Stators and rotors

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gar

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I have not paid close attention to the above discussion, but there is a considerable discussion in "Alternating-Current Machinery" by Bailey and Gault, McGraw-Hill, 1951, on wound-rotor induction motors used for frequency conversion. Part of the discussion was on control of a 40,000 HP motor for a wind tunnel at Wright-Patterson.

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Ingenieur

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Sahib said:
Then rotational speed N of rotor field with respect to rotor N=sxfs/P, where P= number of pole pairs same as stator's. But fs=ns*P. Substituting we get
N=ns-nr. Here ns-nr is rotational speed of rotor field with respect to rotor. But rotor itself rotates at a speed of nr. So rotational speed of rotor field with respect to motor frame is (ns-nr)+nr=ns same as that of stator field.
Click to expand...

that makes no sense
n = s fs / P
fs = ns P
n = s (ns P) / P = s ns
not ns - nr


The speed of rotation and the induced rotor i freq are 2 different things
the freq is different stat vs rot
otherwise it could not be used as a freq convertor
a common applicaton prior to the advent of ss vfd's
 
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winnie

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Ingenieur is describing a technique of using a wound rotor induction motor as a frequency changing device, where the output from the rotor is the altered frequency.

A more modern application which operates in the reverse direction is the 'doubly fed induction machine' where the stator is connected directly to the mains and the rotor is powered by a variable speed drive.

http://www.mondragon.edu/es/eps/inv...1/archivos-de-thesis/Tutorial DFIM Part 1.pdf

My understanding is that this lets a much smaller VSD be used to control a given load, akin to controlling the excitation of a wound rotor synchronous machine, but allowing the rotor to spin at other than synchronous speed.

-Jon
 
M

Meterman Eng

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Sahib, I agree totally. It is the current (and induced voltage) in the cage bars that varies at a different frequency. I did not technically say otherwise, and hopefully did not open a can of worms. The slip-speed changes the frequency of the voltage and current in the rotor. Still, the rotor and stator magnetic fields spin about the axis of the motor at the same frequency.
 
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Ingenieur

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3 ph
60 Hz
s 2.5%
8 pole (4 pole pair)

ns = 60 120 / 4 = 1800 rpm

nr = (1 - 0.025) 1800 = 1755 rpm

rotating mag field cuts rotor bars
= 1800 - 1755 = 45 rpm or 0.75 times / sec

rotor induced emf freq (i is the same)
= 0.75 rev/sec 4 pp cycles/rev = 3 cycles/sec or 3 Hz
almost a DC field

Or slip x elec ns x pp = 0.025 x 30 x 4 = 3 Hz
 
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M

Meterman Eng

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Really, it depends on how you look at it (no really, the frame of reference). Yes, a 3 Hz current wave is induced in the cage bar. This creates a 3 Hz magnetic field (the less slip, the less voltage and frequency induced in the rotor) that would rotate around/through the squirrel-cage (along the bars).. When you spin the cage and look at it from the end, there is an outward-pointing rotatating magnetic field that is rotating in synchoronous with a speed of
nr = 120 * applied frequency/Poles, lagging the applied voltage. It is this rotor magnetic field and the stator's magnetic field that couple and induce torque on the motor. Right...? All said, what is really important is spinning reserves.
 
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Ingenieur

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A synchrnous machine has an externally excited dc field in the rotor
it's freq is 0 (obviously)

the fact that it spins and induces a current in the stator/armature with an elec freq
= pole pairs x rpm / 120
does not mean the rotor i's freq is anything but 0 does it?

Agreed, reference frame matters, but if we don't understand what is actually going on we can't appreciate the beauty of the machine
if we look at it from the stator the rotor field has a freq
from the rotor it has none
 
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GoldDigger

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Sahib said:
But s=(ns-nr)/ns. Substitute and see that n=ns-nr.
Click to expand...

I do not see how you take ns and nr to be field rotation speeds for rotor and stator and get to multiply s by ns to get n.
Maybe I do not understand the variables, or maybe it is just too obvious for me to understand.
 
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Ingenieur

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Sahib said:
But s=(ns-nr)/ns. Substitute and see that n=ns-nr.
Click to expand...

what does that even mean?
1800 - 1755 = 45
so?
that is the speed of the rotor i (or elec n - mech n delta)
45/60 = 0.75 rev/sec
0.75 x 4 pp = 3 Hz

the rotor and stator i freqs are not the same
 
Sahib

Sahib

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Ingenieur said:
what does that even mean?
1800 - 1755 = 45
so?
that is the speed of the rotor i (or elec n - mech n delta)
45/60 = 0.75 rev/sec
0.75 x 4 pp = 3 Hz

the rotor and stator i freqs are not the same
Click to expand...
Are you saying there is no rotor rotational field?
 
Sahib

Sahib

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Ingenieur said:
is the reference frame the stator or rotor?
is the motor mntd vertically or horizontally?
Click to expand...

For your first question, the answer is already in post18: rotor field rotation with respect to rotor. But if viewed from motor frame, its rotation speed same as stator's. For your second question I think mounting does not matter.
 
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Ingenieur

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Sahib said:
For your first question, the answer is already in post18: rotor field rotation with respect to rotor. But if viewed from motor frame, its rotation speed same as stator's. For your second question I think mounting does not matter.
Click to expand...

Are you saying in an induction motor, operating with 2.5% slip, that the electrical (vs mechanical) speed of the rotor and stator are the same?
 
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