Hello guys,
Here is another nice question from behind the dikes:
If asynchronous motor has a rotor bar problem then will it be harmfull to the drive train on the instant of starting the motor?
What seems to be obvious is this:
The pole pass frequency will be 100 hz (here in Europe) that is for sure.
The drive train will be subjected to torsional vibration, at first the frequency will be 100 hz and the ppf will drop proportionally when speed ramps up, that is obvious too.
But how excessive these torsional vibrations can be at start?
Torsional resonance frequency of drive train is not yet known.
This is not theory: a 4 MW disc coupling collapsed!
I seek the advice of the profs in cyberspace as all handbooks let me down on this subject!
Who lends me a hand in estimating the ppf amplitude at starting?
Regards,
Arie Mol

Salient pole syncronous motor started by armotisseur windings are notorious for exciting torsional resonance. They generate torque oscilation at pole pass frequency (twice slip frequency) which is the same as an induction motor, but for different reasons (the reason for the sync motor has to do the torque produced by high reluctance pole pieces going in and out of the field).

During start, this torque oscillation pattern of the induction motor with broken bar (twice slip frequency), just like the torque oscillation pattern of the sync salient pole motor, is likely to excite any torsional resonance below 2*LF because it sweeps from 2*LF down to almost 0 and passes thru every frequency in between.

Attached is a very rough way of estimating the fractional variation. If we make the gross simplyfing assumption that current and flux in the unbroken bars are unaffected, then the torque variation for one broken bar is 2/Nbar where Nbar is the number of bars. For example with 80 bars, the torque variation is a fraction 2/80 = 1/40.

This fraction would multiply the locked rotor torque at start. Locked rotor torque might be in the ballpark 2.5 times full load torque. So 1/40 * 2.5 ~ 6% (the torque variation immediately after start expressed as percent of full load torque). If you only had 40 bars in the motor, then you might expect 12% (again expressed as percent full load torque). If two broken bars out of 40, then perhaps closer to 24%.

Again this is based on a simple model. Maybe others can suggest a better form.

Also I know the baker on-line instrument (and pdma?) can estimate torque vs time from the input power - maybe someone has measurements on a broken bar motor?

SimpleTorqueFromBrokenBar.xls (22 Kb, 13 downloads)

quote:

Originally posted by arie mol:
Who lends me a hand in estimating the ppf amplitude at starting?

It is a real challenge! I do not know the numerical answers. But there will be 3 torques involved:
a) The “forward” torque. This torque would be there even if the rotor were perfectly symmetrical.
b) Reverse torque: This torque appears only when the rotor is not symmetrical (if it has broken bar). In wound motors it appears when one rotor phase is open. There is a talk about the Gorges phenomenon. The torque of a motor with an open bar seems to increase up to the ½ speed. At speeds above ½ of synchronous, the torque will decrease. There is a noticeable bump on the speed – torque curve at ½ of the speed (with broken bars).
c) Then there is an oscillating torque, that oscillates at a double of the slip frequency.
d) Instead of one torque we have 3 of them.
e) In other words, the ppt frequency changes as the motor accelerates.
f) The oscillating torque makes a mess of the otherwise clear picture.
I wish I had the numerical data!
jank

The torque of the motor with broken bars is an interesting subject. I have been searching for an exact solution for a while, so far without any success. In the following ppt file I am trying to show the currents and the speed-torque curve during the acceleration of the motor to full speed. Also shown are the currents and torques in the generator region up to the slip of –0.5.
The derivation of the equations was based on this fact: The stator field induces voltages into each bar depending on the slip and the strength of the stator field. If there are broken bars, the voltage is induced into the bar, but the current cannot flow. If there are 40 bars on the rotor and two are broken, we no longer have a symmetrical 40-phase system. But we can break this nonsymmetrical system into 2 symmetrical ones in a similar way we do it in the 3-phase system. The “forward” system is there even when there are no broken bars. If the rotor spins with slip s, this system spins with slip –s relative to the rotor, so –s+s=0, zero speed (relative to stator rotating field), in other words the same speed as the rotating field created by stator. Hence, those two fields can lock like permanent magnets and create a steady torque, which is called in the ppt file the “forward torque” (red).
The other rotor system spins in the opposite direction relative to rotor than the forward system. It means it spins at slip speed +s relative to rotor and 2s relative to stator Since it does not spin the same speed as the stator field it cannot create a steady torque. By interacting with the stator rotating field it creates an oscillating torque (“reverse torque” – green in ppt).
At the same time, this reverse field induces currents into the stator. Those currents are of different frequencies from the stator frequency. It is assumed that the system feeding the motor is much larger than the motor itself (in terms of kVA). It means the currents of different frequencies can circulate through the feeding system with little or no impedance presented to the currents.
The rotor reverse current can lock with those stator induced currents and create a steady torque (“stator reverse torque”). This is a motor “inside out”.
The currents and torques (calculated under simplified conditions) are shown in the attached ppt file. Note:
a) The bump on the current curves (total and forward) in the area of ½ speed [s=0.5] (black and red).
b) The reverse current is zero at s=0.5 and s=0 (blue).
c) Decrease of the total torque (black) around ½ speed.
d) The stator reverse torque (blue) goes from the motor mode to generator mode at ½ speed.
e) The reverse torque (green) is the oscillating torque. It means that the total torque may be anything shown by the black curve +/- the green curve. The black curve is just an average.
I do not pretend that the curves are accurate, but they show in principle what is happening.
jank

This message has been edited. Last edited by: jank,

torques_with_broken_bars.ppt (346 Kb, 16 downloads)