Static Unbalance. Both bearing reactions occur in the same plane and in the. After balancing no bearing reaction remains theoretically.

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1 BALANCING

2 Static Unbalance A disk-shaft system on rigid rails An unbalanced disk-shaft system mounted on bearings and rotated For the unbalanced rotating system: Dynamic Bearing Reactions + Inertia Forces =0 -(F A + F B )i + mr G 2 i = 0 same direction. Balancing is performed on Both bearing reactions occur in the same plane and in the same direction. Balancing is performed on the plane of unbalance. After balancing no bearing reaction remains theoretically.

3 Dynamic Unbalance A long rotor mounted in bearings at A and B If m1=m2 and r1=r2 the rotor is statically balanced but dynamically unbalanced. Both bearing reactions occur in the same plane but in opposite directions. Balancing may be performed on the planes of unbalances or any other two convenient planes. After balancing, no bearing reactions remain theoretically.

4 Unbalance Types STATIC (Misalignment of axes) DYNAMIC (Tilt of axes) STATIC and DYNAMIC (GENERAL) (Called as DYNAMIC, in industry) (Both misalignment and tilt of axes)

5 Analysis of Static Unbalance F=0 m 1 R m 2 R m 3 R m c R c 2 =0 m c R c is the correction quantity. It is added to the plane of unbalances.

6 Analysis of Dynamic Unbalance Graphical Vectorial Scalar True moment polygon is 90 cw away

7 Industrial Balancing Machines

8 Balance Quality Grade Maximum Permissible Spesific Unbalance G M RA G. m. mr 2 r G Mr G 2 Mr G 2 =mr 2 Mr G =mr=u (Amont of Unbalance) r e=r G =mr/m=u/m (Spesific Unbalance) Empirically, e =constant G (Balance quality grade) ( ) For example: N=3000 rpm, G=6.3 give e=20 m as max permissible eccentiricity. Maximum Speed Permissible spesific unbalance (u/m) OR Mass center s eccentricity (e) 1) As the rotor mass increases permissible unbalance also increases (u=mr G ) 2) As the rotor speed increases permissible unbalance decreases (e =constant ) For small masses and high speeds, higher quality balance grade is required.

9 Balancing A Single-Cylinder Engine m A F C F = F A,B + F C Secondary component does not changed Primary component reduced 50% (Altough a new vertical component added) Eliminated

10 Polar Diagram of Inertia Forces Before adding m C : After adding m C :

11 Imaginary-Mass Approach 1) Two imaginary rotary masses are used for each reciprocating mass. 2) Each mass is equal to half the equivalent reciprocating mass. 3) These imaginary masses rotate about the crank center in opposite directions and with equal angular velocities. A simple model to analyse and recover the unbalance of reciprocating piston mass. 4) The total inertia force in y-direction is zero. 5) The inertia force in x-direction is equal to the inertia force due to reciprocating pisto mass.

12 Balance of Multicylinder Engines 2-Cylinder 2-Stroke Engine First Harmonics: Angles between imaginary masses : 360 /2 = 180 S 1 = 0 M 1 0

13 2-Cylinder 2-Stroke Engine Second Harmonics: Angles between imaginary masses: 720 /2 = 360 S 2 0 M 2 0

14 4-Cylinder 4-Stroke Engine Firing Order: First Harmonics: Angles between imaginary masses : 720 /4 = 180 S 1 = 0 M 1 = 0

15 4-Cylinder 4-Stroke Engine Firing Order: Second Harmonics: Angles between imaginary masses : 1440 /4 = 360 S 2 0 M 2 0

16 4-Cylinder 4-Stroke Engine Firing Order: First Harmonics: Angles between imaginary masses : 720 /4 = 180 S 1 = 0 M 1 0

17 4-Cylinder 4-Stroke Engine Firing Order: Second Harmonics: Angles between imaginary masses : 1440 /4 = 360 S 2 0 M 2 0

18 Balancing Linkages The position of the mass center of whole linkage, is tried to remain stationary.

19 Balancing Linkages (cont d) rs3 rs4 rs2 Vector loop-closure eq n: One of the vectors, for example may be drawn from this eq n: e.g.

20 Balancing Linkages (cont d) This eq n shows that the center of mass may be made stationary at the position, ( 1 and 3) if the following coefficients of the time-dependent terms vanish: r j 3 j r 4 4 j 3 0 m4a4e m3a3 e 0 r r j 2 2 m a e m r m a e Simplification a e r a e 3 j 3 j j 2 2 j 3 m2a2e m3a 3 e r3 3 r 0 These two equations must be satisfied for total force balance

21 Balancing Linkages (cont d) These equations yield the two sets of conditions: r4 and 3 r2 m and 2a2 m3a 3 2 r 3 3 A study of these conditions show that the mass and its location can be specified in advance for any single link (3); and then full balance can be obtained by rearranging the masses of the other two links (2 and 4). Without disturbing 3rd link, counterweights are added to rotating links 2 and 4. In this procedure, m a m a r m a m a m a * * * i i i i i i i i i relations must be satisfied. Unbalanced linkage Counterweight

22 Balancing Linkages (cont d) Therefore gives, * * mi ai ( miai ) ( mi ai ) 2( miai )( mi ai ) cos( i i and tan * 1 i m a sin m a sin i i i i i i i i cos i mi ai cos i m a * m i If it is required to find, m m m * 0 i i i condition must be satisfied.