Linköpings Tekniska Högskola
IKP
Fluid and Mechanical Engineering Systems
EXAMINATION
TMHP51
2003-12-19
Page 1(6)
Sample of Exames TMHP51: 2006-11-06
EXAMINATION IN
Hydraulic servo systems, (TMHP51)
Date:
Friday 19 December 2003, at 8 am - 12 am
Allowed educational aids:
Tables: Standard Mathematical Tables or similar
Handbooks: Tefyma
Formularies:
Hydraulik och Pneumatik, LiTH/IKP
Mekanisk Värmeteori och Strömningslära
Pocket calculator
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:
You will find the solutions of this examination on the notice board
in the A building, entrance 15, C-corridor to the right, (outside the
laboratories of div. of Fluid and Mechanical Engineering Systems).
Results:
Results will be announced on the notice board (see above) on
12 January, 2004.
GOOD LUCK!
19 December, 2003
Karl-Erik Rydberg,
Professor, (tel 073-649 0670)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1.
2(6)
EXAMINATION
TMHP 51
2003-12-19
a) Cavitation in orifices
In hydraulic systems cavitation will occur downstream orifices with high pressure
drops.
For an orifice with a constant inlet pressure of p1 = 200 bar cavitation will start
when the outlet pressure is reduced to p2 = 72 bar. Assume that the inlet pressure
is changed to p1 = 280 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
serial connection. Describe with equations, which of two serial 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) Servo valve with bushing
The main stage in an advanced servo valve consists of a spool and a valve
bushing. Describe qualitatively the benefits of using a bushing compared to a
valve without bushing it the pressure level is high.
(3p)
c) Flow forces on a spool valve in constant pressure system
The figure below shows a 4-port critical centre servo valve connected to a
constant pressure controlled pump. A diagram shows the measured flow forces
versus valve displacement, xv.
pp
qp
Dpv
xv
qv
Fs
65 N
np
pT = 0
Fs
xv
0
0,6*xvmax
xvmax
Assume that the max pump flow (qpmax) is increased to the same level as the max
flow capacity of the valve with the same pressure drop, ∆pv as for the measured
case. Show in a diagram and calculate the maximum flow forces for that case.
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
3(6)
EXAMINATION
TMHP 51
2003-12-19
a) Characteristics of two-stage and single-stage servo valves
Describe qualitatively in which way a two-stage servo valve can show up a
better steady state characteristic than a single-stage valve (direct drive valve).
(2p)
b) Pressure gain (sensitivity) for a servo valve: The figure below shows an
electrical controlled servo valve, which is connected to a test stand and equipped
with transducers for measurements of valve command signal, pressures and flow.
The servo valve is a 4-way one with critical center.
Measurement signals
Potentiometer for
zero point adjust.
Uc
p2
p1
iv
Servo amplifier
ps = constant
Describe with diagram how the pressure gain (Kp) can be measured and show in
the diagram how the characteristics will be changed because of wear in the
valve.
(3p)
c) Bandwidth and steady state errors for a position and a speed servo:
A schematic figure of a linear position servo and an angular speed servo is shown
below. 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 coeff. are Kqi and Kc respectively. The two systems have the
same hydraulic resonance-frequency (ωh) and damping (δh).
xp
Ap
Kf
Uc
V1
V2
Mt
FL
Kf
Ksa
__
Uc s
nm
V1
Dm
Ksa
Jt
TL
V2
Compare qualitatively, (with equations and bode-diagrams) these two systems
with respect to their bandwidth (ωb) and steady state errors.
(5p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
4(6)
EXAMINATION
TMHP 51
2003-12-19
Linear position servo with a valve controlled symmetric cylinder and
velocity feedback
FL
ö
Vt
Kce æç
÷
ç1 + 4 b K s÷
2
Ap è
e ce ø
Threshold
uc
+
-
+
-
Ksa
ir
iv
ein
Kqi
1
Ap 1 + s +
wv
-
.
1
xp 1 xp
s 2 + 2 dh s +
s
1
wh2 wh
Kfv
Velocity feedback
Kf
Position feedback
The figure shows an electro-hydraulic linear position servo with a valve
controlled cylinder. The regulator is proportional with the gain Greg = Ksa. The
position feedback gain is Kf. In order to increase the system accuracy a negative
velocity feedback signal with the gain, Kfv has been implemented. The servo
valve is a 4-port critical center valve with high bandwidth and its nullcoefficients are Kqi0 and Kc0. The valve has a threshold value, which is iT (= εin).
The oil volumes between the valves and the piston are assumed to be equal, V1 =
V2 = Vt/2 and its bulk modulus is β e. The piston area is Ap and the cylinder can
be assumed as loss free. The loading mass of the cylinder is Mt and the external
force is FL.
Parameter values in the actual operating point:
Mt = 500 kg
Ap = 1,96.10-3 m2
-3
3
.
Vt = 1,0 10 m
ps = 14 MPa
Kc0 = 8,0.10-12 m5/(Ns)
Kf = 20 V/m
Ksa = 0,25 A/V
iT = 0,4 mA
β e = 1200 MPa
Kqi0 = 0,025 m3/(As)
Kfv = 1,5 Vs/m
a) Steady state loop gain (Kv) and bandwidth (ωb) of the servo
Derive an expression for the steady state loop gain, Kv, and calculate its value
as well as the bandwidth, ωb, of the servo.
(6p)
b) Influence from the valve threshold on the steady state stiffness
Describe (with equations) how the threshold of the valve (iT = εin) will influence
− ∆FL
, of the closed loop system.
the steady state stiffness,
∆X p
s →0
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
5(6)
EXAMINATION
TMHP 51
2003-12-19
Electro-hydraulic angular position servo with acceleration feedback
Kf
Uc
+
+
Kac
Greg
V1
ps
..
qm
qm
Jt
pL
TL
V2
The figure shows an electro-hydraulic angular position servo with a valve
controlled motor. The regulator is of proportional type with the gain factor Greg =
Ksa = 0,04 A/V. In order to increase the damping an acceleration feedback is
implemented with the gain Kac. . The servo valve is a 4-port critical center valve
with high bandwidth and its null-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 the effective bulk modulus is β e = 1000 MPa. The motor displacement is Dm
= 19.10-6 m3/rad and the inertia on the motor shaft is Jt = 0,5 kgm2. The leakage
coefficient of the motor is Ctm = 8,0.10-13 m5/Ns and the viscous friction
coefficient is Bm = 0.
a) Increased hydraulic damping from the acceleration feedback
Calculate the acceleration feedback gain, Kac, so that the total hydraulic
damping will reach a value of δh' = 0,4.
(7p)
b) Acceleration feedback and its influence on the steady state stiffness
Show with equations that the acceleration feedback will not influence the steady
state stiffness of the closed loop system.
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5.
6(6)
EXAMINATION
TMHP 51
2003-12-19
Hydraulically operated boom with lumped masses
The figure shows a valve controlled cylinder used for operation of a mechanical
arm. The total mass of the moving arm is ML = 1000 kg. The distance from the
gravity center of the mass to the joint (θ) is L = 2,5 m. The lever length for the
hydraulic cylinder is e, which will varies according to xp. xp = 0 gives emax = 0,8
m and xpmax = 1,0 m gives emin = 0,4 m. The piston area is Ap = 3,3⋅10-3 m2 and
its pressurised volume is VL and this volume varies according to the piston
position inside the interval 1,0 litre ≤ VL ≤ 4,3 litre. The effective bulk modulus
is β e = 1000 MPa. The pressure on the piston rod side is assumed as constant, pR
= constant. The mass of the cylinder housing is M0 = 80 kg and the mechanical
spring coefficient for the connection is KL= 4⋅107 N/m.
L
ML
q
e
xp
Ap
VL pL
KL
pR = constant
M0
xv
a) Equivalent cylinder mass
Neglect the mass M0 and derive an expression (via the inertia, ML⋅L2) for the
equivalent mass (Mt), which will load the piston rod.
(2p)
b) Hydraulic resonance frequency and dampning
Calculate the hydraulic resonance frequency (ωh) for xp = 0 and xp = xpmax
respectively, when KL in included and M0 neglected. Then, describe
qualitatively with equations how the product of ωh⋅δh will vary according to the
piston position, xp.
(5p)
c) Influence of the mass M0
For the above system the mechanical frequencies can be expressed as:
KL
KL
ωa =
, ω1 =
M0 + Mt
M0
Show in principle with a bodediagram (amplitude- and phase shift) how the
transfer function G(s) = sXp/Xv will be influenced by the size of the mass M0, M0
<< Mt and M0 = Mt respectively. (Mt = equivalent cylinder mass according to
ML). Assume that KL is much higher than the hydraulic spring coefficient Kh.
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2003-12-19
SOLUTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices:
The critical pressure drop for cavitation depends upon the upstream pressure p1 as
∆p cav = C cav p1 or p 2cav = p1 (1 − C cav ) . p1 = 200 bar and p2 = 72 bar gives
200 − 72
= 0,64 . p1 = 280 bar Þ p2cav = 280(1 - 0,64) = 101 bar
C cav =
200
Two orifices in series, with the pressures p1 → p2 → p3 = 0 gives
∆p1cav
p
∆p1cav = C cav p1 and ∆p 2 cav = C cav p 2 . Þ
= 1 . Since p1 > p2, ∆p1cav >
∆p 2cav
p2
∆p2cav, so the first orifice can takes the highest pressure drop without
cavitation.
(4p)
b) Servo valve with bushing
Using a bushing means that the gap between spool and bushing is kept constant,
independent of the pressure level. Therefore, the leakage in the valve will be more
constant than for a valve without bushing. For a valve with bushing the tolerances
between spool land and valve ports can be produced with higher accuracy, which
means better linearity for the flow characteristics.
(3p)
c Flow forces on a spool valve in constant pressure system
Fs
108 N
pp
qp
Dpv
xv
qv
65 N
np
pT = 0
Fs
xv
0
0,6*xvmax
xvmax
Steady state flow forces: Fs = 2C q ⋅ w ⋅ xv ⋅ ∆p ⋅ cos δ . Const. ∆pv means Fs ∼ xv as
shown in the above diagram. qpmax = qvmax (xv = xvmax) at constant ∆pv Þ
1,0
Fs max =
64 = 108 N
0,6
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
a) Characteristics of two-stage and single-stage servo valves
A single-stage valve design means that the spool position will be disturbed by
flow and friction forces because of the limited forces from the control magnet. A
two-stage valve has much higher control forces on the spool, which means more
linear characteristics and lower hysteresis.
(2p)
b) Pressure gain (sensitivity) for a servo valve
pL
ps
Measurement signals
Potentiometer for
zero point adjust.
Uc
Servo amplifier
Kpi0
p2
p1
Worn
valve
1
iv
ps = constant
iv
-ps
The pressure gain (Kp) is measured by plotting the load pressure difference (p1 p2) versus the input signal to the valve, iv, as shown in the diagram above. The
diagram also shows the effect of wear in the valve.
(3p)
c) Bandwidth and steady state errors for a position and a speed servo
xp
Ap
Kf
Uc
V1
V2
Mt
FL
Kf
Ksa
__
nm
V1
Uc s
Dm
Ksa
Jt
TL
V2
The two systems have the same hydraulic frequency (ωh) and damping (δh).
Bandwidth: In principle the two systems have the same open loop gain,
Au ( s ) =
Kv
æs
ö
2δ
ç 2 + h s + 1÷ s
çω
÷
è h ωh
ø
2
Am
Phase shift [degrees]
Kv = d = 0,2
h
wh
Amplitude [dB]
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2003-12-19
Frequency [w/wh]
If the two systems are designed for the same amplitude margin (Am) the
bandwidth ωb = Kv (∼ωh⋅δh) will be the same.
Steady state error: The steady state stiffness is different. Position servo gives:
Ap2
Dm2
− ∆FL
− ∆TL
= Kv
and the speed servo:
= Kv
. The steady
K ce
K ce s
∆X p
∆nn s →0
s →0
state error goes to zero for the speed servo but not for the position servo.
(5p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
SOLUTIONS FOR EXAMINATION
TMHP 51
2003-12-19
3(5)
Linear position servo with a valve controlled symmetric cylinder and velocity
feedback
FL
ö
Vt
Kce æç
s÷
1+
Ap2 çè 4 be K ce ÷ø
Threshold
uc
+
-
+
-
Ksa
ir
iv
ein
Kqi
1
Ap 1 + s +
wv
-
.
1
xp 1 xp
s
+ 2 dh s + 1
2
wh
wh
s2
Kfv
Velocity feedback
Kf
Position feedback
Parameter values in the actual operating point:
Ap = 1,96.10-3 m2, Mt = 500 kg, βe = 1200 MPa, Vt = 1,0.10-3 m3, ps = 14 MPa,
Kqi0 = 0,025 m3/(As), Kc0 = 8,0.10-12 m5/(Ns), Kf = 20 V/m, Kfv = 1,5 Vs/m,
Ksa = 0,25 A/V, iT = 0,4 mA.
a) Steady state loop gain (Kv) and bandwidth (ωb) of the servo
High response valve gives that
1
= 1,0 . Neglecting the threshold,
1+ s /ωv
reduction of the velocity feedback results in:
1 / K vfv
K qi 0
1
where K vfv = 1 + K fv K sa
=
2
2
2δ
2δ h
Ap
s
s
+ h s + K vfv
+
s +1
2
2
ωh ωh
ω h K vfv ω h K vfv
The steady state position loop gain will be as K v = K sa
K qi 0
A p K vfv
K f . Parameter
values gives Kvfv = 5,78 and Kv = 11 1/s.
Bandwidth, ωb: ω < ωh Þ Au ( s ) =
Kv
Þ ωb = Kv = 11 rad/s.
s
(6p)
b) Influence from the valve threshold on the steady state stiffness
− ∆FL
Steady state stiffness,
∆X p
= Kv
s →0
Ap2
K ce
Þ ∆X pF =
− ∆FL
K v Ap2 / K ce
s →0
The relation between position error ∆Xp, and the threshold iT (= εin) is:
iT
∆X pT =
. The total control error is ∆X ptot = ∆X pF + ∆X pT .
K f K sa
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
SOLUTIONS FOR EXAMINATION
TMHP 51
2003-12-19
4(5)
Electro-hydraulic angular position servo with acceleration feedback
Kf
-
Uc
+
+
Kac
Greg
V1
ps
..
qm
qm
Jt
pL
TL
V2
Parameter values: Greg = Ksa = 0,04 A/V, Kqi0 = 0,013 m3/As, Kc0 = 1,0.10-12
m5/Ns, V1 = V2 = 0,5 litre, βe = 1000 MPa, Dm = 6,4.10-6 m3/rad, Jt = 0,5 kgm2,
Ctm = 8,0.10-13 m5/Ns and Bm = 0.
a) Increased hydraulic damping from the acceleration feedback
Calculate the acceleration feedback gain, Kac, that gives δh' = 0,4.
Assuming a fast valve (Gv(s) = 1) and if TL is neglected the acceleration feedback
will change the hydraulic transfer function from:
Gh ( s) =
1
1
to G h ( s ) =
2
2δ h
s
K qi ö
æ 2δ
s
+
s +1
÷s + 1
+ çç h + K ac K sa
2
2
ωh ωh
Dm ÷ø
ωh è ωh
2
where δ h' = δ h +
ωh =
2 Dm
ω h K ac K sa K qi 0
'
and K ac = (δ h − δ h )
ω h K sa K qi 0
2
Dm
4 β e Dm2
K + C tm
= 18 rad/s and δ h = c 0
(V1 + V2 )J t
Dm
βe Jt
= 0,20
(V1 + V2 )
2 ⋅ 6,4 ⋅ 10 −6
Þ Kac = (0,4 − 0,20)
2,7⋅10-4 V/s2.
18 ⋅ 0,04 ⋅ 0,013
(Examination in English: Dm = 19.10-6 m3/rad Þ
−6
Kac = (0,4 − 0,067) 2 ⋅ 19 ⋅ 10
54 ⋅ 0,04 ⋅ 0,013
4,5⋅10-4 V/s2.)
(7p)
b) Acceleration feedback and its influence on the steady state stiffness
The acceleration feedback will only influence the hydraulic damping δh around
the resonance frequency ωh. Therefore, the steady state stiffness will become as:
− ∆TL
∆θ m
= Kv
s →0
Dm2
with no influence from the acc. feedback.
K c 0 + C tm
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
Hydraulically operated boom with lumped masses
L
ML
q
e
xp
pR = constant
Ap
VL pL
M0
KL
xv
Parameter values: ML = 1000 kg, L = 2,5 m, xp = 0 gives emax = 0,8 m, xpmax = 1,0 m gives emin =
0,4 m, Ap = 3,3⋅10-3 m2, 1,0 litre ≤ VL ≤ 4,3 litre, βe = 1000 MPa, pR = constant, M0 = 80 kg and
KL= 4⋅107 N/m.
a) Equivalent cylinder mass
..
Inertia : J t = M L L2
..
..
Torque : Tθ = M L L2 θ = p L A p e
. Wtih θ =
2
Xp
æ L ö ..
Þ p L Ap = M L ç ÷ X p .
e
èeø
Identification gives the equivalent cylinder mass: M t = M L (L / e )
2
(2p)
b) Hydraulic resonance frequency and dampning
Ke
e
=
Mt
L
M0 = 0 gives ω h =
2
Ke
where 1 = V L 2 + 1 Þ K e = K L β e A p 2
ML
K e β e Ap K L
K LV L + β e A p
xp = 0 gives emax = 0,8 m and VL = 1,0 litre: ω h =
0,8 8,56 ⋅ 10 6
= 30 rad/s.
2,5
1000
xp = 1 m gives emin = 0,4 m and VL = 4,3 litre: ω h =
KL > Kh Þ δ hω h ≈
K ce L
2 Ap e
0,4 2,38 ⋅ 10 6
= 7,8 rad/s.
2,5
1000
2
βe
β e M L e β e Ap
K
Þ δ hω h ≈ ce
VL L VL M L
2 VL0 + Ap x p
(5p)
c) Influence of the mass M0
Given frequencies: ω a =
KL
, ω1 =
M0 + Mt
G ( s) =
system, which means
sX p
Xv
=
KL
. M0 = 0 leads to a 1 DOF load
M0
Kv
æ s 2 2δ h
ö
ç 2 +
s + 1÷÷ .
çω
è h ωh
ø
β e A p2
and M0 = Mt
KL > Kh =
VL
Amplitude
101
10-3
100
gives that ωh < ωa < ω1 Þ
101
102
Frequency [rad/s]
0
Phase
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2003-12-19
-100
-200
100
101
102
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1.
2(6)
EXAMINATION
TMHP 51
2002-12-20
a) Cavitation in orifices: The figure below shows measurements made on a
circular orifice with a diameter of d = 1,0 mm. The supply pressure have been
kept constant at the levels 50, 100, 150 and 200 bar and the downstream pressure
has been varied from supply pressure down to ”zero-pressure”. The rings shows
when cavitation starts. The oil density is ρ = 870 kg/m3.
Assume that the flow through the orifice is turbulent and calculate approximately
and show in a diagram the flow coefficient Cq versus the pressure difference ps pd (supply pressure - downstream pressure) for ps = 200 bar. (Calculate in 5
points)
(3p)
b) Reduction of flow forces and radial friction forces. Describe with a
schematic figure how the spool of the 4-port valve below will be modified for
reduction of steady state flow forces and radial forces.
A
T
B
P
T
(3p)
c) Flow forces in spool valves: The figure below shows a 4-port critical centre
servo valve connected to a constant pressure controlled pump. The valving orifices
are matched and symmetrical and its nominal flow is qvN = 40 litre/min with a
total pressure drop of ∆pvN = 7,0 MPa over the valve. The max. flow of the pump
is qpmax = 50 litre/min and its pressure level is adjusted to pp = 21 MPa.
qv
pp qp
xv
pT = 0
Calculate the valve displacement ratio xv/xvmax where the flow forces reach its
maximum value.
Show in principle the steady state flow forces on the servo valve versus the valve
displacement in the interval 0 ≤ xv ≤ xvmax.
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
3(6)
EXAMINATION
TMHP 51
2002-12-20
a) A single stage servo valve:
A 4-port servo valve is shown schematically in the figure below. A conventional
linear magnet is used to control a spring-loaded spool. Assume that the valve has
to be used in a position servo with proportional controller. Describe in which way
the valve can cause trouble in the proposed application. Assume that the valve is
equipped with a spool position feedback. How will that influence the valve
behaviour in the proposed application?
Proportional
magnet
A
B
P
T
(3p)
b) Stiffness of a position servo with valve controlled cylinder: The figure
below shows a block diagram of a position servo where a symmetric cylinder is
controlled by a 4-way critical center valve. The flow/pressure coefficient, Kce is
dominated by the Kc-value of the servo valve.
.
1
ö
Vt
DFL Kce æç
÷
+
s
1
ç
Ap2 è 4 be K ce ÷ø -
s 2 + 2 dh s +
1
wh2 wh
Kqi
1
Ap 1 + s
wv
Saturation
xp
1
s
DXp
Threshold
imax
ein
Ksa
Show in principle in a bode-diagram how the closed loop stiffness S c =
Kf
− ∆FL
∆X p
varies according to the frequency when the bandwidth of the valve ωv is higher
than the resonance frequency ωh.
Describe with equations how the steady state stiffness S c s →0 will be influenced
by an increased value of Kc and also describe qualitatively how the dynamic
stiffness at the frequency ωh will be influenced by the Kc-value.
(4p)
c) Threshold and saturation in the servo valve: The block-diagram in task b
includes threshold and saturation of the servo valve.
Describe with equations how the threshold (εin) will influence the position error
(∆Xp) in the servo system. Show qualitatively in a diagram how saturation will
affect the step response of the servo.
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
4(6)
EXAMINATION
TMHP 51
2002-12-20
Linear position servo with velocity feedback
Kf
Uc
+
+
Kfv
Ksa
Threshold
xp
.
xp
Bp = 0
Ap
V1
V2
Mt
FL
ein
The figure shows an electro-hydraulic linear position servo with a valve
controlled cylinder. The regulator is proportional with the gain Greg = Ksa. The
position feedback gain is Kf. In order to increase the system accuracy a negative
velocity feedback signal with the gain, Kfv has been implemented. The servo
valve is a 4-port critical center valve with high bandwidth and its null-coefficients
are Kqi0 and Kc0. The oil volumes between the valves and the piston are assumed
to be equal, V1 = V2 and its bulk modulus is βe. The piston area is Ap and the
cylinder can be assumed as loss free. The loading mass of the cylinder is Mt and
the external force is FL.
Parameter values:
Ap = 1,96.10-3 m2
Vt = 1,0.10-3 m3
Kc0 = 8,0.10-12 m5/Ns
Mt = 1250 kg
ps = 21 MPa
Kfx = 20 V/m
βe = 1200 MPa
Kqi0 = 2,0⋅10-3 m3/As
Ksa = 0,56 A/V
a) The velocity feegback gain, Kfv:
With velocity feedback the undamped resonance frequency has been calculated to
ωh' = 296 rad/s.
Show with equations how the velocity feedback will influence the resonance
frequency (ωh) and calculate the feedback gain Kfv for the actual operating case.
(5p)
b) The threshold-value (εin) of the servo valve:
Without velocity feedback the steady state position error caused by the valve
threshold has been measured to ∆Xp = 1,0 mm.
Which value of the position error (∆Xp) can be expected with velocity feedback
if the steady state gain (Kv) is the same in both cases (ωh' = 296 rad/s with
velocity feedback)?
(5p)
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Fluid and Mechanical Engineering Systems
4.
5(6)
EXAMINATION
TMHP 51
2002-12-20
Electro-hydraulic speed servo with pump controlled motor
ep
wp
uc +
Ksa
____
uf - s
+
Dm
Dp
Accelerationssignal
Jt
TL
Varvtalsgivare
pm
Kac.s
.
qm
p2 V2
i
-
p1 V1
Kf
The figure shows a electro-hydraulic angular speed servo with a pump controlled
motor. The regulator is of integrator type with the gain factor Ksa. The pump
ε
K pi
displacement controller has the transfer function p =
.
s
i
1+
ωs
The displacement setting coeff. is Kpi = 20 A-1 and the break frequency ωs = 50
rad/s. The pump angular speed is ωp = 157 rad/s and the pump displacement is Dp
= 5,6.10-6 m3/rad. The volumes between pump and motor are V1 = V2 = 0,5 litre
and the effective bulk modulus is β e = 1000 MPa. The motor displacement is Dm
= 19.10-6 m3/rad and the inertia on the motor shaft is Jt = 0,5 kgm2. The total
leakage coefficient of the transmission is Ct = 5,0.10-12 m5/Ns and the viscous
friction coeff. is Bm = 0. The low pressure in the transmission is constant = pm.
a) Influence from acceleration feedback and the bandwidth of the pump
controller:
Show with equations how the acceleration feedback (Kac) will influence the
hydraulic resonance frequency (ωh) and the damping (δh) of the angular speed
servo.
Describe qualitatively how the break frequency of the pump controller (ωs) will
influence the damping of the hydraulic resonance frequency.
(6p)
b) Calculation of the acceleration feedback gain, Kac:
Calculate the acceleration gain factor (Kac) so that the hydraulic damping of the
system reaches the value δ 'h = 0,30 at the frequency ωh, and when the dynamics
of the pump controller is included.
(4p)
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5.
6(6)
EXAMINATION
TMHP 51
2002-12-20
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
KL
be Ap
Ap be
p1 V1
V2 p2
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
s 2 2δ a
+
s +1
ω a2 ω a
The transfer function of the mechanics is G LX ( s ) = 2
, (ωa < ω1)
2δ 1
s
+
s +1
ω12 ω1
a) The transfer function Gh(s) = sXp/i of the hydraulic system
Derive an expression of Gh(s) = sXp/i and show how the transfer function can be
simplified if the hydraulic spring coefficient Kh = 4β eAp2/Vt is much less than the
mechanical spring coefficient KL (ωh << ωa).
(5p)
b) The system loop gain if the mass M1 is neglected:
Assume that M1 << M2 and derive an expression for the position servo loop gain
Au(s), with definitions of ωh and δh.
(5p)
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1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2002-12-20
SOLUTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices:
Data: d = 1,0 mm, ps = 200 bar and ρ = 870 kg/m3.
2
q = C q A0
ρ
∆p gives C q =
q
A0 2∆p / ρ
The diagram gives Cq = 0,79 for pd
= 150 bar.
Cq
1,0
0,8
pd < 75 bar gives
C q ∝ 1 / ∆p
0,6
.
ps - pd
0
100
200 [bar]
(3p)
b) Reduction of flow forces and radial friction forces.
A
B
See compendium for spool design.
T
P
T
(3p)
c) Flow forces in spool valves: Data: qvN = 40 litre/min with ∆pvN = 7,0 MPa,
qpmax = 50 litre/min and pp = 21 MPa.
Calculate xv/xvmax for max. flow forces: Max. flow forces appear when the valve
flow is equal to the max. pump flow, qv = qpmax. q vN = C q wxv max ∆p vN / ρ and
q v = q p max = C q wx v p p / ρ gives
xv
xv max
=
q p max
q vN
∆p vN
= 0,72 .
pp
Fs
Fsmax
qv
pp qp
xv
pT = 0
0
xv/xvmax
0,72
1,0
Constant ∆p means Fs ∼ xv and constant flow means Fs ∼ 1/xv.
(4p)
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a) A single stage servo valve:
Proportional
magnet
A
B
P
T
The valve design means that the spool position will be disturbed by flow forces
and friction forces. In a position servo acts as a positive load pressure feedback,
which will reduce the hydraulic damping in the system.
By using a position feedback for the spool control the disturbance from flow
forces can be reduced, which gives higher hydraulic damping.
(3p)
b) Stiffness of a position servo with valve controlled cylinder:
ö
Vt
DFL Kce æç
÷
ç1 + 4 b K s÷ 2
Ap è
e ce ø
.
1
s 2 + 2 dh s +
1
wh2 wh
Kqi
1
Ap 1 + s
wv
Saturation
xp
DXp
1
s
Threshold
imax
ein
Ksa
Kf
Bode-diagram of the closed loop stiffness S c = −∆FL / ∆X p .
2
Amplitude, (Sc/Ks)
10
1
10
0
10
−1
10
0
10
1
2
10
3
10
10
Frequency [rad/s]
200
150
Phase
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2002-12-20
100
50
0
0
10
1
2
10
3
10
10
Frequency [rad/s]
= Kv
A p2
where
K ce
K v = K f K sa K qi / A p . Increased Kc-value (Kce) means reduced steady state
stiffness. However, if the hydraulic damping is just proportional to the hydraulic
damping (δh) the dip in dynamic stiffness at the frequency ωh will be reduced by
the increased Kc-value to the same amount as the reduction in steady state
stiffness, which means nearly constant absolute value of the dynamic stiffness.
(4p)
The block-diagram gives the steady state stiffness S c
s→0
c) Threshold and saturation in the servo valve:
Threshold: The block-diagram gives ∆X p K f K sa = εi n or ∆X p =
εin
K f K sa
.
Xp
Saturation gives constant velocity
Saturation and step response:
0
Time
(3p)
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3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2002-12-20
Linear position servo with velocity feedback
xp
.
xp
Kf
Uc
+
Kfv
-
Ksa
+
Bp = 0
Ap
V1
Threshold
Mt
V2
FL
ein
Parameter values:
A1 = 1,96.10-3 m2
Vt = 1,0.10-3 m3
Kc0 = 8,0.10-12 m5/Ns
Mt = 1250 kg
ps = 21 MPa
Kfx = 20 V/m
β e = 1200 MPa
Kqi0 = 2,0⋅10-3 m3/As
Ksa = 0,56 A/V
a) The velocity feegback gain, Kfv:
With velocity feedback the undamped resonance frequency is ωh' = 296 rad/s.
Show with equations how the velocity feedback will influence the resonance
frequency (ωh) and for the actual operating case.
s2
ω h2
+
2δ h
ωh
s +1 →
s2
ω h2
Calculate Kfv. ω h =
+
2δ h
ωh
4 β e A p2
Vt M t
s + 1 + K fv K sa
K qi
Ap
and ω h' = ω h ⋅ 1 + K fv K sa
K qi
Ap
.
'
= 121,5 rad/s and ωh = 296 rad/s gives
K qi
K qi
ω h'
= 2,44, which gives 1 + K fv K sa
= 5,94 and
= 1 + K fv K sa
Ap
ωh
Ap
K fv = (5,94 − 1)
Ap
K sa K qi
= 8,6 Vs/m.
(5p)
b) Position error from threshold-value (εin) of the servo valve:
Without velocity feedback the valve threshold gives the error ∆Xp = 1,0 mm.
K qi
Kf .
Without velocity feedback: K v = K sa
Ap
With velocity feedback: K vv = K sav
K qi
A p K vfv
K f where K vfv = 1 + K fv K sav
From task a Kvfv = 6. Position error caused by threshold is ∆X p =
Equal gain means Kv = Kvv ⇒ K sa =
K qi
Ap
εin
K f K sa
.
∆X p 1,0
K sav
=
and ∆X pv =
= 0,17 mm.
K vfv
6
K vfv
(5p)
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4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2002-12-20
Electro-hydraulic speed servo with pump controlled motor
p1 V1
ep
wp
uc +
Ksa
____
uf
s
+
Dm
Dp
qm
p2 V2
i
-
Varvtalsgivare
pm
Accelerationssignal
Kac.s
TL
Jt
.
Kf
The pump displacement controller has the transfer function
εp
i
=
K pi
1+ s / ω s
.
Parameters: Kpi = 20 A-1, ωs = 50 rad/s, ωp = 157 rad/s, Dp = 5,6.10-6 m3/rad, V1
= V2 = 0,5 litre, β e = 1000 MPa, Dm = 19.10-6 m3/rad, Jt = 0,5 kgm2, Ct = 5,0.1012 m5/Ns, B = 0 and p = constant.
m
m
a) Influence from acceleration feedback and the bandwidth of the pump
controller: If TL is neglected the block-diagram will be as follows:
Ksa
____
uc
+
-
s
Kf
+
i
-
Kac.s
Speed feedback
Acc. Feedback gives that
s2
ω h2
+
1
s 2 + 2 dh s +
1
wh2 wh
KpiDpwp
(1+s/ws) Dm
2δ h
ωh
.
qm
Acceleration feedback
s +1 →
 2δ h K ac K pi D pω p 

s + 1
+
+
ω h2  ω h (1 + s / ω s )Dm 
s2
'
That means no influence on ωh and the damping is: δ h = δ h +
ω h K ac K pi D pω p
2 (1 + s / ω s )Dm
If the bandwidth of the pump (ωs) controller is lower than ωh the damping effect
is reduced.
(6p)
b) Calculation of the acceleration feedback gain, Kac for δ 'h = 0,30 at ωh:
ω = ωh gives δ h'
C
δh = t
2 Dm
K ac
βeJt
V1
ω =ω h
= δh +
ωh
K ac K pi D pω p
2
1 + (ω h / ω s ) Dm
= 0,132 and ω h =
2
β e Dm2
V1 J t
2 D 1 + (ω h / ω s )
=  δ h'
− δ h  m
 ω =ω h

ω h K pi D pω p
= 0,30 , where
= 38 rad/s. ⇒
2
4,77 ⋅ 10 −5
⇒ Kac = (0,30 − 0,132)
1,2⋅10-5 A/s2.
0,668
(4p)
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Fluid and Mechanical Engineering Systems
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2002-12-20
Linear position servo with valve controlled cylinder and two masses
xp
Ap
KL
Ap
p1 C1
M2
M1
C2 p2
Position
transducer
BL
xL
Kf
Servo
uc + amplifier
Ksa
uf -
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
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 )
where
and Vt/2 = V1 = V2.
Vt
Kh = 4β eAp2/Vt << KL (ωh << ωa) ⇒ G LX ( s ) ω <ω = 1 och Gh ( s) =
a
K qi / Ap
s

2δ
 2 + h s + 1
ω

 h ωh

2
(5p)
b) The system loop gain if the mass M1 is neglected:
M1 << M2 leads to a 1 DOF load system, which means G LX ( s ) = 1 . According to
task a), Gh(s) = sXp/i is expressed as Gh ( s) =
ωh =
K qi / Ap
s

2δ
 2 + h s + 1
ω

 h ωh

2
where
Ke
K M ω
Vt
1
1
, δ h = ce 2 2 h and
=
+
2
M2
2
Ap
K e 4 β e Ap K L
The position servo loop gain is : Au (s) =
K sa K qi K f / Ap

 s 2 2δ h
 2 +
s + 1s
ω

 h ωh
(5p)
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Fluid and Mechanical Engineering Systems
1.
2(6)
EXAMINATION
TMHP 51
2001-12-19
a) Cavitation in orifices: I hydraulic systems cavitation will occur downstream
orifices with large pressure drops. The figure below shows measurements made
on a circular orifice with a diameter of d = 1,0 mm. The supply pressure have
been kept constant at the levels 50, 100, 150 and 200 bar and the downstream
pressure has been varied from supply pressure down to ”zero-pressure”. The rings
shows when cavitation starts. The oil density is ρ = 870 kg/m3.
Assume that the flow through the orifice is turbulent and calculate approximately the flow coefficient Cq.
The pressure difference required to initiate cavitation can be described as: ∆pcav =
Ccav⋅pu where pu is the upstream pressure. Calculate Ccav for the actual orifice and
study if Ccav will vary depending on the pressure pu.
(4p)
b) Viscosity and bulk modulus are important physical parameters for hydraulic
fluids. Compare qualitatively water and mineral oil with respect to these
parameters and their influence on system behaviour.
(3p)
c) Flow forces in spool valves: The figure below shown a 4-way critical center
servo valve connected to a pressure controlled pump. A diagram shows the flow
forces versus valve displacement, xv (with constant load pressure difference) when
the maximum flow capacity of the valve is bigger than the pump capacity.
Show with equations and diagram the influence on flow forces (Fs) from an
increased pump pressure (pp) and an increased pump shaft speed (np).
xv
pp
qp
Fs
Load
np
pT = 0
Fs
0
xv
xvmax
(3p)
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2.
3(6)
EXAMINATION
TMHP 51
2001-12-19
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)
b) Flow forces in a direct controlled servo valve: The figure shows a block
diagram of a position servo with a single-stage servo valve with a mechanical
stiffness for the magnet armature (Ka) and a proportional controller. The feedback
of the load pressure pL describes the influence from flow forces on the valve
control. Describe with equations how the flow forces will influence the
stability of the servo system.
Kfp
uc
+
-
Ksa
i K xa K
m
a
+
+
xv
1
Kq +
Ka+Kfx
-
1
V
Kce + t s PL
4be
Ap +
FL
-
1
Mt s
.
xp
1_ xp
s
Ap
Kf
(2p)
c) The figure below shows a valve controlled and a pump controlled motor. Both
systems are used as rotating position servos with proportional control. Volumes,
bulk modulus, motor displacement and load inertia are equal in both systems.
Servo valve and pump controller has high bandwidth.
Assume that in the most critical operating point both systems have the same
flow/pressure-coefficient (Kce = Ct) and the same amplitude margin, Am = 6 dB.
Show with equation which of the system as will given the highest bandwidth,
ωb. (The bandwidth of the closed loop system is asked for).
-
Uc Ksa
+
Kf
Uc Ksa
V1
ep
+
Dm
qm
V2
Kf
-
Jt
TL
wp
p1 V1
Dm
Dp
pm =
konst
qm
Jt
TL
p2 V2
(5p)
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3.
EXAMINATION
TMHP 51
2001-12-19
4(6)
Position servo with an asymmetric cylinder
The figure shows an electro-hydraulic position servo with a 3-port valve and an
asymmetric cylinder loaded by a mass and an external force. The servo has
proportional position control with a feedback gain of Kfx and a controller gain of
Kreg. The servo valve is zero-lapped and it has symmetric and matched orifices
and the null-coefficients are Kqi0 and Kc0. The valve has a high bandwidth but has
a threshold, iT. The supply pressure, ps is constant. The piston area is A1, the area
on the piston rod side is A1/2 and the cylinder volume on the piston side is V1.
The cylinder is assumed to have no leakage and friction. The bandwidth of the
closed loop servo system is ωb (with the amplitude –3 dB).
Parameter values:
A1 = 1,96.10-3 m2
V1 = 0,25.10-3 m3
Kc0 = 8,0.10-12 m5/Ns
iT = 0,4 mA
Mt = 200 kg
ps = 14 MPa
Kfx = 25 V/m
βe = 1200 MPa
Kqi0 = 0,013 m3/As
ωb = 18 rad/s
a) Steady state position error: Calculate the maximum position error (∆Xp)
which can be caused by a load disturbance ∆FL = 800 N and together with
threshold of the valve, iT.
(6p)
b) Lag-compensation of the control loop: Assume that the proportional
position controller (Kreg) is extended by a lag filter with the transfer function
1 + s / ω LC
. The steady state gain α = 2,4. Calculate for this case
G LC ( s ) = α
1 + s ⋅ α / ω LC
the position error caused by a load disturbance only, according to task a).
(2p)
c) Influence from a servo valve with slow response: Discuss in general how
the bandwidth of the servo system will be influenced if the valve response is
slower that the dynamics of the actuator and load.
(2p)
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4.
5(6)
EXAMINATION
TMHP 51
2001-12-19
Electro-hydraulic force servo
Kff
Force feedback
Au(s)
uc
-
Ksa
+
i
Gv Kqi
+
-
1
FL
-
Ap
V
Kce + t s PL
4be
Fh +
1
Mt s
.
xp
Ap
The figure shows a block diagram for a linear force servo with a valve controlled
cylinder loaded by a mass, Mt. The servo cylinder is symmetric with the piston
area Ap. The controller is just proportional with the transfer function Greg = Ksa.
The servo valve is a 4-way critical center valve. The dynamics of the valve is
described by the function Gv.
Data:
Ap = 2,4·10-3 m2
Kc0 = 4,0·10-12 m5/Ns
Mt = 1500 kg
ps = 14 MPa
βe = 1,2·109 Pa
Kqi0 = 0,020 m3/s/A
Vt = 1,0·10-3 m3
Gv = 1/(1 + s/ωv)
a) The control loop gain, Au(s): Assume that FL = 0 and derive an expression of
the loop gain Au(s).
Calculate the hydraulic damping δh and the resonance frequency ωh.
(7p)
b) Mechanical spring on the load and its influence on the resonance
frequency:
Fix reference
be
Ap
P1 V1
Ap
xp
be
Mt
V2 P2
KL
Force
transducer
xv
Servo
uc + amplifier
i
Ksa
u
f
Kff
Ps = const.
Assume that the force servo is connected to a mechanical spring (see figure) with
the coefficient KL = 2,8⋅107 N/m. Show with equations how KL will influence
the hydraulic resonance frequency ωh and calculate the actual frequency.
(3p)
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Fluid and Mechanical Engineering Systems
5.
6(6)
EXAMINATION
TMHP 51
2001-12-19
Angular position servo with valve controlled hydraulic motor
The figure shows an angular position servo with 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 coefficient BL).
The controller is of proportional type (Greg = Ksa) and in order to increase the
hydraulic stiffness a velocity feedback is introduced (Kfv⋅s). The servo valve can
be assumed as having very fast response. The figure also shows a block diagram
describing the hydraulic system and the load.
Kfv.s
Kf
KL
J1
Dm
qm
uc
Kq
Xv ___
Dm +
-
+
Ksa
-
J2
BL
qL
i
+
Dm
__________
Vt
Kce + ___
s
4be
PL
Dm
1 G
________
Lq(s)
(J1 + J2) s
.
1 qm
qm ___
s
s 2 2δ a
+
⋅ s +1
ω a2 ω a
G LΘ ( s ) = 2
2δ
s
+ 1 ⋅ s +1
2
ω1 ω1
a) Derive the linearised and laplace transformed transfer function Gh ( s) =
θm
.
Xv
Draw in principle a bode-diagram for Gh(s) under the assumption that the
hydraulic resonance frequency is the lowest frequency, ωh < ωa < ω1.
(6p)
b) Velocity feedback and its influence on the position servo frequencies:
Assume that the hydraulic torsional constant Kh = 4βtDm2/Vt is much lower than
the mechanical torsional constant KL (ωh < ωa). Show the influence the velocity
feedback has on the hydraulic resonance frequency ωh and discuss if this is
positive or negative for the stability of the position servo.
(4p)
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1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2001-12-19
SOLUTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices:
Data: d = 1,0 mm, pu = 50, 100, 150 and 200 bar, ρ = 870 kg/m3.
Orifice flow equation: q = C q A0
⇒ Cq =
q
C cav =
ρ
∆p
The diagram gives Cq = 0,7.
A0 2∆p / ρ
Calc. Ccav: C cav =
2
∆p cav
200 − 74
= 0,63 , pu = 50 bar ⇒
. pu = 200 bar ⇒ C cav =
pu
200
50 − 14
= 0,72 . Ccav increases with decreased pu.
50
(4p)
b) Viscosity and bulk modulus: The viscosity of water and mineral oil will
increase with reduced temperature. However a mineral oil is much more sensitive
to temp. than water. The low viscosity of water gives high leakage flow and pore
lubrication properties. The bulk modulus depends upon fluid density. Therefore,
water has higher bulk modulus than mineral oil. This gives higher pressure
transients for flow disturbances in water hydraulic systems, but also lower
compression work and faster response.
(3p)
c) Flow forces in spool valves:
Show with equations and diagram the influence on flow forces (Fs) from an
increased pump pressure (pp) and an increased pump shaft speed (np).
ρ cos(δ ) 2
Flow force equation: Fs = 2C q wxv ∆p cos(δ ) or Fs =
q . Constant ∆p
C q wxv
means Fs ∼ xv and constant flow means Fs ∼ 1/xv.
Fs
xv
pp
qp
Dp increases
Load
q increases
np
pT = 0
Fs
xv
0
xvmax
(3p)
1
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2001-12-19
a) Valve wear and its influence on valve coefficients:
Wear on the orifice edges in a zero-lapped servo valve will increase the leakage
flow around neutral spool position (xv = 0). That means increased Kc-value,
which leads to an decreased pressure gain, Kp.
For a position servo with proportional controller gain the steady state stiffness is
A p2
Ap
S s →0 = K v
= K f K sa K qi 0
= K f K sa A p K p 0 . Decreased Kp0 means
Kc
K c0
decreased stiffness.
(3p)
b) Flow forces in a direct controlled servo valve:
Describe with equations how the flow forces will influence the stability of
the servo system. Study how the pressure feedback will influence the Kce-value.
Kfp
+
xv
1
Kq +
Ka+Kfx
1
Kce +
Vt
s PL
4be
K fp K q
⇒ K ce* = K ce −
K a + K fx
the reduction of Kce means reduced damping and stability.
. Since δ h ∝ K ce*
(2p)
c) Bandwidth for different servo systems:
-
Uc Ksa
+
Kf
Uc Ksa
V1
ep
+
Dm
qm
V2
Kf
-
Jt
TL
wp
p1 V 1
Dm
Dp
pm =
konst
qm
Jt
TL
p2 V2
Kv
For both systems the open loop gain is Au (s ) =
. With this

 s 2 2δ h
 2 +
s + 1 s
ω

 h ωh
transfer function Am = 6 dB gives Kv = δh⋅ωh and the bandwidth is ωb = Kv. To
find the highest bandwidth, δh⋅ωh for the two systems have to be compared.
K
Valve controlled motor (V1 = V2): δ hω h = ce
Dm
C
Pump controlled motor (V1 = V2): δ hω h = t
2 Dm
βe Jt
2V1
2 β e Dm2
β
= K ce e
V1 J t
V1
βe Jt
β e Dm2
V1
V1 J t
=
K ce β e
2 V1
This comparison gives (Kv)valve = 2(Kv)pump ⇒ (ωb)valve = 2(ωb)pump.
(5p)
2
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2001-12-19
Position servo with an asymmetric cylinder
A1 = 1,96.10-3 m2
V1 = 0,25.10-3 m3
Kc0 = 8,0.10-12 m5/Ns
iT = 0,4 mA
β e = 1200 MPa
Kqi0 = 0,013 m3/As
ωb = 18 rad/s
Mt = 200 kg
ps = 14 MPa
Kfx = 25 V/m
a) Steady state position error (∆Xp) for ∆FL = 800 N and threshold
Position error from stiffnes: − ∆FL
∆X p
= Kv
s→o
A12 . K = ω ⇒ ∆X
v
b
p
K c0
Position error from threshold: ∆X p K fx K reg = iT ⇒ ∆X p
Find Kreg? ω b = K v = K fx K reg
∆X p = ∆X p
F
+ ∆X p
T
=
K qi 0
A1
, K reg =
ω b A1
K fx K qi 0
T
⇒ ∆X p
=
T
F
=
∆FL K c 0
.
ω b A12
iT
.
K fx K reg
=
iT K qi 0
ω b A1
.
∆FL K c 0 iT K qi 0
⇒ ∆Xp=9,3⋅10-5+1,47⋅10-4= 0,24 mm
+
2
ω b A1
ω b A1
(6p)
b) Lag-compensation of the control loop: Assume that the proportional
position controller (Kreg) is extended by a lag filter with the transfer function
1 + s / ω LC
. The steady state gain α = 2,4.
G LC ( s ) = α
1 + s ⋅ α / ω LC
Lag-compensation shall be active for frequencies lower than ωb. That means
∆FL K c 0
increased steady state gain KvLag = α⋅Kv. ⇒ ∆X p =
= 3,86 ⋅ 10 −5 m .
2
F
αω b A1
(2p)
c) Influence from a servo valve with slow response:
Low bandwidth of the valve compared to actuator and load dynamics means ωv <
ωh. Design for stability gives Kv = ωv. So if ωv < δh ωh, Kv must be reduced
compared to a fast valve. This means reduced bandwidth as well as stiffness
for the closed servo system.
(2p)
3
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2001-12-19
Electro-hydraulic force servo
Ap = 2,4·10-3 m2
β e = 1,2·109 Pa
Kc0 = 4,0·10-12 m5/Ns
Kqi0 = 0,020 m3/s/A
Mt = 1500 kg
Vt = 1,0·10-3 m3
ps = 14 MPa
Gv = 1/(1 + s/ωv)
a) The control loop gain, Au(s) for FL = 0:
Data:
Kff
Force feedback
Au(s)
Block Diagram:
uc
-
i
Ksa
+
Gv Kqi
Reduction of lower loop gives: G h ( s ) =
G h (s) =
M t / Ap ⋅ s
s2
ω h2
+
2δ h
ωh
+
-
1
Kce +
Vt
s PL
4be
1
V
K ce + t s
Ap
4β e
+
Ap
Mts
=
M t / Ap ⋅ s

V
 K ce + t
4β e

. The loop gain is: Au ( s) = K ff K sa Gv K qi
βeM t
Vt
and ω h =
Fh
Ap
Mt s
s +1
K
where δ h = ce
Ap
⇒
Ap
M s
s  t2 + 1
 Ap
⇒
M t / Ap ⋅ s
s
2
ω
2
h
+
2δ h
ωh
s +1
4 β e A p2
Vt M t
Numerically: δh = 0,071 and ωh = 136 rad/s
(7p)
b) Mechanical spring on the load and its influence on the resonance
frequency:
Fix reference
be
Ap
P1 V1
Ap
xp
be
Mt
V2 P2
KL
Force
transducer
xv
Servo
uc + amplifier
i
Ksa
u
f
Kff
Ps = const.
The mechanical spring is in serial connection with the hydraulic springs in the
4 K L β e A p2
Ke
Vt
1
1
cylinder. This gives: ω h =
, where
=
+
⇒ Ke =
Mt
K e K L 4 β e A p2
4 β e A p2 + K LVt
KL = 2,8⋅107 N/m gives Ke = 1,4⋅107 N/m and ωh = 96 rad/s
(3p)
4
LINKÖPINGS UNIVERSITET
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Fluid and Mechanical Engineering Systems
Angular position servo with valve controlled hydraulic motor
Kfv.s
Kf
KL
J1
Dm
J2
BL
qm
uc
Kq
Xv ___
Dm +
-
-
Ksa
+
Dm
__________
Vt
Kce + ___ s
4be
qL
i
+
PL
1 qm
qm ___
s
1 G (s)
________
Lq
(J1 + J2) s
Dm
s2
.
+
2δ a
⋅ s +1
ωa
G LΘ ( s ) =
2δ
s
+ 1 ⋅ s +1
2
ω1 ω1
ω
2
a
2
a) Derive Gh ( s) = θ m / X v .
The block diagram gives: Gh ( s ) =
ωh =
4 β e A p2
(J 1 + J 2 )Vt
, δh =
K ce
Dm
θm
=
Xv
K q / D m ⋅ G Lθ ( s )
s

2δ
 2 + h s + G Lθ ( s )  ⋅ s
 ωh ωh

2
where
β e (J 1 + J 2 )
Vt
Bode-diagram for Gh(s) when, ωh < ωa < ω1.
101
-90
100
-140
Phase
Amplitude
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2001-12-19
10-1
10-2
wh
10-3
100
-240
w1
wa
102
Frequency [rad/s]
-190
-290
100
102
Frequency [rad/s]
(6p)
b) Velocity feedback and its influence on the position servo frequencies:
Kh = 4β tDm2/Vt << KL (ωh < ωa) implies that GLθ(s) = 1. The block diagram will
uc
+
-
Ksa +
iv
-
be as:
Kfv.s
Kf
s2
ω h2
+
2δ h
ωh
s +1 →
s2
ω h2
+
2δ h
ωh
1
s 2 + 2 dh s +
1
wh2 wh
Kqi
Dm
s + 1 + K fv
1
__
s
qm
Velocity feedback
Angular position feedback
K qi
Dm
and ω h* = ω h ⋅ 1 + K fv
K qi
Dm
. ωh will
be increased by the velocity feedback and if ωh comes very close to the load
dynamics the stability will be affected.
(4p)
5
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1.
2(6)
EXAMINATION
TMHP 51
2000-12-16
a) Orifices: Show in a diagram the flow coefficient (Cq) versus the Reynolds
number ( Re ) for the following type of orifices:
- sharp-edged orifice (with ideally zero length)
- short tube orifice
- long tube orifice
If an orifice in a hydraulic system is exposed for a too high pressure drop cavitation will appear. Assume that a sharp-edged orifice has a constant supply pressure
of p1 = 70 bar. Cavitation is indicated when the outlet pressure is reduced to p2 =
25 bar. Assume that the supply pressure is changed to p1 = 140 bar. On which
value of outlet pressure p2 will than cavitation occur?
(4p)
b) Hydraulic oils: Describe qualitatively how pressure and temperature will
influence the viscosity of a mineral oil.
(2p)
c) Flow forces: The figure below shown a 4-way critical center servo valve
connected in two different ways to a pressure controlled pump. The maximum
pump flow is 150 litre/min and the pressure level is adjusted to 21 MPa. The
pump is assumed to have an ideal characteristic and leakage and pipe losses are
neglected. The servo valve has symmetrical and matched orifices. When the valve
is connected as shown in the left system below, the maximum flow capacity is
qvmax = 200 liter/min (when the load pressure drop is zero and xv = xvmax).
xv
pp
qp
pT = 0
xv
Load
pp
qp
Load
pT = 0
Calculate the relative spool displacement (xv/xvmax), which will give the highest
steady state flow forces in the two different systems. Assume that the load
pressure drop is zero(0).
Show in a diagram (in principle) how the flow forces vary with the spool
displacement inside the interval 0 ≤ xv ≤ xvmax for the two cases.
Data: Flow coefficient Cq = 0,67, oil density ρ = 860 kg/m3, jet angle δ = 69o.
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
3(6)
EXAMINATION
TMHP 51
2000-12-16
a) Describe qualitatively in which respects a two-stage servo valve can show a
better steady state characteristics (qL versus i) than a one-stage (direct controlled)
servo valve.
(2p)
b) The figure below shows an electrical controlled servo valve, which is
connected to a test stand and equipped with transducers for measurements of
valve command signal, pressures and flow. The servo valve is a 4-way one with
critical center.
Measurement signals
Potentiometer for
zero point adjust.
Uc
p2
p1
Servo amplifier
iv
ps = constant
Your task is to adjust the zero point of the valve and determine the pressure gain
(Kp). Describe how you want to solve the task.
(3p)
c) 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 coeff. 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).
xp
Ap
Kf
Uc
V1
Mt
V2
Kf
FL
K
__sa
Uc s
nm
V1
Dm
Ksa
Jt
TL
V2
FL
ö
Vt
Kce æç
s÷
1+
Ap2 çè 4 be K ce ÷ø
uc
+
-
Greg
iv
-
Kqi
Ap
+
Kf
.
1
xp 1 xp
s 2 + 2 dh s +
s
1
wh2 wh
Kf
(5p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4(6)
EXAMINATION
TMHP 51
2000-12-16
3.
xL
xp
Kf
-
Uc
V1
Ap V2
p1
p2
KL
Mt
FL
Ksa iv
+
Kqi
iv ___
Ap +
-
Ap
__________
Vt
Kc + ___ s
4be
PL
Ap
1 Mt s 2
+1
Mt s K L
sXp
The figure above shows a linear position servo with a valve controlled cylinder
loaded by a mass and an external force. A block diagram of the hydraulic part of
the system is also shown. The piston rod has a weakness, which is modelled by a
spring constant, KL. The controller is of proportional type (Greg = Ksa). The servo
valve is a 4-way zero-lapped valve. The cylinder is assumed to have no leakage
and friction and also the mass of the piston and rod is neglected.
Data:
Ap = 2,4·10-3 m2
Kc0 = 4,0·10-12 m5/Ns
Mt = 1500 kg
KL = 3,0·107 N/m
Kf = 25 V/m
βe = 1,3·109 Pa
Kqi0 = 0,020 m3/s/A
Vt = 1,0·10-3 m3
ps = 14 MPa
a) Show how the hydraulic resonance frequency of the system (ωh) and the
damping (δh) depends upon the mechanical spring constant KL and calculate ωh
and δh when KL is included. (Use the given block diagram.)
(5p)
b) Determine the servo amplifier gain Ksa so the amplitude margin of the servo
will be Am = 6 dB.
(2p)
c) Assume that the valve supply pressure is increased to ps = 28 MPa. Now the
pressure gain of the valve will increase to Kp = 2,0⋅1010 Pa/A, which is two times
higher that for ps = 14 MPa. Show with equations and calculate how the closed
loop steady state stiffness increases with the increased Kp-value and when Ksa is
the same as in task b).
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5(6)
EXAMINATION
TMHP 51
2000-12-16
4.
FL
ö
Vt
Kce æç
÷
ç1 + 4 b K s÷
2
Ap è
e ce ø
Threshold
uc
+
-
+
-
Ksa
ir
iv
ein
Kqi
1
Ap 1 + s +
wv
-
.
1
xp 1 xp
s 2 + 2 dh s +
s
1
wh2 wh
Kfv
Velocity feedback
Kf
Position feedback
The figure shows an electro-hydraulic linear position servo with a valve
controlled cylinder. The controller is just proportional with the transfer function
Greg = Ksa. The position feedback gain is Kf. In order to increase the system
accuracy a negative velocity feedback is introduced and the feedback gain is Kfv.
The servo valve is a 4-way critical center valve with high bandwidth and the valve
coefficients is Kqi0 and Kc0. The cylinder volumes are equal, V1 = V2 and the
effective bulk modulus is βe. The piston area is Ap and the losses in the cylinder
can be neglected. The cylinder is loaded, by the mass Mt and the external force FL.
a) Show, with help of the block diagram, how the hydraulic resonance frequency
(ωh) and the damping (δh) will be influenced by the velocity feedback (neglect the
threshold).
(6p)
b) Describe how the steady state position error will be affected by the threshold
of the servo valve.
(2p)
c) Why is it important that the bandwidth is higher with use of velocity feedback
and without this feedback?
(2p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5.
6(6)
EXAMINATION
TMHP 51
2000-12-16
The figure shows an angular velocity servo with 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 coefficient BL).
The controller is of integrating type (Greg = Ksa/s) and in order to increase the
hydraulic damping an acceleration feedback is introduced. The servo valve can be
assumed as having very fast response.
Kacc .s
Kf
KL
Dm
.
J1
BL
qm
uc
Ksa
____
+
s
-
.
J2
TL
qL
i
+
The following block diagram describe the hydraulic system and the load:
TL
Kq
Xv ___
Dm +
-
Dm
__________
Vt
Kce + ___
s
4be
1+
where
GLT(s) =
s2
PL
G LT(s)
Dm
+
s2
s
2δ1 ω1
2δa
+
s+1
2
ωa
ωa
;
.
qm
1
________
G Lq(s)
(J1 + J2) s
GLθ(s) =
ω2a
s2
+
2δa
s+1
ωa
2δ1
+
s+1
2
ω1
ω1
(ωa < ω1)
a) Assume that TL = 0 and that the hydraulic torsional const. Kh = 4βtDm2/Vt is
much lower than the mechanical torsional constant KL (ωh < ωa) in the above
block diagram. Derive, under these circumstances, the transfer function
⋅
θ
Gh ( s ) = m and show expressions for the hydraulic resonance frequency (ωh) and
Xv
the damping (δh, without acceleration feedback).
(5p)
b) Draw a block diagram for the system (with the hydraulic system and load
reduced as in task a). Show how the acceleration feedback will influence the
hydraulic damping (δh) and if the steady state stiffness of the closed loop system
will be affected or not.
(5p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2000-12-16
SOLUTIONS FOR EXAMINATION
1.
a) Orifices: Cq versus the square root of Reynolds number ( Re )
Cq
Sharp-edged orifice
Short tube
Long tube
Sqrt(Re)
Critical pressure drop when cavitation occurs downstream orifices
The critical pressure drop for cavitation depends upon the upstream pressure p1 as
∆p kav = C 2 p1 or p 2 kav = p1 (1 − C 2 ) . p1 = 70 bar and p2 = 25 bar gives C2 = (7025)/70 = 0,64.
p1 = 140 bar Þ p2kav = 140(1 – 0,64) = 50 bar
(4p)
b) Viscosity of hydraulic oils: The viscosity for a mineral oil increases with
increased pressure and decreases with increased temperature. The influence from
the temperature is much stronger than from pressure.
(2p)
c) Flow forces on valves in a system: qpmax = 150 litre/min and pp = 21 MPa.
Max flow capacity of the valve qvmax = 200 litre/min (when ∆pvtot = 21 MPa or
10,5 MPa per orifice). Cq = 0,67, ρ = 860 kg/m3 and δ = 69o.
xv
pp
qp
xv
Load
pp
qp
Load
pT = 0
pT = 0
Single metering from pump to load and ∆pL = 0: xv/xvmax = qpmax/qvmax = 0,75
Double metering from pump to load and ∆pL = 0:
21
q v 2 max = 2
q v max = 566 litre/min , gives xv2/xv2max = qpmax/qv2max = 0,265
10,5
Fs
1(p-A)
2(p-A)
xv
0
xvmax
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2000-12-16
a) Characteristics (qL versus i) for one- and two-stage servo valve resp.
Two-stage servo valves shows better linearity and lower hysteresis than one-stage
valves. The reason is that the available control force on the main spool is higher in
a two-stage than in a one-stage valve and therefore, friction and flow forces will
not disturb the spool position so much.
(2p)
b) Test set-up for a servo valve
Measurement signals
Potentiometer for
zero point adjust.
Uc
p2
p1
Servo amplifier
iv
ps = constant
The zero point of the valve is adjusted by a potentiometer and when the pressure
p1 and p2 are equal the valve is in zero position when the valve input signal is
zero. The pressure gain (Kp) is measured by plotting the load pressure difference
(p1 - p2) versus the input signal to the valve.
(3p)
c) Steady state stiffness for a position and a velocity servo
The controller gain for the position servo is Greg = Ksa and for the velocity servo
Greg = Ksa/s. The servo valves are both zero-lapped and the coeff. are Kqi and Kc.
xp
Ap
Kf
Uc
V1
Mt
V2
Kf
FL
Ksa
__
nm
V1
Uc s
Dm
Ksa
Jt
TL
V2
FL
ö
Vt
Kce æç
s÷
1+
Ap2 çè 4 be K ce ÷ø
uc
+
-
Greg
iv
-
Kqi
Ap
+
Kf
.
1
xp 1 xp
s 2 + 2 dh s +
s
1
wh2 wh
Kf
− ∆FL
Pos. servo with prop.-contr., Uc=0 gives:
∆X p
= Kv
s →0
A p2
K ce
=
K sa K qi K f A p
K ce
Velocity servo with integrating controller: Ap=Dm, FL=TL and Greg=Ksa/s gives:
K
s 2 2δ h
s +1+ v
+
2
K sa K qi K f Dm
s
Dm2
− ∆FL
− ∆TL ω h ω h
.
K
=
=
→∞
=
v
K ce s s →0
K ce s
∆X p
∆nm
K ce æ
s ö
s →0
s →0
ç1 + ÷
Dm2 çè ω1 ÷ø
(5p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2000-12-16
Linear position servo with valve controlled cylinder and mass load
xL
xp
Kf
-
Uc
Ksa
V1
Ap V2
p1
p2
KL
Mt
FL
Kqi
iv ___
Ap +
iv
+
Data: Ap = 2,4·10-3 m2
Kqi0 = 0,020 m3/s/A
KL = 3,0·107 N/m
-
Ap
__________
Vt
Kc + ___ s
4be
βe = 1,3·109 Pa
Mt = 1500 kg
ps = 14 MPa
PL
Ap
1 Mt s 2
+1
Mt s K L
sXp
Kc0 = 4,0·10-12 m5/Ns
Vt = 1,0·10-3 m3
Kf = 25 V/m
a) Derive the equation and calculate ωh and δh when KL is included
The block diag. gives: sX p
iv
ωh =
K qi é M t 2 ù
s + 1ú
ê
Vt
1
1
Ap ë K L
û
=
+
=
2
K e 4β e Ap K L
é Vt
KcM t
1 ù
2
+
+
+
1
M
s
s
ê
ú
t
2
A p2
ëê 4 β e A p K L ûú
ω K M
Ke
och δ h = h c 0 2 t Numerical: Ke = 1,5⋅107 N/m, which gives
Mt
2 Ap
ωh = 100 rad/s och δh = 0,052
(5p)
b) Determine Ksa for a servo amplitude margine of Am = 6 dB
Am = 6 dB gives K v = K sa K f
K qi 0
Ap
= δ hω h Þ K sa =
Ap
K f K qi 0
δ hω h = 0,025 A/V
(2p)
c) Steady state stiffness of the closed loop system at increased ps
The supply pressure is increased to ps = 28 MPa Þ Kp = 1,0⋅1010 Pa/A, which is
two times higher than for ps = 14 MPa.
A p2 K sa K qi 0 K f A p
− ∆FL
= Kv
=
= K p K sa K f A p . ps = 28 MPa ger:
Kc
K c0
∆X L s →0
− ∆FL
∆X L
= K p K sa K f A p = 1,0 ⋅ 1010 ⋅ 0,025 ⋅ 25 ⋅ 2,4 ⋅ 10 −3 = 1,5 ⋅ 10 7 N / m
s →0
As the steady state stiffness is proportional to Kp the stiffness will be doubled if
Kp is doubled.
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2000-12-16
Position servo with negative velocity feedback
FL
ö
Vt
Kce æç
s÷
1+
Ap2 çè 4 be K ce ÷ø
Threshold
uc
+
-
+
ir
Ksa
-
iv
ein
Kqi
1
Ap 1 + s +
wv
.
-
1
xp 1 xp
s 2 + 2 dh s +
s
1
wh2 wh
Kfv
Velocity feedback
Kf
Position feedback
a) Influence from the velocity feedback on ωh and δh
-
Ksa
UF
Kqi
1
Ap 1 + s +
wv
iv
.
1
xp
s 2 + 2 dh s +
1
wh2 wh
Velocity feedback
Kfv
Reduce the block diagram when
⋅
the valve is fast (
1
1+
⋅
Xp
UF
=
s
ωv
= 1 ), which gives
1 / (1 + K fv K sa K qi / Ap )
2δ h'
s
+
s +1
ω h' 2 ω h'
2
Xp
UF
=
1
K fv K sa K qi
2δ
s
+ h s +1+
2
Ap
ωh ωh
2
ger ω ' = ω 1 + K fv K sa K qi , δ ' = δ
h
h
h
h
Ap
1
1+
K fv K sa K qi
Ap
(6p)
b) Influence from valve threshold on the steady state position error
The relation between position error ∆Xp and the threshold εiN is: ∆X p =
εi N
.
K f K sa
The position error is proportional to the threshold.
(2p)
c) Requirements on high bandwidth of the valve with velocity feedback
The velocity feedback will increase the hydraulic resonance frequency (see task
a). To be able to fully use the velocity feedback the valve must be faster than
the actuator and load dynamics. Therefore, the bandwidth of the valve must
be higher than ω h' .
(2p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2000-12-16
Velocity servo loaded by two inertia J1 and J2
Kacc .s
Kf
KL
Dm
J1
.
qm
uc
Ksa
____
s
+
-
J2
.
BL
TL
qL
i
+
TL
Kq
Xv ___
Dm +
-
Dm
__________
Vt
Kce + ___
s
4be
PL
G LT(s)
Dm
+
.
qm
1
________
G Lq(s)
(J1 + J2) s
1+
G LT(s ) =
s
s
2δ1 ω1
2
2δa
s
+
s+1
2
ωa
ωa
;
G Lθ(s ) =
2
2
ωa
2
s
ω21
+
2δa
s+1
ωa
+
2δ1
s+1
ω1
(ωa < ω1)
a) Derive the transfer function of the hydraulic system as well as ωh and δh
TL = 0 and Kh = 4βtDm2/Vt << KL (ωh < ωa) gives G Lθ ( s ) = 1 . Block diag. gives
⋅
K q / Dm
θ
Gh ( s) = m = 2
where ω h =
2δ h
Xv
s
+
s +1
ω h2 ω h
4 β e A p2
(J 1 + J 2 )Vt
, δh =
K ce
Dm
β e (J 1 + J 2 )
Vt
(5p)
b) Draw a block diagram of the system and show the influence from the
acceleration feedback
A valve with fast response and TL = 0 gives the block diagram:
uc
+
-
+
Ksa
iv
____
s
-
Kqi
Dm
Kacc.s
Kf
1
s 2 + 2 dh s +
1
wh2 wh
.
qm
Acceleration feedback
Speed feedback
The acceleration feedback will influence the hydraulic damping:
K qi ö
K qi
ωh
s 2 2δ h
s 2 æ 2δ h
'
÷
ç
δ
δ
1
+
1
Þ
=
+
K
+
s
+
→
+
+
K
s
h
h
acc
acc
2
Dm
Dm ÷ø
ω h2 ω h
ω h2 çè ω h
Acceleration feedback will only influence the stiffness of the closed loop
system at the resonance frequency ωh.
(5p)
LINKÖPINGS Universitet
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1.
EXAMINATION
TMHP 51
2004-12-15
2(6)
a) Cavitation in orifices
Three equal orifices, with the restrictor area As = 1,13·10-6 m2, are connected
according to the figure below. The inlet pressure to the first orifice p1 and the
outlet pressure of the two orifices in parallel is p2 = 1,0 MPa. The pressure
between the serial connected orifices is pm. For each orifice the critical pressure
drop for cavitation can be calculated as: ∆p cav = 0,62 ⋅ pin , where pin is the inlet
pressure.
Calculate the max allowed inlet pressure p1 for cavitation free flow.
Leading: The pressure drop over the first orifice is much bigger than over the two
orifices in parallel.
(4p)
b) Flow forces on a valve in a constant pressure system
The figure shows a servo valve supplied by a constant pressure controlled pump.
The diagram shows how the flow forces varying according to the valve spool
displacement xv. Max flow forces is found at xv = 0,6⋅xvmax when ∆pv = 14 MPa
and it’s value is Fsmax = 65 N.
Derive an expression for Fsmax as a function of qv and ∆pv. Calculate Fsmax when
∆pv is increased to 28 MPa and the maximum pump flow is constant.
(4p)
c) Bulk modulus for water and mineral oil
The bulk modulus is an important physical parameter for hydraulic fluids.
Compare qualitatively water and mineral oil with respect to this parameter and
its influence on the dynamics of a servo system.
(2p)
LINKÖPINGS Universitet
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
3(6)
EXAMINATION
TMHP 51
2004-12-15
a) Characteristics of single- and two-stage servo valves
The figure below shows a single-stage and two-stage servo valve.
Compare qualitatively the valves performances concerning linearity, influence
from flow forces and bandwidth.
(2p)
b) Pressure sensitivity for a servo valve
The diagram shows a measured pressure sensitivity characteristic of a critical
center 4-port servo valve with a supply pressure of ps = 14 MPa.
Describe in a diagram and show with equations how the pressure sensitivity of
the valve (Kp0) will be changed according to increased supply pressure, ps = 28
MPa and if wear in the valve double the Kc0-value of the valve.
(4p)
c) Steady state stiffness of a valve controlled position servo
The figure below shows a position servo with a valve controlled cylinder. The
servo controller has proportional gain (Ksa).
Assume that the valve in task b (above) is used in the position servo.
Show with equations how the closed loop steady state stiffness, |-∆FL/∆Xp|s→0
will be influenced by increased ps and valve wear respectively.
(4p)
LINKÖPINGS Universitet
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
4(6)
EXAMINATION
TMHP 51
2004-12-15
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 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.
Both cylinder volumes are assumed to be equal, V1 = V2 = Vt/2 and the effective
bulk modulus is β e. The piston area is Ap and the cylinder can be regarded as loss
less. The cylinder is loaded by the mass Mt and an external force FL.
a) Bandwidth of the systems with equal amplitude margin, Am
Compare theoretically the bandwidth (ωb) of the two systems when the steady
state loop gain (Kv) in both cases are adjusted to give equal amplitude margin, Am
= 6 dB.
(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 Universitet
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
5(6)
EXAMINATION
TMHP 51
2004-12-15
Elektro-hydraulic velocity servo with acceleration feedback
Kf
Uc
+
Kac
Ksa
s
+
V1
ps
..
qm
.
qm
Jt
pL
TL
V2
The figure shows an electro-hydraulic angular velocity servo with a valve
controlled motor. The regulator is of integrating type with Greg = Ksa/s and the
gain factor Ksa = 0,04 A/V. In order to increase the damping an acceleration
feedback is implemented with the gain Kac. . The servo valve is a 4-port critical
center valve with high bandwidth and its null-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 the effective bulk modulus is β e = 1000 MPa. The motor
displacement is Dm = 6,4.10-6 m3/rad and the inertia on the motor shaft is Jt = 0,5
kgm2. The leakage coefficient of the motor is Ctm = 8,0.10-13 m5/Ns and the
viscous friction coefficient is Bm = 0.
a) Increased hydraulic damping from the acceleration feedback
Calculate the acceleration feedback gain, Kac, so that the total hydraulic
damping will reach a value of δh' = 0,5 at the frequency ωh.
(6p)
b) Acceleration feedback and its influence on the dynamic stiffness
Show with equations and bode-diagram (amplitude curve) how the acceleration
feedback will influence the closed loop stiffness at the resonance frequency ωh.
(4p)
LINKÖPINGS Universitet
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5.
EXAMINATION
TMHP 51
2004-12-15
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.
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) Incerased damping between the two inertia loads
Assume that the viscouse friction coefficient (BL) between the inertias increases
about 10 times.
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)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
SOLUTIONS FOR EXAMINATION
TMHP 51
2004-12-15
1(5)
SOLUTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices
Outlet pressure p2 = 1,0 MPa. The critical pressure drop for
cavitation depends upon the upstream pressure p1 as ∆pcav = 0,62 ⋅ pin .
Flow eq., q = Cq ⋅ As 2 ( p1 − pm ) = 2Cq ⋅ As 2 ( pm − p2 ) → p1 − pm = 4( pm − p 2 )
δ
δ
p + 4 p2
which gives: pm = 1
. Assuming that the first orifice has a much higher
5
pressure drop than the second ones means that the first orifice is the most critical.
( p1 − pm )cav =  p1 − p1 + 4 p2  = 0,62 ⋅ p1 → p1cav = 4 p2 , p1cav = 4,44 MPa.
5
0,9

 cav
(4p)
b) Flow forces on a valve in a constant pressure system
Steady state flow forces: Fs = 2C q ⋅ w ⋅ xv ⋅ ∆pv ⋅ cos δ . ∆pv = pp and pL = 0 gives
the valve flow as qv = C q w ⋅ xv
∆pv
ρ
. Finally Fs max = 2 ρ cos δ ⋅ qv max ∆pv max .
Constant qpmax = qvmax and increased ∆pv to 28 MPa ⇒ Fs max = 65
28
= 92 N
14
(4p)
c) Bulk modulus for water and mineral oil
The bulk modulus (βe) of water is about 2 times higher than for mineral oil. In a
servo system the resonance frequency and damping for an actuator volume and
load is given by the equations. ωh =
β e Ap2
Vp M t
, δh =
K ce
2 Ap
βeM t
Vp
. This means that
water makes the system stiffer with increased resonance frequency and damping.
(2p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2004-12-15
a) Characteristics of single-stage and two-stage servo valves
A single-stage valve design means that the spool position will be disturbed by
flow and friction forces because of the limited forces from the control magnet. A
two-stage valve has much higher control forces on the spool, which means more
linear characteristic, less influence from flow forces and higher bandwidth.
(2p)
b) Pressure gain (sensitivity) for a servo valve
Pressure sensitivity of a critical center 4-port servo valve with ps = 14 MPa.
The definition of pressure sensitivity is: Kp0 = Kq0/Kc0, where K q 0 = C q w
ps
ρ
28
= K p 014 2 . Increased
14
K c0
1
= K p 014 .
K c0w
2
ps = 28 MPa means increased Kp0 -value K p 028 = K p 014
Kc0 = value gives reduced pressure gain: K p 0 w = K p 0
(4p)
c) Steady state stiffness of a valve controlled position servo
− ∆FL
Steady state stiffness:
∆X p
K v = K sa
K q0
Ap
K f and Kce = Kc0 gives:
Increased ps (ps → p
new
s
Increased Kc0 (Kc0 → K
− ∆FL
):
∆X p
new
c0
new
s →0
− ∆FL
):
∆X p
− ∆FL
∆X p
− ∆FL
=
∆X p
new
s →0
= Kv
s →0
Ap2
K ce
= K sa K p 0 Ap K f .
s →0
old
psnew
.
p sold
s →0
− ∆FL
=
∆X p
old
s →0
K coold
.
K conew
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2004-12-15
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
uc +
u -
Ps = const.
f
Ksa
Position
transducer
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/2, βe, Ap, Mt and FL.
a) Bandwidth of the systems with equal amplitude margin, Am
Am = 6 dB → Steady state loop gain Kv = δh·ωh and bandwidth ωb = Kv .
Valve controlled cylinder: ωh =
the bandwidth:
Vt M t
ωbv = δ hωh = 2 K c 0
Pump controlled cylinder: ωh =
the bandwidth:
4 β e Ap2
ωbp = δ hωh = Ctp
, δh =
K c0
Ap
βeM t
Ctp
2β e M t
, which gives
Vt
Vt
, which gives
βe
Vt
2 β e Ap2
Vt M t
, δh =
2 Ap
βe
Vt
Bandwidth: Ctp = Kc0 ⇒ ωbv = 2ωbp.
(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: ∆X pT =
K f K sa
Valve controlled cylinder: K vv = K sav
Pump controlled cylinder: K vp = K sap
K qi 0
Ap
Kf .
kpN p
Ap
Kf .
Am = 6 dB for both systems means that Kvp < Kvv and with kpNp0 = Kqi0 it will be
stated that Ksap < Ksav.
Finally, the pump controlled system will give the highest error: ∆X pTp > ∆X pTv .
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
SOLUTIONS FOR EXAMINATION
TMHP 51
2004-12-15
4(5)
Elektro-hydraulic velocity servo with acceleration feedback
Kf
Kac
-
-
Ksa
s
Uc
+
+
..
qm
V1
ps
.
qm
Jt
pL
TL
V2
Parameter values: Ksa= 0,04 A/V, Kqi0 = 0,013 m3/As, Kc0 = 1,0.10-12 m5/Ns, V1
= V2 = 0,5 litre, βe = 1000 MPa, Dm = 6,4.10-6 m3/rad, Jt = 0,5 kgm2, Ctm =
8,0.10-13 m5/Ns and Bm = 0.
a) Increased hydraulic damping from the acceleration feedback
Calculate the acceleration feedback gain, Kac, that gives δh' = 0,5.
Assuming a fast valve (Gv(s) = 1) and if TL is neglected the acceleration feedback
will change the hydraulic transfer function from:
1
1
Gh ( s) = 2
to Gh ( s ) =
2
2δ
s
K qi 0 
 2δ
s
+ h s +1
s + 1
+  h + K ac
2
2
ωh ωh
Dm 
ωh  ωh
where δ h' = δ h +
ωh =
ωh K ac K qi 0
2
Dm
(
'
and K ac = δ h − δ h
)ω2DK
m
h
qi 0
4 β e Dm2
K + C tm
= 18 rad/s and δ h = c 0
(V1 + V2 )J t
Dm
βe Jt
(V1 + V2 )
= 0,20
2 ⋅ 6,4 ⋅ 10 −6
= 1,64⋅10-5 As2/rad.
⇒ Kac = (0,5 − 0,20)
18 ⋅ 0,013
(6p)
b) Acceleration feedback and its influence on the dynamic stiffness
The acceleration feedback will only influence the damping δh ( δ h → δ h' ) around
the resonance frequency ωh. Therefore, the dynamic stiffness will become as:

  s 2 2δ h'
 s
 ⋅  2 +

1
s + 1
+
2 
K
ω ωh
D

Sc ≈ Kv m ⋅  v   h
Kce


s
s ⋅ 1 + ' 
 2δ hωh 
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5.
SOLUTIONS FOR EXAMINATION
TMHP 51
2004-12-15
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 =
4βe Ap2
(J1 + J2 )Vt
, δh =
Kce βe (J1 + J2 )
Dm
Vt
where J1 + J2 ⇒ J1 if Kh >>KL
According to the simulation results ωh is much higher than ωa for the load.
(7p)
b) Incerased damping between the two inertia loads
Increased BL means that the inertia J2 will be well damped. In the frequency
domain variation in amplitude will just take place around ωh.
(3p)
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 (ωh) and damping (δh) in the system.
(2p)
c) Flow forces in a 4-port servo valve
xv
pp
qp
Fs
Load
85N
np
pT = 0
Fs
xv
0
0.7*xvmax
xvmax
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
Ap
Ap
V1
Kqi
V2
Kce
Threshold
uc +
Ksa i
u f
xp
Mt
Ps = const.
FL
V1
Ap
V2
xp
Mt
FL
Pm
Position
transducer
Kf
ein
Ap
qp
Kpi
uc+
uf -
Ksa
i
Ct
ep
Position
transducer
Kf
Threshold
ein
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
(ωh) 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 = εin.
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
Position transducer
xp
Ap
V1
Mt
p1
Kf
uf uc Ksa
+
Ap
V2
p2
pR = const.
i
Ps = const.
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 ωb (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
βe = 1200 MPa
Kqi0 = 0,013 m3/As
a) Kce-value for a bandwidth of ωb = 25 rad/s
Calculate the required Kce-value so that the closed loop servo can reach a
bandwidth of ωb = 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(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 = 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
Kf
Kfv
Uc
+
+
Ksav
xp
.
xp
Bp = 0
Ap
V2
V1
Kvfv = 1 + Kfv Ksav
ö
FL Kce æ
Vt
ç1 +
s÷
Ap2 çè 4 be K ce ÷ø -
Duc = 0
FL
Mt
Kqi
Ap
.
-
1/ Kvfv
xp
s 2 + 2 dh s +
1
Kvfv wh2 Kvfv wh
Kqi
Ap
iv
Ksav
Kf
1
s
xp
Position feedback
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 β 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:
Mt = 900 kg
βe = 1000 MPa
Ap = 1,96.10-3 m2
-2
.
-3
3
3
Vt = 1,0 10 m
Kqi0 = 1.0⋅10 m /As
Kc0 = 1,0.10-11 m5/Ns
Ksav = 0,80 A/V
ωh' = 320 rad/s
Kf = 20 V/m
a) Hydraulic damping with velocity feedback
For the servo system, including velocity feedback, has the un-damped resonance
frequency been measured to ωh' = 320 rad/s.
Calculate the hydraulic damping δh' 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
∆X p
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:
εp
K εi
, with the displacement setting coefficient Kεi = 20 A-1 and the brake
=
i 1+ s
ωs
frequency ωs = 100 rad/s. The pump shaft speed is ωp = 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 β 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.
ep
wp
uc +
Ksa
____
uf - s
p1 V1
Dm
Dp
Au(s)
KL
qm
p2 V2
i
.
J1
Speed
transducer
pm
.
J2
TL
qL
Kf
a) Transmission dynamics versus load dynamics
Calculate the transmission hydraulic resonance frequency (ωh) 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 (δh) 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 (ωs) 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 = 1 −
pdcav
= 0,625 → pdcav = 0,375 ⋅ pu (1).
pu
pu = 160 bar in eq. (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
is given by the equations. ωh =
β e Ap2
Vp M t
, δh =
K ce
2 Ap
βeM t
Vp
. The high bulk modulus of
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
xv
pp
qp
Fs
Load
85N
np
pT = 0
xv
Fs
0
0.7*xvmax
xvmax
Pump pressure, pp = 210 bar, max pump flow, qpmax = 74 litre/min and pL = 0.
Steady state flow forces: Fs = 2C q ⋅ w ⋅ xv ⋅ ∆pv ⋅ cos δ . ∆pv = pp and pL = 0 gives the
valve flow as qv = C q w ⋅ xv
pp
ρ
(1). Finally Fs max = 2 ρ cos δ ⋅ qv max p p (2).
Nominal valve flow: qv = C q w ⋅ xv
pp
ρ
⇒ qvN = q p
x v max
xv
∆pvN
= 61 litre/min.
pp
qvmax = qpmax and increased pp to 350 bar in eq. (2) ⇒ Fs max = 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
Ap
Ap
V1
Kqi
V2
f
Ap
FL
V1
Kce
Kf
ein
Ap
V2
xp
Mt
FL
Pm
Position
transducer
Threshold
uc +
Ksa i
u -
xp
Mt
qp
ep
Kpi
uc+
uf -
Ps = const.
Ct
Ksa
i
Position
transducer
Kf
Threshold
ein
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: ∆X p ⋅ K f ⋅ K sa = ∆iT ⇒ ∆X p =
∆iT
K f ⋅ K sa
Am = 6 dB → Steady state loop gain: Kv = δh·ωh
Valve control: (δ h ⋅ ωh )4 − v = 2 K ce
1 β
. Pump control: (δ h ⋅ ωh ) p = Ct e
Vt
2 Vt
(K v )4 − v
(K v ) p
K sap
∆X p 4 − v
1
K sa 4 − v
=
= = 0,25
= 4.
K sa 4 − v 4
∆X pp
K sap
=
K sa K qi K f / Ap
K sa K pi K f / Ap
=4 ⇒
βe
(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
Position transducer
xp
Ap
V1
Ap
Mt
V2
p1
Kf
p2
pR = const.
uf -
uc
+
i
Ksa
Ps = const.
Ap = 1,96.10-3 m2
ps = 21 MPa
Vt = 1,0.10-3 m3
βe = 1200 MPa
Kf = 25 V/m
Mt = 800 kg
Kqi0 = 0,013 m3/As
a) Kce-value for a bandwidth of ωb = 25 rad/s
Am = 6 dB → Steady state loop gain Kv = δh·ωh and bandwidth ωb = Kv = 25 1/s .
4 β e Ap2
K ce β e M t
, which gives
Vt M t
Ap
Vt
β
ωV
the bandwidth: ωb = δ hωh = 2 K ce e ⇒ K ce = b t = 1,04·10-11 m5/Ns
Vt
2β e
Valve controlled cylinder: ωh =
, δh =
(5p)
b) Feed forward loop to reduce the velocity error
Block diagram showing the implementation of the feed forward loop.
Kff s
uc
+
-
+
Feed Forward
+
Ksa
i
Kqi
Ap
1
2
+ dh s + 1
wh2 wh
s2
.
xp
1
s
xp
Kf
The feed forward block includes a derivation (s) of the position command signal (uc =
Ax·sin(ωt), 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: ωb = K v = K sa
K sa
K qi
Ap
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
xp
.
xp
Kf
Kfv
-
Uc
Bp = 0
Ap
V1
Ksav
+
+
Kvfv = 1 + Kfv Ksav
Mt
V2
FL
ö
FL Kce æ
Vt
ç1 +
s÷
Ap2 çè 4 be K ce ÷ø -
.
-
Duc = 0
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
Kqi
Ap
1/ Kvfv
xp
2 dh +
+
s
1
Kvfv wh2 Kvfv wh
s2
Kqi
Ap
iv
Ksav
Kf
1
s
xp
Position feedback
βe = 1000 MPa
Kc0 = 1,0.10-11 m5/Ns
ωh' = 320 rad/s
a) Hydraulic damping with velocity feedback
With velocity feedback the un-damped resonance frequency is ωh' = 320 rad/s.
The basic resonance frequency (ωh) is: ωh =
4 β e Ap2
Vt M t
= 131 rad/s. The block diagram
2
above gives ω = ωh ⋅ K vfv and δ = δ h / K vfv . K vfv
'
h
δ h' =
K co
Ap
βeM t
Vt
'
h
⋅
 ω' 
=  h  ⇒ Kvfv = 6,0 and finally
 ωh 
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:
 s2

K
2δ h

 ⋅ K vfv ⋅ s + K f K sav qi
1
+
s
+
2

Ap
− ∆FL  K vfvωh K vfvωh

. The steady state part is
Sc =
=
∆X p

K ce 
Vt
1 +
s
Ap2  4 β e K ce 
− ∆FL
∆X p
− ∆FL
∆X p
=
s →0
=
s →0
K vfv K sav K qi K f
Ap K vfv
K sav K qi 0 K f Ap
Kc0
⋅
Ap2
K ce
. 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
p1 V1
ep
wp
uc +
Ksa
____
uf - s
Dm
Dp
.
TL
J2
qL
Speed
transducer
pm
Au(s)
.
qm
p2 V 2
i
KL
J1
Kf
ωp = 157 rad/s, Dp = 5,6.10-6 m3/rad, V1 = V2 = 0,6 litre, βe = 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.
a) Transmission dynamics versus load dynamics
The hydraulic frequency for the transmission is ωh =
β e Dm2
Vt (J 1 + J 2 )
. Numerically:
800 ⋅ 106 ⋅ (19 ⋅ 10 − 6 )
ωh =
= 22 rad/s. The load dynamics include two frequencies, ωa
0,6 ⋅ 10 − 3 ⋅ (0,5 + 0,5)
2
KL
2000
. ωa =
= 63 rad/s. ωh is dominant.
J2
0,5
and ω1. The lowest one is ωa =
Au
Bode diagram:
wh
1
wa
Frequency [rad/s]
(5p)
b) Acceleration feedback for increased damping
Implementation of acceleration feedback.
p1 V1
ep
wp
uc +
Ksa
____
+
uf
s
Dm
Dp
.
J1
KL
qm
i
-
Acceleration signal
Speed
transducer
pm
.
J2
qL
Kac.s
Kf
Influence on hydraulic damping Gh ( s ) =
1
 2δ

K
1
+  h + K ac εi
ω p D p  s + 1
ω  ωh
Dm 1 + s / ωs

s
2
2
h
ωh 

K
1
 K ac εi
ω p D p 
Dm 1 + s / ω s
2 

Since ωs = 100 rad/s, is higher than both ωh and ωa its influence on the hydraulic
damping is marginal.
δ h' = δ h +
(5p)