A sub panel must have the neutral and ground isolated. Panels come with a very long, rather thick (about 1/4 x 20) green bonding screw that connects the neutral bar to the can in the case of a primary panel. You don't get a neutral from your utility, you create one with that bonding screw.
Sub panels should be fed with 3 insulated conductors of appropriate size, and a ground that need not be insulated (but can be, if you want). So the first part of your question is, yes, that sub panel must be grounded, but the grounding conductor should be attached to the can using a ground lug, not by landing it to the neutral bar.
Sub panels must also have a fused disconnect, which means they need to be fed from a breaker, and there can't be anything else on that breaker. Don't double tap. You have two options here:
- Increase the size of the existing sub panel
- Put in a new double pole breaker in the existing sub panel and use that to feed your new sub panel. Land the circuits you had to pull out of the existing sub panel into the new one.
As others have noted, you need to watch your loads. If all you need are a few convenience receptacles or a lighting circuit, you should probably be o.k. (hard to tell with what you've given).
Either way, get an amprobe and look at what each incoming phase is pulling in your existing sub panel prior to doing anything. Make sure everything is on when you do. If it's only pulling 25 - 30A on average, you should be o.k. to add a small 8 circuit sub panel. Since it's directly in the line of sight with the existing sub panel, the new sub panel need not have a main breaker since the means of disconnect is right next to it.
A six to eight circuit sub panel runs about $80 without breakers, they typically start at 50A, but you don't have to feed them with 50A. You could feed it from a 30A breaker if all you want are convenience receptacles and lights.
Here is an amprobe being used:
(source: amprobe.com)
Do that on your sub panel first (one phase at a time) just to be sure you have room to add more. If not, you need to replace your existing sub panel, and an electrician is really your best bet there.
Another good thing to do is measure the draw of the circuits you'll have to move to the sub panel in order to make room for the breaker that will feed it. Obviously, you want to move the circuits drawing the least to the new panel in the end. Some re-arranging might be needed to make that happen.
Since this is a garage, take care what you connect to the sub panel. If you are going to be powering something like a compressor (or anything else with a decent sized motor), carefully consider the locked rotor amperage when determining the load. It will be printed on the motor.
Finally, if any of this sounds overwhelming, call an electrician. If you get into any kind of trouble, call an electrician.
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
You can keep the existing wiring and breaker
Range loads are computed from their wattage, based on the rules in NEC's Table 220.55. In particular, since 422.10(A) paragraph 4 explicitly permits range loads to be sized as per Table 220.55:
we can then apply the logic from Note 4 to Table 220.55 to our situation:
In our case, given that we have a 6.4kW, 240V cooktop, we treat this as the branch-circuit size, without derating, giving us something in the 27A range. As a result, since we're below the 8.75kW threshold set by NEC 210.19(A)(3), we can use that as our branch circuit load, giving us a 30A branch circuit and breaker with 10AWG wire. Since it's a 220-240V appliance and not a 110/220 or 120/240V one, we also know that it has no use for a neutral, which means that's not a concern when wiring this, either. In your case, since you are dealing with a conduit run for the existing branch circuit, the new cooktop needs to be wired the same way the old one was: black to black, red to red, and green to the box, with the white wire in the box left capped off as it is right now.