Sugar displacing water

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menschmaschine

menschmaschine

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It's been a while since I naturally carbed a batch and I don't see doing it anytime soon. Can anyone tell me or point me in the right direction?...

How much water is displaced per "unit" of corn sugar? For instance, if you poured 1 oz. (by weight) of corn sugar into 1 cup of water, how much total solution would you have? I prefer metric measurements, but old timey Imperial measurements are fine too.;)

What I'm thinking is to calculate a quantity of corn sugar in a quantity of water and using something like this to inject a number of ml's of this sugar solution into bottles for limited bottling of a batch. So, if I had a batch of beer that I was kegging, but wanted to naturally carb a few bottles, I could use something like what's in the photo below instead of carb tabs. I want to be as precise as possible so as not to over or under-carb.

21pu33NrgEL._SS500_.jpg
 
Couldn't you just take the amount of sugar you need for the whole batch and figure how much would go into each bottle. 4oz is about 141 grams. Assuming 48 bottles that'd be just about 3grams per bottle.

Take however many bottles worth of sugar you need and dissolve it into as little water as it takes. Then measure the liquid volume that it takes up and inject whatever fraction of it you need.

There must be a more scientific method to figure this out, but this is what I'd do if I didn't use carb drops for my extra bottles. Actually if its easy to figure this out it seems like a good way to be more precise. At that point you could use a pressure cooker to keep sterile priming solutions on hand and just inject however much you need.
 
What Bokonon said. Why don't you just do it by weight? It seems simpler and potentially more accurate. BTW 4 ounces is 116g, 5 ounces is 145g. In any event if you divide the number of 12 oz bottles for a 5 gallon batch (50) into the standard priming sugar dose for that batch (5 oz/145g) it comes out to just under 3g. (145/50 = 2.9) Just adjust that for your preferred carbonation level.
 
I do it this way: let's say I have 40 bottles to carbonate.
I calculate, how much priming sugar do I need, I dissolve it in a small amount of water, and then I top up the water until I reach some even number, let's say 160 ml so that I could take nice and easy 4 ml to each of 40 bottles.
 
The answer is almost none once the sugar dissolves. The molecules of sugar and water fit around each other, so the total volume is less than the sum of the parts.
 
Thanks david_42. I guess everyone else is right too. However, I calculate my priming sugar additions with a priming calculator, which accounts for temperature. But let's say I wanted to naturally carb two 0.5 liter bottles of an ordinary bitter. The low carb level would only put me at only around 1.5 grams of sugar per bottle. A total of 3 grams. If I dissolve that into say 10 ml of water and put 5 ml into each bottle I should be good. Sound about right? I just wanted to try to calculate this precisely, scientifically. That's why I put it in this forum.;)
 
This is what I have done in the past.
I make a stock solution of 157g Sugar (3/4 cup) into 2 cups boiling water.
Here is the formula if anyone is interested:
(Should be around 2.5 volumes CO2.)
At this level I had 7.85g/l if it was poured into a keg.
At room temp this should be 2.5 volumes.

64oz Bottle 25ml
32oz Bottle 12.5ml
16oz Bottle 7ml
12oz Bottle 5ml
 
menschmaschine said:
How much water is displaced per "unit" of corn sugar? For instance, if you poured 1 oz. (by weight) of corn sugar into 1 cup of water, how much total solution would you have? I prefer metric measurements, but old timey Imperial measurements are fine too.;)
Click to expand...

You are referring to the density of corn sugar. See this.

If I were you, I would do this...

Say we have a 19L batch to which we want to add 120 g sugar to the batch to prime. If you make a 50% solution (120 g sugar and 120 g water), you will end up with around 195 mL of solution. Since a bottle is 12 oz = 0.355 L, that means each bottle contains 0.355L/19L = 1.87% of the total beer. So, you need to put 1.87% of the priming solution into each bottle. This ends up being 0.0187*195 mL = 3.65 mL of priming solution per bottle.
 
rocketman768 said:
You are referring to the density of corn sugar. See this.

If I were you, I would do this...

Say we have a 19L batch to which we want to add 120 g sugar to the batch to prime. If you make a 50% solution (120 g sugar and 120 g water), you will end up with around 195 mL of solution. Since a bottle is 12 oz = 0.355 L, that means each bottle contains 0.355L/19L = 1.87% of the total beer. So, you need to put 1.87% of the priming solution into each bottle. This ends up being 0.0187*195 mL = 3.65 mL of priming solution per bottle.
Click to expand...

OK, that's helpful. So if I have this right...

Total ml of Solution = mlWater + (gramsDextrose * (1 / DensityDextrose))

The density of Dextrose is 1.54 @ 25°C.

So, if I wanted to bottle 1 liter (2 half-liter bottles) of an ordinary bitter for ~1.2 volumes CO2 at 70°F and used 10 ml of water for the solution and 3.0 grams of sugar, the calculation would be:

Total ml of Solution = 10 + (3.0 * (1/1.54))
Total ml of Solution = 11.95 ml

So I could put ~6 ml of this solution in each bottle.
 
menschmaschine said:
OK, that's helpful. So if I have this right...

Total ml of Solution = mlWater + (gramsDextrose * (1 / DensityDextrose))

The density of Dextrose is 1.54 @ 25°C.

So, if I wanted to bottle 1 liter (2 half-liter bottles) of an ordinary bitter for ~1.2 volumes CO2 at 70°F and used 10 ml of water for the solution and 3.0 grams of sugar, the calculation would be:

Total ml of Solution = 10 + (3.0 * (1/1.54))
Total ml of Solution = 11.95 ml

So I could put ~6 ml of this solution in each bottle.
Click to expand...

Looks right to me. Though, you better have a good scale for measuring out just 3 grams of something.
 
I don't think so...look at David42's post early itt. You can't just add the seperate volumes.

EDIT: Would it be easier to just mix a solution to a certain SG and calculate how much sugar content per mL of liquid there is at that SG...and then just calculate your volume of sugar solution needed? That way instead of measuring small masses of sugar you would be measuring slightly larger volumes of liquid and the SG of that liquid would have been measured by your hydro. Might be less prone to error in measurements depending on your equipment.

EDIT2: rocketman's post 3 posts above seems right but then it appears to me mensch made a leap from it to adding the volumes which you can't really do (although like many things beer...it'll work to some degree).
 
SpanishCastleAle said:
EDIT2: rocketman's post 3 posts above seems right but then it appears to me mensch made a leap from it to adding the volumes which you can't really do (although like many things beer...it'll work to some degree).
Click to expand...

I'm not just adding volumes, per se. I'm using the volume the sugar will add to the water based on its density to calculate the total volume. I'm not saying I'm definitely right, but based on the link posted by rocketman and finding the actual density of dextrose (as opposed to sucrose like in rocketman's link), that's what I came up with. I suppose I could sacrifice 3 grams of corn sugar and try it out.:)
 
I'm not just adding volumes, per se. I'm using the volume the sugar will add to the water based on its density to calculate the total volume.
Click to expand...
Yea but density is just mass per unit volume. Calculating the volume from the density and mass still just gives you a volume...a volume that can't really be added to the water volume to yield the true volume. Just seems like a juggling of the same numbers. It'll be close but I assumed you were trying to be as exact as possible.
 
menschmaschine said:
OK, that's helpful. So if I have this right...

Total ml of Solution = mlWater + (gramsDextrose * (1 / DensityDextrose))

The density of Dextrose is 1.54 @ 25°C.

So, if I wanted to bottle 1 liter (2 half-liter bottles) of an ordinary bitter for ~1.2 volumes CO2 at 70°F and used 10 ml of water for the solution and 3.0 grams of sugar, the calculation would be:

Total ml of Solution = 10 + (3.0 * (1/1.54))
Total ml of Solution = 11.95 ml

So I could put ~6 ml of this solution in each bottle.
Click to expand...


That is definitely wrong.

You best bet is just to do the experiment yourself. If you have a graduated cylinder and a gram scale, it should be easy.
 
It appears the equation to be used would be:

V(t) = n(a)V(a) + n(b)V(b)

where:

V(t) = total volume
n(a) = moles dextrose (180g = 1 mole)
V(a) = partial molar volume of dextrose
n(b) = moles water (18.016g = 1 mole)
V(b) = partial molar volume of water

Partial molar volume is dependent on temperature and pressure. That's where I'm stuck... figuring out reasonably accurate partial molar volumes.
 
SpanishCastleAle said:
I don't think so...look at David42's post early itt. You can't just add the seperate volumes.

EDIT: Would it be easier to just mix a solution to a certain SG and calculate how much sugar content per mL of liquid there is at that SG...and then just calculate your volume of sugar solution needed? That way instead of measuring small masses of sugar you would be measuring slightly larger volumes of liquid and the SG of that liquid would have been measured by your hydro. Might be less prone to error in measurements depending on your equipment.

EDIT2: rocketman's post 3 posts above seems right but then it appears to me mensch made a leap from it to adding the volumes which you can't really do (although like many things beer...it'll work to some degree).
Click to expand...

Guys, look at the link in my post. It acknowledges that you can't just add the volumes, but it also quite plainly states that for sucrose, the difference is negligible.

I just did this experiment...

Added 30 mL water to a conical graduated cylinder with cap. Verified the mass of the water was 30.0 g +- 0.1 g. Measured 10.0 g sucrose, poured into conical cylinder with water. Shook for 1 minute to ensure dissolution. Verified sucrose was completely dissolved since there was no sucrose sediment at the bottom of the cylinder. Measured volume of solution was between 36 and 37 mL. Now, the density of sucrose is about 1.587 g/mL. So, by just adding volumes, you would expect that you would get 30 + 10/1.587 = 36.3 mL of solution, which entirely agrees with my result.

Conclusion: you can't just add two volumes together for any arbitrary solution, but, at least for sucrose, it makes a very good approximation at room temperature when the solution is not close to saturation.
 
rocketman768 said:
Guys, look at the link in my post. It acknowledges that you can't just add the volumes, but it also quite plainly states that for sucrose, the difference is negligible.

I just did this experiment...

Added 30 mL water to a conical graduated cylinder with cap. Verified the mass of the water was 30.0 g +- 0.1 g. Measured 10.0 g sucrose, poured into conical cylinder with water. Shook for 1 minute to ensure dissolution. Verified sucrose was completely dissolved since there was no sucrose sediment at the bottom of the cylinder. Measured volume of solution was between 36 and 37 mL. Now, the density of sucrose is about 1.587 g/mL. So, by just adding volumes, you would expect that you would get 30 + 10/1.587 = 36.3 mL of solution, which entirely agrees with my result.

Conclusion: you can't just add two volumes together for any arbitrary solution, but, at least for sucrose, it makes a very good approximation at room temperature when the solution is not close to saturation.
Click to expand...

Thanks for doing that, rocketman. I'll try that with dextrose when I get a chance.
 
rocketman768 said:
Guys, look at the link in my post. It acknowledges that you can't just add the volumes, but it also quite plainly states that for sucrose, the difference is negligible.

I just did this experiment...

Added 30 mL water to a conical graduated cylinder with cap. Verified the mass of the water was 30.0 g +- 0.1 g. Measured 10.0 g sucrose, poured into conical cylinder with water. Shook for 1 minute to ensure dissolution. Verified sucrose was completely dissolved since there was no sucrose sediment at the bottom of the cylinder. Measured volume of solution was between 36 and 37 mL. Now, the density of sucrose is about 1.587 g/mL. So, by just adding volumes, you would expect that you would get 30 + 10/1.587 = 36.3 mL of solution, which entirely agrees with my result.

Conclusion: you can't just add two volumes together for any arbitrary solution, but, at least for sucrose, it makes a very good approximation at room temperature when the solution is not close to saturation.
Click to expand...



:rockin:


....
 
Not sure if anyone has the correct answer yet, but here's my crack at it:
dextrose will add 46 points per pound per gallon (for my example), if you add one lb of dextrose to one gallon of water (weighing ~8.35 lbs), you get a mixture that weights 9.35 lbs and a SG of 1.046. One gallon of solution with an SG of 1.046 will weight 8.734 lbs/gal; but we have a volume of solution that now weighs 9.35 lbs, so the volume of the new solution is actually 9.35(lbs)/8.734(lbs/gal)=1.07 gal, IOW the volume of the gallon of water increase by about one cup in volume, even though the volume of a pound of dextrose is roughly three cups.
 
shek said:
Not sure if anyone has the correct answer yet, but here's my crack at it:
dextrose will add 46 points per pound per gallon (for my example), if you add one lb of dextrose to one gallon of water (weighing ~8.35 lbs), you get a mixture that weights 9.35 lbs and a SG of 1.046. One gallon of solution with an SG of 1.046 will weight 8.734 lbs/gal; but we have a volume of solution that now weighs 9.35 lbs, so the volume of the new solution is actually 9.35(lbs)/8.734(lbs/gal)=1.07 gal, IOW the volume of the gallon of water increase by about one cup in volume, even though the volume of a pound of dextrose is roughly three cups.
Click to expand...

The volume of a pound of dextrose is 0.078 gallons which agrees with your calculation. The density of dextrose is 1.54 g/mL, and a pound is 453.6 g. So, the volume of a pound of dextrose is 453.6g / (1.54g/mL) = 294.5 mL = 1.24 cups = 0.078 gal. Maybe you say it's 3 cups because that would appear to be how much room a pound of granulated dextrose takes up in a measuring cup? Don't forget there's a ton of airspace between those granules.
 
You cannot do this calculation with ppg since the "g" in ppg is the volume of the sugar solution and not the volume of the water.

The solution is very easy for sucrose since the Plato/Balling/Brix tables were done with that.

Plato = 100 * m_sucrose / (m_water + m_sucrose)

sg = plato_to_sg(Plato) <---- you'll have to look this up or use one of the many calculators out there.

V_sugar_solution = (m_water + m_sucrose) / sg

Do all this in metric and you'll be fine. It's likely the same for glucose but keep in mind that the dextrose we commonly use is dextrose monohydrate which is 91% glucose and 9% water.

Mensch, don't you have a gram scale for salt additions to the mash? Working by weight is much simpler and you can still mark the syringe accordingly after you tested how much volume the desired sugar solution occupies. That's how I have done this before when I primed a few bottles with a syringe and sugar solution.

Kai
 
rocketman768 said:
The volume of a pound of dextrose is 0.078 gallons which agrees with your calculation. The density of dextrose is 1.54 g/mL, and a pound is 453.6 g. So, the volume of a pound of dextrose is 453.6g / (1.54g/mL) = 294.5 mL = 1.24 cups = 0.078 gal. Maybe you say it's 3 cups because that would appear to be how much room a pound of granulated dextrose takes up in a measuring cup? Don't forget there's a ton of airspace between those granules.
Click to expand...
I meant that it was 3 cups if you were simply measureing the dextrose with a cup. But when you mix it into a solution it only adds about 1 cup of volume. I'm guessing if you had a brick of dextrose that weighed one pound it would be about 1 cup in volume.

Kaiser, you are correct. The solution will weigh 8.734 pounds, and since you added 1 pound of dextrose, that means that you had 7.734 pounds of water to begin with, which is 0.926 Gallons, meaning that the volume increase from the water you started with to your new solution was .074 (a little more than a cup worth of volume). Maybe my original calc was wrong, but this one is def right.
 
In addition to that, because the relationship between plato and sg is not linear, the volume per weigt contributed by the dissolved sugar depends on the concentration.

Kai
 
I really wish I had just started with Plato and metric...soooooo much better.

Regarding the equation:
V_sugar_solution = (m_water + m_sucrose) / sg

The unit is mL in case anyone was wondering (should be fairly obv). That equation threw me at first because I didn't think the dimensions added up. So for all us non-metric slackers: the reason why there can be a volume on one side and a mass on the other is because we are really dividing by density (i.e. mass/volume), not really SG per se. But Density and SG are the same actual number in metric so it works...beautifully.

US units I cast thee out!
 
Kaiser said:
Mensch, don't you have a gram scale for salt additions to the mash?
Click to expand...

Thanks, Kai. No, I do not yet have a scale for measuring mineral additions. My digital scale is good for grains and hops, but not precise enough for minerals, etc.

Assuming your "m_" is mass, I'm getting 11.85ml for the total solution volume of 3 grams sugar and 10ml water.
 
This thread is proof that the beer science forum is a good thing!

wowsers, my head hurts, but I think I understood it all! :mug:
 
Now, let's do this thing the right way as meticulously as possible...

Let's assume everything is at 20C. We have 10 mL of water and 3 g glucose we wish to put in solution. How much total volume will there be?

Since water has a density of 0.9982071 kg/L at 20C, and since glucose has about a 92% yield, the plato degree "P" of the solution is
3*.92/(3*.92+9.98) * 100 = 21.66

Since the relationship between plato and apparent 20C/20C gravity "g" is
P = 135.997g^3 - 630.272g^2 + 1111.14g -616.868,
we find that P = 21.66 corresponds to g = 1.0904 by root-finding techniques for this cubic equation.

Now, 1.0904 is only the apparent 20C/20C density. Since the real density of water at 20C is as I listed above, the actual density of the solution is
1.0904*.9982071 kg/L = 1.088 kg/L = 1.088 g/mL.

Since the mass of solution is 12.98 g, the volume of the solution is 12.98 / 1.088 = 11.93 mL...if you really want to get anal.
 

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