Examen Vibraciones Nivel II

VIBRATION ANALYSIS

CERTIFICATION

Chapter 1 1.

Which of the following statements is false? a.

Predictive Maintenance has the purpose of identifying and monitoring machine flaws.

b.

Predictive Maintenance help to planning in a periodic way the machine repairing,

c.

Predictive Maintenance minimizing the production losses due to shutting down of driving equipment.

d.

Predictive Maintenance keep high the equipment reliability.

Answer: 2.

Which of the following statements is considered as a Predictive Maintenance disadvantage? a.

Allows visualization of flaw evolution.

b.

Optimizes the Predictive Maintenance administration.

c.

Facilitates flaw analysis.

d.

It is successfully only with personal well trained.

Answer: 3.

The surveillance and vibration analysis are the most important tools in Predictive Maintenance, and it is based in: a.

All the machines have a normal vibration level, due to the manufacturing tolerances.

b.

All the machines have a normal vibration level, due to manufacturing and mounting tolerances.

c.

All the machines have a normal vibration level, due to manufacturing, mounting and operation tolerances.

d.

All the machines have a normal vibration level, due to manufacturing, mounting and operation not allowable tolerances.

Answer:

4.

If there is an important increment of the normal vibration level of a machine, then it is possible to say that: a.

There is an increment of the flaws severity.

b.

The measurement method has been changed.

c.

The measurement instrument is uncalibrated.

d.

All the previous ones.

Answer: 5.

Which of the following statements is false? a.

The recurring and transient mechanical impacts generate energy peaks that excite natural frequencies of the elements.

b.

The impacts are of low frequency and the natural frequencies are of very high frequency.

c.

The natural frequencies are of very high frequency and they are caused by friction of two surfaces.

d.

It is not advisable measuring at very high frequency to detect incipient defects in bearings and gears.

Answer: 6.

Which of the following lub oil monitoring techniques is not subjective? a.

Visual.

b.

Odors.

c.

Viscosity.

d.

The filter life.

Answer: 7.

Ferrography is a technique used to analyze …. a.

With a microscope; shapes and sizes of the lub oil metallic particles and comparing results with identification patterns.

b.

The amount of metallic particles (PPM) contained in the lub oil due to friction among components of the machine.

c.

Metallic particles bigger than 0.5 mm (19 mils).

d.

Metallic particles smaller than 0.1 microns.

Answer: 8.

Thermography, is a technique that is based in…… a.

the amount of heat generated from the surface of a body.

b.

The body temperature is direct relationship to the infrared wave longitude size that they emit..

c.

Use instruments that allows one to see and measure the infrared energy and turns it into visible images.

d.

Measure the infrared wave longitude size that any body emit from the surface.

Answer: 9.

Which is the main requirement Predictive Maintenance personnel skill?: a.

Analytical, good observer, tidy and patient.

b.

Self taught.

c.

Experience in planning.

d.

Experience in operation and maintenance of machines.

Answer: 10.

Which is the correct order for the Predictive Maintenance Program start up? a.

Selection of the machines, Collection of the initial data, Planning, Programming, Execution and Control.

b.

Selection of the machines, Planning, Collection of the initial data, Programming, Execution and Control.

c.

Selection of the machines, Programming, Planning, Collection of the initial data, Execution and Control.

d.

Selection of the machines, Planning, Programming, Collection of the initial data, Execution and Control.

Answer:

Chapter 2 11.

Which of the following statements is false? a.

Due to an excitation force, the system (rotor-bearing) responds with three vectorial forces having a magnitude that will depend on the structural characteristics of the system, namely stiffness, inertial mass and damping.

b.

The stiffness force is 180° out of phase from the inertial force.

c.

The inertial force varies with the mass system and with the vibration acceleration.

d.

The damping force varies with vibration displacement and with the system damping.

Answer: 12.

Convert to velocity (in / sec RMS) and acceleration (g´s peak), 1.5 mils peak-peak at 3,585 CPM. a.

0.199 in / sec RMS y 0.274 g´s peak.

b.

0.398 in / sec RMS y 0.548 g´s peak.

c.

0.282 in / sec RMS y 0.387 g´s peak.

d.

None of the previous ones.

Answer: Velocity = 2¶fD= 2¶x(3,585/60)x(1.5/2)x0.707/1,000 = 0.199 in / sec RMS Acceleration=(2¶f)2D = ((2¶x(3,585/60)) 2x(1.5/2)/1,000)/386.1 = 0.274 g´s peak D: Displacement peak (Mils). f: Frequency (CPS). A: Acceleration peak (in/sec 2) (1 g = 386.1 in/sec2) 13.

Convert to displacement (mils RMS) and velocity (in / sec peak), 4.5 g´s peak at 1,780 CPM. a. 50.005 mils RMS y 6.541 in/sec peak. b. 35.354 mils RMS y 9.321 in/sec peak. c. 100.01 mils RMS y 0.387 g´s peak.

d. None of the previous ones. Answer: Velocity = 2¶fD Acceleration = 2¶fV = (2¶f)2D Velocity = Acceleration/2¶f = 4.5 x 386.1/(2¶x1,780/60) = 9.321 in/sec peak Displacement =(Velocity x 0.707/(2¶x1,780/60))x1,000 = 35.354 mils RMS 14.

For a rotor running with a speed of 25 RPM, calculate the maximum time to collect a complete 4 revolution time waveform. a.

0.16 sec.

b.

2.4 sec.

c.

4.8 sec.

d.

9.6 sec.

Answer: Calculation of a revolution period T T = 1/f = 1/(25/60) = 2.4 sec. Time of four (4) revolutions = 4 x T = 4 x 2.4 = 9.6 sec. 15.

The exit voltage of a seismic transducer is 0.35 volts, measured with a voltmeter. Which will be the vibration velocity peak, if the transducer sensibility is 1,080 mV/(in / sec) peak? a.

0.032 in/sec peak.

b.

0.324 in/sec peak.

c.

0.378 in/sec peak.

d.

None of the previous ones.

Answer: Velocity = 0.35 x 1,000 / 1,080 = 0.324 in/sec peak 16.

Why there are limits of frequencies in the mounting methods of an accelerometer in order to take reliable vibrations data. Because it depend of: a.

The size of the machine to measure.

b.

The accelerometer natural frequency excited by resonant frequencies.

c. d.

The machine operation speed. The excitation of the new natural frequencies that appear according to the accelerometer mounting method.

Answer: 17.

Calculate the number of samples of the time waveform to obtain a spectrum of 800 lines. a.

2048

b.

1024

c.

800

d.

512

Answer: The number of samples = 2.56 x # of lines The number of samples = 2.56 x 800 = 2,048 18.

Calculate the maximum time for taking a spectrum of 800 lines having a frequency range of 20,000 CPM with 4 averages and 67% overlap. a.

9.6 sec.

b.

6.432 sec.

c.

4.776 sec.

d.

3.192 sec.

Answer: T MAX

=

60 x # lines_____ Range of frequencies

T MAX = 60 x 800 = 2.4 sec. 20,000 T TOTAL = T MAX x [1 + (# Average - 1)(1 – (overlap/100))] T TOTAL = 2.4 x [1 + (4 - 1)(1 – (67/100))] = 4.776 sec. 19.

Calculate the dynamic range of an instrument to visualize two vibration peaks in an acceleration spectrum; 0.123 in/sec peak to 180,000 CPM and 0.01 g´s to 165,000 CPM. a.

55.57 dB.

b.

52.56 dB.

c.

21.80 dB

d.

None of the previous ones.

Answer: Acceleration = (2¶f)V = (2¶x(180,000/60)x(0.123))/386.1 = 6.005 g´s peak Dynamic Range = 20 log (6.005/0.01) = 55.57 dB 20.

Calculate the bandwidth of an instrument that uses Hanning window, to analyze the sidebands at pole pass frequency, of a induction motor of four poles; speed of the rotor 1,790 RPM, Line frequency 3,600 CPM and frequency range 12,000 CPM. a.

45.0

b.

22.5

c.

11.25

d.

5.625

Answer: Separation of frequencies is: 7,200 – 7,160 = 40 CPM Resolution of Frequency < 40 / 3 = 13.33 CPM According to Table 2.7, the lowest frequencies Resolution at 13.33 CPM is 7.50 CPM and it corresponds to 1,600 lines, therefore the Bandwidth will be: Bandwidth = Range of frequencies x Window Factor = 12,000 x 1.5 = 11.25 CPM #lines FFT 1,600 Then the Bandwidth is 11.25 CPM and the desired Frequency Resolution is achieved with 1,600 lines. Chapter 3 21.

In induction motors, which of the following statements is false?. a.

When the electric current is applied on the stator a rotational magnetic field occurs in the air gap.

b.

The rotating magnetic field of the stator induces a current in the rotor bars (in short circuit with end rings) that is proportional to the speed of the magnetic field that cuts the bars of the rotor.

c.

The current induced in the rotor bars produces its own magnetic field that act with the stator magnetic field and generate a force at the rotor bars.

d.

If there are any broken rotor bars, the resistance will decrease and the driving force will be less, consequently there will not be a couple but an unbalanced force that will cause vibrations.

Answer: 22.

The test of idler motor, is good to evaluate incipient problems?. a.

Yes, because it does not have any influence on the driven machine.

b.

Yes, because we get a spectrum free of interferences of other machines.

c.

No, it is only good to evaluate severe electromagnetic problems, because the forces that are generated are proportional to the square of the current.

d.

Yes, but the vibrations should be taken with instruments of higher dynamic range.

Answer: 23.

In an induction motor; What does type of flaw cause vibrations at 2FL frequency accompanied by sidebands at multiple of number of poles x slip frequency.? a.

The short circuits in the stator isolation sheets produce in the stator, located heating and distortion.

b.

Rotor bars in short circuit.

c.

Rotor eccentricity.

d.

The electrical motor stator support looseness or weakness.

Answer: 24.

In an induction motor; What does type of flaw cause vibrations at 2FL frequency without sidebands at multiple of number of poles x slip frequency.? a.

The short circuits in the stator isolation sheets produce in the stator, located heating and distortion.

b.

Rotor bars in short circuit.

c.

Rotor eccentricity.

d.

The electrical motor stator support looseness or weakness.

Answer: 25.

¿What is the difference between the eccentricity rotor vibration spectra of a 2 poles induction motor and another of 4 poles?. a.

There is not any difference.

b.

The difference are the sidebands around the line frequency (FL) , they are at their pole pass frequencies of each motor type.

c.

The difference are the sidebands around (2 FL), they are at their pole pass frequencies of each motor type.

d.

The difference are the sidebands around FL and 2 FL, they are at their pole pass frequencies of each motor type.

Answer: The difference are the sidebands (iFp = p.s) around (2FL) i: 1, 2, 3, 4, … Fp: Pole pass frequency. p: Number of poles. s: Slip frequency. 26.

What is the difference between the eccentricity stator vibration spectra of a 2 poles induction motor and another of 4 poles? a. There is not any difference. b. The difference are the sidebands around the line frequency (FL) , they are at their pole pass frequencies of each motor type. c. The difference are the sidebands around (2 FL), they are at their pole pass frequencies of each motor type. d. The difference are the sidebands around FL and 2 FL, they are at their pole pass frequencies of each motor type.

Answer: There is not any difference, the vibration frequency is at 2F L because they originate in the stator and do not depend neither on the rotor speed nor on the slip frequency. 27.

What does type of flaw cause vibrations at 2F L Frequency, accompanied by sidebands at 1/3 F L?.

a. Broken connections, corrosion originated false contacts and worn contact surfaces. b. Rotor eccentricity with false contacts and worn contact surfaces in connections. c. Stator eccentricity with false contacts and worn contact surfaces in connections. d. Rotor and stator eccentricity with false contacts and worn contact surfaces in connections. Answer: 28.

¿Describe the low frequency spectrum of rotor broken bars, of an electric motor that has 27 bars, 8 poles and 10 CPM of slip frequency?. a.

Spectrum with rotor speed harmonics (X = 890 CPM).

b.

Spectrum with synchronous speed harmonics (X = 900 CPM) accompanied by sidebands at the pole pass frequency (#p x s = 8x10 = 80 CPM).

c.

Spectrum with rotor speed harmonics (X = 890 CPM) accompanied by sidebands at the pole pass frequency (#p x s = 8x10 = 80 CPM).

d.

Spectrum with line frequency FL harmonics (X = 3600 CPM) accompanied by sidebands at the pole pass frequency (#p x s = 8x10 = 80 CPM).

Answer: 29.

Describe the high frequency spectrum of rotor broken bars, of an electric motor that has 27 bars, 8 poles and 10 CPM of slip frequency?. a.

Spectrum with one or two Rotor Bar Pass Frequency harmonics (RBPF = 27x890 = 24,030 CPM) accompanied by sidebands at 2F L (7,200 CPM).

b.

Spectrum with one or two Rotor Bar Pass Frequency harmonics (RBPF = 27x900 = 24,300 CPM) accompanied by sidebands at 2FL (7,200 CPM).

c.

Spectrum with one or two Rotor Bar Pass Frequency harmonics (RBPF = 27x900 = 24,300 CPM) accompanied by sidebands at FL (3,600 CPM).

d.

Spectrum with one or two Rotor Bar Pass Frequency harmonics (RBPF = 27x890 = 24,030 CPM) accompanied by sidebands at F L (3,600 CPM).

Answer:

Low and high broken rotor bar frequency spectrum of an electric motor that has 27 bars, 8 poles and 10 CPM of slip frequency 30.

Which is the slip frequency of the figure n° 1 spectrum of a two poles induction motor with rotor broken bars.

Figure n° 1 a. 229.6 CPM. b. 114.8 CPM. c. 57.4 CPM. d. 28.7 CPM. Answer: Chapter 4 31.

If a two stage pump, is hydraulically balanced on the axial or on the longitudinal position of the shaft; What is the function of the thrust bearing?. a. The thrust bearing has the function of; support the axial loads generated in the pump starting.

b. The thrust bearing has the function of; support the axial loads generated in the pump shut down. c. The thrust bearing has the function of; support the axial loads generated during the load pump variations. d. All the above. Answer: 32.

¿Why the pump efficiency decreases to higher or lower liquid flow with respect to the Best Efficiency Point (BEP)?. a. Because the fluid will leave the impeller at the ideal angle, if and only if the pumping of liquid flow is done at the best efficiency point (BEP). b. Because if the liquid flow is bigger or smaller than the corresponding to the Best Efficiency Point, then the entrance angle of the fluid to the impeller will not be ideal and additional turbulences will take place. c. Because if the liquid flow is bigger than the corresponding to the Best Efficiency Point, then turbulences will take place and if the liquid flow is smaller, then cavitation will take place. d. Because if the liquid flow is bigger than the corresponding to the Best Efficiency Point, then recirculation will take place and if the liquid flow is smaller, then cavitation will take place.

Answer: 33.

The hydraulic phenomena of a centrifugal pump when it operate out of the Best Efficiency Point, cause. a.

Misalignment induced hydraulically.

b.

Premature mechanical seal failure.

c.

Excessive hydraulic noise (tinkling).

d.

Premature bearing failure.

Answer:

34.

What is the range of vibration frequencies of cavitation?. a.

The range is between 10 to 50 CPM in the high side of the impeller RPM frequency.

b.

The range is between 10 to 50 CPM in the low side of the impeller RPM frequency.

c. d.

The range is between 12,000 to 120,000 CPM. The implosions excite frequencies in the range of 300,000 to 760,000 CPM.

Answer: 35.

The cavitation damage is the erosion on the pump´s components on ………….. of the pump impeller. a.

The non-visible or underside of the discharge vane side.

b.

The visible or the pressure side of the vane's leading edge or underside of the discharge vane side.

c.

The non-visible or underside of the suction vane side.

d.

The visible or the pressure side of the vane's leading edge or underside of the suction vane side.

Answer: 36.

Cavitation occurs when; a. There is insufficient NPSH (net positive suction head) available. b. In turbulent liquid flow conditions (not laminate). c. In turbulent liquid flow conditions caused by 90° pipe elbows just before the pump suction. d. All the above.

Answer: 37.

What is the range of vibration frequencies of recirculation?. a. The range is between 12,000 to 120,000 CPM. b. The range is between 10 to 50 CPM in the high side of the impeller RPM frequency. c. The range is between 10 to 50 CPM in the low side of the impeller RPM frequency. d. It has fixed frequency; Frequency = # blades x RPM.

Answer:

38.

What is the range of vibration frequencies of turbulence?. a. The range of frequencies is small (10 to 50 CPM) with amplitudes that are not constants and located close to the impeller RPM. b. The range is between 10 to 50 CPM in the high side of the impeller RPM frequency. c. The range is between 10 to 50 CPM in the low side of the impeller RPM frequency. d. It has fixed frequency; Frequency = # blades x RPM.

Answer: 39.

The blowers resonance failures occur when: a.

The pipe natural frequency coincides with the blade pass frequency.

b.

The volute natural frequency coincides with blade pass frequency.

c.

The rotor natural frequency coincides with its turn frequency.

d.

All the above.

Answer: 40. Which of the following statements is false?. a.

Many centrifugal compressors work at high speeds above their first natural frequency.

b.

If actual flow is bigger than the flow compressor design it will work in an unstable condition and surge will occur.

c.

The efficiency of a compressor depends on the blade profiles.

d.

If actual flow is less than the flow compressor design it will work in an unstable condition and surge will occur.

Answer: Chapter 5 41. Which of the following statements is false?. a.

Vee belts and sheaves cause high vibration if and only if are combined with other problems.

b.

Vibrations measured in the perpendicularly and in the parallel direction to Vee belt tension are the most significant measurements for the detection of transmission failures.

c.

If the wear in sheave grooves exists, then the Vee belt slip increases and the vibration spectrum shows sheave frequency harmonics.

d.

Eccentric sheaves cause high radial vibration at the sheave eccentric speed frequency. The most important vibration is parallel to the Vee belt tension.

Answer: 42. The premature thrust bearing failure of Vee belt transmission occur …: a.

When any of the multiple Vee belts sets is not well tightened. Under this circumstance, wearing of sheaves grooves and of Vee belts will be different.

b.

When there is misalignment between sheaves.

c.

When the sheave has excessive end play.

d.

All the above.

Answer: 43. If there is Vee belts resonance, then it is possible minimize the vibrations in the following way: a.

Increasing the Vee belt tension.

b.

Decreasing the Vee belt tension.

c.

Increasing the Vee belt number.

d.

Decreasing the Vee belt number.

Answer: 44. Calculate the RPM of the driven sheave, if a belt transmission system has a driving sheave of 12 inches in diameter that rotates with a speed of 3580 RPM, a driven sheave of 18 inches and a distance between sheave centers of 50 inches. a.

5,370 RPM.

b.

39.78 Hz.

c.

2,386.67 Hz.

d.

89.5 Hz.

Answer: Driver sheave RPM: M = NxD/E = 3,580x12/18 = 2,386.67 RPM RPS = 2,386.67/60 = 39.78 Hz

45. Calculate the Vee belt length, of question 44. a.

145.1 inches.

b.

148.7 inches.

c.

164.1 inches.

d.

184.2 inches.

Answer: G = sin-1 [(E+D)/2L] = sin-1 [(18+12)/2x50] = 17.46 J = √[L2 -((18-D)/2)2] = √[502 -((18-12)/2)2] = 49.91 H = π[1-(G/90)] = π[1-(17.46/90)] = 2.53 I = π [1+(G/90)] = π [1+(17.46/90)] = 3.75 Belt length (K) = [(HxD/2)+(IxE/2)+2J] K : [(2.53x12/2)+(3.75x18/2)+2x49.91] = 148.77 inches. 46. Calculate the Vee belt speed, of question 44. a.

930 RPM.

b.

907 RPM.

c.

822 RPM.

d.

732 RPM.

Answer: Vee belt speed: NxDx π/K = 3,580x12x π/148.77 = 907.18 RPM 47. Enumerate the belt frequencies in the spectrum, of question 44. ANSWERS DESCRIPCIÓN Vee Belt RPM |RPM1 – RPM2| RPM2 RPM2 +/- Vee belt RPM RPM1 RPM1 +/- Vee belt RPM RPM1 + RPM2 Answer:

a CPM

b CPM

c CPM

d CPM

930 1,790 5,370 2,387 +/- 930

907 1,193 2,387 2,387 +/- 907

822 1,790 5370 2,387 +/- 822

732 1,193 2,387 2,387 +/- 732

3,580 3,580 3,580 3,580 3,580 +/- 930 3,580 +/- 907 3,580 +/- 822 3,580 +/- 732 8,950

5,967

8,950

5,967

48. Which is the spectrum of an unbalanced 1778 RPM rotor mounted on an antifriction bearing with an excessive radial play?. a. b.

Up to five rotor speed sub harmonics; 335 CPM, 711 CPM, 1067 CPM, 1422 CPM and 1778 CPM. Several rotor speed harmonicas.

c.

Only one peak vibration at the rotor speed frequency in the vertical position.

d.

Only one peak vibration at the rotor speed frequency in the horizontal and vertical position.

Answer:

49. Which of the following statements is false?. a.

Mechanical looseness is not a cause of vibration; it is a reaction to other problems in the machine.

b.

The distortion caused by level lack between the skid and the foundation of the machine, will cause vibrations at the rotor RPM.

c.

The lack of stiffness on pedestals could be caused by looseness in their housing bearing bolts.

d.

The increase of stiffness on pedestals could be caused by cracks in their structure.

Answer:

50. If the bearing vibration spectrum shows multiple shaft RPM harmonics and erratic phase angle, then the failure could be: a.

Friction between rotor and stator caused by excessive wear, if and only if there are multiple subharmonics at 1/2xRPM or 1/3xRPM.

b.

Looseness combined with misalignment..

c.

Inner race bearing wear.

d.

Coupling locked up.

Answer: Chapter 6 51. Which of the following statements is false? a.

The gear and pinion teeth profiles are designed to avoid slipping.

b.

Only, the slipping between gear and pinion teeth excite natural frequencies.

c.

The common factors between the number of teeth in gear and pinion cause gearmesh frequency subharmonics.

d.

Excessive backlash cause vibrations at; 1xGMF and 2xGMF.

Answer: 52. The gears transmission generate vibrations whose frequencies depend on the following: a.

Eccentric gears.

b.

Shaft deflection.

c.

Number of common factors.

d.

All the above.

Answer: 53. A gear tooth with a flaw will cause knocking or impacts every time that it meshes, its vibration frequencies are; a.

1xRPM of the gear tooth with a flaw and at 1xGMF, the impacts excite natural frequencies that are modulated by the gear speed frequency.

b.

1xGMF, the impacts excite natural frequencies that are modulated by the gear speed frequency.

c.

1xRPM of the gear tooth with a flaw and at 1xGMF accompanied by sidebands at 1xRPM of the gear, furthermore, the impacts excite natural frequencies that are modulated by the gear speed frequency.

d.

1xRPM of the gear and pinion, 1xGMF accompanied by sidebands at 1xRPM of the gear and pinion, furthermore, the impacts excite natural frequencies that are modulated by 1xRPM of the gear and pinion.

Answer: 54. What is the vibration difference between; An eccentric gear and a looseness gear caused by excessive wear bearing? a.

Both, show vibrations at the gear mesh frequency, the eccentric gear is accompanied by sidebands and the looseness gear is not accompanied by sidebands.

b.

Both, show vibrations at the gear mesh frequency, the looseness gear is accompanied by sidebands and the eccentric gear is not accompanied by sidebands.

c.

Both, show vibrations at the gear mesh frequency; the eccentric gear show sidebands on the high side.

d.

Both, show vibrations at the gear mesh frequency; the looseness gear show sidebands on the high side.

Answer: 55. What is the gear mesh frequency of a gear transmission, made up of a pinion of 14 teeth that rotates at 3,560 RPM and of a gear of 35 teeth with one broken tooth?. a.

48,940 CPM.

b.

2,076.67 Hz.

c.

124,600 CPM.

d.

49,840 CPM.

Answer: GMF = 3,560 x 14 = 49,840 CPM 56. Find the common factor of question 55.

a.

1 and 7.

b.

1, 2 and 7.

c.

1, 5 and 7.

d.

2 and 5.

Answer: The factors of 14 are = 1 x 2 x 7 The factors of 35 are = 1 x 5 x 7 Then the common factors of 14 and 35 are: 1 and 7. 57. Describe the frequency spectrum of question 55. a.

1424, 3560, 7120, 14240, 21360, 28480 CPM, 35600, 42720, 49840 +/- i x 1424 CPM.

b.

1424, 3560, 7120, 14240, 21360, 28480 CPM, 35600, 42720, 49840 +/- i x 3560 CPM.

c.

3560, 8900, 17800, 35600, 53400, 71200, 89000, 106800, 124600, +/- i x 3560 CPM.

d.

3560, 8900, 17800, 35600, 53400, 71200, 89000, 106800, 124600, +/- i x 8900 CPM.

Answer: The common factor 7, affects the gear mesh frequency and take place harmonics of GMF/7, such us; GMF/7 (7120 CPM), 2xGMF/7 (14240 CPM), 3xGMF/7 (21360 CPM), 4xGMF/7 (28480 CPM), 5xGMF/7 (35600 CPM) and 6xGMF/7 (42720 CPM), GMF accompanied by sidebands at gear RPM, because it has a broken tooth, GMF (49840 +/-i x 1424 CPM).

58. How many revolutions should the pinion rotate so that both number 1 tooth on pinion and gear mesh twice, of question 55?. a.

2.

b.

5.

c.

7.

d.

None of the previous ones.

Answer: The Pinion: 5 revolutions. 59. How many revolutions should the gear rotate so that both number 1 tooth on pinion and gear mesh twice, of question 55?. a.

2.

b.

5.

c.

7.

d.

None of the previous ones.

Answer: The gear: 2 revolutions. 60. What is the life expectation of gear transmission, of question 55? a.

70%.

b.

50%

c.

20%

d.

14.29%

Answer: The Common Factor: 7, defines the teeth wear and affects the gear and pinion life expectancy in inverse way = (1/7)x100 = 14.29% Chapter 7 61. Which of the following statements is false? a.

Mechanical impacts that are repetitive and transient generate energy that excites natural frequencies.

b.

The friction between two surfaces, cause impacts.

c.

The impacts cause peaks of energy of very high frequency.

d.

The impacts frequency are of very high frequency.

Answer: 62. Which one of the following statements is not a consideration for mensuration signals at high Frequency. a.

Measurement of high frequencies is very sensitive to the type of sensor and to the mounting methods.

b.

A slight slip of the sensor will cause change of high frequency readings, that will take to false conclusions.

c.

Before to take the readings, it is necessary to know the accelerometer and engine resonant frequencies.

d.

The signals have a frequency higher than 300,000 CPM, hand held probes should not be used to measure them.

Answer: 63. Calculate the BPFO (ball pass frequency outer race) frequencies of failure of a antifriction bearing that has 9 balls with 0.5 inches of diameter; 2.6 inches of bearing primitive diameter; 35° of contact angle. Its interior race rotates at 1,785 RPM and the external race at 600 RPM. a.

17,381 CPM

b.

6,172 CPM

c.

3,157.46 CPM

d.

4,492 CPM

Answer: BPFO = N | RPMo – RPMi | (1 – B cosΦ) = 2

P

BPFO = 9 | 600 –1,785| (1 –0.5cos 35°) = 4,492 CPM 2

2.6

64. Calculate the BPFI (ball pass frequency inner race) frequencies of failure of a antifriction bearing, of question 63. a.

28,046 CPM

b.

6,172 CPM

c.

3,157.46 CPM

d.

4,492 CPM

Answer: BPFI = N | RPMo – RPMi | (1 + B cosΦ) = 2

P

BPFI = 9 | 600 – 1,785 | (1 + 0.5 cos 35°) = 6,172 CPM 2

2.6

65. Calculate the BSF (ball spin frequency) frequencies of failure of a antifriction bearing, of question 63. a.

58,983 CPM

b.

6,172 CPM

c.

3,157.46 CPM

d.

4,492 CPM

Answer: BSF = P | RPMo – RPMi | (1 + B2 cos2Φ) = 2B

P2

BSF = 2.6 | 600 – 1,785 | (1 + 0.52 cos2 35°) = 3,157.46 CPM 2x0.5

2.62

66. Calculate the FTF (fundamental train frequency) frequencies of failure of a antifriction bearing, of question 63. a.

1,099 CPM

b.

1,331.3 CPM

c.

3,157.46 CPM

d.

4,492 CPM

Answer: FTF = RPMi (1 - B cosΦ ) + RPMo (1 + B cosΦ) = 2

P

2

P

FTF = 1,785 (1 – 0.5 cos 35° ) + 600 (1 + 0.5 cos 35°) = 1,099 CPM 2

2.6

2

2.6

67. Which one of the next steps of the signal process followed to obtain the modulated frequencies spectrum HFD or Spike Energy, is false. a.

The signals from the accelerometer are processed by a special circuit with a high pass corner frequency (usually 300,000 CPM).

b.

The band pass filtered high frequency signal passed through a rectifier that passes only positive content.

c.

The rectified signal is passed through a peak-to-peak detector. that detects and retains the peak amplitude of the signal, that is to say an enveloping is traced through all the positive peaks of the time wave form. The enveloping gives a new time wave form of high frequency.

d.

The low frequency time waveform is digitized and processed by the Fast Fourier Transform algorithm to generate a spectrum; its component frequencies are the impact frequencies..

Answer: 68. Sort the very high frequency signals processing. I.

The low frequency time waveform is digitized and processed by the Fast Fourier Transform algorithm to generate a spectrum.

II.

The rectified signal is passed through a peak-to-peak detector. that detects and retains the peak amplitude of the signal.

III.

The band pass filtered high frequency signal passed through a rectifier that passes only positive content.

IV.

The signals are processed by a special circuit with a high pass corner frequency.

V.

Vibration signals are measured with an accelerometer.

b.

I, II, III, IV y V

c.

I, IV, III, II y V

d.

V, II, III, IV y I

e.

V, IV, III, II y I

Answer: 69.

Industrial stethoscopes are: a.

Highly reliable in the early detection of bearing flaws.

b.

Subjective and not reliable.

c.

Useful because they can be easily compared.

d.

Useful because the noises taken place by flaw bearings are easily identified from other noises such as turbulences, frictions, etc.

Answer: 70. If there is looseness between the shaft and the inner race and between the outer race and the housing bearing, then we can not affirm: a.

Cause multiple RPM harmonics due to They cause harmonic due to the no-linearity of the vibrating system.

b.

The phase angle is unstable among measures, because the rotor changes its position from one start to another.

c.

The vibration due to looseness is directional.

d.

In this condition, the balancing process requires of high quality instruments.

Answer: Chapter 8 71.

What is the balancing procedure and in how many planes should be balance, a symmetrical rotor speed is 1,780 RPM, weigh 50 kilos, has a diameter of 60 cm and a wide (among balancing planes) of 25 cm? Show the following vibrations, filtered at rotor RPM: V1: 16 at 340° and V2: 17 at 343°

Figure n° 2 b.

Use the balancing procedure of one plane and the correction weights will be in one plane.

c.

Use the balancing procedure of one plane and the correction weights will be in two planes located in the same angle.

d.

Use the balancing procedure of one plane and the correction weights will be in two planes located in different angles.

e.

Use the balancing procedure of two planes and the correction weights will be in two planes in the same angle.

Answer: The rotor should be balance using a single plane procedure and the correction weights will be in two planes located in the same angle, because the phase angles (340° and 343°) of both supports are similar. 72. Calculate the test weight or trial weight, of question 71. a.

2.35 gr located in the same angle in the planes I and II.

b.

4.71 gr located in the same angle in the plane I.

c.

4.71 gr located in the same angle in the planes II.

d.

4.71 gr located in the same angle in the planes I and II.

Answer: Trial Weight = PP (gr) = 35782656 W / (RPM 2 D) Trial Weight = 35782656 x 25 / (1780 2 x 60) = 4.71 gr Trial Weight per plane = 4.71 / 2 = 2.355 gr

73. Calculate the correction weight and its location, when the test weights calculated in the question 72 are installed in the rotor, the following vibrations are obtained at the rotor RPM; V1+T: 12 at 220° y V2+T: 12 at 217° a.

3.1 gr at 25.28° clockwise from the trial weight position.

b.

3.1 gr at 25.28° counterclockwise from the trial weight position.

c.

1.55 gr at 25.28° counterclockwise from the trial weight position, in planes I and II.

d.

1.55 gr at 25.28° clockwise from the trial weight position, in planes I and II.

Answer: Draw in a polar graph and find the vector magnitude, VT = 24.31

Location: Shift 25.28° clockwise from the trial weight position. Correction Weight (P.C.) = PP x V 1 / VT Correction Weight (P.C.) = 4.71 x 16 / 24.31 = 3.1 gr Place the correction weights in the planes I and II, in order to not increase the dynamic unbalance. Correction Weights = 1.55 gr at 25.28° clockwise from the trial weight position, in planes I and II.

74. With regard to the correction weights obtained in the question 73, find the equivalent correction weights and their location, if the rotor has 8 available points where to place the correction weights (the point 8 is the initial test weight position). a. Plane I and II: Point 7; 0.94 gr and Point 8; 0.74 gr. b. Plane I and II: Point 1; 0.74 gr and Point 8; 0.94 gr. c. Plane I and II: Point 1; 0.94 gr and Point 8; 0.74 gr. d. Plane I and II: Point 7; 0.74 gr and Point 8; 0.94 gr. Answer:

Split the test weight Point 1:

1.87 gr.

Point 8:

1.48 gr.

Place the correction weights in the planes I and II, in order to not increase the dynamic unbalance: Plane I y II: Point 1; 0.94 gr and Point 8; 0.74 gr. 75. Calculate the trial weight of an overhung rotor fan rotates with 3575 RPM, weight 75 kilograms, its diameter is 50 cm, distance between balancing planes is 20 cm and it has the following vibrations, filtered at rotor RPM: V1 : 10 at 220° and V2 : 14 at 330°

Figure n° 3 a.

2.1 gr.

b.

4.2 gr.

c.

8.4 gr.

d.

16.8 gr.

Answer: Trial Weight = PP (gr) = 35782656 W / (RPM 2 D) Trial Weight = 35782656 x 75 / (3575 2 x 50) = 4.2 gr 76. When the trial weight (calculated in the question 75) is placed on the rotor (plane closest to the in bearing), the following vibrations are obtained, filtered at rotor RPM: V1+T1 : 4 at 145° and

V2+T1 : 14 at 330°

Calculate the correction weight of the static unbalance and its location on the plane closest to the in bearing. a.

4.3 gr at 23.32° counterclockwise from the trial weight position.

b.

4.3 gr at 23.32° clockwise from the trial weight position.

c.

8.6 gr at 23.32° counterclockwise from the trial weight position.

d.

8.6 gr at 23.32° clockwise from the trial weight position.

Answer:

Draw in a polar graph and find the vector magnitude, VT1 = 9.77 Location: Shift 23.32° clockwise from the trial weight position of Plane I. Correction Weight (P.C.) = PP x V 1 / VT1 Correction Weight (P.C.) = 4.2 x 10 / 9.77 = 4.3 gr Static Correction Weight = 4.3 gr a 23.32° clockwise from the trial weight position of Plane I. 77. Calculate the dynamic correction weights and its location, when placing the trial weight on the rotor fan (farthest plane from the in bearing) and opposed to it, a similar weight on the plane closest to the in bearing, the following vibrations are obtained; V1+T2 : Low amplitude and V2+T2 : 18 at 120° a. Plane I: 1.9 gr at 16.92° and Plano II: 1.9 gr at 196.92° clockwise from the trial weight position in the Plane II. b. Plane I: 1.9 gr at 196.92° and Plano II: 1.9 gr at 16.92° clockwise from the trial weight position in the Plane II. c. Plane I: 1.9 gr at 196.92° and Plano II: 1.9 gr at 16.92° counterclockwise from the trial weight position in the Plane II. d. Plane I: 1.9 gr at 16.92° and Plano II: 1.9 gr at 196.92° counterclockwise from the trial weight position in the Plane II. Answer:

Draw in a polar graph and find the vector magnitude, VT2 = 30.95

Correction Weight (P.C.) = PP x V 2 / VT2 Correction Weight (P.C.) = 4.2 x 14 / 30.95 = 1.9 gr Location: Shift 16.92° counterclockwise from the trial weight position of Plane II. Dynamic Correction Weights: Plano II: 1.9 gr at 16.92° Plane I: 1.9 gr at 196.92° 78. Calculate the combination of Static and Dynamic Correction weights (found in questions 76 and 77) on the Planes I and II: a. Plane I: 2.42 gr at 28.34° and Plane II: 1.9 gr at 16.92° b. Plane I: 2.42 gr at 16.92° and Plane II: 1.9 gr at 28.34° c. Plane I: 1.9 gr at 28.34° and Plane II: 2.42 gr at 16.92° d. Plane I: 1.9 gr at 16.92° and Plane II: 2.42 gr at 28.34° Answer: Plane I: 4.3 gr at 23.32° and 1.9 gr at 196.92° = 2.42 gr at 28.34° Plane II: 1.9 gr at 16.92°

79. Calculate the maximum trial weight allowable for a rotor gas turbine, diametral correction weight 50 inches and maximum residual unbalance of 4 gr - inch. a.

12.5 gr

b.

0.16 gr

c.

0.08 gr

d.

6.2 gr

Answer: 80. Which of the following statements is true? a.

In order to balance, a 6 stages rotor pump using the Static and Couple Derivation Method, are required vibration readings in at least 6 supports.

b.

The Static and Couple Derivation Method means to balance the static and the dynamic unbalance simultaneously.

c.

The Static and Couple Derivation Method means to balance the static unbalance first and after the dynamic unbalance.

d.

The Static and Couple Derivation Method means to balance the dynamic unbalance first and after the static unbalance.

Answer: Chapter 9 81. Which of the following statements is false? a.

Misalignment between two shafts produces rotating forces at the bearings.

b.

The phase angle between the vibration readings horizontal or vertical filtered at rotor RPM, in the supports close to the coupling are 180°.

c.

When angular misalignment exist, the axial vibration has a frequency at 1RPM and when offset misalignment exist, the radial vibration has a frequency at 2RPM.

d.

The misalignment generate an orbit “banana” shape pattern.

Answer: 82. There are many reasons for the existence of a misalignment condition, such us; a.

Bearings failures.

b.

Pipe strain in pumps.

c.

Very rigid pedestals.

d.

All the above.

Answer: 83. Why is necessary to verify, the shafts driver and driven end play? a.

In order to compare with the end play that the coupling can absorb.

b.

In order to adjust the distance among the coupling hubs, thus avoid the axial efforts.

c.

In order to be able to center the rotors, inside their end plays.

d.

All the above.

Answer: 84. If the vibration readings indicate a misalignment condition and the shafts were checked and the misalignment is insignificant, then it is possible that: a.

The machine level is not accurate.

b.

Exist worn bearings.

c.

The coupling misalignment maximum allowable is very small.

d.

All the above.

Answer: 85. If the frequencies spectrum of looseness is at 1X and 2X and the frequencies spectrum of deflection is also at 1X and 2X, then. How they can be identify? a.

If it is looseness, then phase angle of the vibrations at 1X and 2X will be 0°.

b.

If it is deflection, then the phase angle of the vibrations at 1X and 2X will be 180°.

c.

If it is looseness, then the phase angle of the vibrations at 1X and 2X will change.

d.

It is not possible identify with vibrations.

Answer: 86. Which of the following statements is true? a.

The couplings are designed to absorb small offset and angular misalignments.

b.

If the coupling is locked up, then insignificant misalignment will cause excessive vibrations.

c.

If the coupling is locked up, then the axial vibration and phase angles readings will change in each start.

d.

All the above.

Answer: 87. In order to align hot working machines, it is not necessary to consider the following statements: a.

The thermal dilation of pedestals.

b.

The thermal efficiency of the machine.

c.

The Integration of differential dilations of each pedestal.

d.

The exponential spread of pedestal temperature.

Answer: 88. The gear misalignment is not caused by: a.

Gearbox housing bearings misaligned.

b.

Loose gear on its shaft or loose bearings in their housing.

c.

Excessive gears backlash.

d.

The conical end shaft is not the same one that the conical of the gear hole.

Answer: 89. Describe the frequencies spectrum of a gear transmission misaligned, if the pinion has 27 teeth and the eccentric gear of 55 teeth rotates at 1,760 RPM?. a. 1760, 3585, (96800 +/- i 3585), (193600 +/- i 3585), (290400 +/- i 3585) b. 1760, 3585, (96800 +/- i 1760), (193600 +/- i 1760), (290400 +/- i 1760) c. 1760, 3585, (47520 +/- i 3585), (95040 +/- i 3585), (142560 +/- 3585) d. 1760, 3585, (47520 +/- i 1760), (95040 +/- i 1760), (142560 +/- 1760) Answer:

90. Which are the vibration characteristics of a machine that has the coupling locked up due to lubrication fails? a. Excessive vibrations due misalignment. b. Excessive vibrations due misalignment (1X, 2X and 3X) even with insignificant misalignment. c. The axial vibration and phase angle readings change in each start. d. All the above. Answer: Chapter 10 91. Describe the frequencies spectrum of Oil Whirling?. a. The vibration frequency is between 40% and 45% of the rotor’s RPM. b. The vibration frequency is 50% of the rotor’s RPM. c. The vibration frequency is between 45% and 49% of the rotor’s RPM. d. None of the previous ones. Answer:

92. Which of the following statements is false?. a.

The Oil Whirling is a vibration that appears only in rotors that use pressure lubricated friction bearings that work with speeds above their first critical speed.

b.

Since the shaft operates in an eccentric balance position with respect to the bearing center, the oil takes the form of a pressurized wedge that is bigger in the biggest gap between the shaft and the bearing.

c.

Close to the shaft, the speed of oil is similar to that of the shaft’s and is similar to zero when close to the bearing.

d.

The oil average speed is similar to RPM/2, but considering friction losses, the oil average speed will be between 45% and 49% of the rotor RPM.

Answer: 93. The Oil Whirling problem can be caused: a.

Always in all the pressure lubricated friction bearings.

b.

Always in rotors that work with speeds above their first critical speed.

c.

Only when the external forces become bigger than the damping force, in rotors that use pressure lubricated friction bearings that work with speeds above their first critical speed.

d.

Only when the oil film force prevail, pushes the shaft and takes it out of its balanced position and the shaft is forced to rotate inside the bearing, carried by oil film wedge.

Answer: 94. It doesn't cause Oil Whirling vibrations.

a.

A bad distribution of rotor weight on its supports, if the static load of the shaft on the bearing is too small then the oil film force will be predominant.

b.

The shaft eccentricity with respect to the center of the bearing.

c.

The increment of pressure and/or of oil viscosity.

d.

External forces that become smaller than the damping force of the lubricant.

Answer: 95. They cause Oil Whirling vibrations. a.

Misalignment.

b.

The distortion of the housing bearing, decreases the horizontal clearances between the shaft and the bearing.

c.

Any friction of the rotor with the stator such us; labyrinth seals, tip blades rotor with the stator, etc.

d.

All the above.

Answer: 96. It doesn't cause Oil Whirling vibrations. a.Any friction of the rotor with the stator. b.Lub oil with water. c. Lub oil with low anti foam additive d.Work speed close to a critical speed. Answer: 97. The frictions that cause Oil Whirling vibrations are the following: a.

Between the stator and rotor labyrinth seals.

b.

Distortion of the stator labyrinth seal, due to misalignment between their holes with their dowels.

c.

Friction, between the rotor and stator blades.

d.

All the above.

Answer: 98. If the compressor RPM is changed (from 9000 to 10000), and the vibration filtered, at 47% also change (from 4230 to 4700), then: a.

The vibration at 47% is due to the Oil whip.

b.

The vibration at 47% is due to resonance.

c.

The vibration at 47% is due to the Oil whirling.

d.

The vibration at 47% is due to dry friction in the bearing.

Answer: 99. If the compressor RPM is changed (from 9000 to 10000), and the vibration filtered remains in 4230 CPM, then: a. The vibration at 47% is due to the Oil whip. b. The vibration at 47% is due to distortion. c. The vibration at 47% is due to the Oil whirling. d. The vibration at 47% is due to dry friction in the bearing. Answer: 100. If the bearings show superficial porosities in the babbit layer due to cavitation, then we should take the following actions: a. Change the lub oil, because it is contaminated with water. b. Clean the lub oil tank. c. Change the worn bearings. d. All the above. Answer: