Lecture 2-Field Balancing

7/30/2019 Lecture 2-Field Balancing

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LECTURE 2 - FIELD BALANCING OF ROTORS

Balancing of rotors in the field is necessary for the following reasons:

(i) the rotor would have been balanced in a commercial balancing machine running at 400 or 500 rpm, but

would be operating at a much higher speed of 3000 rpm. This may necessitate balancing afresh.

(ii) A heavy rotor which would have been operating at site for a long time and would have under gone

structural changes during shut down necessitating rebalancing at site.

(iii) Some machines such as fans and blowers may develop unbalance as a result of erosion. All these

cases necessitate the field balancing of rotors.

.2.1 Single Plane Balancing:

Quite often we come across rotors whose diameter is large in relation to its length.(more than 2).These

units can be treated as single plane objects. The resulting unbalance can be assumed to be in the same

plane. As a consequence the balance weight can be assumed to be located in the same plane. Let us

assume that the rotor is held overhanging from the supports shown in Fig2.1. Let us assume that the

dynamic response at the support is linearly proportional to the unbalanced force. It can be measured using

a vibration transducer. Let the reading be a . Let us choose a known weight and fix it at a known

location on the rotor(hereafterwards called zero degree position). Let the reading be b . It may be noted

that a and b are complex quantities since the magnitudes and phases of these responses need not be the

same(Fig8.2). Let us take the same known mass and place it at a location 180o

away from the earlier

position.(hereafterwards called 180 degree position). Let the reading be c . They are shown vectorially in

Fig.2.2.3. OA represents a , OB represents b and OC represents c . Let AD be equal to

OA . It is easy to visualise that the quadrilateral OCDB is a parallelogram. Since OA is equal to AD and

AC is equal to AB. AC and AB are the effects of the same known mass and hence their magnitudes must

be the same. In the triangle OCB, since OA is the meridion, one can express AB as

( ) ( )AB OC OB OA c b a= + + = + 2 2 2 2 2 22 2/ / (2.1)

If the figure drawn is correct, using cosine rule one can write

cos( )( )

1

2 2 2 2 2 2

2 2=

+ =

+ OA OB AB

OA OB

a b AB

ab(2.2)

Mathematically the two admissible solutions are 1 and 2. Similarly


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cos 2

2 2 2

2=

+ a c AC

ac(2.3)

To find out which one of the two is correct ,we can take the fourth reading by shifting the known mass to

90 degree position .If1 is correct, OE should be the reading and if2 is correct, OF will be the correct

reading. With these four readings the system can be balanced.

Example 2.1: Compute the magnitude and location of the balance weight needed to balance a rotor in the

field using the single plane balancing technique. Trial mass tried is weighing 100N at a radius of 1 meter.

Vibration reading at the support = 10mm/sec. Known mass at 0o

location = 12 mm/sec. Known mass at

180o

location = 14mm/sec. Known mass at 90o

location = 2.2mm/sec. Fig.2.3 gives the relative positions.

( )AB = + = =196 144 2 100 70 8 36/ .

cos ( ) / ( ) / /12 2

70 2 100 144 70 240 174 240= + = + =a b ab

Hence 1 =43.53o

From Eq(8.5) cos 2=(100+196-70)/(280).

Hence 2=36.18

cos =(144-70-100)/(28.3610)= -0.155

hence =99o

.

hence OE2=102+(8.36)2-2108.36cos 99=4.85

Hence OE=2.2

This is the recorded measurement also. Otherwise it would have been OF. The correction weight needed is

10/8.36 times the trial mass and is to be placed at a radius of 1meter at a location 99 from zero degree

position in the counterclockwise sense.

In the foregoing discussion it was indicated that four trials were needed to locate the phase and magnitude

of the correct weight. If the phase angles can be directly measured, one of trials can be saved.

.2.2 Two Plane Balancing:

If the rotor has a length comparable to the diameter( more than 0.5 times the diameter) it is not correct to

assume that the balancing can be done by adding masses in one unique plane. Then two plane balancing

becomes a necessity. This is usually done by the seven run method. The format for the seven run method

is indicated in Table(2.1)


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Table2.1 Format for the seven run method

run trial weight near end amplitude far end amplitude

1 nil AL1 AR1

2 near end 0o

BL1 BR1

3 180o

CL1 CR1

4 90o

DL1 DR1

5 far end 0o

BL2 BR2

6 180o

CL2 CR2

7 90o

DL2 DR2

Let the two planes in which the trial weights are added be referred to as near end and far end planes and

the two locations where the measurements are made (which are usually the left and the right bearings) be

referred to as the left and the right side.(Fig 2.4)The readings obtained are tabulated in Table2.1.

The following information as contained in Table2.2 is to be extracted from Table2.1. From the logic

adopted for single plane balancing one can visualise L0 and R0 are the same as AL1 and AR1. The readings

BL1,CL1 and DL1 will help in locating L1 and BR1, CR1. And DR1 will locate R1. Similarly the other six

readings will be used to determine L2 and R2.

Table 2.2 Six vectors for two plane balancing

Run trial wt. Near end

amplitude

phase far end

amplitude

phase

1 0 L0 0 R0 0

2 W plane 1 L1 1 R1 1

3 W plane 2 L2 2 R2 2

These are shown in Fig8.7. It is seen from this figure that the effect of trial weight in plane1 on the first

bearing is A (represented as a complex quantity ),on the second bearingA

. Similarly the effect of the

trial mass in plane 2 on the right bearing is B and on the left bearingB . It may be noted that and

have to be real numbers whose magnitudes have to be less than 1. This is because they must be equal to

the ratios of distances of planes of balance from the left and right side supports which in all cases will be


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less than 1. If the final solution to the problem is in the form of times the trial weight in plane 1 and

times the trial weight in plane2 (both and complex numbers) the solution can easily be written as

L A Bo + + = 0

R A Bo + + = 0 (2.4)

Hence

( )( )

=

=

+

L B

R B

A B

A B

L R B

AB

o

o o o

1(2.5)

Similarly

( )( )

=

L R AAB

o o

1(2.6)

The main advantage of the seven run method is that no phase angle measurement is needed. The technique

is easy to use and no sophisticated measuring equipment other than the vibrometer is needed. This will be

explained through an example.

Example 8.2 The readings obtained by the seven run method are shown in Table 2.3 Compute the balancing

mass needed in both the planes for complete balance.

Table 2.3 readings by seven run method

run near end (mm/sec) far end (mm/sec)

1 10.0 16.4

2 15.9 18.3

3 5.4 14.5

4 8.9 16.8

5 8.5 19.4

6 14.3 17.9

7 5.8 7.6

Let us assume that Lo=10 00

. Obviously L1= 15.9 and L2=8.5. Similarly Ro=16.4, R1=18.3 and R2=19.4.

The phase angles corresponding to L1, L2, Ro,R1 and R2 have to be determined. Referring to Fig8.8 one can

observe

A =+

=5 4 15 9

210 6 4272

2 2

2. .

.


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cos. .

.1

2 2 215 9 10 6 4272

2 15 9 10=

+

Hence 1=11.6

Now L1= 15.9 11.6o

Hence A=L1-Lo=6.4272 30o

.

Similarly

A =+

=14 5 18 3

216 4 1 9

2 2

2. .

. .

cos. . .

. .1

2 2 218 3 16 4 1 9

2 18 3 16 4=

+

Hence 1=0 .

A=1.9 30o

=1.9/6.4272=.3

Amust have the same phase angle as A since the ratio between them must be a number less than 1.

But

A R Roo o o= = = 1 18 3 30 16 4 30 1 9 30. . .

Since A has a phase angle 30 R1 and R2 must have the same phase angle. Referring to Fig 8.8 likewise we

observe

B =+

=8 53 14 3

210 6 21

2 2

2. .

.

B =+

=19 4 17 9

216 4 8 91

2 2

2. .

. .

Hence =6.21/8.91=.7

Likewise cos. .

.2

2 2 28 53 10 6 21

20 8 53=

+

2=38.12o

B o o o= = 8 53 38 15 10 0 6 21 122. . .

Bo o= = 19 4 16 4 30 8 91 1222. . .

Hen ce Bo

= 19 4 57 1 1. .


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=0.21

Hence

( )

=

=

R L

A

o o o

10 82 32 6. .

( )

=

=

L R

B

o o o

12 73 185 87. .

C PROG. 12C--------------------------------------------------------------------C PROGRAM TO COMPUTE THE UNBALANCED FORCE FOR TWO PLANE BALANCINGC METHOD WITH 7 TRIAL RUNSC Ref:-'TWO PLANE IN-SITU BALANCING' ,V.RAMAMURTI,K.ANANTARAMAN,C JSV(1989),134(2),pp 343-352C----------------------------------------------------------------------C READINGS WITH LOADS ON PLANE 1C AL1,AR1---READINGS ON THE LEFT & RIGHT BEARINGS WITH NO WTS.C BL1,BR1---READINGS WITH TRIAL WEIGHT AT 0 DEGREE POSITIONC CL1,CR1---READINGS WITH TRIAL WIEGHT AT 180 DEGREE POSITIONC DL1,DR1---READINGS WITH TRIAL WEIGHTS AT 90 DEGREE POSITIONC (4TH RUN IS FOR CHECK)C READINGS WITH LOADING IN PLANE 2

C BL2,BR2---READINGS WITH TRIAL WEIGHT AT 0 DEGREE POSITIONC CL2,CR2---READINGS WITH TRIAL WIEGHT AT 180 DEGREE POSITIONC DL2,DR2---READINGS WITH TRIAL WEIGHTS AT 90 DEGREE POSITIONC NOTATIONS ARE ALL IN THE ANTICLOCKWISE SENSEC----------------------------------------------------------------------

IMPLICIT REAL*8(A-H,O-Z)REAL L0,LAMBDA,MU,LAM_ANG,MU_ANGCHARACTER*32 INPFILE,OUTFILECOMMON PIPI=4*ATAN(1.0)con=PI/180WRITE(*,*)'ENTER THE INPUT FILE NAME'READ(*,*)INPFILE

WRITE(*,*)'ENTER THE OUTPUT FILE NAME'READ(*,*)OUTFILEOPEN(15,FILE=INPFILE)OPEN(16,FILE=OUTFILE)READ(15,*)ICHOICEIF(ICHOICE.EQ.1)THENWRITE(16,*)'----------------------------------------------------'WRITE(16,*)' SINGLE PLANE BALANCING:'WRITE(16,*)'----------------------------------------------------'READ(15,*)AL1


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READ(15,*)BL1READ(15,*)CL1READ(15,*)DL1ENDIFIF(ICHOICE.EQ.2)THENWRITE(16,*)'----------------------------------------------------'WRITE(16,*)' TWO PLANE BALANCING:'WRITE(16,*)'----------------------------------------------------'READ(15,*)AL1,AR1READ(15,*)BL1,BR1READ(15,*)CL1,CR1READ(15,*)DL1,DR1READ(15,*)BL2,BR2READ(15,*)CL2,CR2READ(15,*)DL2,DR2ENDIF

C----------------------------------------------------------------------C gamma_A is the inclination of A with L0(OR AL1)C gamma_aA IS THE INCLINATION OF aA WITH R0(OR AR1)C gamma_R IS THE INCLINATION OF R WITH L0

C gamma_B IS THE INCLINATION OF B WITH R0C----------------------------------------------------------------------

CALL SOLVE(AL1,BL1,CL1,A,gamma_A)CALL RUN4(AL1,BL1,CL1,DL1,A,gamma_A)IF(ICHOICE.EQ.1)THENL0=AL1LAMBDA=L0/Aunbal_ang=(PI-gamma_A)/conWRITE(16,6)unbal_angWRITE(16,2)WRITE(16,5)LAMBDAGOTO 10ENDIFCALL SOLVE(AR1,BR1,CR1,aA,gamma_aA)CALL RUN4(AR1,BR1,CR1,DR1,aA,gamma_aA)gamma_R=gamma_A-gamma_aA

CALL SOLVE(AR1,BR2,CR2,B,gamma_B)CALL RUN4(AR1,BR2,CR2,DR2,B,gamma_B)

C CONVERTING THE gamma_B W.R.TO L0gamma_B=gamma_B+gamma_R

CALL SOLVE(AL1,BL2,CL2,bB,gamma_bB)CALL RUN4(AL1,BL2,CL2,DL2,bB,gamma_bB)

C gamma_B &gamma_bB should be the same

beta=bB/Balpha=aA/AL0=AL1R0=AR1

C----------------------------------------------------------------------C THE SYSTEM WILL BE IN BALANCE WITH LAMBDA TIMES THE TRIAL MASSC IN PLANE1 & MU TIMES THE MASS IN PLANE2C NOTE:-L0,R0,A,B ARE VECTORSC LAMBDA=(-L0+R0*beta)/(A*(1-alpha*beta))C MU=(-R0+L0*alpha)/(B*(1-alpha*beta))


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C----------------------------------------------------------------------C=-L0+R0*beta*COS(gamma_R)S=R0*beta*SIN(gamma_R)CALL POLAR(C,S,R1,theta_Lam)LAMBDA=R1/(A*(1-alpha*beta))lam_ang=theta_lam-gamma_AC1=L0*alpha-R0*COS(gamma_R)S1=-R0*SIN(gamma_R)CALL POLAR(C1,S1,RR1,theta_mu)MU=RR1/(B*(1-alpha*beta))mu_ang=theta_mu-gamma_bB

A_INC1=(lam_ang-gamma_A)/CONA_INC2=(mu_ang-gamma_A)/CONIF(A_INC1.GE.360)A_INC1=A_INC1-360IF(A_INC2.GE.360)A_INC2=A_INC2-360WRITE(16,1)A_INC1WRITE(16,3)A_INC2WRITE(16,2)WRITE(16,7)LAMBDA

WRITE(16,4)MU10 WRITE(16,*)'----------------------------------------------------'1 FORMAT(2X,'PHASE ANGLE OF LAMBDA W.R.TO THE 0 degree POSITION

* =',F11.5)2 FORMAT(2X,'THE WEIGHTS TO BE ADDED TO REMOVE THE UNBALANCE IS:')3 FORMAT(2X,'PHASE ANGLE OF MU W.R.TO THE 0 degree POSITION

* =',F11.5/)4 FORMAT(2X,'MU =',F11.5,'*TRIAL MASS IN PLANE1'/)5 FORMAT(2X,'LAMBDA=',F11.5,'*TRIAL MASS')6 FORMAT(2X,'PHASE ANGLE OF LAMBDA W.R.TO 0 degree POSITION IS:',

* F11.5/)7 FORMAT(2X,'LAMBDA=',F11.5,'*TRIAL MASS IN PLANE 2')

STOPEND

C----------------------------------------------------------------------SUBROUTINE SOLVE(A0,A1,A2,A,theta)

C----------------------------------------------------------------------IMPLICIT REAL*8(A-H,O-Z)COMMON PIA=DSQRT((A1**2+A2**2)/2-A0**2)theta=ACOS((A0**2+A**2-A1**2)/(2*A0*A))RETURNEND

C----------------------------------------------------------------------SUBROUTINE POLAR(C,S,R,theta)

C----------------------------------------------------------------------IMPLICIT REAL*8(A-H,O-Z)

COMMON PIR=DSQRT(C**2+S**2)theta=ATAN(S/C)IF(C.LT.0)THETA=PI+thetaRETURNEND

C----------------------------------------------------------------------SUBROUTINE RUN4(OA,OB,OC,OD,AB,theta)

C----------------------------------------------------------------------IMPLICIT REAL*8(A-H,O-Z)


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COMMON PItheta1=theta-PI/2OD1=DSQRT(OA**2+AB**2-2*OA*AB*cos(theta1))error1=abs(OD1-OD)theta2=theta+PI/2OD2=dsqrt(OA**2+AB**2-2*OA*AB*cos(theta2))ERROR2=ABS(OD2-OD)IF(ERROR1.LE.ERROR2)THENtheta=PI-thetaELSEtheta=PI+thetaENDIFRETURNEND

EXAMPLE 8.2INPUT FILE:-210 16.415.9 18.3

5.4 14.58.9 16.88.5 19.414.3 17.95.8 7.6OUTPUT FILE:-----------------------------------------------------

TWO PLANE BALANCING:----------------------------------------------------PHASE ANGLE OF LAMBDA W.R.TO THE 0 degree POSITION = 32.16496PHASE ANGLE OF MU W.R.TO THE 0 degree POSITION = 63.71981

THE WEIGHTS TO BE ADDED TO REMOVE THE UNBALANCE IS:LAMBDA= 1.09265*TRIAL MASS IN PLANE 2MU = 1.96269*TRIAL MASS IN PLANE1

----------------------------------------------------

Fig 2.1 Over hanging disc

90o

O

E

B

F

C

A

Fig 2.2 Single plane balancing


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R1

L2

L1

R2

L0

2

1

(2-

0)

(1-

0)

R0Fig 2.5 Two plane balancing

l4

l1

l2

l3

Fig 2.4 Two plane balancing =l1/l

2,=l

3/l

4

Measuring

plane

Measuringplane

Near

end

plane

Far

end

plane

90o

O

E

B

C

A

Fig 2.3 Single plane balancing

14

10

2.2

12


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38.12o 580

L0=10

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Lecture 2-Field Balancing