Updated: May 5
Dwelling Unit Load Calculations - Part 2
In Part 2 we are going to dig into the basics of load calculations for the Service and/or Feeder calculations as they pertain to cooking equipment within dwelling units.
At this point, it would be again appropriate, much like in Part 1 of this series, to introduce an extremely important section in the National Electrical Code, as it pertains to Electrical Cooking Appliances, which is Sec. 220.55. This section is necessary to understand load calculations for cooking equipment. This section also is what is utilized for applying demand factors, where applicable, to the load calculations for standard method calculations supply dwelling units.
"220.55 Electric Cooking Appliances in Dwelling Units and Household Cooking Appliances Used in Instructional Programs.
The load for household electric ranges, wall-mounted ovens, counter-mounted cooking units, and other household cooking appliances individually rated in excess of 13∕ 4 kW shall be permitted to be calculated in accordance with Table 220.55. Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for loads calculated under this section.
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." - 2020 National Electrical Code (Fair Use Extract)
The above code language directly permits the electrician to use Table 220.55 when calculating the loads for electric ranges, wall-mounted ovens, counter-mounted cooking units (cooktops), and other household cooking appliances that are individually rated more than 1,750 watts (1 3/4kW). This will come in handy as we begin our process with load calculations for the same 12 kW Electric Range we used in Part 1 when we sized the branch circuits.
What is the service and/or feeder demand loaf for a 12 kW Electric Range?
Visit 220.55 and notice it gives "permission" to use Table 220.55 to perform the calculation to determine the loads.
Looking at Table 220.55 (Figure 1), the electrician will notice the heading of the table, more specifically the portion that is in parenthetical text. It says "(Column C to be used in all cases except as otherwise permitted in Note 3.)". Since Note 3 doesn't apply to this question we are working in Column C.
Column C says "Not over 12 kW Rating", which we are not over 12 kW.
The number of appliances, in the left column of Table 220.55, is (1) for our 12kW Range, we follow the horizontal plane of the table to the right and we select the Maximum Demand to be utilized.
The resulting calculated load or demand load is 8 kW, which we promptly convert to 8000 watts by taking 8kw x 1000 = 8000 watts. Also as stated in 220.55, kilovolt-amperes (VA) are equivalent to kilowatts, also making watts equivalent to volt-amperes. Now, I understand that the original electric range was 12 kW but based on demand allowances that have been reduced to 8 kW at this point.
Figure 1 - The extract used under the Fair Use Law*
The result is simple in the case of a single 12 kW Range. The demand load is now 8 kW and represents the cooking demand loads used when calculating the service or feeder rating. Now, if it was all just that simple. Well, it's not so let's kick things up a bit.
What is the service and/or feeder demand loaf for a 15 kW Electric Range?
In this situation, we now have (1) Range that is above the values given in Column C of Table 220.55. Again we are going to rely on the applicable notes below the table to help us figure this out. In this case, since there is only (1) range we are going to apply Note 1.
"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." - 2020 National Electrical Code (Fair Use Extract)
Considering our range is 15 kW, based on note 1 of Table 220.55, we need to determine how many kW our 15 kW range exceeds 12 kW. Well, 15 kW - 12 kW = 3 kW and the note said for every 1 kW that exceeds 12 kW, you increase the maximum demand as shown in Column C by 5 %. So, 3 kW x 5% = 15%.
At this point some folks get confused. Since the note said to increase the maximum demand in column C, you would be using the 8 kW for a single range as that is the maximum demand given in Column C. This would be 8 kW x 15% = 1.2 kW.
Now, note 1 says to add that increase to the maximum demand in column C which would be 8 kW + 1.2 kW = 9.2 kW or 9,200 Watts (9,200 VA or Volt-Amperes).
The conclusion is the demand load for (1) 15 kW range, which will be added to our dwelling unit load calculations (standard method) is 9,200 volt-amperes.
What is the service and/or feeder demand load for (1) 3 kW Counter-Mounted Cooking Unit and (1) 8 kW Wall-Mounted Oven?
Addressing this question is a little different because now we have (2) cooking appliances and they may align differently under Table 220.55. In fact, they will be starkly different.
Since the 3 kW counter-mounted cooking unit (cooktop) falls under column A of Table 220.55 and the 8 kW wall-mounted oven falls under column B of the same table the electrician now needs to look at Note 3 as permitted in the heading of Table 220.55.
Let's be clear, we have (2) cooking appliances, and the bold heading in the parenthetical text of Table 220.55 states we are to use column C in all cases except where permitted in Note 3. So let's look at Note 3.
"3. Over 13∕ 4 kW through 83∕ 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 13∕ 4 kW but not more than 83∕ 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."- 2020 National Electrical Code (Fair Use Extract)
The phrase " In lieu" means to look at what the kW maximum demand would be using column C but if the cooking appliance values fall within columns A and/or B you can use the multipliers in those columns to determine the final kW and whichever is lessor, in the comparison to column C is permitted to use. This will demand that you do two calculations and compare the results to column C's listed kW value so let's do that.
The 3 kW Counter-Mounted Cooking Unit falls in column A and has a multiplier of 80% (.80). The 8 kW Wall-Mounted Oven Unit falls in column B, which has a multiplier of 80% as well. In note 3 it says, "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."
3 kW Counter-Mounted Cooking Unit = 3 kW x 80% = 2.4 kW or 2,400 Watts (2,400 VA).
8 kW Wall-Mounted Cooking Unit = 8 kW x 80% = 6.4 kW or 6,400 Watts (6,400 VA)
Combine the Loads = 2,400 VA + 6,400 VA = 8,800 VA or 8,8 kW or 8,800 Watts as they all means the same thing in Article 220.
At this point, the electrical designer would compare the results of columns A and B with column C. The value for (2) appliances in column C is 11 kW. Since the 11 kW is greater than the 8.8 kW the selection would be to use the lessor value of 8.8 kW or as we like to say 8,800 VA (volt-amperes).
Pretty Simple Right?
In the next part (Part 3) of this series, the discussion will shift to load calculations that pertain to the service and feeder loads for multifamily dwellings. The principles are the same but the number of cooking appliances increases and it also introduces us to Note 2 where we may encounter cooking appliances that are different ratings that exceed 12 kW.
Paul Abernathy, CMECP® | CEO & President
Electrical Code Academy, Inc. | www.FastTraxSystem.com Office: 214-945-0653
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