Linköpings Tekniska Högskola
IEI
Fluid and Mechanical Engineering Systems
EXAMINATION
TMHP51/TEN1
2008-12-13
Page 1(6)
EXAMINATION IN
Hydraulic Servo Systems, TMHP51 / TEN1
Date:
Saturday 13 December 2008, at 8 am - 12 am
Room:
??
Allowed educational aids: Tables: Standard Mathematical Tables or similar
Handbooks: Tefyma
Formularies:
Formula Book for Hydraulics and Pneumatics, LiTH/IEI
Mekanisk Värmeteori och Strömningslära
Pocket calculator
Number of questions
in the examination:
5
On the front page and on all following pages the student must write:
AID-number, TMHP51/TEN1, YYMMDD, page number
Responsible teacher:
Tel.no. during exam:
Will visit at:
Course administrator:
Karl-Erik Rydberg
28 11 87, mob 073-806 18 39
9:30 and 11:00 o’clock
Rita Enquist, tel.nr. 28 11 89, [email protected]
Score:
Maximal score on each question is 10.
To get the mark 3, you will need 20 points
To get the mark 4, you will need 30 points
To get the mark 5, you will need 40 points
Solutions:
Results:
GOOD LUCK!
13 December 2008
Karl-Erik Rydberg
Professor
You will find the solutions of this examination on the notice board
in the A building, entrance 17, C-corridor to the right.
Results will be announced ??.
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
1.
EXAMINATION
TMHP 51
2008-12-13
2(6)
a) Cavitation in orifices
Two orifices are used to create a middle-pressure pm, as shown in the figure below. The
first orifice has the area A1 = 1,5·10-6 m2 and the second area is A2. Inlet pressure to the
first orifice is p1 = 100 bar and the outlet pressure after the orifices is p2 = 14 bar. For
each orifice the critical pressure drop for cavitation can be calculated as:
Δpcav = 0,63 ⋅ pin , where pin is the inlet pressure.
Calculate the pressure pm, which gives cavitation free flow for the two orifices.
Calculate the orifice area A2 for that pm.
(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 vary according to the valve spool displacement xv.
Max flow force occurs at xv = 0,6⋅xvmax when Δpvmax = 21 MPa and it’s value is Fsmax =
65 N. The pump shaft speed in np = 1500 rpm.
Derive an expression for Fsmax as a function of qv and Δpv.
Calculate Fsmax when the pump speed is increased to np = 2000 rpm and the pump
pressure is increased so that Δpvmax = 25 MPa. (Assume constant pump efficiency).
(4p)
c) Hydraulic fluids and it’s influence on servo system performance
Describe qualitatively how a reduction of the bulk modulus and the viscosity of a
hydraulic fluid will influence the dynamic behaviour of a servo system, like stiffness
and oscillation amplitude.
(2p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
2.
EXAMINATION
TMHP 51
2008-12-13
3(6)
a) Efficiency of servo systems with different supply units
The figure shows two valve controlled servo systems with different supply units. In the
first system the pump displacement is fixed and the second system has a constant pressure controlled variable displacement pump. The two systems have the same pump
pressure ps = 21 MPa and the load pressure in one operation point is pL = 14 MPa. Max
pump flow is assumed to be equal to max valve flow (pL = 0).
Calculate the hydraulic system efficiency, ηhs = Ph,out/Ph,in, for the two systems at the
given operation point, (all components are assumed to have ideal characteristic).
Describe qualitatively how a change in pL will influence ηhs for the two systems.
(4p)
b) Selection of a servo valve
You have to select a servo valve for a linear position servo with proportional control.
The system has to reach some piston velocity, have low steady state control error, high
stiffness and high bandwidth.
Describe the most important valve parameters, in order to fulfil the system spec.
(3p)
c) Valve controlled position servo with an orifice in parallel to the piston
The figure shows schematically a valve controlled position servo and a block diagram
for the hydraulic part of the system (without Cs). A fixed orifice with the flow/pressurecoefficient Cs, has to be implemented in parallel to the piston (see figure).
Show in a block diagram (as above) how the orifice (Cs) shall be included. Derive an
expression for the hydraulic damping (δh), as a function of Cs.
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
3.
EXAMINATION
TMHP 51
2008-12-13
4(6)
Valve controlled position servo and velocity servo respectively
The figure shows a system used as a position servo and a velocity servo respectively.
The position servo has a proportional controller (Greg = Ksa) and the velocity servo has
an integrating controller (Ksa/s). The two applications are equipped with the same servo
valve, which is zero-lapped with the coefficients Kqi and Kc. The internal leakage flow
in the motor can be neglected compared to the servo valve. In the most critical operation
point the resonance frequency is ωh = 125 rad/s and the hydraulic damping is δh = 0,16.
a) Loop gain and bandwidth of the systems with equal amplitude margin, Am
Calculate for the two systems the steady state loop gain (Kv), which gives the
amplitude margin Am = 6 dB. Calculate the bandwidth (ωb) of the two systems.
Show in a bode-diagram how the amplitude margin for the velocity servo will be
influenced by increased motor speed.
(4p)
b) Stiffness of the closed loop system
The figure describes a general block diagram for the stiffness of a closed loop system.
Derive from the block diagram the stiffness Sc =
− ΔTL
− ΔTL
and Sc =
for the
•
Δθ m
Δθm
position- and velocity servo respectively. Show in a bode-diagram how the amplitude
of the velocity servo stiffness will be changed when the controlled motor speed is
increased.
(4p)
c) Influence of threshold in servo valve
Assume that the servo valve has a threshold value iTH = 0,01·imax.
Describe with equations if the threshold results in different control errors in the two
systems when the steady state loop gain in both cases are adjusted to give equal
amplitude margin, Am = 6 dB.
(2p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
4.
EXAMINATION
TMHP 51
2008-12-13
5(6)
Linear position servo with velocity and acceleration feedback
The figure illustrates a block diagram for an electro-hydraulic position servo with a
symmetric servo valve and cylinder, loaded by the mass Mt. The controller has three
different gains, one proportional gain KP of the position error, the gain KD for the
negative velocity feedback and KAC for the negative acceleration feedback. Servo valve
bandwidth is much higher than ωh.
For the basic system (without velocity and acceleration feedback) the hydraulic
resonance frequency is ωh = 100 rad/s and the damping is δh = 0,15.
The total velocity feedback gain is adjusted for the value, Kfv·KD·Kqi/Ap = 5,0 [-].
For the acceleration feedback the stationary gain is Kfv·KAC·Kqi/Ap = 8,0·10-3 [s].
a) Velocity and acc- feedback influence on frequency and damping
Show with equations how the velocity and acceleration feedback influence the system
resonance frequency (ωh) and damping (δh).
Calculate the new hydraulic resonance frequency ωh' and damping δh' according to the
velocity and acceleration feedback.
Lead: Reduce the given block-diagram in order to include the velocity and acceleration feedbacks in the
transfer function for the actuator/load dynamics.
(6p)
b) The feedbacks influence on the steady state stiffness of the position servo
Calculate the increase of the position servo steady state stiffness, Sc =
− ΔFL
ΔX p
s →0
according to the velocity and acceleration feedbacks, compared to the case without
these feedbacks. Assume that the steady state loop gain (Kv) in both cases are adjusted
for equal amplitude margin, Am = 6 dB.
Lead: Kv = δh·ωh.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechanical Engineering Systems
5.
EXAMINATION
TMHP 51
2008-12-13
6(6)
Angular position servo with pump controlled motor and two inertia loads
The figure shows an electro-hydraulic angular position servo with a pump controlled
motor. The controller is proportional with the gain factor Ksa. The dynamics of the
pump displacement controller is represented by a first order low pass filter with the
break frequency ωs=150 rad/s.
The dynamics of the system, including the load is described in the following block
diagram:
The total inertia on the motor shaft is, Jt = J1 + J2 = 1,0 kgm2 and if the load dynamics
is assumed as very stiff the value for the hydraulic resonance frequency is ωh = 30 rad/s
and its damping is δh = 0,20.
a) The load dynamics influence on system stability margin
Show with equations and bode diagram, which requirement the load transfer function,
GLθ must fulfil to not influence the stability margin of the system.
Calculate an approximative value of the mechanical torsion spring coefficient KL
[Nm/rad] to fulfil the requirement above, when J1 = J2 = 0,5 kgm2.
(The load dynamics shall be approximately 5 times faster than ωh.)
(6p)
b) Slow pump displacement controller
Describe qualitatively how the closed loop system bandwidth will be changed if the
pump displacement controller is slower than the system dynamics (ωs < ωh).
How will the load dynamic stiffness now influence the closed loop system bandwidth
and stability?
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
SOLUTIONS FOR EXAMINATION
TMHP 51
2008-12-13
1(5)
SOLUTION SUGGESTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices
. The critical pressure drop for cavitation depends
upon the upstream pressure pin as Δpcav = 0,63 ⋅ pin .
Orifice A1: ( p1 − pm )cav = 0,63 ⋅ p1 gives pm, cav = (1 − 0,63) p1 = 37 bar
Orifice A2: ( pm − p2 )cav = 0,63 ⋅ pm gives pm, cav = p2 /(1 − 0,63) = 37,8 bar
Requirements on pm for cavitation free flow: 37 < pm < 38 bar, pm = 37,5 bar is OK.
p − pm
A2 = ? Flow eq., q = Cq ⋅ A1 2 ( p1 − pm ) = Cq ⋅ A2 2 ( pm − p2 ) → A2 = A1 1
pm − p2
ρ
ρ
A1 = 1,5·10-6 m2 gives: A2 = 1,5 ⋅10 −6
100 − 37,5
= 2,45·10-6 m2.
37,5 − 14
(4p)
b) Flow forces on a valve in a constant pressure system
Steady state flow forces: Fs = 2C q ⋅ w ⋅ xv ⋅ Δpv ⋅ cos δ . Δpv = pp and pL = 0 gives the
valve flow as qv = C q w ⋅ xv
Δpv
ρ
. Finally Fs max = 2 ρ cos δ ⋅ qv max Δpv max . Constant
Pump flow increases to qpmax = qp·2000/1500 = qvmax and increased Δpv to 25 MPa ⇒
2000 25
Fs max = 65
= 94,6 N
1500 21
(4p)
c) Hydraulic fluids and it’s influence on servo system preformance
Bulk modulus, β e: Reduction of βe gives lower frequency, since ωh ∝ β e and lower
damping, δ h ∝ β e , which means reduced stiffness and higher oscillation amplitude.
Viscosity, η: Lowering of the viscosity gives higher leakage flow and therefore higher
Kce-value, which means reduced stiffness and reduced oscillation amplitude.
(2p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2008-12-13
a) Efficiency of servo systems
ηhs = Ph,out/Ph,in
Fixed pump: qs = qLmax gives qs = Cq ⋅ w ⋅ xv 0
1
ρ
ps → ηhs =
q L pL
=
qs ps
.
ps − p L ⋅ pL
ps ⋅ ps
.
21 − 14 ⋅ 14
= 0,385.
21 ⋅ 21
p
14
Variable pump: qs = qL gives ηhs = L =
= 0,67.
ps 21
ps = 21 MPa and pL = 14 MPa → ηhs =
For the fixed pump the efficiency 0,385 is a max value, so a change in load pressure
from the point pL = 0,67ps will give a lower efficiency value. The variable pump has an
efficiency proportional to the load pressure.
(4p)
b) Selection of a servo valve
Select a servo valve for a linear position servo with proportional control; for some piston
velocity, low steady state control error, high stiffness and high bandwidth.
The piston velocity is given by the nominal flow of the valve (qN at Δpv = 70 bar). Low
steady state error requires low threshold value of the valve. High stiffness requires a
valve with high pressure gain (Kp). High bandwidth requires a fast valve, ωv > ωh .
(3p)
c) Valve controlled servo cylinder with hydro-mech load pressure feedback
Damping: Including Cs in the block between flow and load pressure gives the new block
1
K c + Cs +
Vt
s
4β e
where K c + C s = K ce δ h ∝ K ce gives δ h =
K c + Cs
Ap
βeM t
Vt
.
(3p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
3.
SOLUTIONS FOR EXAMINATION
TMHP 51
2008-12-13
3(5)
Valve controlled position servo and velocity servo respectively
Parameter values: ωh = 125 rad/s and δh = 0,16
a) Loop gain and bandwidth of the systems with equal amplitude margin, Am
The loop gain is equal for the two systems, as Au ( s) =
K sa K qi K f / Dm
⎛ s 2 2δ
⎞
s ⋅ ⎜⎜ 2 + h s + 1⎟⎟
⎝ ωh ωh
⎠
where
K sa K qi K f / Dm = K v . Am = 6 dB gives K v = δ h ⋅ ωh = 125 ⋅ 0,16 = 20 1/s.
System bandwidth: The system loop gain gives that ωb = Kv, so ωb = 20 rad/s.
Influence on the amplitude margin for the velocity servo by increased motor speed.
(4p)
b) Striffness of the closed loop system
Closed loop stiffness of position and velocity servo:
⎛ s
⎞ ⎛ s 2 2δ h
⎞
⎜
⎟⎟ ⋅ ⎜⎜ 2 +
+
1
s + 1⎟⎟
⎜
2
⎠
Position servo: Sc = − ΔTL ≈ K v Dm ⋅ ⎝ K v ⎠ ⎝ ωh ωh
Δθ m
⎛
⎞
K ce
s
⎜⎜1 +
⎟⎟
⎝ 2δ hωh ⎠
⎛ s
⎞ ⎛ s 2 2δ
⎞
s3
2δ h 2 s
⎜⎜
+ 1⎟⎟ ⋅ ⎜⎜ 2 + h s + 1⎟⎟
s +
+
+1
2
2
ω
ω
Velocity servo:
D K vωh K vωh
Dm ⎝ K v
− ΔTL
Kv
h
⎠ ⎝ h
⎠
Sc =
= Kv
⋅
≈ Kv
⋅
•
K
K
⎛
⎞
⎛
⎞
V
s
ce
ce
t
Δθ m
⎟⎟
s ⋅ ⎜⎜1 +
s ⎟⎟
s ⋅ ⎜⎜1 +
⎝ 4 β e K ce ⎠
⎝ 2δ hωh ⎠
2
m
(4p)
c) Influence of threshold in servo valve
In the position servo the control error ΔӨm is proportional to ΔiTH but in the velocity
servo ΔsӨm is proportional to sΔiTH, which means that ΔsӨm → 0 when s → 0.
(2p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
4.
SOLUTIONS FOR EXAMINATION
TMHP 51
2008-12-13
4(5)
Linear position servo with velocity and accleration feedback
ωh = 100 rad/s and δh = 0,15. Kfv·KD·Kqi/Ap = 5,0 [-]. Kfv·KAC·Kqi/Ap = 8,0·10-3 [s].
a) Velocity and acc- feedback influence on frequency and damping
Velocity and acceleration feedback: The second order transfer function in the block
1
1
diagram changes as
→
⎞
⎛ s 2 ⎛ 2δ h
⎛ s 2 2δ h
K ⎞
K ⎞
⎜
⎜⎜ 2 +
s + 1⎟⎟
+ ⎜⎜
+ K fv K AC qi ⎟⎟ s + 1 + K fv K D qi ⎟
2
⎜ ωh
Ap ⎠
Ap ⎟⎠
⎠
⎝ ωh ωh
⎝ ωh
⎝
The transfer function on standard form gives:
1 / K vfv
1 / K vfv
, where Kvfv = 1+ Kfv·KD·Kqi/Ap
=
2
⎛ s2
⎞ ⎛ s
⎞
2δ h'
⎛ 2δ h
K fAC ⎞
⎜
⎟ s + 1⎟ ⎜⎜ ' 2 + ' s + 1⎟⎟
+⎜
+
ωh
⎜ ωh2 K vfv ⎜ ωh K vfv K vfv ⎟
⎟ ⎝ ωh
⎠
⎠
⎝
⎝
⎠
K fAC ⎞
ωh' ⎛⎜ 2δ h
'
'
⎟.
ω
=
ω
K
+
and KfAC = Kfv·KAC·Kqi/Ap. → h
h
vfv and δ h =
2 ⎜⎝ ωh K vfv K vfv ⎟⎠
Kvfv = 6, KfAC = 8,0·10-3 s, ωh = 100 rad/s and δh = 0,15 gives:
ωh' = ωh K vfv = 245 rad/s and the damping δ h' =
K fAC ⎞
ωh' ⎛⎜ 2δ h
⎟ = 0,225.
+
2 ⎜⎝ ωh K vfv K vfv ⎟⎠
(6p)
b) The feedbacks influence on the steady state stiffness of the position servo
− ΔFL
General definition of steady state stiffness: Sc =
ΔX p
s →0
Ap2
= Kv
K ce
Block diagram and its reduction of feedbacks gives the steady state loop gain is:
K
K
K v' = K P' ⋅ qi ⋅ f . Without inner feedbacks, K v = δ hωh ⇒ Am = 6dB , where
Ap K vfv
Kv = K P ⋅
K qi
Ap
⋅ K f . With velocity and acceleration feedbacks: K v' = δ h' ωh' ⇒ Am = 6dB
K v = δ hωh = 15 1/s och K v' = δ h' ωh' = 55 1/s.
− ΔFL
Velocity and acc feedbacks gives: Sc =
ΔX p
Sc , va
Sc
=
s →0
'
v
K K vfv
Kv
=
= K K vfv
'
v
s →0
Ap2
K ce
, which gives:
55 ⋅ 6
= 22. The steady state stiffness increases 22 times.
15
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechanical Engineering Systems
5.
SOLUTIONS FOR EXAMINATION
TMHP 51
2008-12-13
5(5)
Angular position servo with pump controlled motor and two inertia loads
Parameter values: ωs=150 rad/s, Jt = J1 + J2 = 1,0 kgm2, ωh = 30 rad/s and δh = 0,20.
a) The load dynamics influence on systemets stability margin
K pi D pω p
GLθ ⋅ K f
K sa
,.
From block diagram the loop gain is: Au ( s ) = s ⋅ (1 + s / ω )D ⋅ ⎛ 2
⎞
2δ h
s
s
m
⎜⎜ 2 +
s + GLθ ⎟⎟
⎝ ωh ωh
⎠
If GLθ ( s ) = 1,0 the load dynamics have no influence on ωh, δh and the steady state loop
gain Kv. Since Am ∝
Kv
the stability margin is not influenced if GLθ ( s ) = 1,0 .
− 2δ hωh
GLθ ( s ) → 1,0 for ω < ωa < ω1. For this application it means that ωh < ωa < ω1.
Loop gain versus ω:
KL that gives ωa = 5ωh? ωh =
ωa =
KL
. ωa = 5ωh →
J2
Kh
= 27rad / s → K h = ωh2 ( J1 + J 2 ) = 900 Nm / rad .
J1 + J 2
Kh
52 K h J 2
KL
gives KL = 11250 Nm/rad.
=5
→ KL =
J2
J1 + J 2
J1 + J 2
(6p)
b) Slow pump displacement controller
K sa K pi D pω p K f
Kv
⋅
=
.
s (1 + s / ωs )Dm s (1 + s / ωs )
Au ( s )
1
Closed loop transfer function: Gc ( s ) =
.
=
1 + Au ( s ) s (1 + s / ωs ) / K v + 1
A reasonable stability margin gives that Kv = ωs and the system bandwidth ωb ≈ ωs.
ωs < ωh gives the dominant loop gain as Au ( s ) =
Closed loop system gain:
If ωs is much lower that ωh the load dynamics can have less stiffness (ωa = 5ωs)without
any influence on system bandwidth and stiffness.
(4p)