Hydraulic actuators, Dispense di Ingegneria Meccanica

Scarica Hydraulic actuators e più Dispense in PDF di Ingegneria Meccanica solo su Docsity! 1 Control and Actuating Devices for Mechanical Systems Prof. Francesco Braghin Hydraulic actuators 2 Hydraulic actuators Controller F(s) Y(s) Actuator Mechanicalsystem Sensor U(s)Yref(s) E(s)+ - Actuator 5 Hydraulic pistons piston rod 1st chamber (pressure p1) flow rate Q2 flow rate Q1 2nd chamber (pressure p2) The force/torque developed by the actuator is proportional to pressure difference. 6 Hydraulic pistons: classification F p A k x= ⋅ − ⋅ 1 1 2 2F p A p A= ⋅ − ⋅ 1 2( )F p p A= − ⋅ single-acting piston  single rod double-acting piston  single rod double-acting piston  double rod 7 Servovalves Open and closed loop control of hydraulic actuators is relatively simple using valves and pumps. Hydraulic actuators: typical applications Hydraulio high pressure pump system Hydraulic fiying control surfaces Ax servo actuator ‘system Manometric Lines Oxygen lines Fuel manifold system Aîr Landing gear conditioning hydraulic retraction system jacks or brakes systems Potable galley Waste Auxiliary power water system water unit fuel feed system 10 11 Hydraulic system forward chamber return chamber piston rod servovalve to the pump from the pump 12 Hydraulic system supply return By acting on the servovalve, the volumetric flow rates (and thus the pressures) can be adjusted so as to control:  the force developed  the piston-rod displacement y TYPICAL “SPOOL VALVE” PISTON ROD 1 2( )F p p A= − ⋅ 15 Governing equations 0 0 pβ ρ ρ ∂ = ⋅ ∂ 0 11 p Tρ ρ α β   = ⋅ + ⋅∆ − ⋅∆    isothermal bulk modulus 00 1 T ρα ρ ∂ = − ⋅ ∂ isobar cubical expansion coefficient Assuming constant oil temperature (isothermal conditions): 0 1 pρ ρ β  ∆ = ⋅ +    Thus 0d d p dt dt ρρ β ∆ = ⋅ 16 Governing equations 3) CONTINUITY EQUATION + EQUATION OF STATE From the combination of these two equations, it is possible to express the continuity equation in terms of volumetric flow rates Assuming we have ρ = ρ0 (low compressibility fluids, see equation before). Thus, 1p β ∆  The contribution is not negligible in case of fast pressure variations and significant volumes of the actuator’s chambers. V d p dtβ ∆ ⋅ in out dV V d pQ Q dt dtβ ∆ − = + ⋅ ( ) ( ) ( ) 0in out Vd d dV d p dVQ Q V V dt dt dt dt dt ρρρ ρ ρ ρ ρ β ∆ − = = + = + 17 Hydraulic piston Let’s apply the above equation to the two chambers of a double-acting double rod piston: Chamber 1: Chamber 2: Cip = internal leakage coefficient of piston (laminar flow) Cep = external leakage coefficient of piston (laminar flow) 1 1 1 1 1 2 1( )ip ep dV V d pQ C p p C p dt dtβ ∆ − ∆ −∆ − ∆ = + ⋅ 2 2 2 2 1 2 2( )ip ep dV V d pQ C p p C p dt dtβ ∆ − + ∆ −∆ − ∆ = + ⋅ 20 Servovalve return supply return Q1 Q2 Q1 Q2 return supply return TYPICAL “SPOOL VALVE” Fluid flow through valve clearances can be analyzed by making reference to the flow through an orifice diaphragm 21 Servovalve 1 1 2 2 3 3Au A u A u= = turbulent flow (high Reynolds numbers) the fluid particles are accelerated between section 1 and 2 A2 = minimum jet area  vena contracta Applying Bernoulli’s equation: Applying the continuity equation for incompressible flow the volumetric flow rate Q is equal to: 2 0c cA C A C= ⋅ → is the contraction coefficient 2 2 2 2 1 1 1 1 2 2 u p u pρ ρ+ = + Assuming no pressure recovery during expansion from section 2 to 3: 1 2 1 3p p p p p∆ = − = − 22 Servovalve 2 2 2 2 1 1 1 1 2 2 u p u pρ ρ+ = + ( ) 2 2 2 22 1 2 2 1 2 2 2 1 2 1 2 1 1 1 2 2 1 p Ap p p u u u u A A A ρ ρ ρ ∆     − = ∆ = − = − ⋅ ⇒ =       −    Thus, the volumetric flow rate (evaluated in the minimum jet area) is equal to: ( )02 2 0 2 0 02 0 1 2 2 1 c c d c C A p pQ A u C A u C A A C A A ρ ρ ∆ ∆ = = = ⋅ = ⋅   −     0 2 2 0 1 ( ) 1 c d c CC A AC A =   −      discharge coefficient (nonlinear function of the orifice section) ( )0 0 0 2 ( , )d pQ C A A Q Q A p ρ ∆ = ⋅ ⇒ = ∆ 25 Hydraulic actuator: complete model FUNDAMENTAL EQUATIONS a) PISTON’S LINEARIZED EQUATION b) SERVOVALVE’S LINEARIZED EQUATION c) PISTON’S EQUATION OF MOTION b) SERVOVALVE’S LINEARIZED EQUATION Combining the two equations: 26 Hydraulic actuator: complete model 0 2 2 ep L L ip L C V dpdyQ C p A dt dtβ   ⇒ = + + + ⋅    L q c LQ K x K p= − 0 2 2 ep L q c L ip L C V dpdyK x K p C p A dt dtβ   − = + + + ⋅    a) PISTON’S LINEARIZED EQUATION 27 Hydraulic actuator: complete model 0 2 2 epL ip c L q CV dp dyC K p K x A dt dtβ   ⋅ + + + = −    *C *0 2 L L q V p C p K x Ay β + = − ⇒ 1st order LTI differential equation 0 2 2 ep L q c L ip L C V dpdyK x K p C p A dt dtβ   − = + + + ⋅    30 Hydraulic actuator: state-space model * 0 0 0 2 2 2 q L L KA Cp y p xV V V β β β = − − +  p L p p c Ay y p m m = − + ⋅  y y=  L y z y p    =       STATE VARIABLES: * 2 ep ip c C C C K= + + 31 * 0 0 0 00 0 1 0 0 0 2 2 2 p p p q c A m m z z xK CA V V V β β β     −       = +       − −          Hydraulic actuator: state-space model L y z y p    =       STATE VARIABLES: [ ] [ ]z A z B u u x = + ≡  32 ( ) ( ) ( )2p p Lm s c s Y s AP s+ = ( ) ( ) ( )*0 2 L q V s C P s K X s AsY s β   + = −    Hydraulic actuator: TF model and applying Laplace transform: *0 2 L L q V p C p K x Ay β + = −  p p Lm y c y p A+ = ⋅  Recalling the two system’s differential equations: * 2 ep ip c C C C K= + + 35 ( )2 * 20 ( )( ) ( ) 2 q yx p p AKY sG s X s Vm s c s s C A s β = =   + + +    Y(s)X(s) ( )yxG s Hydraulic actuator: TF model PL(s) Y(s) As X(s) ( )2p p A m s c s+*0 1 2 V s C β   +    qK + - … it is possible to get the transfer function between the input displacement of the control valve and the output displacement of the piston rod: 36 ( )2p p L m s c s P Y A + = Hydraulic actuator: TF model ( ) ( ) ( ) 2 2 2 * 20 2 p p p p q L p p m s c s m s c s AK P Y X A A Vm s c s s C A s β + + = = ⋅   + + +    ( ) ( ) 2 2 * 20 ( )( ) ( ) 2 q p pL px p p K m s c sP sG s X s Vm s c s s C A s β + = =   + + +    … it is also possible to get the transfer function between the input displacement of the control valve and the output pressure drop across the piston rod: