Linköping Tekniska Högskola
IEI/Flumes
EXAMINATION Page 1(7)
TMHP51/TEN1, 2013-08-30
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:
2013-08-30, 08:00
TR4
5
Magnus Sethson, ([email protected])
013-288945, 0734-619890
09:00, 11:00
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 uploaded on the course site.
Results will be announced on 2013-09-09 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.
I hope you had a nice summer and are able to put that energy into this exam. Good Luck!
Magnus Sethson
Senior Lecturer
August 27, 2013
On the front page and on all following pages the student must write:
AID-number, TMHP51/TEN1, YYMMDD, page number
Linköping Tekniska Högskola
IEI/Flumes
1
EXAMINATION Page 2(7)
TMHP51/TEN1, 2013-08-30
Cavitation in Orifices
[A, 4 points] A cavitation test is carried out using a 200bar pressure setting on the supply system. The
result is seen in figure 1. Visual cavitation is observed at a downstream pressure of 73.4bar when the
flow is 6.8l/min. The upstream pressure is increased to 270bar by the supply system. At what pressure
can now cavitation be assumed? What is the flow then?
Cavitation Test
8
Cavitation (73.4bar,6.8l/min)
Flow [l/min]
6
4
2
200
150
100 90 80 70 60 50
Downstream Pressure [bar]
0
Figure 1: The result of a cavitation test.
[B, 2 points] What maximum supply pressure can be assumed when at maximum flow of 10l/min in
figure 1?
[C, 4 points] Suppose the arrangement of orifices is changed so that another identical orifice is added
in series with the first. At a supply pressure of 300bar, what cavitation downstream pressure after the
two orifices can be expected?
Linköping Tekniska Högskola
IEI/Flumes
2
EXAMINATION Page 3(7)
TMHP51/TEN1, 2013-08-30
Servo Valves Characteristics
[A, 2 points] Servo valve bandwidth is one of the main characteristics for the hydraulic servo system’s
response. In figure 2 is the data sheet characteristics of an MOOG G761-3005 valve shown. The valve is
a two-stage valve without hydraulic pilot-stage. What is the bandwidth (in Hz) for a ±40% amplitude
response when only considering a maximum 3dB drop in amplitude ratio?
Frequency response of 15gpm servovalve MOOG G761-3005
3
150
±40%
±100%
100
−3
−6
−9
50
Phase Lag [deg]
Amp. Ratio [dB]
0
−12
−15
20
30
40 50
Frequency [Hz]
100
150 200
0
300
Figure 2: The frequency repsonse for a 15 gallon-per-minute servo-valve MOOG G761-3005.
[B, 2 points] Using figure 2, what is the bandwidth (in Hz) for a ±100% amplitude response when
considering both a maximum 6dB drop in amplitude ratio and a maximum allowed phase lag of 45 .
[C, 4 points] A servo valve like in figure 2 is going to be used in position control of a system. The
resonant frequency defined by the system sti↵ness k = 16MN/m and total mass of m = 32kg sets the
hydraulic frequency. Also, in general a valve phase lag of 60 can be tolerated in such a system. And at
the same time an amplitude margin of at least 6dB is required. The lowest of these three frequencies will
set the practical application of the system. What is it? Assume motions of moderate amplitude (±40%)
[D, 2 points] The servo-valve in figure 2 is going to be used in a test system for generating a sinusoidal
flow through the test object. The requirements are ±12gpm (gallon-per-minute) with a frequency of
30Hz. The generated flow will lag behind the reference signal feed into the servo valve amplifier. How
long time in milli-seconds will the lag be? (Such a measure can be used for feed-forward control in critical
applications.)
Linköping Tekniska Högskola
IEI/Flumes
3
EXAMINATION Page 4(7)
TMHP51/TEN1, 2013-08-30
Hydraulic Servo Systems
[A, 4 points] A servo valve is used for producing a pressure response according to the diagram in figure
3. The dynamics may be approximated by a first order system with an initial delay. What is the system
bandwidth, !s in Hz for both the 50bar and 120bar response? Assume 1ms delay for the 50bar response
and a 2ms delay for the 120bar
120
Response
120bar
50bar
Pressure Response [bar]
100
80
60
40
20
0
0
0.005
0.010
0.015
0.020
Time [s]
0.025
0.030
0.035
Figure 3: The diagram shows two step responses from a pressure control application. The reference
pressures are set to 50bar and 120bar.
Linköping Tekniska Högskola
IEI/Flumes
EXAMINATION Page 5(7)
TMHP51/TEN1, 2013-08-30
[B, 2 points] Figure 4 show schematically a valve controlled position servo. The system is quite old.
At centre position of the piston the chamber volumes are equal, V1 = V2 = V2t = 0.46litre. The cylinder
piston area, Ap is 15.9cm2 . The moving mass mt is 42kg.
The system has been tested with the piston in centred position using step responses. It has been seen
from these tests that the damping is quite good, h = 0.31. New oil where used during the tests with a
bulk modulus of e = 1GPa. From this, calculate the hydraulic resonance frequency !h and the servo
valve e↵ective leakage coefficient Kce .
!h =
h
=
s
Kce
2Ap
e Ap
mt
s
2
✓
e mt
1
1
+
V1
V2
✓
◆
1
1
+
V1
V2
◆
(1)
(2)
Figure 4: A classical servo system with a symetric cylinder, a controller and a servo-valve.
[C, 2 points] Most of the time the system in figure 4 is operated in an o↵-centre position, V1 = 0.25Vt
and V2 = 0.75Vt . What are the dynamic system characteristics ( h , !h ) then?
[D, 2 points] After some time it is discovered that the system characteristics is changed due to the
rather old components. Investigations find a problem with supply system tank. Its poor design adds
quite a lot of air into the system. As a consequence the sti↵ness of the oil is reduced by 20%. In the
o↵-centre position described above, what are the air-a↵ected dynamic system characteristics ( h , !h )
now? Assume oil density unchanged.
Linköping Tekniska Högskola
IEI/Flumes
4
EXAMINATION Page 6(7)
TMHP51/TEN1, 2013-08-30
System Models
[A, 6 points] In figure 5 is a servo system seen with its symmetrical cylinder, feedback and servo-valve.
Draw the fully reduced block-diagram of the closed-loop system. Indicate the feedback controller and
the electromagnet as separate blocks. Also indicate the intermediate signals iv and xv along with
the input and output signals Xr , Xp and FL . Assume the hydraulic damping and frequency are
represented by h and !h . The oil is represented by e and the valve by Kce and Kq . Vt = V1 + V2
Figure 5: A servo system with a symetrical cylinder, feedback loop and a servo valve. The feedback loop
is represented by a position transducer with a transducer gain Kf , a controller Ksa and a electromagnet
Kix . The damping is represented by Bp . The external load force is FL .
[B, 4 points] Kix represents the amplifier stage of the electric power stage for the electromagnet. The
force applied by the electromagnet is opposed by the flow-forces inside the valve. For larger velocities
of the piston, ẋp , these forces can be significant. Compare two cases, one where the valve is operating
close to zero valve position, (xv ⇡ 0), to another case where the piston is moving at maximum speed,
ẋp = 0.6m/s. What flow force can be expected in this later case? Assume Cq = 0.67, = 69 , maximum
xv = 0.4mm and the diameter of the valve is 6mm. The supply pressure Ps is 210bar and during this
maximum speed motion a pressure di↵erence across the servo-valve’s outlet ports (marked A and B) of
60bar have been measured.
Linköping Tekniska Högskola
IEI/Flumes
5
EXAMINATION Page 7(7)
TMHP51/TEN1, 2013-08-30
Feedback Systems
[A, 10 points] Draw a reduced and simplified block diagram for the closed-loop system in figure 6.
Also, calculate the hydraulic characteristics !h and h for the given parameter values. Finally, indicate
in a sketch the principle look of the open-loop system Bode-diagram and indicate !h and the amplitude
margin Am by value. There is a substantial internal leakage in the cylinder, Cip = 2.2 ⇥ 10 10 . The
external damping Bp can be neglected. V1 = V2 = 9.0l, Ap = 65cm2 , mt = 180kg, Kce = 8 ⇥ 10 11 , Kq =
0.28, e = 1.03GP a. The mechanical linkage forms a regulator Greg that implements the control law
xv = 0.4(xref xp ).
Figure 6: A servo system with mechanical linkage as feedback system and a symetric cylinder. Input to
the system is an externally applied motion to the xref point.
9/7/13 6:36 PM
/Users/magse/exams/TM.../TMHP51_130830.m
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%Solution to exam in TMHP51 2013−08−30
format compact;
%%
disp ’Task 1A’;
C2=(200e3−73.4e3)/200e3
P2=270e3*(1−C2)
XwCq=6.8/sqrt(200e3−73.4e3) % this combines all of valve size, opening, Cq and density
q=XwCq*sqrt(270e3−P2)
disp ’Task 1B’;
P1=10^2/(C2*XwCq^2)
disp ’Task 1C’;
%assume cavitation in first orifice
q=XwCq*sqrt(300e3*C2)
% have second orifice already cavitated then?
%upstream pressure of second orifice = downstream pressure of first orifice:
P2=300e3*(1−C2)
% downstream pressure of second orifice according to restriction:
P3=P2 − q^2/XwCq^2
% NEGATIVE!!! means that cavitation takes place first in second orifice,
% contrary to assumption above
% for the second orifice:
P2=300e3/(1+C2)
P3=P2*(1−C2)
%%
disp ’Task 2A’;
% from diagram: 71Hz
71
disp ’Task 2B’;
% from diagram: 34.5Hz < 67.5Hz −> 34.5Hz
34.5
disp ’Task 2C’;
% 6dB −> 79Hz, 60deg phase −> 140Hz
f=sqrt(16e6/32)/(2*pi)
fmin = 79
disp ’Task 2D’;
% from diagram, phase lag = 37grad
tlag=1000/30*37/360
%%
disp ’Task 3A’;
f120=1/(0.0113−0.002)
f50 =1/(0.0086−0.001)
disp ’Task 3B’;
betae=1e9;
Ap=15.9e−4;
mt=42;
V1=0.46e−3;
V2=0.46e−3;
Vt=V1+V2;
dh=0.31
w=sqrt((betae*Ap^2)/mt*(1/V1+1/V2)) % in rad/s
Kce=dh*2*Ap/sqrt(betae*mt*(1/V1+1/V2))
disp ’Task 3C’;
V1=0.25*Vt;
V2=0.75*Vt;
w=sqrt((betae*Ap^2)/mt*(1/V1+1/V2)) % in rad/s
dh=Kce/2/Ap*sqrt(betae*mt*(1/V1+1/V2))
disp ’Task 3D’;
9/7/13 6:36 PM
/Users/magse/exams/TM.../TMHP51_130830.m
betae=0.8e9;
w=sqrt((betae*Ap^2)/mt*(1/V1+1/V2)) % in rad/s
dh=Kce/2/Ap*sqrt(betae*mt*(1/V1+1/V2))
%%
disp ’Task 4A’;
% see formula book
disp ’Task 4B’;
Cq=0.67;
d=69;
x=0.4e−3;
dv=0.006;
Ps=210e5;
Pab=60e5;
w=dv*pi;
dP=0.5*(Ps−Pab);
Fs=2*abs(2*Cq*w*x*(dP)*cos(d/180*pi)) % 2 orifices, inlet and outlet
%%
disp ’Task 5A’;
betae=1.03e9;
Ap=65e−4;
mt=180;
V1=9e−3;
V2=9e−3;
wh=sqrt((betae*Ap^2)/mt*(1/V1+1/V2)) % rad/s
Kce=2.2e−10+8e−11;
dh=Kce/2/Ap*sqrt(betae*mt*(1/V1+1/V2))
Kq=0.28;
Kreg=0.4;
Kv=Kq/Ap*Kreg
Am=−20*log10(abs(Kv/(−2*dh*wh)))
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9/7/13 6:35 PM
MATLAB Command Window
1 of 2
>> what
MATLAB Code files in the current folder /Users/magse/exams/TMHP51/TMHP51_130830
TMHP51_130830
>> TMHP51_130830
Task 1A
C2 =
0.6330
P2 =
99090
XwCq =
0.0191
q =
7.9009
Task 1B
P1 =
4.3253e+05
Task 1C
q =
8.3283
P2 =
110100
P3 =
−7.9800e+04
P2 =
1.8371e+05
P3 =
6.7422e+04
Task 2A
ans =
71
Task 2B
ans =
34.5000
Task 2C
f =
112.5395
fmin =
79
Task 2D
tlag =
3.4259
Task 3A
f120 =
107.5269
f50 =
131.5789
Task 3B
dh =
0.3100
w =
511.5741
Kce =
7.2950e−11
Task 3C
w =
590.7149
dh =
0.3580
Task 3D
w =
528.3515
dh =
0.3202
Task 4A
Task 4B
Fs =
54.3108
Task 5A
wh =
231.7872
dh =
9/7/13 6:35 PM
0.1481
Kv =
17.2308
Am =
12.0088
>>
MATLAB Command Window
2 of 2