I helped an electrician at work the other day, do some calculations for a power run for our new press.
He was figuring it off of 125% of listed FLA.
I showed him section of 670
670.4 Supply Conductors and Over current Protection.
(A) Size. The size of the supply conductor shall be such as to have an ampacity not less than 125 percent of the full-load current rating of all resistance heating loads plus 125 percent of the full-load current rating of the highest rated motor plus the sum of the full-load current ratings of all other connected motors and apparatus based on their duty cycle that may be in operation at the same time.
(B) Over current Protection. A machine shall be considered as an individual unit and therefore shall be provided with a disconnecting means. The disconnecting means shall be permitted to be supplied by branch circuits protected by either fuses or circuit breakers. The disconnecting means shall not be required to incorporate over current protection. Where furnished as part of the machine, over current protection shall consist of a single circuit breaker or set of fuses, the machine shall bear the marking required in 670.3, and the supply conductors shall be considered either as feeders or taps as covered by 240.21.
The rating or setting of the over current protective device for the circuit supplying the machine shall not be greater than the sum of the largest rating or setting of the branch-circuit short-circuit and ground-fault protective device provided with the machine, plus 125 percent of the full-load current rating of all resistance heating loads, plus the sum of the full-load currents of all other motors and apparatus that could be in operation at the same time.
Well the largest motor was 120A, second and third motor both 93.1 A each. The heaters are 17,500w; 21,600w; 22,100w, and 1,200w. So I came up with calculations as fallows.
Largest motor 102A*125%= 150.0A
Second motor + 93.1 A
Third motor + 93.1 A
Total motor load = 336.2 A
Total heater W = 62,400 W
Heater Voltage is 480V 3phase so = 480V*sqarroot of 3
So total heater Amps = 75A
Then 125% 0f 75A = 93.82A
Total = motors +heaters = 336.2A + 93.8A = 430A
The name plate says FLA of 442A so all good
Now we would rather not run over 500kcm so we parallel them. This gives us 4/0 wire.
4/0 THHN = 260A 260A+260A = 520A
Number of Current-Carrying Conductors Percent of Values in Tables 310.16 through
310.19 As Adjusted for Ambient
Temperature if Necessary
4–6 = 80%
Now the question is do we count the two sets of wires as current carrying conductors when they are paroled? I can not find this answer any where.
520A * 80% = 416A Not enough.
Also what about This?
310.4 Conductors in Parallel.
Aluminum, copper-clad aluminum, or copper conductors of size 1/0 AWG and larger, comprising each phase, neutral, or grounded circuit conductor, shall be permitted to be connected in parallel (electrically joined at both ends to form a single conductor).
Exception No. 1: As permitted in 620.12(A)(1) .
Exception No. 2: Conductors in sizes smaller than 1/0 AWG shall be permitted to be run in parallel to supply control power to indicating instruments, contactors, relays, solenoids, and similar control devices provided
(a) They are contained within the same raceway or cable,
(b) The ampacity of each individual conductor is sufficient to carry the entire load current shared by the parallel conductors, and
(c) The over current protection is such that the ampacity of each individual conductor will not be exceeded if one or more of the parallel conductors become inadvertently disconnected
Does this mean that the over current protection would have to be 260A on each wire?
We have never done it this way. Have we been wrong? :o
Any help would be appreciated
Sorry about the length