Linköpings Tekniska Högskola
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP51/TEN1
2010-12-17
Page 1(6)
EXAMINATION IN
Hydraulic Servo Systems, TMHP51 / TEN1
Date:
Tuesday 17 December 2010, at 8 am - 12 am
Room:
??
Allowed educational aids: Tables: Standard Mathematical Tables or similar
Handbooks: Tefyma
Formularies:
Formula Book for Hydraulics and Pneumatics, LiTH/IEI
Mekanisk Värmeteori och Strömningslära
Pocket calculator
Number of questions
in the examination:
5
On the front page and on all following pages the student must write:
AID-number, TMHP51/TEN1, YYMMDD, page number
Responsible teacher:
Tel.no. during exam:
Will visit at:
Course administrator:
Karl-Erik Rydberg
013-28 11 87
9:30 and 11:00 o’clock
Rita Enquist, tel.nr. 28 11 89, [email protected]
Score:
Maximal score on each question is 10.
To get the mark 3, you will need 20 points
To get the mark 4, you will need 30 points
To get the mark 5, you will need 40 points
Solutions:
Results:
GOOD LUCK!
17 December 2010
Karl-Erik Rydberg
Professor
You will find the solutions of this examination on the notice board
in the A building, entrance 17, C-corridor to the right.
Results will be announced ??.
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
1.
EXAMINATION
TMHP 51
2010-12-17
2(6)
a) Cavitation in orifices
In hydraulic systems cavitation will occur down-streams orifices with high pressure
drops. For an orifice with a constant inlet pressure of p1 = 150 bar cavitation will start
when the outlet pressure is reduced to p2 = 54 bar. Assume that the inlet pressure is
changed to p1 = 210 bar. At which value of p2 can cavitation now be expected?
A way to avoid cavitation is to split the pressure drop over two or more orifices in series
connection. Describe with equations, which of the two series connected orifices can
take the highest pressure drop without any risk for cavitation. (Assume that the outlet
pressure for the second orifice is zero).
(4p)
b) Flow forces on servo valve in a constant pressure system
The figure shows a servo valve, supplied by a pressure controlled pump with Dp = 35
cm3/rev. In the diagram, valve flow force (Fs) versus valve displacement, xv is shown.
Max flow force is reached at xv = 0,6⋅xvmax and the value is Fsmax = 70 N.
Assume that the pump is changed to one with bigger displacement (same np and pp) so
that max pump flow is equal to max valve flow at max pressure drop, (pp – pT)max.
Calculate the pump displacement, Dp1 and max flow force, Fs1max for that system.
xv
pp
qp
Fs
Load
np
pT = 0
Fs
0
xv
xvmax
(3p)
c) Servo valve pressure gain and it’s influence on a position servo stiffness
The figure shows the pressure gain (pL versus iv) for a critical center servo valve and a
schematic block-diagram of a closed loop position servo.
Describe, with equations how the valve pressure gain and the steady state stiffness are
affected by increased supply pressure, ps.
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
2.
EXAMINATION
TMHP 51
2010-12-17
3(6)
a) Direct and pilot operated servo valves
The figure shows a direct controlled 1-stage valve and a 2-stage valve with an electrohydraulic pilot-stage. Both valves have feedback control of the main spool position.
Compare qualitatively the two valves in respect of control accuracy, threshold,
bandwidth and flow capacity.
(4p)
b) The servo valve bandwidth and it’s amplitude dependency
The diagram below shows how the bandwidth of a servo valve varies at different input
signal amplitudes (in % of max amplitude).
Describe qualitatively the reason of the bandwidth variation at input signal amplitude
for direct-driven and pilot-operated servo valves respectively.
(2p)
c) Hydraulic damping in servo systems
The figure below shows schematically a valve controlled hydraulic servo and it’s block
diagram. However, the orifice between valve and cylinder (flow/pressure-coeff. Cs) is
not included in the block diagram. Without Cs the hydraulic damping is, δh0 = 0,18.
Add the coefficient Cs to the block-diagram above and show with equations it’s
influence on the hydraulic damping, δhC when the cylinder is assumed as loss free.
Calculate δhC when Cs = 1,5·Kc.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
3.
EXAMINATION
TMHP 51
2010-12-17
4(6)
Position servos with valve and pump controlled symmetric cylinder
Ap
Ap
xp
Mt
Ap
Ap
FL
xp
Mt
FL
Pm
ep
Position
transducer
xv
Dp
Servo
uc + amplifier i
Ksa
u f
Kf
Ps = const.
uc +
u f
Ksa
Position
transducer
Kf
i
The figure shows a valve controlled and a pump controlled symmetric cylinder. Both
systems are used as linear position servos with proportional control, Greg = Ksa. The
position feedback gain is Kf.
The servo valve, in the valve controlled system, is a 4-port critical center valve and its
null-coefficients are Kqi0 and Kc0. The bandwidth of the valve is high. In the pump
controlled system just the pressure in one of the cylinder chambers is controlled and the
low pressure is kept constant (pm). The pump displacement controller is assumed to be
fast and the maximum flow capacity of the pump is equal to the maximum flow of the
servo valve (Kqi0·imax = kpNp0·imax). The leakage flow coefficient of the pump is Ctp =
Kc0.
The same cylinder is used in the two systems. The piston area is Ap, the total volume is
Vt and it’s effective bulk modulus is β e. The cylinder can be regarded as loss less. The
cylinder is loaded by the mass Mt and an external force FL.
a) Closed loop steady state stiffness at equal amplitude margin, Am
The steady state loop gain (Kv) are in the two systems adjusted to give the same
amplitude margin, Am = 6 dB in the most critical operation point respectively.
Compare, during these circumstances, the closed loop steady state stiffness
Sc s → 0 =
− ΔFL
ΔX p
of the two systems.
s →0
(6p)
b) Influence of threshold in servo valve and pump controller
Assume that the servo valve and the pump controller have the same threshold value iT =
0,01·imax.
Describe with equations if the threshold results in different position errors in the two
systems when the steady state loop gain in both cases are adjusted to give equal
amplitude margin, Am = 6 dB.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
4.
EXAMINATION
TMHP 51
2010-12-17
5(6)
Pump controlled angular velocity servo with PI-controller
The figure shows the system and the block diagram for an electro-hydraulic velocity
servo with a pump controlled motor and PI-controller, GPI(s) = KP + KI/s. The pump
control unit is fast compared to the motor/load dynamics, (ωps > ωh). The static gain
factors in the hydraulic system have the values: Kps·Dp·ωp/Dm = 1225 rad/(As) and Kf =
0,10 Vs/rad. For the motor/load dynamics the values are, ωh = 73 rad/s and δh = 0,22.
a) Tuning of PI-controller for the amplitude margin Am = 6 dB
Determine the controller parameters Kp in [A/V] and KI in [A/(Vs)], so that the control
loop (Au(s)) amplitude margin is Am = 6 dB and the controller shall give integrating
action up to a frequency of, ω ≈ 2,5·ωh.
(6p)
b) Controller tuning for slow pump controller (ωps < ωh)
I the system above the bandwidth of the pump-controller is, ωps ≈ 200 rad/s. Assume
that this bandwidth is reduced to ωps = 40 rad/s.
Tune the controller parameters (Kp and KI), so that the control loop (Au(s)) amplitude
margin is Am = 6 dB.
Determine the bandwidth, ωb of the closed loop speed control system.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
5.
EXAMINATION
TMHP 51
2010-12-17
6(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. Specify the expressions for the hydraulic resonance frequency
and damping (ωh och δh).
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) Suppression of speed oscillations
Describe qualitatively where in the system the damping must be improved in order to
minimize the speed oscillations in the system above.
Show how the transfer function Gh(s) will be influenced in the frequency domain
(amplitude in a bode-diagram) by this action.
(3p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2010-12-17
SOLUTION SUGGESTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices:
The critical pressure drop for cavitation depends upon the upstream pressure p1 as
Δp cav = C cav p1 or p 2cav = p1 (1 − C cav ) . p1 = 150 bar and p2 = 54 bar gives,
150 − 54
= 0,64 . p1 = 210 bar ⇒ p2cav = 210(1 - 0,64) = 76 bar
Ccav =
150
Two orifices in series, with the pressures p1 → p2 → p3 = 0 gives Δp1cav = C cav p1 and
Δp1cav
p
Δp 2 cav = C cav p 2 . ⇒
= 1 . Since p1 > p2, Δp1cav > Δp2cav, so the first orifice can
Δp 2cav
p2
takes the highest pressure drop without cavitation.
(4p)
b) Flow forces on servo valve in a constant pressure system
Data: Dp = 35 cm3/rev and xv = 0,6⋅xvmax gives Fsmax = 70 N.
xv
pp
qp
Fs
Load
np
Fs
pT = 0
xv
xvmax
0
Calculate the new pump displacement, Dp1 and max flow force, Fs1max for qpmax = qvmax.
xv max
Constant pump pressure and speed gives, q p1 max = D p1n pηvp =
D p n pηvp , and
0,6 ⋅ xv max
1
1
D p1 =
D p = 58,3 cm3/rev. Constant Δpv → Fs ∝ xv and Fs1 max =
Fs max = 117 N.
0,6
0,6
(3p)
c) Servo valve pressure gain and it’s influence on a position servo stiffness
Describe, with equations how the valve pressure gain and the steady state stiffness are
affected by increased supply pressure, ps.
Block-diagram gives: S c
Theoretically K p =
s →0
− ΔFL
=
ΔX p
= Kv
s →0
Ap2
K ce
= K sa K p A p K f , where K p =
Kq
K ce
.
2( p s − p L 0 )
. Increased ps gives increased Kp and stiffness.
xv 0
(3p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2010-12-17
a) Direct- and pilot operated servo valves
Direct controlled 1-stage valve and a 2-stage valve with an electro-hydraulic pilot-stage.
Control accuracy: 2-stage valve gives the best control accuracy, because of highest
control force.
Threshold: 1-stage valve has bigger threshold than 2-stage, because of spool friction.
Bandwidth: 2-stage valve has the highest bandwidth, because of high pilot stage gain.
Flow capacity: 2-stage valve can handled much higher flow, because of higher control
power.
(4p)
b) The servo valve bandwidth and it’s amplitude dependency
Bandwidth, 10% ampl. gives, ωb ≈ 60 Hz and 90%, ωb ≈ 22 Hz. Reduced bandwidth at
high amplitude is caused by saturation in the valve controller. Using a electro-hydraulic
pilot stage is the common way to increase the valve bandwidth, because this concept
gives high control power.
(2p)
c) Hydraulic damping in servo systems
Data: Without Cs, δh0 = 0,18. Calculate δhC when Cs = 1,5·Kc.
Cs in block-diagram:
Calculate δhC when Cs = 1,5·Kc:
δ h0 =
Kc
Ap
βe M t
Vt
and δ hC =
K c + Cs
Ap
βe M t
Vt
, which gives δ hC = δ h 0
K c + Cs
.
Kc
Numerically: Cs = 1,5·Kc gives, δ hC = 0,18 ⋅ 2,5 = 0,45.
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2010-12-17
Position servos with valve and pump controlled symmetric cylinder
Ap
xp
Ap
Mt
Ap
xp
Ap
Mt
FL
FL
Pm
ep
Position
transducer
xv
Dp
Servo
uc + amplifier i
Ksa
u f
Position
transducer
Kf
uc +
u -
Ps = const.
f
Ksa
Kf
i
Parameter values in the actual operating point:
Controller: Greg = Ksa, feedback gain Kf, valve parameters: Kqi0, Kc0, pump param.:
kpNp0·imax = Kqi0·imax, Ctp = Kc0, cyl.:V1 + V2 = Vt, βe, Ap, Mt and FL.
a) Closed loop steady state stiffness at equal amplitude margin, Am
Steady state stiffness Sc s → 0 =
− ΔFL
ΔX p
= Kv ⋅
s →0
Ap2
K ce
Am = 6 dB → Steady state loop gain Kv = δhmin·ωhmin , (most critical operation point).
Valve controlled cylinder: ωh min =
steady state loop gain:
Vt M t
K v , p = δ h minωh min =
β e Ap2
Vt M t
Kc0
Ap
βeM t
Ctp
βe M t
2 Ap
Vt
, δ h min =
K v , 4v = δ h min ωh min = 2 K c 0
Pump controlled cylinder: ωh min =
state loop gain:
4 β e Ap2
, which gives the
Vt
βe
Vt
, δ h min =
, which gives the steady
Ctp β e
2 Vt
Loop gain ratio: Ctp = Kc0 ⇒ Κv,4v = 4·Κv,p. Stiffness ratio:
Sc v
Sc
p
=
Kv,v
= 4.
Kv, p
(6p)
b) Influence of threshold in servo valve and pump controller
Servo valve and the pump controller have the same threshold: iT = 0,01·imax.
iT
Position error according to threshold: ΔX pT =
K f K sa
K qi 0
Kf .
Ap
k N
= K sap p p K f .
Ap
Valve controlled cylinder: K v , v = K sav
Pump controlled cylinder: K v , p
Am = 6 dB for both systems means that Kv,p < Kv,v and with kpNp0 = Kqi0 it will be stated
that Ksap < Ksav.
Finally, the pump controlled system will give the highest error: ΔX pTp > ΔX pTv .
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
4.
SOLUTIONS FOR EXAMINATION
TMHP 51
2010-12-17
4(5)
Pump controlled angular velocity servo
Kps·Dp·ωp/Dm = 1225 rad/(As), Kf = 0,10 Vs/rad, ωh = 73 rad/s and δh = 0,22, Am = 6 dB.
PI-controller: GPI(s) = KP + KI/s, where Kp is in [A/V] and KI in [A/(Vs)].
a) Tuning of PI-controller for the amplitude margin Am = 6 dB
Kv
.
⎞
⎛s
2δ h
s ⋅ ⎜⎜ 2 +
s + 1⎟⎟
ω
ω
h
⎠
⎝ h
Dω
K
Am = 6 dB → K v = ωh ⋅ δ h = 73 ⋅ 0,22 = 16 1/s. GPI ( s ) K ps p p K f = v , where
Dm
s
K
(1 + ( K P / K I ) ⋅ s )
(1 + s / ω PI )
. GPI ( s ) s =1 = K I , which gives
GPI ( s ) = K p + I = K I
= KI
s
s
s
Dω
K v Dm
16
=
K v = K I K ps p p K f and K I =
= 0,131 A/(Vs).
K ps D pω p K f 1225 ⋅ 0.1
Dm
Gps = 1,0 gives the target loop gain as, Au ( s) =
2
Integration up to ω ≈ 2,5·ωh gives, ωPI = K I = 2,5 ⋅ ωh and K P =
KP
KI
= 7,18·10-4 A/V.
2,5 ⋅ ωh
(6p)
b) Controller tuning for slow pump controller (ωps < ωh)
ωps < ωh means that the equation, GPI ( s ) K ps G ps ( s )
which gives, K I
ω PI =
D pω p
K
K f = v must be satisfied,
s
Dm
K
(1 + s / ω PI )
= I . This is fulfilled if ωPI = ωps.
s ⋅ (1 + s / ω ps )
s
KI
= ω ps and KP = KI/ωps where KI = 0,131 A/(Vs) as in task a) and the
KP
proportional gain is KP = 0,131/ωps (increased KP compared to task a).
Bandwidth: ωb = Kv for the given Au(s), which gives, ωb = 16 rad/s,
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
5.
SOLUTIONS FOR EXAMINATION
TMHP 51
2010-12-17
5(5)
Hydraulic system with a valve controlled motor loaded by two masses
a) Transfer function and bode diagram for the system
Derive the transfer function Gh(s) = sθm/Xv.
K q / Dm ⋅ GLθ ( s )
sθ
The block diagram gives: Gh ( s ) = m = 2
where
Xv ⎛ s
⎞
2δ h
⎜⎜ 2 +
s + GLθ ( s ) ⎟⎟
⎝ ωh ωh
⎠
ωh =
K β (J + J )
4βe Dm2
, δh = ce e 1 2
(J1 + J2 )Vt
Dm
Vt
According to the simulation results ωh is much lower than ωa for the load. For
K /D
sθ
frequencies up to ωh the transfer function is: Gh ( s ) = m ≈ 2 q m
⎞
Xv ⎛ s
2δ
⎜⎜ 2 + h s + 1⎟⎟
⎝ ωh ωh
⎠
(7p)
b) Suppression of speed oscillations
The hydraulic damping, δh =
Kce βe (J1 + J2 )
must be improved in order to minimize the
Dm
Vt
speed oscillations in the system above. This will be done by increased Kce.
(3p)