Is NFPA's NEC Calculation Correct?

[jar546]

An industrial multi-building facility has its service at the rear of its main building and then provides 480Y/277-volt feeders to additional buildings behind the main building in order to segregate certain processes. The facility supplies its remote buildings through a partially enclosed access corridor that extends from the main switchboard rearward along a path that provides convenient access to services within 15 m (50 ft) of each additional building supplied. Two building feeders share a common raceway for approximately 45 m (150 ft) and run in the access corridor along with process steam and control and communications cabling. The steam raises the ambient temperature around the power raceway to as much as 35°C. At a tee fitting, the individual building feeders then run to each of the two buildings involved. The feeder neutrals are not connected to the equipment grounding conductors in the remote buildings. All distribution equipment terminations are listed as being suitable for 75°C connections.

Each of the two buildings has the following loads:

• Lighting, 11,600 VA, comprised of electric-discharge luminaires connected at 277 V
• Receptacles, 22 125-volt, 20-ampere receptacles on general-purpose branch circuits, supplied by separately derived systems in each of the buildings
• 1 Air compressor, 460 volt, three phase, 5 hp
• 1 Grinder, 460 volt, three phase, 1.5 hp
• 3 Welders, AC transformer type (nameplate: 23 amperes, 480 volts, 60 percent duty cycle)
• 3 Industrial Process Dryers, 480 volt, three phase, 15 kW each (assume continuous use throughout certain shifts)

Determine the overcurrent protection and conductor size for the feeders in the common raceway, assuming the use of XHHW-2 insulation (90°C):

Calculated Load {Note: For reasonable precision, volt-ampere calculations are carried to three significant figures only; where loads are converted to amperes, the results are rounded to the nearest ampere [see 220.5(B)]}.

This is what they come up with as an example for us to follow:

• Noncontinuous Loads: 38,886 VA
• Continuous Loads: 56,600 VA
• Total Load: 109,636 VA (with continuous loads adjusted)
• Current: 132 A
• Conductor Size: 1/0 AWG copper with XHHW-2 insulation.
• Overcurrent Protection: 150 A circuit breaker. - Would a 175A also be acceptable?

 

[wwhitney]

Noncontinuous Loads: 38,886 VA
Continuous Loads: 56,600 VA

First step is to check these totals. The continuous loads total is easy to confirm, as it appears the only loads being treated as continuous are the lighting and the 3 industrial process dryers, and those add up to 11,600 + 3 * 15,000 = 56,600 VA as above. That leaves the non-continuous loads.

For the receptacles, I assume 22 means 22 duplex receptacles, which are 180VA each for 22 * 180 = 3,960 VA.

On the motors, the Table 430.250 full load currents are 7.6A for the 5HP and 3A for the 1.5 HP. We need to convert to VA, and use a 1.25 factor for the largest motor, so these contribute 480V * sqrt(3) * (7.6*1.25 + 3) = 10,392 VA.

On the welders, I understand them to be single phase. I'm not particularly familiar with Article 630, but if they are nonmotor generator arc welders, then the current gets a 0.78 factor to account for the 60% duty cycle. 3 welders get what I'll call a group factor of 2.85 under 630.11(B). That makes the load 23A * 480V * 0.78 * 2.85 =24,541 VA.

Adding up these non-continuous loads gives 38,893 VA, close enough to the 38,886 VA above. So I'll stick with the number in the OP.

[I find it a bit odd that 630.11(B) doesn't specify that the group factor is per ungrounded conductor and based on the number of welders connected to that ungrounded conductor. To balance the load above, you'd need to put each welder on a different pair of ungrounded conductors, meaning each ungrounded conductor would only be connected to 2 welders. So for any given conductor, I don't see how the presence of the third welder not connected to that conductor would have any bearing on the load on that conductor. Thus I would argue that it would be better to use a 3.0 group factor, making the load 1,292 VA higher.]

Cheers, Wayne

 

[wwhitney]

Two building feeders share a common raceway for approximately 45 m (150 ft) and run in the access corridor along with process steam and control and communications cabling. The steam raises the ambient temperature around the power raceway to as much as 35°C.

Next is to determine the derating factors that apply. Two 3 phase feeders in the same raceway is 6 current carrying conductors, so that provides an 80% ampacity adjustment factor. The conductors are specified to be a type with 90C rated insulation; as the table ampacities are based on 30C ambient, the ampacities get a temperature correction factor of 0.96. Multiplying those together gives 0.768.

• Noncontinuous Loads: 38,886 VA
• Continuous Loads: 56,600 VA
• Total Load: 109,636 VA (with continuous loads adjusted)
• Current: 132 A
• Conductor Size: 1 AWG copper with XHHW-2 insulation.
• Overcurrent Protection: 150 A circuit breaker. - Would a 175A also be acceptable?

At this point in the calculation, I find it easier to work with the currents, so the non-continuous loads represent a current of 38,886VA/480V/sqrt(3) = 47A, and the continuous loads represent a current of 56,600VA/480V/sqrt(3) = 68A. Thus the total load is 115A, while some of the checks require us to use an extra 25% of the continuous load, or 132A.

For the "run of the feeder" check, we need to use the derating factors, but get to start with the 90C ampacity, and don't need any extra 25% on the continuous load. That is, we need a conductor with 90C ampacity at least 115A / 0.768 = 150A. #1 Cu is just shy at a 90C ampacity of 145A; we need to use 1/0, with a 90C ampacity of 170A.

For the "termination check," we don't need to use the derating factors, but we are stuck with the 75C ampacity, and we do need the extra 25% on the continuous load. So we need a conductor whose 75C ampacity is at least 132A. Again #1 Cu is just shy at a 75C ampacity of 130A; we need to use 1/0, with a 75C ampacity of 150A.

The smallest standard size OCPD we may use for the load is the next size up from 132A, or a 150A OCPD. To use a 150A OCPD, 240.4(B) says that we need a conductor with an ampacity of at least 126A. If we select 1/0 Cu, the ampacity is the minimum of the 75C termination ampacity of 150A (no derating) and the derated 90C ampacity of 0.768*170A = 131A, meaning the ampacity is 131A, greater than 126A. Note that sometimes this check will require us to bump up the conductor size, as in the "run of the feeder" check we only required a final ampacity equal to the load, 115A.

So the 1 AWG in the OP is not correct, it should be 1/0 Cu.

Cheers, Wayne

 

[jar546]

So the 1 AWG in the OP is not correct, it should be 1/0 Cu.

You are correct. This is right from the NEC appendix and I made a typo. The rest is correct according to the NEC. I will correct the OP.

 

[jar546]

This is what it says about the feeder neutral conductor:

Feeder Neutral Conductor

(see 220.61)
Because 210.11(B) does not apply to these buildings, the load cannot be assumed to be evenly distributed across phases. Therefore the maximum imbalance must be assumed to be the full lighting load in this case, or 11,600 VA. (11,600 VA ÷ 277 V = 42 A.) The ability of the neutral-to-return fault current [see 250.32(B) Exception No. 2] is not a factor in this calculation.
Because the neutral runs between the main switchboard and the building panelboard, likely terminating on a busbar at both locations, and not on overcurrent devices, the effects of continuous loading can be disregarded in evaluating its terminations [see 215.2(A)(1) Exception No. 3]. That calculation is (11,600 VA ÷ 277 V = 42 A), to be evaluated under the 75°C column of Table 310.16. The minimum size of the neutral might seem to be 8 AWG, but that size would not be sufficient to be depended upon in the event of a line-to-neutral fault [see 215.2(B), second paragraph]. Therefore, because the minimum size equipment grounding conductor for a 150 ampere circuit wired with 2/0 AWG conductors, as covered in Table 250.122, is 6 AWG, that is the minimum neutral size required for this feeder.