Calculating half frequency whirl ?

[VibFrank]

Hy,
a longer question..
We measure the transient bearing loads of a vertical rotor with low bearing load. We also see, that a half frequency whirl occurs. If I take the load and calculate with this load the linear stiffness and damping coefficients of the bearing and then the damped natural frequencies of the machine, I really get a (low) damped natural frequency near half frequency and the mode shape looks like a stiff shaft, rotating in the bearings.
I do not understand this. In my understanding, this nonlinear behaviour (halffrequency whirl) cannot be simulated with linear calculated stiffness and damping coefficients. But the linear calculation shows them and even at (nearly) correct halffrequency. Why ? Accident ?
Thanks a lot.

 

[electricpete]

I believe that the frequencies of whirl and other instabilities in a sleeve bearing machine can be calculated by eigenvalue analysis of a sufficiently detailed linearized system model (including cross coupling).

Coming up with the correct bearing coefficients is one of the more difficult parts (as far as I understand.... I have never done that calculation)

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[electricpete]

There is some discussion in the article “FUNDAMENTALS OF FLUID FILM JOURNAL BEARING OPERATION AND MODELING” Here”

http://turbolab.tamu.edu/pubs/Turbo34/T34pg155.pdf

As shown in Figure 19, the fluid film can be represented by
springs and dampers. The static load W, such as gravity, establishes
the journal’s steady-state equilibrium position. Then, some dynamic
force, such as rotor unbalance forces, pushes the journal away from
its equilibrium and causes it to whirl on an elliptical orbit (Figure
20). To have an acceptable vibration level, the orbit’s size must be
relatively small compared to the bearing clearance. When this is the
situation, the vibration is said to be in the linear range and the film
dynamic forces are directly proportional to the displacements and associated velocities
This relationship is given by....

where the Kij and Cij are called linearized stiffness and damping
coefficients, respectively. In other words, at an instantaneous
journal position, the horizontal and vertical forces due to the oil
film can be obtained by expanding Equation (7)....

where the negative sign implies that the force is acting on the rotor.
To justify the use of linearized dynamic properties, one should
understand the other situation where the vibration levels are relatively
large. Figure 21 shows the classic example of an unstable
shaft where the orbit nearly fills up the entire bearing clearance.
Here the rotor has almost reached the so-called “limit cycle.” Since
this motion is large relative to the bearing clearance, linearized
coefficients are inadequate to represent the film dynamic forces.
Therefore, they cannot be used to predict the actual amplitudes for
such large vibrations. However, the strength of the linearized coefficients
is their ability to predict whether or not such unstable
vibrations will occur. This ability, combined with their accuracy in
predicting vibration amplitudes within the range of interest [up to
40 percent of the clearance according to Lund (1987)], enables
modern rotordynamics to be firmly based on their use.

I didn’t see them talk about the frequency of the whirl but I’m pretty sure this type of analysis would show the frequency of the whirl. Note it would not be exactly 1/2 frequency but slightly less.


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[electricpete]

I believe that the frequencies of whirl and other instabilities in a sleeve bearing machine can be calculated by eigenvalue analysis of a sufficiently detailed linearized system model (including cross coupling).

I should have mentioned that not all eigenfrequencies would correspond to whirl frequencies. Only those eigenvalues on or to the right of the imaginary axis. Damped whirl frequencies would die down in the absence of excitation at those frequencies.

The eigenvalue to on or to the right of the imaginary axis would not be possible in normal damped mass spring systems. But I believe the assymetric cross-coupled stiffnesses can cause it for sleeve bearing machines. There is more discussion of this on page 10 of 22 of the link above.

At least that's how I understand it. Anyone feel free to correct me.




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[WCFoiles]

Using linear coefficients does determine the onset frequency and speed. With a rigid mode shape for the rotor, the rotor model does not have to be too detailed for obvious reasons. However, not all sub-synchronous instabilities have rigid shapes.

With a vertical rotor the support flexibility can influence both the frequency and the stability status (stable or not). Support flexibility can lower the stability threshold or onset speed of an instability. A dynamically asymmetric support has been shown to help in some situations.

This low frequency resonance may (should) be present even if the system is stable. For low log dec calculations consider system changes like the bearing. If you have a plain circular geometry bearing, this is almost certain to have stability issues if unloaded (vertical rotor no side loads).

Regards,

Bill

 

[electricpete]

So maybe worthwhile to talk about three types of possibilities:

1 – An unstable eigenvalue can vibrate in the absence of any excitation at that frequency. I believe oil whirl is an example of this.

2 – A stable eigenvalue would only vibrate only in the presence of excitation (possibly broadband) at that frequency. An example would be a resonance or critical speed.

3 – Bill mentioned onset-type vibration which don’t occur until a certain speed is reached. I believe in this case typically the whirling frequency is the rotor critical, and the onset machine speed for this whirling speed is somewhere just above twice rotor critical. As machine speed increases above this onset speed, the whirl frequency does not increase further but instead remains constant at rotor critical speed. Oil whip would be an example of this.

Again i am open to comments/corrections.


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[WCFoiles]

Either whirl or whip (if you differentiate) have an onset rotor speed (or may just be stable). If one calculates this for an unloaded (vertical) machine, that onset speed could be nearly 0 rpm.

If one has made a distinction between whirl (rigid body mode for rotor - stiff rotor at this condition) and whip (flexible rotor mode), then upon reaching the onset speed the rotor goes unstable and vibrates at the eigenfrequency, below running speed (almost always). Increase the speed with whip and the whirl frequency increases (note different use of whirl here). With oil whirl increase the speed and the whirl frequency has small or no increase, because of the flexible mode the bearings approximate nodes (if not close to nodes speed increase may result in frequency increase somewhat).

Regards,

Bill

 

[electricpete]

On the left side of page 11 of 22 of the link above is a campbell diagram for whirl and whip.

They show whirl frequency increases with speed. Whip frequency (once above the speed threshhold) occurs at a relatively constant frequency (independent of speed) which corresponds to a shaft critical. Those are the dominant features I have heard described for whirl and whip.

In this particular case they also show a relatively stiff rotor shape for whirl and a flexible rotor shape with nodes at the end for whip which matches your terminology.

Since these vibrations apparently occur in the absence of exciting forces at those frequencies, wouldn’t they correspond to unstable or marginally unstable (on the j*w axis) eigenvalues of the system?


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[WCFoiles]

What one looks for in the analysis is positive real parts to the eigenvalue (right half plane). In practice one generally sets a minimally acceptable log dec for eigenvalues in the left half plane (nominally stable eigenvalues) to be conservative, e.g. API standards.

If one has a complete set of eigenvectors and some are on the imaginary axis these have neutral stability (they wouln't be BIBO stable - bounded input,bounded output). However, in theory (probably not in practice) one can have generalized eigenvalues on the imaginary axis; these are unstable, but not with the usual exponential growth.

Generalized eigenvalues on the imaginary axis grow polynomial in time. Ordinary right half plane ones grow exponentially with time for linear models.

Regards,

Bill

 

[electricpete]

I am used to the fact that typical m/k/c systmes have roots in the left half of the plane, and if c=0 then roots on the imaginary axis. I was a little surprised that the addition of the assymetric cross stiffness (unique to fluid film bearings but not typical of other damped mass/spring systems) can push the eigenvalue toward instability and can even create unstable systems (roots in right half of the plane).

Here's my feeble attempt to demonstrate the destabilizing effect of asymmetric cross-coupling by examining a system with no damping (c=0). In absence of cross-stiffness terms, it is marginally stable (roots on the imaginary axis). As we increase asymmetric cross stiffness (relative to direct stiffness), it moves further and further into the right half of the complex plane. It may not represent a realistic system (due to the absence of damping), but it illustrates that the system gets more unstable as asymmetric cross stiffness increases in relation to direct stiffness.

The state vectors is X = [x,y]'
M is [m 0 ; 0 m]
K is [Kd Kc ; -Kc, Kd]
where m, Kd =Kdirect, Kc=Kcross are all positive constants, negative cross term is explicitly shown with a minus.

F = M*s^2*X +K*X = 0
M*s^2*X = - K*X
s^2*X = - Minverse * K*X
0 = [ -Minverse*K - s^2*Identity] * X
non-trivial solution for X exists only when det[ ] =0
det[ -Minverse*K - s^2*Identity] =0
multiply through by -1
det[Minverse*K + s^2*Identity] =0
det [ [1/m 0 ; 0 1/m] * [Kd Kc ; -Kc Kd] + s^2*Identity] = 0

det [ [ Kd/m Kc/m ; -Kc/m Kd/m] + s^2*Identity] = 0

det [ Kd/m+s^2 Kc/m ; -Kc/m Kd/m+s^2]

(Kd/m + s^2) ^2 + (Kc/m)^2 = 0

(Kd/m + s^2) ^2 = - (Kc/m)^2
take sqrt
(Kd/m + s^2) = j *(+/- Kc/m) (pure imaginary but two possible signs)

s^2 = j *(+/-Kc/m) - Kd/m

s = sqrt{ j *(+/- Kc/m) - Kd/m }

The quantity inside the curly brackets (call it "{ }") has an angle within (Pi/2, Pi) for the + sign selection and (-Pi/2,-Pi) for the sign negative selection. "(,)" indicates the endpoints are excluded.

s will be the square root of { } and will have an angle half as big as { }
s will lie within (Pi/4, Pi/2) for the + selection and (-Pi/4, -Pi/2).
Since again it does not include the endpoint, s lies in the right half of the plane.

If Kc is 0, { } is on the negative real axis and s lies on the j*w axis.
If Kc is very small in comparison to Kd, { } is just above the real axis and s is just to the right of the j*w axis.
As Kc gets larger, s moves further into the unstable region.

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[Zackman]

I've read your posts and the linked to article, please help me understand a little better. My apologies if I'm regressing this thread.

I've always thought that whirl could not occur at frequencies of 0.5X and above without some type of outside motivational force such as injection of pressurized fluid. Otherwise, the circumferential flow, due to the rotating shaft, around the bearing must be below 0.5X due to frictional losses.

What am I missing?

 

[electricpete]

I'm no expert but I agree with your comments.

it is my belief as well that oil whirl and oil whip cannot have a whirl frequency above 0.5x running speed. (although there may be other unstabilities that occur above 0.5x).

Here's my view (similar to yours)

The oil velocity at the stationary bearing wall is 0. The oil velocity at the shaft is 1x running speed. The average oil velocity is an average between these two. Since bearing wall is a little bigger than shaft wall, it will be a little less than 0.5 times running speed (let's say 0.48 times running speed).

If the shaft is whirling at anywhere less than average oil velocity (<~0.48x) , there is a rotating choke point at the narrowest gap. Since average oil velocity is above whirl velocity, there is net forward flow of oil through this gap, oil pressure builds up behind the gap and continues to push the shaft in the forward whirl direction.

If the shaft is whirling at anywhere above the average oil velocity (>~0.48x) , there is again a rotating choke point at the narrowest gap. Since net flow of oil is now backward through this gap, oil pressure builds up in front of the gap and tends to oppose the whirl of the shaft. I don't see how unexcited whirl can continue in this situation.

I'm interested to hear any more comments.

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[VibFrank]

Hy,

electripete:
I read your calculation und I think I understood it (one question: isn't there another squareroot at 180 degrees of "your" roots ? That one would be stable..)

I tried to make your calculation with a damped system, but I failed. Have you done such a calculation or do you know of such analyses ? perhaps a link or literature ?

Thanks a lot