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How to select a Servo or Proportional valve Introduction

Criteria of valve sizing Following criteria's have to be considered for sizing a valve! • Which

kind of actuator shall be taken • Differential cylinder • Synchronising cylinder • Hydromotor

• The

conditions of the actuators • Dimension of the actuator • Maximum required flow and pressure • The natural frequency of the actuator • What kind of controler is needed ( position, force, speed )

• Possible MOOG Training

disturbances and environmental conditions Seite 2

Valve sizing 6 Steps • Determination of the needed pressure at the cylinder • Determination of the Pressure Drop of the Servovalve • Determination of the needed maximal Flow • Determination of the natural frequency of the actuator • Selection of the Servovalve • Calculation of the gain loop and therefore the accuracy of the controller

MOOG Training

Seite 3

Example

Vmax

d

A

F D

P

Type of Piston Annular (piston) area Max velocity (load) ΣForce (load) Mass Installation position MOOG Training

: : : : : :

T

Synchronising cylinders (D=63 / d=45 / H=400) A = 15,26 cm² Vmax = 1,5sec / 300 mm Fr = 9000 N (1*) 900 Kg Horizontal / valve direct mounted on the cylinder 1* : Simplified => Only the Mass

Seite 4

Step 1 Determination of the needed pressure on the actuator The needed pressure „P“ relates on

F

A Surface of the piston „A“

Formula

Pload = F / A MOOG Training

Loaded force „F“

Pload = 9000N / 15,26 cm² Pload = 590 N/cm² Equivalent to Pload = 59 bar => 60 bar (1bar = 10N/cm²) Pload = 590000 Pa (Pascal) Seite 5

Step 2 Determination of the pressure drop at the Servovalve 1/3 - 2/3 Rule This rules describes the best condition •1/3 Pressure drop at the valve •2/3 Pressure drop at the actuator

Formula 2/3 of „P“

1/3 of „P“

PT

MOOG Training

∆ PVa = Pload / 2 ∆ PVe = 60 bar / 2 ∆ PVe = 30 bar Seite 6

Step 3 Determination of the needed maximal Flow The maximum flow „Q“ relates on

A

Vmax

Surface of the piston „A“

Formula

Q = vmax * A

Requested max speed (*1)

Q = (30cm/1,5 sec) * 15,26 cm² Q = 20 cm/sec * 15,26 cm² Q = 305 cm³ / sec Equivalent to

Q = 18,3 L/min MOOG Training

*1: The requested speed have to be calculated with and with out load

Seite 7

Step 4 Determination of the natural frequency of the actuator Natural frequency “ωo“ or „f“ relates on

m spring constant “C”

the mass “m”

Each actuator has to be considered as a hydraulic spring-mass system All fluids are compressible and therefore they are reacting like a spring.

MOOG Training

m

The load of the cylinder is the mass

Principle graph

Seite 8

Step 4 Determination of the natural frequency of the actuator Calculation of the natural angular frequency „s –1“ (*1)

Formula

ωo=

C m

ω

4 * 1,4 * 10 7 kg * 15,26cm² o= cm * s ² * 40cm * 900 Kg ωo ≈ 154 s-1

Calculation of the natural frequency „f“

ω f= Hz 2π

154 f= Hz = 24,5 Hz 2π * s natural frequency „f“ for valve selection

f1/3 = 24,5 Hz / 3 = 8,71 Hz MOOG Training *1: calculation based on : Synchronising cylinder / valve installed direct on the cylinder

Seite 9

Step 5 Selection of the servovalve Principle criteria's of valve sizing for position and speed controler The smallest possible valve for the application is the best valve The natural frequency of the valve should be 2,5 to 3 times higher then the natural frequency of the valve The trapped oil between valve and actuator should be as small as possible Oil Volume

The natural frequency of the sensor should be 10 times higher

MOOG Training

Seite 10

Step 5 Selection of the servovalve Principle criteria’s of valve sizing for force/pressure controller The smallest possible valve for the application is the best valve The faster the valve the better the damping which is necessary for a stable closed loop controller The higher the oil capacity (volume) the better Oil Volume

Smaller flow pressure gain The natural frequency of the sensor should be 10 times higher MOOG Training

Seite 11

Step 5 Selection of the servovalve / valve sizing The flow rate of servo valves is determined in the catalogues at 70 or 10 bars. There are two ways to select the needed valve size and type. Selection from the Flow / Pressure Diagram

Calculation of the needed flow at 70 and 10bars

Formel Q = QN *

Q = 18,3 L/min

Detla p Detla p N 10bar 30bar

Q= 10,6 L/min at 10 Bar

Q = 18,3 L/min

70bar 30bar

Q= 28 L/min at 70 Bar Example shown in the D63x catalogue MOOG Training

Seite 12

Step 5 Selection of the servovalve / valve sizing

Take the catalogue of the D63 Series and select out the correct valve size

MOOG Training

Seite 13

Step 5 Selection of the servovalve / criteria's of spool selection Position controller loop 4 Way valve

accurate axis cur (< +/- 1%)

Delta P (%) 70%

High pressure gain (80% P at 2% signal)

2% -4

4 Input signal (%)

At differential cylinders : according to the cylinder surface ratio adopt spools Ak MOOG Training

Ar Seite 14

Step 5 Selection of the servovalve / criteria's of spool selection Force or pressure controller loop 4 Way valve

accurate axis cur (< +/- 3%)

Delta P (%)

Lower pressure gain -4

4 Input signal (%)

At differential cylinders : according to the cylinder surface ratio adopt spools Ak MOOG Training

Ar Seite 15

Step 5 Selection of the servovalve / conclusion

We selected out ... a 4 Way valve with 40 L/ min at 70 bar with a dynamic of 75 Hz (90% Signal at –90° phi) with axis cut of (< +/- 1%) and a high pressure gain (80% P at 2% signal)

MOOG Training

Seite 16

Typkey / of D633

MOOG Training

Seite 17

Step 6 Calculation of the gain loop and therefore the accuracy of the controller fSV w

xw x

controller

KV

servo valve

fH hydr. actuator

fS sensor

w = command signal x = control value xW = error signal KV = loop gain

MOOG Training

Seite 18

Additional items to consider

Cavitation Fluid Temperature Changing mass Human and machinery safety regulations Installation position of the cylinder Environmental conditions like shocks, vibrations and electrical conditions (EMC) MOOG Training

Seite 19

View more...
Criteria of valve sizing Following criteria's have to be considered for sizing a valve! • Which

kind of actuator shall be taken • Differential cylinder • Synchronising cylinder • Hydromotor

• The

conditions of the actuators • Dimension of the actuator • Maximum required flow and pressure • The natural frequency of the actuator • What kind of controler is needed ( position, force, speed )

• Possible MOOG Training

disturbances and environmental conditions Seite 2

Valve sizing 6 Steps • Determination of the needed pressure at the cylinder • Determination of the Pressure Drop of the Servovalve • Determination of the needed maximal Flow • Determination of the natural frequency of the actuator • Selection of the Servovalve • Calculation of the gain loop and therefore the accuracy of the controller

MOOG Training

Seite 3

Example

Vmax

d

A

F D

P

Type of Piston Annular (piston) area Max velocity (load) ΣForce (load) Mass Installation position MOOG Training

: : : : : :

T

Synchronising cylinders (D=63 / d=45 / H=400) A = 15,26 cm² Vmax = 1,5sec / 300 mm Fr = 9000 N (1*) 900 Kg Horizontal / valve direct mounted on the cylinder 1* : Simplified => Only the Mass

Seite 4

Step 1 Determination of the needed pressure on the actuator The needed pressure „P“ relates on

F

A Surface of the piston „A“

Formula

Pload = F / A MOOG Training

Loaded force „F“

Pload = 9000N / 15,26 cm² Pload = 590 N/cm² Equivalent to Pload = 59 bar => 60 bar (1bar = 10N/cm²) Pload = 590000 Pa (Pascal) Seite 5

Step 2 Determination of the pressure drop at the Servovalve 1/3 - 2/3 Rule This rules describes the best condition •1/3 Pressure drop at the valve •2/3 Pressure drop at the actuator

Formula 2/3 of „P“

1/3 of „P“

PT

MOOG Training

∆ PVa = Pload / 2 ∆ PVe = 60 bar / 2 ∆ PVe = 30 bar Seite 6

Step 3 Determination of the needed maximal Flow The maximum flow „Q“ relates on

A

Vmax

Surface of the piston „A“

Formula

Q = vmax * A

Requested max speed (*1)

Q = (30cm/1,5 sec) * 15,26 cm² Q = 20 cm/sec * 15,26 cm² Q = 305 cm³ / sec Equivalent to

Q = 18,3 L/min MOOG Training

*1: The requested speed have to be calculated with and with out load

Seite 7

Step 4 Determination of the natural frequency of the actuator Natural frequency “ωo“ or „f“ relates on

m spring constant “C”

the mass “m”

Each actuator has to be considered as a hydraulic spring-mass system All fluids are compressible and therefore they are reacting like a spring.

MOOG Training

m

The load of the cylinder is the mass

Principle graph

Seite 8

Step 4 Determination of the natural frequency of the actuator Calculation of the natural angular frequency „s –1“ (*1)

Formula

ωo=

C m

ω

4 * 1,4 * 10 7 kg * 15,26cm² o= cm * s ² * 40cm * 900 Kg ωo ≈ 154 s-1

Calculation of the natural frequency „f“

ω f= Hz 2π

154 f= Hz = 24,5 Hz 2π * s natural frequency „f“ for valve selection

f1/3 = 24,5 Hz / 3 = 8,71 Hz MOOG Training *1: calculation based on : Synchronising cylinder / valve installed direct on the cylinder

Seite 9

Step 5 Selection of the servovalve Principle criteria's of valve sizing for position and speed controler The smallest possible valve for the application is the best valve The natural frequency of the valve should be 2,5 to 3 times higher then the natural frequency of the valve The trapped oil between valve and actuator should be as small as possible Oil Volume

The natural frequency of the sensor should be 10 times higher

MOOG Training

Seite 10

Step 5 Selection of the servovalve Principle criteria’s of valve sizing for force/pressure controller The smallest possible valve for the application is the best valve The faster the valve the better the damping which is necessary for a stable closed loop controller The higher the oil capacity (volume) the better Oil Volume

Smaller flow pressure gain The natural frequency of the sensor should be 10 times higher MOOG Training

Seite 11

Step 5 Selection of the servovalve / valve sizing The flow rate of servo valves is determined in the catalogues at 70 or 10 bars. There are two ways to select the needed valve size and type. Selection from the Flow / Pressure Diagram

Calculation of the needed flow at 70 and 10bars

Formel Q = QN *

Q = 18,3 L/min

Detla p Detla p N 10bar 30bar

Q= 10,6 L/min at 10 Bar

Q = 18,3 L/min

70bar 30bar

Q= 28 L/min at 70 Bar Example shown in the D63x catalogue MOOG Training

Seite 12

Step 5 Selection of the servovalve / valve sizing

Take the catalogue of the D63 Series and select out the correct valve size

MOOG Training

Seite 13

Step 5 Selection of the servovalve / criteria's of spool selection Position controller loop 4 Way valve

accurate axis cur (< +/- 1%)

Delta P (%) 70%

High pressure gain (80% P at 2% signal)

2% -4

4 Input signal (%)

At differential cylinders : according to the cylinder surface ratio adopt spools Ak MOOG Training

Ar Seite 14

Step 5 Selection of the servovalve / criteria's of spool selection Force or pressure controller loop 4 Way valve

accurate axis cur (< +/- 3%)

Delta P (%)

Lower pressure gain -4

4 Input signal (%)

At differential cylinders : according to the cylinder surface ratio adopt spools Ak MOOG Training

Ar Seite 15

Step 5 Selection of the servovalve / conclusion

We selected out ... a 4 Way valve with 40 L/ min at 70 bar with a dynamic of 75 Hz (90% Signal at –90° phi) with axis cut of (< +/- 1%) and a high pressure gain (80% P at 2% signal)

MOOG Training

Seite 16

Typkey / of D633

MOOG Training

Seite 17

Step 6 Calculation of the gain loop and therefore the accuracy of the controller fSV w

xw x

controller

KV

servo valve

fH hydr. actuator

fS sensor

w = command signal x = control value xW = error signal KV = loop gain

MOOG Training

Seite 18

Additional items to consider

Cavitation Fluid Temperature Changing mass Human and machinery safety regulations Installation position of the cylinder Environmental conditions like shocks, vibrations and electrical conditions (EMC) MOOG Training

Seite 19

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