Grain Spec: Confused on Brix/Plato vs SG calculations

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malweth

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If an example grain spec has:

%FG = 80% (Fine Grain Extract Yield)
%M = 4% (Moisture)

G = 1 lb (mass of grain)
V = 1 gal (vol of water)
E = 100% (Mash efficiency - to make calculations easier)​

The mass of sugars M should be given by:

M = G * (%FG - %FG * %M / 100) / 100 * E

M = 1lb * (80% - 80% * 4% / 100) / 100 * 1 = 0.768 lb​

Into 1 gallon, the degrees plato (P) should be:

P = M / (8.34 lb/gal * V + M)

P = 0.768 lb / (8.34 lb/gal * 1 gal + 0.768 lb) * 100 = 8.43°P

Now, given that 1 lb sugar in 1 gal yields 1.046, the SG should be:

SG = (%FG - %Meff)/100 * G * 46 pt*gal/lb * V * E / 1000 + 1

SG = 0.768 * 1 lb * 46 pt/(lb*gal) * 1 gal * 1 / 1000 + 1 = 1.0353​

Now... the PDF at http://www.ams.usda.gov/AMSv1.0/getfile?dDocName=STELPRDC5091778 has a conversion table between Brix and SG. The interesting rows:

Code:
°P  / SG @ 20°C / lb dissolved solids per Gal
8.4 / 1.03344   / 0.722
8.5 / 1.03385   / 0.731

8.8 / 1.03508   / 0.758
8.9 / 1.03549   / 0.767

It seems to me like, while the calculated numbers are close, the degrees plato are off from the lbs sugar per gallon.

Am I calculating °P wrong? It should be (sugar wt / total wt) and there are ~8.34 lbs per gallon water (8 lbs / gal gets closer, but is less correct).

For distilled water: http://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html
 
If an example grain spec has:

%FG = 80% (Fine Grain Extract Yield)
%M = 4% (Moisture)

G = 1 lb (mass of grain)
V = 1 gal (vol of water)
E = 100% (Mash efficiency - to make calculations easier)​

The mass of sugars M should be given by:

M = G * (%FG - %FG * %M / 100) / 100 * E

M = 1lb * (80% - 80% * 4% / 100) / 100 * 1 = 0.768 lb​
To get to the grain they analyzed 0.96 lb of grain absorbed 0.04 lb moisture so the dry basis for 1 lb of this grain is 0.96 lbs. Assuming the yield is the dry basis yield the extract would be
M = (1 - .04)*0.8 = 0.768 lb (same answer - simpler calculation)

Into 1 gallon, the degrees plato (P) should be:

P = M / (8.34 lb/gal * V + M)

P = 0.768 lb / (8.34 lb/gal * 1 gal + 0.768 lb) * 100 = 8.43°P


Now, given that 1 lb sugar in 1 gal yields 1.046, the SG should be:

SG = (%FG - %Meff)/100 * G * 46 pt*gal/lb * V * E / 1000 + 1

SG = 0.768 * 1 lb * 46 pt/(lb*gal) * 1 gal * 1 / 1000 + 1 = 1.0353​

The official interconversion between SG (20/20 Apparent) and Plato is given by the ASBC polynomial
°P = ((135.997*S-630.272)*S+ 1111.14)*S -616.868

For 8.43 °P the specific gravity (obtained by setting the polynomial to 8.43 and solving for S) is sg(8.43) =1.03356
and the density is 8.34*sg(8.43) = 8.61989 lbs/gal
Thus you have (8.34 + 0.768)/(8.34*sg(8.43)) = 1.05663 gal wort which contains 0.768 pounds of extract for
0.768 / ( (8.34 + 0.768)/(8.34*sg(8.43)) ) = 0.726842 lbs extract/gal


Now... the PDF at http://www.ams.usda.gov/AMSv1.0/getfile?dDocName=STELPRDC5091778 has a conversion table between Brix and SG. The interesting rows:

Code:
°P  / SG @ 20°C / lb dissolved solids per Gal
8.4 / 1.03344   / 0.722
8.5 / 1.03385   / 0.731

8.8 / 1.03508   / 0.758
8.9 / 1.03549   / 0.767

Now lets put 0.763924 lbs of sugar in 1 gal of water which actually weighs 8.33041 lbs at 20 °C. The concentration is 0.763924/(8.33041 + 0.763924) = 0.084 i.e. exactly 8.4 °P.

sg(8.4) = 1.03344 by inversion of the ASBC polynomial - exact agreement with the table.
The volume of this wort would be
(8.33041 + 0.763924)/(8.33041*sg(8.4)) = 1.05638 gal
and the extract per gal
0.763924 / ( (8.33041 + 0.763924)/(8.33041*sg(8.4)) ) = 0.723152
which disagrees with the table by 0.001. Possible reasons for this is that the table uses older definitions of the gallon and the temperature scale. The density of water is 0.998203 at 20 °C on the ITS 90 temperature scale.


It seems to me like, while the calculated numbers are close, the degrees plato are off from the lbs sugar per gallon.

Am I calculating °P wrong? It should be (sugar wt / total wt) and there are ~8.34 lbs per gallon water (8 lbs / gal gets closer, but is less correct).

You'll get much closer if you use the ASBC polynomial instead of the 'points'. Things are almost, but not quite linear to the extent that it takes a 3rd order polynomial to adequately model the relationship between SG and °P. As for the extract per gallon - I've shown how to calculate that so you don't need to rely on the table.
 
Wow, great posts by both of you. Thanks for helping me understand this!
 
I have three lab grade Hydrometers that measure the oldest measure of sugar dissolved in water, the Balling scale. Is there a conversion to Brix, Plato, and SG? I understand it is close to Brix, but Adolf Brix changed his scale for better accuracy.
 
Balling was first, Brix second, Plato third and the ASBC the last in the history of estimation of the sugar content from specific gravity. Thus you might assume that the ASBC tables are the most accurate and the Balling tables the least and there may be some truth in that. Balling apparently only worked to 3 decimal places and Brix and Plato to 6. The rationale behind the Normaleichungskomission (Plato) was that there were known error in the 5th and 6th decimal places in the Brix tables.

Other than the differences in inherent accuracy the differences between the tables stem from different definitions of specific gravity. Balling compared the density of the sample at 17.5°C to the density of water at 4 °C. Whether this was apparent or true is moot as the difference would not be seen in the 3rd decimal place. Brix measured specific gravity 4/20 and Plato used 4/20. I do not know whether Brix measured true or apparent SG but Plato measured true. The ASBC has taken the Brix data and converted from 4/20 true to 20/20 apparent. You could work out the conversion factor between 4/17.5 and 20/20 and thus figure out how much the specific gravity difference for a particular level of Plato/Brix would be and convert that to error in °P. It's going to be less than 0.1 °P so unless you are reading to less than that I think you are safe in considering ° Balling and °P the same.
 
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