.Centrifugal Forces. ( ) Geometric Centerline. :Principal Inertia Axis. shaft material. Balancing

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Transcript:

مقدمة الى الموازنة الديناميكية : - -

. - - - - - -. - - -. - -. ١

١. مقدمة حول الموازنة... -.Centrifugal Forces. () Geometric Centerline :Principal Inertia Axis shaft material.. Balancing. ٢

- : Manufacturing Faults.. ( ). : Assembly Faults. Rotor Keys. :. wear. ( ). - : :.. :.. :.. :.. ٣

:Personnel Safety. :.... ٢. انواع اللااتزان.Principal Inertia Axis : Static nbalance - Principal Inertia Axis Shaft Axis ( ). ).Force nbalance. ( Couple nbalance - Shaft Axis Principal Inertia Axis 180 Principal Inertia Axis ( )..180 ٤

Shaft Axis Principal Inertia Axis Dynamic nbalance -.. ).. ( ٣. الحدود المسموح بها لقيم اللااتزان ودرجات الجودة Balancing.. Residual nbalance. Residual Permissible nbalance -. E per : E per per per = (g.mm/kg) (3.1) M g.mm per kg M E per. Principal Inertia Axis ٥

E per E per : E per ω = constant (3.2) mm/sec -.. Balance quality grades are standardized in ISO 1940. Balance quality grade G Vibration velocity in mm per second (E per ω) G 100 100 G 40 40 G 16 16 G 6.3 6.3 G 2.5 2.5 G 1 1 G 0.4 0.4 The smaller the number, the smoother the operation ω= 2π N/60, where N is the speed in RPM Rotor types General examples Crankshaft drives of large Diesel engines Complete engines for trucks and locomotives Crankshaft drives for engines of trucks and locomotives Parts of crushing machinery Parts of agricultural machinery Fly-wheels Fans Aircraft gas turbine rotors Electrical armatures Process plant machinery Pump impellers Machine-tool drives Turbo compressors Small electric armatures Turbine-driven pumps Grinding machine drives Textile bobbins Automotive turbochargers Gyroscopes Disk-drives Spindles for high-precision applications - ٦

- : :Balance Quality Grades ISO 1940 - G630 G E per ω = 630 630 mm/sec :rad/sec. E per Grade ω = (3.3). 2.5 - Mass :Experimental Balancing Production.. Field or in-site Balancing. Coupling or Joints :Bearing Forces...Hard Bearing Balancing Machines.. ٧

- (3.1).M E per per per. Single Plane Balancing -- E M per per per = (g.mm) (3.4) R per : mm R Permissuble Mass per = (gram) (3.5) R Two-Plane Balancing -- Left Correction Plane Right Correction Plane per : CG :. a h L b d h R.d. b : ٨

per, L per, R per h b h b R = per L = per 70% Left Correction Plane per, L per, R h L h d b b h d b b R = per L = per per per 30% per,r per,l (3.6).Narrow Rotor. b d. : d/b (3.7) Narrow or Overhung Rotor. d b d CG Left Correction Plane C Overhung Rotor Right Correction Plane.d h R b Static Correction Plane Right Correction Plane b : Static. Correction Plane.. :. ٩

per, static per, couple per d = 2 2C per 3d = 2 4b Left Correction Plane b Right Correction Plane C Static Correction Plane d Narrow Rotor (3.8). (3.8) : 1 = + 2 cuople, L = L = cuople, R ( ) static L R R static static (3.9) R.. per,couple L per,static ٤. مكاي ن الموازنة وطرق الموازنة. Soft Bearing. Hard Bearing Soft-Bearing Balancing Machines - Motion Measuring Balancing Machines. Inertia Forces ١٠

.(-). -).ω r m ( (-) M ω r m C k (-) M : ١١

(4.1) Mx + Cx + kx = F Sin ωt 0 x F 0 = m ω 2 r.m : Steady-state Solution x XSin( ωt φ) = (3.11) φ X X = φ = tan 2 ( k M ω ) + ( C ω) 1 F 0 Cω 2 k M ω 2 2 (3.12) : k C C ωn =, ζ = = M C 2M ω c n : (3.12) X = k φ = tan 1 0 2 2 2 ω ω 1 + 2ζ ω n ω n ω 2ζ ω n ω 1 ωn 2 F (3.13) (ω/ω n ) ω n (3.13). 180 (ω/ω n ). : m.r = i t mr e ω ١٢

: x x = α Influence (3.14) α.coefficient Trial ( ) x Calibrators Masses. : x 0 0 ١. نفترض ان اللااتزان الاول ي الموج ود ف ي المح ور ه و : x = α 0 0 (3.15) trial :.٢ x = α + ( ) 1 0 trial (3.16) : α.٣ α ( x 1 x 0 ) = (3.17) trial.α x 0 ١٣

2 1 : (-) x = α + α i i 1 1 i 2 2 2 () α ij.i x i 1 (3.18) 2 1 i.i j Left Correction Plane (Plane 1) Right Correction Plane (Plane 2) Left Support (Support 1) Right Support (Support 2) a b d (-) (3.18) x 1 α11 α12 1 1 [ α ] x = = α α 2 21 22 2 2 : (3.19) : 10. x 10, x 20 20 : x = α + α 10 11 10 12 20 x = α + α (3.20) 20 21 10 22 20 ١٤

t 1 :. x = α + + α ( t ) ( t ) 11 11 10 1 12 20 (3.21) x = α + + α 21 21 10 1 22 20 : x = α + α + ( t ) ( t ) 12 11 10 12 20 2 x = α + α + 22 21 10 22 20 2 t 2 (3.22). : (3.20 3.22). α α α α 11 12 21 22 = = = = ( x x ) 11 10 t 1 ( x x ) 12 10 t 2 ( x x ) 21 20 t 1 ( x x ) 22 20 t 2 (3.23) : x [ ] 1 10 α 10 = x (3.24) 20 20.Computerized System Multi-plane Balancing.(number of correction planes x number of measurement planes) ١٥

Hard Bearing Balancing Machines - ( ) (-).. (ω/ω n ) (3.13) k F و F R L.. :(- ) F = 0 FL + FR + F1 + F2 = 0 M L = 0 af + ( a + b) F + df = 0 1 2 R (3.25) : F 2 F 1 ( d a b ) FR ( a + b ) FL F1 = b afl ( d a) FR F2 = b (3.26) : i Fi = 2 ω (g.mm) (3.27). ١٦

.. : [1] "Theory of Vibration with Application", William T. Thomson, London 1988. [2] "Theory of Machines and Mechanisms", Joseph E. Shigley and John J. icker, New York 1995. [3] "A Computerized System for Rotor Balancing", M. S. Abdelsattar, MSc Thesis, niversity of Basra 1999. [4] " ISO 1940/1, Balance Quality Requirements of Rigid Rotors", International Organization for Standardization. [5] "Balancing Quality Requirements of Rigid Rotors", IRD Balancing Tech. Paper 1. ١٧