Electric Brewing Question

[meshbackhats]

I was just curious if you could power a 5500 watt element using an SSR with an AC adapter for a laptop:
Something like this
http://www.ebay.com/itm/FOR-SONY-VA...Accessories_PowerSupplies&hash=item27d660a4cd

which would switch the SSR- something like this
http://www.ebay.com/itm/AC-250V-100...262?pt=LH_DefaultDomain_0&hash=item35dc236e86

My thinking is that the laptop switch Mode PSU combined with that SSR which is rated for high voltage as being able to handle high current. I don't have access to 240v as I'm living in an apt and thought something like this might work- taking the 15amp outlet and having the SSR do the heavy lifting.

I am sure I am missing something....

Thanks in advance.

 

[Bartp]

You want to apply the 19.5v DC to switch the SSR on? Yes you could but I'm not sure why you would want to do that. SSR is usually used in conjunction with a controller like a PID. SSR turns on or off the power being supplied to other devices, like a switch. If you want to plug the element straight without controlling the element, than SSR is not needed.

I've put some information on how I wired up my panel for use with 120v 15a, check it out, it might help you. www.AleDrinkers.com

 

[meshbackhats]

Thanks for the reply-
I would use a PID to control the element.

It looks like in your design (every design that I have seen) the 5500 watt element is only capable of putting out of 1375w as it is only being pushed by a 120v circuit.

I guess my question is more along the lines of-
Can a 120V circuit power a 5500 watt element at full power?

powering the SSR with the Laptop Power Supply

 

[GotPushrods]

I guess my question is more along the lines of-
Can a 120V circuit power a 5500 watt element at full power?

The term "5500 watt element" is only useful when paired with the intended voltage.

If this was intended as a 240v element, it would imply an element resistance of 10.47 Ω. You can actually check this with a multimeter. This is the only constant of the element. Your wattage will be proportional to the voltage you give it.

At 120v, the element will run at 120v² / 10.47 Ω = 1375 watts.

 

[Bartp]

What GotPushrods said. Wattage is function of voltage multiplied by amperage. For instance, we know that the element produces 5500w at 240v. Therefore we can extrapolate that this element needs 22.9 amps of current (5500/240).

Your 15amp 120v breaker at most can only provide 1800w (15*120). To reach 5500w on a 120v line you'll need to support 45.8amps; Meaning you would need to run a 50a breaker (Do not run 50a breaker!)

This element is designed to work on a double pole 240v circuit. It requires two hot lines. Your 15a 120v breaker is single pole, and has only one hot line.

The reason why it will produce only 1375w (quarter of power instead of half) is because of resistance, or Ohms.

Ohms law is v = I * R (where v = voltage, I = amperage, R = resistance)

For this element we know that at 240v, we need 22.9A (240 = 22.9 * R) so we can figure out that resistance of the element is 10.48 ohms (like GotPushrods said) The resistance value does not change if we lower voltage or amperage.

when we take the same formula but replace it 240v with 120v we get 120v = I * 10.48R. so we can extrapolate that at this voltage and resistance, we're capable of only 11.45 amps. Now since we know voltage and amperage, we can calculate wattage (w = I * v), 120v*11.45 = ~1375w.

Hope this clarifies it.

 

[meshbackhats]

GotPushrods & Bartp

Great info.

Thanks for taking the time to explain that to me!
Very helpful.