CHAPTER 10 EXPERIMENTAL ESTIMATION OF DYNAMIC PARAMETERS OF BEARINGS

Transcription

1 CHAPTE 10 EXPEIMENTAL ESTIMATION OF DYNAMIC PAAMETES OF BEAINGS One o the important actors governing the vibration characteristics o rotating machiner is the eective dnamic stiness o the supports as seen b the rotor as shown in Figure The dnamic stiness o the support is determined b the combined eects o leibilit o the bearing, the bearing pedestal assembl (bearing housing) and the oundations on which the pedestal is mounted. For the case o turbo generator rotors mounted on oil-ilm bearings might be three times more leible as compared to pedestals and oundations. Figure 10.1(a) A simpliied representation o a rotor-bearingoundation sstem (b) Single-rotor-degree o reedom idealisation In the case o aeroengine compressor shats are mounted on rolling element bearings, the oundation o bearing is ar more leible. The theoretical models available or predicting the rotor support stiness are insuicientl accurate. It is or this reason that designers o high-speed rotating must rel on empiricall derived values (i.e. eperimental) or support stiness and damping in their design calculations. Following methods are available which is classiied in terms o tpe o orcing applied (i) Static orce method (ii) Dnamic orce method o Use o an electromagnetic vibration or eciter Comple receptance method Direct comple impedance derivation Multi-requenc testing

2 o Use o centriugal orces Imbalance mass attached to the journal Imbalance mass attached to an independent vibrator shat o Transient methods Measurement on the running machine o Forces inherent in the sstem (residual imbalance and random) 10.1 Static orce method It is possible to determine all our stiness coeicients (i.e. K,, K K and K ) o the bearing oil ilm b application o static loads onl. Unortunatel this method o loading does not enable the oil-ilm damping coeicient to be determined. Figure 9. (a) A bearing model (b) Stead state locus curve o the shat center The eact operating position o the shat center on a particular bearing depends upon the Sommereld number. Because the bearing oil-ilm coeicient are speciic to a particular location o shat center on the static locus as shown in Figure 10.. A static load must irst be applied in order to establish operation at the required point on the locus. The net step is to appl incremental loads in both the horizontal and vertical directions, which will cause changes in the journal horizontal and vertical displacement relative to the bearing bush (or more precisel with respect to its static equilibrium position). B relating the measured changes in displacements to the changes in the static load it is possible to determine our-stiness coeicients on the bearing oil ilm. We have, increments in orces as 399

3 = k + k ; and = k + k (10.1) where and are the journal displacement in and directions, respectivel (with respect to static equilibrium position). I the displacement in the direction is made to zero b application o suitable loads and then k = / ; k = / (10.) Similarl i the displacement in the -direction is made zero then k = / ; k = / (10.3) Determination o the oil-ilm coeicient in this wa necessitates a test rig, which is capable o appling loads to the journal in both the horizontal and vertical directions. The method is somewhat tedious in the eperimental stage since evaluation o the required loads to ensure zero change in displacement in one or other direction is dependent on the application o trial loads. Alternative method (i) Instead o appling loads in both and directions, to ensue zero displacements in one o these directions, it is easier to simpl appl a load in one direction onl and measure resulting displacements in both directions. Equations (10.1) can be written as { } = [ K ]{ d} (10.4) with k k = = = k k { } ; [ K ] ; { d} I the [K] matri is inverted then equation (10.4) can be written as with { d} [ α ]{ } [ K ] [ α ] = (10.5) 1 α α = = α α 400

4 where the quantities α, α, etc. are called the oil-ilm inluence coeicients. I the orce in the direction is zero then α = ; α = (10.6) Similarl, when the orce in the direction is zero, we have α = ; α = (10.7) The bearing stiness coeicient ma be obtained b inverting the inluence coeicient matri i.e. [ K ] [ α ] 1 =. This method still requires a test rig which is capable o providing loads on the bearing in both and directions (speciall in the horizontal direction, ). Alternative method (ii): I there is no acilit on the test rig or appling loads transverse to the normal stead-state load direction o the bearing, it is still possible to obtain approimate value o the stiness coeicients. Figure 10.3 Shit in the journal center position due to a horizontal load In Figure 10.3, e is the eccentricit, φ is the altitude angle, A is the stead state position or a vertical load w, B is the additional imaginar orce F is applied to change its stead state running position to B, is the resultant o w and F, d θ is the angle o with respect to vertical line i.e. w, d φ is the 401

5 change in altitude angle due to additional The inluence coeicient can be obtained as F, ( )( ) e + d e is the new eccentricit ater application o F. ( ) sin ( ) PB P PB SA e + de φ + dφ esinφ α = = = = F F F F = e + d sinφ cos dφ + cosφ sin dφ esinφ e F Since or small displacement, we have ( d ) e coeicient can be simpliied to e + e, sin d φ = dφ and cos d φ = 1. The inluence ( ) e sinφ + dφ cosφ esin φ e( dφ)cosφ α (10.8) F F A urther simpliication can be made i the resultant is considered to be o verticall same magnitude as the original load w, ecept that it has been turned through an angle dθ. We ma write dφ dθ = tan dθ = F w (10.9) On substituting equation (10.9) into (10.8), it gives ecosφ F ecosφ d α = (10.10) F w w w Similarl it ma be shown that esinφ d α = = (10.11) w w Since vertical load is eas to appl, one can get α = and F α =. Stiness coeicients can F be obtained as [ ] [ ] 1 k α =. 40

6 Eample 10.1: Under particular operating conditions, the theoretical values o the stiness coeicients or a hdrodnamic bearing are ound to be; K =30 MN/m, K =6.7 MN/m, K =-0.96 MN/m, K =11.7 MN/m. A testing is being designed so that these values can be conirmed eperimentall. What increment in horizontal (F ) and vertical (F ) loads must the rig is capable o providing in order to provide (a) a displacement increment o 1 µm in the horizontal direction whilst that in the vertical direction is maintained zero and (b) a displacement increment o 1 µm in the vertical direction whilst that in horizontal direction is maintained zero. Solution: From equation (10.1) static orces required in the and directions to a given displacement can be obtained. For case (a) ollowing orces are required = 30 1 = 360 N and = = N For case (b) ollowing orces are required = = 30.4 N and = = N The MATLAB code INPUT FILE % Input ile name is input_qus_1_1.m % For the irst condition 1 =1*10^-6; % displacement in horizontal direction when F is applied in m =0; % displacement in horizontal direction when F is applied in m % For the second condition 1 =0; % displacement in vertical direction when F is applied in m =1*10^-6; % displacement in vertical direction when F is applied in m %F = load in horizontal direction in N. %F = load in vertical direction in N. K =30*10^6; % dnamic stiness coeicient or a bearing in N/m K =6.7*10^6; % dnamic stiness coeicient or a bearing in N/m K =-.96*10^6; % dnamic stiness coeicient or a bearing in N/m K =11.7*10^6; % dnamic stiness coeicient or a bearing in N/m MAIN FILE clear all; input_qus_1_1; =[ 1 ; 1 ]; k=[k k ;k k ]; =k*; print ('The loads to be applied in the irst condition'); print ('\n 1 ='); print (numstr ( (1,1))); print (' N\n'); print ('\n 1 ='); print (numstr ( (,1))); print (' N\n'); print ('\nthe loads to be applied in the second condition'); print ('\n ='); print (numstr ( (1,))); print (' N\n'); print ('\n ='); print (numstr ( (,))); print (' N\n'); 403

7 OUTPUT The loads to be applied in the irst condition 1 =360 N, 1 = N The loads to be applied in the second condition =30.4 N, =140.4 N Eample 10.: The test rig described in Eample 10.1 is used to measure the hdrodnamic bearing stiness coeicients b appling irst o all a horizontal load o 360 N, which is then removed and replaced b a vertical load o 30 N. The horizontal load produces displacement o 10.3 µm and 3.3 µm in the horizontal and vertical directions respectivel, whilst the vertical load produces respective displacements o 18.3 µm and 19.7 µm. Calculate the value o stiness coeicients based on these measurements. Solution: For the horizontal load o 360 N alone rom equation (10.4), we have α = = m/n ; α 3.3 = = m/n For the vertical load o 30 N alone rom equation (10.5), we have α 18.3 = = m/n ; α = = m/n 30 From equation (10.3), we can obtain stiness coeicients as [ K ] MN/m k k = k k = = = The MATLAB code INPUT FILE % Input ile name is input_qus_1_.m 1 =10.3*10^-6; % displacement in horizontal direction when F is applied (in meter) 1 =3.3*10^-6; % displacement in vertical direction when F is applied (in meter) F 1 =360; % load in horizontal direction (in N) F 1 =0; % load in vertical direction (in N) =-18.3*10^-6; % displacement in horizontal direction when F is applied (in meter) =19.7*10^-6; % displacement in vertical direction when F is applied (in meter) F =0; % load in horizontal direction (in N) F =30; % load in vertical direction (in N) MAIN FILE clear all; input_qus_1_; =[F 1 F ; F 1 F ]; X=[ 1 ; 1 ]; 404

8 a=x/; k=a^-1; print ('Dnamic stiness coeicients are'); print ('\nk ='); print (numstr (k (1,1))); print (' N/m\n'); print ('\nk ='); print (numstr (k (1,))); print (' N/m\n'); print ('\nk ='); print (numstr (k (,1))); print (' N/m\n'); print ('\nk ='); print (numstr (k (,))); print (' N/m\n'); OUTPUT Dnamic stiness coeicients are k = N/m, k = N/m, k = N/m k = N/m Eercise 10.1 For the estimation o bearing stiness coeicients b the static load method, the static load o 400 N is applied in the vertical and horizontal directions, one at a time. When the load is applied in the horizontal direction, it produces displacements o µm and 0 µm in the vertical and horizontal directions respectivel, whilst the vertical load produces respective displacements o 4 µm and 1 µm. Obtained bearing stiness coeicients rom the above measurements. Answer: The stiness coeicients are k = MN/m, k = MN/m, k =5.581 MN/m and k =3.56 MN/m. Eercise 10. A test rig is used to measure the hdrodnamic bearing stiness coeicients b appling irst o all a horizontal load o 400 N. It produces displacements o 10 µm and 4 µm in the horizontal and vertical directions, respectivel. Then in second case onl a vertical load o 300 N is applied. It produces displacements o -0 µm and 0 µm in the horizontal and vertical directions, respectivel. Calculate the value o the stiness coeicients based on these measurements. 10. Use o Electromagnetic Vibrator In order to ull analse the behavior o a bearing under dnamic loading it is necessar to cause the journal to vibrate within the bearing bush under the action o a known eciting orce as shown in Figure 10.4(a). Alternativel, the bearing bush can be allowed to loat reel on the journal as shown in Figure 10.4(b), which is mounted on a slave bearings and the orcing is applied to the bush. B measuring the resulting sstem vibrations and relating these to the orce, it is possible to determine the eective oil-ilm stiness and damping coeicient. B varing the amplitude, requenc and shape o the electrical signal input to the vibratior it is possible to eercise ull control over the orcing applied to the sstem. 405

9 Fluid Fied non-loating bearing housing ( 0 and 0) b b Journal (t) (t) Figure 10.4(a). A ied bearing and a rotating journal loating on the luid Fluid Journal Floating bearing bush ( 0 and 0) b (t) b (t) Figure 10.4(b). A ied rotating shat and a non-rotating bearing loating on the luid Comple eceptance Method The method involves appling a sinusoidall varing orce to the journal in the horizontal direction, whilst the orcing in the vertical direction is zero, and measuring the resulting displacement amplitudes in the horizontal and vertical directions together with their respective phase relative to the eciting orce. It is then necessar to repeat the procedure with the orcing applied onl in the vertical 406

10 direction. The knowledge o orce amplitude and measured displacement quantities, then enables the eight oil-ilm coeicient to be derived. The orce transmitted across the oil-ilm ma be represented in the orm = k + k + c + c and = k + k + c + c (10.1) Assuming sinusoidal variations o and (i.e. unction), equation (10.1) gives jωt = Xe etc., where ω is requenc o orcing ( k + jω c ) + ( k + jωc ) = and = ( k + j c ) + ( k j c ) ω + ω (10.13) which can be written in matri orm as Z = Z Z Z (10.14) where Z is a comple stiness coeicient given b Z = k + jωc = (10.15) where [ ] [ ] 1 = Z is called the comple receptance matri. For the case o orcing in horizontal direction onl equation (10.15) gives = and = (10.16) where and are the measured displacement in the horizontal and vertical directions at a particular time and is the orce in the horizontal direction at that instant. For the case when orcing is in the vertical direction the other reacceptance terms are derived as = and = (10.17) 407

11 On inverting [ ] elements o [ Z ] can be obtained. The elements o [ Z ] contain all eight bearing stiness and damping coeicients as deined in equation (10.14) i.e. Z = k + jωc where ω is the requenc o orcing unction. The problems related to this method can be easil solved in the MATLAB and is illustrated in the ollowing eample. Eample 10.3 A bearing is orced in the horizontal direction b a orce F = 150 sin 00t N. The resulting vibrations are = sin(00t-0.) m in the horizontal direction and = sin(00t-0.3) m in the vertical direction. When the same orcing is applied in the vertical direction the horizontal and vertical displacements take the respective orms = sin(00t ) m and = sin(00t-0.3) m. Determine dnamic coeicients o the bearing. Solution: We have two sets o measurements (i) For F = 150 sin 00 t N and F = 0 6 = 7 10 sin(00t 0.) m and = t sin(00 0.3) m which can be written in comple plane as For F j00t = 150e alone, we have and (ii) F = 0 and F = 150sin 00 t N e e 6 j(00-0.) 6 j(00-0.3) 7 10 t t = and = 0 10 (A) t 6 = sin( ) m and = t sin(00 0.3) m which can be written in comple plane as For F e j00t = 150 alone, we have e e 6 j( ) 6 j(00-0.3) t + t = and = 6 10 (B) 408

12 Bearing dnamic coeicients are deined as F = k + k + c + c and F = k + k + c + c (C) On substituting the irst set o measurement rom equation (A) into equation (C), we have j00t 6 j(00-0.) 150 ( j00 ) ( k j00 ) t c e k + c e = 6 j(00t-0.3) 0 ( k j00 c ) ( k j00 c ) e (D) Similarl on substituting the second set o measurement rom equation (B) into equation (C), we have 6 j( ) 0 ( j00 ) ( k j00 ) t c + k + c e = j00t 6 j(00t-0.3) 150 e ( k j00 c ) ( k j00 c ) e (E) Let the dnamic stiness is deined as Z = k + jωc with ω = 00 rad/sec First set o equations rom equations (D) and (E), we have 7 10 e Z e Z = 150e (F) 6 j(00t-0.) 6 j(00t-0.3) j00t and 8 10 e Z e Z = 0 (G) 6 j(00t-0.15) 6 j(00t-0.3) Equation (G) gives 6 10 e Z Z Z e Z 8 10 e 6 j(00t-0.3) 0.45 j = or j(00t-0.15) = (H) On substituting equation (H) into equation (F), we get 409

13 Z j00t 150e 150 = = 6 j(00t-0.3) 6 j(00t-0.) j 6 0.j 6 10 e e 0 10 e e or 7 7 Z = j (I) On substituting equation (I) into equation (H), we get 7 8 Z = j (J) Similarl rom irst set o equation (E), we have 6 j(00t 0..3) ( 0 10 e ) Z = Z Z =.8571e Z 7 10 e 0.1 j 6 j(00t 0.) (K) On substituting equation (K) into second equation o (E), we get Z j00t 150e = = 6 j(00t 0.3) 6 j(00t+ 0.15) j0.3 j e e 6 e.86 e (L) In simpliication o equations (K) and (L), we get Z = j and Z = j (M) Stiness and damping coeicients can be obtained b separating real and imaginar part o the dnamic stiness coeicients rom equations (I), (J) and (M), as k k k k = N/m; = N/m; = N/m; = N/m c = N/ m-sec ; c = 9400 N/m-sec ; c = N/m-sec; c = N/m-sec MATLAB solution: INPUT FILE % The name o the input ile is input_qus_5.m 410

14 w=00; % Frequenc o the applied orce % For the given 1st condition F 1 =150; % Amplitude o the applied horizontal orce (in N) F 1 =0; % Amplitude o the applied vertical orce (in N) X 1 =0*10^-6; % amplitude o the horizontal vibration (in meter) α 1= -0.; % Phase angle o horizontal vibration (in radian) Y 1 =0*10^-6; %Amplitude o the vertical vibration (in meter) α =-.3; % Phase angle o vertical vibration (in radian) % For the given nd condition F =0; % Amplitude o the applied horizontal orce (in N) F =150; % Amplitude o the applied vertical orce (in N) X =8*10^-6; % amplitude o the horizontal vibration (in meter β 1 =0.15; % Phase angle o horizontal vibration (in radian) Y =6*10^-6; %Amplitude o the vertical vibration (in meter) β =-0.3; % Phase angle o vertical vibration (in radian) 1 =X 1 *(cos(α 1 )+i*sin(α 1 )); 1 =Y 1 *(cos(α )+i*sin(α )); =X *(cos(β 1 )+i*sin(β 1 )); =Y *(cos(β )+i*sin(β )); MAIN FILE clear all; input_qus_1_5; F=[F 1 F ; F 1 F ]; X=[ 1 ; 1 ]; K=X\F; print ('Dnamic coeicients o the bearing are'); print ('\nk ='); print (numstr (real (K (1,1)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (1,)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (,1)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (,)))); print (' N/m\n'); print ('\nc ='); print (numstr (imag (K (1,1)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (1,)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (,1)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (,)))); print (' N/m-sec\n'); OUTPUT The dnamic coeicients o the bearing are K = N/m K = N/m K = N/m K = N/m C = N/m^ C = N/m^ C = N/m^ C = N/m^ The above problem we can be solved b an alternative method: INPUT FILE % The input ile name is input_altr_qus_1_5.m w=00; t=pi/(4*w); % time o operation in second % For the irst condition F 1 =150*sin(00*t); % applied orce in horizontal direction (in N) F 1 =0; % applied orce in vertical direction (in N) A 1 =-.; % phase o the vibration in the horizontal direction or the irst condition (in radian) B 1 =-.3; % phase o the vibration in the vertical direction or the irst condition (in radian) 411

15 M 1 =0*10^-6; % amplitude o the vibration in the horizontal direction or the irst condition (in meter) M 1 =0*10^-6; % amplitude o the vibration in the vertical direction or the irst condition (in meter) % or the second condition F =0; % applied orce in horizontal direction (in N) F =150*sin(00*t); % applied orce in vertical direction(in N) A =.15; % phase o the vibration in the horizontal direction or the second condition (in radian) B =-.3; % phase o the vibration in the vertical direction or the second condition (in radian) M =8*10^-6; % amplitude o the vibration in the horizontal direction or the second condition (in meter) M =6*10^-6; % amplitude o the vibration in the vertical direction or the second condition (in meter) 1 =M 1 *(sin(w*t)*cos(a 1 )+j*sin(a 1 )*cos(w*t)); % displacement in horizontal direction when F is applied (in meter) =M *(sin(w*t)*cos(a )+j*sin(a )*cos(w*t)); % displacement in horizontal direction when F is applied (in meter) 1 =M 1 *(sin(w*t)*cos(b 1 )+j*sin(b 1 )*cos(w*t)); % displacement in vertical direction when F is applied (in meter) =M *(sin(w*t)*cos(b )+j*sin(b )*cos(w*t)); % displacement in vertical direction when F is applied (in meter) MAIN FILE clear all; input_altr_qus_1_5; F=[F 1 F ; F 1 F ]; X=[ 1 ; 1 ]; K=X\F; print ('Dnamic coeicients o the bearing are'); print ('\nk ='); print (numstr (real (K (1,1)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (1,)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (,1)))); print (' N/m\n'); print ('\nk ='); print (numstr (real (K (,)))); print (' N/m\n'); print ('\nc ='); print (numstr (imag (K (1,1)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (1,)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (,1)))); print (' N/m-sec\n'); print ('\nc ='); print (numstr (imag (K (,)))); print (' N/m-sec\n'); OUTPUT The dnamic coeicients o the bearing are k = N/m k = N/m k = N/m k = N/m c = N/m-sec c = N/m-sec c = N/m-sec c = N/m-sec Eample 10.4 A bearing is orced in the horizontal direction b a orce F = 150 sin 00t N. The resulting vibrations are = sin(00t-0.) meters in the horizontal direction and = sin(00t-0.3) meters in the vertical direction. When the same orcing is applied in the vertical direction the horizontal and vertical displacements take the respective orms = sin(00t +0.15) meters and = sin (00t-0.3) meters. Determine elements o comple receptance matri or the bearing. Solution: The ollowing measurement were done 41

16 Case I: For F = 150sin 00 t N and F = 0, we have = t = t sin(00 0.) m and 0 10 sin(00 0.3) m Case II: For F = 150sin 00 t N and F = 0, we have = t + = t sin( ) m and 6 10 sin(00 0.3) m For a orce F leading a displacement X b θ is shown in Figure 8.5. F cosθ F sinθ F θ Figure 8.5 Phase between the displacement and orce vectors From Figure 8.5 the receptance can be epressed as X X = F cosθ + j F sinθ where X and F are displacement and orce amplitudes. The displacement is lagging behind orce b θ angle. F = 150 N 0. rad X = m Figure 10.6 A tpical orce and displacement vectors From Figure 10.6, we have 6 X = = = = F F cosθ + j F sinθ 150 cos 0. + j 150sin ( j ) 10 m/n 413

17 = = = ( ) 150cos0.3 + j150sin j m/n = = = ( + ) 150cos0.3 + j 150sin j m/n = = = ( ) 150cos j 150sin j m/n Hence, the receptance matri can be written as (45.74 j 9.7) ( j 7.97) m ] = = (16.56 j 41.94) ( j 51.) + MN MATLAB Solution: INPUT FILE % The name o the input ile is input_qus_1_6.m w=00; % Frequenc o the applied orce % For the given 1st condition F 1 =150; % Amplitude o the applied horizontal orce (in N) F 1 =0; % Amplitude o the applied vertical orce (in N) X 1 =7*10^-6; % amplitude o the horizontal vibration (in meter) α 1= -0.; % Phase angle o horizontal vibration (in radian) Y 1 =0*10^-6; %Amplitude o the vertical vibration (in meter) α =-.3; % Phase angle o vertical vibration ( in radian) % For the given nd condition F =0; % Amplitude o the applied horizontal orce (in N) F =150; % Amplitude o the applied vertical orce (in N) X =8*10^-6; % amplitude o the horizontal vibration (in meter β 1 =0.15; % Phase angle o horizontal vibration (in radian) Y =6*10^-6; %Amplitude o the vertical vibration (in meter) β =-0.3; % Phase angle o vertical vibration (in radian) 1 =X 1 *(cos(α 1 )+i*sin(α 1 )); 1 =Y 1 *(cos(α )+i*sin(α )); =X *(cos(β 1 )+i*sin(β 1 )); =Y *(cos(β )+i*sin(β )); MAIN FILE clear all; input_qus_1_6; F=[F 1 F ; F 1 F ]; X=[ 1 ; 1 ]; K=X/F; print ('The elements o comple receptance matri or the bearing are'); print ('\n ='); print (numstr (K (1,1))); print (' N/m\n'); print ('\n ='); print (numstr (K (1,))); print (' N/m\n'); print ('\n ='); print (numstr (K (,1))); print (' N/m\n'); print ('\n ='); print (numstr (K (,))); print (' N/m\n'); 414

18 OUTPUT The elements o comple receptance matri or the bearing are =4.5736e e-009i m/n =5.734e e-009i m/n =1.656e e-008i m/n =1.6559e e-008i m/n Eercise 10.3 A bearing is orced in the horizontal direction b a orce F = 00 sin150t N. The resulting journal vibrations are sin(150t 0.35) = m (in the horizontal direction) and sin(150t 0.4) = m (in the vertical direction). When the same orce is applied in the vertical direction the horizontal and vertical displacements take the respective orms sin(150t 0.3) sin(150t 0.38) = + and =. Determine elements o the comple impedance matri or the bearing. MATLAB Solution: INPUT FILE % The name o the input ile is input_qus_8_4.m w=150; % Frequenc o the applied orce % For the given 1st condition F 1 =00; % Amplitude o the applied horizontal orce (in N) F 1 =0; % Amplitude o the applied vertical orce (in N) X 1 =1*10^-6; % amplitude o the horizontal vibration (in meter) α 1 =-0.35; % Phase angle o horizontal vibration (in radian) Y 1 =0*10^-6; %Amplitude o the vertical vibration (in meter) α =-0.4; % Phase angle o vertical vibration (in radian) % For the given nd condition F =0; % Amplitude o the applied horizontal orce (in N) F =00; % amplitude o the applied vertical orce (in N) X =13*10^-6; % Amplitude o the horizontal vibration (in meter) β 1 =0.3; % Phase angle o the horizontal vibration (in radian) Y =5*10^-6; % amplitude o the vertical vibration (in meter) β =-0.38; % Phase angle o the vertical vibration (in radian) MAIN FILE clear all; input_qus_1_4; 1 =X 1 *(cos(α 1 )+i*sin(α 1 )); 1 =Y 1 *(cos(α )+i*sin(α )); =X *(cos(β 1 )+i*sin(β 1 )); =Y *(cos(β )+i*sin(β )); F=[F 1 F ; F 1 F ]; X=[ 1 ; 1 ]; K=X\F; print ('The elements o comple impedance matri or the bearing are'); print ('\nk ='); print (numstr (K (1,1))); print (' N/m\n'); print ('\nk ='); print (numstr (K (1,))); print (' N/m\n'); print ('\nk ='); print (numstr (K (,1))); print (' N/m\n'); print ('\nk ='); print (numstr (K (,))); print (' N/m\n'); OUTPUT The elements o comple impedance matri or the bearing are K = i N/m 415

19 K = i N/m K = i N/m K = i N/m Eercise For the bearing dnamic parameter estimation, how man minimum numbers o independent sets o orce-response measurements are required? Justi our answer. (Assume there is no residual imbalance in rotor.) 10.. Direct Comple Impedance Derivation It is possible to determine the comple stiness coeicients Z, Z etc. in equation (10.14) directl without resorting to the use o receptances. This ma be done provided that orcing in both the horizontal and vertical directions be able provide simultaneousl, and with independent control over each input with respect to its amplitude and relative phase. The sstem orce-displacement relationship is given b equation (10.14). In the present approach it is to ensure that one o the resulting sstem displacement vectors, sa Y, is zero. This can be made to be the case b correctl adjusting the amplitude o the orce in the -direction and its phase relative to that in the -direction. Suitable values or these quantities can be ound relativel easil b trial and error. The irst line o equation (10.14) thus gives Z = (10.18) which will allow the value o Z to be determined directl provided that the amplitude and phase o the horizontal displacement relative to the horizontal orce had been measured. I the amplitude and phase o the orce in the -direction were also noted then the value o Z could also be determined as Z = (10.19) In the above case, it is the phase o the -direction displacement amplitude relative to the orce in the -direction that is signiicant. Similarl b adjusting orcing amplitudes and relative phases so as to ensure a zero horizontal displacement,, then the values o Z and Z could also be determined. This method requires more complicated eperimental procedure. It ma be more costl in terms o equipment, since two vibrators and additional control units are required. Some eperimental measurement considerations are discussed now. Choice o orcing requenc (or orcing requenc range) is an important parameter to choose. It depends upon the sstem resonance. 416

20 I the sstem is ecited close to its resonant requenc then a response o suitable magnitude ma be obtained or a lower orce amplitude input. (since the bearing impedance changes with journal vibration non-linearit eect will pla a major roll). The advantage o eciting the sstem at a requenc in the region o its resonant requenc, that is that phase lag will be generall greater than zero. (between response and orce). With this or small inaccuracies in their (phase) measurement are less likel to substantiall alter the magnitude o coeicients, which are derived. This is not the case hen the lag angle is ver small or when it is close to In these cases (i.e. close to 0 0 and 0 90 ) ill conditioning o the equation o motion results in signiicant changes in the magnitude o the derived coeicient or even a change o onl 3-4 degrees o phase (which ma be about the accurac to which phase is measured). This coeicient derived using data generated well awa rom the critical speed ma well be considerabl in accurate (o the order o 100% in some cases). Also since at critical speeds it is observed that the orbit o the shat center is elliptical in nature and that leads to well conditioning o regression matri Multi Frequenc Testing It has advantage that the certaint o eciting all sstem modes within the prescribe requenc range, and inherent high noise rejection. The method involves orcing the sstem in both and directions, at all requencies within the range o interest, simultaneousl. The aim is to arrive at more accurate values o the coeicients, which are assumed to be independent o requenc, b means o some averaging procedure. When all (several) requencies are ecited simultaneousl, the knowledge o bearing behavior at man dierent requencies should enable more accurate results to be obtained. Also it saves the laborator time. Fourier analsis can be used to convert measured input and output signals rom the time domain to the requenc domain. ecent advances in laborator instrumentation, or eample, the emergence o spectrum analsers (FFT analsis) capable o carring out the Fourier transorm, have helped the technique to evolve. In theor, an shape o input signal with multi-requenc content can be used to orce the sstem. For eample an impulse signal (Figure 10.7) is actuall composed o signals at all requencies in coeistence. Because o the likel concentration o the signal at the low-requenc end o the spectrum however, a impulse in practice provides useul signals over onl a relativel small requenc range. For higher requenc the signal/ noise ratio becomes too low. An alternative is a white noise signal, which contains all requencies within its spectrum Band-limited white noise, sometimes reerred to as coloured noise, contains all requencies within a prescribed range. One wa o producing such a signal is with pseudo random binar sequences (PBS) where the requenc range that is present is chosen to ecite appropriate modes in the sstem under test. Unortunatel, both with impulse and PBS signals there is a danger o saturating the sstem so that amplitudes at some requencies are so large that non-linearit are encountered and the test becomes invalid. These disadvantages can be overcome b using a signal 417

21 mode up o equal-amplitude sinusoidal signals whose requenc are those which one wishes to ecite within a particular requenc range. One signal o this tpe is Schroeder phased harmonics. Figure 10.7 Impulse and white noise signals in the time and requenc domains I the sstem response to multi-requenc signal is recorded, bearing properties ma be obtained as ollows. The displacement in the and directions occurring at a requenc ω are written in the orm = and = (10.0) jωt jωt Xe Ye Thus j ω t j ω t j ω t j ω t (10.1) = j ω Xe ; = j ωye ; = ω Xe and = ω Ye The orcing unction ma similarl be deined as j t = Fe ω and j t = Fe ω (10.) Equations o motion o the journal in the and directions are k k c c = M and k k c c = M (10.3) where M is the mass o journal and Substituting equations (10.0) to (10.) into equation (10.3), we get k, c etc. are oil ilm stiness and damping coeicients. ( ) ( ) ( ) ( ) Z ω Z ω X F + Mω X = Z ω Z ω Y F + Mω Y (10.4) 418

22 where Z = k + jωc. On separating the real and imaginar parts, we rearrange equation (10.4) as k k c c r i r i r i r r X ωx Y ωy F F k k Mω X Mω Y i r i r i r = i i X ω X Y ωy F F c c Mω X Mω Y (10.5) Equation (10.5) ma be written or ω = ω0, ω0, 3 ω0,,nω0 (Total o n times in all). Values o ω and quantities in the irst and last matrices o equation (3) are determined b the Fourier transormation o time-domain signals. All o these equations (10.5) ma be grouped as a single matri equation as [ ] n 6[ Z ] 6 = [ A] n D (10.6) The contents o the [ Z ] matri might be best obtained b means o a least squares estimator. This involves recognizing that measurements obtained in the laborator will be inaccurate and so there are no values o the coeicient in the [ Z ] matri which will satis all lines o equation (10.6). A residual matri is developed which deines the errors between the let hand and right hand sides o equation (10.6) i.e. [ E] = [ A] n [ D] n 6 [ Z ] 6 n (10.7) The contents o the [ Z ] matri are deined as being those values, which ensure that the sums o the squares o the elements in the [ E ] matri are minimized. On multipling equation (10.7) b [ ] T n we get T T [ D] [ D][ Z ] = [ D] [ A] or [ Z ] [ D] [ D] T 1 T [ ] [ D] [ A] D 6, = (10.8) Since measured terms used to make up the [ D ] and [ A ] matrices are obtained via Fourier transormation o the output and input signals. The noise occurring at a requenc greater than n ω0 is automaticall iltered out o the analsis. 419

23 10.3 Use o Centriugal Forces One o the simplest was o eciting a journal in a sinusoidal manner is b means o centriugal orcing, simpl b attaching imbalance masses o known magnitude to a rotating shat. Advantage with this method is that there is no need or costl electromagnetic eciter. In present method the processing o measured data in the time domain. Out o three methods in two methods an imbalance o known magnitude attached to the journal (and are based on the assumption that inherent rotor imbalance is insigniicantl small). The third method involves use o a separate imbalance mass shat, which can rotate at requencies independent o the journal rotational requenc. (i) Imbalance mass attached to the journal This means o eciting the journal can be used to determine the bearing oil-ilm damping coeicients when the bearing stiness coeicients are alread known. (or eample b static orce method). The eperimental involves measurement o the horizontal and vertical displacement amplitudes o the journal relative to the bearing, and o the bearing or pedestals itsel relative to space (ied oundation). In addition, measurements o the corresponding phase lag angles o each o these displacements behind the imbalance orce vector are also made. Figure 10.8 A rotor-bearing sstem with an unbalance In addition to imbalance orce as shown in Figure 10.8, the rotor also has oil-ilm orces acting on it, these being transmitted to rotor b shat. Governing equations o the rotor can be written as and ( ) k k c c = M + p ( ) k k c c = M + (10.9) p 40

24 where and are displacements o journal relative to bearing (or pedestal), p and p are the displacement o pedestal (or bearing) relative to ied oundation, M is the central rotor mass (smmetric) and is the known imbalance orce on the rotor. We can write j t = Fe ω ; j t = F e ω so that j t = Xe ω ; j t = Ye ω ; j t p = X pe ω ; j t p = Ype ω (10.30) where F, F, X, Y, X p and Y p are in general comple quantit and contain the amplitude and phase inormation and ω is the rotational speed o the rotor. On substituting equation (10.30) into equation (10.9), we get ( p ) ( p ) Z ( ω ) Z ( ω ) X F Mω X X + + = Z ( ) ( ) ω Z ω Y F + Mω Y + Y (10.31) On separating real and imaginar terms, we get Mω k k ωc ωc X i Mω X pi k k Mω ωc ωc F M Y Y r + ω p r = ωc ωc k Mω k X r F + Mω X pr ωc ωc k k Mω Yi Mω Y pi (10.3) Quantities M, F, F, ω, X, Y, X and Y are either known or are measured during the course o p p the eperiment. k, c, are unknown (eight or the present case). Equation (30) has our equations so i our stiness coeicient are known (b static orce method) the remaining our damping coeicient can be obtained rom this. Alternativel, i bearing dnamic parameters are speed-independent then measurement at least two speed will be suicient to obtain all eight coeicients or b changing F and rotating the rotor at same speed (or speed-dependent bearing parameters) all eight coeicients can be obtained. Imbalanced mass attached to an independent vibrator shat 41

25 In previous method equation (30) has eight unknowns ( k, c, ) and it has our equations. I a test rig capable o providing ecitation b means o imbalance orcing, where the orcing requenc could be varied without upsetting the journal rotational requenc. Thus a second equation (30) could be obtained, resulting in eight simultaneous equations in all, b using a dierent value o ω without upsetting the bearing Sommereld number. Two sets o measurements can be taken or two dierent rotational requenc o secondar shat, most convenient would be position should not disturb (as shown in Figure 10.9b). ± ω so that the stead state (a) Ecitation unit arrangement Fluid Fied non-loating bearing housing ( 0 and 0) b b Journal Anti-snchronous ecitation (b) Basic principle o the ecitation unit Figure 10.9 An anti-snchronous ecitation b an auiliar unbalance unit 4

26 Eample 10.5 For estimation o bearing dnamic coeicients the ollowing measurements were made: (i) X 1 and Y 1 or simultaneous application o F and application F and 1 F 1 and (ii) X and Y or simultaneous F ; where X and Y are displacements and F is orce and in general the all are comple in nature. I shapes o both the orbits o the shat center are circular in shape, whether it would be possible to estimate all bearing dnamic coeicients rom these two measurements. Solution: Consider a single bearing and use a comple stiness, requenc to describe the equation o motion in requenc domain, as = ω + ω, at a single Z k m j c Z Z = Z Z (A) Using two unbalance runs with corresponding responses 1, 1, and and right hand sides,, 1 1 and, equation (A) ma be written as Z Z = 1 1 Z Z (B) 1 1 The solution o equation (B) is obtained as Z Z 1 1 Z Z = ( ) (C) For circular orbits 1 = j1 and = j (or negative, depends on the deinition o aes, and the direction o rotation). Then the denominator o equation (C) becomes = j ) ( j ) 0 (D) 1 1 1( 1 = and hence, equation (D) is ill-conditioned or circular orbits. Having a third unbalance run does not help. For three unbalances equation (B) ma be written as Z Z Z Z = (E) 43

27 The least squares solution involves the ollowing inversion = ( ) = ( )( ) ( ) (F) I = i j, then the denominator o the equation (F) becomes i ( )( ) ( ) ( 1 3 ) ( 1 3 ) ( 1 3 ) = + + j + + j + + = 0 (G) and the circular orbits are ill-conditioned. There is another possibilit when ill-conditioning ma occur, namel when 1 = α1 and = α or an value o α, where α is a constant. Then the denominator o equation (C) becomes zero, leading to ill-conditioning. This means that a change in orbit rom one unbalance to the net is required. The ill-conditioning due to a circular orbit ma be avoided b taking measurements in both the clockwise and anticlockwise directions o rotation o the rotor. For this case 1 = j1 and = j. Then the denominator o equation (C) becomes = j ) ( j ) 0 (H) 1 1 1( 1 and hence, equation (C) becomes well-conditioned. Eercise 10.5 The eight bearing stiness and damping coeicients are to be determined b using the method described above. Eperimental measurements o journal vibration amplitude and phase lag angle are given in the Table 10.1; pedestal vibrations are ound to be negligible. Determine the values o oil-ilm coeicients implied b these measurements, and the maimum change in the direct crosscoupling terms introduced b an error o +4 in the measurement o the phase recorded as 4.5. Table 10.1 Some test data used to calculate bearing stiness and damping coeicients 44

28 Forward ecitation everse ecitation Horizontal vibration amplitude 66.4 µm 46.6µm Horizontal phase lag Vertical vibration amplitude 55.5 µm 38.4 µm Vertical phase lag Force amplitude 1.0 KN 1.0 KN Forcing requenc 1.6 Hz 1.6 Hz Journal mass 150 kg 150 kg MATLAB Solution: INPUT FILE % Name o this input ile is input_qus_1_7.m X 1 =66.4*1.0e-6; % horizontal vibration amplitude (in meter) A 1 =4.5; % horizontal phase lag (in degree) Y 1 =55.5*1.0e-6; % vertical vibration amplitude (in meter) B 1 =9.9; % vertical phase lag (in degree) F 1 =1*1.0e+3; % orce amplitude (in N) n 1 =1.6; % orcing requenc(in Hz) M=150; % journal mass (in Kg) % For the reverse ecitation condition. X =46.6*1.0e-6; % horizontal vibration amplitude (in meter) A =-0.9; % horizontal phase lag (in degree) Y =38.4*1.0e-6; % vertical vibration amplitude (in meter) B =-111; % vertical phase lag (in degree) F =1*1.0e+3; % orce amplitude (in N) n =1.6; % orcing requenc (in Hz) M=150; % journal mass (in Kg) MAIN FILE clear all; input_qus_1_7; w 1 =*pi*n1; w =*pi*n; a 1 = A 1 *(pi/180); b 1 = B 1 *(pi/180); a= A *(pi/180); b = B *(pi/180); p=[-x 1 *sin(a 1 ) Y 1 *cos(b 1 ) 0 0 w 1 *X 1 *cos(a 1 ) w 1 * Y 1 *sin(b 1 ) 0 0; X 1 *sin(a 1 ) Y 1 *cos(b 1 ) 0 0 w 1 * X 1 *cos(a 1 ) w 1 *Y 1 *sin(b 1 ); X 1 *cos(a 1 ) Y 1 *sin(b 1 ) 0 0 w 1 * X 1 *sin(a 1 ) - w 1 *Y 1 *cos(b 1 ) 0 0; 0 0 X 1 *cos(a 1 ) Y 1 *sin(b 1 ) 0 0 w 1 * X 1 *sin(a 1 ) - w 1 *Y 1 *cos(b 1 ); - X *sin(a) Y *cos(b ) 0 0 w * X *cos(a) w *Y *sin(b ) 0 0; X *sin(a) Y *cos(b ) 0 0 w * X *cos(a) w *Y *sin(b ); X *cos(a) Y *sin(b ) 0 0 w * X *sin(a) - w *Y *cos(b ) 0 0; 0 0 X *cos(a) Y *sin(b ) 0 0 w * X *sin(a) - w *Y *cos(b )]; =[-M* w 1^* X 1 *sin(a 1 ); F 1 +M* w 1^*Y 1 *cos(b 1 ); F 1 +M* w 1^* X 1 *cos(a 1 ); M* w 1^*Y 1 *sin(b 1 ); -M* w ^*X *sin(a); F +M* w ^*Y *cos(b ); F +M* w ^*X *cos(a); M* w ^*Y *sin(b )]; k=p\; disp ('The bearing coeicients are'); print ('\nk ='); print (numstr (k (1,1))); print (' N/m\n'); print ('\nk ='); print (numstr (k (,1))); print (' N/m\n'); print ('\nk ='); print (numstr (k (3,1))); print (' N/m\n'); print ('\nk ='); print (numstr (k (4,1))); print (' N/m\n'); print ('\nc ='); print (numstr (k (5,1))); print (' N/m\n'); print ('\nc ='); print (numstr (k (6,1))); print (' N/m\n'); print ('\nc ='); print (numstr (k (7,1))); print (' N/m\n'); 45

29 print ('\nc ='); print (numstr (k (8,1))); print (' N/m\n'); OUTPUT The bearing stiness and dnamic coeicients are K = N/m K = N/m K = N/m K = N/m C = N/m C = N/m C = N/m C = N/m 10.4 Transient Methods In this method it is possible to take measurement on running machines. In this method the sstem consists o a smmetrical rigid rotor mounted in two identical journal bearings. Transient vibration o the rotor in the bearings is caused b appling a orce impulse (as shown in Figure 10.10) to the rotor center o gravit. In practice this is provided b striking the rotor with a calibrated hammer whose head mass is known. This means o ecitation results in an impulse, which lasts or a inite period o time (tpicall a raction o second). I an accelerometer is mounted in the hammerhead, it is possible to determine the instantaneous orce, which is applied to the rotor. The electrical output rom the hammer will then indicate the vibration o the applied orce with time. An impulse can be considered as made up o a number o sine waves o dierent requencies, all occurring simultaneousl. B varing the hammer head mass, the stiness o the hammer impact orce (tip) and the initial hammer head velocit, it is possible to var the amplitude, requenc content and duration o the applied impulse. EOM o the journal would be k k C C = M and k k C C = M (10.33) Figure An impulse in the time and requenc domains Since orcing ma be considered to be sinusoidal, albeit at several dierent requencies, an one component will be o the orm in the horizontal and vertical directions 46

30 j t = Fe ω and = (10.34) j t Fe ω The corresponding horizontal and vertical displacements and will be o the orm so that j t = Xe ω and = j ; = ; = j ; and = j t Ye ω (10.35) ω ω ω ω (10.36) On substituting in EOM ields ( k Mω + jωc ) ( k + jωc ) ( k jωc ) ( k Mω jωc ) F X = F + + Y (10.37) which can written as where ( ω jω ) ( jω ) ( jω ) ( ω + jω ) X 1 k M + C k + C F = Y D k C k M C F ( ω jω )( ω jω ) ( jω )( jω ) D = k M + C k M + C k + C k + C (10.38) Equation (36) is similar to the case o electromagnetic eciter method (irst method) ecept in present case the inertia orce has now also been allowed or. I orcing is applied in one direction (or eample the hammer strikes the rotor in horizontal direction) then it is possible to deine the reacceptance as : (rom equation 36) ( ω + jω ) k M C X = = and D F ( jω ) k + C Y = = (10.39) D F Similarl i hammer strikes the rotor in the -direction then ( jω ) k + C X = = and D F ( ω + jω ) k M C Y = = (10.40) D F 47

31 The reacceptance terms deined in equations (10.39) and (10.40) are clearl unctions o requenc and so take a dierent value depending on the vibration requenc being considered. The reacceptance terms are in general comple because displacement and orce are not in-phase. The method o determining the oil-ilm stiness and damping coeicient makes use onl o the modulus o the reacceptance terms, however, doesnot use the data describing phase. In eperiment the right hand side o equations (10.39)-(10.40) eist at man dierent requencies simultaneousl, and the corresponding receptance terms must be determined or each o these requencies. The reacceptance will be jωt e dt Fourier transorm o ( ω) = = (10.41) Fourier transorm o jωt e dt ( t) ( t) This ma be obtained in eperiment b spectrum analzer and it will displa reacceptance as shown in Figure Figure A tpical variation o magnitude o receptance terms The above reacceptances have been obtained rom right hand side o equations (10.39)-(10.40) using two independent orcing. Now our aim is to obtain k etc. so that when it is substituted back in let hand side o equation (10.39)-(10.40) it should give the value o the right hand side o equation (10.39)-(10.40). These processes can be repeated until appropriate values are ound which results in the dierence between let hand side and right hand side o equations (10.39)-(10.40) being minimized, or all requencies under consideration. The least squares error criteria ma be used so that to minimize a scalar quantit s [ ij ω) ( ω ] = i j ω theor ij ) ep ( (10.4) 48

32 Using ilter snchronous the imbalance response must be subtracted as shown in Figure Figure 10.1 Eect o residual unbalance on the impulse response Step unction: This step unction can be generated b giving gradual static load to the rotor and suddenl releasing the load at well deined upper limit o the static load as shown in Figure Figure A step unction orcing and corresponding response eerences [1] Goodwin, M.J., 1991, Eperimental Techniques or Bearing Impedance Measurement, ASME Journal o Engineering or Industr, Vol. 113, No. 3, [] Mitchell, J.., Holmes,. and Ballegooen, H.V., , Eperimental Determination o a Bearing Oil Film Stiness, in the 4 th Lubrication and Wear Convention, IMechE, Vol. 180, No. 3K, [3] Parkins, D.W., 1979, Theoretical and Eperimental Determination o the Dnamic Characteristics o a Hdrodnamic Journal Bearing, ASME Journal o Lubrication Technolog, Vol. 101, No., [4] Swanson, E.E. and Kirk,.G., 1997, Surve o Eperimental Data or Fied Geometr Hdrodnamic Journal Bearings, ASME Journal o Tribolog, Vol. 119, No. 4, [5] Tiwari,., Lees, A.W. and Friswell, M.I., 004, Identiication O Dnamic Bearing Parameters: A eview, Shock & Vibration Digest, Vol. 36, No., [6] Tripp, H. and Murph, B. T., 1984, Eccentricit measurements on a tilting pad bearing, Trans ASLE, Vol.8, No.,

33 [7] Woodcock, J.S. and Holmes,., , The Determination and Application o the Dnamic Properties o a Turbo-otor Bearing Oil Film, Proceedings o IMechE, Vol. 184, No. 3L,