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Worked Examples Solutions

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CAT II Worked examples solutions Page 91

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PRINCIPLES OF VIBRATION

1. The peak amplitude of the waveform below is 6

The Peak to Peak amplitude is 12

The RMS amplitude is 6 x 0.707 = 4.24

2. The peak amplitude of the waveform below is 9

The Peak to Peak amplitude is 16

The RMS amplitude can be calculated from the analog signal, digitally or from the spectrum. This is not a sine wave so the RMS is not 0.707 x pk

3. The period of the waveform below is 0.1 seconds

The frequency in Hz is: 1 / 0.1 = 10 Hz

The frequency in CPM is: 10 Hz x 60 = 600 CPM

4. The definition of the period is: The time it takes to complete one complete cycle of vibration, usually measured in seconds or milliseconds.

5. The definition of Hz is: How many cycles per second

6. The definition of CPM is: How many cycles per minute

7. The relationship between the period and the frequency in Hz is

Period (seconds) = 1 / Frequency (Hz)

Frequency (Hz) = 1 / Period (seconds)

8. One cycle of vibration is 360 degrees. It is the same as one revolution of a shaft.

9. The most common convention for measuring phase begins with the reference (the arrow) and the measures to the right to the next peak of the wave.

Page 92 CAT II Worked examples solutions


10. Phase is a measure of the relative timing of two events, therefore, in order to measure phase you need “two things.” Absolute phase is measured between a vibration sensor and a phase reference such as a tachometer, strobe light or Keyphasor™. These phase references utilize reflective tape, or keyways on the shaft. Absolute phase must be used for balancing. Relative phase is measured directly between two vibration sensors such as accelerometers or proximity probes. A benefit of relative phase measurements is that they can be easier to setup. Since no reference is necessary in the shaft, it is not necessary to shut the machine down to install a reflective tape or to set up a tachometer.

11. Just like a clock, look at the angle that the line is pointing in. Up is 0, right is 90, down is 180 and left is 270 degrees.

12. What are the phase angles of the two graphs below? 90 and 270 degrees. Start from the tach pulse and measure to the next peak to the right. The distance between two peaks in a sine wave (or between two tach pulses) is one cycle which is 360 degrees.

They are 180 degrees out of phase. When one point is moving up, the other is moving down.

You can use a Keyphasor or a strobe as a reference instead of an optical or laser tachometer. A relative phase measurement can be made using a two channel data collector and two vibration sensors



13. Phase is used for rotor balancing (although a method for balancing exists that does not require phase).

Phase helps one understand how different test points move relative to one another as it can be used for structural analysis.

Phase is used in resonance testing.

Phase can be used to detect common machine faults such as unbalance, misalignment, rotating looseness (where phase is erratic), bent shafts and cocked bearings.

14. Displacement for low frequencies

Velocity for mid frequencies

Acceleration for high frequencies.

These can be measured directly using the following three types of sensors proximity probes (eddy current probes), velocity probes (velocimeters or velometers) and accelerometers respectively.

15. It is possible to collect data in acceleration and convert it to velocity or displacement via a process called integration.

16. When converting from acceleration to velocity, the phase: b. Changes by 90 degrees

17. How far something is moving acceleration

How fast something is moving displacement

How quickly something is speeding up or slowing down velocity

18. in/sec

mm/sec2 acceleration

g’s displacement

mils velocity

microns

mm/sec

19. Please match each unit to the one that is typically used by convention

a. in/sec pk

b. mm/sec2 RMS

c. g’s pk for imperial and RMS for metric

d. mils pk-pk



e. microns pk-pk

f. mm/sec RMS

20. High frequencies (above 2000 Hz) - Acceleration

Mid frequencies (10 – 2000 Hz) - Velocity

Low frequencies (below 500 Hz) - Displacement

21. First note the key at the bottom of the formula sheet to determine which unit each of the variables is in and to select the correct formulas (metric or imperial). Note that the metric formulas are in the box on the right and in the key at the bottom.

F = frequency in CPM. The frequency given is 100 Hz, so first convert this to CPM using the formula: CPM = Hz x 60.

F = 100 Hz x 60 = 6000 RPM

Note in the key that Acceleration “A” is given as g’sRMS. You were given the acceleration as 1 gRMS so you can say: A = 1 gRMS. You want to convert to mm/secRMS – note in the key that “V” is given as mm/secRMS.

Therefore, you have “F” and you have “A” and you want to calculate “V”. Find a formula in the box on the right that has “V = ” on the left and F and A on the right side of the equals sign. The top right formula works.

V = 93712 A / F

Enter the values for A and F and solve the equation: V = 93712 x 1 / 6000 = 15.62 mm/secRMS

Now convert to microns pk-pk.

Looking at the key it says that “D” is in microns pk-pk. Now find an equation with “D =” on the left side and F and A on the right side. You could also look for an equation with F and V on the right side but you may not want to do this in case you made a mistake when you calculated V in the last part of the problem – that said the equation with F and V on the right side looks easier to solve, so let’s solve that one. You could also solve both to check your answer.

D = 27009 V / F

D = 27009 x 15.62 / 6000 = 70.3 µmpk-pk

Convert 0.5 in/secpk at 1800 RPM to mm/secRMS

This involves two steps. The first is to convert from imperial units (inches) to metric units (mm). The second step is to convert from pk to RMS. Use the formula: in x 25.4 = mm: 0.5 x 25.4 = 12.7 mm/secpk

To convert from pk to RMS use the formula: RMS = 0.707 x pk: 0.707 x 12.7 mm/secpk = 8.98 mm/secRMS

Convert 1 in/secpk at 3600 RPM to in/secRMS



When dealing with a single frequency, the vibration is a sine wave, therefore you can use the formula RMS = 0.707 x pk: 1 in/secpk x 0.707 = 0.707 in/secRMS

22. Convert: 1gRMS at 10 RPM to in/secpk

Use the top right formula in the left box where A = 1 gRMS (given) and F = 10 RPM (given) and according to the key, V = in/secpk which is what you want to solve for.

V = 5217 A / F

V = 5217 x 1 / 10 = 521.7 in/secpk

Typical vibration readings in in/secpk have values around 0.2 or 0.5, so 521 is far, far bigger than this. Assuming there are no forcing frequencies of interest anywhere near 10 RPM, you want to filter this vibration out so that you can focus on the vibration of interest which has far, far lower values. Typically a 10 Hz high pass filter or “low cutoff frequency” is used for this (this is a setting in the software or data collector). It is like taking a photo of your friend standing next to a skyscraper and having to get the whole skyscraper in the photo even though you are only interested in photographing your friend.

23. Please describe the ISO RMS overall level

a. 2 – 1000 Hz

b. It is a simple measurement that tells you how much energy is in the vibration

c. Simple, inexpensive

d. Pros: It is easy to measure with inexpensive simple equipment. Because it is only one number it is easy to understand. If the number goes up you would suspect there is a problem. Alarm charts are available to help one determine what levels are acceptable and not acceptable.

Cons: One number doesn’t tell you what the problem is. What do you do with the information? Rolling element bearing defects might not raise the RMS overall level – yet they are a key failure mode you want to measure. On the other hand, the RMS level might go up when there is no mechanical problem such as in the case of turbulence or flow noise in a pump.

e. Analog, digital or from the spectrum. Many modern data collectors and software packages offer all three options. It is important to choose one of the options and stick with it on a particular measurement point as the different calculations may lead to different values.

24. What is the crest factor? It is peak / RMS

It can tell you if you have impacting. Impacting is associated with defects in bearings, gears, cavitation, rubs etc.

If the peak amplitude of a sine wave is 1, the RMS value is 0.707. The crest factor is then 1 / 0.707 = 1.41

Peak / RMS = 5.2 / 1.4 = 3.7



25.

The period is the time to complete one cycle of vibration.

Period (seconds) = 1 / Frequency (Hz). Period = 1/4 = 0.25 seconds. All you really need to draw is one cycle of vibration in 0.25 seconds but you can draw as many cycles as you want.

Note that in the spectrum you need to select whether one wishes to display the data in CPM, Hz or Orders or pk, pk-pk or RMS. You can also select amplitude units of velocity, acceleration or displacement.

In this question you were asked for in/secRMS therefore multiply 0.707 x 4 in/secpk to get 2.8 in/secRMS

You were asked to place the frequency axis in CPM, therefore multiply 4 Hz x 60 to get 240 CPM.

Note that the graph needs to be labeled appropriately so everyone knows what the units are.

26. Instead of using units of frequency such as CPM or Hz, you instead relate everything to the motor shaft speed where the motor shaft speed is “1x”

What does 1x refer to? The motor shaft.

It makes it easier to calculate forcing frequencies. If the motor is running at 1787 RPM and it has 7 blades on a cooling fan, the cooling fan blade pass rate is 7 x 1787 = 12,509 RPM. In orders, you can say the motor shaft is 1x. There are 7 blades then the blade pass frequency is 7 x 1x = 7x. This is much easier!

Order normalization makes it easier to “line up” and compare spectra taken at different times where the running speed may have varied a bit from test to test. Below is a graph from a pump running at 10 RPM with 6 vanes. The motor shaft speed is labeled “A” and the pump vanes are labeled “B”.



Next time you take a reading, the motor is running at 11 RPM. Because this is not a very big difference in speed it is still OK to compare these readings. The original reading and the new reading are included in the next graph.

In the graph below, the peaks from the new test are at 11 and 66 RPM for the motor speed and vane pass.

We want to compare the levels of these peaks (trend), but they do not line up. To solve this problem we can convert the graph to “orders.” This is called order normalization. In orders, the motor shaft is 1x and the pump vanes are 6x. This is the same for both tests. You simply need to tell your software which peak is 1x.

Orders are also useful in helping you determining where the vibration is coming from. In vibration analysis 6x is very different from 6.1x. 6x is called a synchronous frequency and 6.1x is non synchronous. Is it possible to have a pump vane pass frequency at 6.1x? No it is not. The pump can have 6 vanes, 5 vanes, 7 vanes etc. but never 6.1 vanes so if you see a peak at 6.1x you know it is NOT the pump vanes. Rolling element bearings create non synchronous peaks, so 6.1x could be a bearing defect, but 6x wouldn’t be.



27. First convert the motor shaft rate to Hz using the formula Hz = CPM/60

Hz = 1800 / 60 = 30 Hz

Next calculate the vane pass rates by multiplying the number of vanes times the shaft rate.

PV1 = 13 x 30 = 390 Hz

PV2 = 19 x 30 = 570 Hz

See if you can also find harmonics of the vane pass frequencies in the plot

28. When working in orders, the motor shaft rate is 1x by definition. There are 23 teeth on the motor gear. The gearmesh frequency is the number of teeth times the shaft rate or 23 teeth x 1x = 23x. This peak is labeled in the graph with the arrow on the left.

Harmonics are multiples e.g. 2 x 23x = 46x and 3 x 23x = 69x. The peak to the right labeled with the arrow is at 46x

29. The belt rate peak is labeled “BR”. In order to calculate the belt rate you need to know the belt length in addition to the pulley diameters. You were not given this information. However, the belt rate will always be sub synchronous (it rotates slower than both of the shafts because each shaft must go around more than once before the belt can go all the way around). The peak labeled BR is the only sub synchronous peak on this graph.

Use the formula: S2 = S1 x (D1/D2) where S1 and S2 are the input and output shaft speeds respectively and D1 and D2 are the input and output pulley diameters respectively.



When working in orders, the motor shaft speed is 1x by definition: S1 = 1 and S2 = 1 x (34/23) = 1.47x

Note that the smaller pulley always turns faster, so the answer above is correct. The fan turns 1.47x faster than the motor.

Harmonics of the fan shaft speed are 1.47 x 2, 1.47 x 3, 1.47 x 4 etc. or you can use a ruler and measure them on the graph as shown above.

30. First calculate the forcing frequencies. By definition, the motor shaft S1 is 1x and the motor shaft rate harmonics are then 2x, 3x, 4x etc. which are labeled on the X axis of the plot.

The motor fan blade rate is the number of blades (6) times the motor shaft speed (1x): 6 x 1x = 6x.

To calculate the pump shaft rate use the formula: S2 = S1 x (T1 / T2), here S1 and S2 are the input and output shaft rates respectively and T1 and T2 are the number of teeth on the input and output gears respectively.

S2 = 1 x 23/33 = 0.7x

Harmonics of the pump shaft rate are multiples of 0.7x or 2 x 0.7x = 1.4x and 3 x 0.7x = 2.1x etc. These are marked with the three left-hand arrows on the plot.

The pump vane rate is the number of vanes (17) times the pump shaft rate (0.7x) or 17 x 0.7x = 11.9x.

31. It is best to draw a diagram of the machine and or write down all of the variables and all of the equations you will need. This will help make things less confusing:

S1 = 1xM = 1x by definition

D1 = 50, D2 = 23, BL = 250, T1 = 73, T2 = 29, #CV = 7

BR = 3.14 x S1 x D1 / BL = 3.14 x 1500 x 50/250 = 0.628x

The belt rate must be sub synchronous so this looks correct

Gin = S1 x D1 / D2 = 1 x 50/23 = 2.17x

The smaller pulley should be turning faster – which it is, so this answer also looks correct.

GMF = Gin x T1 = 2.17 x 73 = 158.7x

1xP = Gin x T1/T2



1xC = 2.17 x 73/29 = 5.47x

The smaller gear turns faster, so this looks correct

CV = #CV x 1xC = 7 x 5.47x = 38.2x

To convert orders to CPM or Hz, multiply all of the answers by the motor shaft rate in CPM (1500) or Hz (1500/60 = 25)

UNDERSTANDING SIGNALS

1. Because they are in phase, these waves add up point by point and you will get a sine wave with the same frequency but an amplitude of 6 mm/secRMS.

What if the second wave is 180 degrees out of phase with the first? These waves will also add up but because they are 180 degrees out of phase they will add up to 0. They are like mirror images of each other, when one is at +3 mm/sec the other is at -3 mm/sec – so they add to zero at each point.

This principle is used in rotor balancing; an equal weight is put opposite to the unbalance weight creating an equal vibration in the opposite direction (180 degrees out of phase with it) – this cancels the effect of the unbalance weight. The same principle is used in noise cancelling headphones – they measure the sound coming in and play back a sound 180 degrees out of phase with it to cancel it.

2. This will cause beating. If the two signals start out in phase, they will gradually go out of phase and then back in phase again repeatedly – like two people running around a track at different speeds. They begin together (in phase) but eventually the other person ends up on the opposite side of the track from the first person (out of phase) then he eventually catches up to the first runner and they are on the same side again (in phase).

a. This happens when they are in phase and the two amplitudes add together: 0.8 + 0.2 = 1 in/secpk

b. This happens when they are out of phase: 0.8 = 0.2 = 0.6 in/secpk

c. Beating

d. The beat frequency is 100.5 Hz – 100 Hz = 0.5 Hz. The beat period is 1/ beat frequency => 1/0.5 = 2 seconds Therefore, you will hear a pulsation every 2 seconds.

3. The period, marked by the horizontal arrow in the time waveform, is 0.01 seconds. Therefore the frequency is 1/ 0.01 sec = 100 Hz.

You will see a peak at 100Hz, but because this wave is repetitive (periodic) but not a sine wave, there will also be harmonics or multiples of 100 Hz (2x 100 Hz, 3x 100 Hz, 4x 100 Hz etc.)



4. First calculate the higher frequency from the period = 0.01 s Frequency (Hz) = 1/ period (s) = 1/0.01 = 100 Hz. This will result in a peak at 100 Hz in the spectrum.

Next calculate the period of the modulating frequency. There are 20 cycles from the lowest amplitude to the lowest amplitude making the period = 20 x 0.01s = 0.2 seconds. The frequency is therefore 1/0.2 = 5 Hz. The modulating frequency will result in 5 Hz sidebands around 100 Hz or peaks at 95 Hz and 105 Hz

Thus, you will see a peak at 100 Hz with sidebands at 95 and 105 Hz

5. There are two primary causes of a raised noise floor in the spectrum (not including bad data or instrumentation problems). They are: single events or impacts and random vibration

Please describe common causes (in machines) of these two patterns of vibration.

Random vibration: loose sensor, loose parts or rattles, flow noise or turbulence in pumps blowers or compressors, etc.

Impacts or single events: Cavitation, late stage bearing wear etc.

Because these two might have very different causes but look the same in the spectrum it is a good idea to also analyze the time waveform.

6. Amplitude modulation results in a pattern of sidebands in the spectrum. It is often associated with these machine components: gears, bearings and AC motors.

7. In the time waveform, amplitude modulation can be confused with beating, however beating will result in a pattern of two closely spaced peaks in the vibration spectrum.

This is because beating is caused by the addition of two sine waves of slightly different frequencies. Amplitude modulation is a sine wave being modulated (multiplied) by another sine wave.



SIGNAL PROCESSING

1. Fmax Low pass

ISO RMS overall reading (10-1,000Hz) High pass

10 Hz low cutoff frequency Band pass

Filtering out data below 5000Hz for Band stop demodulation reading

2. To remove low frequency noise from the measurement in order to prevent it from getting amplified during the integration process.

3. Hanning Sensor calibration

Flat top Bump test

Rectangular Normal route test

4. The time waveform is measured and the spectrum is calculated using an algorithm called the Fast Fourier Transform (FFT).

5. If the Fmax is 1000 Hz then the sample rate is: Fs = 2.56 x Fmax => 2.56 x Fmax = 2560 Hz and the time between samples in the time waveform is: Ts = 1/Fs => Ts = 1/2560 = 0.0004 seconds.

6. If Fmax = 1000 Hz and there are 800 lines of resolution:

a. N = 2.56 x LOR = 2.56 x 800 = 2048

b. T = LOR/Fmax = 800/1000 = 0.8 seconds

c. Convert RPM to Hz: Hz = RPM / 60 = 1200 / 60 = 20 Hz. If it rotates 20 times per second then it takes 1/20 (= 0.05s) of a second to rotate once (period (S) = 1/frequency(Hz)).

The time waveform length (T) from the last question is 0.8 seconds. Divide this by the time for 1 revolution (0.05) => 0.8 / 0.05 = 16 revolutions.

d. R = Fmax / LOR = 1000 / 800 = 1.25 Hz

e. Bandwidth (BW) = Resolution (R) x WF = 1.25 x 1.5 = 1.875 Hz

7. The Hanning window is used to solve a problem called leakage



8. Please match the average type to the application:

Bump test Linear averaging

Normal route test Peak hold averaging

Special gearbox test Negative averaging

Bump test on a machine that is running Exponential averaging

Give more weight to newer data in averaging process Time synchronous averaging

Find the highest vibration amplitudes on a machine when tested over a period of time

9. How many averages are typically used with linear averaging on typical machines? 4 – 12

What overlap %? 67% or 50%

10. There are approximately 5.8 impacts in 0.2 seconds. Each impact is therefore 0.2 / 5.8 = 0.034 seconds. This is the period.

The frequency in Hz = 1 / period = 1/0.034 = 29 Hz

Hz x 60 = CPM = 29 x 60 = 1740 CPM

1740 RPM could be 1x which would make this an impacting at 1x – typically associated with rotating looseness or a rub. Note also that the waveform is asymmetrical – it goes down further than it goes up – like it is hitting something at the top.

DATA ACQUISITION

1. 200 mV/mil means that the probe will output 0.2 Volts for each mil of displacement. If the probe moves 5 mils it will output 5 x 0.2 V = 1 Volt.

2. The DC voltage output from a proximity probe relates to the gap or average distance between the probe and its target and the AC voltage relates to dynamic movement or vibration.

3. A proximity probe pointed at a keyway in the shaft can be used to measure shaft speed and phase. When used in this way it is called a Keyphasor®.

4. When analyzing data from a pair of proximity probes mounted 90 degrees apart from each other on a turbine journal bearing, the graph one typically analyzes is called an orbit.

5. The plot is produced from DC voltages from the proximity probes.

The outer edge of the plot has a value of 1. It means the shaft is touching the bearing. The bottom center point of the plot shows where the shaft is when the machine is not running. It is at the bottom of the bearing.

The arrow in the top left corner of the plot indicates the direction of rotation of the shaft; clockwise in this case.



How is this plot measured? During a start up or coast down, the DC voltages from the two proximity probes are plotted in this graph. It shows the average position of the shaft in the bearing.

What do the numbers along the line towards the bottom of the plot refer to? The shaft speed as it speeds up.

6. Unbalance, misalignment, looseness or rubs, oil whirl, oil whip

7. Oil analysis or wear particle analysis can be used to find early warning of wear in the babbit. Temperature may increase at a late stage of wear and can be detected using IR technology or installed temperature probes.

8. A large journal bearing can go from perfect health to catastrophic failure in approximately how much time? In seconds. Therefore one typically employs a continuous monitoring protection or shut down system to monitor these bearings.

9. For each “g” of vibration the sensor puts out 100 mV when it is operating in its linear range. Therefore, 5 g’s would be 5 x 100 mV = 500 mV

10. A high speed compressor with high amplitude 10 mV/g high frequency vibration

A low speed bearing 100 mV/g

A typical 1800 RPM centrifugal pump and motor 500 mV/g

Higher acceleration levels require a less sensitive sensor.

Lower acceleration levels require a more sensitive sensor.

A typical “off the shelf” general purpose accelerometer is 100 mV/g.

11. It is an accelerometer with a built in integrator. Therefore the electrical output of the sensor is in velocity. These are often used with inexpensive online systems or vibration switches where alarms are set based on velocity.

12. No sensor can measure perfectly at every frequency, this is why there is more than one type of sensor on the market. The frequency response of the sensor (which is the same as its calibration curve) shows how accurately the sensor measures at each frequency. How the sensor is mounted also effects how accurately it measures at each frequency. It is important to understand these concepts when choosing a sensor and a mounting technique.

13. The more rigidly the sensor is mounted the higher the frequency it can measure accurately. A flat magnet on a flat pad is a more rigid connection and therefore has a better high frequency response.

14. Repeatability – everyone mounts the sensor on the same test point.

15. One solution is to monitor bearings 2, 3, 6, 7 and 8. If a triaxial sensor is being used then triaxial readings are best. If not then take one radial reading at each bearing (either vertical or horizontal) and take an axial reading at bearings 2, 3, 6 and 7 if possible.



Because the motor is small, (less than 30 in in diameter), it is OK to take only one bearing. The coupled end (2) is preferred because it will aid in detecting motor to gear misalignment – this is also why an axial reading can be taken here. The free end of the motor might have a cover on it so there may not be a good test point.

Bearings 3 and 6 on the gearbox are taken for the same reason – i.e. because they are coupled to the other components. The distance between bearings 3 and 6 is large enough to warrant taking readings on both. Because bearing 4 is close to 3 and bearing 5 is close to 6, it is unnecessary to take readings at 4 and 5.

Because the diameter of the pump is large (> 30 in) it is a good idea to test both bearings (7 and 8).

16. Nameplate information: To verify that this is the correct machine, that it has not been replaced with a similar machine.

Test points clearly named and identified on machine drawing: So everyone knows where to take the readings. It also helps when analyzing the graphs to see where they came from and to see which test point to select in the data collector. If using test pads this tells everyone where to replace the pads when they fall off.

Test instructions: To be sure the machine is always tested in the same way - speed, load, etc.

Schematic of the machine: So you know what components are in the machine, what forcing frequencies to look for and what faults to try to diagnose. This should include the bearing types and bearing numbers if available.

By documenting the above, anyone should be able to test the machine correctly and the program should not fail when the “expert” leaves.

FAULT DETECTION

1. The pump shaft rate is marked with the arrow. The peak to the left of this, marked with a box is 1/2x and the peaks marked with triangles are harmonics of 1/2x. These are called “sub harmonics” – not to be confused with “sub synchronous”. Sub harmonics are sub multiples of the frequency of interest. They can be halves (as in this case), thirds or quarters.

Sub harmonics are often associated with rotating looseness or rubs

2. If the motor speed is 1800 RPM then the electrical line frequency is 60 Hz (if it was 50 Hz the motor speed would be 1500 RPM for a 4 pole induction motor). Therefore 2xLF is 2 x 60Hz = 120Hz. The motor speed in Hz is 1800 / 60 = 30Hz. This would make 120 Hz appear at 4x (120 / 30 = 4). However, there is always slip in an induction motor, i.e. the motor likely runs at 1740 RPM (29Hz) or somewhere near there. This means that 2xLF will be slightly higher than 4x or the peak marked ‘D’.

The amplitude of this peak can go up if the stator is eccentric. This is usually caused by soft foot. Loose stator windings will also result in an increase as would unbalanced line phase voltage.

3. The peaks labeled A and B are non synchronous. They appear to be at about 3.3x and 6.6x. B is a harmonic of A. These are most likely bearing tones, probably from an outer race defect (assuming the inner race of the bearing is rotating).



Typical bearings have between 8 – 12 balls and their defect frequencies often appear somewhere between 3 – 12x. One can use a simplified formula to estimate the frequencies where BPFO (outer race) is approximately 0.4 x #balls and BPFI (the inner race) is 0.6 x #balls (in orders). Therefore, a bearing with 8 balls would have an outer race defect frequency at around 3.2x.

It would be nice to know if these peaks gradually appeared in the spectrum over time as the defect occurred and got worse. Trending is always preferred.

Another possibility is that this is external vibration from another machine. One could check vibration spectra from nearby machines to see if any of the forcing frequencies match.

4. The peaks marked ‘B’ are non synchronous and they are in a range (between approximately 3- 12x) where one typically finds defects in rolling element bearings with between 8 to 12 balls.

The peaks marked ‘A’ are a distance of exactly 1x to the right and left of the first ‘B’ arrow. If the first ‘B’ peak is at 3.24x then the ‘A’ arrows are at 2.24x and 4.24x. These are called “shaft rate sidebands” and they are caused by amplitude modulation.

This pattern is common for an inner race defect when the inner race of the bearing is spinning and for an outer race defect when the outer race is spinning. Essentially what causes this is the defect rotating in and out of the load zone each time the shaft turns around.

5. These peaks are all non synchronous. Notice how easy it is to determine that the peaks are non synchronous when the data is displayed in orders? The middle of the three arrows close together appear to be at about 5.9x. The arrow to the far right is a harmonic of this at (5.9 x 2 =) 11.8x. Non synchronous peaks in this range are often caused by bearing defects. The arrows to the left and right of the one at 5.9x are equally spaced from 5.9x at a distance of about 0.27x. These are sidebands, but not shaft rate sidebands (shaft rate sidebands would be at 4.9x and 6.9x). These are cage rate (FTF) sidebands and 0.27x represents the rotational rate of the cage. A defect on a ball or roller will move in and out of the load zone each time the cage goes around. This causes amplitude modulation which results in sidebands in the spectrum.

This is therefore likely a defect on a ball or roller.

6. The 1xH will be much higher than 1xV taken on the same bearing and the phase difference between them will not be 90 degrees (as it would be for unbalance). The relationship between 1xH and 1xV should be the same on both bearings of the same machine component (say, both motor bearings).

The difference between them is in how they are solved. Looseness implies that there is something that can be tightened, such as hold down bolts, to solve the problem.

Weakness implies that the foundation itself will need to be stiffened or the machine held in place with braces.



7. The pump vanes will be at 13x and 19x. They are marked with arrows in the plot below:

The motor bar related peak(s) are marked below on the spectrum. The motor bar peak will be at the number of bars times the shaft rate and it will have 2xLF sidebands. There are typically between 30 and 90 bars, so look between 30x and 90x. This is an 1800 RPM machine (30Hz) so 2xLF will be slightly higher than 4x. The vertical grid lines in the plot are 5x apart.

How many poles does this motor have? 4

If the motor is running at 1760 RPM, what is the slip frequency? 1800 – 1760 = 40 RPM

What is the pole pass frequency? The number of poles (4) times the slip frequency (40 RPM) = 160 RPM

8. Log or dB scales allow you to see small amplitude peaks in the presence of large amplitude peaks. The small amplitude peaks may be your bearings whereas large peaks may be 1x and vane or blade pass frequencies.

Think about trying to photograph your friend standing next to a large building, where the top of the building needs to be in the photo. In a linear scale this will only work if the building is less than 10x taller than the person. If it is greater than this you can see the top of the building but your friend is cut off at the bottom – or you can see your friend but not the top of the building. When using a linear scale, one must re-scale the graph like this to see both. This takes time!

In a log or dB scale, the building can be 1,000 times taller and you can still see both the building and your friend clearly in the picture.

Because the log or linear scale allows you to view all of the important data in one graph, it also means that you can use a fixed graph scale instead of auto-scaling the graphs. This offers another HUGE benefit.

When all of your graphs are always scaled the same way it gets very easy for your eyes and your mind to pick out patterns and to compare one plot to another – say from different test points, test dates or different axes on the same machine. This drastically speeds up the analysis process.



The benefit of dB over log is that the vertical scale in a dB plot is “linear” and the log part is contained in the unit itself. This means that you can look at a peak in the spectrum and estimate its amplitude without clicking on it. You cannot do this with a log scale because the scale is graduated (it goes up by factors of 10).

In a dB scale just remember that an increase of 6 dB is a doubling of amplitude. When using the imperial version of VdB (velocity decibels) a good fixed scale to use is 60 – 120 VdB.

BALANCING

1. A vector describes a magnitude (amplitude) and a direction (phase angle).

2. 4 mils at 90 degrees is the horizontal vector at the bottom pointing to the right.

3. To add 3 mils @ 0 degrees to this, place the tail of the 3 mil vector to the tip of the 4 mil vector and draw the vector vertically (in the direction of 0 degrees). Next, draw a vector from the tail of the 4 mil vector to the tip of the 3 mil vector – this is the solution. You can measure the length of this vector with a ruler and the angle “a” with a protractor

To calculate this mathematically, it is a right triangle (90 degrees angle between the 4 mil and 3 mil vectors) so you can use x2 + y2 = z2

32+ 42 = z2

Z = 5 mils

The angle “a” is tan-1 (3/4) = 36.9 degrees

5 mils

4 mils

3 mils

90 a

180

270

0



4. The “O” vector describes the 1x amplitude and phase angle for the original run. This is the machine running with its original amount of unbalance.

A trial weight is added to the rotor and a second measurement is taken. The “O+T” vector is the second measurement and it represents the vibration and phase angle with the original unbalance and the trial weight unbalance on the shaft.

The “T” vector is calculated by subtracting “O” from “O+T”. “T” represents the effect that the trial weight had on the rotor. With this information you can now calculate how much weight to add and where to add it in order to get the effect you want. The effect you want is to create a vector equal and opposite to “O”.

You can think of a trim run like playing golf. The goal is to hit the ball in the hole (to create a vector equal and opposite to “O”) but maybe you have never hit a golf ball before so you do not know how hard to hit it. You can just give it a try (add the trial weight and see what happens)

Based on your trial hit, you can now take the ball back to the tee (remove the trial weight and add the balance weight) and hit it again. This time you have a better idea of how hard to hit it and in which direction to get it to go where you want it to go.

You can also take your second swing from where the ball landed after the first swing (how one would normally play golf) this is the same as leaving the trial weight on and adding additional balance weight.

Even though in either case you have a better idea of how to hit the second shot, it doesn’t mean it is going to go in the hole. In balancing, you only have to get close or meet your target. Therefore you will also do trim runs (these are like putts on the green). If you don’t get it in the hole (or close enough) after two putts then you either start from the beginning or say it is good enough and stop.

5. When balancing a rotor in-situ, if the trial weight is too small it will not influence the rotor enough for you to calculate the balance solution. If it is too large it can damage the machine. The trial weight should generate a force equal to approximately 10% of the rotor weight and it should change the original phase and amplitude readings by approximately 30%.

6. Wide rotors often suffer from dynamic or couple unbalance and they therefore require a two plane balance solution.

7. The amount of unbalance remaining in a rotor after it has been balanced is called its residual unbalance.

8. There are two general types of balance standards. The first type is based on vibration amplitude and the second type is based on residual unbalance – which is the amount of unbalance left in the rotor after it has been balanced (note that rotors are never perfectly balanced, there is always some residual unbalance).

High vibration levels cause problems. If you are printing computer chips and you have an unbalanced fan near the assembly line, the vibration from the fan can travel through the floor and affect the product. Therefore you might want to set limits on the vibration level in the fan.

The forces produced by an unbalance weight are proportional to the residual unbalance, the radius of this unbalance from the center of rotation and the rotational speed squared. The



forces produced by the unbalance can cause damage or fatigue to the machine and its foundation. Therefore you might want to set limits on the forces (or residual unbalance).

Both types of standards are useful and have their place.

SETTING ALARMS

1. According to the chart, the upper limit is 6.3 mm/secRMS for a start-up or acceptance test on a new piece of equipment. A machine on isolators is deemed to be flexibly mounted.

2. 3.5 mm/secRMS places the vibration in Zone B (between 3.2 and 5.1 mm/secRMS for this pump) therefore unrestricted long term operation is allowable. However, you should report that the pump is operating far from its best efficiency point.

The upper limit of Zone B for this pump is 5.1 mm/secRMS. Multiply this by 0.25 (25%) and add the value to the baseline (3.0 mm/secRMS) to set the alarm. 0.25 x 5.1 + 3.0 = 4.275 mm/secRMS

In order to use the alarm that you calculated in the last question (4.3 mm/secRMS), based on the baseline, the pump needs to be tested under the same conditions each time, e.g. the same BEP. For the test where BEP was again at 90% and the RMS level was 4.4 mm/secRMS, the machine would be in alarm.

For the test where it was operated at 50% of BEP, you have two options The better option would be to retest the pump at the correct BEP and compare the value to the alarm you generated. If this cannot be done you can compare the measured level to the alarm chart directly. In this case, 4.4 mm/secRMS would be in Zone B (between 3.2 and 5.1).

3. The piping vibration is the only one that exceeds the C/D threshold. It is possible that there is a resonance condition in the pipes. You would recommend doing a pump test on the pipes or changing the compressor speed to verify if this is a resonance problem.

4. This is called a “mask” or “envelope” alarm. One of the main problems with it is that bearing defects will usually result in peaks that are much smaller in amplitude than other forcing frequencies such as 1x and vane or blade pass. If the alarm is set high enough to avoid false alarms it may be too high to catch the bearing defects. The peaks marked with arrows are likely bearing tones that are not crossing the alarm.



Band alarms, like the one shown below, have the same capability of detecting any peak that goes above the alarm level, but they also have a second type of alarm based on the amount of energy in each band. Although a bearing tone may never get high enough to cross the peak alarm, it will add energy to the band and might trip the band alarm. That makes this type of alarm superior to the mask alarm.

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