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lighting dimmers for element control?

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And dimmers are usually the exact same thing as the element. Resistance. And all that heat when you tune the dimmer down has to go somewhere. I'd think there would be a hell of a heatsink on a 2000W dimmer.


Dimmers are not resistance elements. There's no way the electrical code would allow you to put up to 2000W heat dissipation into a small metal box inside your wall. They simply to change the firing angle of a triac or thyristor to chop the duty cycle of the input current. This accomplishes the same thing without wasting electricity and creating a fire hazard.

As long as the dimmer is rated for the voltage required and the power, I see no reason why it wouldn't work. What it will do is change the wattage that the element puts out on a continual basis, rather than an on/off switching of the element. It would be harder to control than the PID and SSR but it could be done I think.
 
You can do it...

But those dimmers are more expensive than a PID, SSR and heat sink combined.

They can only control 2000W

You cannot buy an ULWD 2000W element for use in a BK

It would work, but I dont know what you would use it for with the above restrictions.
 
i've installed several dimmers of that size in establishments before and the inspecter passed it just fine so i'm pretty sure it's not a code violation... they're rather large and take up a 2 gang box and have a large heat sink... http://electricalsupplies4less.stores.yahoo.net/lutatsiinrod3.html i just had a couple laying around the shop for FREE so i was wondering if it was worth the time and effort to see if they worked on heating elements... just didn't want to hook it all up to the dimmer and then act like a flourescent acts when on a dimmer... they flicker. and elements do operate by resistance soooooo technically it should work
 
Seems like we all keep repeating ourselves in this thread.:D

Post Whores!
 
I am interested in element controls because I am taking a different path with my electric BK than most.

I have a ceramic cooktop range that is too inefficient to do 5 gallon AG batches. So, in addition to my electric cooktop I am going to use a 5500 ULWD element in my BK wired to 120 vac to produce 1375 watts. This element, in addition to the ceramic cooktop, will make my 5 gallon batches much more manageable.

Additionally, once my basement is arranged in such a way that it is feasible to brew in, then I already have my 5500 watt element in place I can use on my 220 vac feed down there.
 
Now, that would work... using it to control essentially a heat stick would work.
 
I made my Electric Kettle using a Simmerstat to control power to my element. The following are pictures of my upgraded version.

If anyone has questions I would be happy to answer them.

TD

iD7101D7E-AB6A-4574-B23E-1D6D5B64AFB2.jpg


iD96B1993-3426-4C71-9581-77E216F4D644.jpg
 
I am putting together a parts list with suppliers of what it took to build my Electric Kettles. If any one is interested let me know and I will send it to you. I will also try and put to together a "How To" on building your Kettle.

Let Me Know
TD
 
I'm curious. Between you and Pol I may end up with an electric brewery before the weather gets nice.
 
I just ordered my PID, SSR, Heatsink and K type thermocouple today. The only parts that are not here or on order are the junction box and the distribution block.
 
I have the equipment list done and will work on the instruction of building the Electric Kettle this weekend.

TD
 
can someone reasure me that if a water heater element is converted from 220 to 110, the wattage of that element changes? rather than just raise the amperage needed for 110? it seems like i've seen a few setups like that, but my boss was arguing that if a 5500 watt water heater element were converted to 110, the wattage would stay the same, u would just trip the breaker
 
Nope. The wattage is based on the constant resistance of the element. If you change the voltage, you change the wattage. Drop the voltage in half, and the power drops to a quarter.

P=E^2 x R

If R is constant and you decrease E (voltage) power will go down considerably more.
 
Just to add my 5 cents; that equation should be P=E^2/R. An R is roughly (but not exactly) constant; all metals increase their resistance with temperature; although the effect is small at low temperatures (less than 2% for nichrome at 100 Centigrade).

To sum up, if E (voltage) is decreased by a factor K, then the power is decreased as K^2; (ignoring the change in resistance).
 
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