Last Motor Problem

[Roger]

Use the below diagram to calculate the size of the feeders, branch circuit conductors, OCPD, and overloads for the given motor design, Assume 75C terminations. Neutral not shown. An engineering error exists in the drawing M6 should be connected to L3.

http://i18.photobucket.com/albums/b149/thermocline/motor3.jpg

[Homer]

I think that the motor design needs to be tweeked a bit to optimize the load balancing. :)

A single terminal change will lower the required feeder ampacity by 5.5A. :D

Homer

[Roger]

During construction a union strike was called. Outsourced labor was brought in by the building owner to the dismay of the GC. When the strike ended and union electricians returned to the job it was discovered that the temporary labor had overlooked a slight error in the motor design drawings by the engineer, by specifing M6 on the wrong phase for proper balancing. The union electricians quickly moved M6 to L3 and requested a drawing correction. The error is rather significant in that it will allow a smaller feeder by one wire size from the original drawing specs.


YOU HAPPY NOW!! :D

[Homer]

YOU HAPPY NOW!! :DI'm always happy.

M1 Table FLA 74.8A, Branch Ampacity = 93.5A, #3 Copper, Max. 125A Inv.Time Breaker, Max. 62.5A Overload. (largest motor)

M2 Table FLA 18.7A, Branch Ampacity = 23.4A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 15A Overload.

M3 Table FLA 18.7A, Branch Ampacity = 23.4A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 15A Overload.

M4 Table FLA 13.2A, Branch Ampacity = 16.5A, #14 Copper, Max. 35A Inv.Time Breaker, Max. 12.5A Overload.

M5 Table FLA 20A, Branch Ampacity = 25A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 20A Overload.

M6 Table FLA 20A, Branch Ampacity = 25A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 20A Overload.

Required Feeder Ampacity = 145.4A, 1/0 Copper, Max 200A Inv.Time Breaker.

Note 1: §240.4(G) allows #14 up to 20A and #12 up to 25A with 75° terminations for motor applications.

Note 2: The neutral conductor is clearly carrying the unbalanced load only in this case as it's a 4-wire wye system. Thus by §310.15(B)(4) the neutral is not a current carrying conductor for derating purposes. This can even be proven mathemetically now that I refreshed my memory while proving something similar to jsmall.

If we read carefully;

§310.15(B)(4) Neutral Conductor

(a) A neutral conductor that only carries the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of §310.15(B)(2)(a).

(b) In a 3-wire circuit consisting of two phase wires and the neutral of a 4-wire, 3 phase, wye connected system, a common conductor carries approximately the same current as the line to neutral load currents of the other conductors and shall be counted when applying the provisions of §310.15(B)(2)(a).

(c) On a 4-wire, 3 phase wye circuit where the major portion of the load consists of nonlinear loads, harmonic currents are present in the neutral conductor; the neutral shall therefore be considered as a current carrying conductor.


Sections (b) and (c) don't apply here since we have a full 4-wire wye feeder (3 Hots, not 2) and no nonlinear loads. Also, (a) applies to this situation since the neutral carries only unbalanced current.

I saw a similar question with a posted solution here in the past that claimed that it wasn't a true neutral is this case. I should have disagreed at the time.

It appears that Mike Holt agrees with my interpretation in his own example (http://www.ecmweb.com/mag/electric_motor_calculations_part_2/).

Homer

[Roger]

It appears that Mike Holt agrees with my....

Yep, his artciles along with others by different resources are what I have been studying. The one you give a link to is the article I formated this problem after....looks like you figured that out. :eek: Not quite confident enough to derive my own but I believe I could put one together now. I'll have to study the other stuff you added but thanks for taking the time to provide it.

I have the same results as you....except on overloads for M1 I get 50 x 1.15 = 57.5 amps max. overload....next size down 50 amps...Using other motors not listing service factors.

Thanks Homer for the education you always give me on these problems I post.

[Homer]

Yes, I was in too big of a hurry to finish. :mad: I didn't look back at the sketch enough times!

I used a 125% factor for all when I should only have used it for the motors with a posted SF of 1.15. Those others needed a factor of 115% as you state.

BTW, remember that the overloads will have a range for setting (including decimals) unless you are using fuses (not recommended).

Homer

[Roger]

BTW, remember that the overloads will have a range for setting (including decimals) unless you are using fuses (not recommended).

Yes thats right these werent fuses so would not be next size down but a range up to a maximum.

[Wgoodrich]

The following I calculated per NEC 2005 rules on the same problem balancing the load to 126.8 amps highest feeder conductor.

The first was Homer's comments. Teh COMMENT part was my calcs following Homer to ease confusion;



M1 Table FLA 74.8A, Branch Ampacity = 93.5A, #3 Copper, Max. 125A Inv.Time Breaker, Max. 62.5A Overload. (largest motor)

COMMENT;
Breaker maximum is 250% of FLA = 187 amps adjust up on branch per 240.6 = max inv. Time breaker of 200 amp breaker for fla of 74.9 and 3 awg copper branch circuit conductor size. Overload minimum 115% x 74.8 = 86.2 amps overload size allowed minimum if motor won’t start maximum overload allowed is 130%x74.8= 97.24 amps overload size.

M2 Table FLA 18.7A, Branch Ampacity = 23.4A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 15A Overload.

COMMENT;

Breaker maximum is 250% of FLA = 46.7 amps adjust up on branch per 240.6 = max inv. Time breaker of 50 amp breaker for fla of 18.7 and 12 awg copper branch circuit conductor size. Overload minimum 125% x 18.7 = 13.8 amps overload size allowed minimum if motor won’t start maximum overload allowed is 140%x 26.2 = amps overload size.


M3 Table FLA 18.7A, Branch Ampacity = 23.4A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 15A Overload.

COMMENT;

Breaker maximum is 250% of FLA = 46.7 amps adjust up on branch per 240.6 = max inv. Time breaker of 50 amp breaker for fla of 18.7 and 12 awg copper branch circuit conductor size. Overload minimum 125% x 18.7 = 13.8 amps overload size allowed minimum if motor won’t start maximum overload allowed is 140%x 26.2 = amps overload size.


M4 Table FLA 13.2A, Branch Ampacity = 16.5A, #14 Copper, Max. 35A Inv.Time Breaker, Max. 12.5A Overload.

COMMENT;

Breaker maximum is 250% of FLA = 33 amps adjust up on branch per 240.6 = max inv. Time breaker of 35 amp breaker for fla of 13.2 and 14 awg copper branch circuit conductor size. Overload minimum 115% x 13.2 = 15.8 amps overload size allowed minimum if motor won’t start maximum overload allowed is 130%x 17.6 = amps overload size.


M5 Table FLA 20A, Branch Ampacity = 25A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 20A Overload.

COMMENT;

Breaker maximum is 250% of FLA = 50 amps adjust up on branch per 240.6 = max inv. Time breaker of 50 amp breaker for fla of 20 and 12 awg copper branch circuit conductor size. Overload minimum 115% x 18.7 = 23 amps overload size allowed minimum if motor won’t start maximum overload allowed is 130%x 26 = amps overload size.


M6 Table FLA 20A, Branch Ampacity = 25A, #12 Copper, Max. 50A Inv.Time Breaker, Max. 20A Overload.

COMMENT;

Breaker maximum is 250% of FLA = 50 amps adjust up on branch per 240.6 = max inv. Time breaker of 50 amp breaker for fla of 20 and 12 awg copper branch circuit conductor size. Overload minimum 115% x 18.7 = 23 amps overload size allowed minimum if motor won’t start maximum overload allowed is 130%x 26 = amps overload size.


Required Feeder Ampacity = 145.4A, 1/0 Copper, Max 200A Inv.Time Breaker.


Largest motor load increased 25% ; 74.9 x 125% = 93.6 + 18.7 + 13.2 + 20 = 145.5 amps

1/0 copper. Max Inverse time breaker; largest overcurrent device allowed = calc of 200 amp + full load current of all other motors on same line. 200 + 18.7 + 13.2 + 20 = 251.9 amps adjust up to next overcurrent device listed per 240.6 = 300 amp inverse time breaker for feeder size 1/0 Copper. Feeder neutral size is max unbalanced load of 126.8 – 112.3 = 14.5 amp feeder neutral. However 215.2 requires the feeder neutral conductor shall not be smaller than the grounding conductor. This makes the minimum feeder neutral conductor to be size equal to grounding in 250.122. being a 8 awg copper feeder neutral conductor.


HOpe this helps

Wg

[Homer]

Hope this helpsNot really.

M1 is a wound rotor induction motor. I stand by my calculation of a max. 125A inv. time breaker for this the largest motor.

That changes the entire feeder calculation.

Also I question the rounding up for the feeder breaker. The wording of §430.52 and §430.62 states that only a branch circuit OCPD may be 'rounded up'. No mention of the feeder. Remember you may have already rounded up the OCPD for the largest motor before adding the FLAs of the rest. I have always rounded down at the final step.

Homer

[Wgoodrich]

Homer you are correct the Feeder Over current device is to round down in 250.6 forgot that one.


Exception No. 1: Where the values for branch-circuit short-circuit and ground-fault protective devices determined by Table 430.52 do not correspond to the standard sizes or ratings of fuses, nonadjustable circuit breakers, thermal protective devices, or possible settings of adjustable circuit breakers, a higher size, rating, or possible setting that does not exceed the next higher standard ampere rating shall be permitted.

Table in 430.52

Home

You are correct with the wound rotor max 150 %. Forgot that one too.

Must be getting rusty.


Thanks for pointing out my misses.

Eating Crow is not good tasting but good learning or reality check in getting out of practice.

Wg