LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
1.
EXAMINATION
TMHP 51
2006-12-14
2(6)
a) Flow characteristics for an hydraulic orifice
The figure shows measured flow versus outlet pressure for a sharp edged orifice
supplied with constant inlet pressure, pu = 200 bars. The orifice flow coefficient
can be assumed as constant, Cq = 0.67, at cavitation free flow. The critical
pressure drop
for cavitation is
Δpcav=125 bar.
The flow
through an
orifice at
cavitation can
be described as a change of the flow coefficient Cq.
Calculate and show in a diagram how the Cq-value varies according to the
downstream pressure, as in the diagram above. How about the oil viscosity and
it’s influence on cavitation in the orifice?
(4p)
b) Design of servo valve spool
The figure shows a drawing of a servo valve spool and its bushing. Why is a
”spool-bushing” used and what is the aim of using circulated tracks on the spool
lands (2 circulated tracks on each land)?
(3p)
c) Flow forces in a 2-port seat valve
A 2-port seat valve is supplied by a constant pressure controlled pump with the
max flow qpmax = 335 litre/min. The figure below shows the valve flow (Q) versus
the valve displacement (y) at two different max pressure drop (100 and 200 bar).
Show qualitatively in a diagram how the steady state flow forces (Fs) varies
according to the displacement (y) and the pressure drops (100 and 200 bar).
(3p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
2.
EXAMINATION
TMHP 51
2006-12-14
3(6)
a) Pressure gain for servo valves with different “lapping”
In position servos normally 4-ports symmetric servo valves with null-lapping
(critical center) or under-lapping are used.
Describe in diagram how the ”lapping” will influence the valve pressure gain
(Kp). How can wear in the valve be recognised on the Kp-value?
(3p)
b) Concepts for hydraulic servo systems
In hydraulic servo systems the power to the load actuator (cylinder or motor) can
be controlled by a servo valve or a variable pump.
Compare qualitatively these two system concepts according to efficiency, servo
bandwidth and possibilities to improve system dynamics by different kind of
feedbacks.
(3p)
c) Valve controlled position servo – Position error
Ap
Ap
V1
Kqi
uc
+
ue K i
sa
uf -
V2
Kce
xp
Mt
FL
Position
transducer
Kf
Ps = const.
The figure shows schematically a valve controlled (4-port ”critical center” valve)
position servo. The servo has a proportional controller (Greg = Ksa). The controller
gain (Ksa) is adjusted for the amplitude margin, Am = 6 dB when the resonance
frequency and damping are ωh = 120 rad/s and δh = 0,15 respectively.
Assume that the input signal, uc is a ramp, adjusted for the steady state velocity
vp = 0,25 m/s. The valve threshold value can be neglected.
Derive an expression and calculate the position error ΔXp at steady state
velocity.
Show in a simplified block diagram a control concept, which will reduce ΔXp at
steady state velocity.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
3.
EXAMINATION
TMHP 51
2006-12-14
4(6)
Position servo with 3-port valve controlled asymmetric cylinder
The figure shows an electro-hydraulic position servo with a 3-port valve and an
asymmetric cylinder loaded by a mass and an external force. The servo has
proportional position control with a feedback gain of Kf and a controller gain of
Kreg. The servo valve is zero-lapped and it has symmetric and matched orifices
and the null-coefficients are Kqi0 and Kc0. The valve has a high bandwidth but has
a threshold, iT. The supply pressure, ps is constant. The piston area is A1, the area
on the piston rod side is A1/2 and the cylinder volume on the piston side is V1.
The cylinder is assumed to have no leakage and friction. The bandwidth of the
closed loop servo system is ωb (with the amplitude –3 dB, for a given operation
point).
Parameter values:
A1 = 1,96.10-3 m2
V1 = 0,25.10-3 m3
Kc0 = 8,0.10-12 m5/Ns
iT = 0,4 mA
Mt = 400 kg
ps = 14 MPa
Kf = 25 V/m
βe = 1200 MPa
Kqi0 = 0,013 m3/As
ωb = 15 rad/s
a) Steady state position error at load disturbance
Calculate the maximum position error (ΔXp) which can be caused by a load
disturbance ΔFL = 1000 N and together with the threshold value of the valve, iT.
The pressure gain of the loaded valve is Kpi0 = 1,0.109 N/(m2A).
(5p)
b) Lag-compensation of the control loop
Assume that the proportional position controller (Ksa) is extended by a lag filter
1 + s / ω LC
with the transfer function G LC ( s ) = α
. The steady state gain α =
1 + s ⋅ α / ω LC
2,0. Calculate for this case the position error caused by a load disturbance and
the threshold, (according to task a).
(3p)
c) Influence from a servo valve with slow response
Discuss in general how the bandwidth of the servo system will be influenced if
the valve response is slower that the dynamics of the actuator and load.
(2p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
4.
5(6)
EXAMINATION
TMHP 51
2006-12-14
Angular velocity servo with pump controlled motor and flexible shaft
ep
wp
uc +
Ksa
____
uf - s
p1 V 1
KL
Dm
Dp
p2 V2
i
Au(s)
Jt
.
qm
TL
Speed
transducer
pm
Kf
The figure illustrates an electro-hydraulic velocity servo with a pump controlled
motor loaded by an inertia, Jt with a flexible shaft (KL). The controller is of
integrating type with the static gain Ksa. The pump displacement setting controller
εp
K εi
=
, with the displacement setting coefficient
has the transfer function:
i 1+ s
ωs
Kεi = 20 A and the brake frequency ωs = 120 rad/s. The variable pump shaft
speed is ωp = 157 rad/s and the displacement is Dp = 8,0.10-6 m3/rad. The
-1
volumes between pump and motor are V1 = V2 = 0,5 litre and the effective bulk
modulus is βe = 800 MPa. The motor displacement is Dm = 25.10-6 m3/rad and
the inertias on the motor shaft is Jt = 1,0 kgm2. The torsion spring coefficient of
the motor shaft is KL = 2000 Nm/rad. The transmission total leakage flow
coefficient is Ct = 7,5.10-12 m5/Ns and the viscous friction coefficient is Bm = 0.
The low pressure side has constant pm.
a) Pump controller and it’s influence on the open loop gain, Au(s) and stability
Show with “s-functions” an expression for the open loop gain Au(s) of the
system.
Calculate the hydraulic resonance frequency ωh' and damping δh' with attention
to the spring coefficient, KL.
Show in a bode-diagram how the brake frequency of the pump controller (ωs)
will influence the stability margin of the servo.
Lead: Springs in serial connections gives ω =
'
h
K e ' Ct ⋅ J t ωh'
,δ h =
⋅
Jt
Dm2
2
(6p)
b) Closed loop stiffness versus frequency
Derive a simplified expression of the closed loop stiffness
− ΔTL
at the
•
Δθm
s →1
frequency ω = 1 rad/s.
Show in a bode-diagram the closed loop stiffness amplitude versus frequency.
(4p)
LINKÖPINGS TEKNISKA HÖGSKOLA
IKP
Fluid och Mekanisk Systemteknik
5.
6(6)
EXAMINATION
TMHP 51
2006-12-14
Servo with valve controlled cylinder loaded by two masses
xp
be Ap
Ap be
p 1 V1
V2 p2
KL
M2
M1
BL
xL
GHL(s) =
sXp Kq
=
Xv Ap s2
xv
ω2h
GLX(s)
+
2 δh
s + GLX(s)
ωh
Ps = const.
The figure shows a valve controlled hydraulic cylinder loaded by two masses (M1
and M2) and the corresponding transfer function, GHL(s), from valve spool posis 2 / ω 2 + 2δ a s / ωa + 1
tion (Xv) to cylinder piston velocity (sXp), where GLX ( s ) = 2 a2
.
s / ω1 + 2δ 1 s / ω1 + 1
Frequencies and damping for the masses are ωa = 30 rad/s, δa = 0,05, ω1 = 67
rad/s and δ1 = 0,11. The hydraulic resonance frequency and damping are, according to GHL(s), ωh = 120 rad/s and δh = 0,15, when the load is assumed to be a
single mass system (Mt = M1 + M2).
a) The load dynamics and it’s influence on the cylinder dynamics
Show with equations how the load dynamics (GLX(s)) will influence the hydraulic
resonance frequency and damping and calculate their equivalente values
( ωh' and δ h' ) for the given case. (GLX can be simplified when the damping terms are small.)
Which will occur for the hydraulic resonance frequency and damping if the
mechanical spring coefficient, KL is increased 50 times?
(6p)
b) Dynamic load pressure feedback for increased hydraulic damping
In order stabilize the system above it is very important to increase the hydraulic
damping (δh). For that purpose a dynamic load pressure feedback has to be
implemented.
Show in a block-diagram how you will implement the feedback and derive an
expression for the hydraulic damping versus frequency.
(High pass filter for dynamic pressure feedback: GHP ( s ) =
s /ω f
1+ s /ω f
)
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
1(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2006-12-14
SOLUTIONS FOR EXAMINATION, TMHP51
1.
a) Flow characteristics for an hydraulic orifice
0,67
0,53
Cq
pu = 200 bar and Δpcav = 125 bar .
The figure shows the orifice flow characteristics and the variation in Cq during cavita2
( pu − pd ) (1). During cavitation, q0 = constant and
tion. Flow equation: qo = C q A0
ρ
the flow coefficient can be calculated as C qcav = C q
Δpcav
where (pu-pd)>= Δpcav.
( pu − p d )
The influence from oil viscosity on the flow char is very small for a sharp edged orifice.
(4p)
b) Design of servo valve spool
The spool-bushing is used to guarantee constant gap between spool and bushing,
independent of pressure. That means reduced valve leakage flow at high pressures.
Circulated tracks on each land gives reduction of the radial forces on the spool, which
means reduced friction forces and more accurate spool control.
(3p)
c) Flow forces in a 2-port seat valve
qpmax = 335 litre/min, pressure drop: 100, 200 bar). Flow forces versus spool position, y
Fs
Fsmax
q=constant
Dp=200 bar
Dp=100 bar
0
Flow force equations: Fs = 2C q wxv Δp cos(δ ) or Fs =
1,5
ρ cos(δ )
C q wx v
y [mm]
3,0
q 2 . Constant Δp means
Fs ∼ xv and constant flow means Fs ∼ 1/xv.
(3p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
2.
2(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2006-12-14
a) Pressure gain for servo valves with different “lapping”
Valve pressure gain is defined as K p = K q / K c . For a 4-port servo valve, null-lapping
gives the highest Kp-value because of low Kc-value. Under-lapping means that Kp is
proportional to the inverse of the under-lap, U. Big under-lap means low Kp. Wear
reduces the Kp-value in all kind of valves.
pL
ps
Valve with big
under-lapp
Criticalcenter valve
Kpi0
Worn criticalcenter valve
1
iv
-ps
(3p)
b) Concepts for hydraulic servo systems
Valve controlled systems: Low efficiency because of high pressure drop over the
valve. High bandwidth because of fast valve response. Fast valve response makes it
possible to improve the actuator-load dynamics.
Pump controlled systems: High efficiency because of low pressure losses in the main
circuit. Low bandwidth because of slow pump controller response. Slow pump
controller makes it difficult to improve the actuator-load dynamics.
(3p)
c) Valve controlled position servo
Ap
V1
Kqi
uc
+
xp
Ap
Mt
V2
FL
Position
transducer
Kce
Am = 6 dB, ωh = 120 rad/s, δh = 0,15.
Kf
ue K i
sa
Ps = const.
uf -
Derive an expression and calculate the position error ΔXp at steady state velocity, vp=0,25 m/s.
Control error: U e = U c − U f ,
K qi
Ue Uc −U f
=
= sX p = vb
and U e ⋅ K sa
Kf
Kf
Ap
Ap
v
Ue
vb = b . Am = 6 dB gives Kv = ωh·δh = 18 1/s.
= ΔX p gives ΔX p =
K f K sa K qi
Kv
Kf
Position error, ΔXp at steady state velocity: ΔXp = 0,25/18 = 0,014 = 14 mm.
Reduction of position error by using feed forvard:
Kff s
uc
+
-
+
Feed Forward
+
Ksa
i
Kqi
Ap
1
s 2 + 2 dh s +
1
wh2 wh
.
xp
1
s
xp
K ff =
Ap
K sa K qi
Kf
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
3.
3(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2006-12-14
Position servo with 3-port valve controlled asymmetric cylinder
Parameter values:
A1 = 1,96.10-3 m2, Mt = 400 kg, βe = 1200 MPa, V1 = 0,25.10-3 m3, ps = 14 MPa, Kqi0 =
0,013 m3/As, Kc0 = 8,0.10-12 m5/Ns, Kf = 25 V/m, ωb = 15 rad/s and iT = 0,4 mA.
a) Steady state position error at load disturbance
Position error (ΔXp) caused by a load disturbance ΔFL = 1000 N and threshold value, iT.
Pressure gain Kpi0 = 1,0.109 N/(m2A). Steady state stiffness gives:
Ap2
− ΔFL
= Kv
= K sa K pi A p K f . According to the given system, the
S c s →0 =
ΔX p
K ce
→0
s
position error is, ΔX pF =
K sa K qi 0 K f
− ΔFL
ω A
. Ksa? ωb = K v =
K sa = b 1
K sa K pi 0 A1 K f
A1
K qi 0 K f
− 1000
= 2,24·10-4 m = 0,224 mm.
9
−3
0,091 ⋅ 1,0 ⋅ 10 ⋅ 1,96 ⋅ 10 ⋅ 25
ΔiT
Position error from threshold: ΔX p ⋅ K f ⋅ K sa = ΔiT ⇒ ΔX pT =
= 0,176 mm.
K f ⋅ K sa
Ksa = 0,091 A/V → ΔX pF =
Total position error: ΔX ptot = ΔX pF + ΔX pT = 0,40 mm.
(5p)
b) Lag-compensation of the control loop
Lag fifter: G LC ( s ) = α
1 + s / ω LC
with the steady state gain α = 2,0. Lag-comp
1 + s ⋅ α / ω LC
means increased Kv-value. KvLC = α·Kv or KsaLC = α·Ksa. The equations in task a) gives
ΔX pF ΔX pT
+
the total position error with lag-compensation as ΔX ptLC =
= 0,20 mm
α
α
(3p)
c) Influence from a servo valve with slow response
If the bandwidth of the servo valve ωv << ωh the valve dynamics are dominant in the
control loop, which means that the bandwidth of the closed loop servo system will be
equal to the valve bandwidth.
(2p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
4.
4(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2006-12-14
Angular velocity servo with pump controlled motor and flexible shaft
ep
wp
uc +
Ksa
____
uf - s
p1 V1
KL
Dm
Dp
.
qm
p2 V2
i
Au(s)
Jt
Speed
transducer
pm
TL
εp
i
=
K εi
s
1+
ωs
Kf
Parameter values:
Kεi = 20 A-1, ωs = 120 rad/s, ωp = 157 rad/s, Dp = 8,0.10-6 m3/rad, V1 = V2 = 0,5 litre,
βe = 800 MPa, Dm = 25.10-6 m3/rad, Jt = 1,0 kgm2, KL = 2000 Nm/rad, Ct = 7,5.10-12
m5/Ns, Bm = 0 and pm = constant.
a) Pump controller and it’s influence on the open loop gain, Au(s) and stability
Open loop gain, Au(s)? The figure gives Au ( s ) =
K εiω p D p
Kf
K sa
⋅
⋅ 2
s (1 + s / ωs )Dm ⎛ s
⎞
2δ '
⎜⎜ ' 2 + 'h s + 1⎟⎟
ωh
⎝ ωh
⎠
Kh K L
β e Dm2
K e ' Ct ⋅ J t ωh'
1
1
1
ω =
,δ h =
and K e =
⋅
with
; Kh =
.
=
+
Jt
Dm2
2
V1
Kh + K L
Ke Kh KL
Kh = 1000 Nm/rad and KL = 2000 Nm/rad gives Ke = 667 Nm/rad.
Ct ⋅ J t ωh'
Ke
'
'
Resonance frequency: ωh =
= 26 rad/s and the damping δ h =
⋅
= 0,156.
Jt
Dm2
2
'
h
ωh' << ωs gives no influence on Au(s) around the hydraulic resonance frequency.
Au
wh
1
ws
Am w [rad/s]
1
(6p)
b) Closed loop stiffness versus frequency
From the figure and Formula book the closed loop stiffness can be derived as:
− ΔTL
= Kv
•
Δθm
s →1
Dm2 1
D2
⋅
= K v m . Versus frequency we have,
Ct s s →1
Ct
(4p)
LINKÖPINGS UNIVERSITET
Department of Mechanical Engineering
Fluid and Mechanical Engineering Systems
5.
5(5)
SOLUTIONS FOR EXAMINATION
TMHP 51
2006-12-14
Servo with valve controlled cylinder loaded by two masses
xp
be Ap
Ap be
p 1 V1
V2 p2
KL
M2
M1
GHL(s) =
BL
xL
sXp Kq
=
Xv Ap s2
ω2h
xv
GLX(s)
+
2 δh
s + GLX(s)
ωh
Ps = const.
s 2 / ωa2 + 2δ a s / ωa + 1
GLX ( s ) = 2 2
s / ω1 + 2δ 1 s / ω1 + 1
ωa = 30 rad/s, δa = 0,05, ω1 = 67 rad/s, δ1 = 0,11, ωh = 120 rad/s and δh = 0,15
a) The load dynamics and it’s influence on the cylinder dynamics
GHL(s) gives: sX p =
Xv
K /A
K q / Ap
= 2 q 'p
s
2δ
s
2δ h
+
+ 'h s + 1
s +1
2
'2
ωh ωh
GLX ωh GLX ωh
2
'
where ωh = G LX ( s ) ⋅ ωh and
δ h' = δ h / GLX ( s ) . The absolute value of GLX(s) at high frequencies (about 200 rad/s)
2
2
is GLX ( s ) iω = 200
⎛ω ⎞
⎛ 200 2 / ωa2 ⎞
⎟ = ⎜⎜ 1 ⎟⎟ = 5,0, which gives ωh' = 5 ⋅ ωh = 268 rad/s
≈ ⎜⎜
2
2 ⎟
⎝ 200 / ω1 ⎠
⎝ ωa ⎠
and δ h' = δ h / 5 = 0,067.
KL increases 50 times and ωa =
KL
gives ωa' = 50 ⋅ ωa = 212 rad/s.
M2
ωa' > ωh means that ωh is dominant, which gives ωh = 120 rad/s and δh = 0,15.
(6p)
b) Dynamic load pressure feedback for increased hydraulic damping
Implementation of load pressure feedback with high pass filter GHP ( s ) =
s/wf
s/wf + 1
uc
+
-
s /ω f
1+ s /ω f
.
Kpf
Dynamic load pressure feedback
Ksa
The block diagram gives δ h' =
i
Kqi
+
-
1
Kce +
Vt
s
4be
PL
Ap
Ap
+
.
xp
s /ω f
K ce'
⋅ δ h where K ce' = K ce + K pf
K sa K qi
K ce
1 + s / ωs
Frequency dependent hydraulic damping:
(4p)