Annex-D, Problem D8, Wound Rotor Load left out of Feeder Calc ?????

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ToolHound said:
Jumper. Thanks.

I looked at what you nicely wrote above.
The part in red...I am wondering about that.
One question about the '81 A' , in red above...one question
about that would be...would that '81 A' need to be included in the 'feeder protection calculation' ? ( i.e., the feeder protection calculation laid out in the solution provided in the book )?

Thanks. ToolHound
Click to expand...

No. That is the minimum conductor ampacity of the wires to the motor.

Feeder OCPD sizing is largest individual motor OCPD + FLC of all other motors..

From the example in question.

Feeder Short-Circuit and Ground-Fault Protection
The rating of the feeder protective device is based on the sum of the
largest branch-circuit protective device (example is 110 A) plus the sum of
the full-load currents of the other motors, or 110 A + 40 A + 40 A = 190
A. The nearest standard fuse that does not exceed this value is 175 A [see
240.6 and 430.62(A)].
Click to expand...
 
Think of it this way, a step down transformer will have a higher current on its secondaries inverse proportional to the current on the primaries, do we add this secondary current to the primary current when sizing conductors or circuits feeding a transformer? Nope
 
hurk27 said:
Correct the conductors from the slip rings (brushes) to the contactors and resistor banks are required to be sized as above but these do not add any load to the supply circuits feeding the motors primary, all these conductors are for is to step the rotor windings through a set of resistor banks shunting the rotor current thus providing speed control.
Click to expand...

jurk27. Got it. Thanks for the help and information.
--ToolHound
 
hurk27 said:
Think of it this way, a step down transformer will have a higher current on its secondaries inverse proportional to the current on the primaries, do we add this secondary current to the primary current when sizing conductors or circuits feeding a transformer? Nope
Click to expand...


Exactly. Got it. Thanks. --ToolHound
 
130415-1125 EDT

Toolhound:

You need to study the theory of AC induction motors. The resistance in the rotor coils, and thus the current in those coils, determines the speed torque characteristics of the motor. Most AC induction motors are made with very low resistance, usually aluminum, rotor conductors. These are effectively a shorted secondary of a transformer.

Yes current is magnetically induced in these conductors. That current comes from the current in the stator coils. As turk27 said, like a transformer. That induced current and its magnetic field interacts with the rotating magnetic field of the stator coils to produce motor output torque. The more load on the motor shaft the greater is the slip and induced current, and output torque up to a breakdown point.

A wound rotor induction motor simply has coils on the rotor that are not shorted on the rotor as in a normal induction motor, but connect thru slip rings to an external adjustable resistance. This allows the rotor resistance to be adjusted without rebuilding the rotor and thus dynamically you can adjust the speed torque curve of the motor.

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I think that you are (mistakenly) thinking that you are ENERGIZING the rotor circuit of a Wound Rotor Induction Motor (WRIM), and you are wondering why the current that is going to the rotor is not factored into the motor circuit feeder conductor sizing, is that right?

Here is where that is flawed: the rotor in a WRIM is NOT being powered separately by the supply, just because it has slip rings and windings, as opposed to a squirrel cage induction motor that just has a rotor cage and no slip rings. What is ALWAYS happening in any induction motor is that the rotor current is induced onto the rotor conductors (rotor bars or windings as the case may be) by the expanding and colapsing magnetic fields of the stator windings, hence the term "induction motor". So ALL of the energy in the rotor circuit comes from the energy that goes into the stator, it is NOT separate. A WRIM is not being POWERED through the slip rings, all you are doing is changing the RESISTANCE of the rotor circuit, which changes the flux and thus the torque that the motor produces. So although you must size the conductors that go from the rotor slip rings to the resistor bank based on the maximum rotor current, ALL of that current still came from the stator, and therefore from the motor feeder conductors you already ran to it.
 
ToolHound said:
Annex-D Problem D8
Motor Circuit Conductors, Overload Protection, and Short-Circuit and Ground-Fault Protection

The entire problem, with solution, is included in the attached Word.doc file.

This problem includes three(3) motors
1 squirrel cage motor
2 wound rotor motors

I'm wondering why the load (current) for the wound rotor secondary circuit is NOT included in the Feeder Short-Circuit and Ground-Fault Protection calculation.

In other words...the feeder calculatoin for the three-motors is provided with the problem. But the load for the wound rotor secondary is left out. Why ? Any guess ?


Because wound rotor/secondary is powered by induction, not by it's own power source, so to speak ?


Thanks, ToolHound
Click to expand...

It's because you only have to motors that require only two feeders.

You might have been confused as to why there were three calculations on conductors sizing?

  1. The first item computed the required conductor size for the squirrel-cage induction motor and,
  2. Item b. computed the conductor sizes for a.) the motor feeder plus b.) the size of the conductor you are required to attach your secondary resistors of the wound-rotor induction motor.
Crystal!
 
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