A simple way to infer a ballpark initial grist weight via known mash efficiency, volume, and OG

[Silver_Is_Money]

First we will exhibit this in units of Lbs. and Gal's and Deg. F., and then we will exhibit it for units of Kg. and Liters and Deg's C.
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Example: If I collect 7 gallons pre boil at 150 degrees F. and its SG (or OG) is 1.043, and my mash efficiency is 73%, what inferred initial weight of grist did I begin with?

1st, lets establish a baseline for gravity points:

1 Lb. of sugar when dissolved in 1 Gallon of water yields a SG of 1.04621
1000(1.04621 - 1) = 46.21 gravity points

Lets assume that for the average grist the course grind efficiency vs. pure sugar is 78%

46.21 x 0.78 = 36.0438

Now lets apply our mash efficiency of 73%

36.0438 x 0.73 = 26.312 gravity points per pound per gallon

This means we should expect 1 Lb. of our recipe/grist when mashed in 1 Gallon of water to yield a SG of 1.026312

We have collected 7 gallons of 1.043 SG Wort at 150 degrees F.
At room temperature we will have 7 x 0.98 = 6.86 gallons

6.86 x 1000(1.043 - 1) = 294.98 gravity points

294.98/26.312 = 11.21 Lbs. of inferred initial grist weight

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The above in Metric units:

Example: If I collect 26.5 Liters pre boil at 65.6 degrees C. and its SG (or OG) is 1.043, and my mash efficiency is 73%, what inferred initial weight of grist did I begin with?

1st, lets establish a baseline for gravity points:

1 Kg. of sugar when dissolved in 1 Liter of water yields a SG of 1.3856
1000(1.3856 - 1) = 385.6 gravity points

Lets assume that for the average grist the course grind efficiency vs. pure sugar is 78%

385.6 x 0.78 = 300.768

Now lets apply our mash efficiency of 73%

300.768 x 0.73 = 219.561 gravity points per Kg. per Liter

This means we should expect 1 Kg. of our recipe/grist when mashed in 1 Liter of water to yield a SG of 1.219561

We have collected 26.5 gallons of 1.043 SG Wort at 65.6 degrees C.
At room temperature we will have 26.5 x 0.98 = 25.97 Liters

25.97 x 1000(1.043 - 1) = 1,116.71 gravity points

1,116.71/219.561 = 5.086 Kg. of inferred initial grist weight

 

[Vale71]

Considering that to calculate efficiency you need to know the grist way I find it a bit of pointless excercise to then calculate backwards to get the same number (grist weight) you started with? Unless you know of a way to calculate efficiency without knowing the initial grist weight to begin with. that is.

 

[Silver_Is_Money]

Lets say I want to make a beer that has a pre-boil SG of 1.043 and I intend to end up with 7 pre-boil gallons, and I know that my typical conversion or mash efficiency for a grist of this type and at this SG is 73%, but I don't know how must grist in Lbs. I should I start with.

 

[Silver_Is_Money]

Or to make this way more complicated lets say that I know a malts Buffering coefficient only in expressed terms (units) of mEq/Liter_pH_SG and I need to know how much weight of grist I must start with in my mash of a known volume whereby to produce a Wort that exhibits this level of buffering coefficient. Or other derivations along these general lines....

 

[Sammy86]

Lets say I want to make a beer that has a pre-boil SG of 1.043 and I intend to end up with 7 pre-boil gallons, and I know that my typical conversion or mash efficiency for a grist of this type and at this SG is 73%, but I don't know how must grist in Lbs. I should I start with.

This would be fine if all recipes used the same amount of grain and had the same OG...but they don't so as stated above what's the point?

 

[Silver_Is_Money]

This would be fine if all recipes used the same amount of grain and had the same OG...but they don't so as stated above what's the point?

Follow the same steps but apply to them the variables of your needs. Perhaps you want to make 5 room temperature gallons of 1.070 SG pre-boil Wort, but don't know how much grist weight you will need to begin with. No problem.

5 x 1000(1.070 - 1) = 350 gravity points

350/26.312 = 13.30 Lbs. of inferred initial grist weight (presuming a conversion/mash efficiency of 73%, which can also be varied to match your efficiency)

 

[Silver_Is_Money]

Lets say your mash efficiency is 85%. Then:

36.0438 x 0.73 = 26.312 gravity points per pound per gallon
Becomes:
36.0438 x 0.85 = 30.637 gravity points per pound per gallon

Substituting this into post #6 above gives us:
350/30.637 = 11.42 Lbs. of inferred initial grist weight required (presuming a conversion/mash efficiency of 85%)

 

[Sammy86]

Don't need to math when then brewing programs do it for me boss...but hey to each their own.

 

[Silver_Is_Money]

Don't need to math when then brewing programs do it for me boss...but hey to each their own.

Now you can write your own program.

 

[VikeMan]

Lets say your mash efficiency is 85%. Then:

36.0438 x 0.73 = 26.312 gravity points per pound per gallon
Becomes:
36.0438 x 0.85 = 30.637 gravity points per pound per gallon

Substituting this into post #6 above gives us:
350/30.637 = 11.42 Lbs. of inferred initial grist weight required (presuming a conversion/mash efficiency of 85%)

I might be misinterpreting your idea, but are you assuming that you can take a standard "user" mash efficiency and apply it to grain bills/gravities of various sizes? If you were not assuming that, I apologize and will go back to sleep.

(Mash efficiency decreases as grist size increases. And there's no standard way to scale that, i.e. it's process and equipment dependent.)

(ETA: When I say there's no standard way to scale it, I don't mean that it can't be done, only that you need to know certain brewhouse (process/equipment) parameters to do the math.)

 

[Silver_Is_Money]

I might be misinterpreting your idea, but are you assuming that you can take a standard "user" mash efficiency and apply it to grain bills/gravities of various sizes? If you were not assuming that, I apologize and will go back to sleep.

(Mash efficiency decreases as grist size increases. And there's no standard way to scale that, i.e. it's process and equipment dependent.)

(ETA: When I say there's no standard way to scale it, I don't mean that it can't be done, only that you need to know certain brewhouse (process/equipment) parameters to do the math.)

I believe post #3 addresses this. Mash efficiency is by no means standard, or intended to be standard. All along a presumption must be made that the end user understands the efficiency of his/her equipment and process at varying OG's. This is a ballpark primer. It can be developed into more. But if you have a general knowledge of your mash efficiency you should have enough info here to infer a ballpark grist weight.

My interest however is actually in buffering coefficients and how to equate (as in normalize) them to Worts of varying SG's.

 

[VikeMan]

I believe post #3 addresses this.

I guess I'm confused. If you're saying that you are (quite correctly) assuming the user's "typical" mash efficiency (e.g. 73%) only applies to the wort gravity at which that mash efficiency has been observed, then the user already knows the amount of grains needed to get that gravity, because they've done it before.

But for other desired OGs, the 73% can't really be applied, because mash efficiency varies with grain bill size.

 

[Silver_Is_Money]

I guess I'm confused. If you're saying that you are (quite correctly) assuming the user's "typical" mash efficiency (e.g. 73%) only applies to the wort gravity at which that mash efficiency has been observed, then the user already knows the amount of grains needed to get that gravity, because they've done it before.

But for other desired OGs, the 73% can't really be applied, because mash efficiency varies with grain bill size.

Don't get too hung up on 73% efficiency. I picked that at random out of thin air whereby to merely present an example. Then in post #7 I altered it, again out of thin air. But I was adding to post #11 when you sent post #12. My interest or intent lies within my addition.

 

[VikeMan]

Don't get too hung up on 73% efficiency. I picked that at random out of thin air whereby to merely present an example. Then in post #7 I altered it, again out of thin air.

Hmmm. I'm not hung up on 73%. I'm hung up on any one "standard" efficiency for any given user. And if you are suggesting (correctly) that the user should substitute the efficiency that they have experienced with that grain bill size/gravity, then again, they already know the grain amount required. What am I missing?

 

[Silver_Is_Money]

Hmmm. I'm not hung up on 73%. I'm hung up on any one "standard" efficiency for any given user. And if you are suggesting (correctly) that the user should substitute the efficiency that they have experienced at that grain bill size/gravity, then again, they already know the grain amount required. What am I missing?

You are presuming that the brewer knows intuitively how much grain to add to get to a desired endpoint, whereas I do not make this presumption. It is not a matter of missing anything. It's a matter of educating. Isn't that what a science forum should do? And when I educate, I try to approach it from a bottom up perspective which presumes that the brewer is not full of knowledge at the onset. As opposed to (for example) AJ hitting us from a top down approach, and then wondering why no one gets it.

 

[VikeMan]

You are presuming that the brewer knows intuitively how much grain to add to get to a desired endpoint, whereas I do not make this presumption. It is not a matter of missing anything.

No, I'm not presuming that. I'm saying that for any batch a brewer has brewed, they know what gravity they got and what the grain bill size was, unless they didn't bother to record it. And they can compute a mash efficiency from that. So far, so good.

But, that mash efficiency cannot be applied to a different desired gravity without being modified by some considerable math that involves some brewhouse parameters.

IOW, there's no such thing a "typical" mash efficiency for any given brewer, unless that brewer is brewing the same thing (or very close to it) over and over, and nothing of a very different gravity.

 

[Silver_Is_Money]

No, I'm not presuming that. I'm saying that for any batch a brewer has brewed, they know what gravity they got and what the grain bill size was, unless they didn't bother to record it. And they can compute a mash efficiency from that. So far, so good.

But, that mash efficiency cannot be applied to a different desired gravity without being modified by some considerable math that involves some brewhouse parameters.

IOW, there's no such thing a "typical" mash efficiency for any given brewer, unless that brewer is brewing the same thing (or very close to it) over and over, and nothing of a very different gravity.

We seem to be merely talking past each other.

Let me add that someone who has not delved into things such as this does not need to be presented with all of the conceived complexities of the subject all at once, as such an approach will more often than not confound the subject more than is initially required and scare people off rather than further engaging their minds. This expands upon my thinking that brewing science is best learned (by most) from the bottom up. Had I not intended this to be ballparking it, I would not have said within the subject line the word "ballpark", or for that matter the word "inferred".

 

[VikeMan]

Ah, okay, if this is meant to be a thought experiment/lecture to help develop young brewing minds, that's admirable. I would just include a disclaimer that the further they get away from their average grist size, the less valid the answer will be.

Carry on!

 

[Silver_Is_Money]

Would something quick and easy along these (highly tentative, admittedly crude, and linear) lines bring a degree of satisfaction to this threads naysayers?

Desired_OG_efficiency ~= Known_efficiency x (1 + (Known_OG - Desired_OG) x 2.5*)
(* = initial rough ballpark guesstimate for this factor is 2.5, and is clearly pending improvement or revision)

Example:

Brewers Known Mash Efficiency = ~73% for beers of OG = ~1.043 pre-boil as per the above

Brewer wants to make a beer of pre-boil OG = 1.095

Brewer needs to know estimated mash efficiency for OG = 1.095, so grist weight can be assessed as per the above method

1.095_efficiency ~= 73 x (1 + (1.043 - 1.095) x 2.5*)

1.095_efficiency ~= 63.51%

Just tossing this out such that the intent for achieving a solution replaces the complaining that does not help with offering a solution.

 

[VikeMan]

Taking the usual caveats as read...

With my equipment and process (including a "no sparge"), the factor would need to be 3.35 to hit the numbers in the example, based on going from a baseline of 73% mash efficiency at 1.043 to 60.4% mash efficiency at 1.095. (I iterated through BrewCipher's mash efficiency predictor until the changes became very small to determine the 60.4% value.)

 

[Silver_Is_Money]

Taking the usual caveats as read...

With my equipment and process (including a "no sparge"), the factor would need to be 3.35 to hit the numbers in the example, based on going from a baseline of 73% mash efficiency at 1.043 to 60.4% mash efficiency at 1.095. (I used BrewCipher's mash efficiency predictor to determine the 60.4% value.)

Thanks much! I made a spreadsheet and incorporated all of the above (post #1 plus post #19) and was in the process of honing in on the factor for this simple method being about 3.5, so my thinking is that your 3.35 is a much better bet than my initial guess of 2.5.

 

[Silver_Is_Money]

If you look at the highly valued an sophisticated charted efficiency curves that @doug293cz worked up for all sorts of mash sparging methods (including no-sparge) the slopes are exponential but relatively close to linear, and also close to having the same slope. Somewhere around 3.35 for my down and dirty method as seen in post #19 is likely a decent linear fit 'compromise' to Doug's far more accurate slopes.

If it is acceptable on this forum to attach zipped spreadsheet files, I will attach my spreadsheet that I mentioned above. I've gone with the 3.35 factor for it.

 

[VikeMan]

One thing to maybe keep in mind when trying to determine the "best" default/avg factor to use... IIRC, Doug's excellent charts are (by necessity) generic, i.e. they don't account for things like mash tun dead space (which in any given system would be a constant, i.e. it doesn't vary with grist size). The upshot is that the calculated relative mash efficiency "penalty" for using the larger grain bill is actually smaller larger than it would be if mash tun dead space or transfer losses etc. were considered.

What would the average unrecoverable loss volume out there be? Beats me. But it's something to think about.

ETA: Or maybe, with BIAB being all the rage, just assume 0 losses and crunch Doug's data.

ETA: I don't know why I typed "smaller" above. "Larger" is correct, and has a slightly different meaning.

 

[doug293cz]

One thing to maybe keep in mind when trying to determine the "best" default/avg factor to use... IIRC, Doug's excellent charts are (by necessity) generic, i.e. they don't account for things like mash tun dead space (which in any given system would be a constant, i.e. it doesn't vary with grist size). The upshot is that the calculated relative mash efficiency "penalty" for using the larger grain bill is actually smaller larger than it would be if mash tun dead space or transfer losses etc. were considered.

What would the average unrecoverable loss volume out there be? Beats me. But it's something to think about.

ETA: Or maybe, with BIAB being all the rage, just assume 0 losses and crunch Doug's data.

ETA: I don't know why I typed "smaller" above. "Larger" is correct, and has a slightly different meaning.

You are correct about the charts being for 0 undrainable MLT volume. My spreadsheet does include a parameter for undrainable volume (and any transfer losses should be lumped with undrainable.) The charts are busy enough without including more sets of curves for different undrainable volumes.

Undrainable volume has the same effect on lauter efficiency as grain absorption, except that it does not vary with grain bill size.

Brew on

 

[bracconiere]

Considering that to calculate efficiency you need to know the grist way I find it a bit of pointless excercise to then calculate backwards to get the same number (grist weight) you started with? Unless you know of a way to calculate efficiency without knowing the initial grist weight to begin with. that is.

lol, i remember learning about trigonometry too!

 

[doug293cz]

First we will exhibit this in units of Lbs. and Gal's and Deg. F., and then we will exhibit it for units of Kg. and Liters and Deg's C.
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------

...

1st, lets establish a baseline for gravity points:

1 Lb. of sugar when dissolved in 1 Gallon of water yields a SG of 1.04621
1000(1.04621 - 1) = 46.21 gravity points

...

This is incorrect. 1 lb of sugar when dissolved in enough water to create 1 gal of solution yields an SG of 1.046173. This is a very common misunderstanding of the definition of potential. Turns out the volume of water required to create 1 gal of solution from 1 lb of sugar is 0.92613 gal @ 68°F (20°C) or 7.715 lb of water (density of water @ 68°F is 8.3304 lb/gal.)

The above is determined starting with the definition of °Plato: °P = 100° * Mass of Sucrose / (Mass of Sucrose + Mass of Water)

You then convert °P to SG using the "ASBC" equation: SG = 1 + °P / (258.6 - (°P / 258.2) * 227.1) [valid at 68°F (20°C)]

Next covert weight of solution to volume: Volume (gal) = (Mass of Sucrose + Mass of Water) / (8.3304 lb/gal * SG)

You can put this in a spreadsheet, and then use "Goal Seek" to adjust the weight/volume of water to get a solution volume of 1 gal.

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The difference between 1.04621 and 1.046173 may be due to errors in the ASBC equation or temp differences, IDK. But, I use 46.173 as the PPG for sucrose.

Brew on

 

[doug293cz]

You can get a pretty accurate estimate of the grain bill weight required for a specific volume of wort at a target OG by using my spreadsheet. Just fill out your system parameters at the top of the sheet, taking a guess at the Grain Bill Weight. You will need to know what your typical conversion efficiency is, and it shouldn't vary with grain bill size. The spreadsheet will take care of estimating the lauter efficiency factor in mash efficiency (mash efficiency = conversion efficiency * lauter efficiency.) It is lauter efficiency that varies with grain bill size.

Once you fill in your parameters, then use "Goal Seek" to get the target pre-boil SG [cell K52] or post-boil SG (i.e. OG) [cell K55 or K59] by using the grain bill weight [cell B5] as the variable parameter.

Brew on

 

[Silver_Is_Money]

This is incorrect. 1 lb of sugar when dissolved in enough water to create 1 gal of solution yields an SG of 1.046173. This is a very common misunderstanding of the definition of potential. Turns out the volume of water required to create 1 gal of solution from 1 lb of sugar is 0.92613 gal @ 68°F (20°C) or 7.715 lb of water (density of water @ 68°F is 8.3304 lb/gal.)

The above is determined starting with the definition of °Plato: °P = 100° * Mass of Sucrose / (Mass of Sucrose + Mass of Water)

You then convert °P to SG using the "ASBC" equation: SG = 1 + °P / (258.6 - (°P / 258.2) * 227.1) [valid at 68°F (20°C)]

Next covert weight of solution to volume: Volume (gal) = (Mass of Sucrose + Mass of Water) / (8.3304 lb/gal * SG)

You can put this is a spreadsheet, and then use "Goal Seek" to adjust the weight/volume of water to get a solution volume of 1 gal.

--------------------------------------

The difference between 1.04621 and 1.046173 may be due to errors in the ASBC equation or temp differences, IDK. But, I use 46.173 as the PPG for sucrose.

Brew on

I'll accept your 1.046173. No one has done more for efficiency study at the amateur level than you have.

 

[doug293cz]

I'll accept your 1.046173. No one has done more for efficiency study at the amateur level than you have.

"Blush"

Brew on

 

[Silver_Is_Money]

Here is the link to the spreadsheet I built using post #1, post #19, 1 lb table sugar in one gallon water = 1.046173, and the 3.35 factor. After going to this link, click the down arrow next to the printer icon in the upper right corner to download it whereby to launch and run it.

https://drive.google.com/file/d/1BpffiG0RSrsItmZ34mW77XCf_q-E7wwN/view?usp=sharing
Edit, made a quick text correction, and provided a new link with the corrected version. The above link now points to the corrected version.

 

[Silver_Is_Money]

The first 3 data entry cells of the spreadsheet anchor (hone, or dial in) its logic to your specific system and process. Choose your "knowns" (the first 3 data entry cells) wisely from direct measured past brewing experience on a given system, then venture into unknown brewing OG territory with data entry cell #4. All of the cells to the right of data entry cell 4 will reflect a projection of how your system should respond when you actually brew a beer of the desired (unknown, data entry cell #4) pre-boil OG, and the last data output cell provides a prediction of the grist weight you should target.

NOTE: You can do a double check on your first 3 data entry values (your anchors) validity simply via duplicating the cell #2 OG entry in cell #4. If the predicted grist weight matches that of the actual grist weight that generated your anchor data, the anchor data is to be considered valid. If not, the anchor data in the first 3 data entry cells must be tweaked until you get a grist weight match.

NOTE 2: If pretty close to everything that you boil winds up in your fermenter you can enter post boil and cooling values (I.E., initial fermenter conditions, pre fermentation) across the board instead of pre-boil values and get respectfully decent output thereby.

 

[VikeMan]

Here is the link to the spreadsheet I built using post #1, post #19, 1 lb table sugar in one gallon water = 1.046173, and the 3.35 factor.

Cool. This doesn't actually affect your spreadsheet the way you have set it up, but one of the points @doug293cz was making is that it's not a pound of sugar in a gallon of water, but rather a pound of sugar in enough water to make a gallon of solution. I'm sure you get that. I mention it only for future readers.

 

[Silver_Is_Money]

The aging chemist in me is well aware of making up solutions to volumes, but I tend to get sloppy with the terminology.

 

[doug293cz]

The aging chemist in me is well aware of making up solutions to volumes, but I tend to get sloppy with the terminology.

You, and so many others. It is just so much easier to say/write it incorrectly.

Brew on

 

[SFC Rudy]

I put John Palmers calculations into a spreadsheet. You can enter your OG, Pre-boil volume, mash efficency and base grain type and it will tell you how much grain you need.

 

[Silver_Is_Money]

Since there has been some confusion as to how to make this "Grist Weight as tuned to your system and process" predicting calculator/spreadsheet function properly, I've neatened it up some and provided instructions within. Link is as seen below.

Download at the 'Google Drive' link seen below (by pressing the "down arrow" seen next to the printer icon in the upper right hand corner) and then launch the downloaded spreadsheet file and run in Excel or LibreOffice Calc.

https://drive.google.com/file/d/1BpffiG0RSrsItmZ34mW77XCf_q-E7wwN/view?usp=sharing