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.
The most important thing to check is probably the size of the main supply fuse on your incoming feed. Between 60 and 100 amps is typical for a domestic supply. If you have a 60 amp fuse, I'd be very wary of trying to run two chargers at 7KW each simultaneously.
The other thing to consider is the actual consumption of the chargers that will be connected. Just because the supply point is rated to 7KW, doesn't mean that the charger will actually draw that.
You don't need to accommodate the full current capacity of all the breakers because you're unlikely to be using everything that you possibly can at any one time, but if your peak load exceed the main fuse, you have a problem.
Personally, I'd be very tempted to have nothing further to do with British Gas and approach a local electrician about the work. If their electrical work is priced like their plumbing, it'll be very expensive. (And I wouldn't be surprised if they subcontract it out anyway).
Best Answer
Straight up, this is a hard sell
First, EVSE's require a dedicated circuit. NEC 210.17.
Second, a 50A circuit can only serve one thing - NEC 210.23. So it cannot serve a hardwired driveway melter and also an EVSE receptacle. So that's right out.
Third, NEC 625.42 says the EVSE can be cord-and-plug connected if all are true:
1) it's part of a listed system
2) The receptacle <= 50A
3) It's cord connected to facilitate a) ready removal for interchange, b) maintenance and repair, or c) repositioning of the EVSE
4) Cord length <= 6 feet
5) Receptacles located to avoid physical damage to cord.
I think item 3 will be a hard sell.
So you're trying to argue that you should get an exception to the dedicated circuit rule (NEC 210.17); this outlet isn't an EVSE outlet -- it's a general purpose outlet that sometimes (wink, nudge) has a level 2 EVSE plugged into it. Hmmmm. That'll be an interesting sell to the AHJ. AHJ==your local inspector; the one that you pull the permit from.
The other prong of your argument, then, is that it's somehow reasonable for the driveway melter to be cord-and-plug-connected. In practice, that means it comes to a junction box, which then has a cord or inlet. This doesn't make a whole lot of sense, but it also absolutely requires the melter to call out a 50A breaker in its installation instructions. If the melter is looking for a 20A-40A breaker, no deal because the breakers don't match.
However, it would sell as a subpanel
If you convert this 50A circuit so it serves a subpanel instead of an EVSE, different deal. The subpanel does not need a "main breaker" if it's in the same building as it's served from.
In the subpanel you have a 50A breaker for the EVSE, and a ??A breaker for the driveway melter. That solves the breaker size problem.
Here's an interesting fact. Nothing in Code prohibits you from misusing a "generator interlock kit" to interlock any two random branch-circuit breakers, just for the chuckles. Only one can be on at a time. Go with either a Siemens/Murray panel or a Square D "QO". Use Siemens' ECSBPK02 interlock ($30), or a Square D's QO2DTI interlock ($25).
Now we open an ugly chapter: Provisioning. The feeder cable must be provisioned to be able to suppor--- Oh, wait. You've got an interlock there, don't you? Well, then. Nevermind.