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Author Topic: Water Heater Control?  (Read 1352 times)
Lin
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« on: March 14, 2011, 08:30:03 PM »

I was wondering if I could put my electric water heater on some sort of rheostat switch rather than the on/off switch I currently have.  The idea being that if I have to be managing power usage, I could turn the amperage on the water heater down instead of completely off and still avoid popping a breaker.
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Sean
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« Reply #1 on: March 14, 2011, 09:19:12 PM »

You are better off getting a smaller element if your tank has a single element, or putting in a switch to go from two elements down to one if your tank has two elements.

You can't really use a rheostat; the amount of power you'd have to dissipate would make it enormous and would require a huge heat sink, and now you're leaving power on the table because you'll be wasting so much of it as heat.  If you managed to cut the current in half, you'd drop the effective power down to 1/4, and your water might never get hot.  For this application you really need a variac, which is a spendy item in those current ratings.  A 5a variac runs around $100, and you'd need at least a 15a unit for even the smallest water heater, which would be closer to $200.
 
-Sean
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Lin
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« Reply #2 on: March 14, 2011, 10:29:38 PM »

Thanks Sean, on/off switch it is.
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« Reply #3 on: March 14, 2011, 10:52:17 PM »

Lin,. you can do what you want by using a diode in series with your switch. This would produce very little waste heat (about 3 watts) at  the diode.

With a three position switch, you could have off, low and high power at very low cost. The low power would be very close to one half of high power.

We have a halogen portable lamp stand that uses this setup on each light, as near as I can tell. This circuit is much more effective as a heater than as a lamp.

For what it's worth.

Tom Caffrey
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Sean
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« Reply #4 on: March 15, 2011, 10:27:26 AM »

... you can do what you want by using a diode in series with your switch. ...

Good point, Tom -- yes, you can get about half power by using a diode to transform the input AC current into half-wave rectified DC.

However, bear in mind that the formula for RMS current in this case makes it about 71% of full-wave current, so while you will have half power (~0.712=~0.5), you will only reduce current by 29%.

By contrast, switching off one of two elements reduces both current and power by half.  Also, it only requires a single SPST switch and no other parts.  Similarly, on a single-element tank, changing from, say, an 1,800-watt element to a 1,200-watt element will reduce both power and current by one third, making it more efficient than the diode method.  However that would not give you the option to select a higher-power setting with just a switch, making recovery time longer even when the extra power is available.

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« Reply #5 on: March 15, 2011, 12:09:43 PM »

I do not think I want to change out the element since I would like to be able to heat water as fast as possible at some times. Since this is only a small, ten gallon single element heater, the diode idea sounds like a good compromise.  I am not sure of what type and size diode would be needed and how to wire it in.  Is it that I would use a two pole, 3 position switch that would have the hot connected to one pole and the other pole connected to that through the diode?
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Oonrahnjay
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« Reply #6 on: March 15, 2011, 12:16:48 PM »

  Good point, Tom -- yes, you can get about half power by using a diode to transform the input AC current into half-wave rectified DC.   (snip)   

Does this work with a single-element, 110V heater?  Thanks,  BH
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« Reply #7 on: March 15, 2011, 12:26:53 PM »

..  I am not sure of what type and size diode would be needed and how to wire it in. ...

You need a silicon "rectifier diode" rated for 200 volts and 15 amps.  That's a big diode, you won't find it at Radio Shack.  Try Newark

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...  Is it that I would use a two pole, 3 position switch that would have the hot connected to one pole and the other pole connected to that through the diode?

No, you want a SPDT switch (one pole, but double-throw).  It does not need a third, center-off position unless you want to be able to turn it off completely from the same switch.

The incoming hot power goes to the center "common" terminal.  The diode goes between the other two terminals, and the outgoing power to the water heater is connected to one of them -- it doesn't matter which.  When the switch is operated to the position where the common terminal connects directly to the output terminal you will get full power, and in the other position you will get half power.

If your water heater draws 12.5 amps (1500-watt unit, like mine) on full power, it will now draw a little under 9 amps on half power, saving you about 3.66 amps.

Does this work with a single-element, 110V heater?

Yes.

-Sean
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« Reply #8 on: March 15, 2011, 02:52:05 PM »

I'm sort of getting there.  The reason I mentioned the 3 position switch was because I do want that single switch to control off, low, and high.  How would I do that?  Thanks
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« Reply #9 on: March 15, 2011, 03:20:43 PM »

Another question.  I may not have been clear, but this is a 120v water heater.  Would I still need a diode rated to 200v?
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« Reply #10 on: March 15, 2011, 09:37:06 PM »

... I do want that single switch to control off, low, and high.  How would I do that? 

Connect it exactly as I described, just use the type of SPDT switch that has a center "off" position.

Another question.  I may not have been clear, but this is a 120v water heater.  Would I still need a diode rated to 200v?

Yes.  Remember that "120 volts" is an RMS (root mean square) value; not peak.  The peak voltage of a 120 VAC system is actually 170 volts, so the diode will need a reverse bias value higher than 170.  200 is a good number for shopping purposes, but if you found one rated at, say, 180 volts that would work in practice.

-Sean
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