CAT III Example Book PDF

Page 24 u l- CAT Ill Act ivity book SIGNAL PROCESSING c...

Page 24 u l- CAT Ill Act ivity book SIGNAL PROCESSING c F-:- Useful information : P= N N lines T=T xN=- s Fs 2.56 X Fmax Fmax T = Time required to collect the waveform c T, = Time between each sample F, = Sampl ing rate = Samples per second c N = Number of samples (1024, 2048, 4096, et c.). Fmax Resolution =. 1,nes C: Bandwidth = Resolution x Window factor C Window factor= 1 (no window/un iform/rectangular) or 1.5 (Hann ing window) Separating frequency~ 2 x Bandwidth~ 2 x Resolution Window Factor C Required spectral lines< 13. Fn1ftx - - - -- - - - -- - - - - - - - - - ~ Low pass ISO RMS ovemll readlog (10 - 1000 High pass 10 Hz low cut-offfrequencr Band pass FIitering out data below 5000 Hz for Band stop demodulation reading 14. To remove low frequency noise from the measurement in order to prevent it from getting amplified during the integration process. 15. Hanning Sensor calibration Flat top ,Bump test Normal route tes~ 16. The time waveform is measured and the spectrum is calculated using an algorithm called the Fast Fourier Transform (FFT). 17. If the Fmax is 1000 Hz then the sample rate is: Fs = 2.56 x Fm.. => 2.56 x Fmax = 2560 Hz and the time between samples in the time waveform is : T, = 1/F, => T, = 1/2560 = 0.0004 seconds. 18. If Fmax = 1000 Hz and there are 800 lines of resolution: N = 2.56 x LOR= 2.56 X 800 = 2048 T=LOR/Fmax = 800/1000 = 0.8 seconds 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 R = FmaJLOR = 1000/800 = 1.25 Hz Bandwidth (BW) = Resolution (R) x WF = 1.25 x 1.5 = 1.875 Hz 19. The Hanning window is used to solve a problem called leakage 20. How many averages are typically used with linear averaging on typical machines? 4. 12 What overlap %? 67% or 50% [email protected] Copyright ~2020 Moblus Institute www.mobiusinstitute.com  Page 112 CAT Ill Worked examples solutions 21. 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 lx - 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 someth ing at the top. - TIME WAVEFORM ANALYSIS 1. 0.2 !◄ ·! ·V\ \l 1\I I~~ ►,~~ ► I \ 0.01 It is difficult to tell from the time waveform alone if this is modulation or beating. If we knew the forcing frequencies of what was being monitored we could perhaps tell which it is, but without that information we cannot. If this is amplitude modulation, the center frequency or frequency being modulated has a period of 0.01. The frequency is FH, = 1/P, = 1/0.01 = 100 Hz. Next count how many cycles are in one cycle of modulation. There appear to be 20, giving us a period of modulation of 0.2 sec. The modulating frequency Is therefore F = 1/0.2 = 5 Hz. Therefore a 100 Hz wave is being modulated at a rate of 5 Hz. The spectrum will have a peak at 100 Hz with 5 Hz sidebands or 95 Hz, 100 Hz and 105 Hz. If th is is beating, one of the frequencies Is 100 Hz, as calculated above. The beat period is 0.2 sec, making the beat frequency= 1/0.2 = 5 Hz. 100 Hz Is the average frequency. The two beating frequencies are 100 + 2.5 and 100 - 2.5 Hz. In the spectrum we wlll simply see two peaks, one at 97.5 Hz and one at 102.5 Hz. The two peaks wlll be close to the same amplitude becalise when they are out of phase and subtract , the waveform Is at a minimum which is close to a value of r zero. lea rn@mob1us1ns111u1 e.com Copyright © 2020 M ob,u s Institut e WWW.ll\Ob1u s1nsti tL1t ~.(O IH  CAT Ill Worked examples solutions Page 113 2. 0.05 :.. ·.. : : -2 -l N ·r~ N : :.... ·~~. ~. ~. ' '. l~ : I :...., ··-·. , _, · -··. - '.................. 0" 01 02 0) First find the center frequency or modulated frequency. The period is 0.005 seconds as shown above, making the frequency 1/0.005 = 200 Hz. The modulating frequency has a period of 0.05 sec making the frequency= 1/0.05 = 20 Hz. Therefore we have a 200 Hz wave being modulated at a frequency of 20 Hz. This will give us a peak at 200 Hz with 20 Hz sidebands or: 180 Hz, 200 Hz, 220 Hz. Because the wave is distorted and asymmetrical (it goes down to -6 but only up to +3) we will also see harmonics of 200 Hz and these harmonics will have 20 Hz sidebands. This pattern is common in gearboxes. 25. 1S ···· · T 1 3. The circle represents one shaft period. If you count, there should be 33 impulses occurring in one shaft rotation, which would correspond to the number of gear teeth. The pulses are marked with small arrows. One of the pulses is larger than the others. This likely corresponds to a damaged tooth. 4. Note that two of the sensors are pointing In different directions from the other one. The phase angles need to be adju sted lo reflect the way the sensors are mounted. If we select the one on t, arnoh rn12u1t111m111 ill 11,CPI\I CoJ>yriKhl (02020 Mobius institute www.,nobiusinstilute.coni ri~ Page 114 CAT Ill Worked examples solutions bearing 2 with the reading of 210 as a reference, we simply add 180 degrees to the sensors pointing in the opposite directions. The adjusted phase readings will then be 210, 210 and 210 meaning each of these bearings is in phase in the axial direction. SYSTEM DYNAMICS 1. Use the formula on the left to solve the problem, where n is the natural frequency, k = stiffness and m = mass. fn = 2rr 1 ffeo W In a simple mass spring system : If the stiffness of the spring increases, the natural frequency goes up. If the stiffness decreases it goes down. If the mass increases it goes down and if the mass decreases it goes up. You can also remember this by thinking of a guitar - if you tighten a string the note goes up. If you loosen it, it goes down. This is "stiffness." The fat string is the low note and the thin one the high note. This is "mass." The three types of damping are viscous, frictional and hysteresis. A bicycle pump produces viscous damping when you press down on it because you are moving a piston through air, which is a fluid. When a mass bounces up and down on a spring, the spring heats up - this is the result of hysteresis - which means internal molecular friction. The molecules inside the spring rub together to produce heat - this results in a loss of kinetic energy in the system or damping. True. This would be considered viscous damping Damping is proportional to velocity. The faster you push down on a bicycle pump or move your hand through water, the more the fluid resists the motion. Force is proportional to acceleration (F = ma) Stiffness is proportional to displacement. If a spring is more rigid, the mass will move less distance when the same force is applied (F = kx). 2. Jt wiJJ v~ry slowly go back to rest as if moving through honey ct:itically damped It will bounce up and down forever over damped It will go back to rest at "normal" speed and then it will stop under damped Jt will bounce up and qown a number of times before coming to rest has no damping Critically damped~= 1 Over damped ~ > 1 Under damped O < ~ < 1 It has no damping~= 0 learn@mobiuslnsritut e.com Copyright ([)2020 Mobius Institute www.mobiusinst irute.com CAT Ill Worked examples solutions Page 115 Damping has a small effect on the natural frequency. 3. The spring mass system has a single frequency that it likes to vibrate at called its natural frequency. When you move your hand slower than that frequency, the mass moves in phase with your hand, and the spring behaves as if it is rigid. The amplitude of displacement of the mass is equal to that of your hand. As you speed up the movement of your hand, the phase of the mass begins to lag that of your hand and the spring becomes more flexible. When you move your hand at exactly the same frequency the mass and spring likes to vibrate, the system is said to be in resonance and the phase lags by 90 degrees and the amplitude of vibration of the mass increases, perhaps significantly, although it depends on the amount of damping. As you continue to speed up past this frequency, the phase lags until it reaches 180 degrees and the amplitude of vibration of the mass goes down. This entire process can be described in a Bode or Nyquist plot. 4. This graph below is called a Nyquist or Polar plot. The numbers 5445, 6035, 7927 are frequencies in CPM A rotor or some other system was sped up from 500 to 7927 RPM. At 500 RPM the phase angle was 180 degrees and the amplitude was very small. At 6035 RPM it passed through a natural frequency. I know this because at that frequency, the amplitude has reached a maximum value of 2 milspk-pk (note the graph radius is defined at the bottom of the chart) and the phase has shifted by 90 degrees (from 180 to 270). As it continues to speed up from here, the phase approaches 360 degrees (for a total lag of 180) and the amplitude gets smaller. Note that the amplitude at any point can be found by drawing a line from the center of the plot to that point. The length of the line is the amplitude and the direction the line is pointing is the phase angle - as marked around the circumference of the plot. s. What is the name of the plot below? Bode plot The curve B because it does not amplify the vibration as much (it doesn't go up as high in the bottom plot) and the phase shifts more gradually (top plot). The peak is also more rounded in the bottom plot. The natural frequency is approximately 6000 RPM as this frequency coincides with the point of maximum amplitude in the bottom plot and it Is where the phase passes through 270 degrees (which is where it crosses the vertical grid line on the top graph) You might also note this is the same data from the prior question but it is In a Bode plot format rather than a Nyquist plot. 6. Heavy spot - "The angular location of the imbalance vector at a specific lateral location on a shaft." [email protected] Copyright t02020 Moblus Institute www.mobluslnsti111te 1corn  CAT Ill Worked examples solut ions --- ,.- ;.--. --,, ,,,, Page 116 ;;;..-.-,, High spot : "The angular location on the shaft directly under the vibration transduce of closest proximity." When balancing a rotor, one attempts to place a trial weight opposite the heavy spot. r at the point ---- -- , ,..--.....-- , proximity probe When using a proximity probe to measure vibration displacem ent and another as a Keyphasor, the phase reading is relative to the high spot. are relative When balancing, you are trying to balance the heavy spot, but your phase readings to the high spot. If these are not the same then you must be aware of it! The high spot lags the heavy spot. The weight passes, then the rotor moves... point to If you draw a line from the center of the plot to 5922 on the curve the line will the high spot has lagged the approxim ately 260 degrees. The plot begins at 180 degrees, so the critical speed. heavy spot by 260 - 180 = 80 degrees and 5922 RPM is just before Three other sources of phase lag : Mechanical, sensor, and electronic. natural 7. A single degree of freedom system (SDOF) is a mass, spring and damper with one by adding frequency. We can define more complicated structures like the diving board below system. together spring mass systems or SDOF's to create a MDOF or multi degree of freedom Each natural frequenc y of an MDOF is defined as an SDOF. of moveme nt In a complex structure like a bell, each natural frequency coincides with a form and other called a mode or mode shape that includes points called nodes that do not move points called anti-nodes that move the most. or as a SDOF Each natural frequenc y can be defined in terms of mass, stiffness and damping mass spring system the same If structure like a diving board was in resonance, would you expect it to vibrate frequenc y has a amount in the vertical and horizontal directions? No, because each natural from the mode shape. The first bending mode in the vertical direction would look very different which mode(s) is second bending mode in the horizonta l direction. How it vibrates depends on excited. -.. ODS, MODAL ANAL VSIS AND RESONANCE not in resonance 1. False. An ODS simply animates the vibration of a structure. If the structure is then it tells you nothing about resonance.. True. An ODS is performe d with the machine running or the structure vibrating that is changing False. A two-channel frequency based ODS cannot be performe d on a machine speed during the test. The beam is probably In Its first bending mode. l