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DYNAMIC BALANCING OF ROTORS By

Dr. Rajiv Tiwari Department of Mechanical Engineering Indian Institute of Technology Guwahati 781039

Under AICTE Sponsored QIP Short Term Course on Theory & Practice of Rotor Dynamics (15-19 Dec 2008) IIT Guwahati Dr. R. Tiwari ([email protected])

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Introduction The unbalance in rotors will not only cause rotor vibrations, but also transmit rotating forces to bearings and to the foundation structure. The force thus transmitted may cause damage to machine parts and its foundation. If the transmitted force is large enough, it might affect even the neighboring machines and structures. Thus, it is necessary to remove the unbalance of a rotor, to as large an extend as possible, for its smooth running. The residual unbalance estimation in rotor-bearing system is an age-old problem. From the state of the art of the unbalance estimation, the unbalance can be obtained with fairly good accuracy [1-5]. Now the trend in the unbalance estimation is to reduce the number of test runs required, especially for the application of large turbo generators where Dr. R. Tiwari ([email protected]) 2 the downtime is very expensive [6,7].

Static balancing: Single plane balancing Dynamic balancing : (i) Two plane balancing: For rigid rotors only. ( ω < ω cr ) (ii) Flexible rotor balancing : If the shaft deflects, and the deflection changes with speed, as it does in the vicinity of (ω > ω cr ) critical speeds .

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Dr. R. Tiwari ([email protected])

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Dr. R. Tiwari ([email protected])

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Dr. R. Tiwari ([email protected])

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Balancing of a flexible rotor Dr. R. Tiwari ([email protected])

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Basic Principles of Static Rigid Rotor Balancing

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Basic Principles of Couple Rigid Rotor Balancing

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Basic Principles of Dynamic Rigid Rotor Balancing

Dr. R. Tiwari ([email protected])

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Balancing of Rigid Rotor Cradle balancing machine : The rotor is placed in the bearings of a cradle as shown in Fig. 1.

+

Figure 1 Craddle balancing machine Dr. R. Tiwari ([email protected])

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The cradle is placed on two springs and can be fulcrum about F1 or F2 to form a simple vibrating system. Two fulcrum can be located at two chosen balance planes (i.e. I and II), where the correction mass to be added. The rotor can be driven by a motor through a belt pulley arrangement. If the spring system is such that the natural frequency of the system is in the range of motor speed, the phase angle or the location of the unbalance mass in either plane can be determined as follows. Fulcrum the cradle in plane I, by fixing F1 and releasing F2. Run the rotor to resonance, observing the maximum amplitude to the right of fulcrum F2. This vibration is due to all the unbalance in plane II, since the unbalance in plane I has no moment about F1. Dr. R. Tiwari ([email protected])

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Use a trial mass at a chosen location and determine the amplitude of vibration.

Figure 2 Plot of vibration amplitudes versus trial mass locations

Make a plot of this amplitude for different location of the same trial mass (see Fig. 2). The trial mass for correction is added at the location where the amplitude of vibration is minimum. Increase or decrease the trial mass at the same locations, until the desired level of balance is achieved. Similar procedure can be repeated by Fixing F2 and releasing F1. This procedure is tedious and sometimes may be time Dr. R. Tiwari ([email protected]) 13 consuming.

A procedure to determine the correction mass and location can be laid down as follows, based on four observations of amplitude : (i) without any addition to the rotor (ii) with a trial mass at

= 0°

(iii) with a trial mass at 180°and (iv) with same trial mass at conveniently chosen location.

= ±90°, where

is measured from a

This procedure has to be repeated for two cases (e.g. when fulcruming at F1 and then for F2 ). (1) Let OA is the amplitude measured with trial run (2) OB is the amplitude measured in trial run by addition of a trial mass Wt at 0°(arbitrary chosen location on rotor). Dr. R. Tiwari ([email protected])

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Hence the vector AB will represent the effect of trial mass Wt. (At this stage we do not know the location of vector OA on the rotor). (3) OC is the vibration measured in the trial run with the trial mass at 180°. So we will have AB = AC with 180°phase difference between them. (Hence AC vector is also the effect of trial mass Wt so the magnitudes AB = AC and phase will be 180°). However we know only OA, OB & OC from test run (1), (2) & (3) respectively & conditions AB = AC with 180°phase. From these information we have to construct or locate points O, A, B & C on a plane.

Dr. R. Tiwari ([email protected])

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Construction Procedure : D E’ C

A

α

~900 B

φ E

O

Figure 3 Construction procedure Dr. R. Tiwari ([email protected])

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Erect a line OAD equal to 2×OA. With O as the center and OB & OC as radii and D as center and OC & OB as radii draw arcs to intersect at B & C (point B and C we will be obtained by above construction). . Draw a circle with BC as diameter and A as center. Construct the parallelogram OBDC. Now AB represent 0° position (i.e. reference line) and AC 180° position on the rotor (AO is actual unbalance). The angular measurement may be clockwise or CCW and is determined from the fourth observation. The observation could be either OE or OE’ (+90° or –90° ). If the value observed is in the vicinity of OE, then the angle to be measured CCW. However it will be CW if OE’ is the reading observed in test (the fourth run also checks the validity of the linearity used in the balancing procedure). Dr. R.W Tiwari ([email protected]) The magnitude of trial mass t is proportional to AB.

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The unbalance OA can be obtained accordingly in the mass term. The location of unbalance is ∠OAB and the direction from figure (i.e. CW or CCW). The test is repeated by making the cradle fulcrum bed at FII and measurements are made in plane I. This procedure is very time consuming and also restricts the mass and size of the rotor. Modern balancing machines use amplitude and phase measurement in two planes for balancing a rotor. Machines are either soft support or hard support machines.

Dr. R. Tiwari ([email protected])

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900

1800

00

B

E O

+ve

2700

00

Unbalance position at shaft location, (ccw direction +ve)

1800

B O

Location of fourth measurement for trial mass at 2700 (ccw dir. +ve)

1800

B

O

2700

Location of fourth measurement for trial mass at 900 (cw dir. +ve)

Unbalance position at shaft location (ccw dir. +ve) 900

00

2700

Unbalance position at shaft location, (ccw direction +ve)

+ve

00

+ve

E’

00

E

A

900

2700

900

00

900

Location of fourth measurement for trial mass at 900 (ccw dir. +ve)

900

1800

00

+ve

00

E’

B O

2700

Location of fourth measurement for trial mass at 2700 (cw dir. +ve)

Dr. R. TiwariFigure ([email protected]) 4

2700

Unbalance position at shaft location (ccw dir. +ve) 19

Example 3.1 In the balancing process we make the following observations: (i) ao = amplitude of vibration of the unbalanced rotor “as is” (ii) a1 = amplitude with an additional one-unit correction at the location 0 deg and (iii) a2 = same as a1 but now at 180 deg. The ideal rotor, unbalanced only with a unit unbalance (and thus not containing the residual unbalance), will have certain amplitude, which we cannot measure. Call that amplitude x. Let the unknown location of the original unbalance be ϕ. Solve x and ϕ in terms of and show that in this answer there is an ambiguity sign. Thus four runs are necessary to solve the problem completely.

Answer: Measurements are (i) amplitude of vibration with residual unbalance U R ∠ϕ (ii) amplitude with unit trial mass at an angle of 0 0 0 (iii) amplitude with unit trial mass at an angle of 180

0

(iv) x = amplitude with 1 at an angle of 0 and without residual imbalance (i.e. U R = 0 ), OA = a0 AB = a1 and AC = a2 Dr. R. Tiwari ([email protected])

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Figure 3.5 shows variation parameters involved in the present problem. From ∆OAB , we have

a02 + OB 2 − a12 cos ϕ = 2a0OB

(A)

and

a02 + OC 2 − a22 cos (π − ϕ ) = 2a0OC Since

OB =OC

(B)

, we have

a 2 + OB 2 − a 2 2 − cosϕ = 0 2a OB 0

(C) Dr. R. Tiwari ([email protected])

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D

C

x

0

x

ϕ

Reference line

B

a1

a2 a0

A

Figure .5 Geometrical constructions for determination of unbalance

On equating equations (A) and (C), we get

2a0OB cos ϕ = −(a02 + OB 2 − a22 ) = (a02 + OB 2 − a12 )

(D)

which gives 2a02 + 2OB 2 − ( a12 + a22 )

Dr. R. Tiwari ([email protected])

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OB = x, since OB

(or OC ) are effect of trial mass of unit magnitude.

Hence equation (D) gives x 2 = ( a12 + a22 ) / 2 − a02

or

x=±

1 2

( a12 + a22 ) − a02

(E)

Equations (A) and (C) gives (noting that OC = OB = x ),

x 2 = 2a0 x cos ϕ − a02 + a12 and

x 2 = −2a0 x cos ϕ − a02 + a22

(F) (G)

On equating equations (F) and (G), we get

cos ϕ = ( − a12 + a22 ) / 2a0 x

(H)

Equation (E) gives the magnitude of the unbalance and equation (H) gives the magnitude of the phase angle, the direction or sense of the phase cannot be Dr. R. Tiwari ([email protected]) 23 obtained from only above measurements.

Example 3.2. A short rotor or flywheel has to be balanced. Observations of the vibration at one of the bearings are made in four runs as follows: Run 1; rotor “as is” Run 2; with 5gm. at 0 deg. Run 3; with 5 gm. at 180 deg. Run 4; with 5gm. at 90 deg.

amplitude 6.0 µm amplitude 5.0 µm amplitude 10.0 µm amplitude 10.5 µm

Find the weight and location of the correction. Take the trial and balancing masses at the same radius.

Dr. R. Tiwari ([email protected])

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E

D

Answer:

C A e Ref

line B e c ren

O OA = AD = 6 cm DB =10 cm OB = 5 cm OE = 10.5 cm AB = 6.3 cm Imbalance position = angle BAO = 71 deg. CCW

Figure 3.6 Geometrical constructions Figure 3.6 shows the geometrical construction of the present problem with lengths of various arcs. From this the net effect of the imbalance is given as AB = 6.3 cm ≡ 5 gm Hence, the residual imbalance is given as OA = AD = 6 cm ≡ 4.762 gm Dr. R. Tiwari ([email protected])

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AB is the reference line. The fourth observation is intersecting at E, hence angle to be measured in the CW direction (i.e.∠BAE ). Hence, the unbalance position is given as ∠BA0 = 710 CCW direction.

The unbalance magnitude and phase can be also obtained from equations (E) and (H), we have x=±

1 2

( a12 + a22 ) − a02 = 4.09 gm

and

cos ϕ = (− a12 + a22 ) / 2a0 x = 0.6065

ϕ = 52.8deg

Dr. R. Tiwari ([email protected])

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The Influence Coefficient Method

Definition of Influence coefficients (i) Only force F1 1

F1

y11

2

y21

y11 = displacement at station 1 due to force F1 at station1 = α 11 F1 y21 = displacement at station 2 due to force F1 at station1 = α 21 F1

Dr. R. Tiwari ([email protected])

(7)

(ii) Only force F2 F2

1

y12

2

y22

y12 = α12 F2 y22 = α12 F2

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(iii) When both F1 and F2 are present F1

F2

y1

y2

y1 = y11 + y12 = α 11 F1 + α 12 F2 y 2 = y21 + y 22 = α 21 F1 + α 22 F2 y1 α 11 α 12 = y2 α 21 α 22

or

F1 F2

Figure 7

Influence coefficients can be obtained by experimentation or by strength of formulae i.e.

α 11

y11 = , F1

α

21

y 21 = F1

e tc.

Dr. R. Tiwari ([email protected])

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The Influence Coefficient Method L L1

R

a

b No trial mass

1

L1

aR

R1 1

R1

L2 2 2

bR

R2 bL

L1 R3 3

3

R1

Trial mass TL

aL

L3

Figure 5 Bearing measurements and influence coefficients for a rigid rotor

Dr. R. Tiwari ([email protected])

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In soft support machines, the resonant frequency of the rotor support system is low and the rotor runs at a speed above the resonance of the support system. Vibratory amplitudes are measured, which are then converted to forces. In hard support system, the support natural frequency is very high and they measure the rotor unbalance forces directly, independent of rotor mass and configuration. The balancing procedure is based on influence coefficient measurement. We choose two convenient planes L and R for trial mass and two measurement planes a and b (can be chosen as bearing locations). Let L1 and R1 be the initial readings of vibration levels (displacement, velocity or acceleration) measured with phase angle 1 and 1 respectively. Dr. R. Tiwari ([email protected])

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Signal from station a shaft

t2

T

t1

Spike due to notch in the shaft surface reference signal

Phase = 0 with respect to notch

Phase lead =

2πt1 radians T

2πt 2 Phase lag = radians T

Figure 6 Dr. R. Tiwari ([email protected])

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The phase angles are measured with the same reference during the test and their relative locations with respect to rotor is initially known. In the second run, place a trial mass TR at a convenient location in plane R and let the observations be L2 and R2 with phase 2 and 2 respectively in the a & b planes. The difference between R2 and R1 will be the effect of trial mass in right plane R on the measurement made in plane b. We can denote this as an influence coefficient α bR (1) α bR = ( R 2 − R 1 ) / T R where ‘ ’ represent vector since displacement has magnitude and phase information. Similarly (2) α

aR

= ( L 2 − L1 ) / T

R

We remove the trial mass from plane R and place TL in plane L and repeat Dr. R. Tiwari ([email protected]) 32 the test to obtain the measured values

α bL = ( R3 − R1 ) / TL

and α aL = ( L3 − L1 ) / TL

(3,4) With the help of equations (1) to (4), we can obtain influence coefficient experimentally. Let the correct balance masses be W R and W L . Since the original unbalance response is R1 and L1 as measured in right and left planes, we can write (5) −R = W α +W α and − L = W α +W α 1

R bR

L bL

1

R aR

L aL

Correction masses will produce vibration equal and opposite to the vibration due to unbalance masses. Hence,

R1 L1

α bR =− α aR

α bL α aL

WR

(6)

WL

These can be calculated either by a graphical method or analytical Dr. R. Tiwari ([email protected]) 33 method of vectors (complex algebra) i.e.

a

b

c

d

−1

1 = ∆

d

−b

−c

a

w h e re

∆ = (ad − cb )

which gives WR =

L1 .α b L − R1 .α a L α b R .α a L − α aR .α b L

and

WL =

R1 .α a R − L1 .α b R α b R .α a L − α aR .α bL

Dr. R. Tiwari ([email protected])

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Experimental set-up for influence coefficient method of balancing Phase mark on shaft Accelerometer

Measurement Plane a

R

L

Accelerometer (or proximity probes on the shaft near to the bearing)

Photo electric Probe

Charge Amplifier

Measurement Plane b

Vibration meter

Phase meter Oscilloscope

Maximum displacement location

Photo sensitive mark

Hardware or virtual instrumentation

Photo electric probe

Figure 8 Experimental set-up for influence coefficient method of balancing Dr. R. Tiwari ([email protected])

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Example 3.3 A rigid rotor machine is exhibiting vibration problems caused by imbalance. The machine is symmetric about its center-line. A trial balance mass of 0.3 kg is sited at end 1 at an angle of 300 relative to some reference position; this causes changes in vibration vectors of 50 µm at 610 at end 1 and 42 µm at 1300 at end 2. Determine the influence coefficients for use in balancing the machine, and calculate the balance mass required at each end of the machine if the measured imbalance vibrations are –30 µm at 2300 at end 1 and –70 µm at 3300 at end 2.

Solution: Given data are Trial mass in plane 1: TR1 = 0.3 kg

at 300 phase , which can be written as

TR1 = 0.3(cos30 + j sin 30) = ( 0.2598 + j 0.15 ) Kg 0 Displacement in plane 1: R2 = 50 m at 61 phase , which can be written as

R2 = ( 24.2404 + j 43.73) Displacement in plane 2:

µm

L2 = 42 m at 1300 phase , which can be written as L2 = ( −26.997 + j 32.173)

µm

Dr. R. Tiwari ([email protected])

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Measured responses due to residual imbalances are In plane 1: R1 = −30 m at 2300 phase ≡ R1 = (19.2836 + j 22.98 ) µm In plane 2:

L1 = −70 m at 3300 phase ≡ L = ( −60.621 + j 35 ) µm 1

We have, influence coefficients as

α11 = α bR =

R2 - R1 = (48.8919 + j 51.63766) × 10−6 TR1

µm/kg

and 12

= α aR =

L2 - L1 = (92.3559 − j 69.1995) × 10−6 µm/kg TR1

It is given that machine is symmetric about centreline.

α 21 = α12 and α 22 = α11 Dr. R. Tiwari ([email protected])

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Measurements, influence coefficients and correction mass are related as − R1 α11 α12 = α 21 α 22 − L1

wR wL

which can be simplified as

wR =

1 1 {L1α12 − R1α 22 } and wL = {R1α 21 − L1α11} ∆ ∆

with ∆ =

11 12

−

12

22

=

2 11

−

2 12

= ( −468.066 + j 16907.73 × 10−6 )

(µm/kg)2

which gives the balancing mass and its angular position as

wR = −3.3519 × 10−3 + j 7.123 × 10−3 ≡ 7.893 × 10-3 kg at 2950 and wL = 3.90356 ×10−3 − j 2.7136 ×10−3 ≡ 4.7541×10-3kg at -350 Dr. R. Tiwari ([email protected])

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Balancing of Flexible Rotors : As long as the rotor experiences no deformations i.e. it remains as a rigid rotor, the balancing procedure discussed earlier is effective. Once the rotor bends while approaching a critical speed, the bend center line whirls around and additional centrifugal forces are set-up and the rigid rotor balancing becomes ineffective. (sometimes rigid rotor balancing worsens bending mode whirl amplitude). Two different techniques are generally employed (i) Modal balancing technique. Bishop, Gladwell & Parkinson and (ii) Influence coefficient method. Tessarrik, Badgley and Rieger. Modal balancing method A practical procedure to balance the rotor by model correction, masses equal in number to the flexible mode shapes, N, known as N-plane method. Dr. R. Tiwari ([email protected])

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Run the rotor in a suitable hard bearing balancing machine, to a safe speed approaching the first critical speed and record the bearing vibrations or forces. Choose an appropriate location for the trial mass. For first critical speed, this should be roughly in the middle for a symmetrical rotor in its axial distribution of mass. Record the readings at the same speed as before. Using the above two readings, the correct mass and location can be determined. (single plane balancing). With this correction mass, it should be possible to run the rotor through the first critical speed without appreciable vibration. Next, run the rotor to a safe speed approaching the second critical speed, if the operating speed is near the second critical or above the second critical Dr. R. Tiwari ([email protected]) 40 speed.

1

2

Rigid body modes

Figure 9

Note the readings. Add a pair of trial masses 180° apart in two planes without a affecting the first mode. (in fact if we try to balance one particular mode it will not affect balancing of other modes). Note the readings at the same speed near the second critical speed. Two readings can be used to determined the correction mass required. Dr. R. Tiwari ([email protected])

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Similarly higher modes can be balanced i.e. up to Nth mode can be balanced by N balancing planes. Instead of N plane correction, Kellenberger suggested that the rotor should be corrected in N+2 planes, so as not to disturb the rigid body balancing. Modal Balancing (formulations) z x bearing axis

y Figure 10

Assume that all unbalance is distributed only in the x-y plane. Let the rotor speed be

and the deflection of the rotor be y(x). Dr. R. Tiwari ([email protected])

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The deflection y(x) can be written in terms of summation of mode shapes as (8) y(x) = φ Y (x) i

i

where Yi (x) is the mode shape in the ith mode and øi is unknown constant. The deflection y(x) can be measured experimentally. For example for simply supported end conditions the mode shapes are I Mode II Mode

y1 ( x) = sin

πx

y 2 ( x) = sin

l

2πx l

i th Mode

iπx y i (([email protected]) x) = sin Dr. R. Tiwari l

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For other end conditions mode shapes can be obtained by free vibration analysis. Modal series for eccentricity can also be written as

a( x) = i

α i Yi ( x)

(9)

It can be written in terms of mode summation since the y(x) is the result of a(x). The main objective is to find out for the eccentricity distribution to be known. Multiplying (8) by the mass per unit length m(x) and the mode shape Y j ( x) and integrate from 0 to 1. l

0

l

φiYi ( x) Y j ( x)dx

m( x) y( x)Y j ( x) = m( x)

(10)

i

0

noting the orthogonality condition of mode shapes l

0

m ( x )Yi ( x )Y j ( x ) dx = 0

for all i ≠ j

Dr. R. Tiwari ([email protected])

(11)

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Equation (10) gives l

l

l

m( x) y( x)Y j ( x)dx = m( x)φ j Y j ( x)dx =φ j m( x)Y j ( x)dx

0

2

0

2

(12)

0

which can be written as φ

j

1 = M

l

j

m ( x ) y ( x )Y j ( x ) dx

(13)

0

with the generalized mass in jth mode (= M j ) is given as

M

i

=

l

m ( x )Y

2 j

( x ) dx

0

The generalized mass M jcan be obtained by knowing m(x)andYj (x); Yj (x) can be obtained by free vibration analysis and theoretically speaking can be found by experiment. Dr. R. Tiwari ([email protected])

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Governing equation for shaft motion is

d2 dx

2 ′ ′ = [ EI ( x ) y ( x ) ] ω m ( x )[ y ( x ) + a ( x ) ] 2

(14)

where is the rotor speed. On substituting for y(x) and a(x) from equation (8) and (9), we get d2 EI ( x ) 2 dx

i

φ i Yi′′( x )

= ω 2 m ( x) i

φ i Yi ( x ) + { α i Yi ( x )}

(15)

Noting the orthogonality condition and multiply both sides by Yi (x) and integrate over the length of shaft, the left hand side of equation (15) gives l

0

d2 dx 2

φ iYi ″ ( x )

EI ( x )

Y j ( x ) dx

i

On performing integration by parts, we get

Dr. R. Tiwari ([email protected])

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[

d EI ( x ) dx

Y j (x)

{

φ i Y i ′′( x )

}]

l 0

l

d dx

− 0

φ i Y i ′′( x )

EI ( x )

Y j′′( x ) dx

i

First term vanishes (since it is zero for all boundary conditions), on taking again integration by parts of the second term, we get

[

− Y j′ ( x ) EI ( x )

{

φ i Y i ′′( x )

}]

l 0

l

+

φ i Y i ′′( x ) Y j′′( x ) dx

EI ( x ) 0

i

First term again vanishes for all boundary conditions. On noting the orthogonality condition equation (11), we get l

= φ

[

EI ( x ) Y j′′( x )

j

]2 dx

= φ jK

j

(16)

0

where the generalized stiffness in jth mode is defined as l

K

j

[

EI ( x ) Y j′′ ( x )

= 0

]

2

dx

Dr. R. Tiwari ([email protected])

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From right hand side of equation (15), noting equation (8) and (9), we have

ω 2 (φ j + α j )M j

(17)

Therefore form (15), noting equations (16) and (17), we have

φ j K j = ω 2 (φ j + α j )M j which can be rearranged as

α

j

=

K

j

/M

ω

j 2

−ω

2

(φ j )

(18)

Once α j is obtained, then distribution of eccentricity a(x) can be found from equation (9). Equation (18) requires m(x), y(x), Yj (x) and pj = Kj / Mj . The m(x) can be accurately found out, y(x) is difficult to obtained, Y j ( x) is obtained by eigen analysis and p j is natural frequency in ith mode = K j M j which can be obtained by eigen value analysis. Dr. R. Tiwari ([email protected])

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Influence coefficient method F1 2

1

α 21

α11

2

′ α 21

ω 2, q number is the measuring planes, generally it is two i.e. at bearing planes. Let the unbalance in each of the balancing planes be U 1 , U 2 , v1 v2 vq

=

α 11 α 21 α q1

α 12 α 22

α1p α2p

U1 U2

α qp

Up

where υ is the vibration measurement at Measurements are taken at number of speeds.

,U

p

(19)

the

measuring

plane.

On writing equation (19) for each of the speeds Dr. R. Tiwari ([email protected])

50

v 21

α α

1 11 1 21

α α

1 12 1 22

α α

1 1 p 1 2 p

v q1 v 12

α α

1 q1 2 11

α α

1 q 2 2 12

α α

1 qp 2 1 p

α

2 q1

α

2 q 2

α

2 qp

v 1n v 1n

α α

n 11 n 21

α α

n 12 n 22

α α

n 1 p n 2 p

v qn

α

n q1

α

n q 2

α

n qp

v 11

v

2 q

=

U

1

U

2

U

3

U

p

or

{ν } = [ α

] {U }

(20)

Once the influence coefficients [α ] are known for all speeds equation (20) can be used to obtained unbalances :

(

{U } = [α ] [α ] T

)

−1

{v} Dr. R. Tiwari ([email protected]t.in)

(21) 51

Influence coefficient matrix can be obtained by attaching trial masses and measuring displacements, from equation (19), we get for a particular speed 1 v11 1 v 21

=

vq11

α1

α1

α1

α

α

α

11 1 21

α1

12 1 22

α q1 2

q1

U 1 + TR

1p 1

U2

2p

α1

(22)

Up

qp

On subtracting equation (22) from first q equation in equation (19), we get 1 v11 − v11 1 v 21 − v 21

v q11 − v q1

=

α1

α1

α1

α

α

α

11 1 21

α1

q1

12 1 22

α1

q2

Equation (23) gives

1p 1 2p

α1

qp

TR 0

(23)

0

1 1 v − v 1 α21 = 21 2 , α1 = , 11 Dr. R. T TR Tiwari ([email protected]) R 1 v11 − v11

,

α

1 q1

=

v q11 − v q1 TR

(24)

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Similarly by attaching a trial mass on plane 2 we get second column of the influence coefficient matrix in equation (24), the above analysis should be done at a constant speed. Similarly we can find the influence coefficient-matrix for different speeds.

Dr. R. Tiwari ([email protected])

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Dr. R. Tiwari ([email protected])

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Dr. R. Tiwari ([email protected])

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Dr. R. Tiwari ([email protected])

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References [1] W. Kellenburger 1972 Transactions of the American Society of Mechanical Engineers, Journal of Engineering for Industry 94, 584-560. Should a flexible rotor be balanced in N or N+2 planes? [2] J. Drechsler 1980 Institution of Mechanical Engineers Conference on Vibrations in Rotating Machinery, Cambridge, UK, 65-70. Processing surplus information in computer aided balancing of large flexible rotors. [3] P. Gnilka 1983 Journal of Vibration 90, 157-172. Modal balancing of flexible rotors without test runs: an experimental investigation. [4] J.M. Krodkiewski, J. Ding and N. Zhang 1994 Journal of Vibration 169, 685-698. Identification of unbalance change using a non-linear mathematical model for rotor bearing systems. [5] M.S. Darlow 1989, Springer – Verlag, Balancing of High-Speed Machinery, Dr. R. Tiwari ([email protected])

57

[6] S. Edwards, A.W. Lees and M.I. Friswell 2000 Journal of Sound and Vibration, 232(5), 963-992. Experimental Identification of Excitation and Support Parameters of a Flexible Rotor-Bearings-Foundation System from a Single Run-Down. [7] R. Tiwari, 2005, Mechanical System and Signal Processing, Conditioning of Regression Matrices for Simultaneous Estimation of the Residual Unbalance and Bearing Dynamic Parameters (in press). [8] IS 5172 : 1969 Specification for Balancing Bench, [9] IS 13274 : 1992/ISO 1925 : 1990 Mechanical vibration – Balancing – Vocabulary. [10] IS 13275 : 1992/ISO 2371 : 1974 Description and evaluation of field balancing equipment. [11] IS 13277 : 1992/ISO 2953 : 1985 Balancing machine - Description and evaluation. Dr. R. Tiwari ([email protected])

58

[12] IS 13278 : 1999 /ISO 3719 : 1994 Mechanical Vibration - Symbols for Balancing Machines and Associated Instrumentation. [13] IS 13280 : 1992/ISO 5406 : 1980 Mechanical balancing of flexible rotors. [14] IS 14280 : 1995/ISO 8821 : 1989 Mechanical vibration - Balancing - Shaft and fitment key convention. [15] IS 14734 : 1999 /ISO 7475 : 1984 Balancing Machines Enclosures and Other Safety Measures. [16] IS 14918 : 2001 Mechanical Vibration - Methods and Criteria for the Mechanical Balancing of Flexible Rotors

Dr. R. Tiwari ([email protected])

59

Thank you

Dr. R. Tiwari ([email protected])

60

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Dr. Rajiv Tiwari Department of Mechanical Engineering Indian Institute of Technology Guwahati 781039

Under AICTE Sponsored QIP Short Term Course on Theory & Practice of Rotor Dynamics (15-19 Dec 2008) IIT Guwahati Dr. R. Tiwari ([email protected])

1

Introduction The unbalance in rotors will not only cause rotor vibrations, but also transmit rotating forces to bearings and to the foundation structure. The force thus transmitted may cause damage to machine parts and its foundation. If the transmitted force is large enough, it might affect even the neighboring machines and structures. Thus, it is necessary to remove the unbalance of a rotor, to as large an extend as possible, for its smooth running. The residual unbalance estimation in rotor-bearing system is an age-old problem. From the state of the art of the unbalance estimation, the unbalance can be obtained with fairly good accuracy [1-5]. Now the trend in the unbalance estimation is to reduce the number of test runs required, especially for the application of large turbo generators where Dr. R. Tiwari ([email protected]) 2 the downtime is very expensive [6,7].

Static balancing: Single plane balancing Dynamic balancing : (i) Two plane balancing: For rigid rotors only. ( ω < ω cr ) (ii) Flexible rotor balancing : If the shaft deflects, and the deflection changes with speed, as it does in the vicinity of (ω > ω cr ) critical speeds .

Dr. R. Tiwari ([email protected])

3

Dr. R. Tiwari ([email protected])

4

Dr. R. Tiwari ([email protected])

5

Dr. R. Tiwari ([email protected])

6

Balancing of a flexible rotor Dr. R. Tiwari ([email protected])

7

Basic Principles of Static Rigid Rotor Balancing

Dr. R. Tiwari ([email protected])

8

Basic Principles of Couple Rigid Rotor Balancing

Dr. R. Tiwari ([email protected])

9

Basic Principles of Dynamic Rigid Rotor Balancing

Dr. R. Tiwari ([email protected])

10

Balancing of Rigid Rotor Cradle balancing machine : The rotor is placed in the bearings of a cradle as shown in Fig. 1.

+

Figure 1 Craddle balancing machine Dr. R. Tiwari ([email protected])

11

The cradle is placed on two springs and can be fulcrum about F1 or F2 to form a simple vibrating system. Two fulcrum can be located at two chosen balance planes (i.e. I and II), where the correction mass to be added. The rotor can be driven by a motor through a belt pulley arrangement. If the spring system is such that the natural frequency of the system is in the range of motor speed, the phase angle or the location of the unbalance mass in either plane can be determined as follows. Fulcrum the cradle in plane I, by fixing F1 and releasing F2. Run the rotor to resonance, observing the maximum amplitude to the right of fulcrum F2. This vibration is due to all the unbalance in plane II, since the unbalance in plane I has no moment about F1. Dr. R. Tiwari ([email protected])

12

Use a trial mass at a chosen location and determine the amplitude of vibration.

Figure 2 Plot of vibration amplitudes versus trial mass locations

Make a plot of this amplitude for different location of the same trial mass (see Fig. 2). The trial mass for correction is added at the location where the amplitude of vibration is minimum. Increase or decrease the trial mass at the same locations, until the desired level of balance is achieved. Similar procedure can be repeated by Fixing F2 and releasing F1. This procedure is tedious and sometimes may be time Dr. R. Tiwari ([email protected]) 13 consuming.

A procedure to determine the correction mass and location can be laid down as follows, based on four observations of amplitude : (i) without any addition to the rotor (ii) with a trial mass at

= 0°

(iii) with a trial mass at 180°and (iv) with same trial mass at conveniently chosen location.

= ±90°, where

is measured from a

This procedure has to be repeated for two cases (e.g. when fulcruming at F1 and then for F2 ). (1) Let OA is the amplitude measured with trial run (2) OB is the amplitude measured in trial run by addition of a trial mass Wt at 0°(arbitrary chosen location on rotor). Dr. R. Tiwari ([email protected])

14

Hence the vector AB will represent the effect of trial mass Wt. (At this stage we do not know the location of vector OA on the rotor). (3) OC is the vibration measured in the trial run with the trial mass at 180°. So we will have AB = AC with 180°phase difference between them. (Hence AC vector is also the effect of trial mass Wt so the magnitudes AB = AC and phase will be 180°). However we know only OA, OB & OC from test run (1), (2) & (3) respectively & conditions AB = AC with 180°phase. From these information we have to construct or locate points O, A, B & C on a plane.

Dr. R. Tiwari ([email protected])

15

Construction Procedure : D E’ C

A

α

~900 B

φ E

O

Figure 3 Construction procedure Dr. R. Tiwari ([email protected])

16

Erect a line OAD equal to 2×OA. With O as the center and OB & OC as radii and D as center and OC & OB as radii draw arcs to intersect at B & C (point B and C we will be obtained by above construction). . Draw a circle with BC as diameter and A as center. Construct the parallelogram OBDC. Now AB represent 0° position (i.e. reference line) and AC 180° position on the rotor (AO is actual unbalance). The angular measurement may be clockwise or CCW and is determined from the fourth observation. The observation could be either OE or OE’ (+90° or –90° ). If the value observed is in the vicinity of OE, then the angle to be measured CCW. However it will be CW if OE’ is the reading observed in test (the fourth run also checks the validity of the linearity used in the balancing procedure). Dr. R.W Tiwari ([email protected]) The magnitude of trial mass t is proportional to AB.

17

The unbalance OA can be obtained accordingly in the mass term. The location of unbalance is ∠OAB and the direction from figure (i.e. CW or CCW). The test is repeated by making the cradle fulcrum bed at FII and measurements are made in plane I. This procedure is very time consuming and also restricts the mass and size of the rotor. Modern balancing machines use amplitude and phase measurement in two planes for balancing a rotor. Machines are either soft support or hard support machines.

Dr. R. Tiwari ([email protected])

18

900

1800

00

B

E O

+ve

2700

00

Unbalance position at shaft location, (ccw direction +ve)

1800

B O

Location of fourth measurement for trial mass at 2700 (ccw dir. +ve)

1800

B

O

2700

Location of fourth measurement for trial mass at 900 (cw dir. +ve)

Unbalance position at shaft location (ccw dir. +ve) 900

00

2700

Unbalance position at shaft location, (ccw direction +ve)

+ve

00

+ve

E’

00

E

A

900

2700

900

00

900

Location of fourth measurement for trial mass at 900 (ccw dir. +ve)

900

1800

00

+ve

00

E’

B O

2700

Location of fourth measurement for trial mass at 2700 (cw dir. +ve)

Dr. R. TiwariFigure ([email protected]) 4

2700

Unbalance position at shaft location (ccw dir. +ve) 19

Example 3.1 In the balancing process we make the following observations: (i) ao = amplitude of vibration of the unbalanced rotor “as is” (ii) a1 = amplitude with an additional one-unit correction at the location 0 deg and (iii) a2 = same as a1 but now at 180 deg. The ideal rotor, unbalanced only with a unit unbalance (and thus not containing the residual unbalance), will have certain amplitude, which we cannot measure. Call that amplitude x. Let the unknown location of the original unbalance be ϕ. Solve x and ϕ in terms of and show that in this answer there is an ambiguity sign. Thus four runs are necessary to solve the problem completely.

Answer: Measurements are (i) amplitude of vibration with residual unbalance U R ∠ϕ (ii) amplitude with unit trial mass at an angle of 0 0 0 (iii) amplitude with unit trial mass at an angle of 180

0

(iv) x = amplitude with 1 at an angle of 0 and without residual imbalance (i.e. U R = 0 ), OA = a0 AB = a1 and AC = a2 Dr. R. Tiwari ([email protected])

20

Figure 3.5 shows variation parameters involved in the present problem. From ∆OAB , we have

a02 + OB 2 − a12 cos ϕ = 2a0OB

(A)

and

a02 + OC 2 − a22 cos (π − ϕ ) = 2a0OC Since

OB =OC

(B)

, we have

a 2 + OB 2 − a 2 2 − cosϕ = 0 2a OB 0

(C) Dr. R. Tiwari ([email protected])

21

D

C

x

0

x

ϕ

Reference line

B

a1

a2 a0

A

Figure .5 Geometrical constructions for determination of unbalance

On equating equations (A) and (C), we get

2a0OB cos ϕ = −(a02 + OB 2 − a22 ) = (a02 + OB 2 − a12 )

(D)

which gives 2a02 + 2OB 2 − ( a12 + a22 )

Dr. R. Tiwari ([email protected])

22

OB = x, since OB

(or OC ) are effect of trial mass of unit magnitude.

Hence equation (D) gives x 2 = ( a12 + a22 ) / 2 − a02

or

x=±

1 2

( a12 + a22 ) − a02

(E)

Equations (A) and (C) gives (noting that OC = OB = x ),

x 2 = 2a0 x cos ϕ − a02 + a12 and

x 2 = −2a0 x cos ϕ − a02 + a22

(F) (G)

On equating equations (F) and (G), we get

cos ϕ = ( − a12 + a22 ) / 2a0 x

(H)

Equation (E) gives the magnitude of the unbalance and equation (H) gives the magnitude of the phase angle, the direction or sense of the phase cannot be Dr. R. Tiwari ([email protected]) 23 obtained from only above measurements.

Example 3.2. A short rotor or flywheel has to be balanced. Observations of the vibration at one of the bearings are made in four runs as follows: Run 1; rotor “as is” Run 2; with 5gm. at 0 deg. Run 3; with 5 gm. at 180 deg. Run 4; with 5gm. at 90 deg.

amplitude 6.0 µm amplitude 5.0 µm amplitude 10.0 µm amplitude 10.5 µm

Find the weight and location of the correction. Take the trial and balancing masses at the same radius.

Dr. R. Tiwari ([email protected])

24

E

D

Answer:

C A e Ref

line B e c ren

O OA = AD = 6 cm DB =10 cm OB = 5 cm OE = 10.5 cm AB = 6.3 cm Imbalance position = angle BAO = 71 deg. CCW

Figure 3.6 Geometrical constructions Figure 3.6 shows the geometrical construction of the present problem with lengths of various arcs. From this the net effect of the imbalance is given as AB = 6.3 cm ≡ 5 gm Hence, the residual imbalance is given as OA = AD = 6 cm ≡ 4.762 gm Dr. R. Tiwari ([email protected])

25

AB is the reference line. The fourth observation is intersecting at E, hence angle to be measured in the CW direction (i.e.∠BAE ). Hence, the unbalance position is given as ∠BA0 = 710 CCW direction.

The unbalance magnitude and phase can be also obtained from equations (E) and (H), we have x=±

1 2

( a12 + a22 ) − a02 = 4.09 gm

and

cos ϕ = (− a12 + a22 ) / 2a0 x = 0.6065

ϕ = 52.8deg

Dr. R. Tiwari ([email protected])

26

The Influence Coefficient Method

Definition of Influence coefficients (i) Only force F1 1

F1

y11

2

y21

y11 = displacement at station 1 due to force F1 at station1 = α 11 F1 y21 = displacement at station 2 due to force F1 at station1 = α 21 F1

Dr. R. Tiwari ([email protected])

(7)

(ii) Only force F2 F2

1

y12

2

y22

y12 = α12 F2 y22 = α12 F2

27

(iii) When both F1 and F2 are present F1

F2

y1

y2

y1 = y11 + y12 = α 11 F1 + α 12 F2 y 2 = y21 + y 22 = α 21 F1 + α 22 F2 y1 α 11 α 12 = y2 α 21 α 22

or

F1 F2

Figure 7

Influence coefficients can be obtained by experimentation or by strength of formulae i.e.

α 11

y11 = , F1

α

21

y 21 = F1

e tc.

Dr. R. Tiwari ([email protected])

28

The Influence Coefficient Method L L1

R

a

b No trial mass

1

L1

aR

R1 1

R1

L2 2 2

bR

R2 bL

L1 R3 3

3

R1

Trial mass TL

aL

L3

Figure 5 Bearing measurements and influence coefficients for a rigid rotor

Dr. R. Tiwari ([email protected])

29

In soft support machines, the resonant frequency of the rotor support system is low and the rotor runs at a speed above the resonance of the support system. Vibratory amplitudes are measured, which are then converted to forces. In hard support system, the support natural frequency is very high and they measure the rotor unbalance forces directly, independent of rotor mass and configuration. The balancing procedure is based on influence coefficient measurement. We choose two convenient planes L and R for trial mass and two measurement planes a and b (can be chosen as bearing locations). Let L1 and R1 be the initial readings of vibration levels (displacement, velocity or acceleration) measured with phase angle 1 and 1 respectively. Dr. R. Tiwari ([email protected])

30

Signal from station a shaft

t2

T

t1

Spike due to notch in the shaft surface reference signal

Phase = 0 with respect to notch

Phase lead =

2πt1 radians T

2πt 2 Phase lag = radians T

Figure 6 Dr. R. Tiwari ([email protected])

31

The phase angles are measured with the same reference during the test and their relative locations with respect to rotor is initially known. In the second run, place a trial mass TR at a convenient location in plane R and let the observations be L2 and R2 with phase 2 and 2 respectively in the a & b planes. The difference between R2 and R1 will be the effect of trial mass in right plane R on the measurement made in plane b. We can denote this as an influence coefficient α bR (1) α bR = ( R 2 − R 1 ) / T R where ‘ ’ represent vector since displacement has magnitude and phase information. Similarly (2) α

aR

= ( L 2 − L1 ) / T

R

We remove the trial mass from plane R and place TL in plane L and repeat Dr. R. Tiwari ([email protected]) 32 the test to obtain the measured values

α bL = ( R3 − R1 ) / TL

and α aL = ( L3 − L1 ) / TL

(3,4) With the help of equations (1) to (4), we can obtain influence coefficient experimentally. Let the correct balance masses be W R and W L . Since the original unbalance response is R1 and L1 as measured in right and left planes, we can write (5) −R = W α +W α and − L = W α +W α 1

R bR

L bL

1

R aR

L aL

Correction masses will produce vibration equal and opposite to the vibration due to unbalance masses. Hence,

R1 L1

α bR =− α aR

α bL α aL

WR

(6)

WL

These can be calculated either by a graphical method or analytical Dr. R. Tiwari ([email protected]) 33 method of vectors (complex algebra) i.e.

a

b

c

d

−1

1 = ∆

d

−b

−c

a

w h e re

∆ = (ad − cb )

which gives WR =

L1 .α b L − R1 .α a L α b R .α a L − α aR .α b L

and

WL =

R1 .α a R − L1 .α b R α b R .α a L − α aR .α bL

Dr. R. Tiwari ([email protected])

34

Experimental set-up for influence coefficient method of balancing Phase mark on shaft Accelerometer

Measurement Plane a

R

L

Accelerometer (or proximity probes on the shaft near to the bearing)

Photo electric Probe

Charge Amplifier

Measurement Plane b

Vibration meter

Phase meter Oscilloscope

Maximum displacement location

Photo sensitive mark

Hardware or virtual instrumentation

Photo electric probe

Figure 8 Experimental set-up for influence coefficient method of balancing Dr. R. Tiwari ([email protected])

35

Example 3.3 A rigid rotor machine is exhibiting vibration problems caused by imbalance. The machine is symmetric about its center-line. A trial balance mass of 0.3 kg is sited at end 1 at an angle of 300 relative to some reference position; this causes changes in vibration vectors of 50 µm at 610 at end 1 and 42 µm at 1300 at end 2. Determine the influence coefficients for use in balancing the machine, and calculate the balance mass required at each end of the machine if the measured imbalance vibrations are –30 µm at 2300 at end 1 and –70 µm at 3300 at end 2.

Solution: Given data are Trial mass in plane 1: TR1 = 0.3 kg

at 300 phase , which can be written as

TR1 = 0.3(cos30 + j sin 30) = ( 0.2598 + j 0.15 ) Kg 0 Displacement in plane 1: R2 = 50 m at 61 phase , which can be written as

R2 = ( 24.2404 + j 43.73) Displacement in plane 2:

µm

L2 = 42 m at 1300 phase , which can be written as L2 = ( −26.997 + j 32.173)

µm

Dr. R. Tiwari ([email protected])

36

Measured responses due to residual imbalances are In plane 1: R1 = −30 m at 2300 phase ≡ R1 = (19.2836 + j 22.98 ) µm In plane 2:

L1 = −70 m at 3300 phase ≡ L = ( −60.621 + j 35 ) µm 1

We have, influence coefficients as

α11 = α bR =

R2 - R1 = (48.8919 + j 51.63766) × 10−6 TR1

µm/kg

and 12

= α aR =

L2 - L1 = (92.3559 − j 69.1995) × 10−6 µm/kg TR1

It is given that machine is symmetric about centreline.

α 21 = α12 and α 22 = α11 Dr. R. Tiwari ([email protected])

37

Measurements, influence coefficients and correction mass are related as − R1 α11 α12 = α 21 α 22 − L1

wR wL

which can be simplified as

wR =

1 1 {L1α12 − R1α 22 } and wL = {R1α 21 − L1α11} ∆ ∆

with ∆ =

11 12

−

12

22

=

2 11

−

2 12

= ( −468.066 + j 16907.73 × 10−6 )

(µm/kg)2

which gives the balancing mass and its angular position as

wR = −3.3519 × 10−3 + j 7.123 × 10−3 ≡ 7.893 × 10-3 kg at 2950 and wL = 3.90356 ×10−3 − j 2.7136 ×10−3 ≡ 4.7541×10-3kg at -350 Dr. R. Tiwari ([email protected])

38

Balancing of Flexible Rotors : As long as the rotor experiences no deformations i.e. it remains as a rigid rotor, the balancing procedure discussed earlier is effective. Once the rotor bends while approaching a critical speed, the bend center line whirls around and additional centrifugal forces are set-up and the rigid rotor balancing becomes ineffective. (sometimes rigid rotor balancing worsens bending mode whirl amplitude). Two different techniques are generally employed (i) Modal balancing technique. Bishop, Gladwell & Parkinson and (ii) Influence coefficient method. Tessarrik, Badgley and Rieger. Modal balancing method A practical procedure to balance the rotor by model correction, masses equal in number to the flexible mode shapes, N, known as N-plane method. Dr. R. Tiwari ([email protected])

39

Run the rotor in a suitable hard bearing balancing machine, to a safe speed approaching the first critical speed and record the bearing vibrations or forces. Choose an appropriate location for the trial mass. For first critical speed, this should be roughly in the middle for a symmetrical rotor in its axial distribution of mass. Record the readings at the same speed as before. Using the above two readings, the correct mass and location can be determined. (single plane balancing). With this correction mass, it should be possible to run the rotor through the first critical speed without appreciable vibration. Next, run the rotor to a safe speed approaching the second critical speed, if the operating speed is near the second critical or above the second critical Dr. R. Tiwari ([email protected]) 40 speed.

1

2

Rigid body modes

Figure 9

Note the readings. Add a pair of trial masses 180° apart in two planes without a affecting the first mode. (in fact if we try to balance one particular mode it will not affect balancing of other modes). Note the readings at the same speed near the second critical speed. Two readings can be used to determined the correction mass required. Dr. R. Tiwari ([email protected])

41

Similarly higher modes can be balanced i.e. up to Nth mode can be balanced by N balancing planes. Instead of N plane correction, Kellenberger suggested that the rotor should be corrected in N+2 planes, so as not to disturb the rigid body balancing. Modal Balancing (formulations) z x bearing axis

y Figure 10

Assume that all unbalance is distributed only in the x-y plane. Let the rotor speed be

and the deflection of the rotor be y(x). Dr. R. Tiwari ([email protected])

42

The deflection y(x) can be written in terms of summation of mode shapes as (8) y(x) = φ Y (x) i

i

where Yi (x) is the mode shape in the ith mode and øi is unknown constant. The deflection y(x) can be measured experimentally. For example for simply supported end conditions the mode shapes are I Mode II Mode

y1 ( x) = sin

πx

y 2 ( x) = sin

l

2πx l

i th Mode

iπx y i (([email protected]) x) = sin Dr. R. Tiwari l

43

For other end conditions mode shapes can be obtained by free vibration analysis. Modal series for eccentricity can also be written as

a( x) = i

α i Yi ( x)

(9)

It can be written in terms of mode summation since the y(x) is the result of a(x). The main objective is to find out for the eccentricity distribution to be known. Multiplying (8) by the mass per unit length m(x) and the mode shape Y j ( x) and integrate from 0 to 1. l

0

l

φiYi ( x) Y j ( x)dx

m( x) y( x)Y j ( x) = m( x)

(10)

i

0

noting the orthogonality condition of mode shapes l

0

m ( x )Yi ( x )Y j ( x ) dx = 0

for all i ≠ j

Dr. R. Tiwari ([email protected])

(11)

44

Equation (10) gives l

l

l

m( x) y( x)Y j ( x)dx = m( x)φ j Y j ( x)dx =φ j m( x)Y j ( x)dx

0

2

0

2

(12)

0

which can be written as φ

j

1 = M

l

j

m ( x ) y ( x )Y j ( x ) dx

(13)

0

with the generalized mass in jth mode (= M j ) is given as

M

i

=

l

m ( x )Y

2 j

( x ) dx

0

The generalized mass M jcan be obtained by knowing m(x)andYj (x); Yj (x) can be obtained by free vibration analysis and theoretically speaking can be found by experiment. Dr. R. Tiwari ([email protected])

45

Governing equation for shaft motion is

d2 dx

2 ′ ′ = [ EI ( x ) y ( x ) ] ω m ( x )[ y ( x ) + a ( x ) ] 2

(14)

where is the rotor speed. On substituting for y(x) and a(x) from equation (8) and (9), we get d2 EI ( x ) 2 dx

i

φ i Yi′′( x )

= ω 2 m ( x) i

φ i Yi ( x ) + { α i Yi ( x )}

(15)

Noting the orthogonality condition and multiply both sides by Yi (x) and integrate over the length of shaft, the left hand side of equation (15) gives l

0

d2 dx 2

φ iYi ″ ( x )

EI ( x )

Y j ( x ) dx

i

On performing integration by parts, we get

Dr. R. Tiwari ([email protected])

46

[

d EI ( x ) dx

Y j (x)

{

φ i Y i ′′( x )

}]

l 0

l

d dx

− 0

φ i Y i ′′( x )

EI ( x )

Y j′′( x ) dx

i

First term vanishes (since it is zero for all boundary conditions), on taking again integration by parts of the second term, we get

[

− Y j′ ( x ) EI ( x )

{

φ i Y i ′′( x )

}]

l 0

l

+

φ i Y i ′′( x ) Y j′′( x ) dx

EI ( x ) 0

i

First term again vanishes for all boundary conditions. On noting the orthogonality condition equation (11), we get l

= φ

[

EI ( x ) Y j′′( x )

j

]2 dx

= φ jK

j

(16)

0

where the generalized stiffness in jth mode is defined as l

K

j

[

EI ( x ) Y j′′ ( x )

= 0

]

2

dx

Dr. R. Tiwari ([email protected])

47

From right hand side of equation (15), noting equation (8) and (9), we have

ω 2 (φ j + α j )M j

(17)

Therefore form (15), noting equations (16) and (17), we have

φ j K j = ω 2 (φ j + α j )M j which can be rearranged as

α

j

=

K

j

/M

ω

j 2

−ω

2

(φ j )

(18)

Once α j is obtained, then distribution of eccentricity a(x) can be found from equation (9). Equation (18) requires m(x), y(x), Yj (x) and pj = Kj / Mj . The m(x) can be accurately found out, y(x) is difficult to obtained, Y j ( x) is obtained by eigen analysis and p j is natural frequency in ith mode = K j M j which can be obtained by eigen value analysis. Dr. R. Tiwari ([email protected])

48

Influence coefficient method F1 2

1

α 21

α11

2

′ α 21

ω 2, q number is the measuring planes, generally it is two i.e. at bearing planes. Let the unbalance in each of the balancing planes be U 1 , U 2 , v1 v2 vq

=

α 11 α 21 α q1

α 12 α 22

α1p α2p

U1 U2

α qp

Up

where υ is the vibration measurement at Measurements are taken at number of speeds.

,U

p

(19)

the

measuring

plane.

On writing equation (19) for each of the speeds Dr. R. Tiwari ([email protected])

50

v 21

α α

1 11 1 21

α α

1 12 1 22

α α

1 1 p 1 2 p

v q1 v 12

α α

1 q1 2 11

α α

1 q 2 2 12

α α

1 qp 2 1 p

α

2 q1

α

2 q 2

α

2 qp

v 1n v 1n

α α

n 11 n 21

α α

n 12 n 22

α α

n 1 p n 2 p

v qn

α

n q1

α

n q 2

α

n qp

v 11

v

2 q

=

U

1

U

2

U

3

U

p

or

{ν } = [ α

] {U }

(20)

Once the influence coefficients [α ] are known for all speeds equation (20) can be used to obtained unbalances :

(

{U } = [α ] [α ] T

)

−1

{v} Dr. R. Tiwari ([email protected]t.in)

(21) 51

Influence coefficient matrix can be obtained by attaching trial masses and measuring displacements, from equation (19), we get for a particular speed 1 v11 1 v 21

=

vq11

α1

α1

α1

α

α

α

11 1 21

α1

12 1 22

α q1 2

q1

U 1 + TR

1p 1

U2

2p

α1

(22)

Up

qp

On subtracting equation (22) from first q equation in equation (19), we get 1 v11 − v11 1 v 21 − v 21

v q11 − v q1

=

α1

α1

α1

α

α

α

11 1 21

α1

q1

12 1 22

α1

q2

Equation (23) gives

1p 1 2p

α1

qp

TR 0

(23)

0

1 1 v − v 1 α21 = 21 2 , α1 = , 11 Dr. R. T TR Tiwari ([email protected]) R 1 v11 − v11

,

α

1 q1

=

v q11 − v q1 TR

(24)

52

Similarly by attaching a trial mass on plane 2 we get second column of the influence coefficient matrix in equation (24), the above analysis should be done at a constant speed. Similarly we can find the influence coefficient-matrix for different speeds.

Dr. R. Tiwari ([email protected])

53

Dr. R. Tiwari ([email protected])

54

Dr. R. Tiwari ([email protected])

55

Dr. R. Tiwari ([email protected])

56

References [1] W. Kellenburger 1972 Transactions of the American Society of Mechanical Engineers, Journal of Engineering for Industry 94, 584-560. Should a flexible rotor be balanced in N or N+2 planes? [2] J. Drechsler 1980 Institution of Mechanical Engineers Conference on Vibrations in Rotating Machinery, Cambridge, UK, 65-70. Processing surplus information in computer aided balancing of large flexible rotors. [3] P. Gnilka 1983 Journal of Vibration 90, 157-172. Modal balancing of flexible rotors without test runs: an experimental investigation. [4] J.M. Krodkiewski, J. Ding and N. Zhang 1994 Journal of Vibration 169, 685-698. Identification of unbalance change using a non-linear mathematical model for rotor bearing systems. [5] M.S. Darlow 1989, Springer – Verlag, Balancing of High-Speed Machinery, Dr. R. Tiwari ([email protected])

57

[6] S. Edwards, A.W. Lees and M.I. Friswell 2000 Journal of Sound and Vibration, 232(5), 963-992. Experimental Identification of Excitation and Support Parameters of a Flexible Rotor-Bearings-Foundation System from a Single Run-Down. [7] R. Tiwari, 2005, Mechanical System and Signal Processing, Conditioning of Regression Matrices for Simultaneous Estimation of the Residual Unbalance and Bearing Dynamic Parameters (in press). [8] IS 5172 : 1969 Specification for Balancing Bench, [9] IS 13274 : 1992/ISO 1925 : 1990 Mechanical vibration – Balancing – Vocabulary. [10] IS 13275 : 1992/ISO 2371 : 1974 Description and evaluation of field balancing equipment. [11] IS 13277 : 1992/ISO 2953 : 1985 Balancing machine - Description and evaluation. Dr. R. Tiwari ([email protected])

58

[12] IS 13278 : 1999 /ISO 3719 : 1994 Mechanical Vibration - Symbols for Balancing Machines and Associated Instrumentation. [13] IS 13280 : 1992/ISO 5406 : 1980 Mechanical balancing of flexible rotors. [14] IS 14280 : 1995/ISO 8821 : 1989 Mechanical vibration - Balancing - Shaft and fitment key convention. [15] IS 14734 : 1999 /ISO 7475 : 1984 Balancing Machines Enclosures and Other Safety Measures. [16] IS 14918 : 2001 Mechanical Vibration - Methods and Criteria for the Mechanical Balancing of Flexible Rotors

Dr. R. Tiwari ([email protected])

59

Thank you

Dr. R. Tiwari ([email protected])

60

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