What box for my pull box and my NEMA-6-50R using #4.

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TheWantonElectrician

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I am trying to wire up two kilns. They are Olympic Electric Kilns (Model 2827H, 11,520 Watts).

I'm trying to follow the manufacturers instructions found here: https://www.olympickilns.com/wp-con...ectric_Kilns_Operating_Manuals_1_revision.pdf

The circuits are over 40 ft, so according to the manufacturer require #4's. I have (4) #4's in a 1'' EMT.

Coming into a pull box, since I'm dealing with #4's and 1'' conduit, I'll need a 8'' minimum length box for a straight pull and a 6'' for an angle pull (assuming no other pulls in the box). Is this correct?

According to the manufacturer, the receptacle and plug can be a NEMA 6-50. For a termination box, I'll need either 2 or 3'' of bending space. What kind of box would allow that? I believe even a 4-11/16'' box would be too small.

Feel like I've opened a can of worms by having to use #4.
 
Their instructions dont seem right. I dont think you can put a 6-50 on a 60A breaker. Next, I think you only need 3 wires and not 4. The ground could most likely be an 8 or 6. But with #4 for the ungrounded, i think you need a large box.

Will these fit in a 3/4" conduit?

For a resistive load like this, i would not upsize a 240V circuit until it was 100' long. Staying with #6 makes things easier. Not sure if mfg info is a mandate or suggestion, especially since I dont believe it is legal.
 
You could use a small (4 x 4 x 8?) box next to, between, or under or over two receptacle boxes, horizontal if fed from the side, vertical if fed from above or below.

The hard part about a single box for everything would be a plate. I couldn't find anything ready-made, but these guys might be able to make something: https://www.kyleswitchplates.com/
 
Seams like you are using single phase 240. Is that correct.
If so you running four current carrying in one raceway or two circuits. Is that correct.
 
Leviton (and probably others) makes a 6-50R which is only 1.00" deep and side wired. If you can avoid entering the box from the back, I suggest putting each of your receptacles in a 4-11/16" box with a 2-gang plaster ring. If you are trying to put both receptacles in one box, you would probably need to fabricate something custom.
 
Re read and slow down a little this time.
2- units. So you are putting four current carrying conductors in a single raceway. Two circuits one for each unit.

I like what Larry said. You could also use a tee and two 4-11 with extension ring and raised covers.
 
If this needs to flush in a finished area consider a 5" box with a 2 gang mud ring. Or if looks are not an issue use an RV recep. enclosure.
 
Thank you all for responding!

240V, single phase, 2 circuits, using the EMT as the EGC.

The customers want their kilns with a cord and plug so they can move them later. It'll also satisfy the "disconnect within sight" requirement that I'm not 100% sure applies to kilns.

I want to go with suemarkp's suggestion of using #6, but I'm pretty sure I need to follow mfg instructions and do #4's.

Fortunately I'll be putting the NEMA 6-50R in two different boxes and entering on the top, so it sounds like the 4-11 could work (not sure about the extension ring and raised cover are necessary).

Any chance the mfg instructions of using #4 vs #6 has to do with the potential of folks running 6/2 NM which has a rating of 55A, whereas my (2) #6 have a rating of 75A downrated to 60A? I guess it's voltage drop, not current drop so I'm likely wrong.

Still open to hearing other thoughts.
 
Aren't these basically a resistive heater? Why is the current higher for 208 volts than 240?
 
Not sure about the resistive heater.

Someone please correct me if I'm wrong but since we're not changing from split phase to three-phase and since the watts stays the same, a decrease in voltage (208V) will require more current than for the 240V.
 
TheWantonElectrician said:
What do you mean by "power"?
Click to expand...
Power = wattage. Volts x amps. P = E x I

If you supply a voltage to a resistive load different from the design voltage, the current will vary proportionately, and the resultant power will vary to the square. If you double the voltage, the current will double, and the power will quadruple.

However, for a given desired power level on a different voltage, the load itself must be manufactured differently. To get the same power on 208v vs 240v, for example, a heater element must have a lower resistance and be thicker to get as hot.
 
Thanks, I didn't know that about heater elements.

I still feel like perhaps I'm missing something. The 11,520 watts stays constant, so reducing from 240v to 208v results in increasing from 48A to roughly 55A. Sounds like the load (heater elements) would need to change as well, but it still just follows ohm's law. I could see if you switch from split phase to three phase (sharing the load on all three), you might divide by 1.73 and then use lower current and certainly the heater elements would need to change to allow the new wiring system.
 
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