Thu Nov 11, 2010 6:28 am
I assume the question is as to by how much the gravity of a wort changes if you add a pound of sugar to it. That depends on the volume of the wort to which you add it and the amount of sugar already in the wort. The steps required to make the calculation follow. Skip down to *** if the rest makes your head spin:
1. You will need the °P for the calculation. If you have the specific gravity of the wort convert it to Plato using the ASBC polynomial
°P = ((135.997*sg -630.272)*sg +1111.14)*sg - 616.868
The Plato value is the sugar content per 100 grams iof wort.
Example: wort of SG 1.040 corresponds to 9.994 °P.
2. Find the density of this wort. You will need the specific gravity for this. If you have Plato hydrometers then you will need to convert °P to apparent sg. Do this by putting numbers into the conversion fomula above until it return the correct °P value, use Excel's solver to do this for you automatically or use the inverse of the Lincoln equation
sg = (668 - sqrt(668^2 - 820*(463 + °P)))/410
Multiply the specific gravity by 0.998203 to get the density
Example: 0.998293*1.040 = 1.03813
3. Find the weight of 1 L of the solution by multiplying the density by 1000. Example: 1 L of 1.040 wort weighs 1038.13 grams.
4. Find the amount of sugar in 1L of the solution by multiplying the weight of 1L by °P/100. Example: 0.09994*1028.13 = 102.75 grams/L
5. Subtract the weight of the sugar from the weight of 1 L of solution. This is the weight of the water in 1 L of solution. Example: 1038.13 - 102.75 = 935.38 grams water
6. Add the amount of your sugar addition to the amount of sugar already present.
Example: Adding 1 pound (454 grams) to a 20L (~ 5 gal) sugar is 102.75 + 454/20 = 125.45 grams
7. Calculate the new strength in °P from °P = 100*sugar/(sugar + water)
Example: 100*125.45/(125.45 + 935.38)=11.826°P
8. Convert to sg
Example: 11.826 °P corresponds to 1.04764 apparent sg meaning that adding 1 pound of sucrose to 20L of 1.040 wort increases gravity by 7.64 "points".
Note that the volume will have increased somewhat because of the sugar addition. Assuming you started with 20 L wort you would have 20*935.38 grams of water and 20*125.45 grams of sugar for 20*(935.38 + 125.45) grams of wort at density 1.0476*0.998203 grams/cc implying a volume of 20.289L. Assuming you would boil this off to get back to exactly 20 L then a somewhat different approach is necessary after step 6.
7a: At this point you know you will have s grams of sugar per liter of wort. Plug this into points = 0.139 + 0.38368*s
Example 0.139 + 0.38368*125.45 = 48.27 so sg = 1.0482
***Another and much simpler approach is to observe that if you make up a wort with 1 pound of sucrose per gallon it will be an 11.48 °P wort which is right in the range of most beers we brew and this corresponds to 46.2 sg points. We pretend that things are linear (which they very nearly are) and say that if we add a pound of sugar to 5 gallons wort we'd be adding 46.2/5 = 9.2 points. Starting from 1.040 that would give us 1.0492 (as opposed to the more "accurate" answer 1.0482). That level of error should be tolerable for most purposes.
The density does depend on the type of sugar but is not something you would be able to detect with a hydrometer.
When you add it also matters according as to whether you intend to boil away the extra volume (add in kettle) or not (add in fermenter). Again, the differences are small - certainly not significant relative to the linear approximation, to the "error" caused by evaporation in the fermenter, errors in volume measurement, hydrometer reading error etc. Commercial brewers measure the original gravity after fermentation is complete.
, I had no idea it was that much, was thinking it would be about half that. Thanks Bug