Hitting a target temperature with a decoction can be tricky if your mash tun doesn't have a heat source. That's because (1) your main mash will get colder during its rests and (2) the decoction will need to raise the temperatures of both your mash and your mash tun.

To account for heat loss over time, you should conduct a simple test with your mash tun: add hot water, measure its temperature, wait an hour and measure its temperature again. You can then determine your mash tun's rate of heat loss by plugging the measured temperatures into the following equation:

dQ = 8.33 x Vw x (Tw1 - Tw2) / (Tw1 - Ta)

where:

dQ = rate of heat loss (BTU/hr/degf)

Vw = volume of water, measured cold (gal)

Tw1 = initial water temperature (degf)

Tw2 = final water temperature (degf)

Ta = ambient temperature (degf)

By assuming dQ is a constant (it'll be close enough at mash temperatures), you can calculate the temperature loss of a given mash rest with this equation:

dT = dQ x (T1 - Ta) x t x (8.33 x Vw / mg + 1) / (8.33 x Vw / mg + 0.4) / (8.33 x Vw + mg)

where:

dT = temperature loss (degf)

t = time of rest (hr)

Vw = volume of mash water, measured cold (gal)

mg = mass of grain (lb)

In reality, Vw and mg will decrease after you pull a decoction. However, the math would require iteration if you needed to know the decoction volume to figure out the decoction volume. To get in the ballpark for the portion of a rest where a decoction is being pulled, you can multiply both values by 0.7. Most of the terms cancel out, resulting in the following equation:

dT = dQ x (T1 - Ta) x t x (8.33 x Vw / mg + 1) / (8.33 x Vw / mg + 0.4) / (8.33 x Vw + mg) / 0.7

The reason why I like these calculations, as opposed to assuming a fixed temperature drop per hour, is because they account for the fact that small mashes cool down faster than big mashes.

To deal with temperature changes in the mash tun, I assume that the inner surfaces of my mash tun are at the mash temperature, the outer surfaces are at ambient temperature and that the temperature gradients between the two are linear. These assumptions aren't perfect, but they're adequate for predicting the initial temperature of a mash rest. As far as the math is concerned, half of the mash tun gets heated up to mash temperature and the other half stays at ambient temperature. Including your mash tun in your calculations, and assuming the decoction will have the same water-to-grain ratio as the mash, will change the heat transfer equation from this:

Vd = Vm x (T2 - T1) / (Td - T1)

to this:

Vd = (cm x mm + cmt x mmt / 2) x (T2 - T1) / (cm x (Td - T1)) / dm

where:

Vd = volume of the decoction (gal)

Vm = volume of the mash before pulling the decoction (gal)

T2 = target mash temperature after returning the decoction (degf)

T1 = main mash temperature before returning the decoction (degf)

Td = temperature of the decoction before returning to the main mash (degf)

cm = specific heat capacity of the mash (BTU/lb/degf)

mm = mass of the mash before pulling the decoction (lb)

cmt = effective specific heat capacity of the mash tun (BTU/lb/degf)

mmt = mass of the mash tun (lb)

dm = average density of the mash (lb/gal)

Specific heat capacity of the mash? Mass of the mash? Average density of the mash? Yeah, there are a few variables you'll probably never measure. Thankfully, you can mathematically manipulate them into brewing parameters that you care about:

Vd = (1 + cmt x mmt x (8.33 x Vw / mg + 1) / (8.33 x Vw / mg + 0.4) / (8.33 x Vw + mg) / 2) x (1.018 x Vw + 0.0837 x mg) x (T2 - T1) / (Ts - T1)

Based on some crude testing, my mash tun seems to have an effective specific heat capacity of 0.33 BTU/lb/degf (I love using pounds as a unit of mass. Don't have judgment). If you use a plastic cooler for a mash tun, that number should work for you too. Anyway, you can see that accounting for temperature losses made the calculations a lot more complex. On that note, I'll leave you with some food for thought: should we worry about evaporative water loss from the boiling decoction?

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