A basic thing with the efficiency and cloning recipes of large breweries.

Homebrew Talk - Beer, Wine, Mead, & Cider Brewing Discussion Forum

Help Support Homebrew Talk - Beer, Wine, Mead, & Cider Brewing Discussion Forum:

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
When first reading this thread it was dismissed, given short shrift.

On a second read was data to support the theory, but I thought it couldn't be right from my experience.

After consideration, yes, in some circumstances this would be so.

I normally mash for 90 minutes, 2 hours for with highly kilned malts, don't mash out and sparge with hotter liquor for about 2 hours. This provides a similar level of efficiency to commercial breweries and I have no difficulties rescaling recipes.

So now I agree, but only if mashing for a short period with mashout and certain mixed grist recipes.
 
OP
OP
Kristoffer84

Kristoffer84

Active Member
Joined
Jul 22, 2021
Messages
28
Reaction score
15
Location
Kiel, Germany
But no matter how people see it, it must be clear for everybody that 100gram of Black Malt more or less in the same recipe no matter how big the Brewhouse effi. is, IS a total difference ;)
 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
I'm now not sure I understand what is being claimed.

If I mixed 1% black malt evenly into 99% pale malt, then mash for 90 minutes, carefully sparge for 2 hours to achieve 90% extraction for a 1040 wort, I don't understand why there will be difference between 5 kg or 500 kg apart from volumes and flow rates.

I can understand there being a difference if the black was not mixed with the base malt and it was mashed for a short period, especially when there was a difference in the lautering in a brewery and at home on a small scale.
 
OP
OP
Kristoffer84

Kristoffer84

Active Member
Joined
Jul 22, 2021
Messages
28
Reaction score
15
Location
Kiel, Germany
I'm now not sure I understand what is being claimed.

If I mixed 1% black malt evenly into 99% pale malt, then mash for 90 minutes, carefully sparge for 2 hours to achieve 90% extraction for a 1040 wort, I don't understand why there will be difference between 5 kg or 500 kg apart from volumes and flow rates.

I can understand there being a difference if the black was not mixed with the base malt and it was mashed for a short period, especially when there was a difference in the lautering in a brewery and at home on a small scale.

Ah okay pretty easy explained: Lets say you have 50% efficiency and there you need lets say 5000g of Malt overall to get to your OG. Now you do the same recipe but with lets says 90% efficiency. Lets say for that brew now, due the higher efficiency, you only need 3000g of Malt to get to the OG. Now take for example 10% Black malt.... there will be a difference from 500g to 300g only cause of efficiency. So one recept will pump in 200 gramm more roastiness into the recipe of same volume ...
 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
Ah okay pretty easy explained: Lets say you have 50% efficiency and there you need lets say 5000g of Malt overall to get to your OG. Now you do the same recipe but with lets says 90% efficiency. Lets say for that brew now, due the higher efficiency, you only need 3000g of Malt to get to the OG. Now take for example 10% Black malt.... there will be a difference from 500g to 300g only cause of efficiency. So one recept will pump in 200 gramm more roastiness into the recipe of same volume ...

So I think you are saying that an inefficient system won't extract all base malt sugars, but will extract all flavor from highly kilned malts.

But when all malt is converted into soluble sugars, why would the sparge liquor prefer to dissolve soluble darker malt than soluble pale ones?

To me, this is as if the base malt was mashed while the black malt added to the boiler, so I can follow this thinking when the darker malts are separately steeped.

Thank you for the the hypothesis and your efforts. My next Stout will have 70% Pale Malt, 20% Flaked Barley and 10% Roast Barley. All the grains will be well mixed and mashed for 2 hour for full conversion. 3/4 of the potential extract will be in the boiler before half of the required volume is collected and the last runnings will be ~1006 and no more than pH 5.6 whether the volume is 10L or 100L.

I will readily agree that there is much more to brewing than might be first thought.
 
OP
OP
Kristoffer84

Kristoffer84

Active Member
Joined
Jul 22, 2021
Messages
28
Reaction score
15
Location
Kiel, Germany
So I think you are saying that an inefficient system won't extract all base malt sugars, but will extract all flavor from highly kilned malts.

But when all malt is converted into soluble sugars, why would the sparge liquor prefer to dissolve soluble darker malt than soluble pale ones?
Yes i am saying that. If ALL the Aromes i cant say since i am not a Lab. BUT i would guess yes... i tested with Black Malt pure several steps and it was a harsh difference ;)
Its about the Aromes... the base malt aromes are the most neutrals in your brew. But 200g Black malt (in 20 Liters) more or less IS tasteable. I tried it, its pretty easy. But 200g Base Malt (Pale Ale Malt) more or less nobody can taste, and i mean nobody :D
 
Last edited:

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
So I think you are saying that an inefficient system won't extract all base malt sugars, but will extract all flavor from highly kilned malts.

But when all malt is converted into soluble sugars, why would the sparge liquor prefer to dissolve soluble darker malt than soluble pale ones?

To me, this is as if the base malt was mashed while the black malt added to the boiler, so I can follow this thinking when the darker malts are separately steeped.

Thank you for the the hypothesis and your efforts. My next Stout will have 70% Pale Malt, 20% Flaked Barley and 10% Roast Barley. All the grains will be well mixed and mashed for 2 hour for full conversion. 3/4 of the potential extract will be in the boiler before half of the required volume is collected and the last runnings will be ~1006 and no more than pH 5.6 whether the volume is 10L or 100L.

I will readily agree that there is much more to brewing than might be first thought.
The difference is that sugar extraction and Aroma extraction follow completely different pathways. Aroma extraction is more like a tea, it happens rather quickly and is not so much depending on 2nd factors like ph, conversion, gelatinization etc. Sugar extraction however, can be messed up in multiple ways which would not affect aroma extraction from roasted malts at all. So now assuming that sugar extraction and flavour extraction are correlating closely is not correct if you ask me.

At the end, the recipe won't be the same if % of the total grist is used when facing different efficencies of sugar extraction on different systems. The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash".

I think it is not such a big topic as most recipes are empirically proven and all our efficiencies are relatively close to each other so that on a practical level, the effect on flavour is not so big. But purely technically speaking, it is not correct to go this percentage route with recipes. At least not when there are grains involved that are just or mainly having an effect on the flavour, and/or if there are simple sugars in the recipe which always have 100% efficiency on their own.
 
Last edited:

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash".

Think about what happens when scaling to a lower efficiency system. In your view, you should scale the base malt, but not the specialty malt. When you increase the base malt, you are also increasing the total amount of water (due to grain absorption), and are also likely increasing the total amount of water to compensate for lautering inefficiencies that drove the lower mash efficiency in the first place.

But the volume of wort going into the kettle hasn't changed. All of that "extra" water is left behind somewhere (grains, deadspaces, transfer hoses). And so that fixed "xyz g of roasted barley" has spread out over a larger water (wort) volume.

To use your tea anaology, steep your teabag in 8 oz of water, then pour into a serving cup. Steep another teabag in 16 oz of water, and put half of that into a second serving cup. Each of these teas could in principle have been "fully steeped," but they won't taste the same or have the same color.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
Think about what happens when scaling to a lower efficiency system. In your view, you should scale the base malt, but not the specialty malt. When you increase the base malt, you are also increasing the total amount of water (due to grain absorption), and are also likely increasing the total amount of water to compensate for lautering inefficiencies that drove the lower mash efficiency in the first place.

But the volume of wort going into the kettle hasn't changed. All of that "extra" water is left behind somewhere (grains, deadspaces, transfer hoses). And so that fixed "xyz g of roasted barley" has spread out over a larger water (wort) volume.

To use your tea anaology, steep your teabag in 8 oz of water, then pour into a serving cup. Steep another teabag in 16 oz of water, and put half of that into a second serving cup. Each of these teas could in principle have been "fully steeped," but they won't taste the same or have the same color.
I was talking about the same amount of final wort here, not different amounts. We do not have half a cup here, we would end up with the exact same amount of wort at the end. The main driver would be probably boil off and that does not really make a difference. If someone has significant losses at other stages, i'd kindly suggest to look at the process first, before thinking about the percentage issue here :D
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
I was talking about the same amount of final wort here, not different amounts. We do not have half a cup here, we would end up with the exact same amount of wort at the end.

When you add (base) grains to compensate for a lower mash efficiency, you have to add more water in order to get the same amount of wort in the kettle. But you have now made more wort. The extra wort (and its water component) is left behind, just like the half of the tea that doesn't get poured into the serving cup in my analogy. The wort that is left behind contains a proportional amount of the color and flavor extracted from the (fixed amount of) roasted barley.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
When you add (base) grains to compensate for a lower mash efficiency, you have to add more water in order to get the same amount of wort in the kettle. But you have now made more wort. The extra wort (and its water component) is left behind, just like the half of the tea that doesn't get poured into the serving cup in my analogy. The wort that is left behind contains a proportional amount of the color and flavor extracted from the (fixed amount of) roasted barley.
I do not think that this really correlates with each other in a way that it compensates for each other, the additional roasted grains at lower efficiancy and the additional water loss with more grains in the mash. It also depends on the system. I do not have as much water left in the grain after the mash as a normal mash ton process would lead to as I BIAB and squeeze it really hard.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
The main driver would be probably boil off and that does not really make a difference. If someone has significant losses at other stages, i'd kindly suggest to look at the process first, before thinking about the percentage issue here :D


I see you edited your post after my response. Boiloff is a red herring. Lets assume identical boiloff in all cases, which will focus the discussion on mash efficiency, where it belongs.

What do you think drives mash efficiency issues? A proper mash should get 100% conversion efficiency, or something very close to it. That leaves lautering efficiency, which is driven by grain "absorption" (a function of crush), non-recoverable mash tun dead spaces, and transfer losses.

As I said before, when you add grains to make up for a low mash efficiency, regardless of the cause of the low efficiency, you have to use more water to get the same amount of wort to the kettle, and are thus making more wort. Period. That's not debatable. Less of your theoretical fixed amount of roasted barley flavor/color makes it to the kettle. That's really not debatable. But if you want to continue the debate, please focus on this paragraph and tell me how it's wrong.

And your suggestion "look at the process" first before thinking about the percentage issue doesn't make any sense to me. If there were no process differences (inefficiencies), there would be absolutely no need to talk about how to scale for efficiency differences in the first place.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
I do not have as much water left in the grain after the mash as a normal mash ton process would lead to as I BIAB and squeeze it really hard.

That's good! Squeezing is giving you higher lauter efficiency, and thus higher mash efficiency. But what if you didn't squeeze? Your mash efficiency would be lower, and you would need to use more grains and more water, making more wort. The extra wort would remain in the grain bag. Your fixed roasted barley contribution would now be diluted, as more of it would be left behind.

Just to be crystal clear, what I'm saying is that if you think critically about volumes and concentrations, the inescapable conclusion is that the following statement cannot be true: "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash". "
 
Last edited:

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
That's good! Squeezing is giving you higher lauter efficiency, and thus higher mash efficiency. But what if you didn't squeeze? Your mash efficiency would be lower, and you would need to use more grains and more water, making more wort. The extra wort would remain in the grain bag. Your fixed roasted barley contribution would now be diluted, as more of it would be left behind.

Just to be crystal clear, what I'm saying is that if you think critically about volumes and concentrations, the inescapable conclusion is that the following statement cannot be true: "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash". "
Not 100% correct, but still closer to reaching the goal of creating a repeatable recipe that translates best to different systems than the normal way of just giving the percentage of the roasted grain of the total grist. Actually water to grain ratio in the mash also plays a big role here as well, as this determines how much of the roast really is left behind per amount of grain in the mash, more water per grain in the mash, less aroma per amount of water meaning less aroma per fixed amount of water left back in the grain..... quite a complex scenario actually.

We could take this even one step further and look at simple sugars which would be given directly into the kettle. Given as a percentage of the total grist, this is a catastrophy with different mash efficiencies :D. In this case the grain loss would not play a role, so it is not completely the same thing as the roast grain problem, but same direction.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
Actually water to grain ratio in the mash also plays a big role here as well, as this determines how much of the roast really is left behind per amount of grain in the mash, more water per grain in the mash, less aroma per amount of water meaning less aroma per fixed amount of water left back in the grain..... quite a complex scenario actually.

It's really not all that complex. Some of us have done a lot a "mash/lauter math" over the years and are pretty comfortable with the volumes and concentrations involved. But, either I'm not following what you're saying here (language barrier) or perhaps you're actually agreeing with me. But let's make sure. Do you agree with the following statements?: "when you add grains to make up for a low mash efficiency, regardless of the cause of the low efficiency, you have to use more water to get the same amount of wort to the kettle, and are thus making more wort. Period. That's not debatable. Less of your theoretical fixed amount of roasted barley flavor/color makes it to the kettle." A simple "yes" or "no" can then shape how this debate progresses (or ends).

We could take this even one step further and look at simple sugars which would be given directly into the kettle. Given as a percentage of the total grist, this is a catastrophy with different mash efficiencies :D. In this case the grain loss would not play a role, so it is not completely the same thing as the roast grain problem, but same direction.

Another red herring. Sugar added to the kettle is completely different. It's not subject to the volumes and concentrations of the mash tun. Nobody (hopefully) would claim that kettle sugar additions should be scaled along with the grain bill when scaling the grain bill for different mash efficiencies.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
It's really not all that complex. Some of us have done a lot a "mash/lauter math" over the years and are pretty comfortable with the volumes and concentrations involved. But, either I'm not following what you're saying here (language barrier) or perhaps you're actually agreeing with me. But let's make sure. Do you agree with the following statements?: "when you add grains to make up for a low mash efficiency, regardless of the cause of the low efficiency, you have to use more water to get the same amount of wort to the kettle, and are thus making more wort. Period. That's not debatable. Less of your theoretical fixed amount of roasted barley flavor/color makes it to the kettle." A simple "yes" or "no" can then shape how this debate progresses (or ends).
Yes, but it even further complicates things. Let's just assume we always have the same grain absorption rate, 1kg water / 1kg of grain is lost during mashing/lautering. If we now look at somebody having a 75% mash efficiency using a grain to water ratio of 1/6 and compare that guy to another person who has the same mash efficiency, but uses a grain to water ratio of 1/5, then we see that the same amount of grain traps a different amount of aroma in it together with the water, as the aroma concentration is higher in the second case.

Looks easy at first view, but is actually quite complex at the end.
 
Last edited:

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
Another red herring. Sugar added to the kettle is completely different. It's not subject to the volumes and concentrations of the mash tun. Nobody (hopefully) would claim that kettle sugar additions should be scaled along with the grain bill when scaling the grain bill for different mash efficiencies.
Forgot the second part.

Yes, of course it shouldn't be scaled, yet the amount of sugar is almost always given in % of total grist so everybody is scaling it in some extend. Does not work this way, yet is the norm (at least in home brewing).
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
Let's just assume we always have the same grain absorption rate, 1kg water / 1kg of grain is lost during mashing/lautering. If we now look at somebody having a 75% mash efficiency using a grain to water ratio of 1/6 and compare that guy to another person who has the same mash efficiency, but uses a grain to water ratio of 1/5, then we see that the same amount of grain traps a different amount of aroma in it together with the water, as the aroma concentration is higher in the second case.

Ok. Assuming your same grain absorption, and your two different liquor to grist ratios, and assuming the same boiloff (per my request in post #52), which then requires the same amount of wort into the kettle, a big difference in this case is a large difference in conversion efficiency. If the first brewer (1/6 ratio) had 100% conversion efficiency, then the second brewer (1/5 ratio) had roughly 83% conversion efficiency (100% x (5/6)), which is really bad. We know this because of the difference in total (recovered + not recovered) wort made in each case, but yielding the same gravity. As I said in post #39, if you botch the mash, all bets are off.

But even if we accept an 83% conversion efficiency as normal/acceptable, the example doesn't support the contention that I was challenging at the beginning of this sidebar (starting with posts 47 and 48) , i.e. "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash", because that fixed amount of roasted barley contribution will be weaker in the first brewer's kettle than in the second (as you said).

So far, I haven't seen any logic, other than flawed tea and kettle sugar addition analogies (flawed because of the volumes and concentrations we see when we actually look under the hood), to support that contention.

If you have a specific example that does support that contention, please lay out the math.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
Ok. Assuming your same grain absorption, and your two different liquor to grist ratios, and assuming the same boiloff (per my request in post #52), which then requires the same amount of wort into the kettle, a big difference in this case is a large difference in conversion efficiency. If the first brewer (1/6 ratio) had 100% conversion efficiency, then the second brewer (1/5 ratio) had roughly 83% conversion efficiency (100% x (5/6)), which is really bad. We know this because of the difference in total (recovered + not recovered) wort made in each case, but yielding the same gravity. As I said in post #39, if you botch the mash, all bets are off.

But even if we accept an 83% conversion efficiency as normal/acceptable, the example doesn't support the contention that I was challenging at the beginning of this sidebar (starting with posts 47 and 48) , i.e. "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash", because that fixed amount of roasted barley contribution will be weaker in the first brewer's kettle than in the second (as you said).

So far, I haven't seen any logic, other than flawed tea and kettle sugar addition analogies (flawed because of the volumes and concentrations we see when we actually look under the hood), to support that contention.

If you have a specific example that does support that contention, please lay out the math.
You have just twisted my example in a way that supports your idea :D

I meant it exactly as I said, same efficiency, both end up with the same amount of extract in the kettle, one person just has more water in the kettle in addition to the extract.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
I meant it exactly as I said, same efficiency, both end up with the same amount of extract in the kettle, one person just has more water in the kettle in addition to the extract.

The way I interpreted your example also had the same mash efficiency. But doing it your way...

So the lower gravity higher preboil volume batch has diluted roasted barley color/flavor as compared to the higher gravity lower preboil volume batch. No disagreement there. Now, to get the same post boil gravity for both batches, the lower gravity higher volume batch with diluted roasted barley color/flavor will have to be boiled longer, thus concentrating the roasted barley color/flavor.

If you believe that your example proves your contention that "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash", please flesh out the numbers in the example, show the math, and show the logical connection to your contention. I won't respond to arm waving, and if that's all that ensues, I'll rest my case on my posts thus far.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
OK. So the firs
The way I interpreted your example also had the same mash efficiency. But doing it your way...

So the lower gravity higher preboil volume batch has diluted roasted barley color/flavor as compared to the higher gravity lower preboil volume batch. No disagreement there. Now, to get the same post boil gravity for both batches, the lower gravity higher volume batch with diluted roasted barley color/flavor will have to be boiled longer, thus concentrating the roasted barley color/flavor.

If you believe that your example proves your contention that "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash", please flesh out the numbers in the example, show the math, and show the logical connection to your contention. I won't respond to arm waving, and if that's all that ensues, I'll rest my case on my posts thus far.

You said it yourself, the boiling concentrates the flavour compounds, so you will end up with relatively more falvour compounds than the batch that had less water to go with from start.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
OK. So the firs


You said it yourself, the boiling concentrates the flavour compounds, so you will end up with relatively more falvour compounds than the batch that had less water to go with from start.

Ok. More arm waving with no math. I'm out.
 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
I'll leave this thread too.

Any recipe with 10% Black Malt should be treated with severe caution. I use Black Malt for color, usually Crisp at never more than 2%. I could use Fawcett , or Simpsons , or maybe one from Ireland Minch , or any of the many others. They are all very different from one another. 2% of one might well have more influence in a particular brew that 10% of another.
 

doug293cz

BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
HBT Supporter
Joined
May 14, 2014
Messages
13,403
Reaction score
10,269
Location
Renton
@VikeMan appears to have the best handle on the fundamentals of what factors might affect SG scaling vs. "flavor scaling" among the recent participants here.

What we need to be able to do the math is to define a new "Steeping Efficiency" which is analogous to conversion efficiency, and each specific process will have its own typical steeping efficiency as well as conversion efficiency. The definition of steeping efficiency would be:

Steeping Efficiency = Amount of Flavor/Color Compounds Released into the Mash / Amount of Flavor/Color Compounds in Grain Bill​

Compare to:

Conversion Efficiency = Weight of Extract Created in Mash / Potential Weight of Extract in Grain Bill​
So we have:

Amount of Extract at End of Mash = Conversion Efficiency * Grain Bill Weight * As Is Grain Potential​
Amount of Flavor/Color at End of Mash = Steeping Efficiency * Grain Bill Weight * Flavor/Color / Weight [F/C density in grain]
Then to get amounts of each pre-boil in the kettle, we multiply both of the amounts by the Lauter Efficiency
Extract in BK = Extract at End of Mash * Lauter Efficiency
Flavor/Color in BK = Flavor/Color at End of Mash * Lauter Efficiency
Extract and Flavor/Color in BK Post-Boil will be the same as Pre-Boil (ignoring the effects of heat on F/C degradation/formation, but I'm gonna just wave my hands about that at the end.)

What we really want to know going into the fermenter is the density of extract and density of flavor/color in the wort, which is just amount / volume. So, at the end of boil we have:


Extract Density = Extract in BK / Post-Boil Volume [we measure this with SG]
Flavor/Color Density = Flavor/Color in BK / Post-Boil Volume
So, to properly scale a recipe for both OG and flavor/color, you need to know the original recipe, and for both the original brewer's process, AND your process, the following:

Conversion Efficiency
Steeping Efficiency
Lauter Efficiency
You start with the original recipe, and the originator's efficiencies, and calculate the Post-Boil Extract Density (OG) and Flavor/Color Density using the equations above, in the order given. Then using the equtions in reverse order, you use your efficiencies to back calculate the scaled recipe. Good luck getting the individual efficiencies from the recipe creator. That's the big hole in trying to clone and scale recipes.

As far as flavor and color formation/degradation during the boil, I'm just going to stipulate that you will have to figure out how to give the wort the same heat load per volume in order to match what the originator got. How's that for a punt?

Brew on :mug:

 

DuncB

Well-Known Member
Joined
Jun 5, 2019
Messages
2,570
Reaction score
1,484
Location
Paremata New Zealand
Translation courtesy of google if it helps " muddy " the water

But this seems to summarise it well from Braukaiser

"
In Abriss der Bierbrauerei, German brewing author Ludwig Narziss defines Sudhausausbeute (German for brewhouse efficiency) as the ratio between the amount of extract in the boil kettle and the amount of grain that was used [Narziss, 2005]:

Sudhausausbeute = (kettle volume in l * kettle extract in % * kettle specific gravity) / grain mass in kg

Note that this is a different approach for defining efficiency. The reference is not the laboratory extract of the grain, but the total weight of the grain. The latter includes the weight of the husks and other insoluble material. Because of that the the Sudhausausbeute is also affected by the potential (or laboratory extract) of the malt used. This is also the definition that German home brewers use for efficiency. Thus care needs to be taken when reading efficiency numbers from German sources. While 75% is a very good efficiency number when based on the total grain weight (most grains laboratory extract is about 80% of their weight) it is only a modest efficiency when seen as based on the laboratory extract of the grain."




Sudhausausbeute
The brewhouse yield is a measure of the effectiveness of the work in the brewhouse . It describes what proportion of the malt went into solution during mashing . All work steps in the brewhouse are included, from crushing to beating the wort after hop boiling but before hot break separation .
The extract content and the volume of the cast-out wort are required as measured values . The extract content is measured using a saccharometer ( areometer , beer spindle ) at its calibration temperature in °P (degrees Plato, corresponds to % vol ). The wort may have to be cooled. The volume of the wort is either read off the scale on the calibrated wort kettle or determined using a calibrated measuring rod.
ATTENTION: The calculation of the brewing yield, which is also called the hot wort yield, should be calculated using the hot wort volume (including the hot trub ) , not the cast-out volume 1) .
The brewhouse yield A S is calculated as follows
diagramme-zeichnungen-tabellen:sha1.png

This calculation is not accurate for various reasons. The volume of the wort is determined at almost boiling temperature, but the measuring devices are usually calibrated to 20°C. The cooking pan, on the other hand, expands at high temperatures. Substances were also introduced into the wort from the hops, which falsify the value of the wort quantity. A correction value of 4% is assumed for all of these factors, and the deflection quantity is therefore multiplied by 0.96.
The value read on the saccharometer indicates the percentage by weight of the extract. These must be converted to percentages by volume by multiplying them by the specific gravity of the wort .
This gives the following corrected formula for the brewhouse yield:
diagramme-zeichnungen-tabellen:sha2.png

The specific weight SG can be roughly calculated (see article SG ) or read more precisely from the Plato table . The product of the saccharometer display, specific gravity and expansion correction factor is also summarized there as a yield factor, so that the formula can be simplified again with the help of the Plato table :
diagramme-zeichnungen-tabellen:sha3.png

Example:
23.0 l of wort with 12.0% original wort were produced from a bed of 4 kg of malt. The brewhouse yield is thus
diagramme-zeichnungen-tabellen:sha4.png

The values for the specific gravity (1.04837) or the yield factor (12.08) can be read from the Plato table for the extract content of 12.0% wt (saccharometer reading).
The value of the brewhouse yield in modern breweries is well over 75%. Depending on the mashing process and malt quality, around 65 to 75% is expected in the home brewing sector.
Other values used to assess the effectiveness of the brewing process are the cold wort yield, the fermenting room yield and the overall yield .
Links
1) Brew Recipe Developer Dokumentation DE



I have to say I know where I stand with the definitions used by brewfather summarised by @doug293cz rather than the above.
 
Last edited:

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
Ok. More arm waving with no math. I'm out.
Thanks! You do not have to understand everything.

But for the rest:

I am talking about the difference between "how much" aroma is lost at the same batch with different grain/water ratios, saying that with the same amount of extract in the kettle at the end, you will end up loosing more or less aroma in the kettle depending on the grain to water ratio.

Example: 10kg grain batch, grain absorption rate 1kg water /1kg grain.

batch 1 uses 1/5 grain to water ratio.

batch two uses 1/10 grain to water ratio.

Aroma spreads equally in the water.

Batch 1 is 45 litre big after mashing, loosing 5 litre with the grain. 5/50 of the aroma is left in the grain.

Batch 2 is 95 litre big after after mashing, loosing also 5 litre with the grain. The aroma left is 5/100 of the total amount of aroma.

Question who lost more aroma?

Right, batch one.
 
Last edited:

DuncB

Well-Known Member
Joined
Jun 5, 2019
Messages
2,570
Reaction score
1,484
Location
Paremata New Zealand
@Miraculix
That does assume a linear extraction of your " aroma ".
It's most likely that it isn't and so it could all be extracted in 10 litres and in that case Batch 2 would have less apparent aroma as it's over twice as dilute.

If I take a teabag and put it in a cup of boiling water and another teabag in a litre of boiling water will the smell ( aroma ) of tea be more in the cup of tea or the litre jug of tea?

But you'd probably be able to make a fair second cup of tea out of the teabag that had been used to make the cup of tea and it would probably still have more aroma than the litre of tea.


It's a puzzle that needs more experimentation.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
@Miraculix
That does assume a linear extraction of your " aroma ".
It's most likely that it isn't and so it could all be extracted in 10 litres and in that case Batch 2 would have less apparent aroma as it's over twice as dilute.

If I take a teabag and put it in a cup of boiling water and another teabag in a litre of boiling water will the smell ( aroma ) of tea be more in the cup of tea or the litre jug of tea?

But you'd probably be able to make a fair second cup of tea out of the teabag that had been used to make the cup of tea and it would probably still have more aroma than the litre of tea.


It's a puzzle that needs more experimentation.
Good Point. In my example I'm assuming that the same total amount of flavour gets extracted in both batches, one just being more diluted by the bigger amount of water. If this is a suitable reflection of reality is actually debatable.

Your point shows that this is way more complex than it looks on first sight, and this was also my initial statement which got a bit derailed during the arguing above, so thanks for bringing it back to that initial statement.

Is just like what op wanted to say and what I wanted to underline with my statements, flavour extraction and sugar extraction follow completely different pathways and hence do not necessarily correlate with each other. This means that scaling them the same way might not be the best idea if a constant amount of flavour is desired. It might be that at the end the amount of roasted grains also needs adjustment, just with another factor in front of it than the one for the base malt.
 

DuncB

Well-Known Member
Joined
Jun 5, 2019
Messages
2,570
Reaction score
1,484
Location
Paremata New Zealand
@Miraculix
Agreed.
Back to the cups of tea I expect that putting a teaspoon of sugar in each of them would be quite different. Same amount of sugar but the weaker second tea might taste sweeter, I'm not sure.
Cloning beers to homebrew scale is hard / impossible. It only takes my system about 4 minutes to raise from 63 to 68 celsius and about the same from there to mash out. I haven't been in a commercial big brewery but I find it hard to believe they'd raise a 5000 litre mash tun by 5 degrees in 4 minutes. Just so many variables .
Let's not mention hop bitterness and aroma with time and temperature across the boil and whirlpool and dry hop.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
Thanks! You do not have to understand everything.

But for the rest:

I am talking about the difference between "how much" aroma is lost at the same batch with different grain/water ratios, saying that with the same amount of extract in the kettle at the end, you will end up loosing more or less aroma in the kettle depending on the grain to water ratio.

Example: 10kg grain batch, grain absorption rate 1kg water /1kg grain.

batch 1 uses 1/5 grain to water ratio.

batch two uses 1/10 grain to water ratio.

Aroma spreads equally in the water.

Batch 1 is 45 litre big after mashing, loosing 5 litre with the grain. 5/50 of the aroma is left in the grain.

Batch 2 is 95 litre big after after mashing, loosing also 5 litre with the grain. The aroma left is 5/100 of the total amount of aroma.

Question who lost more aroma?

Right, batch one.

Ok, I'm back in, at least for now. You provided some numbers. Thanks for that. BTW, you are right... I don't "have to" understand "everything," but it's a goal.

You have neatly demonstrated that using more water in the mash will result in more efficient extraction of "aroma." But this is exactly what normally happens with sugars and dextrins. I don't know why you assumed that these two brewers would have had the same mash efficiency. Given your scenario, where the only water loss is to grain absorption, and it's equal for both batches (as you specified), in real life, the batch with more water would have had a higher mash efficiency, assuming that conversion wasn't botched. As I have stated twice before... if you botch conversion, all bets are off. But you chose an example that would not typically happen with normal batches.

To finish your math for you, if we assume 100% conversion for batch 1, then the recovered wort got 90% of the total potential sugars/dextrins, because lauter efficiency was 90% (i.e. 45/50).

Batch 2's recovered wort also got 90% of the total potential sugars/dextrins. But batch 2's lauter efficiency was 95% (95/100). So, we can compute batch 2's conversion efficiency:
90% Mash Efficiency / 95% lauter efficiency = 94.7% conversion efficiency. Botched conversion. Incomplete mash.

I will stipulate here (again) and for ever more that if conversion is not complete, then the correlation between mash efficiency and "aroma" efficiency may become weaker, if you will stipulate that the claim that "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash" is in no way supported by any example/math model presented so far, and in fact is soundly thrashed by the model @doug293cz presented in post #64.

If you haven't critically read post #64, please do. Going forward, I'll be happy to discuss anything relating to @doug293cz's model.
 

Miraculix

Well-Known Member
Joined
Jun 4, 2017
Messages
6,329
Reaction score
5,001
Location
Bremen
Ok, I'm back in, at least for now. You provided some numbers. Thanks for that. BTW, you are right... I don't "have to" understand "everything," but it's a goal.

You have neatly demonstrated that using more water in the mash will result in more efficient extraction of "aroma." But this is exactly what normally happens with sugars and dextrins. I don't know why you assumed that these two brewers would have had the same mash efficiency. Given your scenario, where the only water loss is to grain absorption, and it's equal for both batches (as you specified), in real life, the batch with more water would have had a higher mash efficiency, assuming that conversion wasn't botched. As I have stated twice before... if you botch conversion, all bets are off. But you chose an example that would not typically happen with normal batches.

To finish your math for you, if we assume 100% conversion for batch 1, then the recovered wort got 90% of the total potential sugars/dextrins, because lauter efficiency was 90% (i.e. 45/50).

Batch 2's recovered wort also got 90% of the total potential sugars/dextrins. But batch 2's lauter efficiency was 95% (95/100). So, we can compute batch 2's conversion efficiency:
90% Mash Efficiency / 95% lauter efficiency = 94.7% conversion efficiency. Botched conversion. Incomplete mash.

I will stipulate here (again) and for ever more that if conversion is not complete, then the correlation between mash efficiency and "aroma" efficiency may become weaker, if you will stipulate that the claim that "The correct way woul be something like "create a wort with an OG of xx and include xyz g of roasted barley in the mash" is in no way supported by any example/math model presented so far, and in fact is soundly thrashed by the model @doug293cz presented in post #64.

If you haven't critically read post #64, please do. Going forward, I'll be happy to discuss anything relating to @doug293cz's model.
I think we do not need to dive furthr here, we do not understand what`s happening here completely so we cannot create a mathematical model that get's close to reality. My point was just to show that your claim, that grain absorption compensates for the adittional added roast in lower efficiency brews is not correct as it is not only linked to the efficiency, but also to the grain to water ratio and probably even more.

My "arm waving" was just a hands on approach, which is often a better approach if it is obvious that there are so many unknowns that trying to force it into a mathematical equation leads to something which has nothing to do any more with reality itself.

I think the reasons for this have already been explained deeply enough.

To sum it up, I quote what I have written above:

... what I wanted to underline with my statements, flavour extraction and sugar extraction follow completely different pathways and hence do not necessarily correlate with each other. This means that scaling them the same way might not be the best idea if a constant amount of flavour is desired. It might be that at the end the amount of roasted grains also needs adjustment, just with another factor in front of it than the one for the base malt.
 

VikeMan

It ain't all burritos and strippers, my friend.
Joined
Aug 24, 2010
Messages
4,675
Reaction score
4,077
I think we do not need to dive furthr here, we do not understand what`s happening here completely so we cannot create a mathematical model that get's close to reality.

I think @doug293cz's proposed model has great promise. The challenge there is coming up with data and/or determining where it's safe (i.e. reasonable) to use estimates/defaults. The world is full of models that get answers "close to reality" without all of the ideal inputs, or even knowing what all of the underlying factors are.

In the meantime, I will continue to use mash efficiency as a proxy for flavor/color efficiency. It fits my real world obeservations to the extent I am subjectively capable of making them, and it is sensitive to lauter efficiency, which as you neatly demonstrated in post #66 must be a feature of any good model going forward.
 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
I'm pleased some common ground has been found. However, never having found such to happen when downscaling from commercial brew length to my own, nor when upscaling a recipe for a commercial brewery, I'm unable to support a theory that such applies to homebrewers in general.
 

doug293cz

BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
HBT Supporter
Joined
May 14, 2014
Messages
13,403
Reaction score
10,269
Location
Renton
I'm pleased some common ground has been found. However, never having found such to happen when downscaling from commercial brew length to my own, nor when upscaling a recipe for a commercial brewery, I'm unable to support a theory that such applies to homebrewers in general.
Using my model from post #64, your observations would mean that, for the systems you use, conversion efficiency and steeping efficiency track pretty closely. There may be cases where this is not the case. As a possible example, I will offer up what @RM-MN has said about his particular brewing process. He does BIAB, and grinds extremely fine. He has reported getting ~100% conversion in ~15 minutes, but that he has to let the mash go ~30 minutes in order to fully extract the flavors of the grain. (@RM-MN feel free to correct me if I have materially misrepresented any of your writings.) In this case conversion and steeping efficiency (rates actually) mismatch in the opposite direction of that postulated to be the general case by @Kristoffer84 .

Brew on :mug:
 

RM-MN

Supporting Member
HBT Supporter
Joined
Nov 26, 2010
Messages
15,118
Reaction score
6,219
Location
Solway
Using my model from post #64, your observations would mean that, for the systems you use, conversion efficiency and steeping efficiency track pretty closely. There may be cases where this is not the case. As a possible example, I will offer up what @RM-MN has said about his particular brewing process. He does BIAB, and grinds extremely fine. He has reported getting ~100% conversion in ~15 minutes, but that he has to let the mash go ~30 minutes in order to fully extract the flavors of the grain. (@RM-MN feel free to correct me if I have materially misrepresented any of your writings.) In this case conversion and steeping efficiency (rates actually) mismatch in the opposite direction of that postulated to be the general case by @Kristoffer84 .

Brew on :mug:
It's actually a bit faster than that but I don't want to do too much encouraging to people who do not mill the grain as fine as I do.
 

sibelman

Supporting Member
HBT Supporter
Joined
Nov 22, 2007
Messages
522
Reaction score
397
Location
Portland, OR
A lot of thought has gone into this discussion. This investigation would best be informed by measurement of aroma and flavor. Will the advent of an inexpensive "e-nose" soon take its place beside the scales, thermometers, hydrometer, refractometer, and pH meter in craft (and home) breweries?

 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
Using my model from post #64, your observations would mean that, for the systems you use, conversion efficiency and steeping efficiency track pretty closely. There may be cases where this is not the case. As a possible example, I will offer up what @RM-MN has said about his particular brewing process. He does BIAB, and grinds extremely fine. He has reported getting ~100% conversion in ~15 minutes, but that he has to let the mash go ~30 minutes in order to fully extract the flavors of the grain. (@RM-MN feel free to correct me if I have materially misrepresented any of your writings.) In this case conversion and steeping efficiency (rates actually) mismatch in the opposite direction of that postulated to be the general case by @Kristoffer84 .

Brew on :mug:

I'm British, mostly use British malts to brew British style beers. British malts are not the same as malts produced in America and some other countries. An example.....

Briess pale malt - protein 11.7% - Diastatic power 85
Briess two-row - protein 12% - Diastatic power 140
Briess six-row - protein 13% - Diastatic power 160

A typical British malt:
Fawcett pale malt - protein 9% - Diastatic power 50

My system is three vessel with grains of base and kilned malts cracked, not ground, with all evenly mixed. Mash lasts 90 minutes, the tun fitted with a Valentine Arm which is adjustable to maintain a floating mash bed for a #2 hour sparge. Only clear wort is transferred to the boil kettle. The majority of hops are whole cone and filter the boiled wort without whirlpool.
Open top fermentation with Yorkshire type yeasts roused for the first 2, occasionally 3 days, then lid fitted and the green beer slowly cooled to cellar temperature (~12C).
Clear beer casked 7 days from pitching (with the better yeasts) and ready to drink after a further 3 days. The beer is better after a week and more and bottling is done after 2 weeks in a barrel.

Aren't the variables here the selected ingredients and how they are processed?

Simpson say Black Malt has a pretty neutral flavour, making it a great choice for full flavoured beers in need of deeper colouring. If you pushed us though, we’d say it has a slightly nutty, cocoa flavour, although it’s very subtle.

Crisp say for their Black Malt up to 3% usage.
  • ROASTED COFFEE
  • LIQUORICE AND CURRANTS
Fawcett spec gives strong burnt coffee flavour, dark brown, 3 to 5% usage.

What could a proportional adjustment do if a recipe includes Black or any other malt if the maltster isn't known and it is brewed a totally different way?
 

doug293cz

BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
HBT Supporter
Joined
May 14, 2014
Messages
13,403
Reaction score
10,269
Location
Renton
...

Aren't the variables here the selected ingredients and how they are processed?

...
Yes, but process includes innumerable variables: crush, mash thickness, mash temp profile, mash times, mash pH, mash-out or not, "completeness" mash prior to lautering, etc. Characterizing them all adequately for just two systems to scale between is a Herculean task. Trying to come up with a "many-to-many" generalized scaling process is intractable. The model for thinking about what's involved is not to hard to understand, but in reality, "steeping efficiency" and rates can vary for individual ingredients, and real world complexity goes up exponentially.

Brew on :mug:
 

cire

Well-Known Member
Joined
Sep 25, 2017
Messages
642
Reaction score
499
Location
UK
Yes, but process includes innumerable variables: crush, mash thickness, mash temp profile, mash times, mash pH, mash-out or not, "completeness" mash prior to lautering, etc. Characterizing them all adequately for just two systems to scale between is a Herculean task. Trying to come up with a "many-to-many" generalized scaling process is intractable. The model for thinking about what's involved is not to hard to understand, but in reality, "steeping efficiency" and rates can vary for individual ingredients, and real world complexity goes up exponentially.

Brew on :mug:

Yes indeed, I agree with all you list, but more happens in the mash than starch extraction and its conversion to fermentable sugars. Many base malts will convert their starch to sugars in a few minutes, even low nitrogen/low diastatic power British malts will in the right circumstances, but some vital ingredients (eg free amino nitrogen) are processed there too, while some byproducts (eg phenols) can be eliminated.. All that can happen with sufficient time, so probably warrant inclusion in any algorithm to accounts for homebrewers who find a disproportional influence of non-base malt ingredients.

I will not claim that no beer should or could be made with a short mash, just that a longer mash will make a different beer even with all other processes and ingredients being identical.
 
Top