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Old 11-02-2010, 03:45 PM   #1
ashyg
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Jan 2009
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Does this sound like a good idea?

Say I use a refractometer/hydrometer to get a reading of SG 1.057 from some fresh pressed apple cider. That means the density of the cider sample is 1.057 times the density of water(1kg/liter), correct?

Fructose is the primary sugar in cider. We want to prime a 5 gallon batch with about 5.25 oz of corn sugar(dextrose/d-glucose) at 60 degrees to get 3 volumes of CO2(seems like a decent carbonation level for a sparkling cider).

Molar mass of dextrose is 180.16 g/mol
Molar mass of fructose is also 180.16 g/mol

meaning we would want to use 5.25 oz of fructose as well.

5.25 oz in kg = 0.15 kg

d = m/v

v = m/d

v = 0.15/1.057
v = 0.14 Liters, or 0.6 cups of cider should contain 5.25 oz of fructose and be suitable for priming a 5 gallon batch... right?

I think I'm missing something here


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Old 11-05-2010, 05:48 AM   #2
pimento
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I think the problem is that you are dividing the mass of your priming sugar by the gravity of the apple juice, but the two aren't actually related.

for the AJ to have a gravity of 1.057 means it has approximately 20 oz of sugar per gallon.

20/5.25 = 3.8095
128 fl oz/3.8095 = 33.6 fl oz of aj to get 5.25 oz of sugar

see this thread for a couple other ways to estimate it.
http://www.homebrewtalk.com/f32/prim...e-juice-30797/



 
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Old 11-05-2010, 07:27 AM   #3
plaidpaint
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I don't know much about the science of cider making, but assuming the apple juice is just a solution of sugar (fructose) and water, which seems to be the assumption here, the amount of sugar in the solution should be 57/1000, or 5.7%.

Thus 1 L (1000g) of juice would have 57 g of fructose. So if you want 150 g of fructose, you would need 2.63 L of juice. I don't know how to make it work accurately with imperial units

I think that's a bit less than 2 oz per liter, so about 8 oz per gallon. And 2.63 L should be a bit more than half a gallon. This doesn't mesh with pimento's response, so if he's ever done anything like this, please take his word for it. I just saw a math problem and couldn't resist

 
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Old 11-05-2010, 09:36 AM   #4
Revvy
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Last month's basic brewing radio was all about alternative priming methods, and the guest (who btw, although he is a minister, from michigan, and is an expert on bottling, is NOT ME, but the coincidence is freaky) offers info on calculating how to prime with strange things. IIRC he went by the amount of carbohydrates in what he was using (including juice) and then used the tables linked with the podcast to calculate the amount needed.

Quote:
October 28, 2010 - Alternate Priming Sugars
Home brewer Drew Filkins shares his technique of using alternative ingredients to put the bubbles in his brew.

Click to Listen-Mp3

Hydrometer readings and sugar content charts from HomeWinemaking.com http://www.home-winemaking.com/winemaking-2b.html
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Old 11-06-2010, 12:54 AM   #5
pimento
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Quote:
Originally Posted by plaidpaint View Post
I don't know much about the science of cider making, but assuming the apple juice is just a solution of sugar (fructose) and water, which seems to be the assumption here, the amount of sugar in the solution should be 57/1000, or 5.7%.

Thus 1 L (1000g) of juice would have 57 g of fructose. So if you want 150 g of fructose, you would need 2.63 L of juice. I don't know how to make it work accurately with imperial units

I think that's a bit less than 2 oz per liter, so about 8 oz per gallon. And 2.63 L should be a bit more than half a gallon. This doesn't mesh with pimento's response, so if he's ever done anything like this, please take his word for it. I just saw a math problem and couldn't resist
I should have put a disclaimer that I haven't actually tried it myself. I was more interested in the math as well and the multiple ways that people came up to estimate it. Thanks for the link revvy.



 
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