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Old 09-24-2013, 05:37 PM   #1
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Default Water heater element Amperage

So I'm sure that the answer is out there someplace. Done a lot of reading and thought I saw it but can't find the answer now. I am in the process of building out a semi automated electric propane hybrid brewing panel. The question I have is this. My power coming in will be 120, and I have a 2000 Watt heating element, I am going to just be using it in a HERMS role. I am planning on using a p-1000 PID with a 40 AMP SSR. In between the PID and the heating element I have a 2 position rotary switch that I will wire in to give me "kill" option. Here is my question. Will I burn the switch out if I the heating element is pull more amperage than the switch is rated to? Or will the heating element just operate at 1/2 the capacity? 20A vs 10A?

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Old 09-24-2013, 05:54 PM   #2
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The amp rating of a switch is the maximum current capacity that it is rated for.
If using a switch rated for less than what your application requires, the switch will overheat and fail.

Your 2000 watt element will draw 17 amps when operated on 120 volts.

Knowing that, your SSR rating is more than adequate at 40A. You should plan on using a rotary kill switch that is rated at least 20 A to be safe.

The element will not operate at half capacity if the switching components are rated at 10 Amps vs 20 Amps.

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Old 09-24-2013, 06:16 PM   #3
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That's what I was thinking but figured I should be safe and ask.

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Old 09-24-2013, 06:27 PM   #4
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Is the element a 120v or 240v unit? My 4500w (IIRC) 240v element runs about 12 amps on 120v.

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Old 09-24-2013, 06:46 PM   #5
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Quote:
Originally Posted by joelrapp
So I'm sure that the answer is out there someplace. Done a lot of reading and thought I saw it but can't find the answer now. I am in the process of building out a semi automated electric propane hybrid brewing panel. The question I have is this. My power coming in will be 120, and I have a 2000 Watt heating element, I am going to just be using it in a HERMS role. I am planning on using a p-1000 PID with a 40 AMP SSR. In between the PID and the heating element I have a 2 position rotary switch that I will wire in to give me "kill" option. Here is my question. Will I burn the switch out if I the heating element is pull more amperage than the switch is rated to? Or will the heating element just operate at 1/2 the capacity? 20A vs 10A?
I'm doing something similar but will have 2 SSRs that 2 PIDs will control. The selector switch will power 2 contactors that will allow the power to go to the SSRs but only one or the other.
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Old 09-24-2013, 06:47 PM   #6
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Quote:
Originally Posted by atoughram View Post
Is the element a 120v or 240v unit? My 4500w (IIRC) 240v element runs about 12 amps on 120v.
FWIW, in theory, a 4500W/240V (12.8ohms) element should draw about 18.8amps with 240V applied.

That same element should draw about 9.4amps with 120V applied, and provide about 1125w of heat.
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Old 09-25-2013, 03:49 PM   #7
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Quote:
Originally Posted by raouliii
FWIW, in theory, a 4500W/240V (12.8ohms) element should draw about 18.8amps with 240V applied. That same element should draw about 9.4amps with 120V applied, and provide about 1125w of heat.
Hey raouliii, how did you do your calculations for running a 240v element on 120v? I know very little about electricity but I am very interested so see if I can run an element like the one you did the numbers for and two march pumps on the same outlet.
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Old 09-25-2013, 04:06 PM   #8
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Hey raouliii, how did you do your calculations for running a 240v element on 120v? I know very little about electricity but I am very interested so see if I can run an element like the one you did the numbers for and two march pumps on the same outlet.
Its ohm's law. Watts/volts=amps. The resistance of the element is what determines the wattage. An element that is meant for 220v can be run on 120v, but it would need to be on a 50a circuit using #8 lead conductors. In other words very inefficient.
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Old 09-25-2013, 04:17 PM   #9
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Hey raouliii, how did you do your calculations for running a 240v element on 120v? I know very little about electricity but I am very interested so see if I can run an element like the one you did the numbers for and two march pumps on the same outlet.
If you know the rated wattage is 4500watts at 240 volts, then you need to calculate the resistance of the element first

R = E squared/ W where R is resistance in ohms, E is volts and W is watts

240 x 240 / 4500 = 12.8 ohms for the element resistance.

When you operate the 240 volt element on 120 volts, calculate watts using:

W = E squared / R where E = 120 and R = 12.8

120 x 120 / 12.8 = 1125 watts

That's the theory.

Now, the easy way to do it is just divide the rated wattage at 240 volt operation by 4 and that will be the wattage when the element is operated at 120 volts
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Old 09-25-2013, 04:31 PM   #10
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If you know the rated wattage is 4500watts at 240 volts, then you need to calculate the resistance of the element first

R = E squared/ W where R is resistance in ohms, E is volts and W is watts

240 x 240 / 4500 = 12.8 ohms for the element resistance.

When you operate the 240 volt element on 120 volts, calculate watts using:

W = E squared / R where E = 120 and R = 12.8

120 x 120 / 12.8 = 1125 watts
Your math seems sound, but I think impedance doesn't change no matter what voltage is used. I can put my meter on ohms and test the impedance of the element and that will never change. Regardless though, if you only get 1125w of heating power from a larger more expensive element, why in the hell would anyone do that? I use 120v in my setup and I use 2000w, if I could get 2500w more heating power just by running a #10 home run and 30a breaker I would do it all day long.
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