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Linköpings universitet TMHP51 IEI / Fluid and Mechanical Engineering Systems ____________________________________________________________________________________ Hydraulic Systems with Load Dynamics Karl-Erik Rydberg 2008-10-15 J1 Dm qm xv KL BL J2 qL Linköpings universitet IEI / FluMeS, K-E Rydberg 1 2008-10-13 Hydraulic systems with dynamic loads The design of hydraulic systems is often based on a simple load model, represented by single lumped parameters. This type of load model can be used if the connection between the hydraulic system and the mechanical load is stiff. However, in many applications, the mechanical system, which the hydraulic power elements are connected to, is weak compared to the stiffness of the hydraulic system. Such weak mechanical structures cause resonance’s which can be lower than the hydraulic natural frequency. If the structural resonance’s dominate the frequency response of the servo system, it is extremely important to take this fact into account in system design. The main reasons are that the stability of the systems and the bandwidths are limited by the lowest natural frequency in the control loop. A simple valve-controlled hydraulic cylinder with a load represented by a mass Mt and an arbitrary load force FL is shown in Figure 1. The four-way valve is assumed to be a servovalve with constant flow gain Kq. Figure 1: Valve-controlled hydraulic cylinder with a mass load For the special case of centered piston, the oil volumes between the piston and the valve are V1 = V2 = Vt/2. If then the load pressure is defined as pL = p1 - p2 the following linearized and laplace transformed equations can be derived: Vt ⎞ ⎛ KqXv = Ap sXp + ⎜ Kce + s⎟ P ⎝ 4β e ⎠ L (1) Mt s 2 Xp = Ap PL − FL (2) Linköpings universitet IEI / FluMeS, K-E Rydberg 2 2008-10-13 These equations resulted in the block diagram in Figure 2. Hydraulic system Mechanical system - Load Kce Xv Kq + 1 _____ Vt ___ s 4βe - PL QL FL Ap 1 ____ Mt s + . Xp 1 _ s Xp Ap Figure 2: Block diagram of a valve-controlled hydraulic cylinder with a mass load The question is now to define a load model, which is simple to connect to the hydraulic system. As can be seen from Figure 2 the transfer function of the mechanical load can be expressed as Gm(s) = 1. QL PL QL = Ap sXp ; (3) Loads with one degree of freedom An hydraulic cylinder acting on a simple mass load is illustrated in Figure 3. This is a load with one degree of freedom. Ap p1 C1 xp Ap Mt C2 p2 FL Figure 3: Symmetric hydraulic cylinder with a mass load For a double acting symmetric cylinder loaded by a mass Mt and an external force FL according to Figure 3, the piston position Xp is given by equation (2) as Xp = Ap PL − F L Mt s 2 2 ; FL = 0 ⇒ Gm(s) = Ap s Mt s 2 = 1 Mt 2 Ap (4) s Linköpings universitet IEI / FluMeS, K-E Rydberg 3 2008-10-13 The load dynamics, from force to piston velocity, is in this case represented by a pure integrator. Introducing the total capacitance of the cylinder CL the transfer function Gm(s) can be rewritten as Gm(s) = QL CL = PL CLMt A2p = s CL s (5) ω2h The capacitance of the cylinder CL is definded as V1 1 1 1 = + ; C1 = och CL C1 C2 βe C2 = V2 βe (6) Centered cylinder piston gives the Capacitance CL and the hydraulic natural frequency ωh as Vt V1 = V2 = 2 Vt CL = 4β e ⇒ ⇒ 4β eA2p Vt Mt ωh = (7) Figure 4 shows a more general load situation where the mass load is completed with a spring gradient KL and a viscous damping coefficient BL. The cylinder is also defective with viscous friction, the coefficient Bp. Ap KL Ap Mt p1 C1 C2 p2 xp Bp FL BL Figure 4: Symmetric hydraulic cylinder with a mass, spring and damping load The load transfer function will be expressed in the same way as in equation (5) which gives Gm(s) = QL = PL CLMt A2p 2 s + CL s CLBe A2p s+ CLKL A2p A2p s KL = Mt 2 Be s + s+1 KL KL where the total viscous friction coefficient is Be = BL + Bp. (8) Linköpings universitet IEI / FluMeS, K-E Rydberg 4 2008-10-13 In real systems, we always have a certain elasticity in the mechanical structure. For a cylinder this weakness can be related to spring systems in the rear mounting end (K1) and in the piston rod (KL) as shown in Figure 5. K1 Ap x1 p1 C1 xp Ap KL C2 p2 Bp x2 Mt FL xL Figure 5: Hydraulic cylinder with elastic mountings Suppose that the piston and rod have no mass and the external force FL = 0. The resulting force equation will now be expressed as Ap P L = Mt s 2 XL = −K1 X1 = KL(X2 − XL) (9) The load flow to the cylinder (QL) is QL = Ap s (X2 − X1 ) = Ap sXp (10) The transfer function of the mechanical load can be written as ⎛ A 2p ⎡ QL 1 ⎞⎟ ⎤ 2⎜ 1 +1 Gm = = M s + t PL M ts ⎢⎣ ⎝ K1 K L ⎠ ⎦⎥ (11) Consider the system shown in Figure 5 and let the cylinder be controlled by a 4-port servo valve with the coefficients Kq and Kc. This system is shown in Figure 6. x1 xp K1 V1 Ap V2 p1 p2 x2 xL KL Mt xv Figure 6: Valve controlled cylinder with elastic mountings According to Figure 2 the block diagram for the system will be developed as in Figure 7. Linköpings universitet IEI / FluMeS, K-E Rydberg 5 2008-10-13 Kce Xv Kq + 1 _____ Vt ___ s 4βe - PL QL . ⎛ 1 1 ⎞ ⎤ Xp 1 ⎡ 2 ⎟ + 1⎥ + ⎢M t s ⎜ ⎝ K1 KL ⎠ ⎦ M ts ⎣ Ap 1_ s Xp Ap Figure 7: Block diagram for a valve controlled cylinder with elastic mountings Reduction of the block diagram in Figure 7 and completing with the transfer function from Xp to XL gives the following diagram. ⎛ 1 1 ⎞ M ts 2 ⎜ + ⎟ +1 ⎝ K1 K L ⎠ s2 δ 'h + 2 +1 ω 'h2 ω 'h Xv K q Ap . Xp 1_ s Xp 1 ⎛ 1 1 ⎞ M ts ⎜ + ⎟ +1 ⎝ K1 K L ⎠ XL 2 Figure 8: Complete block diagram for a valve controlled cylinder with elastic mountings where ω 'h = Ke Mt ; 1 Vt 1 1 K ce M t ω 'h ' = + + and δ = ⋅ 2 h Ap 2 Ke 4β e A 2p K 1 K L If can be noted that the effective spring gradient Ke is derived from the series connection 2 4 βe A p and the two mechanical springs K1, KL. between the hydraulic spring gradient Vt 2. Loads with two degrees of freedom a) Mechanical flexibility between two masses on a piston rod Assume that the load consists of two masses (M1 and M2), of which the first one is fixed mounted to the piston and the second mass is connected by a spring (KL) and a viscous damper (BL) to the first mass. Ap p1 C1 xp Ap C2 p2 KL M2 M1 BL xL Figure 9: Hydraulic cylinder with a load with two degrees of freedom FL Linköpings universitet IEI / FluMeS, K-E Rydberg 6 2008-10-13 The force balance equations for the masses M1 and M2 are for FL = 0 M1 s 2 Xp + M2 s 2 XL = Ap PL (12) M2 s 2 XL = −BLs (XL − Xp ) − KL(XL − Xp ) (13) The piston position Xp can be expressed as ⎛⎜ M2 2 BL ⎞⎟ A P ⎝ KL s + KL s + 1⎠ p L Xp = M1 M2 BL ⎤ 2 (M1 + M2 ) s2 ⎡⎢ + s + 1⎥⎥ s ⎢⎣ KL(M1 + M2 ) KL ⎦ (14) With the load flow QL = ApsXp we obtain the transfer function BL ⎞ ⎛ M2 CL ⎜⎝ K s 2 + s + 1⎟⎠ QL KL L Gm(s) = = PL CL M1 M2 BL ⎤ 2 (M1 + M2 ) s ⎡⎢⎢ s + s + 1⎥⎥ 2 ( ) K L ⎦ ⎣ KL M1 + M2 Ap (15) Equation (15) can also be written in the following form 2δa ⎛ s2 ⎞ ⎜⎜ ⎟⎟ + s + 1 ωa ⎝ ω2a ⎠ A2p Gm(s) = (M1 + M2 ) s ⎛ s2 2δ1 ⎞ ⎟⎟ ⎜⎜ + s + 1 ω1 ⎠ ⎝ ω21 KL , M2 ωa = where δa = BL 2 1 , KLM2 KL(M1 + M2 ) = ωa M1 M2 ω1 = δ1 = BL 2 M1 + M2 = δa KLM1 M2 (16) 1+ 1+ M2 M1 M2 M1 From equation (16) it can be seen that the two coupled masses cause one natural frequency of the fundamental mode ω1 and one natural frequency of nodes of vibration ωa. For these frequencies, it is always evident that ωa < ω1 and the damping ratio δa < δ1. A Bode diagram for equation (16) is shown in Figure 10. Linköpings universitet IEI / FluMeS, K-E Rydberg 7 2008-10-13 Amplitude 101 100 10-1 10-2 100 101 102 Frequency [rad/s] Phase 50 0 -50 -100 100 101 Frequency [rad/s] 102 Figure 10: Frequency response according to equation (16). Solid line is valid for M1 = M2 and dashed line for M1 = M2/2 Figure 10 explains how the amplitude and the phase curve change when the mass M1 is reduced in proportion to M2. If M1 instead increases in proportion to M2 the frequency response will be influenced in the way as expressed in Figure 11. Amplitude 100 10-1 10-2 100 101 102 Frequency [rad/s] Phase 50 0 -50 -100 100 101 Frequency [rad/s] 102 Figure 11: Frequency response according to equation (16). Solid line is valid for M1 = M2 and dashed line for M1 = 2M2 Linköpings universitet IEI / FluMeS, K-E Rydberg 8 2008-10-13 It is convenient to combine equation (16) with the hydraulic part of the system. If the external force FL = 0, the valve cylinder combination (compare with Figure 1) and the load with two masses give the block diagram illustrated in Figure 12. Kq Xv ___ Ap + - Ap __________ Vt Kce + ___ s 4βe PL Ap s2 GLX(s ) = where 1 __________ GLX(s) (M1 + M2) s ω2a s2 ω21 + 2 δa s+1 ωa + 2 δ1 s+1 ω1 . Xp Figure 12: Block diagram for a valve-controlled cylinder with a load of two masses (FL = 0) From Figure 12 the transfer function for the hole system, with valve opening Xv as input signal and piston velocity dXp/dt as output signal, can be derived as GHL(s ) = GLX(s) sXp Kq = Xv Ap s 2 2 where ωh = (17) 2 δh + s + GLX(s ) ωh ω2h 4β eAp Vt (M1 + M2 ) ; δh = Kce Ap β e(M1 + M2 ) Vt (18) The frequency response for equation (17), when the hydraulic natural frequency ωh is lower than the structural resonance’s (ωa and ω1) is shown in Figure 13 (see next page). The dashed lines in the diagrams illustrated the response when the mass M2 is increased 5 times compared to the situation of the solid lines. The hydraulic spring rate Kh (=4βeAp2/Vt) has the same value for both cases. In this system, the frequency response will be dominated by the hydraulic system. In the low frequency range is GLX(s) ≈ 1 and the system can be treated as a one-mass system with the load mass Mt = M1 + M2. Linköpings universitet IEI / FluMeS, K-E Rydberg 9 2008-10-13 101 Amplitude 100 10-1 10-2 10-3 100 ωa ω1 ωh 101 Frequency [rad/s] 102 101 Frequency [rad/s] 102 0 Phase -50 -100 -150 -200 100 Figure 13: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = 1. Dashed line: M1 = 1 and M2 =5. Kh = 0,1.KL for both curves If we now consider that the structural resonance’s are lower than the hydraulic natural frequency, the system response will change drastically, which can be seen in Figure 14. Amplitude 101 100 ωa 10-1 100 ω1 ω'h 101 Frequency [rad/s] 102 101 Frequency [rad/s] 102 Phase 100 0 -100 -200 100 Figure 14: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = M0. Dashed line: M1 = M0 and M2 =5M0. Kh = 10.KL for both curves Linköpings universitet IEI / FluMeS, K-E Rydberg 10 2008-10-13 In Figure 14 the hydraulic spring rate Kh is 10 times greater than the mechanical spring rate KL which means that ωh will be the highest frequency. If we look at the amplitude of the transfer function GLX (see Figure 13) with frequencies greater than ω1, this equation will reach a constant value 2 M2 ⎛ ω1 ⎞⎟ ⏐GLX(s)⏐ω>ω1 = ⎜⎝ = 1 + ωa ⎠ M1 (19) Since the hydraulic resonance occur at a frequency higher than the structural resonance’s, the value of ωh have to be influenced by the function GLX written as equation (19). For this case, the hydraulic frequency and damping, here named ω´h and δ´h, may be changed from the original expression (equation 18) to ω,h = ωh δ,h = δh / M2 1+ = M1 1+ M2 Kce = M1 Ap Kh = M1 4β eA2p Vt M1 (20) β eM1 Vt This equation shows that the dynamics of the hydraulic system only is dependent on the mass M1. Increasing of the mass M2, of course, will lower the mechanical resonance’s but the hydraulic frequency ω´h is not influenced by change of this mass. The reason is that the movement of the mass M2 is approximately zero at such high frequencies as ω´h. The hydraulic natural frequency is solely determined by the mass M1 which is stretched between the hydraulic spring in the cylinder Kh and the load spring KL. However, the low value of KL compared to Kh means that KL cannot be seen in equation (20). If the system is changed so that only the hydraulic spring constant Kh (see eq 20) increases, the variation in amplitude and phase shift according to the mechanical structure will be reduced. This situation is shown in Figure 15. From equation (17) it can also be seen that, if the structural resonance’s (GLX(s)) are dominant, the transfer function from valve opening to piston speed approaches GHL(s) = sXp/Xv ≈ Kq/Ap (21) Linköpings universitet IEI / FluMeS, K-E Rydberg 11 2008-10-13 Amplitude 101 100 10-1 10-2 100 101 102 Frequency [rad/s] Phase 100 0 -100 -200 100 101 102 Frequency [rad/s] Figure 15: Frequency response for GHL(s) according to equation (17). Solid line: Kh = Kh0. Dashed line: Kh = 5.Kh0. M1 = M2/2 for both curves Another situation where ωh increases and makes the hydraulic system stiffer will arise if the mass M1 is reduced. Figure 16 shows how the frequency response develops when M1 is reduced from M1 = M10 to M1 = 0,2.M10. Since M1 also influences the mechanical resonance’s this rise of ωh will not cause a reduction of the structural resonance’s as in the case of increased hydraulic stiffness. Amplitude 101 100 10-1 10-2 100 10 1 102 Frequency [rad/s] Phase 100 0 -100 -200 100 10 1 102 Frequency [rad/s] Figure 16: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M10. Dashed line: M1 = 0.2.M10. Kh and KL are the same for both curves Linköpings universitet IEI / FluMeS, K-E Rydberg 12 2008-10-13 The conclusion to be drawn from these examples is that the structural resonance’s are reduced by a stiff hydraulic system if ωh is always higher than the mechanical frequencies. This stiffness can be achieved by low hydraulic capacitance for the valve cylinder combination and/or feedback control. b) Two-mass system with mechanical flexibility in the rear mounting end A further example of a hydraulic cylinder loaded by a two-mass system is shown in Figure 17. K1 M1 Ap Ap p1 C1 xp B1 x1 M2 C2 p2 FL x2 Figure 17: Hydraulic cylinder with flexible rear flange and a load with two degrees of freedom The transfer function for this mechanical system is ⎛ s2 2δa ⎞ ⎜⎜ ⎟⎟ + s + 1 ωa ⎝ ω2a ⎠ QL A2p Gm(s) = = PL M2 s ⎛ s2 2δ1 ⎞ ⎜⎜ + s + 1⎟⎟ 2 ω1 ⎝ ω1 ⎠ where K1 , M1 + M2 ωa = δa = B1 2 K1 = ωa M1 ω1 = 1 K1 (M1 + M2 ) , δ1 = (22) B1 2 1+ 1 = δa K1 M1 M2 M1 1+ M2 M1 According to equation (17) a valve controlled cylinder yields GHL(s ) = sXp Kq = Xv Ap s 2 G LX(s ) = (23) 2 δh + s + GLX(s ) ωh ω2h s2 where GLX(s) ω2a 2 s ω21 + + 2 δa s+1 ωa 2 δ1 s+1 ω1 ; ωh = 4β eA2p Kce ; δh = Vt M2 Ap β eM2 Vt Linköpings universitet IEI / FluMeS, K-E Rydberg 13 2008-10-13 The Bode diagram for equation (23) is shown in figure 18. In this figure the hydraulic resonance ωh is the dominant frequency, lower than ωa. If the hydraulic spring rate Kh is changed, the value of ωh will also be changed which is illustrated in the figure. Amplitude 10 1 10 -3 10 0 10 1 10 2 Frequency [rad/s] Phase 0 -100 -200 10 0 10 1 Frequency [rad/s] 10 2 Figure 18: Frequency response for GHL(s) according to equation (23). Solid line: Kh = 0,04 KL. Dashed line: Kh = 0,1 KL. M1 = 1, M2 = 2 and KL = 1000 (same for both curves) If the hydraulic cylinder is stiffer than the mechanical structure (Kh > KL), we will have a system with a behaviour similar to that described in Figure 15. An increase of Kh is shown in Figure 19 and we can see that the structural resonance’s are reduced by the large value of ωh. Amplitude 10 1 10 0 10 -1 10 -2 10 0 10 1 10 2 Frequency [rad/s] Phase 100 0 -100 -200 10 0 10 1 10 2 Frequency [rad/s] Figure 19: Frequency response for GHL(s) according to equation (23). Solid line: Kh = 4 KL. Dashed line: Kh = 10 KL. M1 = 1, M2 = 2 and KL = 100 (same for both curves) Linköpings universitet IEI / FluMeS, K-E Rydberg 14 2008-10-13 With this stiff hydraulic system, the real hydraulic frequency and damping, ω´h and δ´h , are changed from the original expression (compare with eq 20) to , ωh M2 1+ = M1 = ωh M2 Kce 1+ = M1 Ap , δh = δh / 2 4β eAp (M1 + M2 ) Vt M1 M2 (24) β eM1 M2 Vt (M1 + M2 ) Figure 20 shows the frequency response when the mass M2 is increased five times. Note that the reduction of ωh increases the variation in amplitude and phase shift for the mechanical structure which can also be seen from the expression of ωa and δa (eq 22). Amplitude 101 100 10-1 10-2 100 101 102 Frequency [rad/s] Phase 100 0 -100 -200 100 101 Frequency [rad/s] 102 Figure 20: Frequency response for GHL(s) according to equation (23). Solid line: M1 = M2 = 1. Dashed line: M1 = 1, M2 = 5. Kh = 4 KL (= 100) for both curves. Linköpings universitet IEI / FluMeS, K-E Rydberg 15 2008-10-13 c) Hydraulic motor with mechanical flexibility between two inertia loads Bm KL J1 Dm BL θm TL J2 θL Figure 21: Hydraulic motor with a load with two degrees of freedom This hydraulic motor system with two inertia loads is identical with the cylinder application illustrated in Figure 9. Consider the system shown in Figure 21 and let the motor be controlled by a 4-port servo valve with the coefficients Kq and Kc. The block diagram of the overall system will be developed as in Figure 22. TL Kq Xv ___ Dm + Dm __________ Vt Kce + ___ s 4βe - PL G LT(s) Dm + . θm 1 ________ G Lθ(s) (J1 + J2) s Figure 22: Block diagram for a valve-controlled motor with two inertia loads and a load torque (TL) From Figure 22 it can be seen that the mechanical structure has influence on the torque disturbance (TL) by the transfer function GLT(s). The transfer function GLθ(s) is similar to GLX(s) in Figure 9. These transfer functions can be expressed as 1+ GLT(s ) = s 2 ω2a where 2δ1 ω1 2δa s+1 ωa KL , J2 ωa = δa = + BL 2 s2 s 1 , KLJ 2 ; GLθ(s ) = s 2 ω21 + 2δa s+1 ωa BL 2 (25) 2δ1 s+1 ω1 KL(J1 + J 2 ) = ωa J 1 J2 ω1 = δ1 = ω2a + J1 + J2 = δa KLJ 1 J 2 1+ 1+ J2 J1 J2 J1 Linköpings universitet IEI / FluMeS, K-E Rydberg 2008-10-13 Appendix Transfer functions for loads with 2 DOF 16 Linköpings universitet IEI / FluMeS, K-E Rydberg 2008-10-13 17