Electric Stove Burners in Series

[cswank]

Hello all,
I am working on an electric boiler setup in my basement, and have a question about the wiring. I have a set of 4 6 inch electric stove burners that I have mounted on my frame. The boiler fits nicely on all four, giving me a total of 6000W. It it probably not the most efficient thing to do, but I wanted to try something different.

So I am pondering how to control them. The total current for all 4 burners is 25 amps (240 V system). Would it be a bonehead idea to wire all four burners together in series, then control both hot leads with a separate SSR? I'm not sure if they were designed to handle 4 times the current that would be running through them. Otherwise I'd either have to split them up into two or four circuits, which starts piling up SSR costs in a hurry.

Thanks,

Craig

 

[CodeRage]

When you put resistive loads in series you have to add up the resistance and then recalculate the current draw and power consumption. If they are all the same size element, every time you double the load you halve the power. So with 4 1500 W elements you are only going to have 375W total.

To get 6000W you need to wire them all in parallel.

 

[cswank]

Thank you very much. Arrgh, I knew I was wrong somehow. So then I think I could wire them up to two 40A SSRs. By the way, I'm going to have an electrician inspect everything before it goes live. I'm obviously not the most competent electricity guy.

 

[Bobby_M]

If each element is 1500 watts, the current draw of each is figured as 1500w / 240v = 6.25 amps each. To figure out the resistance of each, it's 240v / 6.25a = 38.4 ohms.

Now, 4 of those in series would be 153.6 ohms. The new current would be 240v / 153.6 ohms = 1.56 amps. So, the new TOTAL power would be 240v x 1.56amps = 374 watts.

The answer is that you'd want to run all of them parallel, just like they were installed in the stove. You can still control with a 30a SSR.

 

[CodeRage]

If each element is 1500 watts, the current draw of each is figured as 1500w / 240v = 6.25 amps each. To figure out the resistance of each, it's 240v / 6.25a = 38.4 ohms.

Now, 4 of those in series would be 153.6 ohms. The new current would be 240v / 153.6 ohms = 1.56 amps. So, the new TOTAL power would be 240v x 1.56amps = 374 watts.

The answer is that you'd want to run all of them parallel, just like they were installed in the stove. You can still control with a 30a SSR.

deja vu?

one 40 amper would be plenty. There are a few preferred methods when working with two live legs. The primer in my sig addresses it a bit.

 

[bohdan987]

parallel is way to go. just connect "hot" wire thru ssr.