Hydraulic Proportional Control_Bosch Rexroth

Design & Control of Proportional and Servo Systems Bosch Rexroth Industrial Hydraulics Rob Decker & Dave Saaski April 2008 Drive for Technology CMA/Flodyne/Hydradyne, Inc.

© Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

History of Hydraulic Control Mechanical No Electric

Flow Controls, Limit Switches and Relay Logic


Using Closed-loop Controllers Examples of Closed Loop Controllers

f Analog f Lower Cost f Computer not required for set-up or adjustment f Examples: • VT-MACAS (AVPC-V or AVPC-mA) - Position or Velocity Control Card • p/Q Cards – Open Loop Flow, Closed Loop Pressure Control Card

f Digital f HACD – Hydraulic Axis Controller: Digital f HNC – Hydraulic Numerical Controller © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

VT-MACAS (AVPC) Analog Velocity Position Control / V or mA

Material Number for Voltage Command = 0811405139 Material Number for milliAmp Command = 0811405140


VT-MACAS (AVPC) Analog Velocity Position Control / V or mA


VT-MACAS (AVPC) Schematic


AVPC / Applications Position Control

Velocity Control


P/Q Cards with Valve Amplifier


Front Plate

P/Q Cards for Valves with OBE


P/Q Card Applications


Controller for the Pressure Difference (Force Control)


HACD

HACD (Hydraulic Axis Controller - Digital) f Multi-loop 32-bit Digital Controller • DeviceNet Available • CANOpen - In development • PROFIBUS - In development f New Generation Upgrade of the DMX


HACD Technology: European + American Product Experience

1997 to 2003 Joint Development


Bus Controllers - HACD f

6 analog Inputs, Voltage or Current (selection via software)

f

2 analog Outputs, 1x voltage or current (selection via software)

f

Digital Feedback SSI or Incremental

f

8 digital inputs (configuration via software)

f

7 digital outputs (configuration via software)

f

Enable Input and OK Output

f

Display and Keys

f

Serial Interface RS 232

f

CAN Bus - DeviceNet and CANOpen protocol


HACD - Setup Program fOne Configuration Software for all Applications fEditor for the Configuration of the Control Structure using predefined functions (programming knowledge not required) fClearly arranged Settings for commands, controller parameters, analog and digital I/O setup fOscilloscope Function also suitable as a data recorder fLanguage Options English / Deutsch © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Bus Controllers - HNC HNC Hydraulic Numerical Controller

f 32 Bit Multi-Axis Controller f User Programmed using NC “G” codes

f Bus Capability f PROFIBUS – Available f CANOpen – Available f SERCOS - Available f DeviceNet - In development © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

HNC Technology: Connectivity + Drive Control Options


HNC - Bus Applications


HNC with SERCOS interface

Robust SERCOS fiber optic interface

SERCOS interface Noise immune fiber optic ring connection Distributed drive architecture Mode: Closed-loop position control (SERCOS cycle time 2 ms) Preferred and freely configurable messages Special control algorithms for interpolating with electro-hydraulic drives HNC100 closes the control loop Digital interfaces for measuring system EnDat absolute, incremental, SSI Command value feedforward via Sercos HNC100 looks and acts like an electric servo to the CNC


HNC with SERCOS Interface CNC control system solutions by Bosch Rexroth

SERCOS (command value and actual value)

MTX

MTC 200

CNC control

CNC control

additional axes can be added

Ecodrive IDIAX

Electrohydraulic actuator controlled by the HNC Control and Power Electronics

Electromagnetic motors

HRS (HNC 100)

Electrohydraulic actuator

SERCOS (command value and actual value)

additional axes can be added

Ecodrive IDIAX

Electrohydraulic actuator controlled by the HNC Control and Power Electronics

Electromagnetic motors

HRS (HNC 100)

Electrohydraulic actuator


HNC Technology: Demonstration and Development Tools The Test Box comes with three Prj.-Files: Demo_ana (analog Feedback) Demo_ink (incremental Feedback) Demo_abs (absolute SSI Feedback)

VT-HNC100DEMO Refer to RD/RE 30133 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

History of Hydraulic Control Proportional Valve, Limit Switches and Relay Logic

Proportional Valve and Linear Displacement Transducer


History of Hydraulic Control Closed Loop Position Control with Servo Solenoid Valve and Linear Displacement Transducer


Open Loop Systems vs. Closed Loop Systems


Closed Loop System Components Setpoint f Potentiometer f Set Point Card f PLC Analog Output Controller f AVPC f p/Q card f DMX f HACD f HNC Amplifier f Amp Card f Amp Cube f OBE Valve f p/Q f Overlap (W) f Zero Lap (V)

Actuator f Motor f Cylinder Measuring Device f Pressure Transducer f Ultrasonic f LVDT f Encoder f Potentiometer


The Three Most Common Types of Control Position Control

Velocity Control

Pressure Control


Physics of Hydraulics Hydraulic Stiffness Example: Apply a 1 ton load to a cylinder. The no-load position is 100 cm Compare the cylinder when filled with : 1. Oil 2. Water 3. Air 4. Steel

No Load

A = 10 cm2 100 cm

PL = 0 bar


Stiffness of Various Materials 1T

1T ΔX = 0.7 cm

ΔX = 0.4 cm

Oil

Water 99.3 cm

99.6 cm

PL = 100 bar

PL = 100 bar

1T 1T ΔX = 0.002 cm Steel

ΔX = 99 cm

99.998 cm

Air 1 cm PL = 100 bar

PL = 100 bar


Solving Stiffness Limitations

In order to maintain a fixed cylinder position independent of load changes, the following must be used: 1. Mechanical stops (metal to metal) or 2.

Closed loop control

The same conditions are true for a constant velocity drive. Again the following solutions must be used: 1. Electric servo drive (Indramat) or 2.

Closed loop hydraulic drive using load sense, load compensator or electro-hydraulic closed loop


Example of Position Control 1. Mechanical-hydraulic Closed loop F

2. Electro-hydraulic Closed loop

Feedback F S

U Position Transducer

Error

Command Controller

Control Valve


Improving the System Stiffness

A four-way valve controls both sides of an actuator.

As a result, the load is held between “two springs”. With this configuration, the spring-constant is higher ! Note:

See Rexroth “Hydraulic Trainer Vol. 2” or “Using Industrial Hydraulics” for more details.


Positioning with Higher Stiffness

1. Mechanical-hydraulic

2. Electro-hydraulic


What Impacts Machine Design ? How does a “real” hydraulic drive differ from an “ideal” drive? Response The “ideal” or “linear” drive converts all input (command) signals into output signals without delays or distortions. Examples of input signals can be: – On / Off, Stop / Go – Analog voltages – PLC program I/O Example: A rod or a lever converts inputs directly into corresponding outputs.

In

Out


Spring – Mass Systems Respond Differently As we will see later, hydraulic drives are “spring-mass” systems. “Spring-mass” systems have two observable properties: 1. Natural frequency 2. Damping

Example: 1. The number of oscillations per second is the natural frequency “fo” 2. After time, the oscillation decays due to damping.

T=

1 fo

T

m


Predicting the Natural Frequency The natural frequency is determined solely by: – –

Spring constant “C” of the drive Mass “M” coupled to the drive

fo =

C M

2π

Why should the natural frequency be high as possible ?

A simple experiment will show: Every machine has mass and is not completely rigid. Consequently all machines are “spring-mass” systems. Take a machine axis with a given fo, and oscillate it between two defined positions, “0” and “10”.


Machine Limits Imposed by Natural Frequency If we move the axis “slowly”", the machine will respond “ideally”, i.e. it moves from “0” to “10”, as commanded.

If we increase the number of cycles per second, the machine output will increase in stroke! This can be dangerous and destructive to the machine! Every machine has a “limit of operation”. If we exceed this limit, the natural frequency of the machine is approached, and the machine starts to resonate at that natural frequency. This MUST be avoided; BAD things will happen. http://timber.ce.wsu.edu/supplements/seismic/frequency.htm © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Benefits of Damping Increasing damping or, How to Control Resonant Response If “damping” is increased in a “spring-mass” system, amplitude increase or “overshoot” can be reduced. The graph that follows shows how our experiment varies if we increase “damping”. To understand the effect, visualize the previous experiment with the moving components immersed in: – Water (d ≈ 0.5) – Honey (d ≈ 2) – Hot tar (d ≈ 20)


Effect of Increased Damping

A damping ratio of d ≈ 0.7 results in the best overall response. Unfortunately, by increasing damping, another problem occurs: A time delay between the input and output, so-called “phase-lag”, increases.


Improving Damping

f Note: A hydraulic machine

f

drive will normally have a damping ratio between 0.05 ~ 0.4, and will respond accordingly. Proportional valves and servo valves are designed with damping ratios of 0.75. They exhibit no overshoot.

fHow to increase “damping”: f Dissipate kinetic energy by converting it into heat, known as “passive damping” (used in car shock absorbers) or; f Actively counter the kinetic energy using a closed loop feedback to cancel oscillations, called “active damping.” Both methods are used in systems today. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

What Creates Damping Note: It would appear that increasing friction is a simple way to increase passive damping. This is a trap! Mechanical friction can be high at low speeds (breakaway friction), and lower at higher speeds (running friction). High breakaway friction deteriorates a system’s performance and must be minimized! Use of low friction PTFE cylinder seals, lubrication, hydrostatic bearings, etc. is normally recommended for this reason. Passive damping is present due to internal leakage

FL1 Δp

= Δp

R

M LA

•Q

Q

Laminar Flow


Passive Damping Sharp-edge Orifice

FL1 Δp

2

Δp

tu =R

rb

•Q

Q

Turbulent Flow

Bypass Valve

Uo

Valve

Underlapped Spool

Leakage caused by cylinder seal leakage, bypass valves, and valve null-flow, resulting from underlapped spools, improves damping.

Disadvantages: – Loss of energy (efficiency) – Decreased static accuracy


Active Damping Example of Active Damping using Pressure Feedback The (simplified) circuit shows an aircraft actuator in closed loop position control. A step-load can cause oscillation, dependent on the system gain. ( B lever ratio in this example) A

Command

B

A

“P”

“T”


Modulator to Provide Active Damping A so-called “modulator” is added to the system. Pressure increases caused by step-load changes allow the lever support to move. Response time and amplitude of the modulator piston is determined by an orifice and a small accumulator. A high damping is created with this type of control (P-DT1 control). Orifice

This idea is 80 years old!

Volume

In some applications today, an electronic equivalent is used, also known as active damping, or state variable control. Modulator

“P”

“T”


Designing for Performance

Components When designing a hydraulic drive, it is essential to consider both static and dynamic characteristics of the drives. For example: If a constant load is applied to an actuator, it will deflect, as seen earlier. The spring constant of the drive is calculated as: E = bulk modulus of the fluid A = area of cylinder Vo = total compressed volume

E • A2 C= Vo © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Stiffness of Actuators

Therefore:

C=

ΔF ΔX

ΔX =

ΔF C

ΔF = load ΔX = deflection

The result is the deflection under load, or the stiffness. If a load is suddenly applied to the drive, or the load is accelerated dynamically, it will respond like a “spring - mass” system, and the dynamic properties will be dominant. Selecting a cylinder should be based on both the static and dynamic requirements:


Defining Drive Requirements

1. Static: - Balance of forces, i.e. the cylinder must have enough force to hold and move the load ΔF = p • A - Rod buckling must be considered. The cylinder attachment design or rod diameter should be changed if required.

2. Dynamic: - The drive should have sufficient dynamic response to accommodate load changes and accelerations. As seen earlier in a bode-plot, a drive will appear to be rigid and have minimal adverse dynamic effects when operated below it’s limit frequency. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Determining Performance Requirements

When designing a hydraulic drive system, it is desirable to have the natural frequency as high as possible, while staying within the allocated budget.

Rule of thumb for “good design”: C M E • A2 fo = , C= 2π Vo A hydraulic designer can only improve the natural frequency of the drive by varying the cylinder size, and minimizing the pipe length between cylinder and the valve. NOTE: M (mass) is not measured in pounds in English units. Rather M(mass) = F (lbs)/A (in./sec2) = (lbs.)/(32.16 (ft./ sec2) x 12 (in./ft)) = (lbs.) / 386 (in./ sec2)


Design Guidelines A quick approach for finding an optimal cylinder: 1. Calculate F = p • A 2. Check for rod buckling Typical Arrangement ΔF

• Single rod cylinder with pipe connections to the valve • fo calculation is difficult

Simplified Approximation A

P

B

T

• Equal area cylinder (added rod results in lower fo) • Eliminate pipe volume by mounting valve directly on the cylinder (improve fo)

ΔF

A

B

P

T


Rule of Thumb Selection Assuming the two changes in the above approximation cancel each other, the value of Vo (cylinder centered) is:

Vo = AA • S/2

AA = Annulus Area S = Stroke

Since:

fo =

C M

Result:

2π

C = C1 + C2 E • A2 C1 = C2 = Vo

fo =

2 • E • AA S/2 • M

2π

Vo = AA • S/2 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Typical Machine Requirements

Therefore:

fo2 • π2 • S • M AA = E

This formula provides an annulus area of the cylinder required for a “desirable” natural frequency. The following are experience based guidelines: fo < 4 Hz Poor dynamics, good static performance only fo ≈ 15 Hz Good frequency for general machine designs fo ≈ 30 Hz Needed for machines requiring high dynamics


Hydraulic Motor Estimation The same rules apply to hydraulic motor drives:

fo2 • 4 • π4 • I q= E

I = Rotating inertia

Note: If a difference of opinion arises, such as the machine designer insisting on using a small cylinder operating at high pressure, but calculations dictate a larger cylinder and lower pressures, you can compare the alternatives with a simulation program, such as HYVOS by BoschRexroth


Machine Considerations MEFF = M • ( m

Valve

M

x 2 y)

x

ATTENTION! In many designs, the mass “M” is not directly driven by the actuator.

In the examples shown here, the “effective” mass or inertia must be calculated. The effective value “seen” by the actuator (reflected inertia) is the (lever-ratio)2 or (gear-ratio)2 !

Therefore:

fo =

C MEFF

y

JEFF N out

N = rpm n=

N in

N out N in

JEFF = J (n)2 x y

MEFF = M • (

x+y 2 y )

M

M

JEFF = (

M

P = pitch =

2 π )

in rev

2 P

2π

r

JEFF = M (r)2 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Machine Stiffness Important Note: If the driven machine structure is flexible (typically the case), the result is a complex pair of 2nd order “spring - mass” systems (4th order and higher in most situations). Manual calculation of these systems is very tedious. Machine designers should always try to make the driven structure as rigid as possible. A goal is to have the natural frequency of the driven structure be 3 times higher than the hydraulic drive’s natural frequency. With this ratio, the machine’s dynamic response will have a minimal influence on the overall drive performance. When the ratio is 10:1, the machine can be considered a rigid structure, and not be a factor in the drive’s design.


Valve Sizing 1. Pressure Drop Pressure drop (Δp) is defined as the difference between the system (pump) pressure and the load pressure. Load pressure can only be measured directly when a meter- in throttle circuit is used. Example 1: pL Q

AP ΔF

ΔPV

ΔF pL = A p ΔpV = pP – pL

pP = 100% © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Pressure Drop with 2 Orifices It is obvious that the pressure drop is not “selected” by the designer. It is a function of the system pressure and the load applied to the actuator.

Example 2: pA

pB ΔF

Δp1

pP = 100%

Q

Δp2

ΔF p L = pA – p B = A A Δp1 = Δp2 Δpv = 2 • Δp1 (Also applies to Hydraulic Motors)


Meter Out and Double Throttles Example 3: pB ΔF

AP = Piston area AA = Annulus area

Q

ΔP1

pP = 100%

Example 4:

pA

pB ΔF

Δp1

Q1

pP • Ap – ΔF ΔpV = pB = AA

Δp2

Δpv = Δp1+Δ p2 for Q1 Δp1 =

pP • Ap – ΔF Ap + AA2 ϕ

pP = 100% © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

ϕ=

Ap AA

How to Find the Proper Valve

f Flow ratings of proportional valves are the rated flow at a NOMINAL pressure drop, and normally have nothing to do with actually sizing a proportional valve.

f First, we need to calculate the ACTUAL ΔpV for the specific system. After calculating the ΔpV, refer to a valve data sheet.

f Example 4:

Flow Pump-pressure Force Cylinder

Q pP F Ap AA ϕ

= = = = = =

450 l/min 210 bar 3600 daN 20cm2 10 cm2 2


f Which Proportional Valve do we choose? Δp1 = Δp1 =

pP • Ap – F Ap +

AA ϕ2

(3000 psi • 3.1 in2) –7920 LB 3.1

in2

+

1.6 in2 0000 22

Δp1 = 394 psi For a 2:1 cylinder (ϕ = 2), we choose a 2:1 valve (QA:QB = 2), Then, Δp1 = Δp2 and: ΔpV = 2 • ΔpV1 = 788 psi (= 54 bar)

⇒

RA 29 115 Page 13-14 4WRZ valve

WRZ 16 is too small WRZ 25 E 220 is 92% open WRZ 25 E 325 is 78 % open


Spool Flow Characteristics

Characteristic curves (measured with spools “E, W6-, EA, W6A” at v = 46 mm2/s and v = 40 ºC) Size 16 150 L/min nominal flow with a 10 bar valve pressure differential

121.5

(460)

105.7

(400)

84.5

(320)

5

P→A →T / or P→B →T /

B 4

A

63.4

(240)

3 2

42.3

(160)

1

21.1

(80) 0

15 20

30

788 PSI

40

50 60 70 Command value in %

80

90

1 2 3 4 5

Δp = 10 bar constant Δp = 20 bar constant Δp = 30 bar constant Δp = 50 bar constant Δp = 100 bar constant

100


Spool Flow Characteristics Characteristic curves (measured with spools “E, W6-, EA, W6A” at v = 46 mm2/s and v = 40 ºC)

Size 25

220 L/min nominal flow with a 10 bar valve pressure differential

211.0

(800)

185.0

(700)

159.0

(600)

132.0

(500)

4

106.0

(400)

3

79.3

(300)

2

52.8

(200)

1

26.4

(100) 0

15 20


30

5

B

788 PSI

A

40

50 60 70 Comman d value in %

80

90

1 2 3 4 5

Δp Δp Δp Δp Δp

= 10 bar constant = 20 bar constant = 30 bar constant = 50 bar constant = 100 bar constant

100

325 L/min nominal flow with a 10 bar valve pressure differential 5

230.0 211.0

(870) (800)

185.0

(700)

159.0 132.0

(600)

106.0

(400)

79.3

(300)

52.8

(200)

26.4

(100)


B

788 PSI 4

A

3

(500)

2

1 2 3 4 5

Δp Δp Δp Δp Δp

= 10 bar constant = 20 bar constant = 30 bar constant = 50 bar constant = 100 bar constant

1

0

15 20

30

40

50 60 70 Co mman d value in %

80

90

100

Δp = Valve pressure differential to DIN 24 311 (input pressure pP minus load pressure pL minus return line pressure pT)


Data Sheet Ratings

fAs shown, the available pressure drop across the valve is a result of pump pressure and load pressure. Data sheets typically give a “nominal flow” at a “nominal pressure drop”. There are two de facto standards used in the hydraulics industry to define the flow through a control valve: f Servo & High-Response Proportional Valves • A pressure drop ΔPv = 1000 PSI (70 bar) is used. When the valve is open 100%, the measured flow is the “nominal flow” at 1000 PSI total pressure drop. f Proportional Valves

Pressure Drop Test Circuit

QNom

pP = 70 bar

• A pressure drop ΔPv = 145 PSI (10 bar) is used. The valve is opened from zero to maximum. The typically non-linear flow characteristic is recorded. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Valve Pressure Drop Rating

fWe use the sharp-edged orifice equation and the valve flow rating to determine the flow at the pressure drop of our system. Q2 = Q1 x √(Δp2/Δp1)

fExample: If a valve is rated for 50 LPM @ 1000 PSI, the flow at 2000 PSI = Q2 = 50 lpm x √(2000/1000) = 71 LPM


Valve Pressure Drop Selection Since proportional valves often not “linearized,” a graph is generated for “flow vs. spool-stroke”. This also eliminates the need to calculate flows at various pressure drops. Instead, many data sheets will give a family of flows for various pressure drops: Δpv = 20 bar (290 PSI) 30 bar (435 PSI) 50 bar (725 PSI) 100 bar (1,450 PSI) Example: 4 WRZ 16 E 150 valve 150 L/min nominal flow with a 10 bar valve pressure differential 5

121.5

(460)

105.7

(400)

84.5

(320)

63.4

(240)

42.3

(160)

21.1

(80) 0

15 20


B 4

A

3 2

QNOM 1

30

40

50 60 70 Co mmand value in %

80

90

1 2 3 4 5


100

Δp = Valve pressure differential to DIN 24 311 (input pressure pP minus load pressure pL minus return line pressure pT) © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Sizing Proportional Valves for Acceleration and Deceleration

f For transfer systems, the highest pressures are normally experienced f

during acceleration of the mass. The natural frequency of a hydraulic system normally determines the minimal allowable acceleration and deceleration. tmin (sec) = (3 x 6)/ ωo tmin ωo C M

fo = ωo / (2 x π)

= minimum time of acceleration (sec) = natural frequency, undamped (radians/sec) = √(C/M) = spring constant = mass (NOT POUNDS)


Simulation “Software”

f Above, we used an “ideal” 1:1 double-rod cylinder in order to simplify calculations and estimate natural frequency and an ideal annulus area:

E • A2 C1 = C2 = Vo

fo =

2 • E • AA S/2 • M

2π

fo2 • π2 • S • M AA = E

f Simulation software such as HYVOS allows extremely accurate f

simulation of the performance of hydraulic systems without the above simplification. But this software can be difficult to access. A simple spreadsheet, though, can quickly calculate many of the most important values necessary for design. Simplifications such as above then become generally unnecessary.


Poor Man’s Simulation “Software” Natural Frequency, 2:1 Cylinder, 1=a side, 2=b side, 2:1 Valve Bulk Mod (lb/in)=

200000

Valve Pressure Drops Required

d-bore1 (in)=

2.00

Weight (lb) =

d-rod1 (in)=

0.00

Mass (lb*ss/in)=

1.81

A1 (si)=

3.14

C (friction)=

1.00

d-bore2 (in)=

2.00

d-rod2 (in)=

1.38

A2 (si)=

1.66

stroke (in)=

700.00

t (stroke) (sec)=

0.6

V-max (in/sec)=

49.7

26,947

A-max (in/ss)=

336.8

w(theor)(rad/sec)=

121.96

F-acceler (lb)=

610.1

w(damp)(rad/sec)=

40.65

F-friction (lb) =

700.0

t (time cont)(sec)=

0.02

P-system (psi)=

1000.0

t(min. accel.) (sec)=

0.15

Qmax=

40.5

42.00 0.63

L-pipe 1 (in)=

48.00

d-pipe 2 (in)=

1.00

L-pipe 2 (in)=

48.00

V-pipe 1 (ci)=

14.72

F (accel) =

610.06

V-pipe 2 (ci)=

37.68

F (friction) (lb)=

700.00

Cmin=

20.0

Ct (lb/in)=

d-pipe 1 (in)=

d-crit (in)=

Stroke (in)=

35.54

Accel.

Const. Vel.

Decel.

P3 (psi)=

408.6

544.8

681.0

P2 (PSI)=

632.7

510.2

387.8

dP1 =

367.3

489.8

612.2

dP2 =

408.6

544.8

681.0

dPt=

775.9

1034.6

1293.2

26,947


Proportional Acceleration Systems

f Remember: f Total pressure drop can be greater than system pressure, especially if the valve ratio does not match the cylinder ratio f A negative pressure in one or the other side of the cylinder means that cavitation will occur. Again, this normally occurs if the ratios do not match. f Infinite acceleration is not possible, so actual velocity and flow will be greater than average velocity and flow. Size valve accordingly. f Reducing volume between the valve and cylinder will increase spring-constant C, and allow for faster acceleration. f Larger bore cylinders also increase spring-constant C. f If natural frequency is too low, there’s always state control. But figure this out before start up!


“Oversizing” and “Undersizing” of Valves

Design Example: Two engineers select the valve and the pump (system) pressure setting for the same application. Engineer #1: calculates PL = 66 bar for load pressure Engineer #2: calculates same as above Engineer #1 sets the pump pressure to 100 bar. Therefore the valve pressure drop is ΔPv = 33 bar. A valve is selected from the catalog that passes the required flow at a pressure drop of 33 bar, when the valve is fully opened.


Valve Selection Example Engineer #2 selects a valve which has a flow rate twice that of Engineer #1’s choice (for example, 200 liter spool instead of 100 liter spool). It will have the same flow with 1/4 of the pressure drop across the valve. Engineer #2 sets his pump pressure to Pp = 74 bar. The valve pressure drop is only 8 bar rather than 33 bar! (see graph)

Design Note: Engineer #2 will have a more efficient system, suitable for “non-dynamic systems”. If dynamic acceleration and decelerations are required, a higher pressure drop will result in better control. The electrical command to the valve will be more accurately followed when a higher pressure drop is used. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Pressure Drop Relationships P=F•v

Q = Flow F = Force v = Velocity P = Power pL = Load pressure Δp = Valve drop

P = pL • Q

200%

Engineer #2

pL = 66 bar (11/12) (1/12) Δpv = 8.5 bar pP = 74.5 bar (12/12) Q, v

Engineer #1

pL

pL = 66 bar (2/3) Δpv = 33 bar (1/3) pP = 100 bar (3/3) 100

Q, V, pL, 50 Δp, P 0

Δp

Q, v pL

8%

2/ 3

Δp

33%

P

P 92%

66%

50

F

2/3

100%

50 F

2/3

11

100%

/12


Effect of Valve Response In open loop applications, a main valve selection criteria is that the valve responds accurately to the command signal. Valve shift times in the data sheet should be compared to the required response times.

Example: 4 WRZ 16 . . .

Transient function with a stepped form of electrical input signal Model 4WRZ... Signal change in %

100

0 –100

75

0 – 75

50

0 – 50

25

0 – 25

0 0

30

60

90

Time in ms

120

150

0

30

60

90

120

Time in ms

In most open loop systems, the dynamic properties of the system dictates the overall system performance. The effects of valve hysteresis and threshold become more critical when “Closing the Loop”. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

150

Closing the Loop An undesirable characteristic of most hydraulic systems is that the relationship between the command signal and the output of a hydraulic drive is not a linear (“proportional”) function. - Valve response to a “stepped” electrical signal is: Transfer function: Stepped electrical input signal Signal change in % 0-100

Size 6 100 90 80

0-75 70 60

Proportional valve type: 4WRA6...

0-50

50 40 30

0-25

20 10 0

40

80

120

160

200

0

40

80

120

160

Time in ms

(A servo valve would respond similarly but up to 10 times faster) © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Cylinder Response Cylinder response to a “stepped” input (flow “Q”) is: Servo Valve i

Cylinder Q

(X) v

X

Two characteristics are observed: 1)

A cylinder with an attached mass responds as a “spring - mass” system

2) An “integration” occurs between the input and output. This results in a “phase-lag” of -90° © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Example of Phase-Lag If the cylinder stroke versus the valve command signal (valve spool stroke) is plotted, the phase-lag can be seen:

0° 90° α

180° 270° 360°

In this example, the valve and cylinder are in open loop. Note: As the valve is operated at increasing frequencies, the phaselag increases due to inertia and spring compliance Important: If -180° total phase-lag occurs in the system, the output will be inverted as compared to the command signal! © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Closing the Loop Mechanically ΔF

Feedback

Command

When a feedback (lever mechanism in this example) is added, the system responds in a proportional manner rather than as an integrator. Response to a step is similar to the response of a valve spool, proportional with some time lag (known as “PT1”). © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Gain of the Closed Loop

If the pivot on the feedback lever is moved to the left, the amount of valve spool opening is increased for the same position error. The system corrects for errors faster, and the positional accuracy improves. If we move the pivot more and more to the left (more gain), the system response will exhibit increasing overshoot, and eventually become unstable.

Example: X

X

X

t Low Gain

t High Gain

t Too High Gain


Closing the Loop Electrically Controller ΔF Feedback Transducer S

Command Amplifier

U Control Valve

This example is essentially the same as the mechanical feedback example shown before. The ultimate system performance (response time and accuracy) is determined by the properties of all components in the system. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Properties of the System Components

1. Cylinder An ideal cylinder would have zero breakaway friction and no internal leakage. In actuality, cylinder friction results in positioning errors (cylinders exhibit stick-slip friction at low speed). Additionally internal cylinder leakage will result in a compensating flow (spool shift) in the valve. These limitations cause position errors.

2. Feedback Transducer An ideal transducer would measure infinitesimal position errors for the controller to correct. An actual transducer has limited resolution, which results in a position measurement error. The response time of some transducers can also cause signal delays, which can limit system response. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Properties of System Components 3. Valve An ideal valve would have zero response time and react to any input signal. “Real” valves have step response times as seen, and require some minimum signal to respond, known as “threshold”. Additionally, a true “zero-overlap” spool, as shown in the examples, is only possible with “poppet” type valves. Since spool valves are normally used, a “zero-overlap” spool has null flow leakage in the center position. A hydraulic designer must therefore select the best available components, within his budget, to achieve the highest static accuracy possible.


Component selection for Overall System Performance

Ultimately the system performance is determined by the quality of all of the components used. What system improvements and tricks can be used to get the most out of a system? What can be done to optimize the system: The goal is to use the highest gain possible to achieve the highest accuracy, and the fastest response time.


Cylinder Selection Guidelines 1. Select a cylinder with: - Lowest friction seal material (PTFE composite, step seals, metal rings, etc); or for highest performance applications, a servo cylinder with hydrostatic bearings. - A cylinder size selected for high natural frequency, as well as for force and buckling. Except for the piping length between the cylinder and valve (best design is the valve mounted directly on the cylinder), the piston area is the only variable for increasing the natural frequency!

fo =

C M

2π

E • A2 C= Vo


Valve Selection Guidelines 2. Select a valve with: - Minimum threshold and hysteresis (highest performance valves use electrical spool feedback) - A pressure drop as high as possible under nominal load conditions (at least 30% of the supply pressure) - Utilizing as much spool stroke as possible, with some reserve margin - Spool center position zero-lapped for position and under-lapped for pressure control - Overlapped proportional valves may only be used with appropriate control compensation features. (DMX, HNC) - Overshoot-free step response - Dynamic response determined by the application (highest available response is not always the best choice) - Special spool, if required © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Example of a Special Spool for Plastic Injection Axes Velocity and Pressure Control


Transducer and Controller Selection Guidelines 3. Select a transducer with: - Resolution five to ten times better than the required accuracy - Good linearity - Minimal time lag (high dynamic response)

4. Select a controller with: - Highest possible resolution - Fastest scan time, if digital - Control algorithms optimized for hydraulic drives: direction-dependent gain, switched integrator for positioning, spool linearization and overlap compensation, following error compensation, P-I-DT1 controller, etc © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Performance and Stability

5. Improve the System Damping Stability criteria defines the maximum system gain (goal for performance) as:

Kvmax < 2 • D • 2π fo Kvmax = Gain [ m/s ] m D = Damping Ratio fo = Natural frequency What are some ways to increase the system gain ? Increase the Damping Ratio “D” by using: 1.

Passive damping

2.

Active damping


Increasing Damping If velocity feedback is added to the position control, the closed loop frequency response can be increased. However, this results in reduced damping. This can cause reduced performance in some hydraulic closed loop systems. If acceleration feedback is added, the system damping will be increased. Load pressure feedback provides similar results when applied correctly. Using these feedbacks to improve performance is known as active damping or statevariable control.

m•a=P•A a~P The modulator shown earlier is an excellent example of improving damping by using pressure feedback !

m = mass a = acceleration P = pressure A = cylinder area © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Observer Control In some systems, a so-called “observer control” is used. The dynamic response of the system is “stored” in an electronic model of the system (analog or digital). From this stored model, calculated values for velocity and/or acceleration (the “state variables”) are derived without using actual transducers.

1) Closed loop with position feedback

Gain fo D

Command Controller

Valve

Gain fo D

X = Position X = Velocity X = Acceleration

F

X

X

Cylinder


Damping with Transducers

2) Closed loop state variable control using position, velocity and acceleration transducers

Command Controller

Gain fo Gain D fo D

F

X

Valve

KX

X

Cylinder

X

Acceleration

KX

Velocity


Damping with Observer Model Values 3) Closed loop using calculated values for velocity and acceleration

Gain fo D Command

Controller

Gain fo D

F

X

Valve

X

Cylinder

KX

XM

Acceleration

KX

XM X - XM

Velocity

Observer Model

XM

Model Corrector


Using Integrators In theory, using an integrator in the controller would result in perfect accuracy (any deviation from the commanded position would result in an increasing opening of the valve spool). However, due to friction and threshold errors, a continuous “integration limit cycle” oscillation can occur.

An integrator also results in an additional -90° of phase lag. This can cause a hydraulic positioning axis to becoming a high power “phase shift oscillator”

In practical use, an integrator is enabled only below a preset minimum velocity, and when within a position error "window" (“switched integrator”).

Along with electronic spool overlap compensation, high accuracy results can be achieved using simple, low cost proportional valves, in closed loop. © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Supply Pressure Considerations 1. Supply Pressure The system pressure should be constant. When high flow is required (high system dynamics), a drop in supply pressure can occur due to acceleration of the oil mass in the lines, and due to the response time of the pump’s pressure control.

Recommendation: An accumulator should be used in the pressure supply, mounted near the control valve.

If the system has a long return line, similar problems may occur. Acceleration of the return oil mass can cause return pressure spikes, which reduces the available working pressure. This will reduce the system performance.

Recommendation: Use return line accumulators near the control valve, as needed.


Supply Pressure Optimization

T

Using accumulators to improve dynamic response.

P

P

M S

Note: The mass of the oil in the return line can have a very significant effect !


Valve Dynamic Selection

There are three general configurations of valve-cylinder dynamics: Case 1:

The valve is high response and the cylinder is low response (fo )

Case 2:

The valve and the cylinder are similar in response

Case 3:

The valve is low response and the cylinder is high response (fo )

focyl fov

= natural frequency of the cylinder = natural frequency of the valve


Relationship of Valve and Cylinder Dynamics State-variable Controller PT1 Controller P Controller PDT1 Controller

0.3 Case 1

0.6

1.0

1.7

Case 2

focyl Case 1: f ≈ 0.3 - Best results are achieved with pressure ov or acceleration feedback

focyl ≈ 1 - A simple PT1 controller offers best results fov Case 3: focyl ≈ 3 - A PD controller will provide best results fov Case 2:


3.0 Case 3

Effect of Controller Parameters The following graphs show step response for all three cases. The parameters are varied per: 1. Increase P Gain, left to right 2. Increase, bottom to top :

Case 1: Acceleration feedback Case 2: Lag time T1 Case 3: Derivative gain


Increased Acceleration Feedback

Effect of Parameters – Case 1

Increased P-Gain


Increased Lag Time T1


Increased P-Gain


Increased Derivative Gain D


Increased P-Gain © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Performance Optimized with Parameters

In the examples shown, it can be seen that a higher gain can be achieved when other compensating parameters are used. By using higher gain, increased accuracy and improved response is achieved, and stability is maintained.


Other Considerations on System Dynamics Note: Case 1: “Fast” valve - “slow” cylinder, is an example of a “classic” servo system, using a servo valve. Optimally, a double-rod, equal area cylinder should be used.

Case 3: “Slow” valve - “fast” cylinder, is a typical application using low cost proportional valves, operating in closed loop. The control performance is not “perfect”, but can usually meet the design requirements. Due to the proportional valve’s spool overlap and nonlinear flow characteristics, as well as the fact that differential cylinders are most often used, the controller must “compensate” for these properties. Controllers such as the HACD and HNC-100 utilize these compensations. Proportional valves can provide a low cost solution with good accuracy, long life, and simple maintenance.


Closed Loop Application Using a Proportional Valve and a Differential Cylinder Encoder Feedback

s in mm

Incremental Signal

Proportional Valve

Cylinder

Amplifier

CNC

m

Feedback Q

v m/min

UCNC

L/min

U

Command

±10V

A

Δs

Drive Gain Kv =

v m/min in Q L/min

Q

UCNC + UCNC in V Electrohydraulic Gain Kv =

Q in UCNC

L/min V

− UCNC in V

Kv

− UCNC in V

Q +Q in L/min

−Q in L/min

v

−Q in L/min

v in m/min v in m/min

Kv

D/A Control

Control

Output stage

21.1

Loop Gain Kv = Kv CNC • Kv El-Hy • Kv Drive • Kv FB in

m min mm

UCNC

Kv Δs + Δs in mm CNC Gain UCNC

Kv =

Δs

V mm

in

(80)

15.9

(60)

10.6

(40)

5.3

(20)

5


B 4

A 3 2 1

10

20

30

40

50 60 70 Command value in %

1 2 3 4 5


80

90

100


Electronic Compensation of Non-linear Flow Characteristics and Valve Spool Overlap Standard

Linearization Compensation Linearization and Overlap

I

I

Amplifier

Output signal

-U

Output signal

Output signal

+U

-U

+U

Resulting signal

-U

+U

Controller signal

I

I

I

+Q

+Q

+Q

-U

Valve

I

+U

-U

+U

-Q

-Q

Positive overlap

Positive overlap

-U

+U -Q Positive overlap


Overall System Characteristics

The following pages show examples of system response, in open and closed loop. System response is shown with step and ramp command inputs, step input opposing force, and the effects of various controller elements.


Elements of an Open Loop System

“State Variables”

Cylinder

PA

X = Position X = Velocity X = Acceleration PB

Valve I V

PS

PT

Command Generator

Valve Driver

Spool Opening


Open Loop Response to a Step Input X

X

Cylinder

PA

X

PB

Valve

I V

PS

PT

Spool Opening

Valve Driver

Command Generator


Open Loop Response to a Force Step

X

X

Cylinder

F PB

PA

Valve

I V

PS

PT

Valve Driver

Command Generator

Spool Opening © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Elements of a Closed Loop System

Cylinder

Transduce r X

Transduce r Interface PA

PB

Valve

I

Command Generator

V

PS

PT

Valve Driver

Controller


Closed Loop Response to a Step Input Transducer Cylinder X Command Feedback

PA

PB

Transducer Interface

X

Valve

I V

PS

PT

Spool Opening

Valve Driver

Controlle r

Command Generator

Valve Signal


Position

Effect of P Gain on Position Control

Transducer

Gain 3 Gain 2 Gain 1 Gain 1 > Gain 2 > Gain 3

Time

Cylinder Transducer Interface

PA

PB

Valve

Velocity

I

V

Valve Driver PS PT Spool Opening

Gain Controller

Time © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Command

Closed Loop Response to a Ramp Input Transducer Cylinder

Command Feedback

X

Transducer Interface PA

PB

X

Valve

I V

PS

PT

Spool Opening

Valve Driver

Controller

Command Generator

Valve Signal


Dynamics and Errors in Closed Loop

Position

Position Error Following Error Command Feedback

Response to a Ramped Position Command

Time Phase Lag

Position

Amplitude Loss

Response to a Sinusoidal Command


Closed Loop Response to a Force Step Transducer Command Feedback

X

X

Cylinder

F Transducer Interface

PA

PB

Valve I V

PS

PT

Valve Driver

Controller

Command Generator


Position Control - Correction for Opposing Force using PI Control Force

Position Error

P Control

Position

Feedback

Command Feedback

Command

Out P Gain

Time

PI Control

Force

Position

Feedback

Command Feedback

Command

Out P Gain + I Rate


Response of a PID Controller to a Step Input

P Term Time

Input Time

Output I Term

D Term

Time

Time

Time


Response of a PID Controller to a Ramp Input P Term Time

Output

Input Time

I Term

D Term

Time

Time

Time


Position Control - Improving Dynamics using PD Control P Control

Position

Feedback

Gain 2 Gain 1

Out

Command P Gain

Gain 1 > Gain 2

Time

Position

PD Control Feedback Gain 2 Gain 1 Gain 1 > Gain 2

Command

Out P Gain + D Lead Time


Position Control - Improving Dynamics using PT1 Control P Control Position

Feedback

Out

Command P Gain

Time

PT1 Control

Position

Feedback

Out

Command P Gain + Lag Time


BoschRexroth Proportional Valves

Which Proportional Valve should I use?


Start here


Low Flow (Direct Operated) Open Loop (No Cylinder Feedback)

f 4WRAB Flows to 25 Lpm, f

minimum cost, limited performance (similar to KDG4V-3 S) 4WRA Greater flow range to 60 Lpm, better performance, more features (Ramp)

4WRAE6

4WRAB6

4WRAE10 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Low Flow (Direct Operated); Open Loop with Better Performance (No Cylinder Feedback)

f 4WRE(E)

f

f Designed for good performance without high cost f Very repeatable f Greater flow capacity f High value 4WRP(E) f Proportional with LVDT f 5 bar/ land or 10 bar delta-p, f E and W spools in housing (no sleeve) f Overlap compensation f High reliability f Most robust OBE available f CE approved

4WREE10

4WRPE10 E Proportional


High Flow (Pilot Operated) Open Loop (No Cylinder Feedback)

f 4WRZ Wide flow range, Cost effective, f

Accurate enough for most open loop needs 4WRL (HPP) with overlap E and W spools, OBE has overlap comp., High dynamics, Most robust OBE, CE approved

4WRLE (HPP) High Performance Proportional

4WRZE


Low Flow (Direct Operated) Closed Loop (Cylinder or system Feedback)

f 4WRP H - “Servo Solenoid” Best

f

Choice, Excellent dynamics for closed loop, Failsafe defined, Outperforms many old servo’s, Very Robust OBE, High reliability, Wide range of nominal flow 2 to 100 Lpm at 70 bar (Higher delta-P for Spool/sleeve valve) 4WRE Wider flow range, spool in housing (normally V), proportional flow rating, lower cost, Performance not as high

4WRPEH 10

4WREE10 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

High Flow (Pilot Operated) Closed Loop - System Feedback

f 4WRLE Servo Solenoid with V-spool,

f

High Flow range, Main spool in housing, Flow rated at 10 bar delta-p, Very Robust OBE 4WRVE (HRV 2-Stage) Higher dynamics, 10 bar drop, Very Robust OBE, 12-pin Connector 4WRLE (Servo Solenoid – 2 stage)

4WRVE (HRV - 2 stage) © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Injection Valves are Preferred

f 4WRLE..Q4 Best choice for Injection valve for Plastic Injection Molding Machines f Can replace Q2-spool f High reliability f Better Performance f Same advantages of Servo Solenoid valves 4WRLE..Q4.-3X/


pQ Valves are Preferred For Closed Loop Pressure and Throttle Function

f 5WRPE (pQ Valve) Can combine

f

“double port” throttle and “3-way” for close loop pressure control, Requires transducer and PID card, 140 Lpm at 11 bar Note: Use P1+B to P2+A to balance flow forces, pmax is 210 bar

0 811 402 107 © Alle Rechte bei Bosch Rexroth AG, auch für den Fall von Schutzrechtsanmeldungen. Jede Verfügungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.

Hydraulic Proportional Control_Bosch Rexroth

Recommend Documents