in lb/gal). If we assume 95% conversion efficiency, then 75% mash efficiency implies a lauter efficiency of 0.75 / 0.95 = 0.79 => 79%. If we look at the 7.5 gal pre-boil volume, then the grain to volume ratio is 26.7 [lb] / 7.5 [gal] = 3.56. Now we need some data on what kind of lauter efficiencies are actually obtainable.

The following chart shows lauter efficiency for various numbers of batch sparge steps for two different grain absorption ratios as a function of grain weight to pre-boil volume ratio. 0.12 gal/lb is typical for a traditional MLT, and 0.06 represents a fairly aggressive squeeze when doing BIAB. An optimal fly sparge can achieve a lauter efficiency about 3 percentage points higher than the triple batch sparge @ 0.12 gal/lb (a bad fly sparge can be worse than a single batch sparge.)

At our initially estimated grain to volume ration of 3.56, the max possible lauter efficiency for a single batch sparge at 0.12 gal/lb is about 69% (reading solid orange line.) Thus we cannot get our initially assumed lauter efficiency of 79%, so we will need to add more grain to make up for lost efficiency. But, adding more grain will reduce our lauter efficiency even more, so our new lauter efficiency assumption need to be less than 69%, so let's go with 65%, which would put our assumed mash efficiency at 0.95 * 0.65 = 0.617 = 61.7%, and now our grain bill goes to 20 / 0.617 = 32.4 lb, and the grain to volume goes to 32.4 / 7.5 = 4.32, which unfortunately puts us off off the chart, and our lauter efficiency is going to be more like 62%.

Who said high gravity beers were easy.

Rather than doing calculations by hand, and iteratively searching for a grain weight that will give you what you are looking for, you can use the spreadsheet here and the "Goal Seek" tool to tell you how much grain you will need for your proposed OG and volume using your specific process parameters. Copy the spreadsheet to your local computer and open in Excel or LibreOffice Calc (Google sheets does not do goal seek.) Fill in cells B5 to B16 (leave B7 blank) with the specifics for your process. Use an educated guess for grain weight (B5). Next go to "Tools" and select "Goal Seek". In the first input box enter "K59", in the second input box enter your target OG, in the third input box enter "B5", now click the "OK" button. The result will be the grain weight you need, given all the assumptions you entered initially. If I use this procedure for our 7.5 gal pre-boil case (boil off 1.0 gal/hr, 1.5 hour boil, 6 gal post-boil, 1.0 qt/lb max mash thickness allowed) the grain weight needed for 1.120 OG is 37.2 lb. The associated lauter efficiency is ~59%, and mash efficiency ~56%.

Brew on

 

[brewman !]

Reiterative brewing gets around all those problems.

Here are some links to reiterative brewing:

https://www.reddit.com/r/grainfathe...cially_the_largest_brew_ive_done_with_the_gf/
https://www.google.com/search?client=firefox-b&q=Reiterated+Mashing:+Multiple+Mashes+for+Massive+Brews
https://homebrewacademy.com/reiterated-mashing-brewing-high-gravity-beers/
https://www.homebrewtalk.com/forum/threads/polygyle-mashing-anyone-done-it.627080/
http://www.nordeastbrewersalliance.org/education/nicks-tips-for-brewing-high-gravity-beers/
https://www.homebrewtalk.com/forum/...s-pumpkin-porter-and-ale-from-one-mash.74927/
http://beerandwinejournal.com/reiterated-mashing-2/
http://beerandwinejournal.com/tripel-iii/#more-4372
http://beerandwinejournal.com/tripel-iii/
http://grainfatherweeklymash.blogspot.com/2016/03/week-62-large-grain-bill-brewing-what.html
http://grainfatherweeklymash.blogspot.com/2016/06/week-74-reiterated-mashing.html?view=magazine
http://www.thebrewingnetwork.com/brew-strong-triple-mashing-2/
http://www.thebrewingnetwork.com/brew-strong-triple-mashing/
http://www.thebrewingnetwork.com/brewing-style-brewing-big-beers/

I'll be using the reiterative mash process for my high gravity beers.

 

[doug293cz]

I have always done full-volume BIAB (no sparge, but squeezed pretty hard). Is it possible to do reiterative mashing that way? Whether full-volume or sparged, is reiterative mashing more efficient than doing two concurrent full-volume mashes with 1/2 the water in each mash?

There is very little efficiency difference between a full volume (two small FV mashes are the same as one big FV mash) mash and a reiterated mash (I can't remember off the top of my head which one is slightly better than the other.) The advantage to the reiterated mash is that the mash thickness is lower (easier to stir, and slightly faster conversion), and it requires a smaller (or fewer) mash vessel.

One option to consider (which is very common with large beers) is to add sugar to the BK in the form of DME, LME, sucrose, corn sugar, candy sugar, etc. This way you your mash efficiency goes up (because you are mashing a smaller beer), and you need less grain.

Brew on

 

[deadwolfbones]

There is very little efficiency difference between a full volume (two small FV mashes are the same as one big FV mash) mash and a reiterated mash (I can't remember off the top of my head which one is slightly better than the other.) The advantage to the reiterated mash is that the mash thickness is lower (easier to stir, and slightly faster conversion), and it requires a smaller (or fewer) mash vessel.

One option to consider (which is very common with large beers) is to add sugar to the BK in the form of DME, LME, sucrose, corn sugar, candy sugar, etc. This way you your mash efficiency goes up (because you are mashing a smaller beer), and you need less grain.

Brew on

Thanks!

I appreciate the big post above, but I've already purchased the ingredients based on my Beersmith recipe and the idea that I was going to do two FV mashes, so I don't really need help formulating the grain bill. I believe the recipe to be solid (I've been doing this for a while, and the base recipe is an award-winner). I just need to figure out how to brew it most efficiently (time-wise) and get the best efficiency (sugar extraction-wise).

If reiterated mashing doesn't provide any advantage over two small FV mashes, either in time or extract efficiency, I'll probably stick to the process I know and am familiar with. That means option 1 or 2, most likely.

I'm definitely open to adding DME if necessary, but I'm also pretty relaxed about the OG. If it's 1.100, fine. If it's 1.130, fine. Obviously I'd like to be on target, but I don't have any experience brewing a beer this big on this system, so I don't know exactly what my efficiency will be like. Regardless, it'll be a big beer and have plenty of flavor.

 

[deadwolfbones]

@doug293cz can you speak to RM-MN's claim in the second post in thread? Would a dunk sparge after a partial volume/full grainbill mash yield better extract efficiency than two smaller FV mashes?

 

[doug293cz]

@doug293cz can you speak to RM-MN's claim in the second post in thread? Would a dunk sparge after a partial volume/full grainbill mash yield better extract efficiency than two smaller FV mashes?

Yes a dunk sparge (in fresh water) will always provide a higher efficiency than a full volume mash (for the same pre-boil volume and grain absorption.) Look at the chart in my previous post. The blue lines represent full volume, no-spage efficiency, and the orange lines represent a single batch sparge. Typical difference is about 8 percentage points in lauter efficiency.

With very large beers, you get into the area of very thick mashes (less than 1.25 qt/lb). In the simulation I did, I used a 1 qt/lb mash thickness, and even that didn't allow for much sparge volume.

You can use the simulator linked to estimate what kind of SG's you will get using your actual process parameters (rather than my assumptions.) You can then make up the difference with sugar in the BK if you wish.

Brew on

 

[deadwolfbones]

So, looking at the chart (and using 0.065 absorption)...

If I have 11.45 lb of grain in 5.2 gallons of water, which is what it would be for the two full-volume/no-sparge small mashes according to the Priceless calculator, my ratio is ~2.2 lb/gal and my efficiency would be about 79%.

If I have 22.9 lb of grain in 6 gallons of water and sparge with 3 gallons, which is what I estimated for the single partial-volume/sparge mash, my ratio is ~3.8 lb/gal and my efficiency would be about 77%.

If I'm reading all of that correctly, it would probably make sense to do the sparge method even though it's ever so slightly less efficient, since it would save me substantial time. Is that right?

 

[doug293cz]

So, looking at the chart (and using 0.065 absorption)...

If I have 11.45 lb of grain in 5.2 gallons of water, which is what it would be for the two full-volume/no-sparge small mashes according to the Priceless calculator, my ratio is ~2.2 lb/gal and my efficiency would be about 79%.

If I have 22.9 lb of grain in 6 gallons of water and sparge with 3 gallons, which is what I estimated for the single partial-volume/sparge mash, my ratio is ~3.8 lb/gal and my efficiency would be about 77%.

If I'm reading all of that correctly, it would probably make sense to do the sparge method even though it's ever so slightly less efficient, since it would save me substantial time. Is that right?

You're using the chart incorrectly. You don't use strike volume to determine the grain to water ratio, you use the pre-boil volume. Makes a huge difference.

If I plug 22.9 lb, 10.4 gal strike water, and 0.065 grain absorption into the simulator, I get a pre-boil volume of 8.91, a grain to wort ratio of 2.57, and a lauter efficiency of ~76%

For a sparged process squeezed to 0.065 gal/lb both after initial draining and sparging, and still using 10.4 gal of total brewing water, the pre-boil volume is still 8.91 gal, and the ratio is still 2.57. For optimal efficiency (equal initial and sparge runnings) you mash with 5.94 gal and sparge with 4.46, with a resultant lauter efficiency of ~85%.

At 100% conversion efficiency, the no-sparge pre-boil SG is about 1.069, and the sparged SG is about 1.077.

Try the simulator - it's fun and educational. It's also less error prone than trying to do these calculations manually or trying to read accurate values off the chart. If you don't have Excel, you can download LibreOffice for free here.

Brew on

 

[deadwolfbones]

Thanks @doug293cz! I got the spreadsheet downloaded and I understand it a lot better now. I think I'll do what it recommends.

The only thing I have to do now is get my hands on my friend's kettle and see what the boil-off rate is on my stove. I'm used to the Robobrew, which struggles to maintain a boil and has a crazy low boil-off (~0.5g/hr). I suppose I could still boil in the Robobrew, but it would be a lot cleaner to do it all in one vessel.

 

[brewman !]

There is very little efficiency difference between a full volume (two small FV mashes are the same as one big FV mash) mash and a reiterated mash (I can't remember off the top of my head which one is slightly better than the other.) The advantage to the reiterated mash is that the mash thickness is lower (easier to stir, and slightly faster conversion), and it requires a smaller (or fewer) mash vessel.

You are wrong. Reiterated mashes will have much higher efficiency, essentially the same as a regular gravity mash. Neither of the other methods will have good efficiency. This is why Heretic is now using this process.

Regular gravity partigyle beers don't often turn out that great. Sure, you utilize the lower gravity wort in a beer, but it is rarely as good as if you mashed the wort from scratch. With reiterative mashing there is no wasted wort and the efficiency is high. Hitting a high gravity is as easy as hitting a regular gravity on a regular beer. The only downside is that it takes extra time.

 

[doug293cz]

You are wrong. Reiterated mashes will have much higher efficiency, essentially the same as a regular gravity mash. Neither of the other methods will have good efficiency. This is why Heretic is now using this process.

Regular gravity partigyle beers don't often turn out that great. Sure, you utilize the lower gravity wort in a beer, but it is rarely as good as if you mashed the wort from scratch. With reiterative mashing there is no wasted wort and the efficiency is high. Hitting a high gravity is as easy as hitting a regular gravity on a regular beer. The only downside is that it takes extra time.

I have analyzed both single mash and double mash processes, and the efficiency results were not much different. Show me your data or calculations that shows an extreme efficiency difference, and I will show you mine that indicates only a small difference. Constraints are:

• Same grain bill
• Same pre-boil volume
• Same grain absorption rate
• Same sparge process

If any of the above are not the same, then all bets are off.

Brew on

 

[brewman !]

I have analyzed both single mash and double mash processes, and the efficiency results were not much different. Show me your data or calculations that shows an extreme efficiency difference, and I will show you mine that indicates only a small difference.

I provided links to discussion of the practice and the testing of it by the likes of Chris Colby, Jamil, John Palmer and John Blichman, the latter who ran a triple mash just to test out the limits of the process. I urge you to listen and read. I suspect you aren't understanding how a reiterative mash works.

Actually your chart proves the point quit nicely. Lets say we are aiming to mash 24 pounds of grain into a preboil volume of 6 gallons. If done all in one mash, the grain to preboil wort is going to be high, ie 4 lbs/gallon. Your table shows a best efficiency of 70% (using the curves for a conventional MLT) However, if reiteratively mashed at 12 pounds per mash, the the grain to wort ratio of each mash will be ~2 (12/6), with a corresponding peak efficiency of ~87%.

I say a grain to water ratio of roughly 2 because the "preboil" volume of the first mash in a reiterative mash should be the final preboil volume plus whatever water is going to be absorbed by the grain in the second mash. So the grain to wort ratio is even lower than 2 for the fisrt mash.

Actually the grain to water ratio in the reiterative mash would be even less than 2 for a comparable pre boil gravity because 24 pounds at 70% efficiency yields 16.8 effective pounds of malt. At a mash efficiency of 87%, that only requies 19.3 pouunds of grain, which when divided into 2 mashes yields a grain to water ratio of ~1.6 pounds per gallon, which, incidentally has an even higher mash efficiency. (>90%)

FYI, 19.3 pounds x 37 points/lb x 87% if mashed reiteritively = 621 points. 24 pounds x 37 points /lb x 70% if mashed conventionally = 621 points. But the original poster wants 720 points total (6 x 120). 24 pounds mashed reiteratively will have a grain/water ratio of 2 pounds per gallon per mash. (12/6). This should easily yield an efficiency of 80%. (Table says 87% for conventional MLT absorption). 24 x 37 x 82% = 728 gravity points, which exceeds the OP's goal.

Compare this to mashing 37 pounds of grain into a pre boil volume of 7.5 gallons and having to do an extra hour of boiling to get the volume down to 6 gallons and I'll chose reiterative mashing every time.

The cost of malt is small compared to the cost of my time. I'm not interested in reiterative mashing for efficiency. I'm interested in it because I have a small mash tun (~7 effective gallons ) and can't mash all the grain at once. Reiterative mashing allows me to brew bigs beers with my existing equipment. The gain in efficiency is just a nice side benefit. But time saving also comes into play if one must do a big boil down to get the required OG when mashing conventionally. I'd much rather mash twice than do a big boil off.

Btw, nice work on the spreadsheet and graph. Palmer mentioned a similar calculation in his discussion of the reiterative mash yield. Your spreadsheet and graph should be included in brewing calculators because it sheds a lot of light on yield limitations versus grain bills.

 

[brewman !]

I have always done full-volume BIAB (no sparge, but squeezed pretty hard). Is it possible to do reiterative mashing that way?

I am not a BIAB brewer, but yes, it should be possible.

Whether full-volume or sparged, is reiterative mashing more efficient than doing two concurrent full-volume mashes with 1/2 the water in each mash?

Yes, massively so. See the efficiency comparison above using Doug293CZ's table for actual figures. In my hypothetical example of wanting to mash 24 pounds into 6 gallons pre boil, it saves nearly 5 pounds of malt (24 - ~19), which is significant. In your proposed situation it appears to save 13 pounds of grain (37-24) and the boiling off of 1.5 gallons of water.

 

[doug293cz]

I provided links to discussion of the practice and the testing of it by the likes of Chris Colby, Jamil, John Palmer and John Blichman, the latter who ran a triple mash just to test out the limits of the process. I urge you to listen and read. I suspect you aren't understanding how a reiterative mash works.
I did listen to the BrewStrong podcasts a few months ago. What I remember them saying is that you didn't lose any efficiency by reiterated mashing, not that you got better efficiency. I do understand how the process works, and have done one reiterated mash BIAB brew myself.

Actually your chart proves the point quit nicely. Lets say we are aiming to mash 24 pounds of grain into a preboil volume of 6 gallons. If done all in one mash, the grain to preboil wort is going to be high, ie 4 lbs/gallon. Your table shows a best efficiency of 70% (using the curves for a conventional MLT) However, if reiteratively mashed at 12 pounds per mash, the the grain to wort ratio of each mash will be ~2 (12/6), with a corresponding peak efficiency of ~87%.

I say a grain to water ratio of roughly 2 because the "preboil" volume of the first mash in a reiterative mash should be the final preboil volume plus whatever water is going to be absorbed by the grain in the second mash. So the grain to wort ratio is even lower than 2 for the fisrt mash.

Actually the grain to water ratio in the reiterative mash would be even less than 2 for a comparable pre boil gravity because 24 pounds at 70% efficiency yields 16.8 effective pounds of malt. At a mash efficiency of 87%, that only requies 19.3 pouunds of grain, which when divided into 2 mashes yields a grain to water ratio of ~1.6 pounds per gallon, which, incidentally has an even higher mash efficiency. (>90%)

FYI, 19.3 pounds x 37 points/lb x 87% if mashed reiteritively = 621 points. 24 pounds x 37 points /lb x 70% if mashed conventionally = 621 points. But the original poster wants 720 points total (6 x 120). 24 pounds mashed reiteratively will have a grain/water ratio of 2 pounds per gallon per mash. (12/6). This should easily yield an efficiency of 80%. (Table says 87% for conventional MLT absorption). 24 x 37 x 82% = 728 gravity points, which exceeds the OP's goal.

Compare this to mashing 37 pounds of grain into a pre boil volume of 7.5 gallons and having to do an extra hour of boiling to get the volume down to 6 gallons and I'll chose reiterative mashing every time.

The cost of malt is small compared to the cost of my time. I'm not interested in reiterative mashing for efficiency. I'm interested in it because I have a small mash tun (~7 effective gallons ) and can't mash all the grain at once. Reiterative mashing allows me to brew bigs beers with my existing equipment. The gain in efficiency is just a nice side benefit. But time saving also comes into play if one must do a big boil down to get the required OG when mashing conventionally. I'd much rather mash twice than do a big boil off.
Sorry, you can't use the chart for the second mash, as the strike liquor is not water, but rather contains a large amount of extract (mostly sugar.) You have to make a correction to for this, which I have done, and will present the results below.

Btw, nice work on the spreadsheet and graph. Palmer mentioned a similar calculation in his discussion of the reiterative mash yield. Your spreadsheet and graph should be included in brewing calculators because it sheds a lot of light on yield limitations versus grain bills.
Thanks. My spreadsheet was used by @pricelessbrewing to create the advanced version of his on-line calculator (he has since found out he can use simpler, less rigorous, equations with essentially no loss of accuracy.) The benefit to my spreadsheet over using the priceless calculator, is that you can use the "Goal Seek" function to do some interesting backwards calculations. You can also look at the cell formulas to see how all the math is done, and once you understand that you have a much better understanding of the mash and lauter processes and what their limitations are.

Ok, lets do the analysis on a 24 lb grain bill (37 pts/lb, 4% moisture), 0.12 gal/lb, 100% conversion efficiency, no-sparge case.

First the single mash case:

We will lose 24 lb * 0.12 gal/lb = 2.88 gal to grain absorption, so we need to start with a strike water volume of 8.88 gal. We create 18.46 lb of extract in the mash, and the wort is 19.91°P (1.0826 SG) We collect 6.0 gal of that wort which contains 10.78 lb of extract, for a mash efficiency of 10.78 / 18.46 = 0.5836 => 58.36%. The grain to pre-boil ratio is 4, and the chart says ~58.5% (can calculate better than read values off a graph.)

Second the double mash case:

In the second mash, we will lose 12 lb * 0.12 gal/lb = 1.44 gal to absorption, so we need to start the second mash with 7.44 gal of wort so we need to collect that volume from the first mash. We lose the same volume in the first mash, so again we start with 8.88 gal of strike water. 12 lbs of grain in the first mash yields 9.23 lb of extract, with a wort at 11.08°P (1.0445 SG.) We collect 7.44 gal of this wort which contains 7.17 lb of extract and 57.57 lb of water. The mash efficiency of the first mash is 7.17 / 9.23 = 0.7767 => 77.67%. The grain to pre-boil ratio is 12 / 7.44 = 1.61, and chart says just under 78%.

In the second mash we start with 7.17 lb of extract and 57.57 lb of water, and create an additional 9.23 lb of extract from the second 12 lb of grain. So, now we have 16.40 lb of extract and 57.7 lb of water (the grain brings in a bit of water in excess of what is consumed during starch hydrolysis, so we gain a small amount of water during the mash.) The wort after the second mash is 22.13°P (1.0926 SG), and we collect 6 gal of this wort. The collected 6 gal contains 12.09 lb of extract, and the mash efficiency is 12.09 / 19.46 = 0.6546 => 65.46%. So we gained about 7 percentage points higher efficiency with the reiterated mash, which is more of a difference than I remembered from when I previously ran this simulation. Either my memory is bad, or I did something wrong the last time I ran a similar simulation.

If I rerun the cases above for a squeezed BIAB with 0.06 gal/lb absorption, the single mash efficiency is 67.86%, and the double mash efficiency is 73.78%

Brew on

 

[brewman !]

First the single mash case:

We will lose 24 lb * 0.12 gal/lb = 2.88 gal to grain absorption, so we need to start with a strike water volume of 8.88 gal.

Agreed.

We create 18.46 lb of extract in the mash, and the wort is 19.91°P (1.0826 SG)

Huh ?

24 pounds x 37 points/lb = 888 points @ 100% efficiency.

The grain to water ratio is 24 pounds / 6 gallons = 4.0. According to your chart, 0.12 gallons per pound @ 0 sparges @ 4.0 = 58.5% efficiency. 888 points x 58.5% = 519.5 gravity points. 519.5 gravity in 6.0 gallons pre boil = 1.086 SG.

We collect 6.0 gal of that wort which contains 10.78 lb of extract, for a mash efficiency of 10.78 / 18.46 = 0.5836 => 58.36%. The grain to pre-boil ratio is 4, and the chart says ~58.5% (can calculate better than read values off a graph.)

Agreed.

Second the double mash case:

In the second mash, we will lose 12 lb * 0.12 gal/lb = 1.44 gal to absorption, so we need to start the second mash with 7.44 gal of wort so we need to collect that volume from the first mash.

Correct

We lose the same volume in the first mash, so again we start with 8.88 gal of strike water. 12 lbs of grain in the first mash yields 9.23 lb of extract, with a wort at 11.08°P (1.0445 SG.)

The first mash "preboil volume" is 7.44 gallons. The grain to water ratio is 12 pounds / 7.44 gallons = 1.61. From the chart @ 0.12 and 0 sparges, the efficiency is 77.8%.

12 x 37 = 444 points @ 100% or 345 points @ 77.8%.

The gravity at the end of the first mash is 345 / 7.44 = 1.0464

We collect 7.44 gal of this wort which contains 7.17 lb of extract and 57.57 lb of water. The mash efficiency of the first mash is 7.17 / 9.23 = 0.7767 => 77.67%. The grain to pre-boil ratio is 12 / 7.44 = 1.61, and chart says just under 78%.

Agreed.

In the second mash we start with 7.17 lb of extract and 57.57 lb of water, and create an additional 9.23 lb of extract from the second 12 lb of grain.

The preboil volume of this mash is 6.0 gallons. 12 pounds / 6.0 gallons = 2.0 Efficiency at 2.0 is 73.8% from the chart. 12 x 37 points = 444 points. 444 x 73.8% = 327.7 points.

Jamil et al say that NONE of the extract from the first mash is lost into the grain of the second mash. Their finding is that the grain in the second mash absorbs water only, no extract. So now we have 345 points from the first mash plus 327.7 points from the second mash = 672.7 gravity points going into the final volume of 6 gallons. This yields a final gravity of 1.112.

672.7 points / 888 points = 75.7% efficiency.

So, now we have 16.40 lb of extract

Not sure where you are getting this number from. Each mash yielded 9.23 pounds of extract for a total of 18.46 pounds of extract.

and 57.7 lb of water

Is this pre mash or pre boil ? 6 gallons x 8.333 pounds per gallon = 50 pounds of water. 7.44 gallons x 8.333 pounds per gallon = 62 pounds of water.

(the grain brings in a bit of water in excess of what is consumed during starch hydrolysis, so we gain a small amount of water during the mash.) The wort after the second mash is 22.13°P (1.0926 SG), and we collect 6 gal of this wort. The collected 6 gal contains 12.09 lb of extract, and the mash efficiency is 12.09 / 19.46 = 0.6546 => 65.46%.

You mean 12.09/ 18.46, not 19.46.

So we gained about 7 percentage points higher efficiency with the reiterated mash, which is more of a difference than I remembered from when I previously ran this simulation. Either my memory is bad, or I did something wrong the last time I ran a similar simulation.

The mash efficiency gain by your yield is 65.5 - 58.5% = 7%
The mash efficiency gain by my yield estimation is 75.7% - 58.5% = 17.2%

However, both of these numbers are MISLEADING because they are NOT FOR THE SAME PRE BOIL GRAVITY as in the single mash case. The single mash case yielded a pre boil gravity of 1.086. Your double mash numbers yielded a pre boil gravity of 1.0926, significantly higher. To reach this gravity with a single mash would require adding more grain, which would further push the grain to water ratio, which would further decrease the efficiency and require even more grain.

Feel free to run the numbers on obtaining a pre boil gravity of 1.0926 using a single mash. I'd be very interested to see how much grain it would take as well as what the efficiency would be. I suspect that the efficiency difference would be even greater, probably something in the order of 10%.

My efficiency yield was based on 100% of the extract from the second batch being additive to the extract yield of the first batch as per the discussion between Palmer, Jamil and Blichman. I am not convinced that this actually occurs so I suspect that the preboil gravity will land somewhere between my calculation and yours, ie 1.0926 and 1.112. But even at a gravity of 1.100 the yield is 600 gravity points from a possible 888 = 67.5%, which is 9% greater than the single mash yielding 1.086, not even accounting that the single mash will require much, much more grain to hit a gravity of 1.100.

Again the better efficiency is just a side benefit. The real benefit is being able to mash with 12 pounds of grain at a time instead of 24 (or 30) and make use of the equipment that I already own.

 

[doug293cz]

...
Huh ?

24 pounds x 37 points/lb = 888 points @ 100% efficiency.

The grain to water ratio is 24 pounds / 6 gallons = 4.0. According to your chart, 0.12 gallons per pound @ 0 sparges @ 4.0 = 58.5% efficiency. 888 points x 58.5% = 519.5 gravity points. 519.5 gravity in 6.0 gallons pre boil = 1.086 SG.
I don't use points in my calculations, but rather do a rigorous mass balance of both water and extract. 37 pts/lb works out to 37 / 46.173 = 80.13% dry basis extract potential (pure maltose/sucrose is 46.17 pts/lb.) The grain is assumed to have 4% moisture, so the dry weight of 24 lbs moist grain is 23.04 lb. So the max extract at 100% conversion yield is 23.04 * 0.8013 = 18.462 lb. We mashed with 8.88 gal of water, which has a weight of 8.88 * 8.3304 = 73.974 lb (water has a density of 8.3304 lb/gal at 68°F, and I adjust all volumes in my calcs to 68°F.) Now the grain was 4% water, or 0.96 lb, and saccharification consumed 0.68 lb, so the total weight of water at end of mash is 73.974 + 0.96 - 0.68 = 74.251 lb. Thus the Plato of the wort in the mash is 100°P * 18.462 / (18.462 + 74.251) = 19.91°P. 19.91°P is equivalent to an SG of 1.0826. Your point calculation does not correct for the moisture content of the grain (a common failing of a lot of brewing software.) If we use the dry weight of 23 lb, the max points are 23 * 37 = 851 pts. At 58.5% mash efficiency the wort will have 851 * 0.585 = 497.4 pts. 497.4 / 6 gal = 83.0 pts/gal => 1.0830. The difference between 1.0826 and 1.0830 is due to rigorous calculation vs. using an approximate calculation (the approx. calc is actually pretty good.)

...

The first mash "preboil volume" is 7.44 gallons. The grain to water ratio is 12 pounds / 7.44 gallons = 1.61. From the chart @ 0.12 and 0 sparges, the efficiency is 77.8%.

12 x 37 = 444 points @ 100% or 345 points @ 77.8%.

The gravity at the end of the first mash is 345 / 7.44 = 1.0464
Again, the difference between my 1.445 and your 1.0464 is you didn't correct for the moisture content of the grain.

...

Jamil et al say that NONE of the extract from the first mash is lost into the grain of the second mash. Their finding is that the grain in the second mash absorbs water only, no extract.
I just gotta say this makes no sense from a fundamental physical chemistry standpoint. I'm gonna have to look for their data. At the moment, I don't believe it. Why would the grain selectively reject dissolved sugar (selectively absorb only water) in the second mash when it doesn't do this in the first mash?

...

Not sure where you are getting this number from. Each mash yielded 9.23 pounds of extract for a total of 18.46 pounds of extract.
But only 7.17 lb of the extract from the first mash made it into the second mash, due to lautering losses. So, the extract in the second mash is 7.17 + 9.23 = 16.40 lb.

Is this pre mash or pre boil ? 6 gallons x 8.333 pounds per gallon = 50 pounds of water. 7.44 gallons x 8.333 pounds per gallon = 62 pounds of water.
7.44 gal of wort at an SG of 1.0445 weighs: 7.44 * 8.3304 * 1.0445 = 64.74 lb of wort. This wort contains 7.17 lb of extract, and 64.74 - 7.17 = 57.57 lb of water. The 57.70 results after correcting for the moisture content of the grain in the 2nd mash and the water consumed by saccharification in the 2nd mash (0.13 lb.)

You mean 12.09/ 18.46, not 19.46.
Yeah, 19.46 was a typo. 12.09 / 18.46 does equal 0.655

The mash efficiency gain by your yield is 65.5 - 58.5% = 7%
The mash efficiency gain by my yield estimation is 75.7% - 58.5% = 17.2%
I stand by my numbers. Your numbers contain an error in the extract weight from the first mash that actually gets into the second mash (at a minimum.)

However, both of these numbers are MISLEADING because they are NOT FOR THE SAME PRE BOIL GRAVITY as in the single mash case. The single mash case yielded a pre boil gravity of 1.086. Your double mash numbers yielded a pre boil gravity of 1.0926, significantly higher. To reach this gravity with a single mash would require adding more grain, which would further push the grain to water ratio, which would further decrease the efficiency and require even more grain.

Feel free to run the numbers on obtaining a pre boil gravity of 1.0926 using a single mash. I'd be very interested to see how much grain it would take as well as what the efficiency would be. I suspect that the efficiency difference would be even greater, probably something in the order of 10%.
I disagree about them being misleading. One of the assumptions for the analysis was using the same grain bill for both cases. If you want to change the assumptions, we can do that. Getting a 1.0926 SG for 6 gal pre-boil with a single no-sparge mash requires 29.5 lb of grain, and yields an efficiency of 53.3%.

My efficiency yield was based on 100% of the extract from the second batch being additive to the extract yield of the first batch as per the discussion between Palmer, Jamil and Blichman. I am not convinced that this actually occurs so I suspect that the preboil gravity will land somewhere between my calculation and yours, ie 1.0926 and 1.112. But even at a gravity of 1.100 the yield is 600 gravity points from a possible 888 = 67.5%, which is 9% greater than the single mash, not even accounting that the single mash will require much more grain to hit a gravity of 1.100.
As stated above, 100% of the extract obtained in the collected wort from the first mash is 7.17 lb, not 9.23 lb.

Again the better efficiency is just a side benefit. The real benefit is being able to mash with 12 pounds of grain instead of 24 (or 30) and make use of the equipment that I already own.
Agree that the smaller mash vessel requirement is the greatest benefit.

Brew on

 

[brewman !]

The first mash "preboil volume" is 7.44 gallons. The grain to water ratio is 12 pounds / 7.44 gallons = 1.61. From the chart @ 0.12 and 0 sparges, the efficiency is 77.8%.

12 x 37 = 444 points @ 100% or 345 points @ 77.8%.

The gravity at the end of the first mash is 345 / 7.44 = 1.0464

Again, the difference between my 1.445 and your 1.0464 is you didn't correct for the moisture content of the grain.

You missed a zero. Your rigorous method came out at 1.0445. My points approximation came out at 1.0464. 1.1 points difference. I'd say my points approximation works well enough.

I just gotta say this makes no sense from a fundamental physical chemistry standpoint. I'm gonna have to look for their data. At the moment, I don't believe it. Why would the grain selectively reject dissolved sugar (selectively absorb only water) in the second mash when it doesn't do this in the first mash?

Think about concentration gradients. The malt contains 4% moisture at the start. The wort is mostly water and the dissolved extract in the wort can freely move to another water molecule. The malt has a very high extract concentration within the malt kernel pieces. The extract concentration in the wort is relatively low. The water moves into the malt and the extract doesn't. The extract moves into the wort.

There is another way to think about this... let's assume the extract concentration within the kernel pieces reaches equilibrium with the extract concentration in the wort during the final running of the final sparge. The lost extract will be 0.12 gallons/pound of wort absorbed by the malt. 0.12 gallons x 24 pounds = 2.88 gallons. If the gravity of the final running during the sparge is 1.050, the total points lost will be 50 x 2.88 = 144 points.

But this isn't quite right either because there would be a "loss" like this if the mash was lautered with plain water. Ie the gravity of the water absorbed by the mash will reach equilibrium with the wort gravity. So the "extra" "loss" is probably the difference between the equilibrium wort concentration during the final stages of the second sparge and the first sparge. Who knows what that is ? 30 points ? 30 points x 2.88 = 86.4 points.

My original statement was "So now we have 345 points from the first mash plus 327.7 points from the second mash = 672.7" Lets assume there is another loss of 86 points dyue to higher extract concentration in the absorbed wort.

The total contribution would then be 672.7 - 86 = 586.3 points. In 6 gallons, that yields a gravity of 1.097

There is another indicator at play here... John Blichman did 3 reiterative mashes and reported no loss of efficiency. "Things were additive", meaning that each mash just kept adding gravity points to the incoming wort. Surely if the process didn't work, John would have observed issues when doing it 3 successive times. Just because we can't explain how it works doesn't mean it doesn't work.

Not sure where you are getting this number from. Each mash yielded 9.23 pounds of extract for a total of 18.46 pounds of extract.

But only 7.17 lb of the extract from the first mash made it into the second mash, due to lautering losses. So, the extract in the second mash is 7.17 + 9.23 = 16.40 lb.

So the no sparge lauter loses (9.23-7.17) / 9.23 = 22% of the extract ! Wow. I know what I'd change in my process. FWIW, the tests that Jamil spoke of fly sparged. I forgot to account for the poor sparge efficiency. Herein lies the difference in our numbers.

However, both of these numbers are MISLEADING because they are NOT FOR THE SAME PRE BOIL GRAVITY as in the single mash case. The single mash case yielded a pre boil gravity of 1.086. Your double mash numbers yielded a pre boil gravity of 1.0926, significantly higher. To reach this gravity with a single mash would require adding more grain, which would further push the grain to water ratio, which would further decrease the efficiency and require even more grain.

Feel free to run the numbers on obtaining a pre boil gravity of 1.0926 using a single mash. I'd be very interested to see how much grain it would take as well as what the efficiency would be. I suspect that the efficiency difference would be even greater, probably something in the order of 10%.

I disagree about them being misleading. One of the assumptions for the analysis was using the same grain bill for both cases. If you want to change the assumptions, we can do that. Getting a 1.0926 SG for 6 gal pre-boil with a single no-sparge mash requires 29.5 lb of grain, and yields an efficiency of 53.3%.

BINGO ! Here is the punchline when comparing a single mash to a reiterative mash.

To get 6 gallons of 1.0926 wort takes 29.5 pounds of grain with a single mash using the no sparge method. This has a malt efficiency of 53.3%.

To get 6 gallons of 1.0926 wort takes 24 pounds of grain with a reiterative mash using the same no sparge method This has a malt efficiency of 65.5%.

The reiterative mash method, even with no sparge, is 12.2% more efficient, needing 5.5 pounds less malt overall. You can also do it in a much smaller vessel.

I'd love to see the OP do a reiterative mash on 24ish pounds of malt and report his numbers. With good sparges, I think he'll be near his 1.120 desired gravity.

 

[deadwolfbones]

I just might. (The grain bill is actually 22.9lb plus 1lb sucrose and 1.5lb maltodextrin, though. And it's a 6 gal batch.)

Since I'm not someone who usually sparges, what's the best way to calculate the necessary mash and sparge volumes for a reiterated mash like this?

The only thing holding me back from going for the reiterated mash whole hog is the time factor. If I can mash the whole grain bill at once, sparge it, and get similar efficiency to the reiterated mash, I don't see an advantage to spending an extra hour and change mashing twice.

 

[Soulshine2]

I'm a full-volume/no-sparge BIAB brewer. I use a Robobrew V3 (no pump).

I recently picked up a 5 gallon whiskey barrel, and I'm going to be making a big-ass pastry stout to fill it with—6 gallons total, estimated OG 1.120.

Since I'll never fit 24 lbs of grain in the Robobrew for a full-volume mash, my initial thought was to do two mashes in sequence and combine in the kettle before boiling. But that would take a long time.

Then I thought, I can ask someone at the brew club to borrow their kettle/bag and do two mashes concurrently and combine in the boil kettle (the Robobrew, in this case). I got someone to loan me their Blichmann 10g kettle and matching bag. Cool!

Then I realized that I could do the entire batch in the 10g kettle if I do a little less than full volume and do a dunk or pourover sparge. Say, 6 gal mash and 3 gal sparge.

I'm not sure which would be the best course of action. The first would take more time but keep me with equipment/processes I know and trust. The second would take less time, but involve using unfamiliar equipment. The third would take less time, but involve using unfamiliar processes.

Anyone have tips on how to proceed, considerations I might have missed, etc?

"I recently picked up a 5 gallon whiskey barrel, and I'm going to be making a big-ass pastry stout to fill it with—6 gallons total"
How are you going to do that?

 

[deadwolfbones]

"I recently picked up a 5 gallon whiskey barrel, and I'm going to be making a big-ass pastry stout to fill it with—6 gallons total"
How are you going to do that?

Ha!

When filling a barrel I find it good to make extra so you can top up after some of the barrel volume evaporates.

I also plan to bottle some of the un-barreled stout for comparison purposes.

 

[brewman !]

Since I'm not someone who usually sparges, what's the best way to calculate the necessary mash and sparge volumes for a reiterated mash like this?

The mash volume is whatever grist/water ratio you want for your mash, plus the volume of water that doesn't touch the grist.

You sparge until you'e got the right amount of wort in the boil kettle. If you sparge too much, you boil off some water.

The only thing holding me back from going for the reiterated mash whole hog is the time factor. If I can mash the whole grain bill at once, sparge it, and get similar efficiency to the reiterated mash, I don't see an advantage to spending an extra hour and change mashing twice.

The efficiencies are nowhere near the same. See my post above.

Let us know how you make out.

 

[brewman !]

How much malt will OP need to hit 1.120 by malt alone, using BIAB with triple sparging ?

Guess: 24 pounds. Grain to water ratio is 4. Efficiency is 78.5% from the chart. 24 lbs x 37 x .785 = 697 points. He needs 720 (6 x 1.120), but he can bump things up with the sugar adjuncts.

Moral of the story... good lautering techniques make a huge difference when trying to produce high gravity worts.

 

[deadwolfbones]

The mash volume is whatever grist/water ratio you want for your mash, plus the volume of water that doesn't touch the grist.

You sparge until you'e got the right amount of wort in the boil kettle. If you sparge too much, you boil off some water.

Yes, of course. I know the fundamental idea. I'm just trying to determine how to calculate it ahead of time so I'm not figuring it out on the fly (and so I prep the right amount of sparge water). It's easy to figure out when doing a single mash & sparge, using Priceless or Doug's spreadsheet, but seems like it would be a bit more complicated when doing the reiterated mash.

 

[brewman !]

With the reiterative mash, work backwards to find your volumes.

You need 6 gallons pre boil into the boil kettle at the end of the 2nd mash. The 2nd mash will absorb 12 pounds x .06 gallons per pound = 0.72 gallons. Thus you need to start the 2nd mash with 6.72 gallons.

The first mash will end with 6.72 gallons. The first mash will absorb 12 pounds x 0.06 gallons per pound = 0.72 gallons. Thus you need to start the first mash with 7.44 gallons.

Add in a bit extra for spills, drips, leaks, etc. If you end up with a bit extra in the boil kettle, boil it off.

 

[doug293cz]

Yes, of course. I know the fundamental idea. I'm just trying to determine how to calculate it ahead of time so I'm not figuring it out on the fly (and so I prep the right amount of sparge water). It's easy to figure out when doing a single mash & sparge, using Priceless or Doug's spreadsheet, but seems like it would be a bit more complicated when doing the reiterated mash.

The basic volume equations are the same for a single mash or reiterated mash, when doing batch sparging:

Total Brewing Water = Pre-Boil Volume Target + Grain Weight * Grain Absorption Rate + MLT Undrainable Volume

Total Brewing Water = Strike Volume + Sparge Volume
MLT Undrainable Volume = Volume of liquid left in bottom of MLT when drained in the same way as during lautering (does not include volume absorbed by grain.)
​

In single mash brewing, you want the initial runnings and sparge runnings to have approximately equal volumes, as this maximizes lauter (and thus mash) efficiency.

Sparge Volume = Pre-Boil Volume Target / 2

Strike Volume = Grain Weight * Grain Absorption Rate + MLT Undrainable Volume + Pre-Boil Volume Target / 2
​

The complication with reiterated mashes is how to allocate sparge water between the two (or more mashes.) Of course it's simple if you do no-sparge, as sparge water is 0, so just use all the brewing water for the strike in the initial mash. In multiple mashing, whatever wort results from the previous mash is use to strike the subsequent mash.

If you want to know how much grain to use to hit a pre-boil SG target at a specific pre-boil volume target, then you need to use a spreadsheet, and the goal seek function to get grain weight. My spreadsheet can do this for a single mash, but not a reiterated mash.

Brew on

 

[doug293cz]

How much malt will OP need to hit 1.120 by malt alone, using BIAB with triple sparging ?

Guess: 24 pounds. Grain to water ratio is 4. Efficiency is 78.5% from the chart. 24 lbs x 37 x .785 = 697 points. He needs 720 (6 x 1.120), but he can bump things up with the sugar adjuncts.

Moral of the story... good lautering techniques make a huge difference when trying to produce high gravity worts.

Define the sparging process for batch sparging reiterated mashes in detail. Do you sparge after all mashes, or only the last? If you sparge more than one mash, how do you allocate total sparge water across the different mashes? How do you know the sparge strategy chosen is optimal?

And, as I said previously, you cannot use the chart for reiterated mashes. The constraints and assumptions are only valid for single mashes with batch sparging.

Brew on

 

[brewman !]

I have always done full-volume BIAB (no sparge, but squeezed pretty hard). Is it possible to do reiterative mashing that way?

Yes, but the efficiency sucks.

Whether full-volume or sparged, is reiterative mashing more efficient than doing two concurrent full-volume mashes with 1/2 the water in each mash?

You mean doing 2 mashes at the same time in separate vessels with half the grain and half the water in each ? The efficiency of that is the same as doing one single mash with all the grain. It is the grain to final wort volume ratio that determines efficiency.

Define the sparging process for batch sparging reiterated mashes in detail.

It is just like you sparge a single mash. Fly, batch, etc. Same thing. Do whatever you normally do.

Do you sparge after all mashes, or only the last?

Each mash is sparged at the end, just like a regular mash. The only difference is that the next mash is started with wort instead of water.

If you sparge more than one mash, how do you allocate total sparge water across the different mashes?

If you start with the right amount of water, you just use the whole amount that you have at the end of the previous mash.

In this 24 pound example, we assume that 0.72 gallons is absorbed in the grain (0.06 x 12) PER MASH if the bag is squeezed. So you start with 7.44 (6 + 2 x 0.72) gallons of treated water. If you are mashing at 1.5 quarts per pound, you put 18 quarts (4.5 gallons) into the mash vessel and mash with it and keep the rest (2.94 gallons) out for sparging at the end of the mash. At the end of the mash, you'll squeeze the grain bag and remove it. You should have collected 6.72 gallons of wort. Remove all but 4.5 gallons. Heat it up to strike temp. Remove the grain from the mash vessel (bag). Put in the new grain. Mash. Sparge with all the collected wort. Squeeze the bag. You should collect 6.0 gallons when done. If you have more, boil it down. If you have less, sparge with a bit of water.

How do you know the sparge strategy chosen is optimal?

Trade off between time, effort and cost. Define optimal.

And, as I said previously, you cannot use the chart for reiterated mashes. The constraints and assumptions are only valid for single mashes with batch sparging.

I emailed John Palmer on the weekend and he assured me that the gravity gain from successive mashes are additive. So the mashes can be considered individually and your chart works for individual mashes.

 

[doug293cz]

.

...

Trade off between time, effort and cost. Define optimal.

Maximum pre-boil SG for a given amount of total brewing water, for a single batch sparge after either or both mashes. Sparge with water only after second mash, sparge with water after both mashes, sparge with water after first mash then sparge with first mash wort after second mash, etc.

I emailed John Palmer on the weekend and he assured me that the gravity gain from successive mashes are additive. So the mashes can be considered individually and your chart works for individual mashes.

I listened to the podcast again, and it is pretty clear that John was talking about conversion efficiency (extract created in mash / max potential extract) not being affected by striking with wort instead of water, i.e. the presence of sugar does not inhibit the saccharification process. All my calculations assume that is the case. My chart is for lauter efficiency (extract in BK / extract created in mash), and if conversion efficiency is 100%, then mash efficiency = lauter efficiency. Lauter efficiency is different when doing reiterated mashing.

Brew on

 

[brewman !]

They were measuring gravity changes between mashes. They were not differentiating between lauter and mash efficiency. Jamil, John Blichman, John Tonsmeier and John Palmer have all brewed reiterative batches and reported that the gravities were "additive".

I'll report back when I do mine.

 

[doug293cz]

They were measuring gravity changes between mashes. They were not differentiating between lauter and mash efficiency. Jamil, John Blichman, John Tonsmeier and John Palmer have all brewed reiterative batches and reported that the gravities were "additive".

I'll report back when I do mine.

Read this (note that Troester uses "brewhouse efficiency" where most homebrewing literature uses "mash efficiency"), and listen to the podcast again. They were measuring the SG of the wort in the MLT at the end of mash. This is what is used to measure conversion efficiency.

Brew on

 

[deadwolfbones]

So let's say I want to do a reiterated/polygyle mash. My recipe, halved, is 11.45 lbs of grain. I want to end up with 7.5 gallons of wort pre-boil, because I'm doing a 3hr boil, my boil-off is 0.5 gal/hr, and I want to end up with 6 gal in the fermenter. I want to sparge each mash, but it's going to be a pretty thin mash (2.1 qt/lb for the first mash) so I probably don't need a huge sparge, right?

This is what I have in mind. Please tell me if it's stupid:

1. First mash: 11.45 lbs of grain + 6 gallons of water.
2. Sparge with 1.5 gallons of water, squeeze bag post-sparge. Grain absorption = ~0.75 gal (0.065/lb x 11.45). End up with 6.75 gal.
3. Second mash: 11.45 lbs of grain + 6.75 gal of wort.
4. Sparge with 1.5 gallons of water, squeeze bag post-sparge. Grain absorption still = ~0.75 gal. End up with 7.5 total gallons of wort.

If it's preferable to have a thicker mash, I could mash with 4 gal the first time (~1.4 qt/lb) and sparge with 2.5 gal each time.