Basics of rotordynamics 2

Transcription

1 Basics of rotordynamics Jeffcott rotor 3 M A O a rigid rotor disk rotates at angular frequency W massless shaft acts as a spring restoring displacements disk can move only in the plane defined by axes 1 and mass centre M and shaft centre A are different points

2 Jeffcott rotor z y O e M A y 1 z 1 j 1 Coordinates (y 1, y ) give the position of the shaft centre A, coordinates (z 1, z ) the position of the mass centre M. The relation between them is z = y + ecosj 1 1 z = y + esinj Rotor disk in the plane defined by the axes 1, The rotor position is defined by three variables, either y 1, y and j or z 1, z and j.

3 Force balance Forces acting on the rotor disk: Inertia forces related to the mass Spring forces from the shaft Viscose damping, for instance, from the surrounding air gravity Equation of motion '' ' mz =-ky -dy '' ' mz =-ky -dy -mg The damping force is assumed to act on point A. This is not generally true put simplifies the model. There would be a third equation governing the angular speed of the disk but the problem is simplified further by assuming a constant speed.

4 Equations of motion Let us choose for the variables y 1, y z = y + ecosj 1 1 z = y + esinj => ( ecosj) '' '' d '' '' = + = - j j - ( ' j ) z1 y1 y1 e sin e cosj dt ( esinj) '' '' d '' '' = + = + j j - ( ' j ) z y y e cos e sinj dt Constant angular speed => ' j = Wt; j = W; j = '' '' 1 ' 1 1 '' ' my + dy + ky = mew coswt my + dy + ky = mew sinwt -mg

5 Steady-state solution Solution in a steady state 1 ( ) = cos( W -e) () sin( W e) y t r t y t = r t- -y g 6. r/ e D =.1 D =.1 D = 1. where r = mew ( k- mw ) + ( dw) e = arctan k mg yg = k dw -mw W/ w n w n = k m; D= dw k n

6 Steady-state solution II y When the motion is observed from coordinate system O set by gravity and bending of the shaft, points A and M move on circular orbits at steady relative positions. These relative positions changes with the speed. At high speeds, the centre of mass tends to move closer to the origin. ( r fi e) y g O O r Wt A e M e y 1 O O O r r r M r r e e M A r M e A W < w n W = w n W > w n A

7 Critical speed The natural frequency of the lateral bending oscillation of the rotor shaft f n = w n /p is called the critical speed of the rotor. The rotors are typically well balanced, but anyhow, a small mass unbalance excitation acts on the rotor. The mechanical damping of the rotor is small and the amplitude of the lateral oscillation increases significantly while operating close to the critical speed. For large electrical machines and high-speed machines, it may happen that the critical speed of a rotor occurs within the normal operating speed range of the machine. An electrical machine cannot usually be run continuously at the critical speed. However, a critical speed can be passed quickly. The damping of the system can be increased, for instance, by using sleeve bearings (sliding bearings) or active magnetic bearings.

8 Eigen-value problem The homogeneous equations of motion for Jeffcott rotor is '' 1 ' 1 1 '' ' my + dy + ky = my + dy + ky = Choosing a common complex variable u = y 1 + jy for the displacements, leads to equation '' mu + du + ku = ' Exponential trial function for the solution ( ) lt ut = ue => ( ) lt ml + dl+ kue = Non-zero solution possible only if ( ml dl k) + + =

9 Eigen-value problem II The characteristic equation has two roots (eigen-values) d d k l1, =- - =- d j wn - d =- d jw m Łmł m The solution is Functions u 1 and u are called the natural modes of vibration j d ( ) ; ( ) d t w t d t -j 1 = = w dt u t e e u t e e ( ) d t j w d - = dt t j + w dt = ( ) + ( ) u t C e e C e e C u t C u t The first function is the forward whirling mode, the second one is the backward whirling mode. d

10 Eigen-value problem III Jeffcott rotor has only two natural modes and those have a common damping factor d and oscillation frequency w d. Real rotors have an infinite number of natural modes. Each of them have their specific damping and frequency, which usually also depends on the rotation speed of the rotor. The speed-dependencies of natural frequencies are presented in Campbell s diagrams. w A mass unbalance is efficient in exciting the forward whirling modes. These are the critical speeds. w 4 w = W -w 3 w w c1 -w 1 W fi w c

11 Force on an eccentric rotor F total We shall study the force acting between rotor and stator when the rotor is performing cylindrical circular whirling motion F tang F radial with respect to the stator. This means that the rotor remains aligned with the stator but the geometrical w centreline of the rotor W travels around the geometrical centreline of the stator in a circular orbit Whirling orbit with a certain frequency, called whirling frequency and with a certain radius, called whirling radius. In conventional analyses of electrical machines, two special cases of whirling motion are typically studied. They are the static and dynamic eccentricity. In the static case, the whirling frequency is zero, in the dynamic one, the whirling frequency is equal to the rotation speed of the rotor. A lateral total force is commonly called an unbalanced magnetic pull.

12 Need for the force analysis Forces related to eccentricity and other kinds of deformation in the machine structures will be discussed quite extensively. The justification for this is related to the fact that speed-controlled electric drives and high-speed drives are becoming more and more common. This means that the critical bending speeds of rotors will more and more often cause problems, especially, for large-power induction motors and highspeed machines. A critical speed means that the natural bending frequency of a rotor is the same as its rotation frequency. In such a case, the centrifugal force associated with only a small mass unbalance can cause a large amplitude vibration. When there is simultaneously a magnetic force acting together with the unbalance force, the rotor motion may become completely unstable. Normally, an electrical machine cannot be run continuously at or close to a critical speed because of too large vibration amplitudes. These speeds must be forbidden in the control algorithms of the drive. In addition to the critical speeds, the varying supply frequency of a speed control excites many other natural frequencies and resonances in the structures of the machine. Most of the practical problems encountered with electrical machines are related to mechanical vibrations. It is important to know the basics of magnetic forces exciting these vibrations.

13 Non-uniform cylindrical air gap The radial air-gap length related to eccentricity can be approximately written (, t) = - cos( - t+ ) d f d e f w a when the eccentricity e is small. We generalise this expression to take also into account possible other deviations in the mechanical dimensions that may modify the shape of the air gap (, t) = - cos( - t+ ) d f d e nf w a By proper choice of the wave number n, we can study, for instance, the effects of an oval stator (n = ). The permeance related to the distorted air-gap length is approximately l Ø e ø l( f, t) =» l ( nf wn an ) e Œ1 + cos - t+ œ 1 - ( nf wn an ) d d cos - t+ º ß n n n n

14 Non-uniform cylindrical air gap The magnetomotive force over the air gap is taken to include the fundamental harmonic, only Bˆ p F( f, t) = cos( pf- wst+ as) l The flux density in the air gap becomes e Bˆ p B( f, t) = lf = l 1 + cos( nf- wnt+ an ) cos( pf- wst+ as) Ł d łl e = Bˆ cos p - t+ + B cos - t+ cos p - t+ ( f w a ) ˆ ( nf w a ) ( f w a ) p s s p n n s s d ( f w a ) = Bˆ cos p - t+ p s s e + Bˆ cos Ø p+ - + t+ + d ( n) f ( w w ) ( a a ) p º s n s n ß e + Bˆ cos Ø p- - - t+ - d ( n) f ( w w ) ( a a ) p º s n s n ß ø ø

15 Eccentricity, i.e. whirling of a circular rotor If we consider the dynamic eccentricity (n = 1, w w = w r ), we have the fundamental plus the two well-known eccentricity harmonics e B= Bˆ cos p - t+ - B cos p 1 - t+ ( f w a ) ˆ Ø( ) f ( w w ) ( a a ) p s s p n d º s r s ß We shall show in the exercises that two flux-density waves the wave numbers of which differ by one interact and produce a radial net force between the stator and rotor. The wave numbers of the eccentricity harmonics differ by one from the fundamental wave number p. As the fundamental is normally the flux-density wave having the biggest amplitude, the eccentricity harmonics, even if their amplitudes were relatively small, typically produce a large forces on the rotor. ø

16 Force on an eccentric rotor We shall show in the exercises that the force associated with the two harmonics with wave numbers m and m+1 is p m+ 1 lr j( f- w + a ) - jøº ( + 1 ) f- w a + 1) øß jf = ˆ ˆ m mt m m m t m m m+ 1 4m F B B e e e df plr = B ˆ ˆ m B m where the complex notation is used to express the x- and y- components of a force. When this equation is applied to the fundamental harmonic interacting with the two eccentricity harmonics, the force becomes p 1 plr F = Bˆ ˆ pbp 1e m ( wp-wp ) t-( ap-ap ) Ø ø mj º 1 1 ß m+ 1 e ( w w ) t ( a a ) -jø º m m+ 1 m m+ 1 ß ø

17 After substituting the amplitudes, frequencies and phases of the eccentricity harmonics the two forces become equal p 1 plr F = Bˆ ˆ pbp 1e m plr 4md Force on an eccentric rotor e Bˆ = ˆ p Bp, wp = ws w, ap = as a d 1 1 w 1 w e Be ˆ ( wp-wp 1) t-( ap-ap 1) Ø ø mj º ß plr e = B ˆ p m d = p Be ˆ ( w t-a ) j p w ( w -w w ) t-( a -a a ) mjøº s sm w s sm w øß The displacement vector of the rotor associated with the whirling motion can be defined j( wwt-aw p= e ) w

18 Force on an eccentric rotor Substituting the displacement vector, the total force becomes p+ 1 p-1 plr j( w -a ) p w w ecc = + = e ˆ t lr F F F Be = ˆ p Bp p md md The force depends linearly on the displacement (eccentricity) and points in the direction of the displacement vector. In the derivation above, many factors were neglected that affect a lot the forces caused by the eccentricity. Such factors are the eddy and circulating currents the eccentricity harmonics induce in the rotor conductors and parallel branches of the stator winding, saturation of the iron core and harmonics associated with the stator and rotor slotting. The following slides try to show the importance of the these factors on the eccentricity force. A small 15 kw four-pole cage induction motor is studied. As a reference, the weight of the rotor is 3 N.

19 Unbalance magnetic bull on an eccentric rotor Fy [N] Fx [N] 1 Unbalanced magnetic pull in a four-pole 4 kw synchronous reluctance machine. The machine runs at the rated load and its rotor is % dynamically eccentric. The force vector rotating at the synchronous speed draws a slightly distorted circle.

20 Force on a whirling rotor of the 4 kw motor The left-hand figure below show the radial (Fr) and tangential (Ft) force components acting on the rotor. The eccentricity is 1 % of the average air gap. The stator winding is connected in series (no parallel paths). In this case, the force is independent from the whirling frequency. The force vector points exactly in the direction of the shortest air gap. The right-hand figure shows the amplitudes of the two eccentricity harmonics p±1. These are also independent from the whirling frequency Force [N] Flux density [T] Whirling frequency [Hz] Whirling frequency [Hz] Fr Ft Bp-1 Bp+1 Four-pole machine; No parallel branches in the stator winding

21 Effect of parallel branches of the stator winding There are now four branches in each stator phase winding connected in parallel. The amplitudes of the two eccentricity harmonics p±1 strongly depend on the whirling frequency. They have clear peaks at the frequencies of ± 5 Hz. The radial force follows the behaviour of the eccentricity harmonics and peaks at the same frequencies. The tangential component of the force is also significant. This means that the direction of the force clearly deviates from the direction of the shortest air gap. Why do the parallel stator branches have such a strong effect on the force? Force [N] Whirling frequency [Hz] Flux density [T] Whirling frequency [Hz] Fr Ft Bp-1 Bp+1 Four-pole machine; Four parallel branches in the stator winding

22 Effect of parallel branches of the stator winding The eccentricity harmonics interact with the fundamental harmonic and produce the force. The machine is supplied from a voltage source and this should keep the fundamental harmonic approximately constant. Thus, the frequency-dependence of the force probably follows from the variation of the eccentricity harmonics. However, as the eccentricity is also constant, the sources of the eccentricity harmonics (the fundamental harmonic and permeance wave) should not change. The peak flux densities are at about the same values as obtained earlier when there was no parallel branches. Is it so that the parallel branches brought something to the system that dampens the eccentricity harmonics at all the frequencies except at ± 5 Hz? We have to study the interaction of the harmonics with the stator winding and its parallel branches. The winding factor of the simple stator winding without parallel branches is x n 1 a q 1 ( + ) 1 sin nqb sin n 1 p = sin n k sin nb p sin( n + 1) m Substituting n = p ±1 = ± 1, m = 4 gives x = for both the eccentricity harmonics. This means that these harmonics do not interact with the series connected stator winding.

23 Effect of parallel branches of the stator winding When there are four parallel branches in a four-pole machine, we have a double layer winding and all the four coil groups of a phase are connected in parallel. The figure below shows the situation except that in the case the 4 kw motor the coils are not chorded and q = We must study the winding factor for a coil group, which is sin 1 a q 1 1 nqb p p xn = sin n, a =, b =, q = 4 k sin nb 48 This winding factor has a non-zero value for both the harmonics, which means that the eccentricity harmonics interact with the coil groups.

24 Effect of parallel branches of the stator winding The two eccentricity harmonics presented in the stator frame of reference are e B + = ˆ p 1 BpcosØ( + ) f- ( w + wn ) + ( a + an ) ø d º p 1 s t s ß e B = Bˆ cosø p t+ - ( ) f ( w w ) ( a a ) p-1 p n n d º s s ß If the supply frequency is 5 Hz, the frequency of the first harmonic becomes zero at a whirling frequency of 5 Hz. This means that at 5 Hz whirling frequency the harmonic is a dc-field and cannot induce any electromotive force or current in the coil groups. At all the other frequencies, the eccentricity harmonic generates some circulating currents in the parallel coil groups. The circulating currents induce their own counter-acting fields that dampen the eccentricity harmonic. The second eccentricity harmonic becomes a dc-field at the +5 Hz whirling frequency. For this harmonic, the damping vanishes at +5 Hz but all the other whirling frequencies cause circulating currents that dampen the second eccentricity harmonic. ø

25 Computed and measured forces 15 kw induction motor; No load; 1 % eccentricity Radial force [N] Tangential force [N] Whirling frequency [Hz] Whirling frequency [Hz] Computed Measured Computed Measured Force in radial direction Force in tangential direction

26 Computed results Rated load; 1 % eccentricity 7 5 Radial force [N] Tangential force [N] Whirling frequency [Hz] Whirling frequency [Hz] Force in radial direction Force in tangential direction

27 Interpretation of the results A force in x,y-plane can be presented as a complex number. The real part represents the x-component, imaginary part the y-component of the force. In this presentation, the force versus frequency curve of the previous slide can be interpreted as a product of a complex frequency response function and complex displacement vector u of whirling motion ( ) ( ) ( ) where F s = K s u s, s= jw The frequency response function seems to have two poles, which are shifted away from the synchronous speed, when the loading (slip) of the motor is increased. The frequency response function could probably be parametrised to have the form c 1 () = c + + s-a1 s-a K s c If reasonable estimates for the parameters c i and a i can be obtained either by using FEA simulations or measurements, a parametric force model has been identified. If the system is linear and time invariant, we should be able to extend the model to also represent transient response.

28 Interpretation of the results II In this case, the frequency-response function K(s) is interpreted as a transfer function and variable s is taken as the Laplace variable. The following second-order state-space model is obtained for the force ( ) q& = aq+ cut () q& = aq + cut () = + + () Ft q q cut 1 The auxiliary variables q 1 and q are some kind of force variables but their physical origins are unknown. Obviously, we have identified a black-box model. Studying the problem of a whirling rotor analytically using Maxwell s equations, it can be shown that the auxiliary variables are related to the harmonics of wave number p±1 produced in the air gap of the machine by the eccentric motion [Holopainen 4]. The variables can be expressed as linear functions of the p±1 harmonic currents induced in the rotor cage. In this way, the physical meaning of the auxiliary variables was gradually revealed.

29 Electromechanical Interaction Jeffcott rotor model Combined equations of motion ( ) mu&& + du& + k - c u+ q + q = f q& = a q + c u p-1 p-1 p-1 p-1 q& = a q + c u p+ 1 p+ 1 p+ 1 p+ 1 p- 1 p+ 1

30 Campbell diagram Non-rotating damping factor: z n =. Modes: 1. Forward whirling. Backward whirling 3. Electromagnetic

31 Decay rates Non-rotating damping factor: z n =. Modes: 1. Forward whirling. Backward whirling 3. Electromagnetic

32 Stability chart A super-critical induction motor A typical ratio for the spring coefficients of induction machines is.15 z n = factor for nonrotating damping

33 Whirl amplification factor for mass unbalance Non-rotating damping factor: z n =.5 Stability limit: z n».3

34 Conclusions 1. A new simple electromechanical rotor model was constructed. Electromechanical interaction affects the vibration characteristics of the machine (stiffness, damping, stability) 3. These effects are significant in large-power induction motors with a flexible shaft, large bearing span, and narrow air gap 4. Previously used models overestimate the electromagnetic stiffness coefficient at most whirling frequencies 5. An unstabilising electromagnetic force is always present at the supercritical speeds 6. Electromechanical interaction reduces the unbalance response close to the critical speed