Code Forum

Jun 1, 1999 12:00 PM, By Fred Hartwell, Senior Editor

Here's exactly how and why you double the range (and dryer) loads on 3-phase distributions, what happens next, and how Example 5(a) integrates this process.

If you're having problems understanding National Electrical Code (NEC) requirements for 3-phase demand calculations, you're not alone. Recently, a reader asked a question on these calculations. Yes, he understood Sec. 220-19 correctly: determine the maximum number of ranges between any two phases; multiply that number by 2; and then refer to Table 220-19 for demand. However, this procedure didn't look like the examples at the end of the NEC. Here, Appendix D, Example 5(a) takes this demand value, divides it by 2 for a per-phase demand, and multiplies that number by 3 for equivalent 3-phase load. When referring to these calculations in Sec. 220-19, our friend also wondered if the NEC intended the word "range" as the key word. Did it mean counter-mounted cooktops and wall-mounted ovens don't apply to the 3-phase procedure? For instance, suppose the 10 units in Example 5(a) consisted of 10 5kW cooktops and 10 4kW wall-mounted ovens.

EC&M panel's response. The procedure in Example D5(a) is correct. You connect ranges, like any other load, between two phase conductors.

Technically, a phase isn't a wire; it's a transformer winding between two wires. This means when you calculate loading on a "phase conductor," you actually figure how much load, in amperes, results from the two phase loads common to that conductor.

In this case, the Code simplifies the problem by just doubling the highest loaded phase. That's the number to use when you go to the demand table. Once you determine the total demand load from this table, then you split it in half to get the calculation for each winding. Ignoring the square root of 3 for the moment, all you're doing is reversing the doubling procedure you started with. So, the total 3-phase load is the arithmetic sum for all three phases of what you just determined, or triple the amount on a single-phase winding. Remember, we're only working with load so far. We haven't addressed ampere loading on a conductor yet. Example D5(a) carries the total load in volt-amperes throughout load calculations. At the end, you divide by the product of voltage and the square root of 3 to get the number of amperes on one of the "phase" conductors.

Still not convinced? Go back to the range load on one of the phases and look at how it relates to a similar load on an adjacent phase. The loads on phase connections A-B and B-C add vectorially on the common conductor "B," for example. For equal loads (A-B and B-C), you do not add them arithmetically. The common conductor only sees one of those phase loads multiplied by the square root of 3. Multiplying a number by 3 (as in the Code example) and then dividing by the square root of 3 at the end is exactly the same as multiplying by the square root of 3 at this point. The Code example simply postpones using the square root of 3 and voltage factors until the end, so you only need to use them once.

For cooktops separate from other components, Note 4 to Table 220-19 allows for a single branch circuit to connect the elements. This reflects a similar allowance in Sec. 210-19(c) Ex. 1. Sec. 220-10, typically requiring that a feeder reflect the loading on the branch circuits it supplies. Here, combine the cooktop and oven in each dwelling and treat the combination as a range. If you wire each appliance on its own circuit, apply Note 3 to Table 220-19.

Editor's Note:

These answers are given by our panel of experts. This author chairs the panel. Other panel members include: Bill Summers, James Stallcup, Dan Leaf, and Joe Ross.

The opinion expressed is that of the panel. If a panelist disagrees with the majority opinion, his explanation is printed following the answer. Although authoritative, the answers printed here are not, and cannot be relied on as formal interpretations of the NEC.