LINKÖPINGS TEKNISKA HÖGSKOLA IKP Fluid och Mekanisk Systemteknik 1. 2(6) EXAMINATION TMHP 51 2005-12-15 a) Cavitation i hydraulic orifices The figure shows measured flow versus outlet pressure for an orifice supplied with constant inlet pressure, pu = 200 bar. Cavitation starts when the outlet pressure is reduced to pd = 75 bar. Assume that the inlet pressure to the orifice is changed to pu = 160 bar. At which level of the outlet pressure pd can now cavitation be expected? Which highest inlet pressure (pu) can be accepted for cavitation free flow if the lowest outlet pressure is pd = 2,0 bar. (4p) b) Water versus mineral oil as fluid in hydraulic servo systems In a hydraulic servo system the mineral oil is changed to a mixture of waterglycol. This means that the fluid viscosity is reduced a factor 10 and the effective bulk modulus increased 1,8 times. Describe qualitatively how the water fluid will influence the resonance frequency ( h) and damping ( h) in the system. (2p) c) Flow forces in a 4-port servo valve #$ (( ,( !" -.&/ !"# 0( ()*+*' !" #$ ' $%&'()*+( #$%&# A 4-port symmetric servo valve is supplied from a constant pressure controlled pump adjusted for pp = 210 bar. Max pump flow is qpmax = 74 litre/min. Flow forces versus valve spool displacement (xv) when the load pressure difference is 0 are shown in a diagram. Calculate the nominal valve flow qvN (total valve pressure diff. pv = 70 bar). Assume that the pump pressure in the system above increases to pp = 350 bar. Calculate the new value for max flow forces, Fsmax (qpmax = 74 litre/min). (4p) LINKÖPINGS TEKNISKA HÖGSKOLA IKP Fluid och Mekanisk Systemteknik 2. 3(6) EXAMINATION TMHP 51 2005-12-15 a) Servo valves with different “lapping” In position servos normally 4-pors symmetric servo valves with null-lapping (critical center) or under-lapping are used. Describe qualitatively how the ”lapping” will influence stability and stiffness of a proportional controlled position servo. (3p) b) Direct controlled servo valve The figure shows a direct controlled servo valve. Compare this valve with a 2stage servo valve (electro-hydraulic pilot stage) according to hysteresis and bandwidth. (2p) c) Valve and pump controlled position servo ?( ?( AB #( ;7 AC ?( !- AB ?( #( ;7 AC !- 6% :,1 :94 )234"2.5/ 89 = :"& 1 8 > < 6."171.0 73&0"/8943 :< !10 6"*+*9.0"7@ ,( :(1 89= 8< > :"& 1 D7 !( 6."171.0 73&0"/8943 :< )234"2.5/ !10 The figure shows schematically a valve controlled (4-port ”critical center” valve) and a pump controlled position servo. Cylinder and load are identical and both servos have proportional controllers (Greg = Ksa). The controller gain (Ksa) is adjusted for the same amplitude margin, Am = 6 dB when the resonance frequency ( h) has its lowest value. The leakage flow coefficients (Kce and Ct) have the same value as well as the flow gain for valve and pump, Kqi = Kpi (qp =i·Kpi). The servo valve and the pump controller have equal threshold values, iT = in. Calculate the position error ratio Xp4-v/ Xpp for the two systems according to the threshold ( iT). (5p) LINKÖPINGS TEKNISKA HÖGSKOLA IKP Fluid och Mekanisk Systemteknik 3. EXAMINATION TMHP 51 2005-12-15 4(6) Position servo with valve controlled cylinders 6."171.0*73&0"/8943 !" $" ;7 &# "# &% "% :< 8< > 89 :"& = $" (E*+*9.0"7@ 1 6"*+*9.0"7@ The figure shows an electro-hydraulic position servo with a 4-port servo valve and two mechanically connected asymmetric cylinders. The load is a single mass Mt. The servo has proportional position control with the feedback gain Kf and the controller gain Ksa. The servo valve is null-lapped and symmetric and its nullcoefficients are Kqi0 and Kc0. The valve bandwidth is high. The supply pressure ps, is constant. Cylinder piston area is Ap and the total pressurised volume between valve and pistons is Vt = V1+V2. The cylinders can be assumed as loss free. The bandwidth of the servo system is b (at the amplitude - 3 dB). The system has the following parameter values: Ap = 1,96.10-3 m2 Mt = 800 kg Vt = 1,0.10-3 m3 ps = 21 MPa Kf = 25 V/m a) Kce-value for a bandwidth of b = 1200 MPa Kqi0 = 0,013 m3/As e = 25 rad/s Calculate the required Kce-value so that the closed loop servo can reach a bandwidth of b = 25 rad/s, with an amplitude margin of Am = 6 dB in the most critical operation point. (5p) b) Feed forward loop to reduce the velocity error Assume that the piston position (xp) has to follow the command signal, uc = Ax·sin( t). In order to reduce the velocity error the command signal shall be fed forward via servo amplifier to the valve to create a signal corresponding to the required velocity profile for the cylinder pistons. Show in a block diagram how you will implement the feed forward loop and calculate the steady state feed forward gain. Assume that Ksa is adjusted for the bandwidth, b = 25 rad/s. (5p) LINKÖPINGS TEKNISKA HÖGSKOLA IKP Fluid och Mekanisk Systemteknik 4. 5(6) EXAMINATION TMHP 51 2005-12-15 Linear position servo with an inner velocity feedback :< :<$ > F9 = > #( "# #( G(*+*' ?( AC AB :"&$ !- ;7 = :$<$*+*B*= :<$ :"&$ . !- :94 + '% )B % ", ?(C )* K ($ & #$ ,- > * :,1 *?( @ BH &()( #( ! C % C '" ! % B &()( &"C &()( &" > $89*+*' :,1 *?( 1$ :"&$ :< B " #( 6."171.0*<44/I&9J The figure shows an electro-hydraulic linear position servo with a valve controlled cylinder. The controller is proportional, with the gain Ksav. The position feedback gain is Kf and the velocity feedback gain is Kfv. The servo valve is of 4port type, null-lapped, high bandwidth and the null coefficients are Kqi0 and Kc0. The cylinder volumes are V1 = V2 = Vt/2 and the bulk modulus is e. The cylinder piston area is Ap and the cylinder losses are very low. The cylinder is loaded by the mass Mt and an external force FL. I service the system has the following parameter values: Ap = 1,96.10-3 m2 Mt = 900 kg e = 1000 MPa -2 . -3 3 3 Vt = 1,0 10 m Kqi0 = 1.0 10 m /As Kc0 = 1,0.10-11 m5/Ns ' Kf = 20 V/m Ksav = 0,80 A/V h = 320 rad/s a) Hydraulic damping with velocity feedback For the servo system, including velocity feedback, has the un-damped resonance frequency been measured to h' = 320 rad/s. Calculate the hydraulic damping ' h at this frequency. The low hydraulic damping can provide oscillation problems. Describe qualitatively a method to reduce the oscillation problems. (5p) b) Steady state stiffness versus velocity feedback gain Derive from the block diagram above the steady state stiffness of the closed loop system, FL Xp as a function of the velocity feedback gain Kvfv and calculate s 0 the stiffness. (5p) LINKÖPINGS TEKNISKA HÖGSKOLA IKP Fluid och Mekanisk Systemteknik 5. 6(6) EXAMINATION TMHP 51 2005-12-15 Velocity servo with pump controlled motor and two inertia loads The figure illustrates an electro-hydraulic velocity servo with a pump controlled motor loaded by two inertias (J1 and J2). The controller is of integrating type with the gain Ksa. The pump displacement setting controller has the transfer function: Ki p , with the displacement setting coefficient K i = 20 A-1 and the brake i 1 s s frequency s = 100 rad/s. The pump shaft speed is p = 157 rad/s and the displacement is Dp = 5,6.10-6 m3/rad. The volumes between pump and motor are V1 = V2 = 0,6 litre and the effective bulk modulus is e = 800 MPa. The motor displacement is Dm = 19.10-6 m3/rad and the inertias on the motor shaft is J1 = J2 = 0,5 kgm2. The torsion spring constant between the masses is KL = 2000 Nm/rad. The transmission leakage flow coefficient is Ct = 4,0.10-12 m5/Ns and the viscous friction coefficient is Bm = 0. The low pressure side has constant pm. !" &" *+ 0 ,)' ---*/ 1 ..) "4 64 ,9 23 2" ?8L"M " /3 "5 65 7 "# 84 !"##$ %&'()$*+#& "3 # 85 :9 /- ,/ a) Transmission dynamics versus load dynamics Calculate the transmission hydraulic resonance frequency ( h) and compare with the load dynamics. Show schematically in a bode-diagram the amplitude of the loop gain Au(s). (5p) b) Acceleration feedback for increased damping In order to increase the hydraulic damping ( h) an acceleration feedback from the hydraulic motor shaft shall be implemented. Show in a block-diagram how you will implement the feedback and show with equations its influence on the hydraulic damping. Is the bandwidth of the pump controller ( s) high enough to not influence the damping? (5p) LINKÖPINGS UNIVERSITET Department of Mechanical Engineering Fluid and Mechanical Engineering Systems 1(5) SOLUTIONS FOR EXAMINATION TMHP 51 2005-12-15 SOLUTIONS FOR EXAMINATION, TMHP51 1. a) Cavitation in hydraulic orifices pu = 200 bar and pd = 75 bar gives cavitation. The diagram gives C2 pdcav pu 1 pu = 160 bar in eq. (1) pdcav 0,625 0,375 pu (1). pdcav = 60 bar. Cavitation free flow and pd = 2 bar pumax = pd/0,375 = 5,3 bar. (4p) b) Water versus mineral oil as fluid in hydraulic servo systems The resonance frequency and damping for an actuator with one control volume and load e is given by the equations. h Ap2 Vp M t , K ce 2 Ap h Mt . The high bulk modulus of Vp e water increases both the resonance frequency and damping. Low viscosity increases the leakage flow and the Kce-value, which increases the damping. (2p) c) Flow forces in a 4-port servo valve #$ (( ,( !" -.&/ !"# 0( #$ !" ()*+*' ' $%&'()*+( #$%&# Pump pressure, pp = 210 bar, max pump flow, qpmax = 74 litre/min and pL = 0. Steady state flow forces: Fs valve flow as qv Cq w xv Nominal valve flow: qv 2C q w xv pp pv cos . pv = pp and pL = 0 gives the (1). Finally Fs max Cq w xv pp qvN qvmax = qpmax and increased pp to 350 bar in eq. (2) qp 2 cos x v max xv Fs max qv max p p (2). pvN = 61 litre/min. pp 85 350 = 110 N 210 (4p) LINKÖPINGS UNIVERSITET Department of Mechanical Engineering Fluid and Mechanical Engineering Systems 2. 2(5) SOLUTIONS FOR EXAMINATION TMHP 51 2005-12-15 a) Servo valves with different “lapping” Null-lapping gives a linear flow gain and low leakage around “zero”, which will improve the stiffness but low hydraulic damping will reduce the stability margin. Under-lapping gives high flow gain and high hydraulic damping around “zero” operation. This will improve the stability but reduce the stiffness. (3p) b) Direct controlled servo valve In comparison with a 2-stage valve the direct controlled valve has low control forces on the main spool. Low control forces give high hysteresis because of friction and flow forces in the valve and the bandwidth will be low because of low acceleration of the main spool. (2p) c) Valve and pump controlled position servo ;8 78 ;8 ?@ 1, ?A ;8 34 78 ;8 ?@ 1, ?A 34 +: +('!,!(" ,%-"'*./&% 09! 0/& #$%&'$()* ./ 5 0'- ! . 6 ./5 .2 6 +'<=</("',> 2 B, !8 +('!,!(" ,%-"'*./&% 08! 02 !!" 98 0'- ! 02 #$%&'$()* !!" Calculate the position error ratio Xp4-v/ Xpp for the two systems according to the threshold ( iT) when the systems have equal parameters and designed for Am = 6 dB in the most critical operation point. Kce = Ct and Kqi = Kpi. Position error from threshold: Am = 6 dB Steady state loop gain: Kv = Valve control: Kv 4 v Kv p X p K f K sa h h 4 v K sa K qi K f / Ap K sa K pi K f / Ap 2 K ce 4 e Vt h· h . Pump control: K sa 4 v K sap iT K f K sa Xp iT 4. X p4 X pp v h h p K sap K sa 4 v 1 Ct e 2 Vt 1 = 0,25 4 (5p) LINKÖPINGS UNIVERSITET Department of Mechanical Engineering Fluid and Mechanical Engineering Systems 3. 3(5) SOLUTIONS FOR EXAMINATION TMHP 51 2005-12-15 Position servo with valve controlled cylinders !"#$%$"&'%()&#*+,-( !" $" $" 4% &# &% "# "% 1. 56'2',"&#%3 +. 0 +, / $ 1#) !#'2',"&#%3 Ap = 1,96.10-3 m2 ps = 21 MPa Mt = 800 kg Kqi0 = 0,013 m3/As a) Kce-value for a bandwidth of b = 25 rad/s Steady state loop gain Kv = Am = 6 dB Valve controlled cylinder: the bandwidth: b h 4 h Ap2 Vt M t 2 K ce h e Vt = 1,0.10-3 m3 = 1200 MPa Kf = 25 V/m e e h· h , K ce Ap bVt h K ce Vt and bandwidth 2 b = Kv = 25 1/s . Mt , which gives Vt e = 1,04·10-11 m5/Ns e (5p) b) Feed forward loop to reduce the velocity error Block diagram showing the implementation of the feed forward loop. '1 .. +, / ! 0 <--*'<"(=)(* / 1#) / $ 18$ '95 : ; ! #" ! ! : ""; "" !; 3 ' 75 : # 75 1. The feed forward block includes a derivation (s) of the position command signal (uc = Ax·sin( t), which means that the feed forward signal represents a velocity signal. The K qi block diagram shows that the gain from command signal to piston velocity is: K sa . Ap If the steady state feed forward gain is set to K ff Ap K sa K qi it represents a true velocity signal in the system. Calculation of Kff: K sa K qi Ap b Kv K sa K qi Ap K f . Kv = 25 1/s and Kf = 25 V/m gives = 1,0, which gives the steady state feed forward gain Kff = 1,0. (5p) LINKÖPINGS UNIVERSITET Department of Mechanical Engineering Fluid and Mechanical Engineering Systems 4. 4(5) SOLUTIONS FOR EXAMINATION TMHP 51 2005-12-15 Linear position servo with velocity feedback !" !" !" %1 %1( # # 23 %(1(565/5$ %1( %&'( 4"5657 )" .0 ./ %&'( *+ ,- ,- %3? * '% (/ $ &+ )"0 () C '$ & #$ +, # 9 5 /: &()( !" 0 &" $ $ ! / &()( %"0 &()( %" !0 # #835657 $ $ Parameter values: Ap = 1,96.10-3 m2 Vt = 1,0.10-3 m3 Kf = 20 V/m Mt = 900 kg Kqi0 = 1.0 10-2 m3/As Ksav = 0,80 A/V %D; 5)" %D; 5)" ;( %&'( %1 / & !" <=&;+;=>51??@A'3B = 1000 MPa Kc0 = 1,0.10-11 m5/Ns ' h = 320 rad/s e a) Hydraulic damping with velocity feedback With velocity feedback the un-damped resonance frequency is 4 The basic resonance frequency ( h) is: ' h above gives K vfv and h ' h h h e Ap2 Vt M t ' h = 320 rad/s. = 131 rad/s. The block diagram / K vfv . K vfv ' h 2 Kvfv = 6,0 and finally h ' h K co Ap Mt Vt e 1 = 0,0625. K vfv A negative dynamic load pressure feedback or acceleration feedback can be used to increase the low hydraulic damping without any influence on the steady state stiffness. (5p) b) Steady state stiffness versus velocity feedback gain From the block diagram above the closed loop stiffness can be derived as: K 2 h s2 s 1 K vfv s K f K sav qi 2 K vfv h K vfv h Ap FL . The steady state part is Sc Xp K ce Vt 1 s 4 e K ce Ap2 FL Xp FL Xp s 0 K vfv K sav K qi K f Ap2 Ap K vfv K ce K sav K qi 0 K f Ap s 0 Kc0 . Numerically the steady state stiffness is: = 3,14·107 N/m. (5p) LINKÖPINGS UNIVERSITET Department of Mechanical Engineering Fluid and Mechanical Engineering Systems 5. 5(5) SOLUTIONS FOR EXAMINATION TMHP 51 2005-12-15 Velocity servo with pump controlled motor and two inertia loads "4 64 $" %" *+ 0 ,)' ---*/ 1 ..) ,9 23 2" !" !"##$ %&'()$*+#& "3 "#$%& :9 85 " ! #3 "5 6 5 7 84 #! ,/ = 157 rad/s, Dp = 5,6.10-6 m3/rad, V1 = V2 = 0,6 litre, e = 800 MPa, Dm = 19.10-6 m3/rad, J1 = J2 = 0,5 kgm2, KL = 2000 Nm/rad and Ct = 4,0.10-12 m5/Ns. p a) Transmission dynamics versus load dynamics The hydraulic frequency for the transmission is Dm2 . Numerically: Vt J 1 J 2 e h 2 800 106 19 10 6 = 22 rad/s. The load dynamics include two frequencies, 0,6 10 3 0,5 0,5 h and 1. The lowest one is KL . J2 a 2000 = 63 rad/s. 0,5 a h is dominant. "# Bode diagram: %' %2 ( )*+,#+-./01*234%5 (5p) b) Acceleration feedback for increased damping Implementation of acceleration feedback. "4 64 $" %" *+ 0 ,)' ---1 0 */ ..) ,9 23 2" !" 84 ! #3 7 1 ;++#<#&'%7=(.)7>('< !"##$ %&'()$*+#& "3 85 " #! ,'+?) ,/ Influence on hydraulic damping Gh ( s ) 1 s 2 2 2 h ' h h h 2 K ac Ki 1 Dm 1 s / p h h K ac Ki 1 Dm 1 s / p Dp s 1 s Dp s Since s = 100 rad/s, is higher than both damping is marginal. h and a its influence on the hydraulic (5p) a
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