Boil Rate and Effects to Wort Concentration

Hello follow brew scientists.
I'm looking for a formula that will help me to find how much the wort sugar concentrates to after boiling. I thought that at first I could use the start volume and the refract number and then mulitply that by the concentration of the boil off but I must be missing something here. Has anyone found a good formula for that.

Ex: 6 gallons at 1.049 boiled down to 5 Gallons would be a 1/6th concentration of the sugars so I took 1.049 x 1.166=1.223 at 5 Gallons. Clearly this is not correct. Do some of the sugars boil off to atmosphere?

This is easy. When calculatung gravities, you work with the number after the decimal point. In your case that would be 49, and not 1.049.

Sugars do not evaporate and stay the same when boiling.

So, if you have 6 gallons x 49 sugars, how much would sugars concentrate if you end up with 5 gallons after boiling?

6 x 49 = 5 x OG
294 = 5 x OG
OG = 294 / 5
OG = 58.8

So your original gravity after boiling will be 1.0588.

Assuming that you posted this to Brew Science because you want to know how things work here's how you do this problem.

Wort, prior to boiling, is considered to be made up of water and 'extract'. Extract is anything dissolved in the wort. The first step in such problems is to find out how much extract you have and how much water. To do that you need to know the specific gravity of the wort (1.049) and its volume (6 gal initially). We really want the weight of the wort and so must correct volume to room temperature if we are to use 8.35 lbs/gal as the weight. In general the weight is
W = 8.35*SG*correction_factor. The correction factor can be derived from tables of the density of water at various temperatures. Wort density and thus weight changes at very like the rate that pure water does (because it is mostly water). Close to boil the correction factor is about 0.96. Assuming that the 6 gal is corrected to room temperature in your example we have

W = 6*1.049*8.35 = 52.6 lbs of wort. To find out how much of that is extract you find the sugar concentration of a 1.059 SG solution. This is done by dividing 59 by a factor close to 4. In this case it is 4.0345 found from the formula f = 3.8694 + 3.3709*(SG -1). The sugar concentration is 59/4.0345 = 12.15 °P or 12.15% w/w and you thus have .125*52.6= 6.38 lbs of extract. Divide this by the amount of grain you used and you have the efficiency of your brew up to this point.

Putting all these steps into one formula we have

W =vol*temp_corr_factor*8.35*SG*1000*(SG-1)/(3.8694 + 3.3709*(SG -1))/100

You will want that formula to put into a spreadsheet for reasons we will get to in a minute.

Boiing removes mostly water (some volatiles like DMS and even a bit of sugar flies off but you are quite safe in assuming that the amounts of these are trivial). You will have a new volume and a new SG and you want to predict what that SG will be based on the volume (you can, of course, measure it). The weight of the sugar is assumed invariant so that

W = vol2*temp_corr_factor2*8.35*SG2*1000*(SG2-1)/(3.8694 + 3.3709*(SG2 -1))/100

is the same as initially. To find the new SG you must solve the equation

vol*temp_corr_factor*8.35*SG*1000*(SG-1)/(3.8694 + 3.3709*(SG -1))/100 = vol2*temp_corr_factor2*8.35*SG2*1000*(SG2-1)/(3.8694 + 3.3709*(SG2 -1))/100

for sg2. To do this put the formula into two Excel spreadsheet cells with separate input cells for SG and volume for each. Create another cell in which the difference calculated by the two copies of the formula is calculated. If vol and vol2 are equal the difference will be 0 if SG2 = SG. In your example you would put 5 gal for vol2. With SG2 = 1.049 the sugar weight will be 5.31 lbs which is different from the original 6.38 by 1.06. So SG2 is clearly larger. To find what it is put trial numbers into SG2 until you zero the difference. A good first guess is one obtained by an approximate method such as the one given in #2. Sticking in 1.0588 the difference is only 0.007 lb. That's pretty close. In fact close enough for most applications.

The easiest way to get the solution is to let Excel's Solver do it for you though you can continue to grope for a solution manually. Solver quickly returns 1.05873 as the correct answer. Unless you are using a digital densitometer you obviously wouldn't notice the difference (and wouldn't anyway because you aren't measuring volumes to 4 decimal places.

Yes this will work thanks for the formula.