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"potential" question

Kim Pedersen

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Hello beerlovers :)

Im sitting and looking at beersmith, and stumbles over the "potential" in the list !
Is there a formula this can be used in ! something like wheight/litres*potential !... I bet its not that simple, but could be fun to know how it works !
 
I'm going to assume you mean potential specific gravity.  The reference is 1 lb of sugar in 1 gallon of water which yields a 1.046 gravity or 46 points per gallon.  BS references everything to it.  It assigns efficiencies for converting different types of grains, sugars, extracts, etc. into sugar bearing solutions.  For example, dissolving extract gives 80% of the sugar you get from pure sugar, 1.046 is 46 points/gallon so 46 ppg*80% solubility=36.9 points per gallon potential or 1.037 Sp. gravity).  This is the gravity BS uses for extracts.  By using the customizing field, BS allows you to tailor the potentials to more closely match what you are getting in real life - of course this assumes you are measuring it and using the information to manage your batches.
 
Like Dave said above the "potential" is the potential amount of sugar that is added by one pound of grain / extract / sugar, etc.  When we start a recipe there is a target Original Gravity (OG), for our purposes lets use 1.050.  If we are measuring the specific gravity the total volume of liquid doesn't really matter. However, if we are trying to mix grain (sugar) and water to get a to a target specific gravity the volume of water and the weight of the sugars do matter.

It is easiest to think (and calculate) using "points".  1 point is 1.001 SG.  Therefore 50 pts is 1.050; again at this point in time the volume is not specified.  To get volume into the equation, we will set the volume to 1 gallon and then describe each type of sugar (grains / extract, etc) in terms of how many points of sugar each pound will add to one gallon of water.  Since there are several less than 100% efficient steps in the process (mashing  & lautering) we express this number as a potential sugar.

So how do we use this number.

In extract brewing the efficiencies are very close (if not exactly) 100%.  Dry Malt Extract (DME) has a potential sugar of 45 points per pound, therefore 1 lb DME in 1 gallon of water will yeild a SG of 1.045 to achieve our target OG in a 5 gallon batch we can use the following equation

                        pounds of DME  =  Target OG  x  # gallons  /    Potential Sugar
                        # DME  =  50 pts  x  5 gal  /  45 pts/gal    =  5.6 lbs of DME

All-grain requires some more information and more calculations.  In all-grain brewing the inefficiencies of mashing and lautering have to be considered.  For our example let's just use 70%.  Typical base grains have a potential of about 36 pts / gallon.  Therefore

                      pounds of grain  =  Target OG  x  # gallons  /  (Potential Sugar x Efficiency)
                      # of grain  =  50 pts  x  # gallons  /  (36 pts/gal  x 0.70)  =  9.9 lbs of grain

Now there are a couple of very important things to notice.  First, the Efficiency is in the denominator of the equation (under the divide line) therefore lower efficiency makes the amount of grain required go up.  The second is the number above the divide line stays the same; 50 pts x 5 gal = 250 pts total (for the 5 gallon batch)  This number becomes very useful for measuring the amount of sugar we have available at any point in the brewing process.

During the mash we will have less than the final 5 gallons and at the beginning of the boil we will have move than the 5 gallons; therefore SG measured at these two time will NOT be the 1.050 we eventually want.  So method I use is to always work in total sugar.  If my target is 50 pts in 5 gallons, I need 250 total points of sugar at each step.  If I measure near the end of the mash and I have 3.5 gallons of water in the mash tun my measured SG should be

                                              SG  =  250 total points  /  3.5 gal    =    71 pts/gal  or  1.071 SG
or if a measure a SG = 1.069 then
                                              Total Points  =  69 pts/gal  x 3.5 gal  =  242 pts

If I measure SG = 1.042 at the beginning of the boil with 6.2 gallons, then I know the expected OG at the end of the boil with 5 gallons is:
      .                                    Total Points  =  42 pts/gal  x  6.2 gallons  =  260 pts
                                            OG  =  260 pts  /  5 gal    =    52 pts  =  1.052 SG 

Now to work efficiency into the equation; assume we started with 10 pounds of grain at a potential of 36 pts per lb
Our total potential sugar is    36 pts/lb  x  10 lb  =  360 total points

at the end of all the processing I a OG of 1.052 in 5 gallons or a total of 260 points

Then efficiency is                        Eff  =  260  /  360  =  72%

There in a nut shell (and ignoring some of the minor issues) is what is going on in the background of BS2

David


 
OMG thanks for these replys :) They are realy good :) Maby im a geek, but I like to know whats going on behind the scenes :) Thanks !
 
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