MV Module 5 Solution. Module 5

Transcription

1 Module 5 Q68. With a neat diagram explain working principle of a vibrometer. D-14-Q5 (a)-10m Ans: A vibrometer or a seismometer is an instrument that measures the displacement of a vibrating body. It can be observed from Fig. 1 that Z/Y 1 when / (range II). Thus the relative displacement between the mass and the base (sensed by the transducer) is essentially the same as the displacement of the base. For an exact analysis, we consider equation (a) = / [ / ] +[/ ].. When the natural frequency of the instrument is low in comparison to vibration frequency, the relative displacement approaches the amplitude of vibrating body irrespective of damping in the instrument.,, = / [ / ] = { } i.e. Thus, when / is large, amplitude recorded is approximately equal to amplitude of vibrating body. In mostvibrometers, dumping is kept as small as possible. Vibrometers are, therefore, known as low natural frequency instruments. The average value of natural frequency, for vibrometers is about 4Hz. 72

2 Q69. Explain principle of accelerometer with sketch & frequency response curve. D-12-Q5 (b)-05m Ans: An instrument used to measures the acceleration of a vibrating body is known as accelerometer. = /.. [ / ] + [/ ] When the natural frequency of the instrument is high in comparison to vibration frequency i.e.if << 1, equation (a) reduces to =, = = Here /,, The expression of. Hence, the amplitude recorded is proportional to the acceleration of vibrating body. Rewriting equation (a) in the form = [ / ] + [/ ] For frequency ratio in the range /., the value of expression in the parenthesis lies between 0.96 and 1.04 for value of varying between Since/ is small, the natural frequency of instrument has to be large which are obtained using small masses and very stiff springs (i.e. short spring). Thus, the instrument is of smaller size. Due to small size and high sensitivity, accelerometers are widely used for vibration measurements. Accelerometers are designed with high natural frequency, so they may be termed High tuned instruments. The natural frequency for most of the good accelerometers is above 10,000 Hz. 73

3 Q.70).Explain vibration based conditions monitoring and fault daigonsis in rotating machine. D-15-Q 6(b)-5m Ans; Conditions monitoring of machine implies determination of condition of machine and its change with time Following techniques are used to monitor the condition of machine: 1. 1.Visual and aural monitoring 2. 2.Operational variables monitoring 3. 3.Temperature monitoring 4. 4.wear clebris monitoring 5. 5.Vibration monitoring Vibration based condition monitoring a. 1.Vibration monitoring is most commonly used m/c condition monitoring. b. 2.vibration signature of m/c are seen to be very much related to the health of a m/c c. 3.Thus measurement of vibration levels of machine component can provide useful information regarding faults like unbalance, misalignment, lack of oil, wear etc. d. 4.fig shiows frequency spectrum of vibration in ball bearings for original new and old ball bearings The increased level of vibrations and additional peak indicates bearing is defected 74

4 Q71). A seismic instrument is mounted on machine at 1000 rpm the natural frequency of seismic instrument is 20 rad/sec. The instruments record relative amplitude of 0.5 mm. Compute displacement, velocity & acceleration of machine. Damping factor of seismic instrument is neglected. D-12-Q3 (b)-10m Ans: Given: N = 100 rpm, = =, =. = =. / The steady-state relative amplitude is, = = As r >1 = [ ] + [ ] [ ] [ ]., =. [. ] [ = ( ) =. =. ] =. and velocity of vibrating machine,.displacement of vibrating machine =. =.. =. / Also, acceleration of vibrating machine, =. =.. =. / 75

5 Q.72). A commercial type vibration pick up has a natural frequency of 5.75 Hz and a damping factor of 0.65, what is lowest frequency beyond which the amplitude can be measured within 2% error. M-13-Q4 (b)-10m Given: =., =., = ±% For vibration pick up, the amplitude of steady state response is, = [ ] +[ ] For percent error, =. on solving the equation (a), for above value of, the roots turn out to be imaginary. Hence, we shall solve the problem using value of the ratio =.. = [ ] +[ ] Squaring on either side and simplifying, we can write the quadratic equation as, +. = The only positive roots of this quadratic equation is, r = 1.56 Hence, for = =. Therefore, =.. =. 76

6 Q.73). A vibrometer is to be designed to limit the maximum error to 1 percent of the true velocity. The natural frequency of the vibrometer is 90 Hz and the suspended mass to be 0.07 kg. Determine the spring stiffness and damping constant. Ans: Given: Maximum error =1% = M-14-Q5 (b)-10m = =. / m =0.07 kg As, = =. =. /, = = [ ] + [ ]. [ = ] The error would be maximum, when maximum. This maximum value can be found by. ( ) = = =. [ = ] = The above equation can be written as a quadratic equation of the form: + + = Where, =, =, =, = For 1% error, the value of R would be, Where, =. =. For =. = =. Solving the quadratic equation,, = ±.,. = ± 77

7 For =. = =. Solving the quadratic equation,, = ±. =. = =. or, =. = =.. = ± C = =.. [ = ] = Any of the above two values can be chosen for the corresponding values of damping coefficient would be =... =. /, =... =. / 78

8 Q.74).An acceleration is constructed by suspending a mass of 0.1 kg from a spring of stiffness 10,000 N/m with negligible damping. When mounted on the foundation of an engine, the peak to peak travel of the mass of the accelerator has been to be 10 mm at an engine speed of 1000 rpm. Determine the maximum displacement, maximum velocity and maximum acceleration. D-14-Q5 (b)-10m =., =,, =.. As, = = =, = =. /., = =.,, = = [ ] + [ ] =.. =. =., = =.. =. /, = =.. =. /.. = 79

9 Q.75).A spring-mass-damper system, having an undamped natural frequency of 100 Hz and a damping constant of 20 Ns/m, is used as an accelerometer to measure the vibration of a machine operating at a speed of 3000 rpm. If the actual acceleration is 10 m/s 2 and the recorded acceleration is 9 m/s 2. Find the mass and the spring constant of the accelerometer. D-15-Q5(a)-10m Given: = = = / C = 20 Ns/m = = = =. = / = / = / As, we know that, = = [ ] + [ ] [. ] + [. ] [ = ] =. Also, = =. =, =. =. Also, = =. =. / 80

10 Q.76) An accelerometer has a suspended mass of 0.01 kg with a damped natural frequency of vibration of 150 Hz. When mounted on an engine undergoing as acceleration of 9.81 m/s 2 at an operating speed of 6000 rpm the acceleration is recorded as 9.5 m/s 2 by the instrument. Find the damping constant (c) and spring stiffness (k) of the accelerometer. M-12-Q5 (b)-10m Ans: : =., =, = = =. / ( ) =., =. / The damped natural frequency is, =, = =. / The steady state amplitude of vibration is, = / [ ] + [ ].. [ ] + [] Now, squaring on both sides,. (. ) = [ ] + []. = [ ] + [] [ ] + [] = = [ ] + [ ] =. =.. + ( ) + =. + ( ). = (a) Also, = = =. 81

11 =. =. =... =... = (b) Putting (b) in eqn (a) +... = + [. ]. = +.. =. +. =. +. = =., =.. =, = =., =. =., =.. Taking, =. Now =. =. =. /, = =.. =. / The damping constant is, =... =... =. / 82

12 Q.77). An accelerometer has natural frequency15 khz, Determine the highest frequency it can measure within 1% accuracy. Assume damping ratio 0.7 D-13-2(b)-10m Ans: Given: Maximum error =1% = =. = ± % =.. Since, we know that, for accelerometer = 83 [ ] + [ ] for, =.. = [ ] + [. ] Squaring on both sides,. = Let =. =. = [ ] + [. ] [ ] +. [ ] +.. {[ ] +. } =. {[ + ] +. } = =,.. +. = Since, both roots are imaginary. Now, for or, =.. = [ ] + [. ] Squaring on both sides,. = Let =. = [ ] + [. ] [ ] +.

13 . = [ ] +.. {[ ] +. } =. {[ + ] +. } = =,... = =. =. taking =. = =. =. also, = =. =. =. 84