Transient Analysis of Three-Phase Wound Rotor (Slip-Ring) Induction Motor under Operating Condition

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Transient Analysis of Three-Phase Wound Rotor (Slip-Ring) Induction Motor under Operating Condition

Obute K.C.

¹

, Enemuoh F.O.

¹

1 – Department of Electrical Engineering Nnamdi Azikiwe University Awka, Anambra State Nigeria Corresponding Author - Obute Kingsley Chibueze

Department of Electrical Engineering

Nnamdi Azikiwe University Awka, Anambra State Nigeria

Abstract:

This work investigates the transient behaviours of a three phase wound rotor type induction motor, running on load. When starting a motor under load condition becomes paramount, obviously a wound rotor (slip ring) induction becomes the best choice of A.C. motor. This is because maximum torque at starting can be achieved by adding external resistance to the rotor circuit through slip-rings. Normally a face-plate type starter is used, and as the resistance is gradually reduced, the motor characteristics at each stage changes from the other (John Bird 2010). Hence, the three phase wound rotor (slip ring) induction motor competes favourably with the squirrel cage induction motor counterpart as the most widely used three-phase induction motor for industrial applications. The world-wide popularity and availability of this motor type has attracted a deep – look and in-depth research including its transient behavior under plugging (operating) condition. This paper unveils, through mathematical modeling, followed by dynamic simulation, the transient performance of this peculiar machine, analyzed based on Park’s transformation technique. That is with direct-quadrature-zero (d-q-o) axis based modeling in stationary reference frame.

Key Words: Transient Condition, d-q-0 reference frame, a-b-c reference frame, arbitrary reference frame, stationary reference frame, and wound rotor induction motor etc.

Introduction

When a circuit possess energy-storing element such as inductance and capacitance, the energy state of the circuit can be disturbed by changing the position of the switch connecting the elements to the source. The circuit then settles down to another steady state after a certain time. (Smarajit Ghosh 2011). Interestingly, the behavior of all electrical machines is only of interest when it is in a steady-state. There exist some cases, however, when the temporary response of a machine to a change in conditions is required. For instance, if a power supply in a machine is switched on, there may be a surge, possibly with oscillation before a steady flow of current is established. This surge (burst of energy) is short-lived and instantnous, and is caused by a sudden change of state of the machine.

Circuits exhibit transient when they contain active components, that can store energy, such as transformers and induction motors. Transient phenomenon is an aperiodic function. The duration for which it lasts is very insignificant as compared with the operating time of the system.

Yet they are very important because, depending upon the severity of these transients, the system may result into shut down of a plant etc. (Wadjwa 2009).

Every electrical system can be considered as made up of linear impedance of elements resistance, inductance and capacitance. For the slip-ring induction motor under discussion, its circuit is normally energized and carries mechanical load, until a fault suddenly occurs. The fault, then corresponds to the closing of a switch (or switches, depending upon the type of fault). The closing of this/these switch(es) charge(s) the circuit, so that a new distribution of currents and voltages is brought about. This redistribution is accompanied in general by a transient period during which the resultant currents and voltages may momentarily be relatively high. It pays to realize that this redistribution of currents and voltages cannot take place instantaneously for the following reason:

a) The electromagnetic energy stored by an inductance (L)(main component of every electrical machine) is 0.5 LI², where I is the instantaneous value of current. Assuming inductance to be constant, the change in magnetic energy requires change in current which an inductor is opposed by an e.m.f of magnitude L ^𝑑𝐼

𝑑𝑡. In order to change the current instatenousely dt = 0 and therefore L ^𝑑𝐼

𝑂 =  (infinity). This suggests that to bring about instantaneous change in current the e.m.f. in the inductor should become infinity which is practically impossible. Hence the charge of energy in an inductor is gradual, and therefore the redistribution of energy following a circuit change takes a finite time.

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b) Another component, the resistance R (also obtainable at the stator/rotor windings) consumes energy. At any time, the principle of energy conservation in an electrical circuit applies. That is the rate of generation of energy is equal to the rate of storage of energy, plus the rate of energy consumption.

In its summarized form, it becomes clear that in any rotating machine;

i) the current cannot change instantaneously through an inductor (winding)

ii) the law of conservation of energy must hold good, are fundamental to the phenomenon of transient in electric machine.

Also from the above explanations, it can be said that in order to have transients in an electrical machine with special reference to three phase wound rotor (slip ring) induction motor, the following requirements should be met.

i) The inductor (windings) should be present

b) A sudden change in the form of a fault or any switching operation should take place.

2- THE TRANSIENT MATHEMATICAL MODELLING OF THE MOTOR USING PARK’S TRANSFORMATION TECHNIQUE

Unlike the steady state mathematical modeling of induction motors, which is not new and has received a considerable attention from researchers, the transient mathematical modeling of induction motors continues to received much attentions. This is as a result of the vital effect the transient behavior of the induction motors has on the overall performance of the system to which it forms a component part. Transient modeling of induction motors prove to be more difficult both in the definition of suitable forms of equations and in the application of appropriate numerical methods needed for the solution of same (Okoro 2003).

The three phase wound rotor (slip-ring) induction motor is modeled using Park’s transformation technique (Direct-quadrature-zero (d-q-o) axis transformation theory).

In electrical engineering, and other associated disciplines, the analyses of three-phase circuits are mostly simplified using d-q-o transformation.

It is a more reliable and accurate technique for investigating the problems of voltage drips, oscillatory torques and harmonics in electrical system, which arise due to high currents drawn by all induction motors when they are started and during the other transient operations.

The three reference frames includes;

i)

Stationary reference frame

ii)

Rotor reference frame

iii)

Synchronous reference frame

These reference frames are used to convert the a-b-c reference frame (input voltage) to the d-q-o reference frame, and/output current (d-q-o reference frame) to the a-b-c reference frame.

Any one of the afore-mentioned three reference frame can be used to analyze the transient dynamic behavior of the three phase motor under consideration, based on the following conditions;

i)

If the stator voltage is unbalanced and the rotor voltages are balanced, the stationary reference frame is useful

ii)

If the rotor voltages are unbalanced and stator voltages are balanced, the rotor reference frame is used

iii)

If the stator and rotor voltages are balanced and continuous, then synchronous reference frame is used 2.1 – TRANSFORMATION OF a-b-c TO d-q-o REFERENCE FRAME

The transformations like d-q-o developed by park can facilitate the computation of the transient solution of the three-phase wound rotor induction motor model by transforming inductances to differential equations with constant inductances.

2.2- Equations of Induction Motor in Arbitrary d-q-o Reference Frame

The relationship between the abc quantities and qdo quantities of a reference frame rotating at an angular speed, w, is shown in figure 1 below;

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The transformation equation from a-b-c to this d-q-o reference frame is given by:

by:

  ^... ¹

 







 









 







 







c b a qdo o

d q

f f f T f f f

Where the variable f can be the phase voltage, currents or flux linkages of the machine. The first quadrant of the reference frame rotating at an arbitrary speed w (t), in figure 1 is shown as demarcated. The transformation angle,  (t), between the q-axis of the reference frame rotating at w and the a-axis of stationary stator winding may be expressed as:

  t

^t_o

w   t dt  ( 0  elect . rad ... 2

   

Likewise, the rotor angle,



r

  ^t

, between the axes of the stator and the rotor a-phases for a rotor rotating with speed

^w

r

  ^t

may be expressed as

  ^t ^w

r

  ^t ^dt

r

⁽ ⁰  ^elect ^. ^rad ^... ³

t o

r



   

The angles  (0) and _r(0) are the initial values of those angles at the beginning of time t.

The qdo transformation matrix,

 T

qdo

  ^ 

is

 

   

^...⁴

2 1 2

1 2

1

2 3 3 sin

sin 2 sin

2 3 3 cos

cos 2 cos

3 2





















     

 



q d o  T

And its inverse is

     



² ₃



^sin



² ₃



¹ ^...⁵

cos

3 1 sin 2 2 3

cos

1 sin

cos

1





















 

 



 Tqdo

2.3 – d-q-o Voltage Equations

In matrix notation, the stator windings a b c voltage can be expressed as abc

S abc S s

abc abc

S

p r i

V   

…6 Applying the transformation to equation (6), we obtain

Cs Cr br

bs

d-axis a-axis W

_r

Ѳ

_r

q-axis

w

Figure 1 - Relationship between abc and arbitrary d-q-o

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     

 

^p



^T

   

^T

  

^r



^T

  

ⁱ ^b

T V

a i

r T p T V T

qdo s qdo abc s qdo S qdo qd

qdo qdo S

abc S abc S qdo S abc qdo

abc s qdo

7 ...

1 1

0



 



















 

    

   



 

    



² ₃



^cos



² ₃



⁰

^{ } ^

^{ }

^ ^{ }

^...⁸

sin

3 0 cos 2 2 3 sin

0 cos

sin

3 1 sin 2 2 3

cos

3 1 sin 2 2 3

cos

1 sin

cos

1 1

1

qdo S qdo qdo s qdo

qdo

p T w

T p

T







 



 

























































Substituting equation (8) back into equation (7b) gives:

 

   

    ^{ }

^ ^{ }^

^{ }

⁰⁰ ¹⁰⁰¹ ^.⁹

0 0 1

3 0 cos 2 2 3 sin

0 cos

sin

1 qdo

S s

qdo s qdo qdo S qdo

qdo

S T w T p r i

V







































   ^ 

 



Rearranging (9) will give:

10 ...

. 0

0 0

0 0 1

0 1 0

qdo S qdo S qdo S os

ds qs qdo

S w p r i

V  







































Where,

dt w d 



^,

and

11  ...



 

















os s os os

ds s ds qs

ds

qs s qs ds qs

i r p V

i r p w

V

i r p w V



2.4 – q-d-o Transformation Of The Rotor

12

abc

...

r abc

r

r i p

v   

The rotor quantities must be transformed on the same d-q-o frame. The transformation angle to the rotor phase quantities is ( - _r). Therefore T ( - _r) is the transformation angle.

 

  ...13

2 1 2

1 2

1

3 sin 2

sin

3 cos 2 3 cos 2 cos

3 2



















 



  



 



  





 



  



 



  





        

 

 



 r r r

r r

r

T r

 

   

14 ...

3 1 sin 2

3 cos 2

3 1 sin 2

3 cos 2

1 sin

cos

1



















 



  



 



  



 



  



 



  





 ^

 

 



 

 



r r

T r

Applying the transformation matrix to equation 14 in the same manner as in that of the stator, the equation will be obtained as: (V. Sarac et al 2013)



w w



p r i a

V_r^qdo _r ^qdo_r ^qdo_r _r^qdo _r^qdo ...15 0

0 0

0 0 1

0 1 0





















   ^.

dt p d

r r

_s^qdo _S



 







 









1 0 0

0 1 0

0

1

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From equation 15a, the park equations are:

 

  b

i r p V

i r p w w V

or or or or

dr dr dr qr r dr

qr qr qr dr r qr

15  ....



 





















2.5 d-q-o Flux Linkage Relation

The stator qdo flux linkages are obtained by applying T_qdo(Ѳ) to the stator abc flux linkages of equation (16), that is:

16

abc

...

r abc Sr abc S abc SS abc

S

 L i  L i



Using appropriate transformation, we obtain:

         



^T

 

^T



^L



^L^T ^T

  

ⁱⁱ



^T^T

 

^L



^L^T



^T

^

ⁱ

^ 

ⁱ ^a

T

qdo r r qds abc Sr qd qdo S qdo abc SS qdo qdo S

qdo r r abc SS qdo s abc SS qdo abc S

17 ...

1 0

1

1 1























Equation (17) gives rise to:

b i L L

i L L L L L

qdo r Sr Sr

qdo S

LS SS SL SS LS

qdo

s ...17

0 0 0

2 0 0 3

0 2 0

3

0 0

2 0 0 3

0 2 0

3





































  .

Similarly, the qdo rotor flux linkage can be derived as,

abc r abc rr abc S abc rS abc

r

 L i  L i



.

Applying the transformation:

         r ^qdo

abc rr r qdo S sr r abc r

r T L T i T L T i

T     ^¹    ^¹

 

 ^T  ^L  ^T ^{ }  ⁱ  ^T

qdo

^

r

^  ⁱ

r^qdo

^a

qdo

s qdo abc rS r qdo qdo

r

 

^¹

 

^¹

... 18

      

This gives:

b i

L L L

L L

i L

L

qdo r

Lr rr Lr

rr Lr

qdo s sr sr

qdo

r ...18

0 0

2 0 0 3

0 2 0

3

0 0 0

2 0 0 3

0 2 0

3



































  ^.

The stator and rotor flux linkage relationships in equations (17b and 18b) an be expressed compactly as

19 ...

0 0 0 0 0

0 0

0

0 0

' ' '

' '

'

' ' '









































or dr qr os ds qs

Lr m Lr m

Lr m

Ls

m m

LS

m m

LS

or dr qr oS dS qS

i i i i i i

L L L L

L L

L

L L

L

L L

L



Where,

22 ...

21 ...

,

20 ...

,

2 '

' '

Lr Lr

dr dr

qr qr

dr dr

qr qr

Nr L L Ns

Ns i i Nr Ns i

i Nr

Nr Ns Nr

Ns

 

 



 



   



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And L_m, the magnetizing inductance on the stator side, is:

... 23

2 3 2

3 2

3

rr r s sr

r

s

L

N L N

N Lss N

Lm   

From equation (11) and equation (15b), For stator,

For rotor

Also, flux linkage of equation 19 can be written as:

 

b

i L

i i L i L

i L i

i L

i L i L L

i i L i L

i L i L L

or lr or

dr ds m dr lr dr

qr qs m qr lr qr

dr ds m ds ls

dr m ds m ls ds

qr qs m qs ls

qr m qs m ls qs

19 ...

' '

'

 







 

















 

 



 



















Substituting the values of flux in the rotor and stator voltage equations (11) and (15b) gives:

 

    

 

 

    

 

^'

' ' '

' '

'

(

or lr or r or

os s os s os

dr ds m dr lr qr r dr

r dr

dr m ds m ls qs ds s ds

qr qs m qr lr dr r qr

r qr

qr qs m qs ls ds qs s qs

i L p i r V

i i L i L p w

w i r V

i L i L L p w i r V

i i L i L p w

w i r V

l i L i L p w i r V



































In terms of

x

^,

s

and



^,

s

stator and rotor voltage equations (.24) can be written as:

os os s os

ds qs qs s ds

qs s qs

s qs

p i r V

p w i r V

p d w i r V













 

' '

' ' '

' '

' ' '

or or

r or

dr qr

r dr

qr dr

r qr

p i r V

p w

w i

r V

p w

w i

r V





















…24

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25 ...

' 0 ' ' 0 '

0

' ' ' '

'

' ' ' '

'

0 0 0

 



 









 



 



  



 



 



  

















r r r b r

dr r qr b

b dr

qr r dr b

r qr

b qr

s s s b s

ds s qs b ds b ds

qs s ds b ds b qs

i w r

v p

i w r

w w w

v p

i w r

w w w

v p

i w r

v p

i w r

w w

v p

i w r

w w

v p



Where

26 ...

' ' ' 0

' ' '

















































or dr qr s ds qs

Lr m Lr m

m Lr m

ls

m m

ls

m m

ls

or dr qr os ds qs

i i i i i i

x o o

o o o

o x x o

o x o

o o x x o o x

o o o

x o o

o x o

o x x o

o o x

o o x x



Where

  w

_b



2.6–q-d-0 Torque Equation in Arbitrary Reference Frame

The sum of the instantaneous input power to all six windings of the stator and rotor is given by

27

'

...

' ' '

cr c ar ar cs cs bs bs as n as

i

V i V i V i V i V i

P     

In terms of q-d-0 quantities, the instantaneous input power is

 ² ²  ^... ²⁸

2

3

_' _' _' _' _'

or or dr dr qr qr os os ds ds qs

qs

i V i V i V i V i V i

V

Pin      

Substituting the values of

V

_qs

, V

_ds

, V

_qr^'

, V

_dr^' from equation (10) and (15a) into the right side of equation (28), we obtain;



   

        ... 29

2 3

' ' '

' '

'

dr qr r dr

r dr r qr

r qr r

dr ds qs ds s ds qs ds qs s qs in

p w

w i r w

w i r i r

i p w i r i p w i r i P



















From equation (29), we obtain 3 kinds of terms: i²r, i

p 

, and

 

i terms. The i²r terms are copper losses;

the i

p 

terms represent the rate of exchange of magnetic field energy between windings. The w



^{i terms}

represent the rate of energy converted to mechanical work. That is

     



^' ^' ^'



. 1 2 3

dr qr qr dr r ds

qs qs ds m

em

w i i w w i i

T  w        

But

p w

_m

 2 w

^b

     

 

...30

2 2

3 '

dr qr qr dr r ds qs qs

dsi i w w i i

wb w

Tem P      



Substituting values of



_qs

, 

_ds

, 

^'_qr

, 

^'_dr from equation (19b) into equation 30, we have

..26

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IJRERD

www.ijrerd.com || Volume 02 – Issue 07 || July 2017 || PP. 19-32

   

 

^'



'

]

qr ds ds qs m

qs m ds qs m ds qs ls

qs dr qs ds m qs ds ls ds qs qs ds

i i i i L

i L i i L i L L

i i i i L i i L i i

















And



qs dr ds qs



m qr dr dr

qr

i 

^'

i  L i i  i i

'



This implies that

 





^' ^'

' '

dr dr dr qs m qr dr dr qr

ds qr qr dr ds

qs qs ds

i i i i L i i

i i













Therefore,



^'dr qr^' ^'qr dr^'



m



dr qr qr ds

 ^... ³¹

ds qs qs

ds

i   i    i   i  L i i  i i



Substituting equation (31) into equation (30) gives;

 

32 ...

2 . 2 3

' '

' ' '

 





 



















m N i i i i P L

m N i P i

Tem

ds qr qs dr m

ds qs qs ds

qr qr dr qr



3–Induction Machine Equation in Stationary Reference Frame

By setting w=0 in equation (21) gives:

qs b qs s s qs

qs b b qs s s qs

os os s s os

ds ds s s ds

qs qs s s qs

w i p r V

w pw i r V

p i r V



































33 ...

Where

  w

_b



Therefore, for stator

34 ...

'

 





 















os s os b os

s ds s s b s ds

qs s qs b s qs

i w r

V p

i w r

V p

i w r

V p



For rotor

Set w=0 in equation (15b):

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IJRERD

35 ...

' '

' ' ' '

'

' ' ' '

 





 





















or r or b or

s dr r s qr b s r qr b s dr

s dr r s dr b s r qr b s qr

i w r

V p

i w r

w w

V p

i w r

w w

V p



36 ...

' ' ' 0

' '

'

' ' '

















































or s dr s qr s

s s ds s qs

Lr m Lr m

m Lr m

ls

m m

ls

m m

ls

s or s dr s qr s os s ds s qs

i i i i i i

x o o

o o o

o x x o o

x o

o o

x x o o x

o o

o x o o

o x

o o x x o

o o

x o o x x



Torque equations,

 

  ^. ^... ³⁷

2 2 3

' '

' ' '

m N i i i i w x P

i w i

P w i Tem P

s ds s qr qs s dr m b

s ds qs s qs s ds b

s qr s dr s dr b















3.1 – Equations for Simulation of Induction Motor in Stationary Reference Frame

Transient Analysis of Three-Phase Wound Rotor (Slip-Ring) Induction Motor under Operating Condition