Linköpings Tekniska Högskola
IEI
Fluid and Mechatronic Systems
EXAMINATION
TMHP51/TEN1
2011-12-17
Page 1(6)
EXAMINATION IN
Hydraulic Servo Systems, TMHP51 / TEN1
Date:
Saturday 17 December 2012, at 2 pm - 6 pm
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
013-28 11 87
15:30 and 17:00 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!
17 December 2011
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 Mechatronic Systems
1.
EXAMINATION
TMHP 51
2011-12-17
2(6)
a) Cavitation in orifices
Two orifices are connected in series, as shown in the figure. The two orifice areas are
A1= 1,75·10-6 m2 and A2 = 1,2·A1. The inlet pressure to the first orifice is p1 and the final
outlet pressure p2 = 5 bar. For each orifice the critical pressure drop for cavitation can
be calculated as: Δpcav = 0,69 ⋅ pin , where pin is the inlet pressure. The orifice flow
coefficient is Cq = 0,67 and the fluid density is ρ = 890 kg/m3.
Calculate the pressure, p1cav, which initiate cavitation in some of the orifices.
Calculate the flow q1 when cavitation starts.
(4p)
b) Flow forces on servo valve in a constant pressure system
The figure shows a direct controlled 4-port servo valve, supplied by the pressure ps = 21
MPa. The valve flow is qv. The spool displacement (xv) is controlled by a proportional
magnet, which create the force Fm. The valve flow force (Fs) will act against the magnet
force. Max magnet force for spool control is Fm,max = 60 N. The flow force must always
be lower than max magnet force to make the valve controllable.
Calculate the maximum valve flow, qv,max that gives max flow force, Fs,max=0,8·Fm,max,
when cos(δ) = 0,358 and ρ = 890 kg/m3.
(3p)
c) Servo valve efficiency in a constant pressure system
The figure shows a valve controlled servo systems where the supply unit is a constant
pressure controlled variable displacement pump. The pump pressure setting is ps and the
pump flow is equal to load flow, qp = qL in full flow range.
Nominal valve flow: qL,N = Cq ⋅ w ⋅ xv max 1 (7,0 ⋅106 )
ρ
Calculate and show in diagram the servo valve efficiency, ηsv = Pv,out/Pv,in versus the
pressure ps in the range 14 <= ps <= 28 MPa, at constant load flow qL = qL,N.
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
2.
EXAMINATION
TMHP 51
2011-12-17
3(6)
a) Pilot operated servo valves (2-stage valve)
The figure shows a 2-stage servo valve with an electro-hydraulic pilot-stage supplied
from the main P-port (ps), (closed X-port in the figure).
Explain qualitatively how the valve bandwidth (ωb) varies according to the electric
control signal amplitude and supply pressure (ps).
(3p)
b) Servo valve pressure gain and it’s influence on a position servo stiffness
The figure shows the pressure gain (pL versus iv) for a symmetric critical centre servo
valve and a schematic block-diagram of a closed loop position servo.
Describe, with equations how the valve pressure gain and the steady state stiffness are
affected if the valve is exposed to heavy ware.
(3p)
c) Hydraulic frequency and damping in servo systems
The figure below shows schematically a valve controlled hydraulic servo and it’s block
diagram. With the cylinder in centre position (V1 = V2) the hydraulic frequency is ωh =
ωhmin = 120 rad/s and the hydraulic damping is, δh = δhmin = 0,15.
Use the block-diagram to derive an expression of the damping δh = f(ωh).
Calculate the hydraulic frequency ωh and the damping δh at the piston position where
the cylinder volumes are V1 = 0,9·Vt and V2 = 0,1·Vt (total volume, Vt = V1 + V2).
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
3.
EXAMINATION
TMHP 51
2011-12-17
4(6)
Angular position and velocity servo with valve controlled motor
The figure shown an angular servo system (same valve and motor) used as a position
and a velocity servo, respectively. The feedback gain in the position servo is Kf,p, and in
the velocity servo Kf,v. The controllers are proportional and integrating respectively.
The block diagram shows the transfer functions from valve input current (ic) to motor
shaft angle (Ө), including the non-linearity’s from threshold and saturation.
Parameters: ΔiTH = 0,01·imax, imax = 0,10 A, Kqi0 = 0,042 m3/As, Dm = 7,96·10-6 m3/rad,
Kf,p = 0,16 V/rad, Kf,v = 0,014 Vs/rad, ωh = 78 rad/s, δhmin = 0,20 (at low motor speed).
a) Max motor speed
Calculate the maximum motor speed (sӨmmax) that can be reached in the two systems.
(2p)
b) Controller gain (Ksa) that gives the amplitude margin Am = 6 dB
Calculate the controller gains Ksa,p (position servo) and Ksa,v (velocity servo) that gives
the two systems the same amplitude margin, Am = 6 dB, when δh = δhmin.
Describe with equations, how the hydraulic damping (δh), in the velocity servo, will
varies according to the motor speed.
(4p)
c) Position and speed error versus valve Threshold at the frequency ω = 1,0 rad/s
Calculate the position and speed error (ΔӨm, ΔsӨm) caused by valve threshold, ΔiTH,
at the frequency ω = 1,0 rad/s and the controller gains according to task b). (If you not
can solve task b), use the controller gains as parameters in your equations).
Explain why the control error in the velocity servo is frequency dependent and define
the steady state control error (theoretical error).
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
4.
EXAMINATION
TMHP 51
2011-12-17
5(6)
Pump controlled angular velocity servo with ID-controller
The figure shows the system and the block diagram for an electro-hydraulic velocity
servo with a pump controlled motor and an ID-controller, GID(s) = KI/s+ KD·s, where
KI is in [A/(Vs)] and KD in [As/V]. The pump control unit is fast compared to the motor/
load dynamics, (ωps > ωh). The static gain factors in the hydraulic system have the
values: Kps·Dp·ωp/Dm = 1225 rad/(As) and Kf = 0,10 Vs/rad. For the motor/load
dynamics the values are, ωh = 65 rad/s and δh = 0,20 (no damping feedback).
a) Connection of ID-controller
Show in a simplified block-diagram how the ID-controller shall be connected, so that
the derivative controller gain, KD acts as a negative acceleration feedback, which
increases the hydraulic damping.
Derive an equation for the hydraulic damping δh,a, with acceleration feedback.
(4p)
b) Tuning of ID-controller for the amplitude margin Am = 6 dB
Adjust the controller derivative gain KD, so that the hydraulic damping reach the value
δh,a = 0,40.
Adjust the controller integrator gain KI, so that the open loop gain (Au(s)) gives the
amplitude margin Am = 6 dB.
(4p)
c) Closed loop stiffness
⋅
Derive an expression of the closed loop stiffness Sc = − ΔTL / Δ θ m
for the low
ω <ω h
frequency range, ω < ωh.
(2p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IEI
Fluid and Mechatronic Systems
5.
EXAMINATION
TMHP 51
2011-12-17
6(6)
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 coefficient BL). The piston position xp is fed back to a
proportional regulator with the gain Greg = Ksa. The servo valve bandwidth is high.
The block diagram from valve input signal (i) to piston velocity (sXp) is:
s2
+
2δ a
s +1
ωa
where G LX ( s) =
2δ
s
+ 1 s +1
2
ω 1 ω1
ω
2
a
2
The load dynamics is described by the function GLX(s) where, ωa < ω1. The load mass
ratio is M2/M1 = 2 and the frequencies and damping are: ωa = 80 rad/s, ω1 = 138 rad/s,
δa = 0,5 and δ1 = 0,87.
a) Load dynamics and its influence on hydraulic frequency and damping
Show in a bode-diagram the amplitude of the load dynamics, |GLX(s| versus frequency
(ω) for the given parameters.
Assume that the hydraulic stiffness is lower than the mechanical stiffness, Kh = 0,25·KL.
Derive, from the given block-diagram, expressions for the resulting frequency ωh' and
damping, δ h' .
(6p)
b) Low mechanical stiffness, KL < Kh
Assume that the mechanical stiffness is so low that ω1 < ωh in the system above.
Discuss qualitatively how the hydraulic system dynamics will be influenced by the low
mechanical stiffness.
How will the hydraulic system dynamics be affected by very low mechanical damping,
δa ≈ 0,1.
(4p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2011-12-17
SOLUTION SUGGESTIONS FOR EXAMINATION, TMHP51
1.
a) Cavitation in orifices
Orifices: A1= 1,75·10-6 m2 and A2 = 1,2·A1
The critical pressure drop for cavitation: Δpcav = 0,69 ⋅ pin . Cq = 0,67, ρ = 890 kg/m3
p1 that starts cavitation? q1 = Cq ⋅ A1 2 ( p1 − pm ) = Cq ⋅ A2 2 ( pm − p2 ) , which gives with
ρ
ρ
A2 = 1,2·A1: p1 − pm = (1,2) ( pm − p2 ) = 1,44 ⋅ ( pm − p2 ) (1). Δpcav = 0,69 ⋅ pin gives
2
( p1 − pm )cav > ( pm − p2 )cav , which means that cavitation starts first after orifices A2.
Eq. (1) gives p1 = 2,44 pm − 1,44 p2 and pm, cav = p2 /(1 − 0,69) = 16,1 bar gives:
p1,cav = 39,4 – 7,2 = 32,2 bar. Flow: q1 = 0,67 ⋅1,75 ⋅10−6
2
(32,2 − 16,1)105 = 7,05·10-5 m3/s.
890
(4p)
b) Flow forces on servo valve in a constant pressure system
Data: ps = 21 MPa, Fm,max = 60 N, Fs,max= 0,8·Fm,max, cos(δ) = 0,358 and ρ = 890 kg/m3.
Calculate qvmax for Fs,max = 0,8·Fm,max
Flow forces and valve flow:, Fs = 2Cq ⋅ w ⋅ xv ⋅ Δp ⋅ cos(δ ) , qv = Cq ⋅ w ⋅ xv
pL = 0 and Δp = ps → Fs = 2 ⋅ ρ ⋅ cos(δ ) ⋅ qv ⋅ ps and qv max =
Fs,max = 0,8·Fm,max, gives qv max =
0,8 ⋅ 60
2 ⋅ 890 ⋅ 0,358 ⋅ 21⋅10
6
1
ρ
( ps − pL ) .
Fs , max
2 ⋅ ρ ⋅ cos(δ ) ⋅ ps
.
= 4,9·10-4 m3/s = 29,4 l/min
(3p)
c) Servo valve efficiency
qp = qL and qL ,N = Cq ⋅ w ⋅ xv max
Calculate ηsv = Pv,out/Pv,in: η sv =
(Δpv = 7,0 MPa) → η sv =
(7,0 ⋅10 ) .
ρ
1
6
qL ⋅ pL
. qp = qL gives ηsv = pL/ps. Assuming qL = qL,N
qs ⋅ ps
ps − Δpv
.14 <= ps <= 28 MPa →
ps
(3p)
LINKÖPINGS UNIVERSITET
IEI
Fluid and Mechatronic Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2011-12-17
a) Pilot operated servo valves (2-stage valve)
2-stage servo valve with electro-hydraulic pilot-stage supplied from the P-port (ps).
Bandwidth versus control signal amplitude: Flow saturation in the pilot stage gives
max velocity for the main spool control. Therefore, the valve bandwidth (ωb) will be
reduced for high signal amplitude.
Bandwidth versus supply pressure: Since, the pilot stage is used as a servo for position
control of the main spool its control loop gain is proportion to K q , p ∝ ps , which
gives ωb ∝
ps
(3p)
b) ) Servo valve pressure gain and it’s influence on a position servo stiffness
Valve pressure gain and its influence on steady state stiffness: The Block-diagram (and
K qi
Ap2
− ΔFL
= K sa
Kf
= K sa Ap K f ⋅ K pi where Kpi is the
Formula Book) gives, Sc =
ΔX p
Ap
K ce
s →0
pressure gain. Heavy valve ware means reduced Kpi and thereby, reduced stiffness.
(3p)
c) Hydraulic frequency and damping in servo systems
Data: ωhmin = 120 rad/s and δhmin = 0,15. V1 = 0,9·Vt and V2 = 0,1·Vt (Vt = V1 + V2)
Expression of δh = f(ωh).: The -1 feedback loop in the block-diagram gives the func.,
Gh ( s ) =
1
V1 ⋅ V2 M t 2 K ce M t
s +
s +1
Vt β e Ap2
Ap2
where ωh =
Vt β e Ap2
V1 ⋅V2 M t
and δ h =
ωh K ce M t
2
⋅
Ap2
V
4
V 11,1
ωh, δh for V1 = 0,9·Vt and V2 = 0,1·Vt V1 =V2 → t = ;V1 = 0,9⋅Vt ,V2 = 0,1⋅Vt → t = ,
V1 ⋅V2 Vt
V1 ⋅V2 Vt
which gives: ωh =
ωh
11,1
⋅ ωh min = 200 rad/s and δ h =
⋅δ
= 0,25.
ωh min h min
4
(4p)
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IEI
Fluid and Mechatronic Systems
3.
SOLUTIONS FOR EXAMINATION
TMHP 51
2011-12-17
3(5)
Angular position and velocity servo with valve controlled motor
Parameters: ΔiTH = 0,01·imax, imax = 0,10 A, Kqi0 = 0,042 m3/As, Dm = 7,96·10-6 m3/rad,
Kf,p = 0,16 V/rad, Kf,v = 0,014 Vs/rad, ωh = 78 rad/s, δhmin = 0,20 (at low motor speed).
a) Max motor speed
Max motor speed (sӨmmax): Block-diagram gives sθ m = i
sθ m max = 0,10 ⋅
K qi 0
K qi
→ sθ m max = imax
Dm
Dm
0,042
= 528 rad/s. The same max speed for the two systems.
7,96 ⋅ 10− 6
(2p)
b) Controller gain (Ksa) that gives the amplitude margin Am = 6 dB
Same Au(s) for the two systems gives for Am = 6 dB, K v = ωh ⋅ δ h, min = 15,6 1/s.
Position servo: K v , p = K sa , p
Velocity servo: K v ,v = K sa ,v
K qi 0
Dm
K qi 0
Dm
K f , p and K sa , p =
K v , p ⋅ Dm
K qi 0 K f , p
=
15,6 ⋅ 7,96 ⋅10−6 = 0,0185 A/V.
0,042 ⋅ 0,16
−6
K ⋅D
K f ,v and K sa ,v = v ,v m = 15,6 ⋅ 7,96 ⋅ 10 = 0,211 A/(Vs).
K qi 0 K f ,v
0,042 ⋅ 0,014
Hydraulic damping in velocity servo: sθ m ∝ q ∝ i ∝ xv ∝ K c ∝ δ h → δ h ∝ sθ m
(4)
c) Position and speed error versus valve Threshold at the frequency ω = 1,0 rad/s
Same valve in the two systems gives same threshold: ΔiTH = 0,01·imax = 0,001 A.
ΔiTH
0,001
Position error versus Threshold: Δθ m =
. Δθ m =
= 0,34 rad.
K f , p K sa , p
0,16 ⋅ 0,0185
Velocity error versus Threshold: Δsθ m =
ΔiTH
0,001
. Δsθ m =
= 0,34 rad/s.
K f , v K sa , v
0,014 ⋅ 0,211
Control error (E) in the velocity servo: Δi = (U c − U f )
K sa
K
sΔi
= ( E ) sa → E =
.
s
s
K sa
Frequency-dependent E means that s→0 → E→0 (steady state control error).
(4p)
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IEI
Fluid and Mechatronic Systems
4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2011-12-17
Pump controlled angular velocity servo
Kps·Dp·ωp/Dm = 1225 rad/(As), Kf = 0,10 Vs/rad, ωh = 65 rad/s and δh = 0,20, Am = 6 dB.
ID-controller, GID(s) = KI/s+ KD·s, where KI is in [A/(Vs)] and KD in [As/V]
a) Connection of ID-controller
Block-diagram of the ID-controller connection, so KD acts as acceleration feedback:
Together with given block-diagram (with
.
Gps=1) the system dynamics becomes as, Gh ( s) =
which gives the hydraulic damping as: δ h ,a
1
⎛s
Dω
⎛
⎜ 2 + ⎜ 2δ h + K D K ps p p K f
⎜ω ⎜ ω
Dm
⎝ h ⎝ h
Dω
ω
= δ h + h K D K ps p p K f (1)
2
Dm
2
⎞
⎞
⎟⎟ s + 1⎟
⎟
⎠
⎠
,
(4p)
b) Tuning of ID-controller for the amplitude margin Am = 6 dB
By reduction of the controller D-action into the system dynamics (just affect damping), the
D pω p
Kv
K
=
K
K
Kf .
,
where
open loop gain became as, Au ( s) =
v
I
ps
Dm
⎛ s 2 2δ h , a
⎞
s ⋅ ⎜⎜ 2 +
s + 1⎟⎟
ωh
⎝ ωh
⎠
Tuning of KD for δh,a = 0,40: (1) → K D =
− δ h )⋅ 2
(0,4 − 0,20) ⋅ 2
= 5,02·10-5 As/V.
=
Dpω p
65 ⋅1225⋅ 0,1
ωh K ps
Kf
Dm
Tuning of KI for Am = 6 dB: K v = K I K ps
Numerical: K I ==
(δ
h ,a
D pω p
Dm
K f = ω h ⋅ δ h ,a → K I =
K ps
ωhδ h,a
Dpω p
Dm
Kf
65 ⋅ 0,4
= 0,212 A/(Vs).
1225⋅ 0,1
(4p)
c) Closed loop stiffness
Steady state stiffness: The block-diagram gives: Sc =
− ΔTL
Δsθ m
=
ω <ω h
K v Dm2
⋅
s Ct
(2p)
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Fluid and Mechatronic Systems
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2011-12-17
Linear position servo with valve controlled cylinder loaded by two masses
s2
+
2δ a
s +1
ωa
2δ
s
+ 1 s +1
2
ω1 ω1
where G ( s ) = ω
LX
2
a
2
Data: M2/M1 = 2, ωa = 80 rad/s, ω1 = 138 rad/s, δa = 0,5 and δ1 = 0,87.
a) Load dynamics and its influence on hydraulic frequency and damping
Bode-diagram for GLX(s):
The block-diagram gives the transfer function Gh(s) = sXp/i as:
4βe Ap2
sX p
K / A ⋅ G (s)
K β (M + M2 )
, δh = ce e 1
= 2 qi p LX
where ωh =
,
⎛s
⎞
(
i
M1 + M2 )Vt
Ap
Vt
2δ h
⎜⎜ 2 +
s + GLX ( s ) ⎟⎟
⎝ ωh ω h
⎠
since ωh is much lower than ωa and |GLX(s)|ω=ωh = 1.0, according to the bode-diagram.
(6p)
Gh ( s ) =
b) Low mechanical stiffness, KL < Kh
Low mechanical stiffness giving ω1 < ωh means that |GLX(s)|ω=ωh > 1,0. By writing the
K qi / Ap
transfer function in standard form as: Gh ( s ) =
, it can be
2
⎛
⎞
s
2δ h
⎜⎜ 2
s + 1⎟⎟
+
G
s
G
s
(
)
(
)
ω
ω
h LX
⎝ h LX
⎠
seen that the hydraulic frequency and damping will be influenced. |GLX(s)| > 1,0, will
increase the frequency and reduce the damping.
Very low mechanical damping, δa ≈ 0,1, will gives heavy oscillations in the system.
(4p)