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I'm aware of this answer to How to brew all-grain indoors. It says

It gives a good rolling boil and doesn't go all floppy when it's hot. It has a 2kW element holds 29 litres

That's nice, about the volume I need, but it looks insulated. We'll be buying new gas stove to our apartment, preferably one with 5 burners. Sadly, most of them only have 3.8 kW on central burner. Of course, on a gas stove I can't use any insulation around my pot, it would be a fire hazard. 5kW stoves are max I could find, but are either expensive or aesthetically lacking.

Given that issues - is 3.8 kW enough? Is there a general way to calculate required power for given volume? Or pre-calculated tables for home brewers?

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This is a fairly straightforward calculation, actually.*

First, calculate the total energy needed to heat up your volume of liquid, using the following equation:

E = Cp * M * dT
  where:
  E = energy required, in kilojoules (kJ)
  Cp = specific heat of liquid (kJ/kg/°C)
  M = mass of liquid (kg)
  dT = temperature change required (°C)

A few assumptions:

  • The specific heat of water (and wort) is ~4.2 kJ/kg/°C
  • I'll assume you are heating from rougly mashing temperature (65°C) to boiling (i.e. dT above will be 100 - 65 = 35°C)
  • As a ballpark estimate, I'll say heat from the burner is delivered to the pot with ~75% efficiency (also accounting for constant heat loss to the environment)

So, the total energy you need to heat 30 kg of wort (since 1L = ~1kg) by 35°C would be:

E = 4.2 kJ/kg/°C * 30 kg * 35°C
E = 4,410 kJ

To find out how much power you need (Watts) you simply divide the total energy needed by the time desired (in seconds), since 1 Watt = 1 Joule per second.

To heat up in 30 minutes (or 1,800 seconds):

P = 4,400 kJ / 1,800 sec
P = ~2.5 kW

Applying the assumed efficiency:

P = 2.5 kW / 0.75
P = ~3.3 kW

So it looks like you'd be fine with this burner, though beware that my assumption about efficiency is entirely made up and may be very different in reality.

* Sort of... See below


After a closer look I'll note that, while the above equation is entirely sound in principle, it doesn't necessarily apply well when considering open gas ranges. My experience is chiefly with closed, steam-heated systems, where heat transfer efficiencies are high and largely predictable.

A few things that make this question difficult:

  • The kilowatt output rating of a gas burner has little to do with how much heat ends up in your pot/liquid. Looking around the internet, I've seen (mostly un-cited) claims that the amount of energy ending up in the cookware ends up being in the 25-40% range (see here, here, here).
  • This efficiency will depend greatly on your particular setup (the size of the pot in comparison to the burner, how high you run the burner, the environment in which you run it).
  • The conditions for loss of heat energy will also vary greatly. Some very interesting discussions can be found here and here. As mentioned therein, the majority of energy lost when heating an open pot of liquid is to evaporation (~80%). The obvious solution is to cover the pot while it's heating and uncover when you achieve a boil. Radiative heat loss will depend on the material of your pot, convective heat loss will depend on the temperature and movement of the ambient atmosphere, and the thickness of the pot itself probably has a small insulating effect.

Let's revisit it from a worst-case perspective: you still need a minimum ~2.5 kW of power to heat the liquid, assuming no heat loss whatsoever. The last paper I link above found a combined heat loss from radiation and convection in a 7 liter pot to be ~100 watts in a shiny metal pot. Scaling this heat loss proportional to the surface area of the given pot up to a 40L pot (~2.8 times more surface area) we might very roughly assume a heat loss of 280 watts.

So, you need 2.5 kW of power and your burner is making 3.8 kW, but we have to assume that only 25% of that is getting into the pot, while the rest is bleeding into the environment. That leaves less than 1 kW making it into your wort. Then, say you're losing .3 kW to radiation and convection (again, assuming you're covering your pot during heat-up). This leads to the conclusion that, in this worst-case-scenario, you'd be short on power by a factor of ~5 (at least in order to do this all in half an hour).

But the real point is that this is a very complex question to answer with the level of specificity you may be hoping for.

| improve this answer | |
  • This is not true. You totally ignored heat loss on the sides of the pot and with steam. Both are important if you are boiling something longer than a few minutes. – Mołot Nov 22 '18 at 17:40
  • Nope: "also accounting for constant heat loss to the environment". It's a rough estimate, for sure, but assuming a constant temperature differential between the pot (boiling) and the environment (ambient) and a static surface area of both pot walls and liquid/atmosphere interface, you can also assume a constant rate of heat loss, and this is functionally equivalent to simply reducing the overall efficiency of heat transferred to the pot. – Franklin P Combs Nov 22 '18 at 17:47
  • Oh, ok... After edit, it looks complete, and even if it does not answer my question fully, it is really helpful. – Mołot Nov 26 '18 at 10:32

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