Residential electric heater

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Rsemp

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Location
Oklahoma
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Electrician
I have two electric heaters, one 20kw with (2) 60 amp two pole breakers and one 15kw with (2) 60 a
two pole breakers. My question is do I run one 240v circuit to each and jumper between breakers or two circuits to each? Also is 6/2 romex acceptable for this? I only have 200amp service availability for this 2600 sq ft home.
Thanks
 
First you better do a load cal on the home especially with electric heat that I assume is in this home

I have seen water heater that took one circuit but you need to check the water heater itself. If it takes one circuit you should not have to jump anything
 
First you better do a load cal on the home especially with electric heat that I assume is in this home

I have seen water heater that took one circuit but you need to check the water heater itself. If it takes one circuit you should not have to jump anything
It’s the two electric heaters I’m talking about not water heater as water heater is gas.
 
What size circuits are you planning to run to these units? The onboard breakers are not for the feeder / branch circuit OCPD.
BTW, 35KW of heat is overkill for 2600 SQ ft.

Roger
 
The circuit and breaker size is what I’m trying to determine. The only other 240v circuits are double oven and clothes dryer. The lightning load will be all LED and (5) ceiling fans. I’m wiring for a friend (I do hold a contractor license) and I’m a little embarrassed to say it’s been too long since I’ve calculated a residential load which is the reason I’m on this site. I’ve looked at several samples on line and the best I can come up with is a calculated load of 191 amps.
 
8/2 nm is only good for 40 amps and 6/2 is only good for 55 amps. That means you may need a number 4 nm for each 60 amp breaker
If the 20kw unit has 2- 9.6kw elements - which is often the case then you may be able to use #6 nm
 
8/2 nm is only good for 40 amps and 6/2 is only good for 55 amps. That means you may need a number 4 nm for each 60 amp breaker
If the 20kw unit has 2- 9.6kw elements - which is often the case then you may be able to use #6 nm

Dennis, we never see these units here in NJ can you show the calculation?
 
A 10 kw unit usually has 9.6 kw of heat along with a fan. 9600/240 =40 amps x 1.25= 50 amps. Then you add the blower which often times comes to 56amp as the minimum circuit ampacity.

In the op's case he would have 2- 9.6 kw units but one fan so I am not sure how they have that configured.
 
The units I have seen with 15 kw always came with a 60 amp and a 30 amp breaker- not 2- 60 amp breakers so I am not sure what that is about either
 
These heaters may have a kit available allowing a single large circuit to feed the subdivided heat coils. I usually find it easier to run multiple smaller circuits especially if the heater already has separate breakers as the expected input. The thing that sucks is too many have an MCA of 56 or 57 amps. That means you need to use some cable other than NM if you wish to use #6. I just used 1" flex conduit and four #6 THHNs with a #10 equipment ground when I did mine (gets you up to 60A per conductor when factoring in bundling 4 current carrying conductors in a raceway).

Find the MCA rating on the furnace breakers to see what size circuits you truly need to run (when mine was installed, the package had about 20 different MCA labels and the HVAC installer was supposed to put the 2 correct ones in the right place -- I see a huge potential for error with that approach). That's a lot of heat for one house, and gas would have to be cheaper to run since it appears they already have gas service unless the pipe is too small or you can't vent it or something.

If you use the Optional calculation for a dwelling service, you can apply a demand factor to the heaters. You need to know how many stages are in each heater. There are most likely at least two since each has 2 breakers, but mine had 3 stages with two circuit feeds. You have two separate heaters so you could have 4 to 6 overall independent heat stages.

If 4 or less separate heating stages, the load calc is 60% of the heat strips. 220.82(C)(4)
If more than 4 separate heating stages, the load calc is 40% of the heat strips. 220.82(C)(5)

** This demand factor only applies to the service, not the branch circuits **

Even with this demand factor, I'd check your total service load calc, as 35KW of heat is a lot of electrical load.
 
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