Linköpings Tekniska Högskola
IEI
Fluid and Mechanical Engineering Systems
EXAMINATION
TMHP51/TEN1
2009-12-15
Page 1(6)
EXAMINATION IN
Hydraulic Servo Systems, TMHP51 / TEN1
Date:
Tuesday 15 December 2009, 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
28 11 87, mob 073-806 18 39
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!
15 December 2009
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 Mechanical Engineering Systems
1.
EXAMINATION
TMHP 51
2009-12-15
2(6)
a) Cavitation in orifices
Three orifices are connected as shown in the figure below. The orifice areas are A1, and
A2 = A3 = 0,50·A1. The inlet pressure to the first orifice is p1 and the final outlet pressure
p2 = 15 bar. For each orifice the critical pressure drop for cavitation can be calculated
as: Δpcav = 0,68 ⋅ pin , where pin is the inlet pressure..
Where will cavitation first take place (which orifice) when the pressure p1 increases?
Calculate the pressure, p1, which initiate cavitation in some of the orifices.
(4p)
b) Servo valve efficiency
The figure shows a valve controlled servo systems where the supply unit is a constant
pressure controlled variable displacement pump. The pump pressure setting is ps = 21
MPa and the pump flow is equal to load flow, qp = qL in full flow range.
Calculate and show in diagram the servo valve efficiency, ηsv = Pv,out/Pv,in versus load
pressure pL in the range 0 <= pL <=·ps.
Show qualitatively in the diagram how the valve efficiency will be affected if the valve
has a constant leakage flow.
(3p)
c) 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.
How will the valve pressure gain be affected by ware?
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
2.
EXAMINATION
TMHP 51
2009-12-15
3(6)
a) Direct- and pilot operated servo valves
Servo valves are manufactured in different concepts from single-stage to pilot operated
valves (two-stage and three-stage valves).
Describe the most important qualities for selection of servo valves, related to power
level and control accuracy requirements in the application.
(2p)
b) Bandwidth of servo valves
The figure below shows step responses of a servo valve at two different input signal
amplitudes, in percentage of max amplitude.
Assume that the servo valve dynamics can be described by a first order low-pass filter
as, Gv = 1 /(1 + τ v ⋅ s ) . Determine the valve bandwidth (ωb, [rad/s]) for input signal amplitude 50% and 100% respectively. Why is the bandwidth reduced at high amplitude?
(3p)
c) Amplitude margin for position servo with valve controlled servo cylinder
The figure shows schematically a valve controlled position servo. At centre position of
the cylinder piston, V1 = V2 = Vt/2 = 0,5 litre the hydraulic resonance frequency is,
ωh = 100 rad/s and the damping δh = 0,20. The steady state loop gain is adjusted to
Kv = 20 1/s, which gives the amplitude margin Am = 6 dB.
Assume that the piston in positioned close to end position where the cylinder volumes
are, V1 = 0,9 l and V2 = 0,1 l. Calculate the servo amplitude margin (Am) in that
operation point, when Kv = 20 1/s.
(5p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
3.
EXAMINATION
TMHP 51
2009-12-15
4(6)
Positionsservo med 3-ports resp 4-ports servoventil
The figure shows hydraulic position servos with 3-ports valve and asymmetric cylinder
and 4-ports valve and symmetric cylinder respectively. The two systems are loaded by
equal masses Mt and the same force FL. The servo systems have proportional position
control with the feedback gain Kf and the controller gains Ksa3 and Ksa4. The servo
valves are zero-lapped with symmetric orifices and their “zero-coefficients” are identical, Kqi0 and Kc0. The valves are fast compared to the cylinder-load dynamics. The
supply pressure ps are constant and equal in the two systems. For the cylinders the piston area, Ah = 2,0·Ap and the max volume in the asymmetric cylinder is related to the
symmetric cylinder as Vhmax = 2,0·(V1 + V2). The cylinders losses can be neglected.
a) Controller gain for equal amplitude margin, Am
Calculate the ratio for the controller gain in the two systems, Ksa4/Ksa3, which gives
the same amplitude margin, Am in the most critical operation point for each system
respectively.
(6p)
b) Servo valve Threshold and Saturation
The figure shows a complete block diagram for the linear position servo with 4-port
valve and symmetric cylinder. The static gain factors have the following values:
Ksa4 = 0,12 A/(V), Kqi/Ap = 6,0 m/(As), Kf = 25 V/m, ΔiTH = 0,5 mA and imax = 0,050 A.
Calculate the steady state position error, ΔXp caused by the threshold, ΔiTH (ΔFL = 0).
Calculate max piston velocity according to current saturation, imax (ΔFL = 0).
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
4.
EXAMINATION
TMHP 51
2009-12-15
5(6)
Pump controlled angular velocity servo
The figure shows the system and the block diagram for an electro-hydraulic velocity
servo with a pump controlled motor and integrating controller (Ksa/s). The pump control
unit is fast compared to the motor/load dynamics, (ωps > ωh). The static gain factors in
the control loop have the values: Ksa = 0,10 A/V, Kps·Dp·ωp/Dm = 1200 rad/(As) and
Kf = 0,10 Vs/rad. For the motor/load dynamics the values are, ωh = 67 rad/s and δh =
0,18.
a) Feedback for increased hydraulic damping
Show in a block diagram how a negative feedback of time derivative of the speed
signal (uf) with the gain factor, KD [A/V] shall be implemented in the control loop.
Show with equations how this feedback will gain the system damping (δh).
Calculate the value of KD, which gives the new damping δh' = 0,40.
(6p)
b) Pump speed and it’s influence on the system amplitude margin, Am
The above parameter values are valid for the pump speed ωp = 155 rad/s. Assume that
the pump speed increases to ωp = 205 rad/s.
Show with equations how the increased pump speed will influence the amplitude
margin, Am of the control loop, when the extra damping feedback is used.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
5.
EXAMINATION
TMHP 51
2009-12-15
6(6)
Linear position servo with valve controlled cylinder and two masses
The figure below shows a valve controlled hydraulic cylinder loaded by the masses M1
and M2. The connection between the masses includes a spring and a damper (spring
constant KL and viscous friction coeff. BL). The piston position xp is fed back to a
proportional regulator with the gain Greg = Ksa.
xp
be Ap
Ap be
p1 V1
V2 p2
KL
Position
transducer
uc +
Servo
amplifier
uf -
Ksa
M2
M1
BL
xL
Kf
i
Au(s)
Ps = const.
The block diagram from valve input signal (i) to piston velocity (sXp) is:
Kqi
i ___
Ap +
-
Ap
__________
Vt
Kce + ___
s
4be
PL
Ap
sXp
1
__________
GLX(s)
(M1 + M2) s
s2
+
2δ a
s +1
ωa
, (ωa < ω1)
The transfer function of the mechanics is G LX ( s ) =
2δ 1
s
+
s +1
ω12 ω1
ω
2
a
2
a) The transfer function Gh(s) = sXp/i of the hydraulic system
Derive an expression of the transfer function Gh(s) = sXp/i and show how the fundamental hydraulic resonance frequency, ωh' and damping, δ h' are varying according to
the gain of the load dynamics, GLX (s ) .
(6p)
b) Negative velocity feedback to improve hydraulic stiffness in the position servo
Assume that ωh < ωa in the system above. Describe qualitatively with equations how
the hydraulic system dynamics will be influenced by implementation of a negative
velocity feedback in the position servo.
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
SOLUTIONS FOR EXAMINATION
TMHP 51
2009-12-15
1(5)
SOLUTION SUGGESTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices
Orifices A1, A2 = A3 = 0,50·A1
The critical pressure drop for cavitation: Δpcav = 0,68 ⋅ pin .
Where will cavitation starts? q1 = Cq ⋅ A1 2 ( p1 − pm ) = Cq ⋅ ( A2 + A3 ) 2 ( pm − p2 ) , which
ρ
ρ
gives with A2 + A3 = A1: p1 − pm = pm − p2 and ( p1 − pm )cav > ( pm − p2 )cav , which means
thar cavitation starts first after orifices A2 and A3.
Which value of p1 starts cavitation? p1 = 2 pm − p2 and pm, cav = p2 /(1 − 0,68) = 46,9 bar
gives: p1,cav = 93,8 – 15 = 78,8 bar.
(4p)
b) Servo valve efficiency
ps = 21 MPa and qp = qL.
qL ⋅ pL
. qp = qL gives ηsv = pL/ps.
qs ⋅ ps
pL
Leakage flow, qlsv gives η sv =
. p → ps means qL → 0 and ηsv → 0.
(1 + qlsv / qL ) ⋅ ps L
Calculate ηsv = Pv,out/Pv,in: η sv =
(4p)
c) Pressure gain for servo valves
All real servo valve characteristic shows hysteresis, because of friction and hysteresis in
the electro magnet material (Threshold gives hysteresis). The pressure gain for a worn
valve will be less than for a new valve, because of increased leakage flow (increased Kcvalue gives reduced Kp-value).
(2p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
2.
SOLUTIONS FOR EXAMINATION
TMHP 51
2009-12-15
2(5)
a) Direct- and pilot operated servo valves
Servo valve selection, related to power level and control accuracy requirements.
Single-stage valves can not be used in high power applications, because of low
controllability (low actuating force for the main spool control). Therefore, multi-stage
(pilot operated) valves must be used in high power applications. Pilot operated valves
also shows much better control accuracy than single-stage (direct operated) valves.
(2p)
b) Bandwidth of servo valves
Calculate valve bandwidth from diagram, when Gv = 1 /(1 + τ v ⋅ s ) .
Bandwidth, ωb = 1/τv [rad/s]. 50% ampl. gives ωb = 200 and 100%, ωb = 148 rad/s.
Reduced bandwidth at high amplitude is caused by saturation in the valve controller.
(3p)
c) Amplitude margin for position servo with valve controlled servo cylinder
Data: V1 = V2 = Vt/2 = 0,5 litre, ωh = 100 rad/s δh = 0,20, Kv = 20 1/s gives Am = 6 dB.
Calculate the amplitude margin (Am) when V1 = 0,9 l and V2 = 0,1 l and Kv = 20 1/s.
V1 = V2 → ωh =
δ h1 =
4 β e Ap2
Vt M t
, δh =
K ce
Ap
βe M t
Vt
and V1 ≠ V2 → ωh1 =
⎛1 1⎞
V
K ce
β e M t ⎜⎜ + ⎟⎟ , which gives ωh1 ⋅ δ h1 = ωh ⋅ δ h ⋅ t
4
2 Ap
⎝ V1 V2 ⎠
and Am = −20 ⋅10 log
Kv
− 2ωh1δ h1
→ Am = −20 ⋅10 log
β e Ap2 ⎛ 1
1⎞
⎜⎜ + ⎟⎟ ,
M t ⎝ V1 V2 ⎠
⎛1 1⎞
⋅ ⎜⎜ + ⎟⎟ = ωh ⋅ δ h ⋅ 2,78
⎝ V1 V2 ⎠
20
= 14,9 dB.
− 2 ⋅ 100 ⋅ 0,2 ⋅ 2,78
(5p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
3.
SOLUTIONS FOR EXAMINATION
TMHP 51
2009-12-15
3(5)
Position servo with 3-ports and 4-ports servo valve respectively
Parameter values:
Ksa3, Ksa4, identical Kqi0 and Kc0, Ah = 2,0·Ap and Vhmax = 2,0·(V1 + V2).
a) Controller gain for equal amplitude margin, Am
Ksa4/Ksa3, which gives the same amplitude margin, Am?
Kv
and K v = C ⋅ δ h ⋅ ωh gives constant Am.
The loop gain is Au ( s ) =
2
⎛s
⎞
2δ h
s ⋅ ⎜⎜ 2 +
s + 1⎟⎟
⎝ ωh ωh
⎠
β
1
β
3-port valve: (δ h ⋅ ωh )3v = K ce e . 4-port valve: (δ h ⋅ ωh )4 v = 2 K ce e .
2
Vh max
Vt
2 Ap K ce β eVh max ⎡ Ah = 2 Ap ⎤
K v 4 K sa 4 ⋅ K qi 0 ⋅ K f / Ap (δ h ⋅ ωh )4 v
K
=
=
=⎢
→ sa 4 =
⎥ = 4.
K v 3 K sa 3 ⋅ K qi 0 ⋅ K f / Ah (δ h ⋅ ωh )3v
K sa 3
0.5 Ah K ce β eVt
⎣Vh max = 2Vt ⎦
(6p)
b) Servo valve Threshold and Saturation in the servo with 4-port valve
Data: Ksa4 = 0,12 A/(V), Kqi/Ap = 6,0 m/(As), Kf = 25 V/m, ΔiTH = 0,5 mA and imax = 0,050 A.
Position error from threshold: ΔX p ⋅ K f ⋅ K sa = ΔiTH ⇒ ΔX p =
Numerically: ΔX p =
0,0005
= 0,00017 m = 0,17 mm.
0.12 ⋅ 25
⋅
⋅
Max velocity from saturation: iv ⋅ K qi / Ap = X p ⇒ X p max =
⋅
Numerically: X p max =
ΔiTH
.
K f ⋅ K sa
imax ⋅ K qi 0
.
Ap
0,050 ⋅
6,0 = 0,30 m/s.
1
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2009-12-15
Pump controlled angular velocity servo
Ksa = 0,10 A/V, Kps·Dp·ωp/Dm = 1200 rad/(As), Kf = 0,10 Vs/rad, ωh = 67 rad/s and δh = 0,18.
a) Feedback for increased hydraulic damping
Implementation of negative feedback of duf/dt = acceleration feedback:
Gps = 1,0 gives
1
2δ h
⎛s
⎞
⎜⎜ 2 +
s + 1⎟⎟
⎝ ωh ωh
⎠
→
1
⎛s
⎞
Dω ⎞
⎛
⎜ 2 + ⎜ 2δ h + K D K ps p p ⎟ s + 1⎟
⎜ω ⎜ ω
⎟
Dm ⎟⎠
⎝ h ⎝ h
⎠
=
1
2δ h'
⎛s
⎞
⎜⎜ 2 +
s + 1⎟⎟
⎝ ωh ωh
⎠
D pω p
ω
where δ h' is the new damping, which can be calculated as: δ h' = δ h + h K f K D K ps
2
Dm
2
2
2
Dm
2
1
KD for δ h' = 0,40: K D = (δ h' − δ h ) 2 ⋅
= 5,47·10-5 A/V.
= (0,4 − 0,18) ⋅
ωh K f K ps D pω p
67 0,1 ⋅ 1200
(6p)
b) Pump speed and it’s influence on the system amplitude margin, Am
How will an increased ωp influence Am, when acceleration feedback is used?
The steady state loop gain is: K v = K sa K ps
D pω p
K f ⇒ K v ∝ ω p and the hydraulic
Dm
damping δ h' = δ h + const ⋅ ω p
Amplitude margin: Am = −20 ⋅10 log
Const ⋅ ω p
Kv
.
= −20 ⋅10 log
− 2ωh (δ h + const ⋅ ω p )
− 2ωhδ h
If const·ωp >> δh the amplitude margin will not be changed by increased ωp.
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2009-12-15
Linear position servo with valve controlled cylinder and two masses
xp
Ap
p1 C1
KL
Ap
M2
M1
C2 p2
BL
xL
Position
transducer
Kf
Servo
uc + amplifier
Ksa
uf -
i
Ps = const.
Au(s)
The block diagram from valve input signal (i) to piston velocity (sXp) is:
Kqi
i ___
Ap +
-
Ap
__________
Vt
Kce + ___ s
4be
s2
PL
Ap
sXp
1
__________
GLX(s)
(M1 + M2) s
+
2δ a
s +1
ωa
G LX ( s ) =
2δ
s
+ 1 s +1
2
ω1 ω1
ω
2
a
2
a) The transfer function Gh(s) = sXp/i of the hydraulic system:
The block diagram gives: Gh ( s ) =
ωh =
4 β e A p2
(M 1 + M 2 )Vt
, δh =
K ce
Ap
sX p
i
=
K qi / A p ⋅ G LX ( s )
⎛ s 2 2δ h
⎞
⎜⎜ 2 +
s + G LX ( s ) ⎟⎟
⎝ωh ωh
⎠
β e (M 1 + M 2 )
Vt
where
and Vt/2 = V1 = V2.
ωh' and damping, δ h' are varying according to the gain of the load dynamics, GLX ( s) as
Gh ( s) =
K qi / Ap
⎛ s 2 2δ h'
⎞
⎜⎜ '2 + ' s + 1⎟⎟
⎝ ωh ωh
⎠
, where ωh' = ωh GLX ( s ) and δ h' =
δh
GLX ( s )
.
(6p)
b) Negative velocity feedback to improve hydraulic stiffness in the position servo
Assume that ωh < ωa in the original system above. Negative velocity feedback with the
feedback gain Kvfv will give the new resonance frequency and damping as:
ωh' = ωh GLX ( s) ⋅ K vfv and δ h' =
δh
GLX ( s ) ⋅ K vfv
.
The velocity feedback gain is always Kvfv > 1,0, which means that the resonance
frequency will increase and the damping decrease. If the the velocity feedback gain the
hydrauliv resonance so it comes very close to ω1, GLX (s ) > 1,0 and the effective damping
will be very low and the system become very oscillative.
(4p)