FLOW CONTROL SERVOVALVE MODEL -...

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FLOW CONTROL SERVOVALVE MODEL

C. ROMAN Imagine, Roanne

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CONTENTS

I Introduction ................................................................................................................. 3

II Description ................................................................................................................... 4

III Modeling ................................................................................................................... 5

III.1 Armature, flapper, flexure tube, wire ...................................................................... 5 III.1.i torque motor ................................................................................................... 8 III.1.ii hydraulic forces on flapper.............................................................................. 8 III.1.iii flexure tube action .......................................................................................... 8 III.1.iv feedback spring force...................................................................................... 8 III.1.v state equation.................................................................................................. 8

III.2 torque motor, electromagnetic model ................................................................... 12

III.3 Hydraulic amplifier .............................................................................................. 21

III.4 valve spool ........................................................................................................... 22

IV Simulation............................................................................................................... 24

IV.1 parameters ........................................................................................................... 26

IV.2 hydraulic characteristics & static performances................................................... 26

IV.3 dynamic performance ........................................................................................... 32

ANNEX .............................................................................................................................. 35

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I Introduction

The aim of this document is to explain the model of a 2-stages servovalve, built with AMESet and AMESim utilities. It is a flow control servovalve, with two stages, double nozzle and mechanical feedback. This kind of system uses many libraries as it deals with electrics, magnetism, hydraulics and mechanics.

Such servovalves are composed of a magnetic torque motor, an armature with flexure tube and feedback spring, an hydraulic amplifier (with nozzles), a valve spool. Each part can be modeled by a specific library. It’s a perfect example to use the new AMESim’s magnetic library.

For specific parts such as armature, AMESet allows you to code your own model. We will try, through this document, to describe the modeling, step by step, and the way to set your configuration.

Then, we will give the characteristics of this servovalve obtained by simulation (hydraulic characteristics, static and dynamic performance) that are usually obtained experimentally. Those characteristics are really similar to the real ones.

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Figure 1 a) and b): scheme of the 2-stages servovalve

MOOG Documents

a)

b)

II Description

• Electrical current in torque motor coils creates magnetic forces on ends of armature.

• Armature and flapper assembly rotates about flexure tube support.

• Flapper closes off one nozzle and diverts flow to that end of spool.

• Spool moves and opens Ps (supplied pressure) to one control port and opens other control port to R (tank pressure).

• Spool pushes ball end of feedback spring creating restoring torque on armature/flapper.

• As feedback torque becomes equal to torque from magnetic forces, armature/flapper moves back to centered position.

• Spool stops at a position where feedback spring torque equals torque due to input current.

• Therefore spool position is proportional to input current.

• With constant pressures, flow to load is proportional to spool position.

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III Modeling Servovalve model is built with the following elements:

• Electromagnetic part: electromagnetic library in industrialization (see technical document in Annex 1).

• Hydraulic part: built with Hydraulic Component Design Library. • Armature, flapper and flexure tube: specific model built with AMESet utility.

III.1 Armature, flapper, flexure tube, wire

With :

• l: half armature length • L: flexure tube length • d1: distance between center of gravity and flexure tube start • d2: distance between center of gravity and nozzles • d3: distance between center of gravity and spool • θ: angle of rotation

To make a complete model of armature, we need few more parameters:

• m : mass of armature • J: moment of inertia • E: Young modulus of flexure tube material • ∅e : exterior diameter of flexure tube • ∅i : interior diameter of flexure tube • k : stiffness of feedback spring • br: damping ratio for rotation • bt: damping ratio for translation

+ l

d1

θ

L

d2

d3

center of gravity flapper

flexure tube

armature

Figure 2 : scheme of the armature and flexure

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Those parameters are sufficient to fully describe the armature. Hypothesis: rigid body, small displacements are considered. At first, we need to know armature displacement in a few points. Armature has 2 degrees of freedom: θ for rotation and xg, center of gravity displacement (z displacement of the gravity center is neglected). We need to express all other displacements with these 2 degrees. Here are the points we need to see: This is a multiport system: there are 8 ports with the following variables:

• Port 1: force (f1) on armature (input), velocity (vz1) (output), displacement (z1) (output)




• Port 5 : flexure tube actions (internal), velocity (vt), displacement (xt) • Port 6 : center of gravity velocity (vg) and displacement (xg), rotation (θ) and rotary

speed (ω) (state variables). • Port 7: right hydraulic force on flapper (fr)(input), flapper displacement and velocity

(output). • Port 8: left hydraulic force on flapper (fl)(input), flapper displacement and velocity

(output). • Port 9: wire displacement (input), fictitious displacement (internal), feedback spring

force (output).

z1 z3

z4 z2

xt

xg

xflapper

xfictitious

Figure 3 : principal displacements

1

2

3

4

5

6

7

9

8

7

We easily find the relations: The relations are identical for their displacements (small displacements).

Now, we need to list the different forces which act on this armature: There is:

• the torque motor • hydraulics forces on the flapper • flexure tube action • feedback spring force

⋅

−−

−−

=

θxg

dtd

dddlll

l

sxfictitiouxflapper

xtzzzz

dtd

312111

0000

4321

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III.1.i torque motor

Torque motor creates magnetic forces on ends of armature. There are 4 forces f1, f2, f3, f4 which corresponds with the displacements z1, z2, z3, z4. The torque motor is: CTorqueMotor = (f2+f3-f4-f1) . l

III.1.ii hydraulic forces on flapper We can separate hydraulic action on flapper by 2 components that we will call fl (left force) and fr (right force). Those forces acts on the flapper. Fnozzle-flapper= fl-fr

III.1.iii flexure tube action

Litterature gives classical relations between warping variables and torque and force (ref: [8] Karnopp, Margolis, Rosenberg).

θθθθτxg

KdKKKdKKxgd

KKKKxt

KKKKxt

LLLL

IEF ⋅⋅+⋅+=⋅⋅=⋅=⋅−

−⋅⋅=122221111211

22211211

22211211

2

23

11

1011

46612

with: E= Young modulus of flexure tube material

I= Moment of inertia of flexure tube. ( )44

64DiDeI −⋅= π

L= Flexure tube length

III.1.iv feedback spring force

This is a simple stiffness: Fw= - kw . ∆x = - kw . ( xfictitious – xw) = - kw . ( (xg – d3 . θ) - xw) xw corresponds to spool position.

III.1.v state equation We already have detailed the different actions on armature. It is now easy to find:

2)(3)4132( dflfrdfwbrlffffJ tube ⋅−+⋅−−⋅−⋅−−+=⋅ τθθ

tubeFfwfrflgxbtgxm −+−+⋅−=⋅

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Replacing some expressions, we have:

( )( ) ( )( ) ( ) 2331 122221 dflfrdxwdxgkwKdKxgKbrCJ TM ⋅−+⋅−⋅−+⋅⋅++⋅−⋅−=⋅ θθθθ

( )( ) ( )( )θθ ⋅⋅++⋅−−⋅−⋅−−+⋅−=⋅ 111211 13 KdKxgKxwdxgkwfrflgxbtgxm In a vectorial writing:

( )( )

( )

00

23

0010

0001

310

3310

111112

1221222

mflfrxwkw

JdflfrdxwkwC

xg

vgmkwK

mdkwKdK

mbt

JdkwK

JdkwKdK

Jbr

gx

gv

TM

−−⋅

⋅−+⋅⋅−

+⋅−−⋅+⋅−−−

⋅+−⋅+⋅+−−

= θ

ω

θ

ω

This is a state equation of the classical form:

uBxAx += A is called state matrix, B is called control matrix. The code of the armature model is based on this state equation. We have tested this model by 2 ways. First, we made a system composed only of this model with no motor torque nor hydraulic components in order to obtain its eigenvalues. Then, we tested it with the motor torque to ensure the gain was good and to see the influence of the motor torque on the previous eigenvalues. First, we needed to set parameters for the armature. We didn’ t have any exact values but, we managed to find a coherent set. File: test_armature.ame

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After linearization, we found 2 pairs of conjugate eigenvalues, which corresponds to 2 frequencies and modal shapes.

Those two modes do not effect on rotation and translation in the same way. The second mode (4034Hz) is much more effective on translation than the first one. To be sure of it, when can have a look to the modal shapes.

m = 5.5 g J = 5.8 10-7 Kg.m² l = 14.5 mm (half armature length) L = 4.8 mm (flexure tube length ) ∅e = 4.46 mm ∅i = 4.36 mm E = 1.2 106 bars (Young modulus) d1 = 3 mm d2 = 8.6 mm d3 = 22 mm kw = 3400 N/m br = 0.001 N/(rad/s) (damper rating for rotation) bt = 10 N/(m/s) (damper rating for translation)

d2

d1

+ θ

L

d3

l

Table 1 : armature parameters Figure 4 : scheme of armature

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The first mode doesn’ t effect much on translation as we can see on figure 5. The second mode shows that translation and rotation are both influenced by this frequency. We can note that translation is much more important for this frequency than for the first mode.

Figure 5 : modal shape for first mode

Figure 6 : modal shape for second mode

Scale factor : 1 Scale factor : 10

Scale factor : 1 Scale factor : 10

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III.2 torque motor, electromagnetic model

The torque motor consists in an armature (the one we studied before), 2 permanent magnets, 2 pole pieces, 4 air gaps. Armature is to rotate, which change air gap lengths. Equilibrium position results from the simultaneous actions of an electromagnetic torque, due to 2 serial coils set on the armature, and a mechanical return torque. This return torque is null when all air gaps are equal. To simplify the problem, we will only consider a half of the problem, and moreover, we will consider that the only reluctance important enough is the air gap ones. Then, to simplify again, we will work with an equivalent air gap instead of 2 air gaps.

N

S S

N

magnet

magnet

Lower pole piece

Upper pole piece

Motor coil

air gap

armature

Figure 7 : scheme of torque motor

Figure 8 : simplified problem

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An air gap is characterized by his pole area, and his length. The equivalent air gap of 2 serial air gaps is an air gap of the same pole area but with length 2 times bigger. We shall take into account the influence of the motor coils, which produce demagnetizing ampere-turns. The aim of the problem is to find the operating point of the system. First, we must evaluate the permeance of the equivalent circuit. For an air gap the permeance is given by the following relation: with: µ0= permeability of free space S= pole area e = air gap You must remember that equivalent air gap is 2 times the normal air gap. Then, you calculate the equivalent permeability of the circuit with the dimensions of the magnet (effective section, lenght). This is the main parameter because it represents the linear characteristic of the system with the relation: But, now we need to evaluate the demagnetizing ampere-turns (Atd), to know exactly in which range the circuit will function. Then, we have the equation of a new linear characteristic of the circuit: The operating point is located at the intersection of the demagnetizing curve and the load line defined by the relation above. This line intersects the H-axis at point H= Atd/la. This point is defined by its coordinates, Bmin and Hmin, by which passes the recoil line. The recoil line is the real characteristic of the system, it means that the operating points will be situated on this line.

∆: B= µ . H

SalaP⋅=µ

eSP ⋅= 0µ

Atd= N . I

∆’ : ( )la

AtdHB −⋅=µ

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In the torque motor supercomponent, you must define your recoil line by specifying the coordinates of the operating point (Hmin, Bmin) and the remanent flux density of the recoil line. It’ s really important for your model to have a good values for your recoil line and the best way is to apply the method above.

Remanent flux density (Br)

Coercitive field (Hc)

Recoil line

∆ ∆’

Atd/la

Hmin

Bmin

Figure 9 : Recoil line determination

Figure 10 : sketch of torque motor

Permanent magnet

Air gap

1

2 4

3

Electrical ports

? Corresponding ports on armature model

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For example, we want to locate your operating point for a system composed of: Magnet: Hc = -30 000 A/m Br = 0.2 T la = 25 mm Sa = 60 mm² Air gap: e = 310 µm S= 15 mm² Serial coil: N = 4200 turns R = 400 Ω Command intensity : 10 mA

86

67 1004.3

10.310210.15

104 −−

−− ⋅=⋅⋅⋅⋅= πP

56

38 1027.1

10601025

1004.3 −−

−− ⋅=⋅

⋅⋅⋅=µ

16801025

01.042003 =⋅

⋅= −laAtd

With these values, we can find some good values for your operating point coordinates as (Hmin=16 000, Bmin=0.13). Supposing you know the recoil permeability of your permanent magnet (for example 3.1 . 10-6 for Alnico 1500), you easily find Brrecoil = Bmin -µr . Hmin = 0.18 (see Annex 2: Alnico 1500 demagnetizing curve). If you don’ t know the recoil permeability, you can see that using Br instead of Brrecoil is not so important. In any case, you would better put values for Bmin and Hmin less than the real ones but not too far, else your model could have problems. Now that AMESim can calculate your torque motor, let’ s see how it looks like. Literature gives a good approximation of that torque with the following relation (Annex 1, ref: [7] M. Lebrun):

( )

⋅⋅++⋅⋅+⋅⋅

−⋅

⋅⋅= ex

EiN

exEiNe

xE

ex

e

SlCTM0

2

0202

2

22

0 11

µ

with x, the displacement of the armature in air gap (note that θ=x/l), E0, magnetomotive force of magnet. We want to linearize this equation, so we transform it in the first order equation: CTM= K1.i + Km.θ With:

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3

200

2

200

1 ,e

ESlKm

eENSlK

⋅⋅⋅=⋅⋅⋅⋅=

µµ

It’ s interesting to see what happens in the state equation when we replace CTM by the linear approximationK1.i+Km.θ. We find:

( )( )

( )

00

23

0010

0001

310

3310

1

111112

1221222

mflfrxwkw

JdflfrdxwkwiK

xg

vgmkwK

mdkwKdK

mbt

JdkwK

JKmdkwKdK

Jbr

gx

gv −−⋅

⋅−+⋅⋅−⋅

+⋅−−⋅+⋅−−−

⋅+−+⋅+⋅+−−

= θ

ω

θ

ω

When you connect the armature submodel and the torque motor, the eigenvalues will be modified as you changed the state matrix. Magnetic stiffness is opposed to others stiffness so that it will reduce eigenvalues. It’ s maybe the most difficult part of the model for setting parameters as we could have seen it earlier in the notice. To test the model, we found some parameters in different catalogues and here are the values we chose: Electrical characteristics: Series coils R = 200 Ω for each coil (total resistance 400 Ω for 2 coils) iR = 10 mA N = 2100 turns per coil ( 4200 turns for 2 series coils) Elements characteristics: Air gap: 310 µm Pole area: 15 mm² Armature length: 29 mm Cross section: 12 mm² The most important is to set parameters for the magnet as we said above. Notices always give the torque motor gain (ref: [4], Moog), which is its main characteristic. We want to set our parameters in order to find a standard value for this gain: 0.03 N.m for i = 10 mA. Nota: CTM = K1 . i + Km . θ (linear model) To examine the motor torque, we built a simple system in which you can change only the intensity and observe the torque ( it means K1), or only the angle ( it means Km). To control intensity and rotation angle, and to observe the resulting torque, we used signal elements. (file: test_torquemotor.ame)

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results: Good values for magnets are: Hmin = -20 000 A/m Brrecoil = 0.15 T Bmin = 0.1 T AMESim find an operating point defined by: H = -9900 A/m B = 0.12 T Which corresponds to a magnetomotive force E0 = 248 A. Analytic calculus (as in paragraph III) gives: K1 = - 2.96 N.m/A Km = 8.18 N.m/rad We obtained the following characteristics:

Figure 11: Benchmark for motor torque

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The simulation finally gives: Km = 7.86 N.m/rad (figure 12 a) ) K1 = -2.73 N.m/A (figure 12 b) ) It means we have a difference of 4% for Km and 8% for K1 with our linear approximation. As we said earlier, motor torque as an influence on armature’ s eigenvalues because torque motor is function of rotation angle (CTM = K1 . i + Km . θ). This test consists in linking torque motor and armature and observing new eigenvalues. (file: test_armature+torque.ame)

a) b)

Figure 12 a) and b) : torque motor constants

Figure 13 : sketch to test armature and torque motor

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As we can see in the state equation, the more the magnetic constant Km grows, the more it will reduce the frequency of eigenvalues. For a certain value of Km, it will annul this frequency, and over this value the system will get unstable. The problem is that the value used for the complete servovalve makes this sytem unstable. But, when we link this system with hydraulic amplifier, it won’ t be unstable anymore, because of hydraulics stabilizing effects. We will test this model with 3 sets of parameters, in order to see the influence on eigenvalues. We want to modify Km, it comes to change the magnets’ magnetomotive force. The best way to illustrate this phenomen, is to translate our recoil line. We will choose 3 sets for Bmin and Br, and see the influence on eigenvalues frequencies and damping.

• Bmin= 0 T, Br = 0.05 T E1= 96.7 A

Recoil line 1

Recoil line 2

Recoil line 3

H . la

B

E3 E2 E1

∆’

Figure 14: Recoil line translation

Table 3 : influence of torque motor on armature’ s eigenvalues

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• Bmin= 0.06T, Br = 0.11T E2= 198.8 A

• Bmin= 0.08 T, Br = 0.142 T E1= 255A

As we can see, the rotation mode is to disappear when permanent magnets get sufficient induction.

Table 4

Table 5

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III.3 Hydraulic amplifier

Let’ s see what happens in the hydraulic amplifier:

• Armature and flapper are rigidly joined and supported by thin-wall flexure tube. • Fluid continuously flows from pressure Ps, through both inlet orifices, through another

orifice, past nozzles into flapper chamber, through drain orifice to return R (tank). • Rocking motion of armature/flapper throttles flow through one nozzle or the other. • This diverts flow to A or B (and builds up pressure if A and B are blocked).

This system can be easily built with AMESim’ s Hydraulic Component Design library. Sketch:

Flapper

Flexure tube

Inlet orifice

Drain orifice

nozzle orifice

T C1 C2

Ps Ps

Figure 15 : Hydraulic amplifier

C1 C2

Ps

Inlet orifice

Drain orifice Tank ( T )

orifice

Armature submodel nozzle

Figure 16 : Sketch for hydraulic amplifier

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nota: There is a model of elastic contact behind the nozzles. Its aim is to absorb an impact between nozzle and flapper. In case of excessive rotation of the armature, it will stop it.

III.4 valve spool The spool is an essential part of the servovalve because it assumes the distribution of pressure between orifices, and by the way, the flow control.

• Spool slides in bushing (sleeve) as pressure in C1 and C2 varies. • Bushing contains rectangular holes (slots) or annular grooves that connect to supply

pressure Ps and return T. • At null position, spool is centered in bushing; spool lobes (lands) just cover Ps and T

openings. • Spool motion to either side of null allows fluid to flow from Ps to one control port,

and from other control port to T.

Ps Ps T T

A B

C1 C2

Feedback spring spool

Bushing

Ps Ps T T

A B

C2 C2

a)

b)

Figure 17 a) and b) : scheme of valve spool

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This time again, you can build the valve spool with an assembly of spools with annular grooves, pistons for the chambers A and B, and a mass with end-stops to take in account the inertia. Here’ s what it looks like:

C1 C2

To feedback spring port

A B

T

Ps

Piston spool

Figure 18: sketch of spool valve

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IV Simulation Complete model: (file: servovalve.ame) sketch:

Figure 19 : Sketch of complete servovalve

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The system is now complete. First, we are going to check the motor gain. To obtain it, we just have to plot (f2+f3-f4-f1).l for a sinusoidal current in input (amplitude: rated current=10 mA).

As you can see, torque motor is well adjusted with the parameters we set in the magnets. This is surely the most difficult to configurate in the complete model.

Tor

que

mot

or (N

m)

Time (s)

Figure 20 : Torque motor for sinusoidal rated current

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IV.1 parameters

Here is the list of parameters we used for the complete model. We kept the same value for motor torque and armature. We will only detail hydraulic amplifier and spool configuration.

IV.2 hydraulic characteristics & static performances

rated flow: simulation conditions: i = rated current (10mA) Ps= 50 bars → 275 bars During 20s.

Supply pressure: 210 bars Nozzle diameter : 0.4 mm Nozzle-flapper opening : 0.06 mm Inlet orifice : 0.2 mm Drain orifice : 0.3 mm Nozzle orifice : 1 mm Spool mass: 10 g Spool diameter: 10 mm Rod diameter: 5.5 mm Spool/bushing radial clearance : 2 µm Spool stroke: 500 µm Spool rounded corner radius: 5 µm Underlap: 10 µm Spool viscous friction: 50 N/(m/s)

Figure 21

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Flow-load function: This is the servovalve control flow when there is load pressure drop, expressed with input current as parameter. Nominal flow to load is given by: where Qv = valve flow to load K = servovalve characteristic constant i = input current ∆Pv = valve pressure drop note: ∆Pv = Ps - R - ∆PL Ps= supply pressure R = return pressure ∆PL = load pressure drop

Load pressure drop % of supply pressure

Con

trol

flow

% ra

ted

VV PiKQ ∆⋅⋅=

LV PRPsiKQ ∆−−⋅⋅=

Figure 22

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flow curve:

This graph represents the flow gain, and show good linearity and low hysteresis(only due to system’ s dynamic). Rated flow is 84 l/min for Ps = 210 bars. Maximum valve flow can reach 135 % rated flow with oversignal.

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Spool lap: In sliding spool valve, the relative axial position relationship between the fixed and movable flow metering at null. The actual configuration of our servovalve is set with an amount of underlap of 10 µm. But, you can specify an amount of underlap or overlap in order to: Underlap (10 Pm):

• increase flow gain at null. • reduce valve pressure gain at null. • increase valve null leakage.

Overlap (10 Pm):

• reduce flow gain at null • reduce null leakage flow • degrade pressure gain (into load)

Flow

con

trol

Input current

Zero lap

underlap

overlap

Figure 24: influence of spool lap on flow rate at null

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Static performances with blocked load Internal leakage:

This characteristic is obtained with blocked load. Internal leakage includes first stage hydraulic amplifier flow, spool null leakage flow, and bushing laminar leakage flow.

• Spool leakage is essentially zero when spool is off-null. • Servovalve internal leakage excluding spool null leakage is called Tare flow.

Hydraulic amplifier flow largely determines servovalve frequency response

• Lower flow degrades response. Spool null leakage is related to maximum valve flow (slot width) and null cut.

null leakage

Tare flow

null

Figure 25 : internal leakage at null

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Pressure gain:

The blocked load differential pressure will change rapidly from one limit to the other as input current causes the valve spool to traverse the null region. Here, the pressure gain at null exceeds 100 % of supply pressure for 1% of rated current.

Figure 26 : Pressure gain

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IV.3 dynamic performance

frequency response(file: bode-servovalve.ame): Frequency response will depend upon signal amplitude, supply pressure…Natural frequency ranges from 45 to 80 Hz. To make a bode diagram, we used a transferometer. You can’ t realize the standard linearization because of numeric problems due to non-linearities.

Figure 27

Figure 28

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Step response:

Step of current: We observe that time response is depending of signal input amplitude. From 50 % of rated current, spool velocity is constant. This phenomenon is due to flapper displacement saturation. We can observe flapper displacement on figure 30.

Figure 29

Figure 30

25 %

50 %

75 %

100 %

34

REFERENCES [1] « Hydrauliques et électrohydraulique » - J. Faisandier – Dunod [2] « Analyse et conception des systèmes hydrauliques » - Imagine [3] « Systèmes dynamiques et automatique linéaire » - Imagine [4] Moog documents [5] « Specification standards for electrohydraulic flow control servovalve » - Moog [6] « Manuel technique des aimants permanents » – Chambre syndicale des producteurs

d’ aciers fins et spéciaux

[7] « Modélisation et simulation d’ asservissements electrohydrauliques» - M. Lebrun [8] « System Dynamics, Modeling & simulation of mechatronic systems » - Karnopp,

Margolis, Rosenberg HTML reports for the different model: test_armature.html test_torquemotor.html test_armature+torque.html bode-servovalve.html servovalve.html

35

ANNEX ANNEX 1 : Torque motor expression. ANNEX 2 : Permanent magnet (Alnico 1500) demagnetizing curve. ANNEX 3: Bond graph of armature. ANNEX 4: Bond graph of torque motor. ANNEX 5: Bond graph of complete servovalve.

36

Torque motor: Here is the demonstration of the analytic formula for torque motor. E0 = magnetomotive force of the permanent magnet R0 = air gap reluctance for neutral reluctance We will consider that the only reluctance important enough is air gap reluctance, as a numerical application could confirm. x = armature displacement in air gap. e = air gap Air gap reluctances are: Magnetomotive force evaluation gives: We obtain the following flow expressions: Torque exerted by the motor is: ie:

R1

R2 N.i

R1

R2 Φ2

Φ2

Φ1

Φ1

ANNEX 1

( )exRS

xeR +⋅=⋅+= 100

1 µ( )e

xRSxeR −⋅=⋅

−= 100

2 µ

1122 φφ ⋅=⋅+⋅− RRiN11220 φφ ⋅+⋅= RRE

( )exRiNE

RiNE

+⋅⋅⋅−=⋅

⋅−=122 0

0

1

01φ

( )exRiNE

RiNE

−⋅⋅⋅+=⋅

⋅+=122 0

0

2

02φ

( )21

22

0φφµ −⋅= S

lC

( )

⋅⋅++⋅⋅+⋅⋅

−⋅

⋅⋅= ex

EiN

exEiNe

xE

ex

e

SlCTM0

2

0202

2

22

0 11

µ

37

ANNEX 2

38

1 4z 1 3z 1 2z

1

122221111211

11

−

⋅+⋅+KdKKKdKK

1 1z

1

flapperx

1 θ

1

Gx

TF : l

TF : l

TF : -l

TF : -l

Magnétique Magnétique Magnétique Magnétique

I : J

I : m

0

TF: -z2 C

0

TF: z3

0

TF: z4

1

fictitiousx

0

C : Kw

1

wx

Fw

Bond graph of the armature

ANNEX 3

39

Air gap

Permanent magnet

Permanent magnet

F x

0

0

1

1

1

1

0

0

0

1

1

1

1

0

C

C

C

C

C

C

C

C

1 1

GY

1

1

C

C

1 1

C

SE 1 port capacitor with derivative causality

coil

Bond graph of torque motor

ANNEX 4

40

ANNEX 5

1

122221111211

11

−

⋅+⋅+KdKKKdKK

flapperx 1

i 1

Gx 1

θ

Pc2 0

Pc1 Pt

SF GY

K1

C : -1/Km

I : J

R :br

I : m

R :bt

C

0

TF : -d2

1

R

0 1

TF

0

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FLOW CONTROL SERVOVALVE MODEL -...

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