Using the 30 A receptacle is a dangerous idea. The internals of the receptacle are designed for a maximum of 30 A, and putting 40 A through it could cause a fire. The receptacle needs to be replaced with a 50 A model (since they don't make 40 A receptacles).
It's fine to have a 40 A breaker feeding a 50 A receptacle. You need to also match the wire size to the breaker rating. A 40 A circuit requires at minimum 8 AWG copper wire, or perhaps thicker for longer runs. 6 AWG might be a good idea. Since you are replacing the receptacle, you should make sure that there is a dedicated safety ground (so four wires are run from the circuit breaker to the receptacle.
Also, I doubt that the old oven died because of the wrong receptacle (assuming that the receptacle is undamaged). However, I have heard of equipment being damaged when not properly grounded (My mom's gas range died prematurely after the installers failed to connect the ground). Double-check the voltages by measuring all 6 pairs of voltages using a voltmeter (Should be 240 between lines (L-L), 120 from L-G and L-N, and 0 from N-G).
Use caution when working with electricity. Don't forget to turn off the breaker (and verify that power is disconnected) before disassembling anything.
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
No.
You should follow the manufacturer's installation instructions, and install a 4-prong cord on the range. You should never replace a 4-prong range receptacle, with a 3-prong.
As @JPhi1618 points out, it looks like your 4-prong receptacle is wired incorrectly. The bare grounding wire should go to the green screw terminal, while the white "neutral" should go to the silver screw terminal.
Also as @JPhi1618 points out. If there's no grounding conductor in the cable feeding the receptacle, you'll have to address that issue separately.