field balancing large rotating machinery january reprinted august reprinted june

7/31/2019 Field Balancing Large Rotating Machinery January Reprinted August Reprinted June

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FACILITIES INSTRUCTIONS,

STANDARDS, & TECHNIQUES

VOLUME 2-2

Previously Published as: Power O&M Bulletin No. 13

Field Balancing Large Rotating Machinery

ENGINEERING DIVISIONFACILITIES ENGINEERING BRANCH

DENVER OFFICE

DENVER, COLORADO

UNITED STATES DEPARTMENT OF THE INTERIOR

Bureau of Reclamation

Revised January 1983by

Darrell TempleMechanical Engineer

Reprinted August 1989

Reprinted June 1994

Internet Version Prepared October 1998


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PREFACEOne of the primary responsibilities of the Facilities Engineering Branch is to provide technical assistance to

field maintenance personnel. Bulletins such as this are written to provide assistance and knowledge for field

personnel to better perform plant operations. Also, it helps to minimize the number of field trips by Denver

Office personnel.

Since the previous edition (No. 13A, 1946) of this bulletin is outdated and more technical than necessary,

effort has been made to simplify the procedures and to update the bulletin.

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CONTENTS

PageTheory of balancing ................................................... 1Instrumentation and measurement............ .................... 2Vector techniques....................... .......................... ....... 4Vector addition....................... ......................... ............ 4Vector subtraction...................................... ................. 5Component vectors........................... ......................... .. 5Static balancing....................................... .................... 6Static balancing - example...................... ..................... 6Dynamic balancing...................................... ................ 9Dynamic balancing - example.......................... ............ 9Analytical static balancing - example........................... 16Analytical dynamic balan cing - example............. 17

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CONTENTS-Continued

FIGURES

Figure Page

I Elevation of rotor having unbalanced mass...... 22 Plan of rotor................. ............. ............. .......... 23 Vibration amplitude measuring points

location ............. ............. ............. ............. ..... 34 Vector OA direction and magnitude............ ..... 45 Vector addition.............. .............. ............. ........ 46 Vector subtraction..................... ............. ........... 57 Component vectors.......... ............. ............. ....... 68 Static balancing - vectors.................................. 89 Dynamic unbalanced rotor................... ............. 910 Dynamic balancing - upper guide bearing......... 1111 Dynamic balancing - lower guide bearing......... 1212 Dynamic balancing - lower guide bearing,

check............................................................... 1413 Dynamic balancing - balance weight

distribution .............. ............. ............. ............. 1514 Conditions at the upper guide bearing............... 16.15 Weight distribution on rotor arms .................... 1716 Deflections at upper guide bearing................. .... 1917 Deflections at lower guide bearing............. ........ 1918 Divided weight for upper rotor arms No. 2

and 3 .............................................................. 21

19 Divided weight for lower rotor arms No. 5

and 6 ................................................................ 21

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FIELD BALANCINGLARGE ROTATING MACHINERY

1. Purpose

This bulletin is intended as a guide to aid plantmaintenance personnel in balancing large rotatingmachinery. Vibration literature1 can be obtained forstudy in more technical detail. The objective ofbalancing is to reduce shaft vibration to a practical

minimum. Reducing shaft vibrations generallyreduces bearing loads and increases the service life.

1.1 - Turbine runners and pump impellers areusually dynamically balanced before they areinstalled and seldom require rebalancing.Normally, the rotating parts of generators andmotors are balanced after installation.Maintenance work and slight shape changes withage - in some instances - can alter the balance ofa generator or motor rotating parts enough torequire rebalancing. This bulletin is concernedwith correcting the unbalanced masses in rotors

and drive shafts.

2. Vibration Sources

Large hydraulic units are subjected to many kinds ofvibrations.

Vibration and unbalance can be caused bymechanical, electrical, or hydraulic problems. Theunbalance of rotating parts is the most common causeof excessive unit vibration. Before attempting tobalance the rotating parts of a unit, determine thepossible sources of vibration and unbalance.

R. P. Kroon, Balancing of Rotating Apparatus II, Journal of

Applied Mechanics, vol. 11, No. 1, March 1944.

E. L. Thearle, Dynamic Balancing of Rotating Machinery in the

Field, translation of the ASME, October 1934.

Mechanical vibration sources are:

Bolted connections in the rotating parts shouldbe checked for tightness

Unit alinement of the rotor or runner Poor bearing conditions or position Foundation rigidity (flexure)

Check for possible electrical sources of vibrationsuch as:

Nonuniform air gap in the motor, generator, orexciter

Short-circuited winding turns in the rotor poles Ellipticity of the rotor

The hydraulic passages of the unit should be checkedfor:

A nonuniform pressure distribution over thesurfaces of the turbine runner or pump impellercan cause hydraulic unbalance

Obstructions in the spiral case or volute Debris between the vanes Incorrect vertical position of the

impeller (relative to distributor)

Excessive cavitation in the unit causes

unbalance.

runner or

hydraulic

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3. Theory of Balancing

Unbalanced masses in the rotating parts create acentrifugal force that causes the unit to vibrate. Field

balancing consists of determining the amplitude(size) and location around the shaft (phase angle) ofthe unbalance and placing weights on the rotor tocounter the unbalance.

3.1- As the rotor rotates (fig. 1 ), the unbalancedmass m tends to pull the rotor toward the bearingson the side the unbalance is located creating anapparent high spot on the shaft. At very lowspeeds, this high spotwill be in phase with theunbalanced (heavy) mass. As speed increases, thehigh spotbegins to lag the heavy spot.

3.2 - For hydroelectric units, the operating speedusually is less than the critical speed and the lagangle will be less than 90.

3.3 - Figure 2 shows the relation between theunbalanced mass and the high spot at normaloperating speed.

3.4 - When balancing, the machine is assumed to belinear; i.e., vibration amplitudes are in proportion tothe forces causing them. For the purpose of this

bulletin, the term vibration amplitude is synonymouswith shaft deflection which is a measure of theunbalance present. The point that the vibrationamplitude is maximum is called the high spot.Consequently, to balance the unit, it is necessary todetermine thevibration amplitude and location of theshaft deflection caused by the unbalanced condition.With this information known, a balance weight isdetermined and positioned which gives a countereffect and balances the unit.

4. Instrumentation and Measurement

Several methods are used to locate and measure theunbalance. The most common method requires dialindicators and a stroboscope. Dial indicators aremounted, with the indicator stems in contact with theshaft, at locations where deflection measurements are

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desired. The shaft is marked and numbered withequally spaced segments. By using the stroboscopespeed control, the scope is adjusted to flash at anumber on the shaft and the dial indicator reading

corresponding to that number is recorded. Dialindicator readings are recorded at each numberedlocation. The largest reading and its location are usedin balancing techniques discussed later in thisbulletin.

4.1 - A more recent method requires the use of adirect writing oscillograph and proximityindicators. The proximity indicators are calibratedto show the shaft deflection magnitude. The shaftis marked by cementing a steel shim at a knownangular location - usually corresponding to No. 1

rotor arm. As the shim passes a proximity device,a characteristic mark will appear on theoscillograph record. This mark then can be usedto locate the angular position of the maximumshaft deflection. This method has the advantagethat - once set up - the deflection at every pointaround the shaft can be quickly and accuratelymeasured and a written record is generatedautomatically.

4.2 - A seismic velocity transducer and vibrationanalyzer also can be used like the proximityindicator and oscillograph. The analyzer indicates

visually the vibration amplitude and phase angle.Some analyzers often include an automatic chartrecorder to provide a written record of thebalancing.

4.3 - Regardless of the method used to obtaindata, the measurements are the same. Shaft runoutreadings are taken simultaneously at (fig. 3):

upper guide bearing

lower guide bearing

turbine guide bearing

The measurement points should be in the samevertical plane. Data from the upper and lowerguide bearings are used to balance the unit. Data

from the turbine guide bearing are monitored toassure that the rotor balancing is not adverselyaffecting the turbine guide bearing.

4.4 - Three trial runs are made to obtain datanecessary to balance a hydraulic unit. The firstrun is made in the as-found condition to determinethe vibration amplitude and location of theexisting shaft deflection.

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For the second run, a trial weight is attached atthe top of a rotor arm. The third run is made withthe trial weight removed from the top of the rotorarm and attached to the bottom of the same rotor

arm.

4.5 - Experience indicates that the trial weightshould be approximately equal to the weight of therotating parts divided by 10,000. Ideally, theweight should be located 180 plus the lag angle- in the direction of rotation - from the high spotto counter the unbalance. Therefore, the massshould be attached to the rotor arm nearest thislocation. In most cases, the lag angle will beunknown and in that event assume it to be 45.Throughout this bulletin, angles are measured in

the direction of unit rotation with respect to rotorarm No. 1 - the reference point.

5. Vector Techniques

The process of determining the required balanceweights and locations can be simplified by usingvectors to represent the unbalance. Definitions andexamples are shown below to explain vectorprocedure.

5.1 - A vector is defined as a quantity that hasboth magnitude and direction. The direction of

vectors is measured with respect to the samereference line.

5.2 - Example 1: Measurements at the uppergenerator guide bearing indicate a shaft deflectionof 0.006 inch at 30 from No. 1 rotor arm. The0.006 inch is the magnitude of vibration amplitudeand the 30 counterclockwise is the location(direction). Using a scale (such as one-fourth inchequals 0.001 inch), draw the deflection as avector. The 0.006 inch at 30 has been scaled asvector OA. Point O is the tail of the vector andpoint A is the head. Note that the arrowheadalways points in the direction that the vector acts,in this case away from point O and towards A atan angle of 30 from the reference line. Thus, thisvector can be represented by the notation OA andis not the same asAO.

6. Vector Addition

A vector can be added to another vector (original) byusing graphical techniques. Draw the vector to be

added using the head of the original vector as thestarting point. A vector drawn from the tail of theoriginal vector to the head of the added vector is thegraphical sum (resultant) of the two vectors.

6.1 - Example 2: A vector OB having amagnitude of 0.005 inch and a direction of 150 isadded to vector OA in figure 4.

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Both vectors are shown infigure 5. To add twovectors, translate vector OB as described aboveand shown in figure 5 as AC. Note ACequalsOB. Complete the triangle by drawing a vector

OCfrom the tail ofOA to the head ofAC. Thisvector OCis the graphical sum ofOA plus OB.The magnitude and direction ofOCcan be scaledas 0.0056 inch and 82, respectively.

Note that the term "graphical sum" has beenstressed. Vectors cannot be added in the usualsense without complicating this discussion. Thus,0.005 plus 0.006 equals 0.011 in ordinaryaddition; however, graphical vector additionyields:

0.005 /150 + 0.006 /30 = 0.0056 /82

The notation 0.005 /150 describes a vectorhaving magnitude 0.005 inch and in a direction150 from some reference line or axis.

7. Vector Subtraction

To subtract vectorB from vectorA, add the negativeof vectorB to vectorA. The negative of a vector isequal in magnitude to the vector but opposite indirection.

7.1 -Example 3: In figure 6, vector OB is to besubtracted from vector OA. The vector - OB isthe negative ofOB. Placing the tail of - OB at thehead ofOA and completing the triangle yields theresultant ofOA - OB which is OD. Note that OA- OB is equal in magnitude but opposite indirection to OB - OA. The magnitude OD is:

0.006 /30 - 0.005 /150 =0.006 /30 + 0.005 /330 = 0.0095 /3

8. Component Vectors

Often it is necessary to reduce a given vector into twocomponent vectors of known directions. Thesevectors will produce the same effect and, whenadded, yield the given vector. To graphically obtainthe magnitudes of the two component vectors,construct a parallelogram where two sides are

parallel to the known directions and also which hasthe given vector as the diagonal. The two sides arethe component vectors.

8.1 -Example 4: It is determined that a 50-poundweight placed 15 from rotor arm No. 1 (at 0)will balance a rotor. However, the closestlocations for installing weights are at 0 and 60(rotor arm No. 2). Therefore, two weights must beinstalled at 0 and 60 that will have the resultanteffect of a 50-pound weight at 15.

8.2 - The 50-pound weight at 15 is scaled as thevector OE(fig. 7).To reduce OE into a pair ofvectors located at 0 and 60(fig. 7)construct aparallelogram with sides at 0 and 60 and OE asthe diagonal at 15. Beginning at the head of givenvector OE, draw a line parallel to the 0 line

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(rotor arm No. 1) and extend it to intersect the 60line (rotor arm No. 2). This intersection is X.Also, beginning at the head of OE, draw a lineparallel to the 60 line and extend it to intersectthe 0 line. This intersection is Y. The pair of

vectors OXand OYare components to the givenvector OE. Therefore, the rotor should balance byplacing a weight equal in magnitude to OXat 60and a weight equal in magnitude to OYat 0. Byscaling, OX and OY are 15 and 42 pounds,respectively.

9. Static Balancing

Static or single plane balancing-can be used tobalance large motors and generators. In such cases,the balance weight has the same effect when placed

on either the top or bottom of the rotor arm. If theresults of the trial runs show only small differences invibration amplitude and phase angle (lag angle),when the weight is changed from the top of the rotorarm to the bottom, static balancing probably will beadequate.

9.1 - The required balance weight and its locationcan be determined either graphically oranalytically. The graphical method will be usedfor static balancing as it only requires draftingaids and does not use trigonometry. An exampleof the analytical method is given later.

9.2 - The graphical method requires triangles,scale, protractor, pencil, and paper. If polarcoordinate paper is used, the protractor can beomitted and the scale replaced by a divider.

10. Static Balancing -Example

Single-plane balancing technique. - The unit rotates

counterclockwise, the rotating parts weigh 200 000

pounds, and the rotor has six arms equally spaced.

Assume a lag angle of 45. Therefore, the trial weight

should be attached to the rotor arm nearest 225

(180 plus 45) to the high spot.

Location for weight = 150 + 225= 375 or 15

Rotor arm No. 1 (reference line) at 0 is the nearest

arm to 15.

w eight o f ro ta t ing parts

T ria l w e ig h t = 10 00 0

200 000= = 20 pounds

10 000

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Since the shaft deflections with the weight on top ofthe rotor arm are similar to the deflections with theweight on the bottom, static balancing may besufficient.

The largest deflection occurs at the upper guidebearing. Because the weight has slightly more effecton the deflection when it is on top of the rotor arm,

the balance weight should be attached there.

The upper guide bearing condition from the as-founddeflection and of the trial weight on top of rotor armNo. 1 are plotted in figure 8.The as-found deflectionis plotted as vector OA (0.009 inch /150) with thearrow pointing in the direction of the deflection.Likewise, the deflection with top trial weight isplotted as vector OB (0.006 /200). Construct avector fromA toB. VectorAB represents the effectof trial weight. The magnitude ofAB and the anglebetween OA and AB are scaled as 0.0069 inch and42, respectively.

By studying the triangle of vectors OA, OB, andABand applying the vector techniques introduced earlier,a better understanding of this method can be made.Note that the vectors are drawn such that:

OA +AB = OB

where:vector OA is the as-found deflection

vectorAB is the effect of trial weightvector OB is the deflection with top trial weight

(resultant)

To obtain a static balanced condition requires that theresultant deflection OB be equal to zero so that:

OA +AB = 0

Since it is not zero, some unbalance still remains.The following procedure can be used to counter theremaining unbalance.

Beginning at the origin O, draw a vector equal andopposite to OA and label its head C(fig. 8).VectorOCrepresents the effect required to balance (reducethe unbalance to zero). For simplification, vector ABis shifted to the origin O and shown as OB'. Aneffect equal to OCis needed to balance the unit. Tomake OB' equal to OC, OB' must be rotated 42 tothe same angular position as OCand increased by theratio OC/OB'.

The magnitude and direction of the effect are directlyproportional to the amount and location of the

balance weight. Therefore, the balance weight alsomust be rotated 42 in the direction of rotation and itsweight increased by the same ratio OC/OB'.

O C 0.00 9Required ratio = = = 1.3

O B ' 0.00 69Required length for OB' = OC

OC = (1.3) OB'= 1.3 (0.0069)= 0.009 inch

Required balance weight = 1.3 (20)= 26 pounds

Since the required location (42) does not coincidewith a rotor arm, the balance weight must bedivided into component weights to place them on the

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two adjacent arms. Draw the required balance weightas the vector, 26 /42 (fig. 8), and divide that vectorinto components corresponding to the positions ofrotor arms No. 1 and 2. By scaling or readingdivisions on the graph paper, the components are:

Weight on rotor arm No. 1 = 9.3 poundsWeight on rotor arm No. 2 = 20 pounds

Remove the 20-pound trial weight from rotor armNo. 1 and place a 9.3-pound weight on rotor arm No.1 and a 20-pound weight on rotor arm No. 2. Theunit balance should be checked by running at normalspeed before the weights are attached permanently tothe rotor arms.

11. Dynamic Balancing

In some cases, weight added at one end of the rotorwill compound vibration. Here, the unbalance massand the added weightform a couple that deflects therotor.

Figure 9 illustrates a rotor having dynamicunbalance. Weight added at the top of the rotor tocounter ml will create a static unbalance on the m2side of the rotor. To correct dynamic unbalance,

weight must be added to both the top and the bottomof the rotor to produce a couple that will counter theeffect of the couple causing the vibration. Thedifficult part of dynamic balancing is determining thecorrect combinations of weights and locations.

11.1 - The required weights and locations canbe determined graphically, analytically, or by acombination of the two. Programs whichdetermine the required combination areavailable for both computers and programmablecalculators. However, the graphical method

presented here should be studied to give someinsight into the forces involved in dynamicbalancing. It may be quite valuable if unusualproblems arise which require innovative actions.An example of an analytical dynamic balancingis at the end of the bulletin.

11.2 - The same equipment used for static balancingis used for the graphical method of dynamicbalancing. It is noted that the graphical method is atrial-and-error method which intuitively may requirejudgment and experience.

12. Dynamic Balancing -Example

The unit to be balanced rotates counterclockwise, therotating parts weigh 250 000 pounds, and the rotorhas six arms equally spaced.

Plot the maximum deflections for the upper andlower guide bearings on separate sheets and label theheads of the vectorsA. Refer to figure 10 andfigure 11. Subscripts t and b refer to upper and

lower bearings, respectively.

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Add a trial weight at the top of the rotor. In theabsence of other information, assume a lag angle of45. To counter the unbalance at the upper guidebearing, the weight should be attached to the rotor

arm nearest 225 (180 plus 45) from the high spot.

Location for weight = 170 + 225= 395 or 35

Rotor arm No.2 at 60 is the arm nearest 35

w eight o f ro ta t ing partsT ria l w e ig h t =

10 00 0

250 000= = 25 pounds

10 000Plot the maximum deflections for the upper . and

lower guide bearings on the respective sheets andlabel the heads of the vectorsB.

Remove the trial weight from the top of rotor armNo. 2 and put it on the bottom of the rotor arm. Thetrial weight now should be located to counter theas-found deflection at the lower guide bearing.

Location for weight = 0 + 225 = 225

Rotor arm No. 5 at 240 is nearest 225

Plot the maximum deflections on the respective sheets

and label the heads of the vectors C.

Construct vectors fromA toB and fromA to Conboth the upper and lower guide bearing sheets. Thevectors OA, OB, and OCare resultant vectors andthe vectorsAB andACare effect vectors. On figure10,AtBt is the effect of top trial weight (on the rotortop) andAtCtis the cross effect of bottom trial weight(on the rotor bottom). On figure 11,AbBb is the crosseffect of top trial weight (on the rotor top) andAbCbis effect of bottom trial weight (on the rotor bottom).

The upper guide bearing has the highest vibrationamplitude (shaft deflection). Using the single-planebalancing technique, find the top balance weightrequired to make the upper guide bearing deflectionzero. Start with OCt rather than OAt. This way thecross effect of the bottom trial weight is accountedfor. Fromfigure 10 and translatingAtB, as OB't:

OCt = 0.009 inch /180

AtBt = OB't = 0.0075 /328

To reduce the deflection to zero,AtBt must be equaland opposite to OCt.

Required angular rotation = 360 - 328= 32 counterclockwise

O CtRequired weight = trial weight

tA Bt

= 25 ( 0.009 ) = 30 p o un ds0.007 5

This change will change the cross effectproportionately at the lower guide bearing. The crosseffect will be increased by the same ratioOCt/AtBt and rotated 32 ccw (counterclockwise). Fromfigure 11, the new cross effectis shown asAbB .x

0.009

)(AbBx = AbBb

0.0075 0 .0 0 9 6 )= (0 .0 018 in ch )( = 0 .0037 in ch0 .0 0 4 7

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.

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AbCx is rotated 32 ccw fromAbCb.

AbBx = 0.0034 inch /281 + 32

= 0.0034 inch /313 as it would appear with

its tail at O.

The total unbalance at the lower guide bearing is the

vector sum of the original unbalance and the new

cross effect.

OBx = OAb +AbBx

Use the single-plane technique to find the balance

weight that, when added to the bottom of the rotor,

will reduce the total unbalance at the lower guide

bearing to zero(fig. 11).

A

OBx = 0.0096 inch /345b Cb = OCb = 0.0047 inch /147

To reduce the unbalance to zero,Ab Cb must be equal

and opposite to OBx . Plot -OBx as OB'x = 0.0096

/165. ShiftAb Cb to the position shown as OC'b.

Required angular rotation = 165 - 147= 18 ccw

O C' xRequired weight = trial weight

O B ' t

= 2 5 ( 0 0 0 9 60 .0047

). = 5 1 p o u n d s

Changing the weight on the bottom of the rotor

changes the cross effect proportionately at the upper

guide bearing. The cross effect will be increased by

the ratio OB 'x/OC 'b a nd rotated 18

counterclockwise. On figure 10, the new cross effect

is shown asAtC.

AtCx = AtCt (0 .0096

)0 .0047

0 .0 0 9 6 )= (0 .0 018 in ch )( = 0 .0037 in ch0 .0 0 4 7

AtCx is rotated 18 ccw fromAtCt

AtCx = 0.0037 inch /231 + 18= 0.0037 inch /249 as it would

appear with its tail at O.

The total unbalance at the upper guide bearing is thesum between the vectors OAt andAtCx.

Total unbalance =AtCx +OAt = OCx

To balance, again use the single plane technique tofind the weight needed to bring the deflection at theupper guide bearing back to zero(fig. 10).

OCx = 0.0094 inch /193AtBt = OB't = 0.0075 inch /328

To bring the unbalance back to zero, AtBt must beequal and opposite OCx. Plot OCx as

OCx' = 0.0094 /13. Recall thatAtBt = OB't.

Required angular rotation = 13 - 328 +360= 45 ccw

O C' xRequired weight = trial weight

O B ' t

0 .0094 ) = 31 pounds= 2 5 (0 ,0075

Now a balance check can be made on the lower guidebearing. Again, the addition of this new weight willchange the cross effect proportionately at the lowerguide bearing. Referring tofigure 12:

0 0 0 9 6AbBxx = AbBb ( . ) / 2 8 1 + 4 5

0 .0 0 4 7

0 0 0 9 6= (0 . 0 0 2 8 in ch ) ( . ) / 3 2 6

0 . 0 0 4 7

= 0 .00 35 in ch / 3 2 6

The total unbalance at the lower guide bearing is thedifference between vectors OBxx and OBx.

Total unbalance = OZ =OBxx - OBx= 0.0009 inch /40

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Draw the desired balance weights as vectors on polarcoordinate paper and resolve them into componentscorresponding to the adjacent rotor arms as shown onfigure 13.

Place the weights on the appropriate rotor arms andobtain another set of deflection readings. Someadjustment may be required due to data or analysisinaccuracies.

13. Analytical Static Balancing -Example

Paragraph 10, Static Balancing - Example, will beused again for analytical static balancing.DataWeight of rotating parts . . . . . . . . 200 000 poundsNumber of rotor arms . . . . . . . . . 6 equally spacedUnit rotation . . . . . . . . . . . . . . . . counterclockwise Trial weight . . . . . . . . . . . . . . . . . . . . . 20 poundsShaft deflections . . . . . . . . . . . . . . . . . . . as foundUpper guide bearing . . . . . . . . . . 0.009 inch /150Lower guide bearing . . . . . . . . . . 0.008 inch /150Turbine guide bearing . . . . . . . . . 0.005 inch /150Shaft deflections - trial weight on top of rotor armNo. 1Upper guide bearing . . . . . . . . . . . 0.006 inch /200Lower guide bearing . . . . . . . . . . . 0.006 inch /200Turbine guide bearing . . . . . . . . . . 0.004 inch /200Shaft deflections - trial weight on bottom of rotorarm No. 1Upper guide bearing . . . . . . . . . . . . 0.007 inch /200 Lower guide bearing . . . . . . . . . . . . 0.006 inch /200Turbine guide bearing . . . . . . . . . 0.004 inch /200The trial weight has more effect on the shaftdeflections when it is on top of the rotor arm; therefore, the balance weight should be attached ontop.Figure 14 shows the as-foundunbalance at the upperguide bearing as well as the effect produced bypositioning the trial weight on top of rotor arm No. 1.

To balance the unit, the balance weight must have an

effect equal to the as-found unbalance and in the

opposite direction.

Required balance weight = weight causing

unbalance

It was assumed that the machine is linear; i.e., the

vibration amplitudes are in proportion to the forces

causing them:

C as - fo un d u nb ala nce=

A e ffe c t o f tr i a l w e i g h t

w eigh t caus ing unba lance=trial w e ig h t

The effect of trial weightA can be determined using

the law of cosines.

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2a = b2 + c2 - 2 bc cos ]0.5a = [b2 + c2 - 2 bc cos

a = [0.0062 + 0.0092 - 2 (0.006)(0.009)

Cos (200 - 150)]0.5

a = [0.000117 - 2(0.000054) 0.64279]0.5

a = [47.6 x 10-6]0.5 = 0.0069 inch

The fact that the required balance weight equals the

weight causing unbalance:

Required balance weight = (trial weight) (c/a)

= 20 (0.009/0.0069)

= 26 pounds

The angle by which the effect must be changed,

equals the angle the weight must be rotated. Solve forusing the law of sines:

a b

sin sin sin = (b/a) sin

= (0.006/0.0069) sin (200 -150V)

sin = (0.006/0.0069))0.76604) = 0.66612

= 41.8

The required location for the weight is 41.8

counterclockwise from rotor arm No. 1, the location

of the trial weight. The required location does notcoincide with a rotor arm so the weight must be

divided for placement on adjacent rotor arms.

The required weights are obtained using the sine law

(fig. 15):

W W1 W2sin sin sin

where:

W= 26 pounds

= 41.8

The angle between rotor arms is 60:thus:

+ = 60

= 60 - = 60 - 41.8

= 18.2

The sum of triangle angles must equal 180 + + = 180= 180 - - = 180 - 18.2 - 41.8 = 120

The weights for rotor arms No 1 and 2 now can bedetermined.

W1 =Wsin

sin=

26 sin 18.2

sin 120

= 9.3 pounds

W1 =

Wsin

sin =

26 sin 41.8

sin 120

= 20.0 pounds

Weights W1 and W2 are attached to the respectiverotor arms.

14. Analytical Dynamic Balancing - Example

Paragraph 12,Dynamic Balancing -Example,will beused again for analytical dynamic balancing.

Data

Weight of rotating parts . . . . . . . . . 250 000 poundsNumber of rotor arms . . . . . . . . . . .6 equally spacedUnit rotation . . . . . . . . . . . . . . . . counterclockwiseTrial weight . . . . . . . . . . . . . . . . . . . . . . . 25 pounds

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Shaft deflections . . . . . . . . . . . . . . . . . . . . . as found Upper guide bearing . . . . . . . . . . . . 0.008 inch /170 Lower guide bearing . . . . . . . . . . . . . 0.007 inch /0Turbine guide bearing . . .. . . . . . . . . . 0.006 inch /0Shaft deflections - trial weight on top of rotor arm No. 2Upper guide bearing . . . . . . . . . . . 0.003 inch /240Lower guide bearing . . . . . . . . . . . . 0.008 inch /340Turbine guide bearing . . . . . .. . . . . 0.007 inch /140 Shaft deflections - trial weight on bottom of rotorarm No. 5Upper guide bearing . . . . . . . . . . . . 0.009 inch /180 Lower guide bearing . . . . . . . .. . . . . 0.004 inch /40Turbine guide bearing . . . . . . . . . . 0.005 inch /180Two more vector operations are needed for analyticaldynamic balancing; i.e., vector division andmultiplication.To divide one vector by another vector, divide themagnitudes and subtract the angles.Example: To divide a vector of 0.006 inch at 30 by a vector of 0.005 at 150, divide magnitudes

0.006 inch/0.005 inch = 1.200

and subtract angles30 - 150 = - 120

To change the negative angle to a positive angle, add360. The result is 1.200 at - 120 or 1.200 at +240.

To multiply one vector by another, multiply themagnitudes and add the angles.

Example: To multiply a vector of 0.004 inch at 180by a vector of 0.010 inch at 45, multiply magnitudes

0.004(0.010) = 0.00004

and add angles180 + 45 = 225

(0.004 inch /180)(0.010 inch /45)= 0.00004 /225

18

The unit was assumed to be linear so the requiredweights Wrt and Wrb and locations will beproportional to the trial weights, that is:

Wrt = Wt

Wrb = Wb

where:

and are vector operators (having bothmagnitude and direction, andtand b refer to top and bottom subscripts,respectively.

To balance the unit, the values of and must be

determined.

Using the notation in the graphical method whereArefers to the as-found deflection, B refers todeflection with top trial weight, and C refers todeflection with bottom trial weight, the vectors are:

OAt = 8 /170 OCt = 9 /180 OBt = 3 /240

OAb = 7 /0 OCb = 4 /40 OBb = 8/340

For convenience, magnitudes are in mils rather thanin decimals (thousandths of an inch).

As shown onfigures 16 and 17:

OBt = OAt +At Bt OBb = OAb +Ab Bb

OCt = OAt +At Ct OCb = OAb +Ab Cb

To balance the unit, the combined effect of thebalance weights must be equal and opposite to theas-found unbalance at both the upper and lowerguide bearings. Since vibration amplitudes areproportional to the forces causing them, AtBt andAbBb are proportional to Wt andAbCb andAt Ct areproportional to Wb. Also, the required balance weightis proportional to the as-found unbalance; i.e., Wrtisproportional to OAt and Wrb is proportional to OAb.

- OAt = AtBt + AtCt- OAb = AbBb + AbCb


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From these equations, the vector operators aredetermined by

(OAt AbCb) - (OAb At Ct)=

(AbBb At Ct) - (AtBt AbCb)(OAb AtCt) - (OAt Ab Cb)

=(AbBb At Ct) - (AtBt AbCb)

Remember, these are vector equations.

OAt = 8 /170

OAb = 7 /0

OBt = OAt + AtBt = 3 /240

OBb = OAb +AbBb = 8 /340OCt = OAt +Ab Cb = 9 /180OCb = OAb +At Ct = 4 /40

The effects of the trial weights are computed usingvector subtraction or can be scaled:

At Bt = OBt - OAt = 3 /240 - 8 /170

Magnitude ofAtBt

= [ (3 sin 240 - 8 sin 1 70)2

+ (3 cos 240 - 8 cos 1 70)2 ]0.5= [(- 3.99)2 + (6.38)2 ]0.5 = 7.52

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Angle of:

-3.99AtBt = arc tan = arc tan - 0.6254

6.38= 328

AtBt = 7.52 /328AbBb = OBb - OAb = 8 /340 - 7 /0

= 2.78 /280.7AtCt = OCt - OAt = 9/180- 8 /170

= 1.79 /231.1AbCb = OCb - OAb = 4 /40 - 7 /0

= 4.70 /146.8Now, use vector multiplication:

OAt AbCb = 8 /170 4.70 /146.8= 8 4.70 /170 + 146.8= 37.60 /316.8

OAb AtCt = 7 /0 1.79 /231.1= 12.53 /231.

OAb AtBt = 7 /0 7.52 /328= 52.64 /328

OAt AbBb = 8 /170 2.78 /280.7= 22.24 /90.7

AbBb AtCt = 2.78 /280.7 1.79 /231.1= 4.98 /151.8AtBt AbCb = 7.52 /328 4.70 /146.8= 35.34 /114.8

Vector subtraction is now used to determine the

numerator and the denominator:

(OAt AbCb) - (OAb AtCt) = 37.60 /316.8 - 12.53 /231.1= 38.73 /335.6

(OAb AtBt) - (OAt AbBb) = 52.64 /328 - 22.24 /90.7= 67.31 /311.9

(AbBb AtCt) - (AtBt AbCb) = 4.98 /151.8 - 35.34 /114.8= 31.51 /289.3

38.73 /335.6 38.73= = /335.6 - 289.3

31.54 /289.3 31.51

= 1.23 /46.3

67.31 /311.9 67.31= = /311.9 - 289.3

31.51 /289.3 31.51

= 2.14 /22.6

Wrt = Wt = 1.23 /46.3 25 /60= 30.75 /106.3

Wrb = Wb = 2.14 /12.6 25 /240= 53.5 /262.6

Because the required locations do not coincide with

rotor arms, the weights must be divided forattachment on the adjacent rotor arms.

The top weight is proportioned between rotor armsNo. 2 and 3 using the sine law(see fig. 18).

Wrt W2 W3= =

sin t sin t Sin t

where:Wrt = 30.75 pounds

t = 46.3

t + t = 60

t = 60 - t = 60 - 46.3 = 13.7t + t + t = 180

t = 180 - ( t + t) = 180 - 60 = 120

sin t sin 13.7W2 = Wrt = 30.75

sin t sin 120

= 8.4 pounds

sin t sin 46.3W3 = Wrt = 30.75

sin t sin 120

= 25.7 pounds

Similarly, the bottom weight is proportioned betweenarms No. 5 and 6(see fig. 19).

Wrb W5 W6= =

sin b sin b Sin b

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where:

Wrb = 53.5 pounds

b = 22.6 b + b = 60 b = 60 - b = 60 - 22.6 = 37.4 b + b + b = 180 b = 180 - ( b + b) = 180 - 60 = 120

sin b sin 37.4 W5 = Wrb = 53.5

sin b sin 120 = 37.5 pounds

sin b sin 22.6 W6 = Wrb = 53.5

sin b sin 120 = 23.7 pounds

The weights are attached to the respective rotor arms.

The literature available has many cases in theanalyses of pump, turbine, and pump-turbinevibrations. As noted, one must first investigate thecause of vibrations and oscillations excited by suchconditions as vortex shedding, draft tube surges,penstock pressure fluctuations, electrical system, andmechanics.

21


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field balancing large rotating machinery january reprinted august reprinted june

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