Detecting Broken Rotor Bars using Vibration Analysis

[NRaaum]

I am looking for a method and explanation to detect broken rotor bars using vibration. Everything I've read indicates to look for Pole pass freq around turning speed or multiples of turning speed. I just don't understand how a broken rotor bar would generate this spectrum. I understand that at slip freq. there would be a vibration since the stator field rotates over the broken rotor bar, but I don't understand how this would be picked up as a sideband.

 

[jbartos]

Suggestion: Try the advanced search in this Forum. There are some postings ready for your reviews.

 

[GusD]

hello NRaaum

I will give it a shot and hopefully somebody can jump in and offer a more theorectical approach to the question.
I don't know if you have already used other methods, other than vibration analyses , to detect Rotor Bar Problems.The reason I say this,itis because when one looks at Motor Current Signature Analyses spectra,the rotor bar fault appears much more clear than on a Vibration Spectra.
On the MCSA the fundamental 2LF is the carrier frequency and the Fp is the modulated Freq.The extent of the problem is directly related to the # of SBands and their Amplitude in relation to the fundamental 2FL.
The vibration spectra shows much the same patterns but the Fp side bands are of much smaller amplitudes in relation to the carrier Freq.in this case running speed or its Harmonics.
A broken rotor bar/bars interrupts the flow of current and consequently distorts the rotor magnetic field.These forces behave like a mechanical umbalance of the rotor causing the Hi 1x running speed vibration.The lower amplitude PPFreq sidebands just ride the wave of the 1x Fundamental.

Hope it helps.



GusD

 

[NRaaum]

Thanks for tips, I have not used MCSA but I will look into it. I understand that the broken rotor bar will distort the rotors magenetic field and cause a hi 1x, but I do not understand how the ppf is picked up around one 1x. I think my lack of understanding may be in how the fft is performed by the analyzer. I am really not familiar with fourier(sp)transforms, but I would think if the sample time was long enough ppf would be picked up at its freq. and not as sidebands. What am I missing?

 

[bmodrow]

This link may shed some more light on the subject for you.

http://www.vibrationschool.com/vibeschool/mans/SpecInter/SpecInter32.htm

 

[electricpete]

Here is your question as I understand it.

You are aware that a given spot on the rotor passes into and out of maximum field at a frequency of pole pass. You want some scientific explanation for the meaning of pole pass frequency sidebands around 1x, 2x etc.

1X +/-FP frequency pattern can correspond to either of two time patterns.
1 - amplitude modulated: cos(2*pi*1x*t)*cos(2*pi*Fp*t) - simple trig manipulation shows this has frequency components at 1X+ Fp and 1X - Fp.
2 - Frequency modulated: cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t>. If you assume m<<1 those same frequency components can be shown algebraically. Let me know if you want the proof... it's a little messy.

So now we can relate the frequency patterns to time patterns, we still need to know the physical mechanism which produces those time patterns. I know of three mathematically and physically possible explanations, although I don't know which are dominant in real life:

A - Torque oscillation due to bar passing in and out of max field results in actual speed oscillation => frequency modulation (item #2). However I have seen several motors with pole pass sidebands and no discernible speed oscillation evident under strobe. I think in presence of broken rotor bar there definitely would be torque oscillation present but it may not have any significant effect due to machine inertia.


B - Torque on the rotor is physically produced by two equal and opposite tangential forces 180 degrees apart on the rotor. In normal rotor there is no net radial force produced by these forces since they are equal and opposite. However the broken bar is uncapable of producing torque. The bar across 180 degrees across from broken bar has a tangential force which is not radially balanced by equal/opoosite tangential force at position of broken bar. Net result is a radial force (in addition to torque). The magnitude of that radial force changes at Fp as that 180-opposite bar moves in and out of the field. The direction of that radial force changes at 1x as the rotor rotates. Taking for example the x-direction component we will have something like cos(2*pi*1x*t)*cos(2*pi*Fp*t). The cos(2*pi*1x*t( part comes from taking horiziontal component of rotating radial force. The cos(2*pi*Fp*t) part comes from the 180-opposite bar passing in and out of the field. This time pattern gives frequency pattern 1X +/-FP. Real life is a little more complicated because the bars adjacent to broken bar carry more current but I think even when you account for this the conclusion is the same.

C - There is an assymetry in the airgap flux. It rotates in position with the broken bar. It changes magnitude according to Fp. Resulting radial force cos(2*pi*1x*t)*cos(2*pi*Fp*t) by similar logic as B above.

I have not heard a definitive explanation on this subject. These are just some thoughts. Interested to hear other thoughts or comments.

 

[jbartos]

Suggestion: Reference:
EPRI, article &quot;Non-invasive Detection of Broken Rotor Bars in Operating Induction Motors&quot;

Fr = motor rotational frequency
LSB1 = lower slip sideband of line frequency
Lf = fine frequency
NP = number of poles
SF = slip frequency

the following formulae may be used:

LSB1 = Lf - (NP * SF)

SF = (2 * Lf/NP) - Fr

Figure 1 in:

http://www.compsys.com/drknow/aplpa...45f2f1731b3a66c7852565a20066ef3a?OpenDocument

shows data from a four pole motor with a rotational frequency of 29.74 Hz and a line frequency of 59.98 Hz. Using the given formulae we obtain:

SF = (2 * 59.98Hz/4) - 29.74 = .25 Hz

LSB1 = 59.98 Hz - (4 * .25 Hz) = 58.98 Hz

Note the LSB1 sideband one hertz below line frequency at 59 Hz.

 

[engtip74]

To electricpete:
I would appreciate it very much if you suggest any reference for the below statement you made, specially the proof of frequency modulation (item#2):

1X +/-FP frequency pattern can correspond to either of two time patterns.
1 - amplitude modulated: cos(2*pi*1x*t)*cos(2*pi*Fp*t) - simple trig manipulation shows this has frequency components at 1X+ Fp and 1X - Fp.
2 - Frequency modulated: cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t>. If you assume m<<1 those same frequency components can be shown algebraically. Let me know if you want the proof... it's a little messy.

Thank you

 

[dgallagher]

Here's a link to an article presented at this years Tubomachinery symposium on Motor Current Signature Analysis with several case histories:

http://turbolab.tamu.edu/pubs/Turbo32/T32chapter15.pdf

 

[jbartos]

Suggestion: Reference:
Herbert Taub, Donald L. Schilling &quot;Principles of Communication Systems,&quot; McGraw-Hill Book Company, 1971,
Chapter 3 Amplitude-Modulation System
Chapter 4 Frequency-Modulation System
4.2 Phase and Frequency Modulation
The waveform or analytical expression for the waveform of the modulating signal is supposed to be known to be able to make a distinction between the frequency modulation and the phase (or angle) modulation.

 

[electricpete]

Proof of item (1):
cos(w1*t)cos(w2*t)=0.5*[cos<(w1+w2)>*t +cos<(w1-w2)*t>]

Proof of Item (2):
Let Wr be the radian frequency of the rotor and Wp be the radian frequency associated with pole pass. Assume we have torque oscillations causing speed oscillations so that position (angle) of rotor is given by theta(t) = cos(Wr*t+m*sin(Wp*t))
where m (known as moduation index in electrical circles) represents magnitude of frequency oscillation as a fraction of Wr. We can use trigonometry to rewrite it as theta(t)=cos(Wrt)*cos(msinWpt)-sin(Wrt)*sin(msinWpt). If m is small (<<1) then we approximate cos(msinWpt)~1 and we approximate sin(msinWpt)~msin(Wpt). Substituting into above equation for theta we get theta(t)~cos(Wrt)-msin(Wpt)sin(Wrt).

The second term above, sin(Wpt)*sin(Wrt), is the amplitude-modulated form described in item (1) above, which gives rise to pole pass sidebands around rotor speed.

Sorry it is tough to write equations in text. Let me know if the above is not clear.

 

[electricpete]

Instead of theta(t) on left hand side of equations above I should have said cos(theta(t)) on left hand side throughout.

It would correspond to having an accelerometer mounted in the horizontal direction. For normal machine with no FM, theta(t)=w*t, the accelerometer sees cos(theta(t))=cos(w*t). For oscillating motor with theta(t) = cos(Wr*t+m*sin(Wp*t)), the accelerometer sees cos(theta(t). That is the quantity on the right hand side which with assumption m<<1 comes out to be cos(Wrt)-msin(Wpt)sin(Wrt).

==============================
I will rewrite the whole thing for clarity:
FM form:
theta(t) = Wr*t+m*sin(Wp*t)
cos(theta(t) = cos(Wr*t+m*sin(Wp*t))

Use trig to rewrite it as cos(theta(t))=cos(Wrt)*cos(msinWpt)-sin(Wrt)*sin(msinWpt).

If m is small (<<1) then we approximate cos(msinWpt)~1 and we approximate sin(msinWpt)~msin(Wpt).

Substituting into above equation for cos(theta) we get
cos(theta(t))~cos(Wrt)-msin(Wpt)sin(Wrt).

 

[jbartos]

Suggestion: electricpete (Electrical) Dec 28, 2003 marked ///\\Proof of item (1):
cos(w1*t)cos(w2*t)=0.5*[cos<(w1+w2)>*t +cos<(w1-w2)*t>]
///t is supposed to be in the cos argument, i.e. cos(w1*t)cos(w2*t)=0.5*[cos<(w1+w2)*t> +cos<(w1-w2)*t>]\\\

 

[jbartos]

Comment on electricpete (Electrical) Dec 23, 2003 marked ///\\
2 - Frequency modulated: cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t>. If you assume m<<1 those same frequency components can be shown algebraically. Let me know if you want the proof... it's a little messy.

///Since m is the frequency modulating function, the expression cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t needs a clarification, namely,
cos(2*pi*Fp*t) can be expanded into series:
cos(2*pi*Fp*t) = 1 - (1/2!)x [cos(2*pi*Fp*t)]^2 + (1/4!)x [cos(2*pi*Fp*t)]^4 - ....
Then,
m x {1 - (1/2!)x(2*pi*Fp*t)^2 + (1/4!)x(2*pi*Fp*t)^4 - ....}=
m - m x (1/2!)x(2*pi*Fp*t)^2 + m x (1/4!)x(2*pi*Fp*t)^4 - ....
It is evident that m modulates the frequency Fp as well as constant 1. This means that there is the frequency modulation and the phase (angle) modulation since there is a stand alone term &quot;m&quot;, which is multiplied by t when substituted back into:
cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t>. Namely,
cos<2*pi*1x[1 + m - m x (1/2!)x(2*pi*Fp*t)^2 + m x (1/4!)x(2*pi*Fp*t)^4 - ....
]*t>.\\

 

[electricpete]

hi jb
Your first comment: yes, there is an obvious typo and the meaning should be clear to anyone.

Your second comment: I think you have an obvious typo of your own. You wrote:
&quot;cos(2*pi*Fp*t) can be expanded into series:
cos(2*pi*Fp*t) = 1 - (1/2!)x [cos(2*pi*Fp*t)]^2 + (1/4!)x [cos(2*pi*Fp*t)]^4 - ....&quot;

It should be instead:
&quot;cos(2*pi*Fp*t) can be expanded into series:
cos(2*pi*Fp*t) = 1 - (1/2!)x [(2*pi*Fp*t)]^2 + (1/4!)x [(2*pi*Fp*t)]^4 - ....&quot;

Beyond the typo's, I don't understand the point of your 2nd message. Do you disagree with the result? Or do you disagree with the assumed math form of the vibration? In the 2nd case, you are welcome to suggest a more appropriate assumed form in the presence of torque oscillating at Fp.

 

[jbartos]

Comment on the previous posting: Thanks for the correction. The first expansion is to be without cos function, only the argument as you indicated. However, the other derivations are correct. I forgot to erase cos in each term.
My point is this:
The modulation index m is applied to frequency Fp as well as to phase (angle) where there is not frequency. It may happen that the phase modulation is negligible with respect to frequency modulation. In that case the m terms pertaining to the phase modulation may be neglected.
I notice that there is an assumption stating &quot;If m is small (<<1) then we approximate cos(msinWpt)~1 and we approximate sin(msinWpt)~msin(Wpt).&quot; This would imply that
cos<2*pi*1x[1 + m - m x (1/2!)x(2*pi*Fp*t)^2 + m x (1/4!)x(2*pi*Fp*t)^4 - ....
]*t>
would become:
cos<2*pi*1x[1 - m x (1/2!)x(2*pi*Fp*t)^2 + m x (1/4!)x(2*pi*Fp*t)^4 - ....
]*t>
thus covering the frequency modulation only.
Beside the above statements, the variable t outside of rectangular bracket in the cos argument:
cos<2*pi*1x[1+m*cos(2*pi*Fp*t)]*t>
also needs a clarification since the typical Wideband Frequency Modulation (FM) has a form:
fc(t)= A x cos[wc x t + (a x kf / ws) x sin(ws x t)]=
= A x cos[wc x t + mf x sin(ws x t)]
where
mf is the modulation index.
See Reference:
B.P. Lathi &quot;Signals, Systems and Communication,&quot; John Wiley & Sons, Inc., 1965.
page 495, Equations 11.55 and 11.56.