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

[ToolHound]

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

 

[jumper]

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

Not seeing your question.:?

Conductor Ampacity
The full-load current value used to determine the minimum required conductor
ampacity is obtained from Table 430.250 [see 430.6(A)] for the
squirrel-cage motor and the primary of the wound-rotor motors. To obtain
the minimum required conductor ampacity, the full-load current is multiplied
by 1.25 [see 430.22 and 430.23(A)].
For the 25-hp motor,
34 A ? 1.25 = 43 A
For the 30-horsepower motors,
40 A ? 1.25 = 50 A
65 A ? 1.25 = 81 A

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)].

 

[ToolHound]

Not seeing your 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)].

Jumper. Thanks.

My question concerns the last part of the problem/solution, as copied here/above in your reply post--which part is the 'feeder protection calculation'. Now, for feeder protection calc...the sum of the largest branch ckt breaker and the other motors' currents....okay....all that makes sense.

But also, at first, I was scratching my head why no consideration is given to some load for the wound rotor/secondary circiut of each of the two 30 hp motors (i.e., why not in the feeder protection calculation add in some juice for the wound rotor/secondary circuit, of each 30 hp motor ?). But...I then sort of change my question to....is the wound rotor/secondary circuit/s powered, not by a separate power source, BUT RATHER BY INDUCTION ? So therefore no load for wound rotor/secondary need be included in the feeder protection calc ? (it gets its juice via induction). Is that a true statement more less you think ?

Thanks, ToolHound

 

[jumper]

Jumper. Thanks.

My question concerns the last part of the problem/solution, as copied here/above--which is the feeder protection calculation. The sum of the largest branch ckt breaker and the other motors' currents....okay....all that makes sense.

But also, at first, I was scratching my head why no consideration is given to some load for the wound rotor/secondary circiut (i.e., why not in the feeder protection calculation add in some juice for the wound rotor/secondary circuit ?). But...I then sort of change my question to....is the wound rotor/secondary circuit powered, not by a separate power source, BUT RATHER BY INDUCTION ? So therefore no load for wound rotor/secondary need be included in the feeder protection calc ? (it gets its juice via induction). Is that a true statement more less you think ?

Thanks, ToolHound

You lost me, I will go back and look, but no, the power source for a motor is not separately derived by induction if that is what your asking. AFAIK an induced voltage, ghost or phantom voltage, will not power anything. Any load will eliminate it.

I am not referring to the induced voltage on a secondary winding of a transformer, that is different.

 

[ToolHound]

You lost me...

Jumper. Thanks.

Ok, I will go at my question this way....

As given in the original problem/solution, from Annex-D, #D8, the nameplate secondary full-load current for each of the two(2) 30-hp motors is...65 A. From where does that 65 A current come ?

If the 65 A comes from somewhere other than from 'induction' then why is the 65 A, for each 30-hp motor, not separately included in the feeder protection calculation (i.e., the feeder protection calculation as laid out in the last step of the problem's solution included in Annex-D)? I'm probably not saying it the clearest way...but something like that.

Thanks, ToolHound

 

[Lectricbota]

Jumper. Thanks.

My question concerns the last part of the problem/solution, as copied here/above in your reply post--which part is the 'feeder protection calculation'. Now, for feeder protection calc...the sum of the largest branch ckt breaker and the other motors' currents....okay....all that makes sense.

But also, at first, I was scratching my head why no consideration is given to some load for the wound rotor/secondary circiut of each of the two 30 hp motors (i.e., why not in the feeder protection calculation add in some juice for the wound rotor/secondary circuit, of each 30 hp motor ?). But...I then sort of change my question to....is the wound rotor/secondary circuit/s powered, not by a separate power source, BUT RATHER BY INDUCTION ? So therefore no load for wound rotor/secondary need be included in the feeder protection calc ? (it gets its juice via induction). Is that a true statement more less you think ?

Thanks, ToolHound

Not sure what you're asking here.:?

 

[ToolHound]

Not sure what you're asking here.:?

Lectricbota. Thanks. My post #5 above was probably slow hitting the thread, and so the information in my post #5 above may be new input from me that you do not see yet. See my post #5 above and that may help say what I am trying to ask.

Thanks, ToolHound

 

[jumper]

65A is the table amps x 1.25

 

[jumper]

Not sure what you're asking here.:?

Me neither. Getting more lost by the minute.

 

[ToolHound]

I am not referring to the induced voltage on a secondary winding of a transformer, that is different.

Jumper. Thanks.

Okay, but the book ( Annex D, D8) refers to, quote,

"nameplate secondary full-load current" (for each 30-hp motor)

Is that 'secondary' the 'wound rotor', and is that a situation of 'induction'?

Thanks, ToolHound

 

[ToolHound]

Me neither. Getting more lost by the minute.

Yikes. I'm re-reading what you've posted, in case I've missed some point.

Thanks, ToolHound.

 

[ToolHound]

65A is the table amps x 1.25

jumper. Thanks.

The motor hp rating is 30 hp.
If I am seeing it correctly, from the table, for 30 hp 460-v, the current is...

40 Amps.

40 X1.25 = 50

Right ?

So the 65 Amps, identified by the book as...
"nameplate secondary full-load current" comes from where?
( I thought 'induction', but apparently I am the only one thinking that, so I must be wrong. ???? ).

Thanks, ToolHound.

 

[Lectricbota]

I see where your question is on the way (b) is worded- now I'm puzzling over that one.

D8 is a little confusing in how it is laid out.

Conductor Ampacity multiplies 65*1.25 to get 81?????

 

[hurk27]

You need to read up on how a wound rotor motor functions, this is why anyone who does not have an understanding of it won't know why you do not add the current developed in the rotor to the motors load.

Wound rotors motors are just a set of poles that are brought out to be shunted between each other through a bank of resistors something like a DC motor, but they have no reference to the circuit feeding the field of the motor, while inversely they do add to the motors load amp but only up to the point of the FLA of the motor, they actually decrease the current the motor pulls as resistance in inserted between the 3 rotor leads, so you only need to include the FLA of the motor as the rotor current is part of but does not add to the FLA.

 

[ToolHound]

Not seeing your question.:?

Conductor Ampacity
The full-load current value used to determine the minimum required conductor
ampacity is obtained from Table 430.250 [see 430.6(A)] for the
squirrel-cage motor and the primary of the wound-rotor motors. To obtain
the minimum required conductor ampacity, the full-load current is multiplied
by 1.25 [see 430.22 and 430.23(A)].
For the 25-hp motor,
34 A ? 1.25 = 43 A
For the 30-horsepower motors,
40 A ? 1.25 = 50 A
65 A ? 1.25 = 81 A

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)].

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

 

[jumper]

jumper. Thanks.

The motor hp rating is 30 hp.
If I am seeing it correctly, from the table, for 30 hp 460-v, the current is...

40 Amps.

40 X1.25 = 50

Right ?

So the 65 Amps, identified by the book as...
"nameplate secondary full-load current" comes from where?
( I thought 'induction', but apparently I am the only one thinking that, so I must be wrong. ???? ).

Thanks, ToolHound.

I screwed up, sorry, the secondary FLC was provided in the example.

I meant to address the needed ampacity of these conductors. 65 x1.25= 81 as shown.

430.23 Wound-Rotor Secondary.
(A) Continuous Duty. For continuous duty, the conductors
connecting the secondary of a wound-rotor ac motor to its
controller shall have an ampacity not less than 125 percent
of the full-load secondary current of the motor.

 

[jumper]

Conductor Ampacity multiplies 65*1.25 to get 81?????

Er, that is what I get on my calculator.

 

[ToolHound]

You need to read up on how a wound rotor motor functions, this is why anyone who does not have an understanding of it won't know why you do not add the current developed in the rotor to the motors load.

Wound rotors motors are just a set of poles that are brought out to be shunted between each other through a bank of resistors something like a DC motor, but they have no reference to the circuit feeding the field of the motor, while inversely they do add to the motors load amp but only up to the point of the FLA of the motor, they actually decrease the current the motor pulls as resistance in inserted between the 3 rotor leads, so you only need to include the FLA of the motor as the rotor current is part of but does not add to the FLA.

hurk27. Thanks.

Ok. So the 65 Amps of 'nameplate secondary full-load current' would rightly not be part of the 'feeder protection calculation', if I am understanding this correctly. ( Now, I was wondering if that 65 Amps comes by way of 'induction', and NOT from any "separate" circuit/source--just by way of induction. But nevertheless, if I understand the point, the point is...the 65 A is not to be included in the feeder protection calculation. ).

And....here I go to get a book on...wound rotor motors.

Thanks, ToolHound

 

[ToolHound]

sorry, the secondary FLC was provided in the example.

I meant to address the needed ampacity of these conductors. 65 x1.25= 81 as shown.

430.23 Wound-Rotor Secondary.
(A) Continuous Duty. For continuous duty, the conductors
connecting the secondary of a wound-rotor ac motor to its
controller shall have an ampacity not less than 125 percent
of the full-load secondary current of the motor.

jumper. Thanks.

You were thinking about the ampacity of the conductors. Got it. That makes sense.

--Thanks, ToolHound

 

[hurk27]

hurk27. Thanks.

Ok. So the 65 Amps of 'nameplate secondary full-load current' would rightly not be part of the 'feeder protection calculation', if I am understanding this correctly. ( Now, I was wondering if that 65 Amps comes by way of 'induction', and NOT from any "separate" circuit/source--just by way of induction. But nevertheless, if I understand the point, the point is...the 65 A is not to be included in the feeder protection calculation. ).

Thanks, ToolHound

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.