RTFM
After reading the user manual for this heater, as per National Electrical Code.
National Electrical Code 2011
ARTICLE 110 Requirements for Electrical Installations
I. General
110.3 Examination, Identification, Installation, and Use of Equipment.
(B) Installation and Use. Listed or labeled equipment shall be installed and used in accordance with any instructions included in the listing or labeling.
It appears that it has subdivided heating elements, which requires 2 separate 240 volt 9 ampere branch circuits. It also states that "the electrical connection must be performed by a qualified tradesperson.".
Connect each lamp circuit to a 230 - 240 Vac, 60Hz, 9 Amp copper wire circuit that is properly grounded. Electrical connection must be performed by a qualified electrical tradesperson.
Installation must conform with the latest edition Electrical Code ANSI/NFPA N0 70 in the
U.S.A. and PART 1 CSA C22.1 in Canada.
Select the Conductors
Since you're dealing with space heating equipment, you'll want to check article 424 of the NEC. In article 424, you'll find that the conductors have to be sized to 125% of the load served.
National Electrical Code 2011
ARTICLE 424 Fixed Electric Space-Heating Equipment
III. Control and Protection of Fixed Electric Space-Heating Equipment
424.22 Overcurrent Protection.
(E) Conductors for Subdivided Loads. Field-wired conductors between the heater and the supplementary overcurrent protective devices shall be sized at not less than 125 percent of the load served. The supplementary overcurrent protective devices specified in 424.22(C) shall protect these conductors in accordance with 240.4.
So you'll have to do a bit of math to figure out the minimum ampacity of the conductors.
9 amperes * 125% = 11.25 amperes
Now that you know the minimum ampacity required, you'll want to check article 310.15 to figure out what size wire you need. A quick look at table 310.15(B)(16), shows that 14 AWG wire is good for 15 amperes at 60°C.
Select the Overcurrent Protection Device
Now that you've selected the wire size, you can choose your overcurrent protection device. In this case, you'll want 2 15 ampere dual pole circuit breakers.
Wiring the Circuits
When it comes to wiring the circuit, you have a couple choices. You can pull two 14/2 with ground cables from the panel to the outlet, or you can pull a single 14/2/2 with ground cable.
Two 14/2 with Ground
Pull two 14/2 with ground cables from the panel to the outlet, and connect the heater as follows:
Don't forget to reidentify the white conductors, by marking them in an approved manner.
One 14/2/2 with Ground
A 14/2/2 with ground cable consists of a black, red, white, white with red stripe, and a bare conductor. The only drawback here, would be that since there are more than 3 current carrying conductors, the conductors have to be derated by 80%. Even with the derating, the conductors are still rated for 12 amperes at 60°C. Which is still above the 11.25 amperes required by the circuit, so there should be no problem using this cable (as long as the run is not overly long).
Again, don't forget to reidentify the white conductors.
Column C
First off you'll notice the text "Column C to be used in all cases except as otherwise permitted in Note 3.)", in the title of table 220.55. This makes it simple. You have 2 appliances, so follow that over in the table, and you'll see 11 in Column C. So there you go, you can just use 11 kW. Done.
11,000 W / 240 V = 45.8333 A
So you'll need a 50 ampere breaker, and wire sized appropriately for the load.
Note 3
Note 3 says:
- Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household
cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or
Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand
factors for each column shall be applied to the appliances for that column, and the results added together.
Perfect, so instead of just using the value from column C you can do math. Let's step through it.
...it shall be permissible to add the nameplate ratings of all household
cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW...
8.4 kW + 5.0 kW = 13.4 kW
...and multiply the sum by the demand factors specified in Column A or
Column B for the given number of appliances...
Let's check the table again... You have 2 appliances, both between 3 1/2 and 8 3/4 kW. So You'll look at column B, and find 65%.
13.4 kW * 65% = 8.71 kW
8710 W / 240 V = 36.2916 A
So using this method you can use a 40 ampere breaker, and appropriately sized wire. However, keep in mind that if you change the equipment, you'll have to do the calculation again. So while you can use this value, you may have to upgrade the circuit later if you change equipment.
Note 4
I'm not exactly sure how note 4 comes into play, but I think it can be used if this is the only equipment on the branch circuit. Just for fun, let's run through that one too.
- Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch-circuit
load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load
for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same
room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.
You're only concerned with the second half of this note, since you have one counter-mounted cooking unit, and one wall-mounted oven, all supplied by a single branch-circuit, and located in the same room. So you can add the nameplate values, and treat it as a single range.
8.4 kW + 5.0 kW = 13.4 kW
So you can treat the units as a single 13.4 kW range. Check the column C again, this time for a single range. You'll find a value of 8 kW. But wait... The column header says "(Not over 12 kW Rating)". Your range is 13.4 kW. That's bigger than 12 kW. Now you'll have to check note 1
- Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum
demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual
ranges exceeds 12 kW.
That's easy enough.
13.4 kW - 8 kW = 5.4 kW
Since .4 is not a "major fraction", you can just use 5 kW. So you'll have to add 5% 5 times.
5% * 5 = 0.25
8000 W * 0.25 = 2000 W
8000 W + 2000 W = 10,000 W
That means you'll have to use 10 kW as your demand.
10,000 W / 240 V = 41.666 A
Which means you can use a 50 ampere breaker, and appropriately size wire.
Best Answer
Sure, you can always upsize the wire.
However, a dramatic upsize introduces two complications.
When dealing with #1 wire, it's not as simple as using a giant orange wire-nut the size of a salt shaker. You have to use a connector such as a Polaris, which is an insulated lug terminal. You could also use uninsulated lug terminals or split bolts, if you don't mind lashing it under unbelievable amounts of electrical tape. You do need to insulate neutrals.
Generally all terminations sized for #1 wire are aluminum friendly; why wouldn't they, since aluminum is normal and expected at these sizes. You just have to take care to use terminations rated for aluminum; a Polaris is fine, however I would not use a split bolt unless the smaller wire is also aluminum. Aluminum is fine to use at these sizes.
But yes, definitely lay the heavier wire. The only case where I wouldn't is if the wire was in conduit, and the run is very short. In that case, the pigtail adapters for the larger wire might be more expensive than just doing the whole run in the smaller wire. It could be changed for the larger wire later.