I have been wondering about how to design a beer that will have a known amount of fruit added to secondary as a purée. I would like to design so as to control the final ABV, not wanting a high %. In order to determine how many pounds of malt I need in the mash to arrive at a predicted OG for the primary fermentation, I am assuming that I would need to have a guess as to the PE of fruit purée. I'm guessing, based on browsing at gotmead.com, that the % weight of the purée that is fermentable sugar is 97%, but really not sure on that. I am planning to measure the SG/Plato of the purée, but I will not know that until the day I make it for adding to the secondary, so I need to make some guesses/assumptions at some point.
I found that for sugar, that there is some confusion, but found this post from another forum posted by Kai.
Begin quote:
Since I always wanted to do that, I just derived the formula for PPG based on the extract potential in %.
Starting point is the equality between the two efficiency calculations that we know. Weight based and PPG based:
100* (Plato * sg * V_liter) / (E * m_kg) = 100 * (GP * V_gal) / (ppg * m_lb)
where:
Plato = wort strength in Plato
sg = wort specific gravity
V_liter = wort volume in liter (V_liter = 3.78 * V_gal)
E = extract potential in %
m_kg = grist weight in kg (m_kg = m_lb * 0.45)
GP = Gravity points (GP = (sg -1) * 1000)
V_gal = wort volume in gal
ppg = extract potential as ppg
m_lb = grist weight in pounds
using the aforementioned substitutions and the simplification that Plato = ( sg 1 ) * 250 I find:
ppg = E * 0.476 / sg
You notice that the ppg number depends on the specific gravity. If you have sugar (E=100%) the ppg for a resulting wort og 1.040 will be 47.6 / 1.04 = 45.77 whereas for a resulting wort of 1.080 it is 44.07.
At home I would have to take the time to calculate this with a more precise sg to Plato conversion, but the trend will be the same: the gravity point gain you get from sugar or malt also depends on the target gravity itself.
In practical brewing this error likely gets absorbed in the efficiency of your system which is known to change with the gravity of the beer anyway and you already account for that.
Kai
End quote.
I am hoping that he will see this thread and maybe offer some help. After reading his post I realize I'm in over my head. Math is not my forte. I was planning to add 2.5 pounds of fruit per gallon and purée the fruit before adding to the secondary. When I attempt to design a recipe, I find that if I use corn sugar as a surrogate for my fruit purée, that adding 2.5 pounds of sugar per gallon will yield an OG of 1.115, and that's the sugar alone. This seems to be far too much and makes me question if I am going about this correctly. My target ABV is 5.8% with a target FG of 1.005. Some calculations show this to mean the OG would be 1.049 including the mash points and the secondary fruit addition points. In my case the fruit is fresh picked blueberries. These are now in the freezer. The is mash is pils and wheat and maybe some flaked wheat or rye. It is going to be a sour beer.
Any help appreciated.
TD
Sent from my iPad using Home Brew
I found that for sugar, that there is some confusion, but found this post from another forum posted by Kai.
Begin quote:
Since I always wanted to do that, I just derived the formula for PPG based on the extract potential in %.
Starting point is the equality between the two efficiency calculations that we know. Weight based and PPG based:
100* (Plato * sg * V_liter) / (E * m_kg) = 100 * (GP * V_gal) / (ppg * m_lb)
where:
Plato = wort strength in Plato
sg = wort specific gravity
V_liter = wort volume in liter (V_liter = 3.78 * V_gal)
E = extract potential in %
m_kg = grist weight in kg (m_kg = m_lb * 0.45)
GP = Gravity points (GP = (sg -1) * 1000)
V_gal = wort volume in gal
ppg = extract potential as ppg
m_lb = grist weight in pounds
using the aforementioned substitutions and the simplification that Plato = ( sg 1 ) * 250 I find:
ppg = E * 0.476 / sg
You notice that the ppg number depends on the specific gravity. If you have sugar (E=100%) the ppg for a resulting wort og 1.040 will be 47.6 / 1.04 = 45.77 whereas for a resulting wort of 1.080 it is 44.07.
At home I would have to take the time to calculate this with a more precise sg to Plato conversion, but the trend will be the same: the gravity point gain you get from sugar or malt also depends on the target gravity itself.
In practical brewing this error likely gets absorbed in the efficiency of your system which is known to change with the gravity of the beer anyway and you already account for that.
Kai
End quote.
I am hoping that he will see this thread and maybe offer some help. After reading his post I realize I'm in over my head. Math is not my forte. I was planning to add 2.5 pounds of fruit per gallon and purée the fruit before adding to the secondary. When I attempt to design a recipe, I find that if I use corn sugar as a surrogate for my fruit purée, that adding 2.5 pounds of sugar per gallon will yield an OG of 1.115, and that's the sugar alone. This seems to be far too much and makes me question if I am going about this correctly. My target ABV is 5.8% with a target FG of 1.005. Some calculations show this to mean the OG would be 1.049 including the mash points and the secondary fruit addition points. In my case the fruit is fresh picked blueberries. These are now in the freezer. The is mash is pils and wheat and maybe some flaked wheat or rye. It is going to be a sour beer.
Any help appreciated.
TD
Sent from my iPad using Home Brew
