Application of 25% Load Factors

[trenictrev]

I am checking a design prepared by a colleague, and I would like to understand the application of 125% that is used when summing the loads under NEC. Details are below in blue, and the numbers I am questioning are highlighted in red.

In this calculation, we have included contingencies on the loads based on potential future variation – we have only a conceptual design right now and do not have the exact loads available. My colleague has also considered a 25% additional “loss in transmission” as part of each load, but my feeling is that this 25% loss in transmission is already covered by the 125% factors applied in the 430.24 sum and/or the 215.2a sizing of the conductor.

Basically, I believe we are double-counting the 25% somewhere and creating an overly conservative design. The actual total load is 17.5 A (including contingencies), but we are ending up with a circuit rated at 32 A.

Question Statement: Can I drop the 25% losses in transmission because it is a duplication of either the 125% in the sum or the 125% in the conductor sizing?

(I can sense the push back that this is a small circuit, so who cares, but actually the overall system combines 50 of these similar loads to a total near 1 MW. I would like to optimize at this level so that the upstream systems can be less costly/complicated).

I appreciate your help applying the 25% factors correctly.

The total load is made up from auxiliary power plus two pairs of motors as follows:

• Continuous Auxiliary Power = 6.4 kW
  • Adding 20% contingency and 25% losses in transmission = 9.6 kW
  • Based on 3-phase 480 VAC and 90% power factor, load current = 12.8 A
• Motor Pair 1 Power = 2.04 kW
  • Adding 50% contingency and 25% losses in transmission = 3.8 kW
  • Based on 3-phase 480 VAC and 90% power factor, load current = 5.1 A (shared to two motors)
• Motor Pair 2 Power = 1.56 kW
  • Adding 50% contingency and 25% losses in transmission = 2.9 kW
  • Based on 3-phase 480 VAC and 90% power factor, load current = 3.9 A (shared to two motors)

Using NEC article 430.24, the total load is as follows:

• Total Load = (2.55 A x 125%) + ((2.55 + (2 x 1.95 A)) x 100%) + (12.8 A x 125%) = 25.6 A

Based on NEC Article 215.2a, conductor size should be 125% of the load, so minimum conductor rating must be 32 A.

 

[Smart $]

Why are you adding in the "25% losses in transmission" to begin with? That would amount to a severe voltage drop. Ampacity requirements keep conductors large enough that voltage drop is not substantial for short circuit runs. The only way you could feasibly have that great of loss under the NEC and still be compliant is with loads at a substantial distance from the source using the minimum ampacity rating for conductors. The NEC recommends keeping circuits to a total 5% voltage drop but does not require such except for fire pump circuits.

Additionally, motor loads are neither continuous or noncontinuous. As a matter of fact, Article 215 has no effect on any circuit with a motor load. Where a circuit has one or more motors and other loads, it is considered a motor feeder. As such, 430.24 and 430.63 apply.