How Many Watts do you need for my Boil Kettle or Hot Liquor Tun??? Math involved.

Here is the math for determining how big an element you need.

Q=m•c•(T2-T1)

Q → energy in joules (J)
m → mass in kilograms (kg)
c → heat capacity in joules per kilogram Kelvin (J/kg•K) ≈ 4184 @ 60ºC
T2 → final temp (K or ºC)
T1 → final temp (K or ºC)

m=d•V

d → density of water (kg/m^3) ≈ 983 @ 60ºC
V → volume of water (m^3)

V= Vg•Vm^3/g = Vg•(1m^3/264.2g)

Vg → Volume in gallons (g)

So combining that we get:

Q = [983(kg/m^3)•Vg(g)/264.2(m^3/g)]•4184(J/kg•K)•[T2-T1] = [983•Vg/264.2]•4184•[T2-T1]

If you are still following all that then we can see how much energy it takes to heat some water...

For 12 gallons from 169ºF to boil it takes:

Q = [983•12/264.2]•4184•[100-76] = 4,483,372 J

Now that we know that, we can see how much power it takes to get that much energy it the water in a given time frame.

P=Q/t

P → power in watts (W = J/s)
t → time in seconds (3600s = 1 hr)

so...

P = 4,483,372J / 3600s = 1,245 watts/hr

If you want to heat the 12 gallons of water to boil from 169º in half an hour:

1,245 watts/hr / 1/2 hr = 2,490 Watts

This means you need to get 2,490 watts into the water to boil it in half an hour.

Now for design purposes you want to take into account different temp ranges. For example, cold water to mash in or strike temperature.

T2 = 164ºF = 74ºC
T1 = 55ºF = 13ºC
Vg = 12g

Q = 11,394,000 J

That same 2,500 watt element would take 2500 = Q/t → t=4,558s or 76 minutes

a 5500 watt element would only take 34 minutes.

Of course, all this assumes 100% efficiency and no losses. Times will be in the ball park though.

Hope this helps!

-Justin