Plato vs SG

[nexy_sm]

Hi all guys,

there is a problem that bothers me today, and the problem is following:

imagine that I mix 1 kg of sucrose and 9l of water. In this way I get 10kg of wort with 10% of sugar by waight. That means that sacharometer should show 10 degrees. But now, if I use the fact that yeald for succrose is 46ppg or 384 HWE, if I mix 1kg of sufar in 9L of water I get SG of 384/9=42.66.

I would be really happy if somebody can explain me this.

Many thanks,
nexy_sm

 

[MachineShopBrewing]

The volume of your solution increases when you add the sugar. Thus your calculation is not accurate as stated.

 

[nexy_sm]

Hm, I looks like I am missing something.
Plato scale gives percentage of the sugar in the total mash weight, which is in my case 10 Kg, and for me looks correct.

 

[nexy_sm]

If only maybe I had a bit problem with understanding:
"As with specific gravity, the scale is just another way of expressing percentages. 5 °Plato indicates that a given volume of water contains 5% sucrose, measured by weight."

So should it be like this: 10 P in 10 L of wort equals to the mass of succrose that would fit into 1 dm^3 or 1 L (10% of 10L) cube or pot?

 

[ajf]

Your Plato calculation is correct but your SG calculation is not.
As MSB said, the volume is incorrect.
If you mix 1 Kg of sugar with 9 l water, you will end up with 10 Kg of solution, but the volume of that solution will be ~ 9.615 l
To use the 46 ppg (which I am familiar with) you have 1 * 2.2046 lbs sugar, made up to 9 * 0.264172 gallons * 46 ppg = 42.653 points.
(1 * 2.2046) / (9 * 0.264172) * 46 = 42.653
Plato is calculated on a weight by weight basis. S.G. is calculated on a weight by volume basis.
-a.

 

[nexy_sm]

Ok, but my calculation of SG and yours yealds give the same result (except I did a bit rounding).
I would start from SG which says if we mix 1kg of sugar, and 9.6L of water we get wort of 1.040. Such wort weights arround 10.6 kg, but 10% doesn't work. But if we say, OK the volume of 9.6L combined with 1kg is 9.6+0.5454 (volume of 1Kg of sucrose, based on density of sucrose of 1.85 g/cm^3) and we are roughly in 10.1 plato.
So for me it looks like if u have wort of 10L and you measure 10 plato, than one tenth of that number is the weight of sugar.

 

[ajf]

Sorry, I made a slight in the calculations. I didn't correctly account for the volume of the solution.
I should have written
To use the 46 ppg (which I am familiar with) you have 1 * 2.2046 lbs sugar, made up to 9.615 * 0.264172 gallons * 46 ppg = 39.93 points.
(1 * 2.2046) / (9.615 * 0.264172) * 46 = 39.9 or a SG of 1.040 (rounded to 3 decimal places).
In other words, mixing 1 KG sugar with 9 L water (at 20C) gives you 10 Plato, but the volume of that solution will not be 10 L. It will have a volume of 9.615 L and an SG of 1.040 (weight of 10 Kg / volume of 9.615)

-a.

 

[nexy_sm]

Hm, I think your calculation doesn't work. You can't calculate specific gravity with 9.615 L, it doesn't make any sense, cause we are mixing 1 Kg into 9 L. The thing here is that there is some coefficient that is missing. I found on the internet that conversion between platos and SG is not linear, so maybe it is not possible to calculate it.

 

[MachineShopBrewing]

Hm, I think your calculation doesn't work. You can't calculate specific gravity with 9.615 L, it doesn't make any sense, cause we are mixing 1 Kg into 9 L. The thing here is that there is some coefficient that is missing. I found on the internet that conversion between platos and SG is not linear, so maybe it is not possible to calculate it.

1 kg into 9L is 9.615L. The added sugar increases the volume. The calculation is not how much sugar into how much water, it is the concentration of sugar of a given volume of liquid.

 

[nexy_sm]

Yes, I understand that. 1 kg into 9 L gives you liquid of 9.615 L volume, but the concentration of sugar inside should be calculated like 1*2.20*46/(9*0.264), and not like ajf suggested 1*2.20*46/(9.615*0.264). Because as he suggested can lead to following calculation: if u mix 1 lb sugar into 1 gal u will not get 1.046 but rather less because in that case u would have to divide not with 1 gal, but with more cause final volume is bigger.

 

[MachineShopBrewing]

See, that is where you are wrong. The calculation is not 1 lb into 1 gallon is 1.046, it is 1 lb into a solution that results in a final volume of 1 gallon = 1.046. That is the distinction that you are missing in your calculations.

 

[ajf]

With all due respect, I think the calculation does work. You are making the fundamental mistake of confusing volume (Liters) with Weight (Kg). If you mix 1 Kg of a substance into 9 L of water, the only circumstance in which you would get 10 L of solution would be if the density (weight /volume) of that substance was equal to the density of water. In the case of sugar, it is considerably denser than water, so mixing 1 Kg sugar with 9 L water will give the solution a weight of 10 Kg, but a volume less than 10 L.
It is relatively simple to calculate the theoretical volume of the solution.
First, mix 1 Kg sugar in 9 L water. This gives 10 Kg of solution at 10 degrees Plato.
Then use http://www.brewersfriend.com/plato-to-sg-conversion-chart/ to look up the specific gravity of the solution. (You could also use other calculators, but the linked chart provides the formula used to apply the conversion.)
You will find that the S.G. of the solution is 1.040
Using the definition of specific gravity (http://en.wikipedia.org/wiki/Specific_gravity), you can calculate the volume of the solution Volume = Weight / specific gravity * density of water i.e 10 Kg / 1.040 * 1.000 = 9.615 Liters
Note, I have not applied temperature compensation to these figures as you did not specify temperatures.

-a.

 

[nexy_sm]

You are right guys, sorry for this mess. I just didn't pay attention to "... to make 1 galon solution". Sorry again and thank you.

 

[hbtasdfg]

the density of the water can change independent of the weight and volume of the sugar addition- this is because the water molecules stay closer together on average when you dissolve something else in them- otherwise gravity would get lower as you added sugar because the water molecules would be pushed farther apart

 

[MachineShopBrewing]

You are right guys, sorry for this mess. I just didn't pay attention to "... to make 1 galon solution". Sorry again and thank you.

No problem. It is an easy mistake to make. Most people make the same mistake when calculating DME additions for making starters.