Electric RIMS wattage, voltage calculations

I'm building an electric RIMS system from an old barrel-style keg. I picked up a 4500W 240V HWD element from Home Depot, and built a 1" copper tube housing around it to let wort flow by.

My understanding from various homebrewing forums I browse is that if I run the element at 120V, I should reasonably expect to get about 1100W of heat from the element (1/2 the voltage equaling 1/4 the wattage), effectively making the element a ULWD type good for not scorching wort, but I'm scratching my head about that. My understanding is that Watts = Voltage X Amperage, which seems to imply that my 4500W element at 240V would be 37 amps at 120V, which could cause all sorts of unpleasantness with my household power circuits.

So I have three questions:

  1. Why would 1/2 the voltage equal 1/4 the wattage,
  2. is this enough wattage to maintain a mashing temperature for up to a 10 gallon batch in a stainless steel kettle, and
  3. would I blow a 15A GFCI circuit by putting the March brew pump on the same line?

Topic rims electric homebrew

Category Mac

1
answered at

For calculating heating element amp draw (as well as other useful things like water heating time), check out the Javascript calculator on my website.

Now for your questions three:

1- Algebra

Watts = Volts * Amperage is correct, but we need to go back a step to Ohm's law (Volts = Amperage * Resistance) to work this out. We can rearrange Ohm's law to I = V/R, where I is Amperage.

We can then add these two formulas together by substituting V/R for I in the Power formula (power is measured in watts):

P = V*I becomes P = V*(V/R) or just P = V*V/R

Solving for Resistance gets us R = V*V/P. Assuming your element is rated 4500 watts at 240V, this means it is offering 12.8 ohms of resistance. That is a constant R in our calculations.

So now we just plug into our modified power formula.

P = 240*240/12.8 = 4500 watts

P = 120*120/12.8 = 1125 watts

2- Edit: I apologize, I misread the question - For a recirculated mash (like a RIMS or HERMS system) I think this will be fine, but I don't have any personal experience.

Now for a boil, 1125 watts is pretty light for a 10 gallon batch. It may be able to maintain a boil at full power, but it will take over 3 hours to get 10 gallons of 70F water to boiling. I use a 5500W element at 240V in my kettle, and run about a 70% duty cycle to maintain a good boil.

3- Unlikely, but possibly if the March has a high startup spike. 1125W at 120V draws 9.38A. The March is rated at only 1.4A. So technically, there should be enough headroom for your March pump.

Hope that's all clear. It's kind of hard to get algebra across in this format.

1
answered at

I can understand your confusion - with potential, current, resistance and power all being defined in terms of each other, it's tricky to know where to start!

An electric element is a resistive load, with constant resistance. Consequently, the power rating is given relative to the voltage across the element. Since we know the power and the voltage, we can work out the constant resistance. For your 4500W element at 240V you have

P = V^2/R, or R = V^2/P = 240*240/4500 = 12.8 ohm.

When running at 240V, the current across the element is I = V/R = 240/12.8 = 18.75 A. (I have a 5500W element, and that draws about 22.5A so this figure seems resonable.)

If you then reduce the voltage to 120V, the current is correspondingly halved also: I = V/R = 120/12.8 = 9.375A.

Calculating power, P = VI = 120*9.375 = 1125W

The key to the 1/4 power is that because of the constant resistance, when you reduce the voltage by half, you also reduce the current flowing through the element also by half, hence the 1/4 power.

I run a 5500W element in a HERMS system, and it's barely on - IIRC about 16-20%. I would think that 1125W is plenty to maintain a 10 gallon mash at temperature. There's no way it's losing 1Kw of energy to the environment!

A march pump draws 1.4A, plus the element at 9.4A gives a total draw of 10.8A. A 15A circuit has a max recommended load of 80% = 12A, so you are below the max. You won't trip the breaker under normal load.

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