training servo hydraulic drive systems 36
TRANSCRIPT
Sizing of electro-hydraulic drive systems with proportional and servo valves
Required knowledge: Hydraulic drives, hydraulic circuits and basic physics of machine dynamics
Version 3.6
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Why is it not possible to size a servo or proportional valve without knowing the controlled system?The servo or proportional valve is an integral part of the controlled system and has to be adopted to the system.
servo valve
aux. valves
PT
inductivity, capacity of pipes
controlled systemm
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What has to be considered for sizing servohydraulic systems?
• Physics of dynamic processes (mass forces)
• Dynamic response of a drive (natural frequency)
• Pipe characteristics (natural frequency)
• Physics of hydraulic resistors (throttle valves)
• Physics of a meter-in & meter-out circuit (cavitation)
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Basic hydraulic-mechanic rules
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Rated Flow
What means rated flow?The rated flow relates to a rated pressure drop, and describes one (arbitrarily selected) working point.
Standard rated pressure drops are 10bar (5bar/edge) or 70bar (35bar/edge)
The rated flow is one defined working point of the valve and, by applying the equation below, any other working point of the valve can be calculated.
NN p
pQQΔΔ
=
Note: Above equation is only valid when the flow speed is < 30m/s, above this flow speed we get flow saturation. This means, even with higher pressure drop no increase of flow will happen! Higher flow speeds lead to cavitation and increased wear of the valve.
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Thumb Rules for Valve Sizing
Max. flow depending on valve size (see valve catalogues)
The listed max. flow can only be achieved, when all components in the loop are designed for the required flow!
Normally this means, that all additional components as well as pipes/drillings in the loop are one size bigger than the proportional or servo valve!
At manifold blocks it is important to have suited drilling diameters (bigger than valve ports) and a low flow resistance design!
valve size (according to ISO standard) port Ø Qmax l/min
ISO 4401-03 7,5 mm 75ISO 4401-05 11,2 mm 180ISO 4401-07 20 mm 550ISO 4401-08 28 mm 1100ISO 4401-08 32 mm 1500ISO 4401-10 50 mm 3600ISO 10372-04 8,2 mm 95
Moog D791 16 mm 360Moog D792 28 mm 1100
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Hydraulic Power Transmission
The smallest possible valve is the best valve for closed loop control!
At 1/3 pressure drop over the valve you get the smallest possible valve for a given application!
Hydraulic Power Transmission at a Throttle Valve
0
10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50 60 70 80 90 100
Differential Pressure in %
Tran
smitt
ed H
ydra
ulic
Pow
er in
%
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Hydraulic actuator with servo valve
hydraulic resistor bridge
v
m
F
The center position of a zero lapped valve spool represents a hydraulic resistor bridge circuit.
At movement only 2 resistors are active, one in meter-in the other in meter-out function.
P
T
A B
<
><
><
><>
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Spool of a Servo Valve
hydraulic resistor bridge
Caution: To stop a hydraulic drive with zero lapped valves in a positionrequires a closed loop system. The spool center position cannot be used for safety circuits.
T TA P B
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Hydraulic – Electric Analogy
throttles are hydraulic resistors
UUx
i
voltage U = pressure pcurrent i = flow QDifference: The hydraulic resistor is nonlinear
PpA/B
T
Q
>< ><
Δ p
Q
characteristic of hydraulic resistor
Δ u
i
characteristic of electric resistor
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Pressure Gain of a Zero Lapped Valve
When flow in A and B is zero (e.g. blocked actuator) there is a flow across the 4 adjustable throttles (= valve metering edges). The slope of the plot of the differential pressure A-B over the valve command signal is the pressure gain.
P
T
A B<
><
><
><>
By moving the valve spool the differential pressure A-B changes.
spool displacement
diff
eren
tial p
ress
ure
A
-B
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Frequency Response of a Servo Valve
[ ]E
A
xxdBA log20= D633
The natural frequency of the valve as well as the damping of the valve can be taken from the frequency response plot.
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Acceleration Force (Mass Force)
Acceleration means a change of the speed of a mass. To change the speed of a mass forces are required.
• acceleration means a force in the direction of movement
• deceleration means a force against the direction of movement
(acceleration, speed and force are vector variables)
Newton's second law: Fm = m * a [1 kg * 1 m/s² = 1 N]
Example: A mass of 500 kg has to be accelerated (or decelerated) within 25 ms from 0 to 1 m/s.
Linear acceleration: a = v/t = 1m/s / 0,025 s = 40 m/s²
Acceleration force: Fm = m * a = 500 kg * 40 m/s² = 20.000 N
At an actuator area of 25 cm² an acceleration pressure of 80 bar is required. Deceleration under the same conditions means deceleration pressure.
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Weight (Mass Force)
The weight (force) can be derived from the mass:
Equation: Fg = m * g
Gravity: g = 9,81 m/s²
Example: Weight (force) of a 500 kg mass
Weight (force): Fg = m * g = 500 kg * 9,81 m/s² = 4.905 N
pressure: p = force / area
At vertical actuators and an actuator area of 25 cm², this means a load pressure of approx. 20 bar caused by the weight.
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Sizing of Proportional- or Servo Valves taking metering edges and dynamic effects into account
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Extending Differential Actuator
Prerequisite for controllability pA & pB > 0 bar
m
+/- Fmv
P T
pBpA
>< ><
Q1 Q2
FG = m * a + FExt
NN p
pQQΔΔ
=
A B
FExt
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Extending Differential Actuator Accelerating
At acceleration the mass force Fm acts against the direction of movement at the big actuator area
m
Fmv
P T
pBpA
>< ><
Q1 Q2
FG = - (m * a) + FExt
NN p
pQQΔΔ
=
A B
FExt
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Critical Case: Extending Differential Actuator decelerates
At deceleration the mass force Fm acts pulling in the direction of the movement on the small actuator ring area
FG = m * a + FExt
m
Fmv
P T
pBpA
>< ><
Q1 Q2
NN p
pQQΔΔ
=
A B
FExt
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Calculation of an extending cylinder decelerating
Question: What is the required pressure in P, to get during deceleration exactly 0 bar in port A (means, just no cavitation)?
Data: deceleration pressure in port B: pB = 80 bartank pressure pT = 0 barsymmetrical valve spoolcylinder area ratio 2:1, means QPA = 2 x QBT
pressure drop B-T ∆pBT = pB – pT = 80 bar – 0 bar = 80 barpressure drop P-A ∆pPA = pP – pA as pA is 0 bar ∆pPA = pP
ppQQ N
N ΔΔ
= barQ
QpBT
BTPA 802
2
∗⎟⎟⎠
⎞⎜⎜⎝
⎛ ∗=ΔBT
BT
PAPA p
QQp Δ∗⎟⎟
⎠
⎞⎜⎜⎝
⎛=Δ
2
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Example for external Mechanics, causing alternating forces
m
F
v
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Very critical case: Hanging (rod top down), extending differential actuator decelerates
Prerequisite for controllability pA & pB > 0 bar
v
FGP T
pBpA
>< ><
Q1 Q2
FG = m * a + Fg + FExt
m
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When the pressure in one of the actuator chambers gets < 0 bar the drive cavitates.
A cavitating drive is not controllable !
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Hydraulic Circuits to prevent Cavitation
approx. 85% of all hydraulic actuators are asymmetric actuators!
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Special Spools
Technically the most elegant and normally the most cost effective solution, is applying an asymmetric valve spool or using a regenerative circuit.
E.g. an asymmetric 2:1 spool. The rated flow of the B-channels (QPB and QBT) is 50% of the rated flow of the A-channels (QPA and QAT).
QPA and QAT = 2 * (QPB and QBT)
(Note: connect always the high flow to the A-port)
2:1 is in line with a frequently used cylinder geometry.
The asymmetric valve spool provides additionally to the cavitation prevention also a better „chucked“ cylinder, what improves the drive stiffness and therefore the controllability.
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Asymmetric Valve Spool
v F
m
A B
TP
Caution:Ports A and B must not be exchanged
Note:The big piston area should be connected to port A of the valve!
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Regenerative Circuit
Disadvantage ?
m
v F
A B
TP
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Regenerative Circuit
(Disadvantages, less force at extending and less drive stiffness)
m
v F
A B
TP
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Regenerative Circuit
Special spools can automatically change over to normal circuit at low signal levels.
There is no generally suitable regenerative spool available. The optimization of the spool has to be done application specific
m
A B
TP
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Additional Selection Criteria for Proportional Valves
• Ambient temperature range (min. / max.)• Shock, vibration at valve location• Electric environment (EM-Fields, HF devices, required protection
class IP??, explosion proof etc.)
• Fluid• Filtration (For filtration sizing, use publications about this subject)
• Human- and machine safety regulations• Special approvals for designated applications
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Sizing Rules for electro-hydraulic Control Systems
Hydraulic drive system characteristicsPipe characteristics
Position- and velocity control systemsPressure- and force control systems
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What effects the accuracy of electro-hydraulic control loops?
MechanicsFrictionLoadMoved massStiffnessBearing backlash
HydraulicsFlow, pressureStatic and dynamic valve characteristicsActuator geometryActuator sealsPipe/hose geometry
ElectronicType of controllerResolution of sensorsResolution of A/D and D/A convertersController sampling timeLinearity, hysteresis, temperature drifts of Sensors & valves
Static and dynamic accuracy of closed loop hydraulic drive systems
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Hydraulic drive system characteristics
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Fluid Compressibility
Compressed volume ΔVfluid = A * Δ s = V0 * Δ p/EfluidV0 = start volumeEfluid = bulk modulus of fluid (dependent on pressure, temperature, solved air)A = piston area
Note: As pipes, actuators and seals are also elastic, their bulk modulus has to be considered for an equivalent system bulk modulus.A realistic value for the equivalent bulk modulus of oil is 7000 to 9000 bar. At operating pressures < 100 bar a lower bulk modulus has to be applied.
6.000
8.000
10.000
12.000
14.000
16.000
18.000
10 40 70 100 130 160 190 220 250 280 310 340 370 400
pressure in bar
Eflu
id in
bar
Efluid is pressure dependentF = Δp*A
s
Δs
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Natural Frequency of Hydraulic Drives
hydraulic spring-mass system (spring = fluid compressibility)
m
A B
TP
mc
A V0, Efluid
0
2
VAEc fluidfluid ∗=
mVAE
mc fluidfluid
H ∗∗∗
==0
22ω
Efluid = Bulk modulus of fluidA = piston areaV0 = trapped volume
mc=ω
Basic equation for the natural frequency of spring-mass-systems
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1,0
1,1
1,2
1,3
1,4
1,5
1,6
1,7
1,8
1,9
2,0
1 50 99
stroke in %
ωΗ/ω
Ηm
in
Natural Frequency depending on Actuator Stroke
The natural frequency of the drive is, at position- and speed control systems, a measure for the dynamic limits (acceleration) and the achievable loop gain (accuracy) of the closed loop system.
Equal area actuatorAsymmetric actuator 2:1
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Pipe characteristics
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Pipe models
LC ∗=
1ω
m(LHI, RHI)
m(LHI, RHI)
Q0
p0
Q1
p1
Equivalent electric pipe modelL R
C
C
Lenght of pipe in m natural frequency in Hz0,5 3151 1572 795 318 20
10 16
Table is calculated with an equivalent fluid bulk modulus of 9000 bar
Equivalent mechanic pipe model
CC
2lEÖl
L ∗=
ρω
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Hydraulic Pipe Effects
Note:Best stability and accuracy results can be achieved when the pressure sensor is mounted directly at the actuator. Reason is the inductivity of the oil between the valve and the actuator as well as the pressure drop in the pipe. A too fast valve will cause pressure oscillations.
Example: Pressure over time in a pipe. L = 5m, every 0,7 m a pressure test point, time scale 20ms
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Pressure Pulse (Joukowsky‘s pulse)
Due to the inductivity of the fluid in a pipe, a pressure pulse is generated at the valve ports when closing the valve. It has to be distinguished between „abrupt“ and „slow“ valve closing:
l = length of pipec = sonic speed in fluid(c = 1366 m/s steel pipe, c = 462 m/s HP-hose)
A valve closing time smaller or equal tcrit means „abrupt“ closing, else „slow“valve closing.
Pressure pulse at „abrupt“ valve closing:∆v = change of fluid velocity
Pressure pulse at „slow“ valve closing : tV = valve closing time
Cavitation at the tank port of the valve can emerge. (The required preload pressure to avoid cavitation can be calculated)
Fluidcrit c
lt ∗=
2
vcp Δ∗∗=Δ ρ
Vtlvp ∗Δ
∗=Δ ρ2
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Pressure Pulse (Joukowsky‘s pulse)
When closing a hydraulic valve “abruptly” a pressure pulse in the P, A and B ports crops up. The pressure pulse is independent from the pipe length! The duration of the pressure pulse is approx. 1.5 ms per meter pipe length.Cavitation at the tank port of the valve can emerge. (the required preload pressure to avoid cavitation see diagram)
Druckstoß (Kompessionmodul 10000 bar)
0
50
100
150
200
250
300
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ölgeschwindigkeit in m/s
Dru
ckst
oß in
bar
vcp Δ∗∗=Δ ρ
or
ÖlEvp ∗Δ=Δ ρ
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Pressure Pulse (Joukowsky‘s pulse)
At „slow“ valve closing a pressure pulse crops up as well.In this case the pressure pulse depends on the pipe length!Cavitation at the tank port of the valve can emerge. (The required preload pressure to avoid cavitation can be calculated with the equation below)
Pressure pulse calculation for „slow“ valve closing:
Note:The pressure pulse when closing a valve, caused by the pipe inductivity (Joukowsky‘s pulse), has to be added to the pressure pulse caused by the moving mass (inertia) of the drive system.
Vtlvp ∗Δ
∗=Δ ρ2
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Layout of dynamic hydraulic drive systems
To achieve a high natural frequency of the drive the valve has to be mounted close to the actuator.Apply hydraulic accumulators close to the valve in P and T, at pilot operated valves also in X and Y.The pipes between the accumulators and the valve have to be extremely short with as big as possible diameter, without any restrictions in the pipe or drillings.Apply in the T pipe an orifice and a pressure preload valve, to avoid cavitation. Pipe between orifice and preload valve as long as possible (high inductivity).
F
m
A B
TP
v
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Note
Due to its inductivity and capacity hydraulic pipes are oscillatory systems with a natural frequency, which act in a hydraulic system as a low pass filter. When sizing dynamic drive systems it is mandatory to take the pipe dynamics into account. Neglecting the dynamic pipeeffects cause a high risk, that the drive system cannot achieve the required performance and damages are very likely.
Mainly the frequently underestimated T-pipes cause severe problems due to cavitation.
The required accumulators have to be sized by experts with a natural frequency above the valves operating frequency. A „normal“accumulator sizing is in most cases not sufficient!
Note: The operating frequency of the pilot stage of pilot operated valves is approximately two times the main stage frequency.
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Position-and velocity control systems
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Electro-hydraulic Position or Velocity Loop
w = command signalx = control value xW = error signalKV = loop gain
w-x
Controller
Sensor
xw
KHy Acyl
KSensorKV
KV = Kp * KHy * 1/Acyl * KSensor
Kp
ωsv, Dsv ωH, DH
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Electro-hydraulic Position or Velocity System, Loop Gain
(KV is calculated according the stability criteria and depends on the natural frequencies and damping factors)
Max. velocity error
KV = loop gainωH = natural frequency of the hydraulic driveDH = damping factor of the hydraulic driveωSV = natural frequency of the servo valveDSV = damping factor of the servo valve (can be calculated from the amplitude ratio)
vmax = maximum speed at 100% open valve (rated flow of valve !)
πω 2SVSV f=
VKvs max=Δ
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛++
= 22
2
22
2V
2.
2.)(
HHH
H
svsvsv
svOL sDssDss
KsTFωω
ωωω
ω
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Electro-hydraulic Position- or Velocity Control Systems
Thumb rules for component sizing at position- and speed control systems:To achieve the max. performance of the hydraulic drive, the natural frequency of the servo valve should, at well damped systems, 3times higher than the natural frequency of the actuator. At poorly damped systems (most cases) the natural frequency of the servo valve should be slightly below the natural frequency of the actuator.
Caution: The natural frequency of pilot operated valves depend on the pilot pressure.
Drives with a natural frequency < 5 Hz are poorly (only with very high effort) controllable. Above 25 Hz natural frequency, hydraulic drives are normally well controllable. The trapped oil volume between valve and actuator should be as small as possible, to achieve a high natural frequency. Design the system for automatic bleeding !
The smallest possible valve for the application is the best possible valve! (Rule: As much as possible electric- as less as possible hydraulic gain in the loop)
The natural frequency of the sensors should be 10times higher than the natural frequency of the drive.
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Pressure- and force control systems
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Pressure or Force Control Loop
Example: hydraulic actuator with 3/3 servo valve
A
TP
pi
Controller
Pconst.
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Electro-hydraulic Pressure or Force Loop
w = command signalx = control valuexW = error signalKV = loop gain
w-x
Controller
Sensor
xw
VQp CH
KV
KP
ωSV, DSV
ÖlH E
VC =
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Stability criteria of a pressure control loop
KP = controller P gain
CH = hydraulic capacity in cm³/bar
ωSV = natural frequency of the servo- or proportional valve
DSV = damping factor of the servo- or proportional valve
VQp = flow-pressure gain of the servo- or proportional valve
Electro-hydraulic Pressure Control Loop
Qp
HSVSVP V
CDK ∗∗∗<
ω2
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Electro-hydraulic Pressure Control Loop
Conclusion from the stability criteria ?Big ωSV, DSV and CH and small VQp increases the loop gain and therefore the static and dynamic performance.
Notes:The application limits are defined by the stability criteria and the valve resolution (threshold).
At pressure and force control loops are significant stability restrictions in case the natural frequencies of attached spring-mass-systems and of the oil in the pipe between valve and actuator are not significantly above the valve‘s natural frequency.
External spring-mass-systems can be excited to resonance !
The natural frequency of the pipe is a low pass filter between valve and cylinder, what means the valve dynamics cannot be transmitted !
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Electro-hydraulic Pressure or Force Control Loop
Hydraulic actuator with servo valve
m
A
TPAt force- and pressure control loops, the natural frequency of an attached spring-mass-system and the natural frequency of the oil in the pipe between valve and actuator is a criteria for the dynamic limits and the possible loop gain of the control loop, in case it is not significantly above the valve natural frequency.
c
Natural frequency of the fluid in the pipe (low pass filter) !
pconst
mc=ωNatural frequency of spring mass system (resonance)
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Electro-hydraulic Pressure Control Loop
Thumb rules for component sizing at pressure- or force control loops:The dominant factors are the natural frequency, damping and the flow-pressure gain of the servo valve as well as the hydraulic capacity of the drive.
The faster the servo valve the better its damping, the smaller the flow-pressure gain and the bigger the hydraulic capacity of the drive is, the better is the control performance. At significant changes of CH during the active control, an automatic adaptation of the controller gain is recommendable.
In contrast to position- and speed control loops has at pressure control loops a bigger trapped volume (hydraulic capacity) advantages. (Caution: Natural frequency of the oil in the pipe is a low pass filter between valve and actuator !)
The smallest possible valve for the application is also for pressure control loops the best possible valve!
(Rule: As much as possible electric- as less as possible hydraulic gain in the loop)
The natural frequency of the sensor should be 10times higher than the natural frequency of the drive.
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Structural Conditions in a Closed Loop System
Backwards branch: Errors in the backwards branch are mirrored 1:1 in the control loop results. This means, the sensor is a key component for the accuracy.
Forward branch: Errors in the forward branch are divided by the controller gain
w-x
Controller
Sensor
xw
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Limits due to Valve Resolution
Additionally to the limits due to the possible loop gain are at all control loops, more frequent at pressure/force control loops, limits due to the valve resolution (threshold).
An electric feedback valve (EFB) has typically a resolution (threshold) of approx. 1:1000. This means the smallest adjustable spool position is 0,1% of the maximum spool position.
At drives, with high speeds (> 0,5 to 1m/s), it is recommendable to use a progressive spool, to improve the flow resolution at smallsignals. The same applies to pressure control loops.
Operating at the valve resolution limits, leads to limit cycling.
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Summary of Valve Sizing
A proportional valve has to be sized for the operating point which requires maximum valve opening.
Normally this operating point is at maximum speed (max. Q) and maximum load (smallest Δp), in some cases at maximum acceleration. In doubt several operating points have to be calculated. A valve is optimal sized, when the maximum valve opening is 70 to 90%.
To determine the operating point, mass forces, flow saturation and cavitation have to be considered.
At position- and speed control loops it is in most cases sufficient to select a valve with a natural frequency close to the natural drive frequency. At pressure- and force control loops the highest possible natural frequency of the valve is in most cases beneficial.
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Summary of Valve Sizing
After sizing the servo valve according to the mentioned rules, the physical limits of the drive (closed loop controlled system) areknown and a proper function will be achieved.
With the calculated values the closed loop controller can be defined. The art of closed loop control is to utilize the physical possibilities of the closed loop controlled system.
A poor performing controlled system cannot be improved even withthe best closed loop controller.
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Summary of Valve Sizing
Due to the complexity of sizing a servo hydraulic drive system it is recommendable to use a sizing
software. Such a software makes it easy to calculate, in case needed, several operating points
of a drive.
Based on the sizing results a more accurate prediction of the dynamic drive performance can be
achieved with a dynamic drive simulation.
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Dynamic Drive Simulation
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Dynamic Drive Simulation