Vibration Readings.

[Paw]

Let me start by saying I am not an expert on vibration analysis, however I need to analize readings taken of our machines. We manufacture large rotory equipment. I took vibration readings at the two bearings using an instrument made by Bently Nevada, Model #TK81. Operating the machine at 1200 RPM & 1800 RPM, the vibration levels in velocity (in/sec.) were measured. I then switched to the displacement mode and took readings (in. pp). At 1800 RPM (30 Hz.) I have a velocity of .469 in./sec. and a displacement of 2.35 in. What is the mathamatical relationship between these two values? All of my readings come out to about a 5x factor between these two values. This doesn't correspond to any formulas I could find on the subject.

 

[manija]

Let's assume that the vibration displacement is a pure sine <br>
x(t)=A * sin(wt), the first derivative gives the velocity<br>
of the vibration that's is :v(t)= A * w * cos(wt)<br>
So the relationship you are looking for is the rotational speed w (measured in rad/sec), the velocity amplitude V= w * the displacement amplitude<br>
For 1800 RPM, w= 30 * PI, so V= 94.2 * D<br>
If D= 2.35 mils (not inches)<br>
V= 94.2 * 0.0023 = 0.216 in/sec <br>
Other aspect that you have to consider is that some equipment measures what is called peak to peak amplitudes and other only peak values, so if you multiply by 2 the velocity you get 2* .216 = .432 (that's almost what you get in your measurement)<br>
On the other hand if your vibration is not a pure sine but a combination things get a little more complex because you have to consider all the components to make the conversion, but if one frequency is predominant a quick conversion is the one described.

 

[VibrationSpecialist]

The mathematical relation stated above by manija is fine. You need to check that you are reading filtered values or direct values. Filtered values is a narrow band values of the overall input. It can be estimated as the single component peak, i.e. the sine wave form at that frequency. This makes the relation valid. Direct or unfiltered signal will not give correct conversion between the Velocity and the displacement. The second thing you need to check is the amplitude type if it is pp (Peak-to-Peak) or p (Aero-to-Peak). <br>
To answer your question:<br>
· at 1800 rpm Velocity of 0.469 in/sec peak (Filtered) is equivalent to 4.97 mils (0.00497 inches) Peak-to-Peak.<br>
· at 1200 rpm Velocity of 0.469 in/sec peak (Filtered) is equivalent to 7.46 mils (0.00746 inches) Peak-to-Peak.<br>
I hope this is good enough.<br>

 

[dgallagher]

Check out the app note &quot;Comparing Vibration Readings&quot; at

http://www.reliabilitydirect.com/appnotes/cvr.html

 

[Hatch]

The bottom line is that your readings appear to be high. The few times I have measured large displacements, I could actually see the movement of the bearing. You have an indicated displacement of 2.35 inches, I would expect that you can see it (motion) with the naked eye. If not, then there may be one of three reasons.

1. The instrument calibration is off.
2. The transducer was rocking or loose or broken.
3. There is a DC (bias)component (ie. something is wrong with your unit or transducer) in the signal.

You should try and identify the peak amplitude at whatever frequency it occurs at (probably 30 or 60 Hertz) and get the velocity and corrosponding displacement reading. The overall reading which you have taken is good only as a general indicator. Filtered readings will be necessary for problem diagnosis. Also if the filtered readings are much lower in amplitude than the overall readings it is another indicator of a DC bias problem. A spectrum would answer all the questions.

 

Am I missing something here? I thought that w (omega) was equal to TWO times pi times the frequency. Therefore, w in this case should be w = 2*pi*30 = 188.5 radians per second. Therefore, a peak velocity of 0.469 inches per second corresponds to a zero-to-peak displacement of x = V/w = 0.469/188.5 = 0.002 488 inches. I'll defer the other discussion to those who know more than I.

 

[VibrationSpecialist]

Dear Bright_Engineering,

You are right that
w = 2 * Pi * Frequency (in Hz)
Or
w = 2 * Pi * rpm/60.

Now for your example, you used 30 Hz, which is equivalent to 1800 rpm.
Doing the simple calculation then
the displacement is 0.002488 in (zero-to-peak)
which is equal to 2.488 mils (zero-to-peak)
and of course
Is equal to 4.976 mils peak-to-peak

Take Care ;-)