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SFD EXPERIMENTAL TESTING & ANALYTICAL METHODS DEVELOPMENT High Load SFD Test Rig Identification of SFD force coefficients Luis San Andrés Mast-Childs Professor May 2011 1 PW-SFD test rig (2010) Static loader Shaker assembly (Y direction) Shaker assembly (X direction) Static loader Shaker in Y direction Shaker in X direction SFD test bearing 2 Test rig description Static loader shaker Y shaker X SFD Y Static loader support rods base X 3 Flow path & main features Oil in, Qin Test rig main features in Journal (D) Oil out, Qt End groove Central groove c L, 2L L L, 2L Bearing Cartridge Journal diameter: 5.0” Film clearance: (A) 5.55mil (B) 5.43mil Film length: (A) 2x1”, (B) 2 x 0.5“ Centering stiffness: variable End groove ISO VG 2 oil Oil out, Qb Oil collector Oil out Base Support rod 4 Test rig cross section – rods installation All dimensions in inches 9 8 4 x Φ 7/8 5.6 BC OD Φ7.50 12 x Φ 7/8 Test rig materials Φ4.75 4.755 Journals, journal base, pedestal, bearing cartridge, Main support rods : AISI 1020 steel Flexural rods: Alloy Steel per ASME Φ11.00 B18.3 5 Eddy current sensors and accelerometers Top view Sensor locations Y Piezoelectric accelerometer θ= 180o and 270o θ= 90o Top Land Eddy current sensor (Proximity probe) Central groove θ= 180o X eddy current sensor (X proximity probe) θ= 0o Journal B X Piezoelectric accelerometer Bottom Land θ= 0o and 90o Top Land Y eddy current sensor (Y proximity probe) θ= 270o Side view: Sensors located in central groove Piezoelectric Accelerometer Bottom Land 6 Pressure sensors θland = 210o and 330o Top view Top Land 0.5 inch PCB (pressure sensors) Central groove 0.5 inch Bottom Land Locations Journal B , θland = 330o and θland = 210o Top Land PCB (Dynamic) 1.5 inch Central groove PCB and Entran Bottom Land Side view: Sensors located at middle plane of film lands BC 7 Pressure sensors 8 Test results for (c) SFD force coefficients – Comparison between short and long open ends dampers 9 Direct damping coefficient [lbf-s/in] compare SFD damping 250 200 CXX ~ CYY Long (L=1 inch) 150 3 C XX1inch 100 C XX 0.5 inch 50 3 LA 1 c 5.55 7.49 A 3 3 0.5 LB c 5.43 B Cxx 1/2 inch Cyy 1/2 inch Cxx 1 inch Cyy 1 inch CXX ~ CYY Short (L=0.5 inch) 0 0 0.5 1 1.5 2 2.5 3 Static eccentricity e s [mil] 3 Ratio of coefficients ~ (L/c3) Long and short SFDs (circular orbits) 10 compare SFD inertia Added Mass coefficients [lb] 50 45 MXX , MYY 40 35 Long (L=1 inch) 30 25 MXX , MYY 20 15 10 M XX1inch 5 M XX 0.5inch 0 0 LA 1 c 5.55 1.96 A 0.5 LB c 5.43 B 0.5 1 1.5 Mxx 1/2 inch Myy 1/2 inch Mxx 1 inch Myy 1 inch 2 2.5 Short (L=0.5 inch) 3 Static eccentricity e s [mil] Ratio of coefficients ~ (L/c) Long and short SFDs (circular orbits) 11 Film and groove dynamic pressures pressure (psi) Pressures at film lands 10 Lands 5 0 5 10 0 1 2 3 4 Frequency=250 Hz time (-) top land (120 deg) bottom land (120 deg) pressure (psi) Pressures at central groove 10 Groove 0 Top Land 1 inch 10 20 PCB (pressure sensors) 0 1 2 time (-) groove (165 deg) groove (285 deg) 3 Central groove 4 1 inch L/D=0.2 x 2 Bottom Land Long open ends SFD. Centered bearing es=0, circular orbit r=0.1cA. Groove pressure PG = 0.72 bar 12 Film and groove peak-peak pressures peak-peak pressures Top land (120) Bottom land (120) Groove (165) P-P pressure (psi) 40 Frequency20-250 Hz Land length=1 in Groove width=0.5 in depth = 3/8 in (75 c) 30 20 PCB (pressure sensors) 10 Top Land groove 0 0 100 Frequency (Hz) 1 inch Central groove 200 1 inch Bottom Land Long open ends SFD. Centered bearing es=0, circular orbit r=0.1cA. Groove pressure PG = 0.72 bar 13 Test results for (d) SFD force coefficients – Comparison between open ends and sealed ends long dampers B journal I Top land A groove A Bottom land II B BC 14 compare SFD damping B Damping coefficients (lb f-s/in) SFD (1 inch land lengths) circular orbits journal I Top land A groove A 450 Sealed ends CYY (B-B) Bottom land II Sealed ends CXX(B-B) CSFD 400 B CXX ~ CYY 350 BC B-B sealed SFD Sealed ends 300 250 open ends CYY 200 CXX ~ CYY open ends CXX 150 Open ends 100 0.0 0.5 1.0 1.5 2.0 2.5 Eccentricity es (mil) Open and sealed ends long SFD (circular orbits) 15 compare SFD inertia Added mass coefficients (lb) SFD (1 inch land lengths) circular orbits 100 90 MXX , MYY Sealed ends MYY(B-B) Sealed ends 80 70 Sealed ends MXX(B-B) 60 Open ends MYY 50 40 MXX , MYY open ends MXX 30 Open ends 20 MSFD 10 0 0.0 0.5 1.0 1.5 2.0 2.5 Eccentricity es (mil) Test data for open and sealed ends (circular orbits)16 Conclusions: Learning from tests and predictions 17 Summary of learning Open ends long damper shows ~ 7 times more damping than short length damper. Inertia coefficients are two times larger. SFD force coefficients are more a function of static eccentricity (max. 40%c) than amplitude of whirl (max 40%c) changing little with ellipticity of orbit (aspect ratios 1:1, 2:1 & 5:1) Piston ring faces orientation affects leakage and force coefficients. Long Sealed SFD shows ~2.6 times more damping than open ends SFD Code benchmarked for long and short SFDs (open and sealed ends). 18 Proposed work (TRC) Linear-Nonlinear Force Coefficients for Squeeze Film Dampers Whirl Orbit Analysis for Identification of SFD force coefficients 19 Types of journal motion x= R x= R e: amplitude of motion rX, rY : amplitudes of motion whirl frequency h Film thickness Y eYo R whirl frequency eXo Y 2rY e eo whirling journal X (a) small amplitude journal motions 2rX X (b) large amplitude journal motions Applications: K,C, M (force coefficients) FX, FY (reaction forces) RBS stability analysis RBS imbalance response & transient load effects 20 SFD predictive code Orbit 1.0 0.8 2r Y 0.6 2 rX 0.4 0.2 Y/c Code & GUI: virtual tool for prediction of SFD forced response (a) Linear force coefficients (K,C,M) (b) Instantaneous reaction forces along orbital path (c) Automated orbit analysis for NL parameter identification -1.0 -0.8 -0.6 -0.4 0.0 -0.2 0.0 -0.2 0.2 0.4 0.6 0.8 -0.4 -0.6 -0.8 -1.0 X/c 21 1.0 Purpose of whirl orbit analysis Orbit 1.0 0.8 0.6 0.4 0.2 Y/c for specified whirl orbit and over specifiedfrequency range: • predict SFD reaction forces vs. time, • conduct Fourier analysis, & • identify SFD linearized force coefficients -1.0 0.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 -0.2 -0.4 -0.6 -0.8 -1.0 X/c 22 SFD example Journal Diameter 5.0 in Total Length 1.0 in Land Clearance 5.0 mil Bearing D Y NO Central Groove Feed holes 3 (120deg) Axial Length 0.5 in Ambient Pressure 0.0 psig Supply Pressure 10 psig Cavitation Pressure -14.70 psig Supply Temperature 77 oF Viscosity at Tsupply 0.43 mReyns Density 49 lb/ft3 Journal Feed hole X Pa, ambient pressure Ps, supply L pressure Section A-A A Open Ends SFD with feed holes 23 whirl orbit induces forces h - with holes - no central groove SFD open ends L=1.0 inch - with holes - no central groove 1.0 Reaction forces 0.8 locus -0.8 -0.6 -0.4 0.0 -0.2 0.0 60.0 approx using Fourier series 0.2 0.4 0.6 0.8 1.0 -0.2 -0.4 40.0 FY eY/c 0.4 0.2 -1.0 80.0 0.6 20.0 -0.6 -0.8 -1.0 eX/c lbf 0.0 -250.0 -200.0 -150.0 -100.0 -50.0 0.0 50.0 100.0 -20.0 -40.0 Eccentric (Off-center) Elliptical orbit -60.0 -80.0 es/c=0.5c r/c=0.25c -100.0 FX Fundamental 1X Force SFD reaction force 24 SFD Forces: predicted and 1X FD open ends L=1.0 inch - with holes - no central groove Frequency 180 Hz 100 lbf 0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 -50 SFD 1X forces do not reproduce NL forces -100 -150 -200 -250 SFD reaction force 1.0 locus 0.8 FX FY SFD open ends L=1.0 inch - with holes - no central groove Journal eccentricity locus Fraction of Period 0.6 eY/c Bearing reaction force es/c=0.5c r/c=0.25c Fundamental 1X Force 50 0.4 0.2 FX_1Fourier -1.0 -0.8 -0.6 -0.4 0.0 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 -0.2 FY_1Fourier -0.4 -0.6 -0.8 -1.0 eX/c 25 SFD reaction forces The SFD instantaneous reaction force superimposes a dynamic force to a static force, i.e., F=Fstatic+Fdyn. The dynamic components of the SFD reaction forces are modeled in a linearized form as Fdyn KSFD z CSFD z + MSFD z where z is a vector of dynamic displacements and (K, C, M)SFD are matrices of stiffness, viscous damping and inertia force coefficients 26 Analysis (I) The dynamic or time varying part of the SFD reaction force is periodic with fundamental period T=2p/. Using Fourier series decomposition, it Fdyn F1e i 2t FII e i 3t FIII e .... To first order effects (fundamental frequency) Fdyn F1e i t where F1 K SFD 2 M SFD i CSFD z1 H z1 H K SFD M SFD i CSFD 2 is the matrix of damper impedances 27 Analysis (II) The code predicts the SFD time varying reaction forces for the orbital path and delivers the fundamental Fourier components of motion and forces, i.e. z and F. Forward and backward whirl orbits ensure linear independence of the two SFD reaction forces. Solution of the system of algebraic equations: F1 F2 H z1 z2 leads to the determination of the impedances: HXX, HXY,HYX, HYY 28 Analysis (III) The analysis stacks impedances for a set of frequencies (k=1,2,….N) from which, by linear curve fits, one determines : 2 SFD SFD K M Re(H ) CSFD Im(H ) 29 SFD Real Impedances vs. frequency SFD open ends L=1.0 inch - with holes - no central groove 2.50E+04 Impedances (real part) lbf/in HXX will give M<0 Re(H) 2.00E+04 1.50E+04 1.00E+04 5.00E+03 0.00E+00 50 100 -5.00E+03 Hyy 200 250 SFD open ends L=1.0 inch - with holes - no central groove H YY fits well model K-M2 -1.00E+04 Hxx 150 Journal eccentricity locus 1.0 locus Whirl frequency (Hz) 0.8 0.6 eY/c 0 0.4 0.2 -1.0 -0.8 -0.6 -0.4 0.0 -0.2 0.0 Hxy -0.4 Hyx -0.8 Frequency range 20-200 Hz 0.2 0.4 0.6 0.8 1.0 -0.2 -0.6 -1.0 eX/c 30 Impedances (imag part) SFD Ima Impedances vs. frequency SFD open ends L=1.0 inch - with holes - no central groove lbf/in 1.00E+05 HXX gives average C Ima(H) 8.00E+04 6.00E+04 4.00E+04 2.00E+04 HYY fits OK model C 0.00E+00 0 50 100 150 SFD open ends L=1.0 inch - with holes - no central groove 200 250 Journal eccentricity locus 1.0 locus -2.00E+04 Hyy Hxy Hyx Frequency range 20-200 Hz 0.6 eY/c Hxx 0.8 Whirl frequency (Hz) 0.4 0.2 -1.0 -0.8 -0.6 -0.4 0.0 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 -0.2 -0.4 -0.6 -0.8 -1.0 eX/c 31 SFD NL-Linear force coefficients Frequency range 20-200 Hz Kxx Kyy Kxy Kyx lbf/in lbf/in lbf/in lbf/in 2.12E+03 -2.11E+02 -6.35E+01 -6.13E+01 Fourier analysis Linearized forces Nonlinear forces 100 50 0 -100 -80 -60 -40 -20 0 Cxx Cyy Cxy Cyx lbf-s/in lbf-s/in lbf-s/in lbf-s/in 68.5 37.7 -0.6 -0.6 Mxx Myy Mxy Myx -200 lbm lbm lbm lbm SFD NL force response-250 -5.0 2.0 0.2 0.2 40 60 80 -50 -100 -150 FY vs FX 1.0 inch - with holes - no central groove locus 20 1.0 Linear force model 0.8 eY/c 0.6 DISSIPATED ENERGY IN A PERIOD or MOTION 0.4 0.2 -1.0 -0.8 -0.6 -0.4 0.0 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 -0.2 -0.637 Non-linear (from time transient response) -0.590 Linear from ALL force coefficients -0.4 -0.6 -0.8 -1.0 eX/c lbf-in 32 Proposed tasks (2011-12) 1. Test ACTUAL short length open ends damper with dynamic loads (20300 Hz) inducing off-centered elliptical orbital motions with amplitude ratios (5:1) to reach 0.8c. 2. Identify SFD force coefficients from test impedances, and correlate coefficients with linear force coefficients and experimental coefficients for smallest whirl amplitude (0.05c). 3. Perform numerical experiments, similar to the physical tests, to extract linearized SFD force coefficients from the nonlinear forces. Quantify goodness of linear-nonlinear representation from an equivalence in mechanical energy dissipation. Y centered journal X circular orbits Y Y off-centered journal X X elliptical orbits 33 Budget (2011-12) Support for graduate student (20 h/week) x $ 1,800 x 12 months $ 21,600 Fringe benefits (0.6%) and medical insurance ($191/month) $ 2,419 Travel to (US) technical conference $ 1,200 Tuition three semesters ($3,802 x 3) $ 10,138 Supplies for test rig $ Total Cost: Y centered journal X circular orbits Y 1,500 $ 37,108 Y off-centered journal X X elliptical orbits 34 Questions (?) 35