I have just brewed a strong Belgian Ale and really need some help in determining what the actual OG is from somebody smart! The main wort was 23l in volume and had an OG of 1.077. 24 hours after fermentation of the wort had commenced, I dissolved 500g of pure glucose in 450mm of Boiled water and cooled. When I tried to ready the gravity of the glucose solution my refractometer couldn't read it (too high) so I created a 1 in 10 solution of the glucose mix and pure water. When I took this reading I got 1.022. I thus assume that the real gravity of the concentrate glucose solution would be 10 x 1.022 = 10.22. So the real question is this, if I now add 450mm of solution with a SG of 10.22 to another solution measuring 23l that had a OG of 1.077 how do I calculate what the effective real OG of the mixed solutions was?
Calcualting Gravity when mixing solutions?
- Thread starter Imperial1
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Did you use anhydrous glucose, or glucose mono-hydrate? There's about a 15% difference in sugar for the same weight.
When you created your diluted solution, did you add one volume of the concentrated solution to 10 volumes of water, or did you add enough water to 1 volume of concentrate to create 10 total volumes?
Your concentrated solution is going to be somewhere around 1.248 SG, not anywhere near 10.22. The actual SG will depend on the answers to the two questions above.
If you add 500 g of anhydrous glucose to 450 ml (g) of water you will have more than 450 ml of solution, since the sugar takes up a significant volume. The solution would have a weight of 950 g, and be 100 * 500 / 950 = 52.63˚Plato. Converting ˚Plato to SG gives 1.248, but I don't know how valid the commonly used conversion equation is for really high gravity solutions. The volume of solution would then be 950 g / 1.248 g/ml = 761 ml.
So, are you adding 450 ml of glucose solution or 761 ml?
If you're just adding 500 g of anhydrous glucose and 450 ml of water to 23 L of 1.077 wort, you can do the calculation of the new SG as follows:
When you created your diluted solution, did you add one volume of the concentrated solution to 10 volumes of water, or did you add enough water to 1 volume of concentrate to create 10 total volumes?
Your concentrated solution is going to be somewhere around 1.248 SG, not anywhere near 10.22. The actual SG will depend on the answers to the two questions above.
If you add 500 g of anhydrous glucose to 450 ml (g) of water you will have more than 450 ml of solution, since the sugar takes up a significant volume. The solution would have a weight of 950 g, and be 100 * 500 / 950 = 52.63˚Plato. Converting ˚Plato to SG gives 1.248, but I don't know how valid the commonly used conversion equation is for really high gravity solutions. The volume of solution would then be 950 g / 1.248 g/ml = 761 ml.
So, are you adding 450 ml of glucose solution or 761 ml?
If you're just adding 500 g of anhydrous glucose and 450 ml of water to 23 L of 1.077 wort, you can do the calculation of the new SG as follows:
23 L * 1.077 g/ml * 1000 ml/L = 24,771 g of wort
1.077 SG => 18.65˚Plato => 18.65 wt% sugar
24,771 g * 0.1865 = 4,619.8 g of sugar
24,771 g - 4,619.8 g = 20,151.2 g of water
4,619.8 g + 500 g = 5,119.8 g of sugar after addition
20,151.2 g + 450 g = 20,601.2 g of water after addition
100 * 5,119.8 g / (5,119.8 g + 20,601.2) = 19.9˚Plato
19.9˚Plato => 1.0825 SG
Brew on 1.077 SG => 18.65˚Plato => 18.65 wt% sugar
24,771 g * 0.1865 = 4,619.8 g of sugar
24,771 g - 4,619.8 g = 20,151.2 g of water
4,619.8 g + 500 g = 5,119.8 g of sugar after addition
20,151.2 g + 450 g = 20,601.2 g of water after addition
100 * 5,119.8 g / (5,119.8 g + 20,601.2) = 19.9˚Plato
19.9˚Plato => 1.0825 SG
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