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Wiring a 1950's Double Oven

7K views 27 replies 7 participants last post by  stickboy1375 
#1 ·
I'm trying to install a 1950's Kelvinator electric double oven (model ER-9D).

The plate below shows a max load of 16140 watts.



Would the amperage be 16140/240 = 67.25 amps? That seems really high. If that is the correct amperage, what breaker(s) will I need? I was planning to wire with 50 feet of new 6/3 to a 50 amp breaker until I calculated the amperage.

I'm also changing the plug on the oven from the original 3-prong to a grounded 4-prong. Below you can see where I'll attach the new 4-prong range replacement cord. The access panel has been removed.



I know the red and black will connect at the bottom on the outside with the white neutral in the middle. Where do I attach the ground? The directions below (from the back of the oven) say to either ground the range to neutral or to a grounding lug at one end of the range back.



I'm assuming I should do the latter (per code), attaching the ground wire directly to the chassis (where?) and removing the ground strap shown in the second photo above. Or is that not a ground strap? Does the photo below show four ground straps (one at each of the access panels)?



And if those aren't enough photos, here's the wiring diagram for fun:



Thanks in advance for any help.

Brian
 
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#2 ·
The code allows you to size cooking equipment circuits to 80% of the nameplate rating. So you can figure the amps as (16140 x 0.8) / 240 = 53.8 A. But you still have to upside the circuit by 125%, so you need a circuit that can handle 67.25 amps:laughing:

6-3 romex and a 50 amp plug won't cut it here. Either 4-3 copper romex or 2-2-2-4 aluminum SER. The strap needs to be removed and a lug bolted to the frame for the ground.
 
#5 ·
Your best option would be to add a cook top unit and supply them both by one branch circuit... this way you can take advantage of derating of the nameplate...


so if you added a 3kw cooktop, both appliances can be treated as one unit, so....

16KW + 3 KW = 19 KW - 12 KW = 7 KW x 5% = 35%

8 KW x 35% = 2.8 KW

8KW + 2.8 KW = 10.8 KW

10800 watts / 240 = 45 amps... :)
 
#13 ·
After doing a careful reading of the NEC, I find you don't need that large of a circuit. The code says the demand for cooking equipment 12 kW and less can be calculated at 8 kW. Then you add 5% of that for every kW over 12.

So... your oven is 4 kW over 12, which means you need to calculate the load as 120% of 8 kW which is 9.6 kW. 9600 watts / 240 volts = 40 amps. So yes, 6-3 romex and a 50 amp plug would work:laughing:

Sorry for the confusion.
 
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