Question parallel conductors

[hhsting]

I have 6-#3 awg conductors plus #6awg gnd between fire pump controller and fire pump that is 40HP, 208V three phase.

I did research minimum size parallel allowed is #1/0 awg and ampacity derating apply for more than 3 current carrying conductors.

However, I am not able to find this:
Can you parallel in one conduit two sets of #3 awg or not?

 

[infinity]

You cannot parallel #3 AWG condcutors but you did say a fire pump and many fire pumps are Wye start/Delta run so there are 6 conductors between the controller and the fire pump.

 

[hhsting]

You cannot parallel #3 AWG condcutors but you did say a fire pump and many fire pumps are Wye start/Delta run so there are 6 conductors between the controller and the fire pump.

So then it may not be paralleled? I dont follow


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[augie47]

You can't parallel smaller than 1/0 regardless of numbers of conduits

 

[infinity]

So then it may not be paralleled? I dont follow

Are you familiar with Wye start/Delta run motors? They have 6 conductors from the controller to the motor. You need to confirm the type of fire pump controller that you have. If it's Why start/Delta run the conductors are not normal parallel conductors which are required to be #1/0 and larger.

 

[roger]

To expand on Rob's last post, the six conductors of a wye start delta run do not carry current at the same time. Three will be energized at start up and drop out after the motor has reached a running state where the other three will take over.

 

[don_resqcapt19]

To expand on Rob's last post, the six conductors of a wye start delta run do not carry current at the same time. Three will be energized at start up and drop out after the motor has reached a running state where the other three will take over.

I don't understand.
The six conductors are the two ends of each of the 3 windings. In the wye start, one end of each winding is connected to the power source and the other ends are shorted together at the stater to create a wye connection.
For the delta run, each end of each coil is connected to the power source to create a delta connection.
In both cases, all six conductors carry current.

 

[infinity]

In both cases, all six conductors carry current.

That's correct. Wye start/Delta run conductors are sized at 72% (58*125%) of the running motor current for that reason. All six carry current it's the configuration that changes through the starter.

 

[roger]

I don't understand.
The six conductors are the two ends of each of the 3 windings. In the wye start, one end of each winding is connected to the power source and the other ends are shorted together at the stater to create a wye connection.
For the delta run, each end of each coil is connected to the power source to create a delta connection.
In both cases, all six conductors carry current.
View attachment 2570539

You are correct, I should have said "full load current"

 

[wwhitney]

In both cases, all six conductors carry current.

Thanks for the explanation. The above would matter for ampacity adjustment.

But for the OP's question, the point is that in neither case are the load end of any two conductors directly connected together, so never do you have two conductors in parallel in the 310.10(G) sense.

Cheers, Wayne

 

[augie47]

Thanks for the explanation. The above would matter for ampacity adjustment.

But for the OP's question, the point is that in neither case are the load end of any two conductors directly connected together, so never do you have two conductors in parallel in the 310.10(G) sense.

Cheers, Wayne

I think that's the "clue". With the wye/delta they are not parallel conductors.

 

[hhsting]

Fire pump Controller is wye delta.

If they are not inparallel then My only question now is full load amps is 115A.

For fire pump supply conductors should it not be 115x1.25=143.75A which should be minimum #1/0 awg each instead of #3 awg?

 

[wwhitney]

Fire pump Controller is wye delta.

If they are not inparallel then My only question now is full load amps is 115A.

For fire pump supply conductors should it not be 115x1.25=143.75A which should be minimum #1/0 awg each instead of #3 awg?

So, the FLA is based on a 3-wire delta-connected load. If you look at the diagram in post 7, where there are blue dots at the end of each coil, a single wire would be connected to both blue dots at each corner of the triangle of coils. That wire would be carrying current from each of the two coils connected to that point (if you like, the "outbound" current from one coil and the "return current" from the other), and those two currents are 60 degrees out of phase with each other. That means the wire is carrying sqrt(3) * the coil current (not twice, due to the phase difference).

In other words, the coil current in 1/sqrt(3) times the FLA. And with 6 conductors going to the motor, in the delta configuration each wire is only carrying a single coil current, there's no combining currents as in a 3-wire delta connection. So each wire would see 115A / sqrt(3) = 66A at full load, and after the 125% factor, that gives 83A. Thus the required ampacity is only 83A (didn't actually check that the appropriate section of Article 430 specifies this, but it must). This is what infinity was alluding to in post #8, 1/sqrt(3) = 58%.

Cheers, Wayne

 

[hhsting]

So, the FLA is based on a 3-wire delta-connected load. If you look at the diagram in post 7, where there are blue dots at the end of each coil, a single wire would be connected to both blue dots at each corner of the triangle of coils. That wire would be carrying current from each of the two coils connected to that point (if you like, the "outbound" current from one coil and the "return current" from the other), and those two currents are 60 degrees out of phase with each other. That means the wire is carrying sqrt(3) * the coil current (not twice, due to the phase difference).

In other words, the coil current in 1/sqrt(3) times the FLA. And with 6 conductors going to the motor, in the delta configuration each wire is only carrying a single coil current, there's no combining currents as in a 3-wire delta connection. So each wire would see 115A / sqrt(3) = 66A at full load, and after the 125% factor, that gives 83A. Thus the required ampacity is only 83A (didn't actually check that the appropriate section of Article 430 specifies this, but it must). This is what infinity was alluding to in post #8, 1/sqrt(3) = 58%.

Cheers, Wayne

NEC 2017 Article 430.22(C) is the correct section then for sizing conductors for this type of controller?

 

[wwhitney]

NEC 2017 Article 430.22(C) is the correct section then for sizing conductors for this type of controller?

Yes, precisely. The 72% figure there, and in post #8, is just 125% / sqrt(3). So I gave you the derivation of that figure.

Cheers, Wayne

 

[infinity]

NEC 2017 Article 430.22(C) is the correct section then for sizing conductors for this type of controller?

Yes if it's Wye start Delta run those are the values you use. 58%*FLC*125%=FLC*72%.

If they are not inparallel then My only question now is full load amps is 115A.

For fire pump supply conductors should it not be 115x1.25=143.75A which should be minimum #1/0 awg each instead of #3 awg?

No #1/0 is not required because you're spreading the load across 6 conductors not 3.

 

[hhsting]

Yes, precisely. The 72% figure there, and in post #8, is just 125% / sqrt(3). So I gave you the derivation of that figure.

Cheers, Wayne

I see. Thanks . I will read through it

 

[hhsting]

I see. Thanks . I will read through it

I see but what about sizing the equipment grounding conductor load side of the controller? I dont see NEC 2017 Article 430?

 

[augie47]

250.122(D)