The two branch circuit breakers will trip if the load on their protected circuit is greater than 32 amperes, and 25 amperes respectively. However, since the main breaker trips at 40 amperes. If both branch circuits are pulling a full load the branch circuit breakers will not trip, but the combination of the loads will trip the main breaker (32 + 25 = 57 > 40 as you have said).
It's a very common situation to have the branch circuit breakers total more than the main breaker, but this is almost never a problem since the circuits usually don't pull a full load. Remember, circuit breakers are there to protect the wires. They shut off the power if you are drawing enough power to damage the wire, so installing a larger main breaker is likely not an option (unless you contact your power company and have the service upgraded, which will include upgrading the feeder cable).
Contact your local power company, and speak to them about a service upgrade. Other than plugging less stuff in, or cycling loads, there is not much you can do yourself in this situation.
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 achieve this with 'home automation'.
What you're trying to do is called 'load shedding'. It's not a common task in a residential setting, but it can be accomplished with home automation technologies.
There are z-wave devices to measure current with an amp clamp, and to control appliances in a split-phase (biphase) set.
You would get one amp-clamp and instal it in the panel, measuring current accross your feeder wires.
Then you install one of those relays at each appliance that you want to be able to shed.
You also will need to purchase or build a controller box that talks to both the amp clamp and the relays. MiCasaVerde is one choice I've worked with personally and I bet it could do what you want.
Then the fun part. You get to write a program that runs on the head-end to constantly measure amperage draw and controlls the relays accordingly.
Be careful that you don't create some loop that rapidly toggles the relays though. At high amperages, I could see that being a nuisance to adjacent units (power blinks) or worse case you could overheat and set one of the relays on fire from the arcs that switching causes.
Good Luck! And remember to share your code and parts list here once you've got this working.