Linköping Tekniska Högskola
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EXAMINATION Page 1(9)
TMHP51/TEN1, 2013-03-26
EXAMINATION
Hydraulic Servo Systems, TMHP51/TEN1
Date:
Room:
Number of questions:
Responsible teacher:
Phone number during exam:
Teacher visit at:
Allowed tables:
Allowed handbooks:
Allowed formularies:
Allowed electronics:
Course administrator:
Score:
Solution:
Results:
Langauge:
2013-03-26, 14:00
T2
5
Magnus Sethson, ([email protected])
013-288945, 0734-619890
15:00, 17:00
Standard Mathematical Tables (or similar)
Tefyma (Ingelstam, Rönngren, Sjöberg), Beta (Råde,
Westergren)
Formula Book for Hydraulics and Pneumatics (LiTH/IEI), Mekanisk Värmeteori och
Strömningslära
Pocket Calculator (No smartphones or computers allowed!)
Rita Enquist (013-281189, [email protected])
Maximum score on each question is 10 points, resulting in an examination maximum of 50 points. To get
the mark 3 you will need 20 points, to get the mark
4 you will need 30 points and to get the mark 5 you
will need 40 points.
The solution will be e-mailed on the course emailing
list after the exam.
Results will be announced on 2013-04-11 at the latest.
You are entitled to answer in either english or
swedish.
Please notice that the questions are not put in an increasing order of difficulty. Plan your work
through the exam.
Good Luck & Happy Easter!
Magnus Sethson
Senior Lecturer
March 26, 2013
On the front page and on all following pages the student must write:
AID-number, TMHP51/TEN1, YYMMDD, page number
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EXAMINATION Page 2(9)
TMHP51/TEN1, 2013-03-26
Cavitation in Orifices
[A, 4 points] An orifice is connected between two cylinders according to figure 1. The fixed orifice area
is described by the orifice opening diameter ds . It may be regarded as an sharp edged orifice. Derive an
expression for the cavitation condition in the system in terms of the loads. The loads are acting vertically.
The cavitation condition pressure drop across an orifice is: ∆Pcav = C1 + C2 Pupstream ≈ C2 Pupstream .
C2 ≈ 0.68. Also, all friction in the cylinders are neglected.
Figure 1: A double cylinder arrangement for speedcontrol of mass m1
Solution: Cavitation will take place when ∆P = p1 −p2 > ∆Pcav = C2 p1 resulting in ( mA11g − mA22g ) >
or (1 − C2 ) mA11g < mA22g .
C2 mA11g
[B, 2 points] Under the conditions described in the previous question, what terminal velocity v1 can
be expected once the mass m1 is released into falling mode. Assume: m1 = 5ton, A1 = 28.27cm2 , m2 =
250kg, A2 = 7.07cm2 , ρ = 760kg/m3 , ds = 5mm, Cq = 0.67
m1 g
= 17.35MPa, mA22g = 3.47MPa. According
A1 q
cq ds 2 π4 ρ2 C2 p1 = 2.32l/s → v1 = 0.82m/s.
Solution: Assume from above:
in place. The flow becomes: q =
to above, cavitation is
[C, 4 points] Suppose the arrangement in the previous figure is replaced by the one in figure 2 in the
purpose of staying out of cavitation within the orifices. What sizes of the sharp edged orifices diameters
d1 and d2 should be used?
Solution: From lectures: di =
r
4q
πCq
p 2P
1
ρ
C2 (1−C2 )i−1
, d1 = 6.6mm, d2 = 8.8mm
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EXAMINATION Page 3(9)
TMHP51/TEN1, 2013-03-26
Figure 2: Another orifice have been introduced to deal with cavitation.
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EXAMINATION Page 4(9)
TMHP51/TEN1, 2013-03-26
Servo Valves Characteristics
[A, 2 points] Servo valve bandwidth is one of the main characteristics for the hydraulic servo system’s
response. In figure 3 is the data sheet characteristics of an MOOG G761-3005 valve shown. What is
the bandwidth (in Hz) for a ±40% amplitude response when only considering a maximum 6dB drop in
amplitude ratio?
Frequency response of 15gpm servovalve MOOG G761-3005
150
−5
±40%
±100%
100
−10
50
Phase Lag [deg]
Amp. Ratio [dB]
0
−15
20
30
40 50
Frequency [Hz]
100
150 200
0
300
Figure 3: The frequency repsonse for a 15 gallon-per-minute servo-valve MOOG G761-3005.
Solution: 110Hz.
[B, 2 points] Using figure 3, what is the bandwidth (in Hz) for a ±100% amplitude response when
considering both a maximum 3dB drop in amplitude ratio and a maximum allowed phase lag of 30◦ .
Solution: 26Hz.
[C, 4 points] A servo valve like in figure 3 is going to be used in position control of a system having a
resonant frequency defined by the system stiffness k = 13MN/m and total mass of m = 22kg. In general
a valve phase lag of 90◦ can be tolerated in such a system. And at the same time an amplitude margin
of 3dB is required. Is this valve well suited for this system or should the designer look for another valve?
Motivate! Assume motions of moderate amplitude (±40%)
Solution: 70Hz ¡ 103Hz ¡ 122Hz. q
No this is not a good valve for this system. It’s required dynamics
k
is bellow the system’s dynamics ω = m
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EXAMINATION Page 5(9)
TMHP51/TEN1, 2013-03-26
[D, 2 points] Servo valves are high precision machinery. One characteristics is the forces upon the
spool due to flow passing the controllable orifices, the flow forces Fs . A spool manoeuvring unit is used
to handle these forces in direct drive design as shown in figure 4. The design uses an electromagnet for
spool displacement operation. Calculate the static flow forces the electromagnet is required to handle
in terms of flow forces under the following conditions: Ds = 12mm, L = 55mm, δ = 69◦ , Pupstream =
210bar, Pdownstream = 95bar, ρ = 780kg/m3 , Cq = 0.67, xv,max = 0.9mm. Assume standard orifice.
Figure 4: The principle dimensions of the valve.
Solution: Use formula book. Fs = 1873N (Area gradient is w = πDs )
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EXAMINATION Page 6(9)
TMHP51/TEN1, 2013-03-26
Hydraulic Servo Systems
[A, 6 points] A servo valve is used for producing a pressure response according to the diagram and
circuit in figure 5. The dynamics may be approximated by a first order low-pass filter. (ωs ∼ 1/τv ).
Unfortunately, the full pressure response is not fast enough for a particular application. The designer is
challenged by a quick solution to the problem. The designer decide to keep the current valve but instead
increase the supply pressure by 45%. What is the original system bandwidth, ωs in Hz? What is the
new one? What is the limiting factor holding the 100% response back from the 50%?
Pressure Response [% Pc/Ps]
100
Response
100%
50%
80
60
40
20
0
0
0.005
0.010
0.015
Time [s]
0.020
0.025
Figure 5: The diagram shows two step responses from a pressure control application. Two step sizes are
applied, 50% and 100%.
Solution: System response is proportional to pressure
time derivative Ṗc in volume
q
q Vc . Ṗc is props,new
ps
ωnew
portional to flow coefficient for the valve, Kq 0 = Cq w ρ . From this follows: ωold =
ps,old . All this
√
combines into: 143Hz 1.45 = 172Hz. The limiting factor at 100% response is the voltage limits of the
servo amplifier, Ksa .
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EXAMINATION Page 7(9)
TMHP51/TEN1, 2013-03-26
[B, 4 points] Figure 6 show schematically a valve controlled position servo. At centre position of the
piston the chamber volumes are equal, V1 = V2 = V2t = 1.1litre. At that operating point the hydraulic
resonance frequency is ωh = 90rad/s and the damping is δh = 0.22. The steady state loop gain is then
adjusted in the servo amplifier to Kv = 14s−1 by the controller.
Now, assume the position of the piston is changed so that V1 = 0.2Vt and V2 = 0.80Vt . The steady
state loop gain is continuously adjusted by the controller using Ksa along the stroke of the cylinder. At
this new position it is adjusted so that Kv = 15.5s−1 . Assume Bp = 0 and Kce unchanged. Is the amplitude margin Am improved? What is the change of the amplitude margin in dB between the two cases?
s
β e Ap 2 1
1
ωh =
+
mt
V1
V2
s
1
1
Kce
βe mt
+
δh =
2Ap
V1
V2
(1)
(2)
Figure 6: A classical servo system with a symetric cylinder and a servo-valve. The feedback loop is
represented by a position transducer, a transducer gain, Kf and a servo amplifier Ksa .
Solution: Amplitude margin is found in the formula book. The centre amplitude margin becomes
Am = 9.0dB
q The scaling
q of both ωh and δh between the two cases will follow the same expression:
1
1
1
1
∗
ωh = ωh 0.2 + 0.8 / 0.5
+ 0.5
= 112.5rad/s. Similar: δh = 0.275. The off-centre position amplitude
margin becomes Am = 12.9dB. The change is an increase of ∆Am = 3.9dB
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EXAMINATION Page 8(9)
TMHP51/TEN1, 2013-03-26
System Models
[A, 6 points] One way to increase the damping in a hydraulic servo system is to include a fixed orifice,
Csm , across the valve ports, as seen in figure 7. Draw open loop block diagram of the system including
the extra orifice. Do not include any controller. In the block diagram use ∆Xv for the input signal,
∆nm (or ∆Θm ) for output and ∆Tm as the disturbance torque. The diagram need to be fully reduced,
having no loops or cross-links.
Figure 7: A classical servo system with a motor and a servo valve. The feedback loop is represented
by a speed transducer, a transducer gain, Kf and a controller Ksa . A damping orifice Csm is included
between the motor/valve ports. The external load torque is Tm and the system damping is represented
by Bm . (nm = Θm )
Solution: See formula book. A note on Kce → Kc + Cim +
Cem
2
+ Csm is required.
[B, 4 points] Some test have been made to establish a value on Csm in figure 7. It turns out that
the Csm orifice is dominating the damping characteristics of the system. The tests conclude that when
applying a static load torque Tm of 123N m to the motor the flow across the orifice becomes 280ml/s.
In the bench test resulting in that measurement no damping Bm where in play. Other parameters are:
V1 = V2 = 0.6l, βe = 1.1GP a, Jt = 1.15kgm2 , Dm = 55cm3 . Also, the motor is assumed ideal in terms
of efficiency. What is the system damping, δh , of the test bench servo?
Solution: See formula book. The damping δh is given by the formula book. Also, the described test
280ml/s
m 2π
. Csm = ∆Q
gives a value of Csm since the pressure across the motor is ∆p = TD
δP = 140bar . δh = 0.37
m
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EXAMINATION Page 9(9)
TMHP51/TEN1, 2013-03-26
Feedback Systems
[A, 10 points] Draw a reduced and simplified block diagram for the closed-loop system in figure 8.
Also, calculate the hydraulic characteristics ωh and δh for the given parameter values. Finally, indicate
in a sketch the principle look of the open-loop system Bode-diagram and indicate ωh and the amplitude
margin Am by value. There is no internal leakage in the cylinder, nor any external damping Bp . Vh =
1.9l, Ah = 65cm2 , Ar = 45cm2 , Ps = 210bar, Mt = 200kg, Kce = 8 × 10−11 , Kq = 0.28, βe = 1.05GP a.
The regulator Greg uses the control law xv = 0.914(xref − xp ). Is this a good design? Motivate!
Solution: ωh = 341rad/s, δh = 0.065, Am = 1dB. The damping is very low and the amplitude margin is just 1dB. The system will become almost unstable in most situations with a resonance frequency
of 54Hz. This is a very bad design.
Figure 8: A servo system with asymetric cylinder.