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Brewfather efficiency / SG predictions seem off

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McKnuckle

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I've been playing with BrewFather and it's quite a nice app in all respects! Easily the best UI and feature set I've seen so far (compared to the usual suspects).

However the efficiency math seems to be off, and I wonder if anyone else has noticed. Basically, if I plug in a given mash efficiency target into Brewfather, it makes both the pre-boil and original gravity estimates too high, usually by 4 or 5 points. To match other software, I have to crank down the Brewfather mash efficiency value by several percentage points.

I am a disciple of resident HBT efficiency guru @doug293cz , and I've incorporated his calculations into my own brewing spreadsheet. I'm not confused by the differences among conversion, lauter, mash, and brewhouse efficiency (just as a disclaimer). And I understand how these calcs use the dry grain weight, which perhaps not all of the popular software is aware of. However, even that, if removed from the equation, doesn't make the two agree.

As an example, take 12 lbs of American 2-row (37 ppg), mash it in 8 gallons of water, and pretend we'll get 100% conversion. Doug's calculations give a theoretical pre-boil gravity of 1.049 at 100% conversion efficiency. This produces a mash efficiency of 92%.

Brewfather doesn't have an input for conversion efficiency, so I have to input mash efficiency instead. If I put in 92% mash efficiency and zero out all losses to rule out other factors, the pre-boil gravity is predicted at 1.053. Naturally, the OG figures are off by at least that much as well.

With the dry grain weight factor removed, the numbers get closer by ~2 gravity points, but they still never match. I can't figure it out and it's driving me nuts!
 

doug293cz

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I get the 1.049 pre-boil SG for 12 lbs @ 37 pts/lb with 8 gal of strike water, but can't get to 92% mash efficiency. What are you assuming for grain absorption and MLT undrainable volume?

Brew on :mug:
 
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McKnuckle

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Hey Doug, thanks for posting. I set both variables to zero for this example. Your spreadsheet shows 91.97% mash eff. in this case. "Desired post boil wort" is 7 gallons, so a 1 gal/hr evaporation rate (irrelevant for a mash efficiency calc, but just in case).
 

doug293cz

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Ok, but 0 gal/lb apparent grain absorption is not feasible in a homebrew setting. Might be possible if you use a hydraulic or pneumatic filter press to lauter the mash. In this case a 0 apparent grain absorption rate translates to a 0.058 gal/lb true absorption rate, since the total wort volume in the mash is 8.699 gal.

I'd suggest putting in a more realistic grain absorption rate, and then see how the comparison with BrewFather works out.

If I am only interested in mash efficiency and pre-boil volume, I set either boil time or boil off rate to 0, then pre-boil volume = post-boil volume

Note: the true absorption rate calculation in the published version of the spreadsheet is incorrect (but that doesn't affect any of the other calculations.)

Brew on :mug:
 
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McKnuckle

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Fine, so a bad example. But the same phenomenon happens when I enter a proper absorption rate. That figure is not relevant to the issue. I've played with this a lot. Check it out online for yourself if you're interested in debugging.
 

doug293cz

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Fine, so a bad example. But the same phenomenon happens when I enter a proper absorption rate. That figure is not relevant to the issue. I've played with this a lot. Check it out online for yourself if you're interested in debugging.
Without access to the code, not possible to debug easily. I see they apparently work backwards from BrewHouse efficiency, which in my opinion is the wrong way to do things.

Brew on :mug:
 
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McKnuckle

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Yeah, unfortunately I realize we can't directly debug it. I thought it might be possible to infer the issue by manipulating some variables, which I've already done, but I'm just not gaining any insight.

They actually start from mash efficiency (and omit brewhouse) if you choose kettle as the target volume instead of fermenter. I have Googled this and actually got a couple of people asking similar questions, but it never went anywhere. Maybe the app doesn't have enough of a user base that's also familiar with other tools.

Anyway thanks for looking.
 

doug293cz

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Yeah, unfortunately I realize we can't directly debug it. I thought it might be possible to infer the issue by manipulating some variables, which I've already done, but I'm just not gaining any insight.

They actually start from mash efficiency (and omit brewhouse) if you choose kettle as the target volume instead of fermenter. I have Googled this and actually got a couple of people asking similar questions, but it never went anywhere. Maybe the app doesn't have enough of a user base that's also familiar with other tools.

Anyway thanks for looking.
Didn't see the option to select post-boil vs fermenter as batch size, but I did find an "experimental" setting to allow entering an assumed conversion efficiency (at menu link at the bottom of the "Settings" page.)

Brew on :mug:
 

doug293cz

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As a BIAB guy, I don't like that BrewFather calculates my strike water volume at mash temp, rather than room temp. Doesn't appear to be a way to change this.

After some thought, BF might be making a common mistake in calculating mash wort SG, which is to use strike volume rather than the actual wort volume, which is more than the strike volume. If we assume 12 lb of 1.037 potential grain in 8 gal of strike water, and use a 4% moisture correction, the incorrect points/gal calculation looks like this:
Pts/gal = 12 [lb] * 0.96 * 37 [pts/lb] / 8 [gal] = 53.28 pts/gal => 1.053 SG
The actual wort volume in this mash would be 8.699 gal so the correct pts/gal calculation looks like this:
Pts/gal = 12 [lb] * 0.96 * 37 [pts/lb] / 8.699 [gal] = 49.0 pts/gal => 1.049 SG
Brew on :mug:
 

schmurf

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There is pretty good support from the developer at their Facebook group, usually with quick answers.
 
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McKnuckle

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Doug, can you explain the 0.699 portion of your math?

The strike volume is 8.0 gallons. The residual water in the grain is 12 * 0.04 = 0.48. That's a total of 8.48 gallons water in that mash.

[Edit] - I answered my own question: It's because 4% is by weight, not volume... So you convert 0.48 lbs (not gallons) to the equivalent amount of water, which is 0.058 gallons. Still not 0.699 though.

About Facebook support; it's too bad as I deleted my account and won't go back there for pretty much anything. Thank you @schmurf for the tip though.
 
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doug293cz

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Doug, can you explain the 0.699 portion of your math?

The strike volume is 8.0 gallons. The residual water in the grain is 12 * 0.04 = 0.48. That's a total of 8.48 gallons water in that mash.

[Edit] - I answered my own question: It's because 4% is by weight, not volume... So you convert 0.48 lbs (not gallons) to the equivalent amount of water, which is 0.058 gallons. Still not 0.699 though.

About Facebook support; it's too bad as I deleted my account and won't go back there for pretty much anything. Thank you @schmurf for the tip though.
There is a common misconception about how the SG form of grain potential is defined. The correct definition is:
The grain potential is the SG that would result from one pound of grain being used to create one gallon of wort, given 100% conversion.​
What a lot of brewers think the definition is:
The grain potential is the SG of the wort created from mashing one pound of grain with one gallon of water.
The problem comes in because if you mash one pound of grain with one gallon of water, you get more than one gallon of wort. The extra volume is due to the volume of the extract in the wort. The wort volume is the strike water volume plus the extract volume (corrected for some minor non-linearity in the mixing equation for density.)

So, in the case we have been discussing, the extra 0.699 gal is the extract contribution to the wort volume.

Brew on :mug:
 

pricelessbrewing

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There is a common misconception about how the SG form of grain potential is defined. The correct definition is:
The grain potential is the SG that would result from one pound of grain being used to create one gallon of wort, given 100% conversion.​
What a lot of brewers think the definition is:
The grain potential is the SG of the wort created from mashing one pound of grain with one gallon of water.
The problem comes in because if you mash one pound of grain with one gallon of water, you get more than one gallon of wort. The extra volume is due to the volume of the extract in the wort. The wort volume is the strike water volume plus the extract volume (corrected for some minor non-linearity in the mixing equation for density.)

So, in the case we have been discussing, the extra 0.699 gal is the extract contribution to the wort volume.

Brew on :mug:
I couldn't count how many times I've tried to explain the 1 gallon of water vs 1 gallon of wort.
 
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McKnuckle

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Hi @doug293cz , I'd like to resurrect this thread because I'm having an offline convo with the developer of Brewfather.

In this example of how to calculate the volume of water contributed to wort from a given weight of grain, I think there's a mistake. Can you please review?

The example has us mashing 12 lbs of grain in 8 gallons of water. 4% of the 12 lbs is water weight. That's 0.48 lbs. If we convert 0.48 lbs to water volume, given that water weighs 8.33 lbs/gallon, we get 0.058 gallons. That's the total amount of water volume from the 12 lbs of grain.

The total volume of the resulting wort should therefore be 8.058 gallons.

You're coming up with 0.699 gallons of water from the grain, leading to 8.699 gallons of wort. I think you're mistakenly counting 0.058 twelve times instead of once. 12 * 0.058 = 0.696 (I assume there's some other small factor leading to 0.699).

Does this sound right, or am I overlooking something? Thanks!
 

jdudek

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Following.

I think the 0.699 is not water from grain but rathervolume of wort including dissolved sugar. 0.699 is the volume difference between strike water and wort post mash.
 
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McKnuckle

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I've thought of that, of course. But if that is the case, I'd like to understand how to calculate it. The coincidence that 12 x 0.058 is so close to 0.699 is odd. But let's see if he chimes in.
 

jdudek

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Here's how I arrive at his figure... or just about

Density of sugar = 1.59 g/cm^3 = 13.16 lbs/gallon

weight of sugar in 12 lbs of grain with 4% moisture and 80% potential = 12lbs * 0.96 * 0.8 = 9.216lbs

Volume of sugar = 9.216 lbs / 13.16 lbs/gallon = 0.693 gallons.


Looks like I'm off by 0.006 from doug's figure. indeed if he chimes in, maybe he can put out his exact calculation.
 

doug293cz

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Here's how to calculate actual wort volume in the mash:

Calculate the dry basis grain weight. The formula for this is:
Dry weight = As is weight * (1 - % moisture / 100%)​
if % moisture = 4% (typical), then dry weight = as is weight * 0.96​
Calculate the maximum potential extract weight from the grain bill:
If grain potential is in PPG, then convert to % by dividing by 46.173, which is the potential of pure sucrose.​
So, a weighted potential of 1.037 (37 PPG) is equivalent to 37 / 46.173 = 0.801 = 80.13%​
Thus 12 lbs of 4% moisture grain with a dry basis potential of 37 PPG contains a potential 12 * 0.96 * 0.8013 = 9.2310 lb of extract​
Calculate the total weight of the wort created at 100% conversion:
Water has a density of 8.3304 lb/gal @ 68°F, so 8 gal of water weighs: 8 gal * 8.3304 lb/gal = 66.6432 lb​
And total weight of wort = 9.2310 lb + 66.6432 lb = 75.8742 lb​
Calculate °Plato of wort, which is just the weight% of extract in the wort:
°P = 100°P * 9.2310 lb / 75.8742 lb = 12.166 °P​
Convert °Plato to Specific Gravity using the formula SG = 1 + °P / (258.6 - (°P / 258.2) * 227.1)
SG = 1 + 12.166 / (258.6 - (12.166 / 258.2) * 227.1) = 1.04908​
Calculate volume of wort using the formula Volume = Weight / (8.3304 lb/gal * SG)
Wort volume = 75.8742 lb / (8.3304 lb/gal * 1.04908) = 8.682​
Finally calculate the "excess" volume due to extract vs. the original strike volume
Volume of extract in the wort = Volume of wort - Strike volume = 8.682 gal - 8.0 gal = 0.682 gal​
Ok, why did I get 0.682 gal this time and 0.699 gal previously? Well, in my spreadsheet I make a couple of small adjustments that aren't included above: I add the weight of the water from grain moisture content (in this case 0.48 lb) to the weight of the strike water, so starting water in the mash is actually 66.6432 + 0.48 = 67.1232 lb. Then I subtract the amount of water consumed by hydrolysis of the starch which works out to about 3.7% of the extract weight, or 9.231 * 0.037 = 0.34 lb. Now the total weight of water at the end of mash is 67.12 - 0.34 = 66.78 lb. If you plug 66.78 into the calculations above, instead of 66.64, the °P becomes 12.144 (a difference of 0.022 °P), the SG becomes 1.04899 vs. 1.04908, and the total wort volume becomes 8.699 gal - a difference of 0.017 gal or about 2 fl oz. The difference in wort volume is about 0.2%

Brew on :mug:
 

doug293cz

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Here's how I arrive at his figure... or just about

Density of sugar = 1.59 g/cm^3 = 13.16 lbs/gallon

weight of sugar in 12 lbs of grain with 4% moisture and 80% potential = 12lbs * 0.96 * 0.8 = 9.216lbs

Volume of sugar = 9.216 lbs / 13.16 lbs/gallon = 0.693 gallons.


Looks like I'm off by 0.006 from doug's figure. indeed if he chimes in, maybe he can put out his exact calculation.
This is another, simpler way to do the calculation. It starts to break down at higher SG's because the water/extract system is not quite an ideal solution, volume in solution is not quite the same as the solid volume. But as you can see, the amount of error is quite small in the overall scheme of things, and way below a homebrewer's ability to accurately measure.

Brew on :mug:
 
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McKnuckle

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Epic post. Thanks Doug for laying this all out!

I weigh my wort to the FV and divide that by OG to get an approximate water volume (compatible with typical brewing software). But of course that's not a super accurate method, just a "close enough for rock and roll" type of thing.
 

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@doug293cz, do you have a source for the definition of grain yield?

"The grain potential is the SG that would result from one pound of grain being used to create one gallon of wort, given 100% conversion."

Googling around, it does seem like pretty much everyone makes the mistake of 1 lbs of malt in 1 gal of water.

Even Palmer says in how to brew:
"One pound of sugar will yield a specific gravity of 1.046 when dissolved in 1 gallon of water."

a few pages later however he works out an example with malt and uses the wort volume
"For our example batch, we will assume that 8.5 pounds of malt was mashed to produce 6 gallons of wort that yielded a gravity of 1.038. The brewer's total sugar extraction for this batch would be 6 gallons multiplied by 38 points/gallon = 230 points. Dividing the total points by the pounds of malt gives us our mash extraction in points/pound e.g. 230/8.5 = 27 ppg."
 

doug293cz

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@doug293cz, do you have a source for the definition of grain yield?

"The grain potential is the SG that would result from one pound of grain being used to create one gallon of wort, given 100% conversion."

Googling around, it does seem like pretty much everyone makes the mistake of 1 lbs of malt in 1 gal of water.

Even Palmer says in how to brew:
"One pound of sugar will yield a specific gravity of 1.046 when dissolved in 1 gallon of water."

a few pages later however he works out an example with malt and uses the wort volume
"For our example batch, we will assume that 8.5 pounds of malt was mashed to produce 6 gallons of wort that yielded a gravity of 1.038. The brewer's total sugar extraction for this batch would be 6 gallons multiplied by 38 points/gallon = 230 points. Dividing the total points by the pounds of malt gives us our mash extraction in points/pound e.g. 230/8.5 = 27 ppg."
Like I said it's a common mistake.

If you dissolve 1 lb of sugar in 1 gal (8.3304 lb) of water, the SG in °P is 100° * 1 lb / (1 lb + 8.3304 lb) = 10.72°P. Look 10.72 up in your favorite Plato to SG converter - it's not 1.0462. I don't mean this to be snarky, but the definition I gave is the only one that's self consistent.

Or you can do this simple experiment: Dissolve 1 lb of sugar in 1 gal of water (or 4 oz in 1 qt or 2 oz in 1 pint) and measure the SG with a good hydrometer. What's your SG?

Brew on :mug:
 
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jdudek

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Agreed @doug293cz , it's the only definition that makes sense after thinking on it a little. I ran the experiment as well. Dissolved 4oz of sugar in 1quart of water, and then again with <1quart of water such that 1 quart of solution was produced. Indeed got 10.4 %brix and 11.3 %brix with a digital refractometer. ignoring the obvious imprecision in my kitchen experiment, 11.3 is just about 1.046 while 10.4 is about 1.043.
 

jdudek

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I tried to summarize and document all this here, mostly for my own understanding but perhaps for reference for others.
hopefully it's useful to someone in the future. appreciate any feedback!
 

cmac62

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Jakub, that is a great write-up. It makes sense. I generally use the calcs to get my grain bill and hops for the style, SRM, ABV, IBU. Man, if I was worrying about .003 difference between the calc estimate and what I actually get I'd stop brewing. LOL. I often forget to take OG readings, although with a refrac I forget less. I love that this hobby is perfect for those like me who "make beer" and the engineer/science types who can geek out on all the minutia. Love me some HB. LOL. :mug:
 
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McKnuckle

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I skimmed most of your blog post, and marked it for future perusal when I have more time. Thanks @jdudek for that!

End of the day, efficiency claims here on HBT, and arguably even among pro brewers, are fraught with dissonance among practitioners of how to correctly calculate it. Sometimes it even seems pedantic, or out of touch with reality, to insist on what is truly correct. It's a bit like people who use words like irregardless. That's not a word. But so many people use it, hence it's become a word. Kinda. Meaning, if everyone uses efficiency figures calculated by software, and all of that software makes a common mistake, does that not become a new standard?

I suppose this is a philosophical digression. After all, efficiency is a human-defined metric, not one emerging in a native form from the universe itself. So if we incidentally re-define it due to widespread misconception, maybe we've created a new standard.

Nevertheless, I will continue to use a method that does things as accurately as possible, even if it's slightly detuned from how the masses operate. And I will never, positively never, say "irregardless."
 

doug293cz

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I skimmed most of your blog post, and marked it for future perusal when I have more time. Thanks @jdudek for that!

End of the day, efficiency claims here on HBT, and arguably even among pro brewers, are fraught with dissonance among practitioners of how to correctly calculate it. Sometimes it even seems pedantic, or out of touch with reality, to insist on what is truly correct. It's a bit like people who use words like irregardless. That's not a word. But so many people use it, hence it's become a word. Kinda. Meaning, if everyone uses efficiency figures calculated by software, and all of that software makes a common mistake, does that not become a new standard?

I suppose this is a philosophical digression. After all, efficiency is a human-defined metric, not one emerging in a native form from the universe itself. So if we incidentally re-define it due to widespread misconception, maybe we've created a new standard.

Nevertheless, I will continue to use a method that does things as accurately as possible, even if it's slightly detuned from how the masses operate. And I will never, positively never, say "irregardless."
I disagree. The definition of efficiency is pretty fundamental.

Efficiency = Amount Obtained / Maximum Available [multiply by 100% to get percent rather than fractional]​

If you put something else in the numerator or denominator, then the result is not efficiency.

You can define a lot of different efficiencies in Brewing:
  • Conversion
  • Lauter
  • Mash ( = Conversion * Lauter )
  • Boil to Fermenter Transfer
  • Fermenter to Package Transfer
  • Brewhouse ( = Mash * Boil to Fermenter Transfer)
  • Packaged ( = Brewhouse * Fermenter to Package Transfer)
Brew on :mug:
 
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McKnuckle

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Yes, the concept is simple, but the math and variables involved in calculating it are quite complex. And the method is not agreed upon, or more accurately, not universally understood and/or executed. That was my point - not the definition of the word itself, which is, as you state, fundamental.
 
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