**Understanding Table 220-55**

The notes to this table are the most important part of the table and they must be read and understood. I will try and explain them in the following pages.

For household ranges that do not use the notes the process is quite simple. Just follow the table for the number of ranges and look up the answer or the demand percentage needed to calculate the load.

Column A is based on the percentage of the nameplate ratings of the cooking appliances that are less than 3 ½ kw. Use the nameplate rating of the appliance and multiply by the demand factor in Column A. For instance, 3 cooktops rated 3kw has a demand factor of 70% per table 220.55.

3 cooktops at 3 kw = 9 w total wattage. Using the 70% demand factor --- 9 kw *.7= 6.3 kw calculated load. Because of the diversity of the load the calculated load, in most cases, will be less than the nameplate ratings.

Column B is very much the same as Column A except that it is applies to ranges 3 ½ kw thru 8 ¾ kw.

Calculate the load for 20 ranges rated 8 kw.

20 ranges x 8 kw = 160kw but there is a demand factor for 20 ranges at 28% based on Column B for 20 ranges.

160 kw x .28 = 44.8 kw

Wow, 160 kw load is allowed to be connected at 44.8 kw. Why? Simply put it is because the ranges will never be on at the same time and with thermostats on ovens and range eyes the load will be much less than the sum of the nameplates.

Just remember that this section is based on the number of ranges and one must multiply the number of ranges by the kw of each unit to get the total kw then multiply by the demand factor.

* *

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.

Obviously, a 12 kw or a 27 kw range cannot be rated 8kw as Table 220.55 seems to show. This is where Note 1 comes into play.

Note 1 states that for any range larger than 12 kw we must increase the demand factor in Column C by 5% for each kw over 12.

Example 1

Let’s look at a 14kw range. A 14kw range is 2 kw larger than 12 kw (14kw-12kw). 5% for each kw over 12 kw means 5%x 2 = 10%. Take our demand factor 8kw (not 12kw) and multiple it by 10%.

8 kw X 10% = .8 kw. Now simply add it to the demand of 8kw.

8 kw +.8 kw = 8.8 kw.

8.8 kw= 8800 watts. Assuming 240v we can divide 8800/240= 36.67amps so a 40 amp circuit would be required for a 14 kw range.

Example 2

17 kw range

17 kw – 12 kw = 5 kw

5% for each kw over 12 kw = 5 x 5% = 25%

8 kw (demand) x 25% = 2 kw

8 kw + 2 kw = 10 kw

10000/240 = 41.7 amps or a 50 amp circuit.

*Note 2*

Over 8 3⁄4 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 8 3⁄4 kW and of different ratings, but none exceeding 27 kW, an average value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (using 12 kW for any range rated less than 12 kW) and dividing by the total number of ranges. Then the maximum demand in Column C shall be increased 5 percent for each kilowatt or major fraction thereof by which this average value exceeds 12 kW.

This one is a bit tricky because it hard to remember that any range rated less than 12 kw and greater than 8 ¾ kw must be used at 12 kw

Example 1

11 kw, 10 kw, 16 kw and 20 kw ranges. Since the 11 kw and 10 kw range are less than 12 kw they must be calculated at 12 kw per Note 2. What we have now is 2 - 12 kw ranges, a 16 kw and a 20 kw range. 12+12+16+20 = 60 kw and divide by 4 (number of ranges) = 60/4 = 15 kw. This 15 kw is an average of the range loads.

Table 220.55 shows a demand factor for 4 ranges at 17 kw. Note 2 states the same thing as Note 1 where the ranges are larger than 12 kw. Since the load is 15 kw the difference between 15 kw and 12 kw is 3 kw.

For each kw over 12 kw the load must be increased.

3 x 5%= 15%

15% x 17 kw = 2.55 kw

17 kw (demand load from table for 4 ranges) + 2.55 = 19.55 kw

Calculate the conductor size for the feeder for these ranges?

19,550/240 = 8.45 amps = 81 amps

At 75C a number 4 copper wire is needed.

Example 2

10 kw, 13 kw, 14 kw, 16 kw, 18 kw--- Again the 10 kw range must be used at 12 kw for this example.

12 + 13 + 14 + 16 + 18 = 73 kw/5 = 14.6 kw

From Table 220.55 -- 5 ranges at 12 kw ha a 20 kw demand. Now we have 14.6 kw as the average so 14.6 kw – 12 kw = 2.6 kw. Since .6 is a major fraction we must round up so now 2.6 kw is now 3 kw.

5% x 3 = 15%

15% x 20 kw (demand) = 3 kw

20 kw + 3 kw = 23 kw

*Note 3*

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.

Example 1-- here we could use Column 3 but we will get a larger number so on tests you will probably always use Columns A or B

6 cooking appliances rated 4 kw, 5 kw, 6 kw, 7 kw, 2@ 8 kw. All fall in the range of over 1 ¾ kw thru 8 ¾ kw so we can use Note 3.

4+5+6+7+8+8 = 38 kw. In Column B for 6 appliances we have 43% demand factor so 38*.43 = 16.34 kw or 16 kw.

Example 2

6 cooking appliances but instead of them all being in Column B some will be in Column A…. 1 kw, 1 kw, 1.5 kw, 4 kw, 5kw, and 6kw.

Now we need to use Column A and B. We use Column A for the 2- 1kw and 1-1.5 kw. 1 + 1+ 1.5 = 3.5kw. For 3 appliances in Column A we have a demand factor of 70%. Thus 3.5 * .7 = 2.45 kw

For the 4kw, 5 kw, and 6 kw we must use Column B 4 +5 + 6 = 15 kw. For 3 appliances in Column B we have 55%.

15 *.55= 8.25 kw. Now we add them together…. 2.45 + 8.25 = 10.7 kw.

*Note 4*

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.

The first sentence of note 4 says we can use Table 220.55—well duh—isn’t that what we have been doing… The second sentence states that we must use the nameplate rating for one oven or one counter mounted cooking unit (cooktop).

Example

An oven rated 5 kw will have a calculated load of 5kw. A cooktop rated 6 kw will have a calculated load of 6 kw. Pretty easy huh. But why isn’t there any demand allowance for these units? When an oven is turned on it will draw the rating of the unit so no demand is allowed. The cooktop does not have an oven to help with the diversity of the load so that unit must also be used at 100% also. It would be unusual to have all the burners going at the same time but it is possible.

The third sentence states that the branch circuit load for one cooktop and one or two wall ovens shall be calculated by adding the kw of all the units together and treating it as one unit.

Example

Calculate the load for one cooktop at 6 kw and 2 ovens rated 5 kw. As note 4 states we will add 6 + 5+ 5 = 16 kw.

From the table in column C we have a demand of 8 kw but we also have to use note 1. 16 kw is 4 kw greater than 12 kw so 5% x 4 = 20%

20% x 8 kw = 1.6 kw

1.6 kw + 8 kw = 9.6 kw

Here is an interesting fact.

16 kw range has the same branch circuit rating as a 9 kw range.

A 16 kw as we have just seen is calculated to be 9.6 kw. 9600/240 = 40 amps

A 9 kw range is rated 8 kw. 8000/240 = 33.3 amps or a 40 amp circuit

Let’s also look at 210.19(A)(3) which states *for ranges of 83⁄4 kW or more rating, the minimum branch-circuit rating shall be 40 amperes*

One last note is Note 5 which is self explanatory.

*5. This table shall also apply to household cooking appliances rated over 1 3⁄4 kW and used in instructional programs*

Last but not least is the calculation for ranges when we have a 3 phase service providing 208 volts.

An apartment has 10 ranges and is feed with a 3 phase 208 Volt system.

The first thing to do is figure out the max number of ranges between any 2 phases. There are 2 ways to do this but one is much simpler. You can draw it out on paper and count the number of ranges between phases but that can be difficult when you have 50 apartments. The other method is simply divide the number of ranges by 3 and if there is a remainder than add another ranges.

In the example above there are 10 ranges. As stated divide the 10 ranges by 3. 10/3= 3 with a remainder of 1 thus we have 4 ranges across 2 phases. If there were 11 ranges don’t get confused by the remainder of 2 because you would still only add 1 range for a total of 4 ranges across 2 phases.

Article 220.55 states *Where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected*

between any two phases.

There are 4 ranges and article 220.55 states to multiply by 2 and then look up that demand in Table 220.55. Column C shows 23 kw or 23,000 VA. This means there is a load of 11,500 per phase (23,000/2) or 34,500 VA for 3 phases (11,500/3). 34,500 va is the number to use when doing a load calculation for an entire apartment.