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· Electrical Simpleton
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Discussion Starter · #1 ·
I will preface my questions by admitting I have no experience with a wye start delta run controller or motor.

I've researched quite a bit and I think I have a loose understanding of the concept. The FP controller is using a closed transition between start and run.

Section 430.21(C) seems to say that the motor conductors between the controller and the motor can be sized not less than 72% of the FLA.

I, most likely, am misunderstanding the concept. The best I can tell, there will be 6 conductors from the controller to the motor. One set for start and the other for run.

The questions;

1. When the motor is up to speed will all 6 of the conductors between the controller and the motor be current carrying conductors or will the "start" conductors be removed via a contactor in the controller?

2. If the start conductors are removed via a contactor wouldn't the run conductors then carry the FLA of the motor?

Pete
 

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I haven't worked on many but all the y start delta run motors I've seen have six conductors feeding the motor off of two contactors. Start contactor pulls in, starts motor, releases at the same time the run contactor pulls in.
 

· Donuts > Fried Eggs
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...When the motor is up to speed will all 6 of the conductors between the controller and the motor be current carrying conductors or will the "start" conductors be removed via a contactor in the controller?
They are not removed. You have three windings with six leads brought out of the motor and those same conductors and windings carry current in both "start" and "run" conditions.

The difference is that in start condition, the windings are arranged in series as a wye, so 480 is divided between two windings, makes for lower voltage across any single winding, and lower current flow.

When the run contactor pulls in those same windings are then re-arranged by the contactor as a delta, so now there's full 480V drop across one winding, more current flow, and more available torque.

As far as 430.21(C) goes, take a look at the FPN that says the windings must be sized to handle 58% of the current. You know that 1.732 we use when multiplying for 3 phase? 58% of that is basically "single phase."

Example:
150A * 1.732 = 260A
260A * 58% = 150A

That is what each single phase lead of that motor will draw when Δ connected, because the remaining current is being supplied by the other two phases.
 

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Pete

If you can disconnect the power to the starter, disconnect the motor. connect power to the secondary of the control transformer, you can safely play with starter. Run it through it paces and see how the controls work without starting the pump.
 

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I haven't worked on many but all the y start delta run motors I've seen have six conductors feeding the motor off of two contactors. Start contactor pulls in, starts motor, releases at the same time the run contactor pulls in.
There are actually 3 contactors, although you only see the 6 leads for the motor coming off of two of them. In fact because he said his is a "Closed Transition" Wye-Delta starter, there are 4 contactors and a resistor bank, but still, only 2 of the contactors have leads coming out of them to the motor.

Side Note:
So here's a puzzler that bothered me for years as an electrician and I could not get a straight answer out of anyone until after I went back to EE school.

When sizing the conductors (and OL heaters by the way), we always use 58% of the motor nameplate current right? So when the motor is in Delta Run, is one contactor and set of conductors carrying 58% of the current, and the other carrying 42%? Because if so, why are both sets of conductors sized for 58% and when placing the OL relay, it doesn't matter which contactor it goes in? Does the motor take 116% of nameplate FLC when running in Delta?

(By the way, rhetorical question aimed at educating the newbies, I already know the answer)
 

· Electrical Simpleton
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3,343 Posts
Discussion Starter · #8 ·
There are actually 3 contactors, although you only see the 6 leads for the motor coming off of two of them. In fact because he said his is a "Closed Transition" Wye-Delta starter, there are 4 contactors and a resistor bank, but still, only 2 of the contactors have leads coming out of them to the motor.

Side Note:
So here's a puzzler that bothered me for years as an electrician and I could not get a straight answer out of anyone until after I went back to EE school.

When sizing the conductors (and OL heaters by the way), we always use 58% of the motor nameplate current right? So when the motor is in Delta Run, is one contactor and set of conductors carrying 58% of the current, and the other carrying 42%? Because if so, why are both sets of conductors sized for 58% and when placing the OL relay, it doesn't matter which contactor it goes in? Does the motor take 116% of nameplate FLC when running in Delta?

(By the way, rhetorical question aimed at educating the newbies, I already know the answer)
Ok... I'll claim ignorance.

Please teach on brother!!

Pete
 
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