Docslide.us tggg Balancing of Rotors Ppt

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DYNAMIC BALANCING OF ROTORS

ByDr. Rajiv Tiwari

Department of Mechanical Engineering

Indian Institute of Technology Guwahati !"#$%

&nder 'I(TE )ponsored *I+ )hort Term (ourse on

Theory & Practice of Rotor Dynamics

,"-"% Dec /##!0

IIT Guwahati

Dr. R. Tiwari ([email protected]) 1

mailto:[email protected]





















Intro!ction

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. f 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

e!tend as possible, for its smooth running. The residual unbalance

estimation in rotor"bearing system is an age"old problem.

#rom the state of the art of the unbalance estimation, the unbalance can be

obtained with fairly good accuracy $1"%&.

'ow the trend in the unbalance estimation is to reduce the number of test

runs reuired, especially for the application of large turbo generators where

the downtime is very e!pDer.nRs.iTviewa$r i (,r *tiw&.ar

[email protected])

+




















tatic balancing-

ingle plane balancing

Dynamic balancing -(i) Two plane balancing- #or rigid rotors only.

( ω

<

ω

)cr

(ii) #le!ible rotor balancing - f the shaftdoes

deflects, and the

deflection changes with speed, as it in the(ω

vicinity of

critical speeds .ω>

)cr

Dr. R. Tiwari ([email protected])






















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






















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












































0alancing of a fle!ible rotor

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






















Basic Princi"#es of Static Ri$i Rotor Ba#ancin$























Basic Princi"#es of Co!"#e Ri$i Rotor Ba#ancin$























Basic Princi"#es of Dynamic Ri$i Rotor Ba#ancin$























4

Ba#ancin$ of Ri$i Rotor

5radle balancing machine -

The1.

rotor is placed in the bearings of a cradle as shown in #ig.

Fi$!re % Cra#e a#ancin$ machine























The cradle is placed on two springs and can be fulcrum about #1 or #+ to

form a simple vibrating system.

Two fulcrum can be located at two chosen balance planes (i.e. and ),

where the correction mass to be added.

The rotor can be driven by a motor through a belt pulley arrangement.

f the spring system is such that the natural freuency of the system is in the

range of motor speed, the

mass in either plane can be

phase angle or the location of the unbalance

determined as follows.

#ulcrum the cradle in plane, by fi!ing #1 and releasing #+. Run the rotor to

resonance, observing the ma!imum amplitude to the right of fulcrum #+.

This vibration is due to all the unbalance in plane ,

plane has no moment about #1.Dr. R. Tiwari ([email protected])

since the unbalance in

1+






















6se a trial

vibration.

mass at a chosen location and determine the amplitude of

Fi$!re ' P#ot of (iration am"#it!es (ers!s tria# mass #ocations

7a8e a plot

(see #ig. +).

of this amplitude for different location of the same trial mass

The trial mass for correction is added at the location where the

amplitude of vibration is minimum.

ncrease or decrease the trial mass at the same locations, until the desired

level of balance is

and releasing #1.

consuming.

achieved. imilar procedure can be repeated by #i!ing #+

This procedure is tedious and sometimes may be timeDr. R. Tiwari ([email protected]) 1






















9 procedure to determine the correction mass and location can be laid

down as follows, based on four observations

(i) without any addition to the rotor

of amplitude -

(ii) with a trial mass at 8 : ;3

(iii) with a trial mass at 1;3

and

mass at 8

location.

: <2;3, where

8

(iv) with same trial

conveniently chosen

is measured from a

This procedure has to be repeated

and then for #+ ).

for two cases (e.g. when fulcruming at #1

(1) =et ;9 is the amplitude measured with trial run

(+) ;0 is the amplitude measured in trial run by addition of a trial mass W t

at ;3(arbitrary chosen location on rotor).Dr. R. Tiwari ([email protected]) 1/























>ence the vector 90 will represent the effect of trial mass W t .

(9t this stage we do not 8now the location of vector ;9 on therotor).

() ;5 is the vibration measured in the trial run with the trial mass at

1;3.

o we will have 90 : 95 with 1;3 phase difference between them.

(>ence 95 vector is also the effect of trial mass W t so the magnitudes 90 :

95 and phase will be 1;3).

>owever we 8now only ;9, ;0 ? ;5 from test run (1), (+) ? ()

respectively ? conditions 90 : 95 with 1;3phase.

#rom these information we have to construct or locate

a plane.

points ;, 9, 0 ? 5 on

Dr. R. Tiwari ([email protected]) 1%






















5onstruction rocedure -D

E’

C

A

A2;; α

B

φE

O

Fi$!re ) Constr!ction "roce!re























Brect a line ;9D eual to +C;9. ith ; as the center and ;0 ? ;5 as radii

and D as center and ;5 ? ;0 as radii draw arcs to intersect at 0 ? 5 (point

0 and 5 we will be obtained by above construction). .

Draw a circle with 05 as diameter and 9 as center. 5onstruct the

parallelogram ;0D5.

;3 line) and 95 1;3 position on'ow 90 represent position (i.e. reference

the rotor (9; is actual unbalance). The angular measurement may be

cloc8wise or 55 and is determined from the fourth observation.

The observation could be either ;B or ;BE (42;3

observed is in the vicinity of ;B, then the angle to be

or F2;3 ). f the value

measured 55.

>owever it will be 5 if ;BE is the reading observed in test (the fourth run

also chec8s the validity of the linearity used in the balancing procedure).

mDar s. sR.W Tiwaisri (prtir woapri@

or iittgio.enrnaelt.itno)

1*The magnitude of trial 90.t































































The unbalance ;9 can be obtained accordingly in the mass term. The

∠OABlocation of unbalance is

55).

and the direction from figure (i.e. 5 or

The test is repeated by ma8ing

plane .

the cradle fulcrum bed at #and

measurements are made in

This procedure is very time

of the rotor.

consuming and also restricts the mass and siGe

7odern balancing machines use amplitude and phase measurement

planes for balancing a rotor.

in two

7achines are either soft support or hard support machines.























2;;

2;;2;;

E 4ve 4ve

180 ; ;

;

180 ; ;

;

A

B E B;;

2;;

+*;;

Unbalance position

at shaft location,

(ccw direction +ve)

;;

+*;;

Unbalan

ce

position

at shaft

location

(ccw dir.

+ve)

OO

Location of fourth

measurement for

trial mass at 900

(ccw dir. +ve)

Location of

fourth

measurement for

trial mass at 900

(cw dir. +ve)2;;

2;;4ve 4ve

E’

+*;;

180 ; ;;

180 ; ;;

E’

+*;;

B;;

B

;;

+*;;

Unbalance position

at shaft location,

(ccw direction +ve)

+*;;

Unbalan

ce

position

at shaft

location

(ccw dir.

+ve)

OO

Location of fourth

measurement for

trial mass at 200

(ccw dir. +ve)

Location of

fourth

measurement for

trial mass at 200

(cw dir. +ve)

Dr. R. Tiwari

F

(r

i

t

$

iw

!

a

r

r

e

i@

*

iitg.ernet.

in)

12











































+,am"#e )-% n the balancing process we ma8e the following observations-

(i) ao : amplitude of vibration of the unbalanced rotor Has isI

(ii) a1 : amplitude with an additional one"unit correction at the location ; deg and

(iii) a2 : same as a1 but now at 1; 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.

5all that amplitude x . =et the un8nown location of the original unbalance be ϕ.

olve x and ϕ in terms of and show that in this answer there is

Thus four runs are necessary to solve the problem completely.

an ambiguity sign.

Answer . 7easurements are

U R ∠ϕ(i) amplitude of vibration with residual unbalance

00

!800

(ii) amplitude with unit trial mass at an angle of

(iii) amplitude with unit trial mass at an angle of

00(iv) x : amplitude with 1 at an angle of and without residual imbalance

(i.e. =

0 ),

AB = a!and AC = a2OA = a0

U R Dr. R. Tiwari ([email protected]) +;
























#igure .% shows variation parameters involved in the present problem. #rom ∆OAB

, we have

a2

+ OB 2

−

a2

0 !

cosϕ = (9)2a0OB

and

2 2 2

a0 + OC − a2

cos (π − ϕ )

=

(0)

2a0OC

, we haveinceOB =OC

a2 +OB 2 −a2

−cosϕ = 0 22a OB

0

(5)

Dr. R. Tiwari ([email protected]) +1






















D

C x

0 x

ϕ

a!

a2

Referenceline

B

a0

A

#igure .% Jeometrical constructions for determination of unbalance

;n euating euations (9) and (5), we get

2 2 2 2 2 2

2a0OB cosϕ = −(a0 + OB − a2 ) = (a0 + OB − a! ) (D)

which gives

2a2 + 2OB 2 − (a2 +

a 2 )

Dr. R. Tiwari ([email protected]) ++0 ! 2






















OB = x, since OB(or OC ) are effect of trial mass

of unit magnitude.

>ence euation (D) gives

! 2 2 22 2 2 2 or x = (a

! + a2) " 2 −

a0

x = ± (a + a ) −

a

2 ! 2 0

(B)

Buations (9) and (5) gives (noting that

= OB = x ),

OC

2 2 2

= 2a0 x cosϕ − a0 +

a!

x(#)

and

x

=

2 2 2

−2a0 x cosϕ − a0 + a2(J)

;n euating euations (#) and (J), we get

(>)2 2

cosϕ = (−a! + a2 ) " 2a0 x

Buation (B) gives the magnitude of the unbalance and euation (>) gives the

magnitude of the phase angle, the dir ection or sense of the phase cannot beDr. R. Tiwari ([email protected]) +

obtained from only above measurements.

























+,am"#e )-'- 9 short rotor or flywheel has to be balanced. ;bservations of

the vibration at one of the bearings are made in four runs as follows-

Run

Run

Run

Run

1K+K

K

/K

rotor Has isI amplitude

amplitude

amplitude

amplitude

.; µm

%.; µm

1;.;

µm

1;.%

µm

with

with

with

%gm. at ; deg.% gm. at 1;deg.

%gm. at 2; deg.

#ind the weight and location of the correction. Ta8e the trial and balancing

masses at the same radius.

Dr. R. Tiwari ([email protected]) +/






















B D

Answer -5

9

0

;

;9 : 9D : cm

D0 :1; cm

;0 : % cm

;B : 1;.% cm 90 : . cm

mbalance position

: angle 09;

: *1 deg. 55

#igure . Jeometrical constructions

#igure . shows the geometrical construction of the present problem with lengthsof various arcs. #rom this the net effect of the imbalance is given as

90 : . cm ≡ % gm

>ence, the residual imbalance is given as;9 : 9D : cm ≡ /.*+ gmDr. R. Tiwari ([email protected]) +%

R e f e r e

n c e l i n

e






















90 is the reference line. The fourth observation is intersecting at B, hence angle to be

measured in the 5 direction (i.e.∠#$%

∠#$0 : *1; 55 direction.

The unbalance magnitude and phase can be also obtained

(>), we have

). >ence, the unbalance position is given as

from euations (B) and

! 2 2 2

x = ± (a + a ) − a = &.09

'm

2 ! 2 0and

2 2

cosϕ = (−a! + a2 ) " 2a0 x = 0.0

ϕ = 2.8 de'

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






















The Inf#!ence Coefficient Metho (*)

Definition of nfluence coefficients

/i0 On#y force

#1

1

F% /ii0 On#y force F'

#+

++ 1

y 11 y +1 y 1+

y ++

α!2 F 2

α!2 F 2

y11 : displacement at station 1 due to y1+

y++

:

:α!! F !force #1 at station1 :

y+1 : displacement at station + due toα 2! F !

force #1 at station1 :

Dr. R. Tiwari ([email protected]) +*






















/iii0 1hen oth F% an F' are "resent

F 1 F 2

y 1 y 2

y!2 a!! F ! + a!2 F 2 y! =

y2

=

y!! + =

y2! y22 a 2! F ! + a 22 F 2+ =

*a!! a!2 y! ! r F ! l

r lor

=l y2 La 2!

Fi$!re 2

a 22 l F 2

nfluence coefficients can be obtained by e!perimentation

formulae i.e.

or by strength of

y ! ! y 2 !a a=

,=

e t

c .! ! 2 !

F ! F !Dr. R. Tiwari ([email protected]) +
























The Inf#!ence Coefficient Metho

l R

l 1 R 1

1'o trial mass

l 1 R 1

l 2

2

abR

R 2

abl

l 1

R 1

R 3

3 Trial mass T l

aal

l 3#igure % 0earing measurements and influence coefficients for a rigid

rotor


+2

b

2

#

a

1

aaR






















n soft support machines, the resonant freuency of the rotor support

system

support

is low and the rotor runs at a speed above the resonance of the

system. Libratory amplitudes are measured, which are then

converted to forces.

n hard support system, the support natural freuency 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.

e choose two convenient planes = and R for trial mass and two

measurement planes a and b (can be chosen as bearing locations).

=et =1 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]) ;






















ignal from station a shaft

T t 1

-pie due to

notch in theshaft surface

reference si'nal

/hase 0 with respect to notch2πt !

T

2

πt 2

/hase lead radians

/hase la' radiansT

Fi$!re 3Dr. R.

Tiwari ([email protected])

1

t 2
























The phase angles are measured with the same reference during the test

and their relative locations with respect to rotor is initially 8nown.

n the second run, place a trial mass TR at a convenient location in plane R

and let the observations be =+ and R+ with phase + and + respectively in

the a ? b planes.

The difference between R+

and R1

will be the effect of trial mass in rightplane R on the measurement made in plane b.

a bR

e can denote this as an

influence coefficient

(1)1

a = ( R − R ) " T bR 2 ! R

where M E represent vector since displacement

information. imilarly

1

has magnitude and phase

(+)a = ( L − L ! ) " T aR 2 R

e remove the trial mass from plane R and place T L in plane =and repeat

+meaDsr. uRr . eTdiwavria(r [email protected])

the test to obtain the


































1 1

abL aaL= ( R − R! ) "

T L

and = ( L − L! ) "

T L

(,/)

ith the help of euations (1) to

e!perimentally.

(/), we can obtain influence coefficient

W =et the correct balance masses be andW L. ince the

in right and left

original

planes,

R

unbalance response is R1 and =1 as

we can write

measured

1 1 1 1

and −

L!

1 11 1 1 1 (%)− R! =W RabR

+W LabL

=W RaaR +W LaaL

5orrection masses will produce vibration eual and opposite to the

vibration due to unbalance masses. >ence,

1 1 1r R!l

*a

−

a ! r W Rl ()bR bL

1= 1 1

**La a lW L l L!

aR aL

These can be calculated either by a graphical method or analytical

(compD

lr

e. R

!.

aTi

lw

ga

eri

b(r

r ti

aw

)ar

ii

.@

ei.itg.ernet.in)

method of vectors

mailto:@ei.itg.ernet.in
























− !

* a* c

b

d

d

− b!

!*

! *= w h e r e ∆ = ( a d − c b ) ∆ * −

c

*

aL L

which gives

1 1 1 1

L .a − R .a

1! b L ! a

1 L

R!.a a R − L

!.a b

1 R

W =a n

d W =1 1 1 1 1 R L

a .a −

a .a

a .a −

a .a

b R a

L

a R b

L

b R a

L

a R b

L

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






















+,"erimenta# set4!" for inf#!ence

hase mar8

on shaft

coefficient metho of a#ancin$

R 9ccelerometer (or

pro!imity probeson the shaft near

to the bearing)

= 9ccelerometer

7easurementlane a

7easurementlane b

>ardware or virtual instrumentationhoto sensitive mar8

7a!imum

displacement

location hoto electric probe

Fi$!re 5 +,"erimenta# set4!" for inf#!ence coefficient metho of a#ancin$


5harge 9mplifier

Librationmeter

hasemeter

;scilloscope

hoto electricrobe






















+,am"#e )-) 9 rigid rotor machine is e!hibiting vibration problems caused by imbalance.

The machine is symmetric about its center"line. 9 trial balance mass of ;. 8g is sited at

end 1 at an angle of ;; relative to some reference positionK this causes changes invibration vectors of %; µm at 1; at end 1 and /+ µm at 1;; at end +.

Determine the influence coefficients for use in balancing the machine, and calculate the

balance mass reuired at each end of the machine if the measured imbalance vibrations

are F; µm at +;; at end 1 and F*; µm at ;; at end +.

Solution. Jiven data are

T = 0. ' at 00 phase , which can be written asTrial mass in plane 1- R!

T R = 0.(cos 0 + 3 sin 0) = (0.298 + 3 0.!) 4'!

0

R2 = 0 5m at ! phase , which can be written asDisplacement in plane 1-

R2 = (2&.2&0& + 3 &.)

L = &2 5m at

!00

µm

p,hwasheich can be written

as

isplacement in plane +- 2

L2= (−2.99 + 3 2.!) µm























7easured responses due to residual imbalances are

R = -65m at 26

6

phase ≡ R = (!9.28 + 3

22.98)

n plane 1- µm! !

6n plane +- L

! = -65m at 6 phase≡ L

! = (-6.2!

+

3 ) µm

e have, influence coefficients as

R2 1 R! = (&8.89!9 + 3 !.)

x!6-a = a = µmN8g

!! bR

T R!

and

L2 1 L! = (92.9 - 3 9.!99) x!6-

a = a µmN8g=!2 aR

T R!

t is given that machine is symmetric about centreline.

a2! = a!2 a22 = a!!

and

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






















7easurements, influence coefficients and correction mass are related as

*a!! a!2 ! r w R l

r - R! l

= **l - L! La2! a22 l w L

which can be simplified as

!

∆

!

∆{ ! 22 } { ! !!} L a - R a R a - L aw =and w

= R ! !2 L ! 2!

- a a = a2 - a2 = (-&8.6 + 3 !96. x!6-) ∆ = a a (µmN8g)+with !! !2 !2 22 !! !2

which gives the balancing mass and its angular position as

- -1 6

w7 = -.!9 x!6 + 3 .!2 x!6 ≡ .89 x!6 ' at 29

and

wL = .96 x!6 - 3 2.! x!6 ≡ &.&!x!6 '

at

1

- -1 6























Ba#ancin$ of F#e,i#e Rotors .

9s long as the rotor e!periences no deformations i.e. it remains as a

rigid rotor, the balancing procedure discussed earlier is effective.

;nce 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 techniues are generally employed

(i) 7odal balancing techniue. 0ishop, Jladwell ? ar8insonand

(ii) nfluence coefficient method. Tessarri8, 0adgley and Rieger.

Moa# a#ancin$ metho

9 practical procedure to balance the rotor by model correction, masses

eual in number to the fle!ible mode shapes, ', 8nown as '"plane

method.Dr. R. Tiwari ([email protected]) 2






















Run the rotor in a suitable hard bearing balancing machine, to a safe speed

approaching the first critical speed and record the bearing vibrations

forces.

or

5hoose an appropriate location for the trial mass.

#or first critical speed, this should be roughly in the middle for a

a

t

symmetrical rotor in its a!ial distribution of mass. Record the readings

the same speed as before.

6sing the above two readings, the correct mass and location can be

determined. (single plane balancing). ith this correction mass, it should be

possible to run the rotor through the first critical speed without appreciable

vibration.

'e!t, 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 criticalDr. R. Tiwari ([email protected]) /;speed.






















Rigid body modes

Fi$!re 6

'ote the readings. 9dd a pair of trial masses 1;3 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).

'ote the readings at the same speed near the second critical speed.

Two readings can be used to determined the correction mass reuired.

Dr. R. Tiwari ([email protected]) /1

1 +






















'thimilarly higher modes can be balanced i.e. up

of

to

'

mode can be

balanced by ' balancing planes. nstead plane correction,

Oellenberger suggested that the rotor should

so as not to disturb the rigid body balancing.

be corrected in '4+ planes,

Moa# Ba#ancin$ /form!#ations0

z

x

bearing a!is

y

Fi$!re %7

9ssume that all unbalance isdistributed

only in the !"y plane.

=et the rotor speed be w and the deflection of the rotor be y(!).

Dr. R. Tiwari ([email protected]) /+






















The deflection y(!) can be written in terms of summation of mode shapes

as

()φ i Y ( x ) y ( x ) = i

Y i ( x) ithwhere is the mode shape in the mode and Pi is un8nown constant.The deflection y(!) can be measured e!perimentally.

#or e!ample for simply

I Moe

supported end conditions the mode shapes

II Moe

are

y ( x) = sin π x 2π x

l y ( x) = sin! 2

l

i th Moe

iπ x y ( x) = sinDr. R. Tiwairi ([email protected])

/

l

























#or other end conditions mode shapes can be obtained by free vibration

analysis. 7odal series for eccentricity can also be written as

a( x) = La i Y i ( x)

i

(2)

t can be written in terms of mode summation since the y ( x ) is the result of

a( x ). The main obQective is to find out for the eccentricity distribution to be

8nown.

7ultiplying () by the mass per unit length m(!) and the mode shape

Y j( x)

l

and integrate from ; to

1.

l * !

m( x) y( x)Y j ( x) = m( x)*LφiY i ( x)*Y j ( x)dx

(1;

)L

i

6 6

noting the orthogonality condition of mode shapes

l

6

m ( x)Y i ( x)Y j ( x)dx = 6 for all i ≠ j (11)//Dr. R. Tiwari ([email protected])

























Buation (1;)

givesl

l l 2 2

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

m( x)φ jY j ( x)dx =φ j

m( x)Y j

( x)dx

(1+)6 6 6

which can be written as

(1)l

6

!

m ( x ) y ( x )Y ( x ) dxφ

=

j j M j

(=

M )

Qthwith the generaliGed mass in mode is given as j

l

6

2 M

=m ( x )Y ( x )

dxi j

m( x) andY j ( x) Y j ( x) can M The generaliGed mass jcan be obtained by 8nowing

be obtained by free vibration analysis and

found by e!periment.

theoretically spea8ing can be

Dr. R. Tiwari ([email protected]) /%






















Joverning euation for shaft motion is

d [ E ( x) y ,,( x)] = ω 2 m( x)[ y ( x)

+ a( x)]

2

(1/)

dx 2

where w is the rotor speed. ;n substituting for y(!) and a(!) from euation

() and (2), we get

*!! *r

!d 2 r !

Y ( x) + {L a Y

( x)}*

* E ( x) L φ = ω 2m ( x) * L

φ

Y ,,( x) * (1%)i i i i i i

dx 2 *L

l

* *Ll i i

'oting the orthogonality condition and multiply both sides by Y i ( x) and

integrate over the length of shaft, the left hand side of euation (1%) gives

l

*o

*

* E

*

!!

!

!r "d 2

L * Y φ Y ( x )( x

)

( x )

dx

*

i i j2dx !l !

L i

;n performing integration by parts, we get


/






















l

*

*! !r

d [ Ei

φ Y ,,

( x ) }]l -

( x )

{

d* Ei ( x )

!φ Y ,,( x )

!

*Y

LL

!

,,( x )

dx

Y ( x ) j i i i i jdx dx *

! *o

l i

Lo

#irst term vanishes

again integration by

(since it is Gero for all boundary conditions), on ta8ing

parts of the second

l

term, we get

*

*!

*

r !

- Y j, ( x ) [ Ei ( x )

{

φ i Y ,,

( x )}] +

! !l

L L* φ i Y i,,( x )

Y

!

j,,( x )

dx

*

Ei

*

( x )io !

li

L o

#irst term again vanishes for all boundary conditions. ;n noting the

orthogonality condition euation (11), we get

l

Ei ( x ) [Y x ) ] dx*2 (1)

=

φ

j,,( = φ ! j j jo

where the generaliGed stiffness in Qth mode is defined as

l

*o

Ei ( x ) [Y j,, ( x ) ] dx

2 !

= jDr. R. Tiwari ([email protected]) /*






















#rom right hand side of euation (1%), noting euation () and (2), we have

ω

(φ

+ a ) M

2

j j j (1*)

Therefore form (1%), noting euations

= ω (φ j + a ) M

(1) and (1*), we have

2

φ j ! j j j

which can be rearranged as2

- ω ! " M j j

a (φ )= (1) j j2

ω

a;nce

j is obtained, then distribution of eccentricity a( x ) can be found from

Y j ( x) and " j = ! j " M jeuation (2). Buation (1) reuires m(!), y(!), .

The m(!) can be accurately found out, y(!) is difficult to obtained, Y j ( x) is

" j

( ! jithobtained by eigen analysis and is natural freuency in mode : M

j

/which can be obtained by eigen value analysis.























Inf#!ence coefficient methoF 1

+

a2

!

ω

sh

aft

1

a

!!

+

a

2!

1

a

!,

!

,

<<

"!

ω ≈

"!

#1 F#e,i#e shaft

Ri$i

a!,,!

ω ≈

"+

Ba#ancin$

+

a2,

,!

Fi$!re %%

"#anes

x

!

z

Meas!rin$ "#anes Dr. R. TiwaFrii($rt!iwr aeri@%ii'tg

.ernet.in)

/2












































5hoose number of balancing planes (where mass can be added or chip

off) where +, ! number

bearing planes.

is the measuring planes, generally it is two i.e. at

=et the unbalance in each of the balancing planes be , , U U ! , U 2 "

11 1

1

1

1*a !! a !2 a ! " ! r U !

!

!

!

!

r #!!

1 !! 1 1

! !! !!# 2 !

!a 2!

a 22 a 2 "

1

!!U

2=!

/%60!! 1 ! ! ! 11!!# !

!a a !

lU

!L

$!

l $ $" "

υwhere is the vibration measurement at the measuring plane.

7easurements are ta8en at number of speeds.

;n writing euation (12) for each of the speeds

Dr. R. Tiwari ([email protected]) %;






















1 1 1 1! ! ! !r

*

*

**

*

*

*

*

**

*

*

*

*

*

*l

l

*

*

**

*

*

*

*

**

*

*

*

*

*

*

. al

*

*

**

*

*

*

*

*

**

*

*

*

*

*

*

a.

.

.

.

.

.

.

a#

1!

1! !

1! 2

1! "

*a

! !

2 !

!

2 2

!

!

2 "a a# *

*

2

1

* 1 1 11

%

! ! !

$ "

2

! "

a a a# *

1$ ! r l

*

*

*

*

1$

1$ 2 1 !

* a1*2 2

! !

2

! 2

2

$2

a a# !

** % 2

1**

*

(+3){v } - : a ; < % =-

o r 1 1 1 1 %* a 2

2

$

2

$ "a a

#* $ !

* * *

n

1* % *1 1 11 *n n n

l * a a a# "

1!

1! !

1! 2

1! "

* a

*

n n

2!

n

2 2

n

$2

n

2 "

n

$ "

a a# !

*1 1 1 1

* an

$

n

$ ! a a# L

;nce the influence coefficients :a ; are 8nown for all speeds euation (+3)can be used to obtained unbalances -

<% = - (:a :a ; )- !< # =


; T (+1)

%1






















nfluence coefficient matri! can be obtained by

measuring displacements,

attaching trial masses and

from euation (12), we get for a particular speed

1! 1! 1!1 1. l

a a

!2

.

.

.

a

! "

!r #!!

l !r %! + T R l! !!

11! 1! 1!! 1! !

!!

!

!

!

!a

!

!!a a!#2! ! % 2

- 2 " !

2! 22

1

!

1

!

!!!

!!

! !1

%

"

! 1! ! 1! (++)!

l#$! a a $ 2 a$"

!l !L

$!

;n subtracting euation (++) from first ! euation in euation (12), we get

.1! 1! 1!

11 la a

!2

.

.

.

a

! "

r # ! - # ! l !r T l! !!!! ! R

1! 1! 1!! 1 1 ! ! !

!a

!

!!!#! - # !

! a a o

!

!

2! 2 - 2 " ! 2! 22

1

!

1

!

! !!! !! 1! 1! ! ! 1! !!

! (+)l#$! - #$ a a

$ 2

a

$"

o!

l !L

$!

#! -1!

1Buation (+) gives ! -

1 ! # ! - # !1 #2 ,

11 #!! 1 #!

$! $! 2! !

$!a2! - 1. 1

aa !

!!

, --, (+/)T

Dr. R. T i Rwari ([email protected] e Rt.in)%+ R




























imilarly by attaching a trial mass on plane + we get second

column of the influence coefficient matri! in euation (+/),

above analysis should be done at a constant speed.

the

imilarly

different

we can find the influence coefficient"matri! for

speeds.























Dr. R. Tiwari ([email protected]) %/






















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












































References

$1& . Oellenburger 12*+ Transa"t#ons o$ t%e Amer#"an &o"#ety o$ 'e"%an#"al

En(#neers, Sournal of Bngineering for ndustry 6*, %/"%3.

fle!ible rotor be balanced in ' or '4+ planes

hould a

$+& S. Drechsler 123 )nst#t*t#on o$ 'e"%an#"al En(#neers Con$eren"e on

+#brat#ons #n Rotat#n( 'a"%#nery , 5ambridge, 6O, %"*3. rocessing

surplus information in computer aided balancing of large fle!ible rotors.

$& . Jnil8a 12 ,o*rnal o$ +#brat#on 67, 1%*"1*+. 7odal balancing of

fle!ible rotors without test runs- an e!perimental investigation.

$/& S.7. Orod8iews8i, S. Ding and '. Uhang 122/ ,o*rnal o$ +#brat#on %36,

%"2. dentification of unbalance change using a non"linear

mathematical model for rotor bearing systems.

$%& 7.. Darlow 122, &r#n(er - +erla(. 0alancing of >igh"peedDr. R. Tiwari ([email protected])

7achinery,%*






















$& . Bdwards, 9.. =ees and 7.. #riswell +333 ,o*rnal o$ &o*n/ an/

+#brat#on, ')'(%), 2"22+. B!perimental dentification of B!citation and

upport arameters of a #le!ible Rotor"0earings"#oundation ystem from

a ingle Run"Down.

$*& R- Ti8ari, +33%, 'e"%an#"al &ystem an/ &#(nal ro"ess#n( , 5onditioning

of Regression 7atrices for imultaneous Bstimation of the Residual

6nbalance and 0earing Dynamic arameters (in press).

$& %1*+ - 122 pecification for 0alancing 0ench,

$2& 1+*/ - 122+N; 12+% - F Locabulary.

1223 7echanical vibration F 0alancing

$13& 1+*% - 122+N; +*1balancing euipment.

- 12*/ Description and evaluation of field

$11& 1+** - 122+N; +2% - 12% 0alancing machine " Description

and evaluation.Dr. R. Tiwari ([email protected]) %






















$1+& 1+* - 1222 N; *12 - 122/ 7echanical Libration " ymbolsfor 0alancing 7achines and 9ssociated nstrumentation.

$1& 1+3 - 122+N; %/3 - 123 7echanical balancing of fle!iblerotors.

$1/& 1/+3 - 122%N; +1 - 122 7echanical vibration " 0alancing" haft and fitment 8ey convention.

$1%& 1/*/ - 1222 N; */*% - 12/ 0alancing 7achines "Bnclosures and ;ther afety 7easures.

$1& 1/21 - +331 7echanical Libration " 7ethodsthe 7echanical 0alancing of #le!ible Rotors

and 5riteria for

Dr. R. Tiwari ([email protected]) %2






















Than8 you





















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