Electric Machines and Drives by Miroslav Chomat - HTML preview

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J =

0

1

= −JT, eJx =

cos( x)

sin( x)

M

1

0

sin( x)

cos( x)

= ( e−Jx) T.

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From Dynamic Modeling to Experimentation of

Induction Motor Powered by Doubly-Fed Induction Generator by Passivity-Based Control

117

[ LrM( LsG + LsM) − L 2 ]

mM I 2

−LmGLrMe−JnGθG

−LmGLmMe−J( nGθG−nMθM)

L− 1( θ) = 1 ⎣

Δ

LmGLrMeJnGθG

LrGLrM I 2

LrGLmMeJnMθM

(9)

−LmGLmMeJ( nGθG−nMθM)

LrG LmMe−JnMθM

[ LrG( LsG + LsM) − L 2 ]

mG I 2

L

1 ⎢

11

L 12

L 13

T

Δ ⎣ L 12

L 22

L 23 ⎦

(10)

T

T

L 11

L 23

L 33

where

Δ = LrG[ LrM( LsG + LsM) − L 2 ]

<

mM

LrML 2 mG

0

(11)

We recall that, due to physical considerations, R > 0, L( θ) = LT( θ) > 0 and L− 1( θ) =

L− 1 T( θ) > 0.

A state–space model of the (6–th order) electrical subsystem is finally obtained replacing (7)

in (6) as

Σ e : ˙ λ + RL( θ) 1 λ = BvrG

(12)

The mechanical dynamics are obtained from Newton’s second law and are given by

Σ

¨

˙

m : Jmθ + Bmθ = τ − τL

(13)

where Jm = diag{JG, JM} > 0 is the mechanical inertia matrix, Bm = diag{BG, BM} ≥ 0

contains the damping coefficients, τL = [ τLG, τLM] T are the external torques, that we will assume constant in the sequel. The generated torques are calculated as usual from

τ =

τG

τ

= 1

λT[ L( θ)] 1 λ .

(14)

M

2 ∂θ

From (7), we obtain the alternative expression

τ = 12 ∂θ iTL( θ) i .

The following equivalent representations of the torques, that are obtained from direct

calculations using (7), (8) and (14), will be used in the sequel

−LmGiT Je−JnGθGi

rG

sG

τ = ⎣

(15)

−LmMiT JeJnMθMi

sG

r M

− nG

˙ λT

J( λ

R

sG M

sG M − LmMeJnMθM ir M)

= ⎢

sG + RsM

(16)

nM ˙ λT Jλ

R

r M

rM

r M

2.1 Modeling of the DFIG-IM in the stator frame of the two machines

It has been shown in (4) and (3) that the DFIG-IM is Blondel–Park transformable using the

following rotating matrix:

e( )

0

0

Rot( σ, θG, θM) = ⎣

0

e( J( σ+ nGθG))

0

(17)

0

0

e( J( σ+ nGθG−nMθM))

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118

Electric Machines and Drives

where σ is an arbitrary angle.

The model of the DFIG-IM in the stator frame of the two machines is given by (see (4) and (3)

for in depth details):

⎧ ⎡

⎤ ⎡

⎤ ⎡

⎤ ⎡

˙

˙

λ

aI

θ

λ

I

rG

2 − nG G J

bI 2

0

rG

2

⎪ ⎢

⎨ ⎣ ˙ λ

˙

˙

⎦ ⎣

⎦ ⎣

sMG ⎦+

aI 2 − nGθG J

eI 2

−cI 2 + nMθMJ

λsMG

= I 2

vrG

˙

˙ θ

λ

0

λ

0

−dI 2

cI 2 − nM M J

(18)

r M

r M

JG ˙ ωG

+ BG

0

ωG

+

f λT

sMG

rG

=

−τLG

JM ˙ ωM

0

BM

ωM

− f λT Jλ

−τ

sMG

r M

LM

or

˙

λ

i

I

rG

rG

2

⎣ ˙ λ

˙

˙

sMG ⎦ + ( R + LG nGθG + LMnMθM)

isM

=

I 2

vrG

˙

0

λ

i

(19)

r M

r M

JG ˙ ωG

+ BG

0

ωG

+

f λT

sMG

rG

=

−τLG

JM ˙ ωM

0

BM

ωM

− f λT Jλ

−τ

sMG

r M

LM

λsMG corresponds to the total leakage flux of the two machines referred to the stators of the

machines.

LsMG represent the total leakage inductance.

with

R

rG I 2

0

0

R =

RrG I 2

( RsG + RsM) I 2 −RrMI 2

0

0

RrM I 2

−J J 0

0

0

0

LG = LMG

−J J 0

et LM = LMM

0

J

J

0

0

0

0

−J −J

with the positive parameters: a = RrGL− 1 , b = R

, c = R

, d = R

,

MG

rG L− 1

sMG

r M L− 1

MM

r M L− 1

sMG

e = ( RrG + RsG + RsM + RrM) L− 1 , f = L− 1 , and the following transformations: sMG

sMG

λrG = LmG eJnGθGλ

eJnGθG v

L

rG,

vrG = LmG

rG

rG

LrG

λrM = LmM eJnMθMλ

eJnGθG i

L

r M,

irG = LrG

rG

r M

LmG

3. Properties of the model

In this section, we derive some passivity and geometric properties of the model that will be

instrumental to carry out our controller design.

3.1 Passivity

An explicit power port representation of the DFIG interconnected to the IM is presented in

Fig.4

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From Dynamic Modeling to Experimentation of

Induction Motor Powered by Doubly-Fed Induction Generator by Passivity-Based Control

119

vrG

irG

τ LG

− ∑

-w

G

G

i

isG

sM

vsM = vsG

M

-τ

w

LM

M

Fig. 4. Explicit power port representation of the DFIG with IM.

Claim 1. The interconnection of the DFIG with the IM presented in the explicit power port

vrG

irG

representation in Fig.4 is a passive system 2 with the passive map

τ

LG

−ωG .

−τLM

ωM

vrG

irG

Proof. Consider the Fig. 4, Σ =

Σ G

Σ

is passive

τLG

−ωG

is passive ?

M

−τLM

ωM

For this purpose, we have to prove that

vT i

rG rG − τLGωG − τLMωM

0. We know that

each machine separately is passive (see [1])

Σ G is a passive system

vTrGirG − τLGωG + iTsMvsM ≥ 0

(20)

Σ M is a passive system

vTsGisG − τLMωM ≥ 0

(21)

where equation (1) has been used in (20) and (21). Let’s consider

d

vTrGirG − τLGωG + iTsMvsM ≥ 0

(22)

Using the energy conserving principle

iTsMvsM = − iT v

sG sG yields

d =

vTrGirG − τLGωG −

iTsGvsG

(23)

From (21) we have

− iTsGvsG ≤ − τLMωM

(24)

Finally (23) and (24) yields

vTrGirG − τLGωG − τLMωM ≥ d ≥ 0

(25)

Hence, the passivity of the DFIG interconnected to the IM is proven

2 Passive systems are defined here with no causality relation assumed among the port variables (13). This, more natural, definition is more suitable for applications where power flow (and not signal behaviour)

is the primary concern.

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120

Electric Machines and Drives

4. Zero dynamics

For the IM speed control, we are interested in the internal behaviour of the system when the

motor torque τM is constant. In addition, for practical considerations, we are interested in the

control of the IM flux norm |λrM|, where | · | is the Euclidean norm.

For the study of the zero dynamics regarding these two outputs, we consider the DFIG-IM

model3 given by (18).

The control vrG is determined to obtain the desired equilibrium points of the DFIG-IM:

¨ θG = ¨ θM = 0,

˙ θG = ˙ θd =

=

G

Constant,

˙ θM = ˙ θdM Constant,

τLG = τLG 0 = Constant,

τLM = τLM 0 = Constant

λT λ

=

r M r M = β 2 d

Constant > 0

(26)

The IM mechanical dynamics show that the desired equilibrium points are obtained if:

τM = τd = τ

M

LM 0 = Constant

Hence:

f λT

=

sMG Jλr M = τM = τd

M

Constant

(27)

The equation (27) can also be expressed by replacing λsMG by its value given by the third line

of the electrical subsystem (18):

f

˙

T

λ

˙ θd

d

r M + cλr M − nM M JλrM

r M = τM

Hence

f

˙ T

T

λ

˙ θd λT λ

˙ λ

˙ θd β 2 = τ

= cte

(28)

d

r M Jλr M − nM M r M r M

= fd rMJλrM − nM M

M = τd

M

The relative degrees of the outputs y 1 = β 2 = λT λ

r M r M and y 2 = τM = f λT

sMG

r M, regarding

the input control vrG, are 2 and 1, respectively.

The zero dynamics of the mechanical subsystem (18) is stable, since the mechanical parameters

are positive.

Following on we will analyze