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EXAMINATION Page 1(13)
TMHP51/TEN1, 2012-12-20
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:
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Solution:
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Langauge:
2012-12-20, 1400
TER1
5
Magnus Sethson, ([email protected])
013-288945, 0734-619890
1430, 1530, 1730
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 2013-01-10 at the latest.
You are entitled to answer in either english or
swedish.
Since this is the first examination with me as the new examiner I would like to take the opportunity
to welcome your comments on the course, both in general terms and details. Please fill in the course
evaluation and do not hesitate to contact me directly. This is very important for me and my colleagues.
Good Luck, Merry Christmas & Happy New Year!
Magnus Sethson
Senior Lecturer
September 7, 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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1
EXAMINATION Page 2(13)
TMHP51/TEN1, 2012-12-20
Cavitation in Orifices
[A, 6 points] An orifice and an adjustment valve is connected in series according to figure 1. The fixed
orifice area is A1 . The tuneable orifice may be regarded as a sharp-edge orifice with an area function
of A2 (x) = w x. Derive an expression for the cavitation condition in the system in terms of the valve
setting x. The cavitation condition pressure drop a cross an orifice is: ∆Pcav = C1 + C2 Pupstream ≈
C2 Pupstream . C2 ≈ 0.68. Pt = 0.
Figure 1: A series of orifices
[B, 2 points] An operator is to manoeuvre the adjustment valve. Most likely will he experience the
characteristic and noisy sound of cavitation. What is the adjustable valve setting then? A1 = 8.4mm2 ,
w = π × 6mm.
[C, 2 points] Suppose the orifice opening area A1 is changed to A1 = 3.4mm2 . Will the operator still
hear the cavitation sound? Valve position x is limited to x ∈ [xmin .. xmax ] → x ∈ [0.3mm .. 1.2mm].
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EXAMINATION Page 3(13)
TMHP51/TEN1, 2012-12-20
Servo Valves Characteristics
[A, 2 points]Servo valves characteristics shows on hysteresis, this is due to friction in the spool-bushing
arrangement and the natural hysteresis of the magnetic circuit actuating the spool motion. Figure 2
show a schematic pressure gain curve for a new servo valve. Since Kc represent the internal leakage flow
it will affect the principle view of the diagram in figure 2. How? Exemplify by redrawing the diagram and
add an pressure gain curve for the used and worn valve. Make sure you mark the two curves correctly;
new valve and worn valve.
1.0
PL/Ps
0.5
0
−0.5
−1.0
−0.10
−0.05
0
xv/xv,max
0.05
0.10
Figure 2: System complemented by orifice A5 .
[B, 2 points] In modern servo valves, that is valves with built-in electronics, some compensation of the
wear over time is done by the software controller for spool displacement control. Simply by changing the
controllers proportional gain in when xv ≈ 0. In what direction? Should the proportional gain at xv ≈ 0
be increased or decreased? An short textual motivation is required.
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EXAMINATION Page 4(13)
TMHP51/TEN1, 2012-12-20
[C, 2 points] Servo valve bandwidth is one of the main characteristics for the hydraulic servo system
response. In figure 3 is the data sheet characteristics of an MOOG G761-3004 valve shown. What is
the bandwidth (in Hz) for a ±40% amplitude response when only considering a maximum 3dB drop in
amplitude ratio? And 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 60◦ .
Frequency response of 10gpm servovalve MOOG G761-3004
200
0
±40%
±100%
−5
100
−10
−15
50
20
30 40 50
100 150 200 300
Frequency [Hz]
Figure 3: The frequency repsonse for a servo-valve MOOG G761-3004.
0
Phoaselag [deg]
Amp. Ratio [dB]
150
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EXAMINATION Page 5(13)
TMHP51/TEN1, 2012-12-20
[D, 4 points] Some years back our division Flumes made some calculations on really large servo valves
for a large hydraulic press. The two main valves had spools in the size of 450mm in diameter. Normally
servo valves have spools in the range of 8mm to 20mm in diameter. These huge valve spool where
independently manoeuvred by a pilot servo hydraulic circuit. It can be assumed that the pilot stage
controlled the motion of the main stage in such a way that the flow through the main stage followed an
mathematical expression like: q(t) = 2.8 sin(0.3t)[m3 /s], where t represent time. What maximum flow
force on the main stage spool could be expected. From measurements it was seen that the largest flow
forces would take place when t = 0.3s. Dimensions according to figure 4. Ds = 450mm, L = 1430mm, δ =
69◦ , Pupstream (t = 0.3s) = 110bar, Pdownstream (t = 0.3s) = 95bar, ρ = 780kg/m3 , Cq = 0.67. Assume
standard orifice.
Figure 4: The principle dimensions of the huge servo valves.
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EXAMINATION Page 6(13)
TMHP51/TEN1, 2012-12-20
Hydraulic Servo Systems
[A, 4 points] A servo valve is tested for its step response according to the circuit in figure 5. The
dynamics may be approximated by a first order low-pass filter. Determine the valve bandwidth ωb in Hz
for input signal amplitudes of 50% and 100% respectively. ωb ∼ 1/τv Why is the bandwidth reduced at
higher amplitudes?
Response
100%
50%
Pressure Response [% Ps]
100
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 servo valve test. Two step sizes are applied,
50% and 100%.
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EXAMINATION Page 7(13)
TMHP51/TEN1, 2012-12-20
[B, 6 points] Figure 6 show schematically a valve controlled position servo. At centre position of the
piston the chamber volumes are equal, V1 = V2 = V2t = 0.7litre. At that operating point the hydraulic
resonance frequency is ωh = 120rad/s and the damping is δh = 0.20. The steady state loop gain is
adjusted in the servo amplifier to Kv = 20s−1 . The amplitude margin is then Am = 7.6dB. Now,
assume the position of the piston is changed in such a way that V1 = 0.15Vt and V2 = 0.85Vt . Calculate
the amplitude margin Am in this new position. The steady state loop gain, Kv = 20s−1 , is unchanged
by having the servo amplifier tuned on Ksa . Assume Bp = 0 and Kce unchanged.
s
β e Ap 2 1
1
+
ωh =
mt
V1
V2
s
1
Kce
1
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 controller Ksa .
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EXAMINATION Page 8(13)
TMHP51/TEN1, 2012-12-20
System Models
[A, 4 points] One way to increase the damping in a hydraulic servo system is to include a fixed orifice,
Cs , across the valve ports, as seen in figure 7. Draw the block diagram of the system including the orifice.
Do not include any controller. The input signal is ∆Xv , output is ∆Xp and the disturbance force is
denoted by ∆FL . You don’t need to draw the intermediate steps, neither does the final block diagram
need to be fully reduced.
Figure 7: 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 controller Ksa . A damping orifice Cs
is included between the chambers.
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EXAMINATION Page 9(13)
TMHP51/TEN1, 2012-12-20
[B, 6 points] Sometimes servo valves are used to control the pressure in an closed volume. This is the
case in some application within the manufacturing industry and also for test equipment. There is no
moving piston in the volume Vt as you can see in figure 8. Neither any external disturbances. The input
to the system is ∆Uc but the output is uncommonly the chamber pressure ∆Pc . Derive a block-diagram
for this system including the valve dynamics modelled by a first order system described by τv . There is
also a servo-amplifier gain Ksa that transforms the input uc into an current input, iv . Take some effort
to simplify the block-diagram as much as possible. The oil parameters are described by ρ for density
and βe for compressibility. The linearized coefficients does not need to be derived.
Figure 8: A pressure control servo valve configuration.
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EXAMINATION Page 10(13)
TMHP51/TEN1, 2012-12-20
Feedback Systems
[A, 10 points] A follower or ”copying” servo is a common element in many systems. Draw a block
diagram for the closed-loop system in figure 9. Also, calculate the hydraulic characteristics of 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. V1 = V2 = Vt /2 = 0.9litre, Ap =
65cm2 , Ps = 210bar, mt = 165kg, Cq = 0.67, w = π × 12mm, Kce = 3.5 × 10−12 , Kq = 0.32, βe = 1GP a.
The mechanical linkage works so that xv = 12 (xref − xp ).
The described system where very common back in the 50s and 60s in Swedish mechanical industry.
The system where used for mass-production of parts, both in metal and wood materials. Often the
machinery worked with a scale-model as reference. Often the model was an upscaled version of the finals
part. A common scale was 8:1. That would change the control-law to xv = 21 (xref − 8xp ). What is the
amplitude margin in this case? Is the system stable?
Figure 9: A follower servo with two separate cylinders of equal dimensions.
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EXAMINATION Page 11(13)
TMHP51/TEN1, 2012-12-20
Suggested Solutions
Cavitation in Orifices
[A] The flow through both orifices are equal; q1 = q2 .
r
2
q1 = Cq A1
(Ps − Pm )
ρ
r
2
q2 = Cq wx
Pm
ρ
Pm = Ps − ∆Pcav = Ps (1 − C2 )
(3)
(4)
(5)
The condition becomes:
xcav
A1
≤
w
r
C2
1 − C2
(6)
[B]
Numerical values in the previous expression.
xcav
8.4 × 10−6
≤
π × 6 × 10−3
r
0.68
= 0.65mm
1 − 0.68
[C] No, he will not hear the sound of cavitation.
r
0.68
3.4 × 10−6
= 0.26mm < 0.3mm
xcav ≤
π × 6 × 10−3 1 − 0.68
(7)
(8)
Servo Valves Charateristics
[A] The hysteresis is not affected, but the pressure gain.
[B] A lowered pressure gain results in an decreased Kq and possible and increased Kce . This gives a
opportunity to increase the feedback gain slightlly due to a lowered loop gain for the spool displacement
control.
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EXAMINATION Page 12(13)
TMHP51/TEN1, 2012-12-20
[C] The values can be found from diagram. 220Hz and 71Hz.
Frequency response of 10gpm servovalve MOOG G761-3004
200
(220,-3)
0
(98,-3)
−5
100
(71,60)
−10
50
(106,60)
−15
20
30 40 50
Phoaselag [deg]
Amp. Ratio [dB]
150
0
100 150 200 300
Frequency [Hz]
[D] The servo-valve is huge, expect unusual numbers.
r
2
(Pupstream − Pdownstream )
q = 2.8 sin(0.3t) = Cq πDs x
ρ
q̇ = 0.3 × 2.8 cos(0.3t)
(9)
(10)
x = 4.28mm, Fs = 4363N + 936N = 5.3kN
Hydraulic Servo Systems
[A] 52ms → 192Hz and 70ms → 143Hz. The reduced bandwidth is due to saturation in the servoamplifier.
Response
100%
50%
Pressure Response [% Ps]
100
80
60
(0.0070,63.0)
40
(0.0052,31.5)
20
0
0
0.005
0.010
0.015
Time [s]
0.020
0.025
[B] The new ωh and δh is defined by the ones at centred piston position like:
r
120
1
1
ωh =
+
= 168
2
0.15 0.85
r
0.20
1
1
δh =
+
= 0.28
2
0.15 0.85
(11)
(12)
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EXAMINATION Page 13(13)
TMHP51/TEN1, 2012-12-20
Kv ≈ 13.5dB
Am = −20 log10 2ωh δh (13)
System Models
[A] This is most likely the easiest task in this exam, ones you realise that Kce = Kc + Cip + Cep /2 from
the formula book transforms into Kce = Kc + Cs . So it is all about drawing a block-diagram from the
transfer function in formula book, replacing Kce → Kc + Cs .
[B] This task might seems difficult, but is rather easy. Write down the model:
r
2
q = Cq wxv
(Ps − Pc )
(14)
ρ
q=
Vc
Ṗc
βe
(15)
This transforms into:
Vt
s Pc → ∆Pc =
Kq ∆Xv − Kc ∆Pc =
βe
"
#
Kq
∆Xv
Vc
βe s + Kc
The first-order system transfer function for the valve spool dynamics will be:
Ksa
∆Xv =
∆Uc
1 + τv s
(16)
(17)
The total transfer function will then be:
Vt
Kq ∆Xv − Kc ∆Pc =
s Pc → ∆Pc =
βe
"
Kq
Vc
s
βe + K c
#
Ksa
∆Uc
1 + τv s
(18)
It is not difficult to write down a block-digram for this open-loop system.
Feedback Systems
[A] ωh = 754rad/s, δh = 0.053, Am = 10.3dB. After the introduction of the scaled-up models the
amplitude margin becomes Am = −7.7dB. That is an unstable system.