Day 427

Another week, another three hours of brewing calculations. Today was all about mixing and the subsequent effect on temperature. The main formula for the morning was

Aa + Bb = Cc


  • A = the volume of one substance (say, grain)
  • a = some quality of A, say temperature
  • B = the volume of a second substance (say, water)
  • b = the same quality as “a”
  • C = the volume of the two substances mixed together
  • c = the resultant new quality of the mixture

We can rearrange this to solve for any of the variables. For instance, if we wanted to solv for “c”, we could rearrange the formula to look like

c = (Aa + Bb)/C

So if we mix 200 kg of grain at 20°C into 300 L of water at 69°C, the new temperature of the mash (“c”) will be:

[(200 kg x 20°C) + (300 kg x 69°C)]/500 kg = 49.4°C

[EDIT: I discovered one week later that I have missed one factor in this formula. See the correct formula and the correct answer on Day 434.]

Many home brewers do not have any external heat source for their mash tuns, such as steam. Instead, when they want to raise the temperature of the mash in order to mash out, they have to add hot water to the mash tun. How much would Mr. Homebrewer have to add to the above mash to raise the temperature from 58°C to 63°C? Can we use the above formula to determine the resultant temperature of the mash? Well, sort of. First we have to do an in-between calculation, because it turns out that grain doesn’t transfer heat nearly as well as water. In fact, grain only transfers 40% of its heat. So we need to determine the Mash Heat Capacity (MCH), which will tell us how much heat is represented by the mash.

MCH = [(volume of grain x 40%) + volume of water]/volume of grain + volume of water

or, for the above example

[(200 x 0.4) + 300]/(200 + 300) = 0.76 = 76%

Since we want to find out the volume of water needed, we can rearrange our old friend Aa + Bb = Cc so it looks like this:

B = [((MCH x C) x (c – a)]/(b – c)

where B is the volume of the water to be added. In this case

B = [(0.76 x 500 L)(63°C – 58°C)]/(95°C – 63°C) = 59.4 L

Those of you are actually following along will notice that I have used 95°C for the temperature of the boiling water instead of 100°C, since by the time you transfer the water from the boiler, it probably loses 5 degrees.

If you are planning a decoction mash–where a portion of mash is removed, heated up and then returned to the mash tun to raise the mash tun temperature, this formula will also work to calculate how much mash has to be removed. As a matter of fact, it’s even easier than calcualting additions of hot water, since you don’t have to calculate the Mash Heat Capacity–you are adding grain and water to grain and water, so the MCH of one cancels out the MCH of the other.

We also learned how to calculate evaporation losses per hour during the boil. This was relatively simple compared to the above:

Step 1  – Calculate the percentage of volume that evaporates: (Finishing volume – starting volume)/starting volume

Step 2 – Determine the evaporation rate per hour: % volume evaporated (from Step 1)/length of boil in hours

Step 3 – Calculate the volume of wort lost per hour: evaporation rate/hr (from Ste 2) x starting volume

So if we do a boil of 1h 15min, start with a volume of 1100 L and end with 1000 L:

  1. (1100 – 1000)/11 = 9% evaporation loss
  2. 9%/1.25 hrs = 7.2 %/hr
  3. 7.2% x 1100 L = 79.2 L per hour evaporation loss

On to History of Beer. Today we had more student presentations, including The History of Molson, E.P. Taylor’s Effect on the Canadian Beer Industry, and Beer in Art. We also had a special guest, beer historian Ian Bowering, who also writes a column for the Great Lakes Brewing News called “The Jolly Giant”.

End of a long day, but assignments are starting to pile up again, so time to burn some midnight oil…

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