 |
12-26-2009, 01:27 PM
|
#1
|
|
Feedback Score: 0 reviews
Join Date: Mar 2008
Posts: 20
|
Electric Stove Burners in Series
|
|
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
|
|
|
12-26-2009, 01:39 PM
|
#2
|
|
Death by Magumba!
Feedback Score: 0 reviews
Join Date: Aug 2007
Location: Melbourne, Fl
Posts: 2,253
Liked 22 Times on 14 Posts Likes Given: 5
|
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.
|
|
|
12-26-2009, 02:06 PM
|
#3
|
|
Feedback Score: 0 reviews
Join Date: Mar 2008
Posts: 20
|
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.
|
|
|
12-26-2009, 02:43 PM
|
#4
|
|
Vendor and Brewer
Feedback Score: 0 reviews
Join Date: Aug 2006
Location: Piscataway, NJ
Posts: 20,661
Liked 460 Times on 325 Posts Likes Given: 8
|
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.
__________________
BrewHardware.com
Sightglass, Refractometer, Ball Valve, Weldless bulkhead, Thermometer, Decals, Stainless Steel Fittings, Compression Fittings, Camlock Quick Disconnects, Scale, RIMS tube, Plate Chiller, Chugger Pump, Super Clear Silicone Tubing, and more!
|
|
|
12-27-2009, 05:13 AM
|
#5
|
|
Death by Magumba!
Feedback Score: 0 reviews
Join Date: Aug 2007
Location: Melbourne, Fl
Posts: 2,253
Liked 22 Times on 14 Posts Likes Given: 5
|
Quote:
Originally Posted by 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.
|
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.
|
|
|
12-30-2009, 01:25 AM
|
#6
|
|
Feedback Score: 0 reviews
Join Date: Dec 2009
Location: Pystomutu, Ukraine
Posts: 198
Liked 1 Times on 1 Posts Likes Given: 1
|
parallel is way to go. just connect "hot" wire thru ssr.
__________________
T.O.P
Brewery
|
|
|
| Thread Tools |
|
|
| Display Modes |
Linear Mode
|
|
|
|