Linköpings universitet
IEI
Fluid and Mechatronic Systems
1.
EXAMINATION
TMHP 51
2010-08-21
(6)
a) Cavitation in orifices
Two orifices are used to create a middle-pressure pm, as shown in the figure below. The
first orifice has the area A1 = 1,5·10-6 m2 and the second area is A2. Inlet pressure to the
first orifice is p1 = 80 bar and the outlet pressure after the orifices is p2 = 8 bar. For
each orifice the critical pressure drop for cavitation can be calculated as:
, where pin is the inlet pressure.
Calculate the pressure pm, which gives cavitation free flow for the two orifices.
Calculate the orifice area A2 for that pm.
(4p)
b) Spool design for servo valves
The figure shows a drawing of a servo valve spool and its bushing. Why is a ”spoolbushing” used and what is the aim of using circulated tracks on the spool lands (2
circulated track on each land)?
(3p)
c) Flow forces in seat valves
A 2-port seat valve is supplied by a constant pressure controlled pump with the max
flow qpmax = 335 litre/min. The figure below shows the valve flow (Q) versus the valve
displacement (y) at two different max pressure drops (100 and 200 bar).
Show qualitatively in a diagram how the steady state flow forces (Fs) varies according
to the displ. (y) at the pressure drops 100 and 350 bar respectively.
(3p)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP 51
2010-08-21
(6)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
2.
EXAMINATION
TMHP 51
2010-08-21
(6)
a) Valve wear and its influence on valve coefficients
Wear on the orifice edges in a zero-lapped servo valve will cause changes in the steady
state characteristics around neutral spool position (xv = 0). Which of the valve
coefficients will be most affected by wear and how is the closed loop stiffness in a
position servo with proportional controller gain influenced by this?
(3p)
c) Hysteresis and pressure gain for servo valves
The figure shows the pressure gain (pL versus current, iv) for a servo valve, with and
without hysteresis.
Describe, why all real servo valve characteristics show hysteresis. Which kind of servo
valve, single-stage or two-stage, will give the highest pressure gain?
(2p)
c) Valve controlled position servo and an angular speed servo
A schematic figure of a linear position servo and an angular speed servo is shown
below. In principle there is also a block diagram. The position servo has a proportional
controller (Greg = Ksa) and the speed servo has an integrating controller (Ksa/s). The
valve is zero-lapped in both applications and the coefficients are Kqi and Kc
respectively.
Compare qualitatively, with equations derived from the block diagram, these two
systems with respect to the steady state stiffness (for the closed loop system).
Linköpings universitet
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP 51
2010-08-21
(6)
(5p)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
3.
EXAMINATION
TMHP 51
2010-08-21
(6)
Position servo with valve controlled cylinders
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 null-coefficients 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 ωb (at the
amplitude - 3 dB).
The system has the following parameter values:
Ap = 1,96.10-3 m2
Mt = 500 kg
Vt = 1,0.10-3 m3
ps = 21 MPa
βe = 1000 MPa
Kqi0 = 0,013 m3/As
Kf = 25 V/m
a) Kce-value for a bandwidth of ωb = 30 rad/s
Calculate the required Kce-value so that the closed loop servo can reach a bandwidth
of ωb = 30 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(ωt).
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, ωb =
30 rad/s.
(5p)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP 51
2010-08-21
(6)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
4.
EXAMINATION
TMHP 51
2010-08-21
(6)
Angular position servo with dynamic pressure feedback
The figure shows an elektro-hydraulic position servo with a valve controlled motor. The
proportional controller has the gain Greg = Ksa = 0,04 A/V. In order to increase the
hydraulic damping a dynamic load pressure feedback with the gain function Gp(s)=
Kpf.s/(1 + s/ωf) is implemented. The servo valve is of 4-port type, zero-lapped, with
high bandwidth and its zero-coefficients are Kqi0 = 0,013 m3/As and Kc0 = 1,0.10-12 m5/
Ns. The volumes between valve and motor are V1 = V2 = 0,5 litre and its bulk modulus
is βe = 1000 MPa. The motor displacement is Dm = 6,4.10-6 m3/rad, the leakage flow
coefficient is Ctm = 8,0.10-13 m5/Ns and the viscous friction coefficient Bm = 0. The
motor shaft inertia is Jt = 0,5 kgm2.
a) Adjustment of the pressure feedback for a given hydraulic damping
Calculate the static gain Kpf for the load pressure feedback, which gives the hydraulic
damping δh = 0,50 at the frequency ωh. The break frequency in the pressure feedback
filter is ωf = ωh/2.
(6p)
b) The pressure feedback and its influence on steady state stiffness
Show with equations that the steady state stiffness of the closed loop system can be
increased by using dynamic load pressure feedback, if the controller gain, Ksa is
adjusted for the same amplitude margin (Am), with as without load pressure feedback.
(4p)
Linköpings universitet
IEI
Fluid and Mechatronic Systems
5.
EXAMINATION
TMHP 51
2010-08-21
(6)
Hydraulic system with a valve controlled motor loaded by two masses
The figure shows a system layout and block diagram for a valve-controlled motor
loaded by the inertia J1 and J2. The shaft between the inertia loads is modelled as a
torsional spring and a damper (torsional constant KL and friction coeff. BL).
A simulation of the system in the time domain with a step input to the valve (xv) gives
the following step response of the motor and the second inertia shafts angular speed:
a) Transfer function and bode diagram for the system
Derive from the block diagram above the linearised and laplace transformed transfer
function Gh(s) = sθm/Xv. The transfer function GLӨ(s) can just be described as a
function, as in the block diagram.
Show in a bode-diagram the principle characteristics of the amplitude and the phase
shift curves for Gh(s) according to the simulation results shown above.
(7p)
b) Increased mechanical stiffness
Assume that the mechanical stiffness of the system, KL increases so its value is much
higher than the hydraulic motor stiffness Kh.
Show how the transfer function Gh(s) will be influenced in the frequency domain
(amplitude in a bode-diagram) and in the time domain.
(3p)