Hanhok Book

8/3/2019 Hanhok Book

1/83

- -

Contents:

Preface 1Linear Transformation in machines 2Transformation of voltage and impedance for invariant power 3Linear Transformation, An example 4

Two winding Transformer 5Three Windings Transformer 6Example -Scott connection of transformers 7The voltage equation of the simple commutator machine 8D.C. Series motor 9D.C. Shunt motor 10D.C. Compound motor 11Example 12Metadyne 13Amplidyne 14Block diagram of D.C. machines 16Industrial Loads 16The equivalent physical-parameters of three phase and two phase machine 17Transformation from three phase axes to two phase axes 18Transformation from two phase rotating axes to two phase stationary axes 19Transformation from three phase rotating axes to two phase stationary axes 20The equivalent impedance of three phase and two phase machine 21Round-Pole Synchronous Machine 22Two-phase winding in Salient-Pole Machine (Reluctance machine) 24Salient-Pole Synchronous Machine 25Synchronous machine torque 27Flux linkages in synchronous machine 27Synchronous machine dynamics (Electromechanical transients) 28Three-phase short circuit (Armature and field resistances neglected) 30Damper windings 31Synchronous Machine with damper winding 33Flux paths for various reactances of a salient pole synchronous machine 36Reactances and Time-constants of synchronous machine 37Terminal short circuit current 39Transformation from three-phase axes to symmetrical component axes 40Transformation from two-phase axes to symmetrical component axes 41Inductance of synchronous machine to zero sequence current 42Inductance to negative sequence 42Study of Short Circuits in Synchronous Machine Using Symmetrical Components 47


2/83

Synchronous machine model with two damper windings in q-axis 50Permanent magnet synchronous motor 51Induction Machine equations transferred to dq stationary axis 52Induction Machine equations transferred to -axis rotating at synchronous speed 53Transformation from stationary axes to forward and backward axes 54Induction Machine Torque 56Equivalent Resistance and Inductance of Squirrel Cage rotor 57Air gap Inductance 60Slot Leakage Inductance 60Measuring synchronous machine parameters 62Synchronous machine parameter estimation using short circuit current test 65Stand Still Frequency Response (SSFR) method 68Induction machine (Parameter estimation tests) 69DC motor speed control 73Induction machine speed control 77 References 82

Preface:This lecture is intended for students who have knowledge of the physical nature of electricalmachines and is concerned with systematic analysis of the performance of electrical machinery in:

- Steady state condition- Transient condition- Unsymmetrical operation- Electrical machine analysis using computer

Some restrictions that are imposed in this lecture:

1- In all cases the rotor is the non-salient member. The equations for the revolving field machines areidentical with those of the revolving armature machines, except the direction of rotation regarded as

positive.

2- All machines are analyzed as motors. Negative terminal power and negative mechanical output power thus indicate generation.

3- Only two-poles machines will be analyzed. To apply the results to multipolar machines it will benecessary to reduce the speed and increase the torque according to the number of pole pairs.

4- The positive direction of rotation of the rotor is taken as counter-clockwise, since this is thedirection of measurement of positive angles.

5- Changes of saturation is ignored.

6- Space harmonics are ignored, so that sinusoidal variation of inductances are considered.


3/83

The voltage components to be induced for each winding are:

1-The terminal voltage. [U]

2- The resistance drop. [iR]

3- The voltage drop due to the changing current in the winding itself. [Ldi/dt]

4- The voltage due to the changing current in the other windings which produce a flux linking thewinding under consideration. [Mdi'/dt]

5- The rotating voltage (in the armature circuit only) generated by the rotation of the winding in thefluxes set up by all currents in coils having an axis which is not parallel to the axis of the armaturecircuit under consideration, irrespective of whether these currents flow in stationary windings or inthe armature itself. [Mi'd /dt]

The sign of mutual inductance: A mutual inductance M af is positive if the flux set up by positivecurrent in one circuit links the other circuit in the same direction as the flux set up by positive currentin that circuit itself.The positive current I f is assumed to magnetize from left to right. The generated e.m.f. in d winding is

positive on the left hand side of the equation, but has negative sign as voltage drops on the right handside.

f af f af aaaaa

f af aaaaa

f af aaaaa

aaaaa

i M pi M pi Li Ru

pi M pi Li Ru

pi M pi Li Ru

pi Li Ru

'

)sin()cos(

)cos(

+

+

+


4/83

Linear Transformation in machines

In electrical machine analysis it is frequently necessary to change the variables to another related setfor the purpose of:

- obtaining the equation of the machine differently connected- simplifying the process of obtaining the solution of a problem more easily and quickly

A transformation may be selected on purely algebraic grounds, e.g. to give a matrix of constant in place of a matrix of functions, or to give a symmetric or a diagonal matrix in place of an asymmetricone. Most of the transformations used have physical parallels. They are derived by physicalconsiderations long before matrix algebra was applied to electrical machine analysis.It is desirable that the two transformations should be so related that the power is the same in terms of

both sets of variables. If this invariance of power under transformation is not maintained, it will benecessary to determine powers and torques only from currents expressed in the original system of reference axes.

Transformation of voltage and impedance for invariant power

The total power input to a number of circuits is given by the sum of the products of terminal voltageand current of each input. If i and u are the current and voltage in old reference, and i' and u' are thetransformed current and voltage in new reference, the power (active and reactive) has to be invariantunder the transformation. When the transformation of both current and voltage are specified, thetransformation of impedance is automatically determined ad below:

[ ]

zC C z uC uvoltageand current complex of termsin

zC C z

i z i zC C Ci z C i z C zi C uC u

zi u

uC uuC i ui ui ui

power iant in for tiontransformabyC i i

Ci i if

ui ui i ui

i u

i

i uuui ui ui u

i

i i

u

uu

t t

t

t t t t t

t t t t t t

t t t

t t t t t t

t

**

'''

'

t

3

2

1

321332211

3

2

1

3

2

1

','

'

''')()'()()('

'''

var

'

][][u

i

i

,

u

=

==

==

=

==

==

==


5/83

Linear TransformationAn example

=

=+

++

=

==

=

==

000

0321

32

21

322

1'

2

2

132

213

3212

2

'2

2

*

02

2

2

2

2

132

213321

00

00

00

00

00

00

)(300

0)(30

00)(3

1

1

111

111

1

1

111

1

1

1

1

111

23

21,

23

21,

1

1

111

,

i

i

i

X

X

X

u

u

u

X

X

X

X X X

X hhX X

hX X h X

Z

hh

hh

X X X

X X X

X X X

hh

hh

Z hh

hh

C

i

i

i

hh

hh

i

i

i

j h j h

hh

hhC if

i

i

i

X X X

X X X

X X X

u

u

u

n

p

n

p

n

p

n

p

t

n

p

c

b

a

c

ba

c

ba


6/83

Two winding Transformer

[ ]

=

++=++=

++

=

=

==

++=

++

=

2

1

2

1

11221212

2112

22211211

222

211211

2221

1211

1

2

2

1

2

1

22211211

2

1

2221

1211

2

1

''

u'

u

''''

'R'M'M'R

)(R (1/k)(1/k)M(1/k)MR

/10

01

R M

MR

/10

01'

/10

01C

'/10

01

R MMR Z

R M

MR

u

u

,

i

i Z

l Ll L M M

M M

p L p p p L

p L p p p L

k p L p

p p L

k ZC C Z

k C

N N

K i

i

k i

i

p L p p p L

i

i

p L p

p p L

t

t

Equivalent circuit of two winding transformer

This circuit is therefore a satisfactory representation of a transformer, except for saturation and coreloss in the case of a transformer with a ferro-magnetic core.


7/83


8/83

ExampleScott connection of transformers

t uut uui

i

ii

i

i

i

i

i

i

i

i

MpMpMp

MpMp

MpMpMp

MpMpMp

MpMp

u

u

u

u

u

L Rwith

i

i

i

i

i

pM L R pM pM

pM L R pM

pM pM L RMp

pM Mp pM L R

pM pM L R

u

u

u

u

u

M M M M M

M M L LM L LM L N N

R N N M M

R N N M

R N

M L L R

N L L

R N

L R

N M

N N N N N N N

wt uuu

wt uuu

wt uuu

i

ii

i

i

p L L R pM pM

p L L R pM pM p L L R pM

pM pM p L L R

pM p L L R

u

uu

u

u

mmC

B

A

C

B

A

C

B

A

l

C

B

A

l

l

lC C

lB B

lA A

C

B

A

C B A

BC CC BB AA

C B A

BC CC BB AA

C B A

m AC

mC B

m B A

C

B

A

l C B

l A

C CC lC C BC

B BC BBlB B

A AAlA A

C

B

A

sin,cos

1001

13/1

13/1

03/2

05.05.00

00086.0

5.0025.025.00

5.0025.025.00

086.00075.0

0

)(0)(5.0)(5.00

0)(00)(86.0

)(5.00)25.0(25.00

)(5.0025.0)25.0(0

0)(86.000)75.0(21

,23

41

,,43

1

21,

23

41

,,43

,

5.0,866.0,

)120cos(

)120cos(

cos

)(00

0)(000)(0

0)(0

000)(

122

1

2

1

2

1

2

1

2

1

22

11

2

1

221

11222

1

2122

211

21

22

1122

21

21

1121

2

1

222222

11111

2

2

1

2

1

==

=

===

++

++++

++++

=

===

=======

===

=======

====+==

=++

++++

++++

=


9/83

The voltage equation of the simple commutator machine

The machine has two pole configuration shown in the figure. There is a pair of brushes "q" in thenormal position relative to the field winding "f" and also a pair of brushes "d" at right angleselectrically. The brushes are shown diagrammatically on the axis along which the m.m.f. due the their current acts (this is not the physical position of the brushes of an actual machine).

Basic commutator machineThe voltage equations can be written down as:

d fd f f f f f

f qf d qd qqqaq

qdq f df d d d ad

piM pi Li Ru

iM piM pi Li Ru

iM piM pi Li Ru

++=+=+++=

'

'

with M dq=L q, M qd=L d, M df =M qf =M The voltage equations can be written in matrix form as u=zi:

++

+=

f

q

d

f f

qqd

qd d

f

q

d

i

i

i

p L RMp

M p L R L

Mp L p L R

u

u

u

0

''

'

The minus sign attached to M qd'id and M qf 'if are justified as follows. The positive currents i d , i f areassumed to magnetize from left to right. Armature conductors in the vicinity of the right hand d brushtherefore, have generated e.m.f.s , aiding the positive terminal voltage u q in magnetizing vertivallyupwards. These e.m.f.s would, therefore be positive on the left hand side of the equation, but havenegative signs as voltage drops on the right hand side.The Impedance matrix Z consists of three parts:

- A resistance matrix R= f

q

d

R R

R

0000

00

- An inductance matrix L given by the coefficient of p as; L=

f

q

d

LM

L

M L

0

00

0

- A matrix formed by the coefficients of ' as; G= 000

0

00

M L

L

d

q

The voltage equation can thus be written as: u=R1+Lpi+G 'i


10/83

D.C. Series motor

The impedance matrix of series motor (with no d circuit) can be written as below (by omitting thecorresponding row and column of d circuit):

[ ]

[ ]

[ ]

2

')()(

')()('f uau11

f uau'

')()(

1

1

f R 0

'aR 11'

111

1

1

1

f R 0

'aR Z

/',f R 0

'aR

f uau

a Mi Mi t i Gi t i T

i M t i pi f La Lt i i f Ra Rt i ut i

i M pi f La Li f Ra Ri z u

u f uaut C u

M p f La L f Ra R

p f L

M pa L ZC t C Z

t C C i f i ai

p f L

M pa L

dt d f i ai

p f L

M pa L

===

+

=

===

+

=

=+

=

For the d.c. machine p may be replaced by zero for steady state conditions, leading to:U=RI-M 'I=RI+Eand T=-MI 2 The negative sign of the torque shows that the armature will rotate in the clockwise direction. So ' isnormally negative and consequently -M 'I=E is positive.For steady state conditions in the a.c. machine the p of the transient impedance matrix must bereplaced by j , leading to:U=RI-M 'I+jLI; with L=L f +L a


11/83

D.C. Shunt motor

[ ]

[ ]

[ ]

[ ]

]')[(0

00

0

R (

')(R 11'

R (

')(R '

1

1

R 0

'R 11

R (1

'

1111

11,

uu

R 0

'R

R (1

R 0

'R

/',R 0

'R

u

u

2a

a

2a

a

a

f 2

a

f

a

a

f 2

a

1

f

a

f

a

f

a

M R R R R

U I pconditions state steady For

iMii

iM iiGiiT

p L R p L R p L L R

M p L L Ruiii

i

ii

p L R p L R p L L R

M p L L RY

p L

M p L

p L R p L R p L L RY

C C uu

uY i

i

p L

M p L

p L R p L R p L L R Z Y

matriximpedancethe Inverting p L

M p L Z

dt d i

i

p L

M p L

f a f a

f a f

a f at

a f f a f a f

a f f f a

f

a

a f f a f a f

a f f

a

f

a f f a f a f

t f

a

f

a

a

f

a f f a f a f

f

a

f

a

f

a

++==

===

+++++++

==+==

+++++++

=

++

+++=

====

++

+++==

++

=

=++

=


12/83


13/83

Example

The brush axis of a separately excited dc motor is displaced by an angle o as shown in the Figure below. Determine the equation of its electromagnetic torque.

[ ]

cos)(2sin5.0

cos

sin

000

0

00

cossin

000

0

00

cos

sin

01

cos0

sin0

0

''

'

a f d qa

f

a

a

d

q

f aat

d

q

f

a

a

a

f

f

q

d

f

q

d

f f

qqd

qd d

f

q

d

iMi L LiT i

i

i

M L

L

iiiGiiT

M L

L

G

i

i

i

ii

i

i

i

i

i

i

p L RMp

M p L R L

Mp L p L R

u

u

u

=

==

=

==

++

+=


14/83

Metadyne

Elementary metadyne

The metadyne, also known as a cross field machine, may be regarded as an elaboration of the D.C.machine. The winding q represents the quadrature axis circuit of the armature. The winding drepresents the direct axis circuit of the armature. The reaction of the quadrature axis current i q on theflux produced by the field current results in torque T qf just as in the case of the D.C. motor. On theother hand, the interaction between the direct axis armature current i d and the flux produced by i q

produces torque T qd.

[ ]

[ ] qf q f f

q

d

af qdqd qd q

f

q

d

qf qd

dq

f qd t

qf qd

dq

f

q

d

f f fd

qf qaqd

df dqd a

f

q

d

M ii

i

i

i

M iM iM iT

i

i

i

M M

M

iiiGiiT

M M

M

G

i

i

i

p L R pM

M p L RM

pM M p L R

u

u

u

==

==

=

++

+=

000

0

00

000

0

00

0

''

'


15/83

Amplidyne

The simple metadyne can be adapted to operate as the amplidyne illustrated in figure below by theaddition of compensating windings S 1 and S 2 in the q and d axes of the stator. The amplidyne isessentially a D.C. generator. The number of field windings are more than one and are regarded ascontrol windings. Amplidyne has following advantages:

- High power gain up to 250,000- Low input power which makes it possible to feed the control windings from electronic switchs- Fast response, the equivalent electromagnetic time constant is of the order of o.1 to 0.2 s.

The bahavior of the amplidyne is defined by the following matrix equation.

Schematic diagram of Amplidyne

+++++

++++=

++

++

+

==

=

+++

++

+

=

p L R pM pM

pM pM pM p L R p L RM M

M M M pM p L R p L R

Z

p L R pM pM

pM p L R pM

p L R pM

pM pM M p L RM

M M pM M p L R

ZC C Z

i

i

i

i

i

i

i

i

i

i

i

i

i

p L R pM pM

pM p L R pM

p L R pM

pM pM M p L RM

M M pM M p L R

u

u

u

u

u

f f f a f s

f a f s sa s saa saaa

f aaa sa sa s saa

f f f s f a

f s s s sa

s s sa

f a sa saaaaa

f a sa saaaaa

t

f

a

a

f

s

s

a

a

f

s

s

a

a

f f f s f a

f s s s sa

s s sa

f a sa saaaaa

f a sa saaaaa

f

s

s

a

a

22

222222221221

12121111111

22

22222

1111

222122221

121112111

2

1

2

1

2

1

2

1

2

1

22

22222

1111

222122221

121112111

2

1

2

1

0

2''

'''2

'

100

010

001

010

001

00

00

000

''

'''

10000

01010

00101

'

100

010

001

010

001

00

00

000

''

'''


16/83


17/83

Block diagram of D.C. machines(An example; D.C. Shunt motor)

p J T T

MiiT

p

Ru

p L Ruii p L Ruu

p R

iM uiiM i p L R

i

i

p L

M p L

L

f a

f

f f

f f f f f f f f

a

a f aa f aaa

f

a

f

a

+==

+=

+=+==

++=+==

++

=

1

/11)(

1/1

)()(uu

R 0

R

u

u

a

f

a

f

a

U a=U f =250V, R f =12 , R a=0.012 , L a=0.35mH, L f =9H, M=0.18H, J=30kg-m2, T L=2375N-m

Industrial Loads

The most common type of industrial load is the Constant torque, such as lifting machines .[T L=A 0]

A load type that occurs most frequently in metal working applications, is the "Constant Horsepower"

load. On applications requiring constant horsepower, the torque requirement is greatest at the lowestspeed and diminishes at higher speeds. Typical applications include machine tools such as drills,lathes, and milling machines. [T L=A-1]

The "Centrifugal type" loads are found in centrifugal pumps, fans and blowers. The torquerequirement is a function of the speed squared and the power is a function of the speed cubed.[T L=A 2]

The "Linear torque" type loads are found in generators. The torque requirement is a function of thespeed and the power is a function of the speed squared. [T L=A1]


18/83

The equivalent physical-parameters of three phase and two phase machine

Three phase and two phase windings

The total power must be unchanged:

Sabc=S 3U abcIabc=2U I UI = 3/2 UabcIabc

U=23

Uabc

I=23

Iabc

N=23

N abc

L= 3/2 L abc (leakage inductance is not considered)

l=23 labc

A=23

Aabc

R = R abc

As a result of this transformation, it is unnecessary to consider three phase machines. The simpler case of the two phase machine can be analyzed, together with the zero-axis quantities if required, andthe results transformed to the three phase equivalents if necessary.


19/83

Transformation from three phase axes to two phase axes

Three phase and two phase phasor diagram

abct

00abc

1

c

b

a

0

c

b

a

c

b

a

00

c ba

iCi,Cii

22

23

21-

22

23

21

-

22

01

32

C,

22

22

22

23

23

0

21

-21

-1

32

22

Cfor

power.of invariancethe

withconsistentaswellassame,the beshouldvoltageandcurrentfor ationstransformthat thedesirableisIt

23

230

21

-21

-1

32

ii

i

23

230

21

-21

-1

32

ii

i

23

i

i

i

23

23

0

21

-21

-1

i

i

i

3/4sin3/2sin0

3/4cos3/2cos1

i

i

i

:values phasetwoand

phasethreetheequatingand,directionsaxisandthealongm.m.f.s phasethreetheResolving

sinI)2/cos(Ii

tcosIi

)3/4cos(Ii ),3/2cos(Ii t,cosIi

==

====

===

==

===

===

t t

t abc

c

b

a

c

b

a

C k C

k k k

C

k k k

N N with

N

N

N

k k k N

N

N

k k k N

N

N

t t

t t

The current i 0 can not be associated physically with a two phase system, where the transformation is a purely mathematical operation. If there is no current on the zero axis, matrix C may be used omittingthe last column.

In the absence of zero components, therefore, iiuu aa 32

,32 == so that iuiu aa // =

The relationship between u a and i a being the same as that between u and i implies that the numericalvalues of the parameters of the two systems are the same.


20/83

Transformation from two phase rotating axes to two phase stationary axes

This transformation has the effect of replacing a moving system by a stationary one. We can resolvethe m.m.f.s along the d and q axes. If zero sequence current exist, it is not transformed, thetransformation matrix has therefore an additional row and an additional column containing a solitaryunity as their common element:

0dq0

dq01-

0

1

00

q

d

Cii

iCi

100

0cossin

0sincos

C

1000cossin

0sincos

Cii

i

1000cossin

0sincos

ii

i

cosisini

sinicosi

sinI)2/cos(Ii

tcosIi

==

===

==

=+=

===

C C

i

i

t t

t

q

d


21/83

Transformation from three phase rotating axes to two phase stationary axes

The transformation matrix can be extracted directly by resolution of the m.m.f.s of the three phasewindings, or transformation from three phase rotating axes a,b,c to two phase rotating axes , andthen from two phase rotating axes , to two phase stationary axes d , q.

abcabc

t C

b

a

i1Cdq0i,dq0iCi

22

22

22

)3/4sin()3/2sin(sin/3)4-cos(/3)2-cos(cos

321C

22

)3/4sin(/3)4-cos(

22

)3/2sin(/3)2-cos(

22

sincos

32

C

0iqidi

22

)3/4sin(/3)4-cos(

22

)3/2sin(/3)2-cos(

22

sincos

32

ci

i

i

=


22/83

The equivalent impedance of three phase and two phase machine

=

==

++

+=

+

=

=

+=

++

+=

0

sin

cos

23

22

22

22

23

23

0

21211

32

0

'

00

05.10

005.1

'

22

23

21

22

23

21

2201

21

21

21

21

21

21

22

22

22

23

23

0

21

211

32

'

22

22

22

23

23

0

21

21

1

32

,

22

23

21

22

23

21

22

01

32

21

t

t

mu

cubuau

u

u

u

ut C u

pla La R

pa L pla La R

pa L pla La R

Z

pa L pla La R pa L pa L

pa L pa L pla La R pa L

pa L pa L pa L pla La R

ZC t C Z

t C C

a Lca M ac M cb M bc M ba M ab M

a Lla Lcc Lbb Laa Lc Rb Ra R

ci bi ai

pcc Lc R pcb M pca M

pbc M pbb Lb R pba M

pac M pab M paa La R

cubuau


23/83

Round-Pole Synchronous Machine

The synchronous machine with a uniform air gap in terms of stationary axes is shown below:

:axesabc back togTransferin

,

00

0

00

:thusisequationtagesteady volfinalThe0,',0conditionstatesteadyfor

23,5.1

00

000

'0'

0'

'

1000

0100

00cossin00sincos

1000

0100

00cossin00sincos

:axesstationary phsetwotoaxesrotating phasetwothengtransferi Now

0sin23

cos23

000

)'cos(sin23

05.10

)'sin(cos23

005.1

:machinesetnewin theequationmatrixtheingthan writ before,mentionesas

axesrotating phasetwotoaxesrotating phasethreethering by transfeobtainedismatriximpedanceIts

00

00

m f sd q sqd

f

q

d

f

m s

s

f

q

d

a

af alaS

f f

laa

S aS

S S a

f

q

d

f

q

d

f

f f af af

laa

af alaa

af alaa

f

X I X I U X I U

I

I

I

R

X X

X

U

U

U

R p

M M L p L Lwith

p L RMp

p L R

M p L R L

Mp L p L R

Z

i

i

ii

Z

u

u

uu

i

i

i

i

p L R pM pM

p L R

pM p L p L R

pM p L p L R

u

u

u

u

==

=

==

=+=

++

++

=

=

+

++++

++

=


24/83

}{ }{ }

sin}sin)2

sin((32

)sinsin)2

sin((32

)sincos(32

,)sincos32

22

)3/4sin(/3)4-cos(

22)3/2sin(/3)2-cos(

22

sincos

32

0

m sd sqa

m f sd sqqd aqd a

q

d

c

b

a

E X I X I u

X I X I X I U U u I I i

I I

I

ii

i

++=

+=+=+=

=

The vector diagram of the machine can be found from the last equation:

To obtain the diagram for a generator, we should have to write I for I.


25/83

Two-phase winding in Salient-Pole Machine (Reluctance machine)

++

=

+

+

++

=

la La Rd Lq L pa Rq Ld L p

q Ld L pq Ld L pa R

Z

q Ld L M M M M

la Lq L M la Ld L M

i

i

i

la La R

q Ld Lq Ld L pa R pM

pM q Ld Lq Ld L

pa R

u

u

u

q

N q L

d

N d L

qd

q

d

00

0]2sin2cos[]sincos)[(

0]sincos)[(]2sin2cos[

sincos)(

sincos)(sincos)(

000

0]2cos)2

()2

[(

0]2cos)2

()2

[

0

2,

2


26/83

Salient-Pole Synchronous Machine(Three-phase rotating axes to two-phase rotating axes to two-phase stationary axes)

t.coefficienconstantwithequationsaldifferentilinear toleadsationtransformtheTherefore

00

000

'0'

0'

1000

0100

00cossin

00sincos

'

1000

0100

00cossin

00sincos

0)(sin)(cos000

')(cos)(sin0)cossin(cossin)(

')(sin)(cos0cossin)()sincos(

23

),(5.1),(5.1

00

0

22

22

0

++

++

==

++

+++++

=

=+=+=

p L RMp

p L R

M p L R L

Mp L p L R

i

i

i

i

Z

u

u

u

u

ii

i

i

pL RMpMp pL R

M pM L L p R L L p

M pM L L p L L p R

uu

u

u

M M L L L L L L

f f

al a

qad

qd a

f

q

d

f

q

d

f f f

al a

qd aqd

qd qd a

f

af aqlaqad lad

}

} }

sin}sin)2

sin((32

)sinsin)2

sin((32

)sincos(32

)sincos3

2

0

22

)3/4sin(/3)4-cos(

22

)3/2sin(/3)2-cos(

22

sincos

32

,

000

00

0,00

0

',0

m E d X d I q X q I au

m X f I d X d I q X q I qU d U au

q I d I ai

I q I d I

ci bi ai

m X f I d X d I qU q X q I d U

f I q I

d I

f Rm X d X

q X

f U qU

d U

a R f I q I

d I

f R M a Rd L

q La R

f U qU

d U

p

+

+


27/83

Salient-Pole Synchronous Machine(Three-phase rotating axes to two-phase stationary axes)

+

++

+

==

==+=++

++

++

==

++

++

+++

+

++

+

++

++

+

++

+

+++

=

f

q

d

f f

al a

qad

qd a

f

q

d

af q g alad g ala

oo

oooo

oo

t

f

c

b

a

f f o

af o

af af

oaf

oaf

o g

laaa

g

ao

g

a

o

af

oaf

g

ao

g

laaa

o

g

a

af

af

o g

a

o g

a

g

laaa

f

c

b

a

i

i

i

i

p L RMp

p L R

M p L R L

Mp L p L R

u

u

u

u

u

M M L L L L L L L L

Z ZC C Z

i

i

i

i

p L R pM pM pM

M

pM

L

L L p R

L

L p

L

L p

M

pM

L

L p

L

L L p R

L

L p

M

pM

L

L p

L

L p

L

L L p R

u

u

u

u

00

00

000

'0'

0'

'

23

,)(5.1,)(5.1

23

000

022

)120sin()120cos(

022

)120sin()120cos(

022

sincos

23

000

022

22

22

0)120sin()120sin(sin

0)120cos()120cos(cos

32

'

)120cos()120cos()(cos

')120sin(

)120cos(

)]1202cos(

)[(

)]2cos(

5.0[

)]1202cos(

5.0[

')120sin(

)120cos(

)]2cos(

5.0[

)]1202cos(

)[(

)]1202cos(

5.0[

')(sin

)(cos

)]1202cos(

5.0[

)]1202cos(

5.0[

)]2cos(

)[(


28/83

Synchronous machine torque

[ ] [ ]

M qi f i q Ld Lqi d i T

M qi f i q Lqi d i d Lqi d i

f i qi d i

M qi q Ld i d Lqi

f i qi d i

M d Lq L

f i qi d i T

TorqueGi t i

power Mechanical i G t i

power Stored Lpi t i

Losses Ri t i

i G t i Lpi t i Ri t i ut i P

i G Lpi Ri u

M d Lq L

G

ff L M q L

M d L

L

f Ra R

a R

R

f i qi d i

p ff L f R Mp

M pq La Rd L

Mpq L pd La R

f uqud u

=

=

=

=+

+

==

+

+=

)(

000

000

'

'

'

000

0

00

,

0

00

0

,

00

00

00

0

''

'

Flux linkages in synchronous machine

af M M

g La Lla Lq L

g La Lla Ld L

f p f i f R f u

pi a Ru

d q pqi a Rqu

qd pd i a Rd u

f i

i qi d i

f L M

Lq L

M d L

f

q

d

f i

i qi d i

p f L f R Mp

p La R

M pq La Rd L

Mpq L pd La R

f u

uqud u

ci bi ai

oo

oo

i qi d i

23

)(5.1

)(5.1

000

000

0000

00000

0

000

0000

0

0

0

2/22/22/2

)120sin()120sin(sin

)120cos()120cos(cos

32

0

=

+

++

+

=

++

+

=

++

=


29/83

Synchronous machine (without damper) dynamics (Electromechanical transients)

P0=P2+P J

P2=P m.sin=E.U.sin /X s

PJ=J. 222

dt d

P dt d

pn

J dt d

j s ==

P0=P m.sin+ 22

dt d

P j

In stable condition PJ=0, P 0=P2=P m.sin0

By increasing the input power P1=P0+P1 1=0+

P0+P1=P m.sin( 0+)+ 202 )(

dt

d P j

+

When 0 P1=Pm.sin( 0+)Pmsin0+ 22 )(dt

d P j

When 0 sin(0+)=sin0+.cos0

P1=P m..cos0+ Pm.sin0 Pm.sin0+ 22 )(dt

d P j

P1/P j= j

m

P P

..cos0+ 22 )(dt

d

P1/P j=2

2 )(

dt

d + j

s

P

P

n j

s

P P

= P1/P j=2

2 )(dt

d + n2 .

= [ ]t P P

n s

cos11

n = Natural Frequency

Using Damper

P0=P2+PD+P J

P2=P m.sin=E.U.sin /X s

PD=dt d

K d

PJ=J. 222

dt d

P dt d

pn

J dt d

j s ==

P0=P m.sin+K d 22

dt d

P dt d

j +

In stable condition PD=0, P J=0, P 0=P2=P m.sin 0


30/83

By increasing the input power P1=P0+P1 1=0+

P0+P1=P m.sin( 0+)+ d 20

20 )()(

dt

d P

dt

d j

+++

When 0 P1=Pm.sin( 0+)Pmsin0+ d 22 )()(dt

d P

dt d

j +

When 0 sin(0+)=sin0+.cos0

P1=P m..cos0+ Pm.sin0 Pm.sin0+ d2

2 )()(dt

d P

dt d

j +

P1/P j= j

m

P P

..cos0+ j

d

P K

2

2 )()(dt

d dt

d +

P1/P j= 22 )(

dt

d + j

d

P

K

dt

d )( + j

s

P

P

n j

d n

j

s D P K

P P

2, == D= s j

d

P P

K 21

P1/P j=2

2 )(dt

d + 2Dn.dt

d )( +n2 .

If D1 =

+

+ + t D Dt D D s

nn e D

D De

D

D D P P )1(

2

2)1(

2

21 22

12

1

12

11

n 21 D =d = Natural Damping Frequency

An example:A 1000 kW, 6.6 kV, 50 Hz, 20 poles synchronous motor:

k d=4 kW/elec. Deg/sec, J=4960 kg m2 k d=4 kW/elec. Deg/sec, J=2480 kg m

2 k d=2 kW/elec. Deg/sec, J=4960 kg m2


31/83


32/83

Damper windings

Synchronous motors, synchronous condensers and salient pole synchronous generators are equippedwith damper windings (also called amortisseur windings).There are two principal types of damper windings: 1) complete; 2) incomplete or open. Both typesconsist of bars placed in slots in the pole faces and connected together at each end. Completedampers are similar to the squirrel cage winding of an induction motor, except that the bars are notuniformly spaced, being omitted between the poles. Damper windings may be classified alsoaccording to their resistance. Low resistance damper windings produce their greatest torque at smallslips; high resistance damper windings, at large slips. The double deck damper windings are alsoused. This consists of a low resistance, high reactance damper, which is effective at small slips, and ahigh resistance, low reactance damper, which is effective at large slips.Field collars can be used together with a single damper winding to give results similar to those given

by a double deck damper.Damper windings are not used on turbo-generators, but the solid steel rotor cores of such machines

provide paths for eddy currents and thus produce the same effects as dampers.

Fig. 1: Complete damper Fig. 2: Incomplete damper Fig. 3: Image of a damper winding

Purposes of damper windings:

To provide starting torque for synchronous motors and condensers. To suppress hunting in generators driven by reciprocating engines and in motors driving loads of

pulsating torque. To damp oscillations started by shocks. To decrease the negative-sequence reactance and negative sequence Voltage during unbalanced

loading. To provide braking torque during unsymmetrical fault, and thus to reduce the accelerating torqueduring the fault. To prevent overheating of the poles from eddy currents induced by negative sequence armaturecurrents. To provide additional synchronizing torque in case the generator is synchronized out of phase or speed. To reduce the stress on the insulation of the field winding during current surges in the armaturecircuit. To suppress harmonic voltages, caused by unbalanced loading, to safe values.


33/83

Damper windings affect the synchronous machine quantities such as:''''

22'''' ,,,,, qd qd T T r X X X

Damper action explained by induction motor theory:A damper winding is similar to the squirrel cage rotor winding of an induction motor, and thearmature winding of a synchronous machine is like the stator winding of an induction motor, thevarious effects of the damper windings on stability can be analyzed approximately by inductionmotor theory.

Fig. 4: Equivalent circuit of induction motor

The power developed would be:S

R I P DS

'23=

If a synchronous machine with dampers hunts or swings, the slip is alternatively positive andnegative. The torque is likewise alternatively positive and negative, and always tends to bring thespeed back to synchronous speed. Low resistance dampers greatly increase the amount of positivesequence damping.

Negative sequence voltage causes a wave of air-gap flux to rotate backward at synchronous speed,the slip with respect to the negative sequence field is (2-S) 2, so the mechanical power output,

'2'2

231

3 DS DS R I S S

R I =

is negative and is equal to half the damper copper loss. In

synchronous generators, negative sequence braking torque results during unsymmetrical short circuit,decreases the accelerating torque.The negative sequence rotor currents have a frequency (2-S)f, or nearly 2f. In a synchronous machinesuch currents flow both in the field winding and in the damper windings if damper windings are

provided, otherwise in the field winding and in the filed pole faces.In order to give high damping torque at small values of slip, the dampers should have a lowresistance. In order to give high negative sequence braking torque, the dampers should have a muchhigher resistance.It must be noticed that incomplete damper windings not noticeably influence the quadrature pulsating

flux produced by negative sequence currents.dc braking:The dc component of armature current induces in the damper circuits, currents of fundamental

frequency, witch give rise to a braking torque similar to that caused by negative sequence armaturecurrents. The dc braking torque decreases the accelerating torque. The dc braking torque decaysexponentially with time constant T a .

Pacceleration =P input -Poutput -P braking -Pdamping


34/83

Synchronous Machine with damper winding

',,'/'',0

'00

0

00

0,',0

'

''

2''0

''

2

'''

2

''

2'

''

2

''

2'

''

2

''

2

'

'

0''

0

''0

''

100

010001

'

'

'

'

'

'

''000

0''0

0''0

'''

''

0

0

'

0,,

.tan

00000

00

'''

''

f I m X d I d X qU q

I q X d U f R f U f I Q I D I

f Rd M d L

q L

Z a R p state steadyUnder

f i

qi d i

p p D L D R

pd M f L f R p

p D L D R

pd M d M

p D L D R

pd M d M p

pQ LQ R

pq M q La R

p D L D R

pd M d L

p p D L D R

pd M d M

pQ LQ R

pq M q L p

p D L D R

pd M d La R

f u

qud u

f i

qi d i

p pQ LQ R

q M

p p D L D R

d M p p D L D R

d M

Qi Di

f i

qi d i

Qi Di

f i

qi d i

pQ LQ R pq M

p D L D R pd M pd M

pd M p f L f R pd M

pq M d M d M pq La Rd Lq M pd M pd M q L pd La R

f u

qud u

Qu Du Alsoq M Qq M d M Df M Dd M fd M

turnsof number samethetoreferred all arecesinducmutual thewhen

Qi Di

f i qi d i

pQ LQ R pQq M p D L D R p Df M p Dd M

p Df M p f L f R p fd M

pQq M Dd M fd M pq La Rd LQq M p Dd M p fd M q L pd La R

Qu Du

f uqud u

+

+

+

+

+

+

+

+

=

+

+

+

=

++

+

+

=

++

+

+

=


35/83

Under transient condition:

ceinductransient subaxisQuadraturelQ Lq M

lQ Lq M la L

Q L

q M q Lq L

ceinductransient subaxis Direct d L

d M lD LlD Llf Llf Ld M

lD Llf Ld M la L

d M f L D L

d M D Ld M d M f Ld L

D L

d M f L

D L

d M d M

D L

d M d Ld L

p D L

d M f L p

D L

d M d M

D L

d M d M p

Q L

q M q L

D L

d M d L

p D L

d M d M

Q L

q M q L p

D L

d M d L

Z Q R D R

p p D L D R

pd M f L f R p

p D L D R

pd M d M

p D L D R

pd M d M p

pQ LQ R

pq M q La R

p D L D R

pd M d L

p p D L D R

pd M d M

pQ LQ R

pq M q L p

p D L D R

pd M d La R

Z

tan'

'

'

2"

tan"

''''

''

2''

2'322'

'2'

2

'

2

'

2"

'

2'0

'

2

''

2

'

2'

'

2'

2'

'

2

'

2

'0

''

2''0

''

2

'''

2

''

2'

''

2

''

2'

''

2

''

2

'

+

+

++

+

=

+

+

+

+

+

+

+

+

=


36/83

Torque

qi d d i q

torqueus Asynchrono

qi Di d M Qi d i q M

torque s Synchronou

f i qi d M qi d i d Lq LT

Qi Di

f i

qi d i

d M d M d Lq M q L

Qi Di f i qi d i T

Qi Di

f i

qi d i

Q Lq M D Ld M d M

d M f Ld M

q M q Ld M d M d L

Q

D

f

q

d

Qi Di

f i

qi d i

pQ LQ R pq M

p D L D R pd M pd M

pd M p f L f R pd M

pq M d M d M pq La Rd Lq M pd M pd M q L pd La R

f uqu

d u

=

=

++

+

+

=

4 4 4 34 4 4 214 4 4 4 34 4 4 4 21

''')(

'

'

'

00000

00000

00000

00

000

'''

'

'

'

'000

0'0

0'0

000

00

'

'

'

'

'

'

''000

0''0

0''0

'''

''

0

0

'


37/83

Flux paths for various reactances of a salient pole synchronous machine


38/83

Reactances and Time-constants of synchronous machine

d-axis synchronous reactance Direct axis synchronous reactance is the sum of the stator leakage reactance and the armaturereaction reactance.

mdlad XXX +=

d-axis transient reactance consists of the leakagereactances of the stator and field windings of thegenerator. If the rotor has laminated pole piecesand yokes, but no damper windings, thetransient reactance is practically equal to thesubtransient reactance.

'

'

'lf md

lf md lad X X

X X X X

++=

d-axis open circuit transient time constant of themachine is related to the change in its opencircuit voltage in response to a step change inexcitation voltage (the open circuit rotor timeconstant).

f f do' /R LT =

d-axis short circuit transient time constantdepends upon the damping characteristics of the

excitation circuit.

/XdX'T'T ddod' =

d-axis subtransient reactance embraces theleakage reactances of the stator and rotor windings of the generator, taking into accountthe effect on the rotor leakage of the damper winding or bars and the solid rotor construction.

''''md

''

lad X X'X'

lf lDlDmd lf

lDlf md

X X X X X

X X X ++

+=

d-axis open-circuit subtransient time constant ++= )]/(

X

X['T' '

md'

md'

'do D

lf

lf lD R X

X X

d-axis short circuit subtransient time constantrelates to the very rapidly decaying componentof the ac short circuit current. It depends uponthe damping characteristics of the rotor circuit,

particularly upon the damper winding.

/X'X''T''T' dddod =

q-axis synchronous reactance

mqlaq XXX +=


39/83


40/83

Terminal short circuit current

In the event of a short circuit, the time variation of the short circuit current is considerably affected bythe specific characteristics of the generators. It begins at time t=0, rises to a peak value, the maximumasymmetrical short circuit current, and then decays to the steady state ac short circuit current. Thedegree of asymmetry and the peak value of the short circuit current are determined by the instant atwhich the short circuit occurs, relative to the voltage waveform, and by the short circuit impedance of the system. If the stator terminals are short circuited at the zero-crossing of the phase-A voltage, theshort circuit currents I sc exhibits the time variations shown in the following Figure:

0

time

S h o r t

C i r c u

i t C u r r e n

t

The short circuit current can be divided into four components:- the continuous component- the transient component- the subtransient component

- the dc componentThe continuous component has a constant peak value of U m/Xd .A relatively slowly decaying component is superimposed on the steady state ac short circuit current.This is mainly determined by the leakage reactances of the stator and field windings, since under short circuit conditions the generator behaves like a short circuited transformer. The peak value of thetransient ac short circuit current is U m/X'd.Superimposed on the transient process for a few cycles is a subtransient process. In synchronousgenerators with damper windings a transient process arises between the damper and excitationcircuits. This produces in the stator circuits the subtransient component of the short circuit current,whose peak value is U m/X" d.Since the effective short circuit impedance of the generator can be considered approximately as an

inductive reactance, with the assumed short circuit conditions the initial value of the dc component isapproximately the peak value of the initial ac short circuit current.The stator resistance, particularly in high power synchronous generators, is relatively small, andnegligible in comparison with the effective reactances. The equation for the short circuit current cantherefore be written in the following form:

)/()(+++= ad

md

t X

U X

exp''

1t]cos

1)exp(-t/T')

X1

-X'1

()')exp(-t/T'X'1

-'X'

1[(UI d

ddd

ddmsc


41/83

Transformation from three-phase axes to symmetrical component axes

A phasor I of any orientation can be regarded as the resultant of three arbitrary components I p, In, andI0 in series. If, as shown in the figure, the component p is extended to a symmetrical three phasesystem with positive phase sequence, the component n to a symmetrical negative phase sequencesystem, and the component 0 to a co-phasal system, the combination of these three components can

be made to represent any asymmetrical three phase system. With this method, it is possible to resolveany asymmetrical three phase system into three symmetrical components. The computing operationshave to be carried out only for one phase.

The following equations represent the cirrents of a three phase system:

===

=

+

+

+

111

21

21

3

3

0111

21

21

3

3*1

121

2111

3

3

23

212,

23

21

012

12111

33

02

0

02

0

00

hh

hh

i ni

pi

hh

hh

t C C hhhhC

j h j hwith

i ni pi

hh

hh

C i Bi Ai

i ni h phi C i Cni cpi C i

i nhi pi h Bi Bni Bpi Bi

i ni pi Ai Ani Api Ai


42/83

Transformation from two-phase axes to symmetrical component axes

In exactly the same way that the three phase quantities are defiend in terms of three balanced systems,

two phase quantities can be defined in terms of two balanced systems, positive sequence two phaseand negative sequence two phase. The zero sequence has no equivalent in a two phase system. The

phasor diagrams are shown below:

The equations are:

=

==

=

=

+=+=+=+=

i

i

j

j

i

i

j

jC C

j jC

i

i

j ji

i

ji jiiii

iiiii

n

p

t n

p

n pn p

n pn p

1

1

22

1

1

2211

2211

22

.2 bymatrixationtransformthedividingour

enecessitatthuswillvoltageandcurrentfor ationtransformidentical power withinvariantobtainTo

*1


43/83

Inductance of synchronous machine to zero sequence current

la X X

a I la jX uqud uaU

t la X m I t m I la Lu f

I p f

L f

R

t m pI la L

f I

t m I

p f

L f

R pd

M

pla Ld M pq Ld L

pd M q L pd L

f u

uqud u

t m I t m I

t m I

t m I

i qi d i

f i

i qi d i

p f L f R pd M

pla Ld

M pq

Ld

L

pd M q L pd L

f u

uqu

d u

=

=

=

+

=

+

=

=++

=

+

=

0

)022

sincos[32

)cos3cos30

)(

sin3

0

0

sin3

0

0

00

000

'0'

0'

0

sin

0

0

3

sin

sin

sin

2/22/22/2

)120sin()120sin(sin

)120cos()120cos(cos

32

0

000

000

'0'

0'

0

Inductance to negative sequence(Synchronous machine without damper winding)

For system studies it is necessary to have a value for the impedance of a synchronous machine to

negative sequence current and voltage. Since harmonics are introduced in the current even if thevoltage is sinusoidal, and in the voltage even if the current is sinusoidal, the impedance to negativesequence is regarded for this purpose as the ratio of the fundamental component of the voltage to thefundamental component of the current. For this purpose the field winding is regarded as closedthrough a negligible resistance (short circuited) and there is no excitation under purely negativesequence condition, it is convenient to eliminate the field winding from the inductance matrix of themachine.

=

==

+

=

q

d

qd

qd

q

d

q f

d d

q f

d d

q

d d

f

d f f

f

q

d

f f d

d qd

d qd

f

q

d

i

i

p L L

L p Li

i

p L LM

L

L p LM

L

u

ui

LM

i R

i

i

i

p L R pM

M p L L

pM L p L

u

u

u

'

'

')(

')(0

0

''

'

'

'

2

2


44/83


45/83

Inductance to negative sequence current(Synchronous machine with damper winding)

2 current;sequencenegativetoReactance

]2

[

)23cos(23cos2

)23cos(2

3cos2

]sincos[32

)2cos(

)2sin(

'

'

23

,

0

)2cos()2sin(

23

)120sin(

)120sin(

sin

2/22/22/2

)120sin()120sin(sin

)120cos()120cos(cos

32

'

'

,)(

)(

)(0

)(0)(

')()(')(

)(')()(

""

2

""

""""

""""

""

""

0

""

""

2"

2

22

2"

2

2

22

222

222

qd

aqd

a

mqd mqd

mqd

mqd

qd a

m

m

qd

qd

q

d

m

m

m

m

m

q

d

q

d

qd

qd

q

d

Q

qqq

D

d F

D

d d

D

d d d d d D f

d Dd f f

f

q

d

D

d F f

D

d d

D

d d

Q

qq

D

d d

D

d d

Q

qq

D

d d

f

q

d

X X X

I X X

jU voltagetheof component l Fandamenta

t I X X

t I X X

t I L L

t I L L

uuu

t I

t I

p L L

L p Lu

u

t witht I t I

t I

t I

t I

i

i

i

i

i

p L L

L p Lu

u

L

M L L

LM

L

LM

M

LM

L LiM L LM LM

i R

i

i

i

p LM

L R p LM

M

LM

M p L

M L

LM

L

p LM

M L

M L p

LM

L

u

u

u

+=

+==

+

++

=

+

++

=+=

++

=

+=++

=

++

+=

=

=

=

=

+

=


46/83

Inductance to negative sequence voltage(Synchronous machine without damper winding)

qd

qd

qd

qd

ma

q

m

d

mqd a

m

m

qd

qd

q

d

q

d

qd

qd

q

d

q

d

qd

qd

q

d

m

m

m

m

m

q

d

m

m

m

X X

X X

X X X

t X X

U icurrent theof component l Fandamenta

t t t X

U

t t t X U

iii

t U

t U

p L L

L p Li

i

u

u

p L L

L p Li

i

i

i

p L L

L p Lu

u

t where

t U

t U

t U

t U

t U

u

u

u

t U

t U

t U

+=+=

+==

++++

+++=+=

++

=

==

+=

++

=++

+=

+=

'

'

'2

'

'

1

'

'

1

'

'

'

'

0

2]

11/[2voltage;sequencenegativetoReactance

cos]11

[2

)]23cos(cos)(2cos1[2

)]23cos(cos)(2cos1[2

]sincos[32

)2cos(

)2sin(

'

'

23

'

'

23'

'

0

)2cos(

)2sin(

23

)120sin(

)120sin(

sin

2/22/22/2

)120sin()120sin(sin

)120cos()120cos(cos

32

)120sin(

)120sin(

sin

voltage phasethreesequence Negative


47/83


48/83

Study of Short Circuits in Synchronous Machine Using Symmetrical Components

Equivalent circuits of the symmetrical components for a synchronous machine are shown below. Theinitial generator voltage; 3 E" as driving voltage; is seen only in positive sequence system. Thenegative sequence and zero sequence systems include no driving voltages. In the following sectionthe formulae required to calculate the initial ac short circuit currents for the various kinds of shortcircuit are deduced for effective voltages derived from the generator.

Equations for the component currents and voltages are studied before:

=

=

C

B

A

hh

hh

n

p

n

p

hh

hh

C

B

A

111

1

1

33

0

01

1

111

33

2

2

2

2


49/83

Three-Phase Short Circuit

112

1

01

0

/",/",/"

,0,0,/"3

00

X hE I X E h I X E I

I I X E I

U U U U U U

C B A

n p

n pC B A

======

======

TWO-Phase (B-C) Short Circuit

)/("2

0),/("3),/("3

0),/("3,

0,,

212

2121

021

X X X E U

I X X E j I X X E j I

I X X E I I U U

I I I U U

A

AC B

n pn p

AC BC B

+=

=+=+=

=+===

===

Single-Phase (A) Short Circuit to Earth

)]/()(["

)]/()(["

)/("3)/("3,0

0,0

021022

1

02102122

021

02100

X X X X X hhX h E U

X X X X hX X hh E U

X X X E I X X X E I I I U U U

I I U

C

B

A

n pn p

C B A

++++=++++=

++=++====++

===


50/83

Two -Phase (B-C) Short Circuit with Earth

)/("3

)/()("3

)/()("3

/,/,/)(

)/("3

0,0

10022102

100221022

10022102

00022020

100221020

X X X X X X X X E U

X X X X X X X X h E j I

X X X X X X X hX E j I

X U I X U I X X X X U I

X X X X X X X X E U U U

I U U

A

C

B

nn p p

n p

AC B

++=++=

++=

==+=

++===

===


51/83

Synchronous machine model with two damper windings in q-axis

qd d qQQq Dqd f qd qd d qt

d d d

qqq

Q

Q

D

f

q

d

Qqq

qQq

Dd d

d f d

la

qqq

d d d

Q

Q

D

f

q

d

Q

Q

D

f

q

d

QQqq

qQQq

D Dd d

d f f d

laa

qqd d qad

qqd d qd a

f

q

d

iiiiM iiM iiM ii L LGiiT

M M L

M M L

G

i

i

i

i

i

i

i

LM M

M LM

LM M

M LM

L

M M L

M M L

i

i

i

i

i

i

i

p L R pM pM

pM p L R pM

p L R pM pM

pM p L R pM

p L R

pM pM M M p L R L

M M pM pM L p L R

u

u

u

u

=++==

=

=

++

++

++

+

=

)()(

0000000

0000000

0000000

0000000

0000000

0000

0000

0000

0000

0000

0000

000000

0000

0000

0000

0000

0000

0000

000000

''0'

''0'

0

0

0

'2

'1

''

'2

'1

'

'0

'2

'1

'

'

'2

'1

'

'0

'2

'1

'

'0

'2

'2

'1

'1

''

'''0


52/83

Permanent magnet synchronous motor

Permanent magnet synchronous motors have the rotor winding replaced by permanent magnets.

Based on the typical B-H line for a permanent magnet, we can write:0

0 F F rc +

= where rc , the

permeance, is the slope of the line. From this equation we can write:

rcrcrc

rcrcrc

Li I N i I N F F

L I N I N F

)()()( 02

00

002

0000

=======

Assuming the PM synchronous motor has damper windings but no field winding, the qd axes linkagefluxes and equivalent circuits are as shown below:

++

++

=

==

'0

'

'

''

''

'

'0

'

'

'

'

''

'

'0

'

'

00

000

00

0

0

0

,

00

000

00

000

00

i

i

i

i

i

pM pM pM

p L R pM

pM p L R pM

M pM M p L RM

pM M pM L p L R

u

u

M Lwhere

i

ii

i

i

LM M

LM

L LM

M L

LM L

Q

D

q

d

d d d

QQq

d D Dd

d qd qqd

d qd qd d

q

d

d rc

Q

D

q

d

rcd d

Qq

rc Dd

qq

rcd d

Q

D

q

d

where is the total linkage flux produced by permanent magnet. This equation suggests thefollowing equivalent circuit for a permanent magnet.

We can now use this equivalent circuit of the permanent magnets on the rotor and the previousequivalent circuits of the synchronous motor todevelop a set of qd0 equivalent circuits for the

permanent magnet synchronous motor.


53/83

Induction Machine equations transferred to dq stationary axis

+

++

+

===

====

=

++

++

==

++

++

==

==

++ +

+

=

+===+=======

Q

D

q

d

s sm

s sm

mmr r r

mmr r r

Q

D

aA DqQd t r

r

Q

D

q

d

s

s

r

r

Q

D

q

d

Q

D

q

d

s s

s s

r r r

r r r

Q

D

Q

D

q

d

s s

s s

r r r

r r r

t

t

Q

D

q

d

Q

D

Q

D

s s

s s

r r

r r

Q

D

AlA sQ Dalar sQ Dr

I

I

I

I

jX R jX

jX R jX

jX X s jX R X s

X s jX X s jX R

U

U s j p

M M withiiiiM GiiT M L

M L

G

i

i

i

i

LM

LM

M L

M L

i

i

i

i

p L RMp

p L RMp

MpM p L R L

M Mp L p L R

u

u

u

u

u

u

p L RMp p L RMp

MpM p L R L

M Mp L p L R

ZC C Z

C

i

i

i

i

i

i

i

i

i

ii

i

pL RMpMp

pL RMpMpMpMp p L R

MpMp p L R

u

uu

u

L L L L L L L L L L R R R R R R

00

00

)1()1(

)1()1(0

0

)1(',

5.1],[

0000

0000

00

00

00

00

00

00

00

00

''

''0

0

0000

''

''

'

1000

0100

00cossin

00sincos

1000

0100

00cossin

00sincos

0cossin

0sincoscossin0

sincos0

5.1,5.1,,


54/83

Induction Machine equations transferred to -axis rotating at synchronous speed

=

==

=

=

++

++

=

++

++

=

=

++

++

=

023

sin23

cos23

cossin

sincos

2

0

0

'

cossin00

sincos00

00cossin

00sincos

cossin00

sincos00

00cossin

00sincos

Z'

:speedssynchronouatrotatingaxistodtransferr becanIt

00

00

''

''0

0

:asdevelopedwasaxisdqstationaryinequationsmachineThe

m

sm

sm

s s

s st

S

S

s

s

s s

S

S

r

r

S S S S S

S S S S S

S r r S r

S S r r r

S

S

S S S S S

S S S S S

S r r S r

S S r r r

s s

s s

s s

s s

s s

s s

s s

s s

Q

D

q

d

s s

s s

r r r

r r r

Q

D

u

t u

t u

t t

t t uC

u

u

sand

f with

i

i

i

i

p L R LMpM

L p L RM Mp

MpMs p L R s L

MsMp s L p L R

u

u

p L R LMpM

L p L RM Mp

MpMs p L R s L

MsMp s L p L R

Z

t t

t t

t t

t t

z

t t

t t

t t

t t

i

i

i

i

p L RMp

p L RMp

MpM p L R L

M Mp L p L R

u

u


55/83

Transformation from stationary axes to forward and backward axes

01

fb0fb00*1

00

f q

bf d

iCi,iCi

200

01

01

22

200

0

011

2

2

200

0

011

2

2

ii

iii

dqdqt

b

f

q

d

b

j

j

C C

j jC

i

i

i

j j

i

i

i

ji j

====

==

+=+=

Circuit Equivalent r jx sr Rm jX

r jx sr Rm jX s jx s R P Z m X r X r x m X s X s x

r jX sr Rm X r jX sr R s jX s R

r jsX r Rm sX r jsX r R s jX s R

P I P U P Z

m sX r X s j r R s jX s R

r X s j r R N U N I

m sX r jsX r R s jX s R

r jsX r R P U P I

N I b I

s jX s Rm jX m X s j r X s j r R

N U p I f I

s jX s Rm jX m jsX r jsX r R

pU

N I

b I P I f I

s jX s Rm jX

m X s j r X s j r R s jX s Rm jX

m jsX r jsX r R

N U

bU P U f U

N I P I b I f I

s jX s Rm jX s jX s Rm jX

m X s j r X s j r Rm jsX r jsX r R

N U P U bU f U

j j

j j


m jX m X sr jX r Rr X sm X sm jX r X sr jX r R

j

j

j

j

ZC t C Z

Q I D I q I d I


m jX m X sr jX r Rr X sm X sm jX r X sr jX r R

QU DU

+++

++=

++=

+=

++=

+

=+

+=

+

+

+

=

++

+

=

++

=

++

=

)/()/(

,

/

2)/)((2))((/

2])2()[(

)2(,

2))((

)2()2(0,

000

)2()2(00

00

00

00

00

)2(0)2(0

00

00

1100

00

0011

00

00

)1()1(

)1()1(

100

100

001

001

21*'

00

00

)1()1(

)1()1(0

0


56/83


57/83

Induction Machine Torque

++

+

==Qi Di qi d i

p s L s R Mp

p s L s R Mp

Mp M pr Lr Rr L

M Mpr L pr Lr R

Qu Du

Qu Duqud u

00

00

''

''0

0

)2(2)(,2)(

)/(*]*[1

Re]**[1

Re

00**1Re,

00**1Re

0000

00

0000

00

1'

00

1100

00

0011

0000

0000

00

00

100

100

001

001

21*'

0000

0000

00

00

sr R

b I bT sr R

f I f T

m jX r jX sr R f I

f I m jX

m X P I f jI m X P I f jI r X f I f jI f T

N I b I m jX r jX

N I b I bT p I f I m jX r jX

P I f I f T

m jX r jX

m jX r jX

N

b

P

f

G

j j

j j m X r X m X r X

j

j

j

j

GC t C G

M r L

M r L

G

+

=

==

=

=

=


58/83

Equivalent Resistance and Inductance of Squirrel Cage rotorA squirrel cage can be imagined as a polyphase winding whose number of pole pairs is equal to thenumber of pole pairs of the rotating field. For practical calculations it is convenient to consider asquirrel cage as a polyphase winding with the number of phases equal to the number of slots.Impedance matrix of the squirrel cage rotor in two-phase axis can be developed as below:

)2

(),2

(

)(2

)(2.

......

......

..)(2

)(2

..)(2

)(22

sin2

])1(cos[

.

.

)cos(

cos

,2

20

20

321

2232221

113

1211

11

)(

2

1

==

++++

+

+

++

++

+

+

+

++++

=

==+

+=

+==

==

lr L

lr Lwhere

p L p L L

R R L L

p L p L

R

p L p L p L

R

p L p L L

R R

p L p L

R

p L p L

R p L

p L p L

R

p L p L L

R R

Z

I I I I I

nt I I

t I I

t I I

slotsrotor of number nwithn p

ijii

nnring bar

ring bar

nnnbar

bar

nbar

bar

ring bar

ring bar

bar

bar

nbar

bar

bar

bar

ring bar

ring bar

ring bar ring ringnbar

mnring

mring

mring


59/83


60/83

=

++

++

+

=

=

'1

'

'

'

'2

1

'

'

'

'2

1

'

'

'2

'2

'2

'2

'''

'''

'

'2

1

'

'

12

1

sin000

cos000

0100

0010

0001

000

00

00

'''

''

sin000

cos000

0100

00cossin

00sincos

sc

D

q

d

Q

D

D

q

d

Q

D

D

q

d

s s

s s

S S

r r r

r r r

Q

D

D

q

d

SC

D

Q

D

D

q

d

i

i

ii

i

i

i

i

i

i

i

i

i

i

p L RMp

p L RMpMp

Mp p L RMp

MpM M p L R L

M MpMp L p L R

u

u

u

u

u

ii

i

i

ii

i

i

i

++

+++

=

==

++

+++

=

==

'2

'2

''

'''

'1

'

'

'2

'2

'''

'''

'1

'

'

1

cossincoscos0

cos)1(sin)1()1(

sin)1(cos)1(

'

)1(',

)(cos)(sin)(cos

)(cos0

')(cos)(sin''

')(sin)(cos'sin000

cos000

0100

0010

0001

sincos000

00100

00010

00001

0

0

0

'

s sM M M

M S S M

M M M r r r

M M M r r r

sc

D

q

d

s s

S S

r r r

r r r

sc

D

q

d

D

jX R jX jX jX jX jX R jX

X s jX X s jX R X s

X s jX jX X s jX R

Z

s j p

i

i

i

i

p L R pM pM pM

pM p L RMp

M pM M p L R L

M pM Mp L p L R

i

i

i

i

Z u

u


61/83

Air gap Inductance

Let us calculate the inductive reactance of the magnetic field in the air gap of a round pole machinewith stator bore diameter D. With a uniform gap having a width all around the periphery, themagnetizing force F; produces a sinusoidally distributed field with an amplitude B m.

In the above equations:f frequencyk packing factor k winding's breadth factor k winding's pitch factor I winding's current

N ph number of turnsP number of pole pairsl core length

Slot Leakage Inductance

The coil of a winding is placed in slots punched in sheet steel laminations assembled into core stacks.The slots may have the following shapes, depending on the rating, voltage and torque characteristicof the machine. Some of them are shown in the figure below:

A) rectangular-shape open; B) rectangular-shape semi-closed; C) trapezoidal-shape semi-closed; D)round-shape semi-closed; E) round-bottom-shape; F) rectangular-shape with round-mouthSlot leakage reactance refers to the leakage due to flux crossing the slots within the slots themselves

and thereby linking the conductors that set up this flux. In the calculation two assumptions are made:

2

20

2

2

0

2

0

0

00

)(2

)(42

22

22/

22

224

,/

P

lD N k k k L

P

lD N k k fk

I E X k k f N E

N k lDkk P

I I P

N k k

P kDl

I P

N k k

F B

I P

N k k F

P kDl B

pha

pha ph

ph

ph

phmm

phmm

=

===

==

==

==


62/83

1) the permeability of the surrounding iron is infinite; 2) the leakage flux crosses the slot in straightlines parallel to the bottom of the slot.

]4

)3

[(]sin2

sin)

3[()

])(3

2623.0[)

)623.0(])2sin2(4

1[)

)(32

[)

3)(2

[)

]3[)()()

1

220

2

0 1

1

1

220

1

1

21

220

1

120

0

22

1

120

21

2

1

120

2

4

2

3

21

2

1

120

211

2

0

2

0

2

21

0

1

120

+=+=

++

+=

+=+=

++=

+++

+=

+=+=

d h

lN h

d hd

hlN L F

d h

R RhlN L E

d h

lN d d h

lN L D

d d h

d h

lN LC

d h

d h

d d h

d h

lN L B

hhd lN LdhhhN d l d hlN L Ah

In a full pitch polyphase winding, the top and bottom coil sides in each and every slot belong to thesame phase. Hence the reactances of all slots are identical. However, if the winding is chorded, therewill be some slots with top and bottom coil sides from the same phases, and other slots with top and

bottom coil sides from different phases. In each of the latter kinds of slots there will be mutualleakage reactances between phases.

])2()([2

]2[

]3)(

[

)(1

)(

]3)(

[)(1

)(

2121

1

2

21

1

1

320

1

220

0 21

120

13021

1

3

1

432

21

120

0

2

310

1

432

21

1202

0 1

3

1

2

21

120

2

310

1

2

21

1201

3

3

3

++=+++=+++=

+++++

=+++++

=

+++

=+++

=

Lq L Lq P L

d h

d d h

d h

lN d h

lN d d h

lN dhd hhN

lN L

d h

d hhh

d d h

lN

dhhhN

d l

d hhh

d d h

lN L

d h

d h

d d h

lN dhhhN

d l

d h

d d h

lN L

h

h

h


63/83

Measuring synchronous machine parameters using finite element/difference method

One of the methods for electrical machine design is the numerical field method, especially the finiteelement method. The finite element design packages for electrical machines have made it easier andsafer to develop machines via computer-aided design. They enable simulations to define machineoperating characteristics and parameters such as voltages, inductances and torque. The programs areused extensively today in commercial and scientific applications in electrical machine design anduniversity research worldwide. Figure below show the results of magnetic field simulations on poleshoes of a synchronous machine under steady state, transient and subtransient conditions.

Measuring synchronous machine leakage reactance (X la) using zero-power-factor tests method

The zero-power-factor saturation curve is a graph of armature terminal voltage against field currentwith armature current held constant at its full load value and power factor angle constant at 90 O lag.Point B (see figure below) lies to the right by a constant distance BD and downward by a constantdistance AD from the corresponding point A on the no-load saturation curve. In this test only two

points are require on the zero-power factor curve: the short circuit point B' and point B at ratedvoltage. The vertical line AD equals X laI. Since I is known, X la can be found.

O.C. = open circuit { 0,),( = I snn f I f E ag = air gap line

Point B ==

0cos,

,

N I I snn N U U

B'C'=field current by { 0,, =U snn N I sc I CA ag

I

ADla X p X

3=

Measuring synchronous reactance (X S) using no-load and short circuit tests

The saturated and unsaturated values of the synchronous reactance at rated voltage can be found fromthe open- and short-circuit tests.

O.C. = open circuit }0,),( === I nn I f E s f , ag = air gap line, S.C.=Short Circuit }0,),( === U nn I f I s f


64/83

I1 is the armature current read from the short circuit characteristic at the filed current corresponding tonominal voltage U N on the open-circuit characteristic, as shown in the figure below. I 2 is the armaturecurrent read from the short circuit characteristic at the filed current corresponding to nominal voltageon the air-gap line.

23

13

I N U

S X

I N U

S X

ag =

=

Measuring direct axis subtransient reactance

Any two phases of the machine are connected in series and a single phase voltage is applied acrossthem. The rotor is at standstill. The rotor position is adjusted with hand to get maximum deflection of the ammeter placed in the field winding circuit. Under these conditions:

2"2""

2max

"

max

"

2

2

d d d

d

d

R Z X

I P

R

I U Z

=

=

=

Measuring quadrature axis subtransient reactance

If the rotor is rotated through half a pole-pitch, then peak of the resultant armature m.m.f. coincidesthe q-axis. In this position the ammeter in the field winding should record minimum (sometimes zero)reading. Under these conditions:


65/83


66/83

Measuring direct and quadrature axis synchronous reactance (slip test)

For the measurement of X p and X d a more feasible procedure is the slip test, in which the rotor isdriven at a speed differing slightly from synchronous speed as determined from the frequency of theapplied armature currents. These currents are modulated at slip frequency by the machine; they havemaximum amplitude when the quadrature axis is in line with the m.m.f. wave and minimumamplitude when the direct axis is in line with the m.m.f. wave. Because of impedance in the supplyline, the armature voltage is usually modulated at slip frequency too, being greatest when the currentis least and vice versa. In the slip test the field winding should be open so that slip frequency currentis not induced in it.

max

min

min

max

0

sin

I U

X

I U

X

I

t U U

q

d

f

sr

smtest

=

=

==

=

Synchronous machine parameter estimation using short circuit current testSynchronous machine parameters and time constants appearing in the short circuit equation can bedetermined graphically with sufficient accuracy from the short circuit current oscillograms recordedafter the sudden application of a symmetrical three phase short circuit at the machine terminals. Asmooth line passing through the crest or peak values of the oscillograms is drown to get the envelopeof short circuit current as shown in the example below:

d d d I I I ++ '" The dc component of the current is reduced to zero and the difference between the transient envelope

and the steady state current (11 kA) is obtained.


67/83


68/83


69/83

Stand Still Frequency Response (SSFR) method(IEEE Standard 115)

Test Procedures:1) Excitation signal selection; step signal, ramp signal, variable frequency sinusoidal signal2) Direct-axis identification3) Standstill test application for measuring the direct axis impedance d a s sL R Z +=

4) Standstill test application for measuring the transfer functiond

f s i

i sG=

5) Quadrature-axis identification6) Standstill test application for measuring the quadrature axis impedance qa s sL R Z +=

7) Measuring armature leakage reactance; L la

8) Selection the order of model

9) Extracting the time constants and machine parameters from the measured data using curvefitting procedure and numerical techniques

Some other researches using methods such as: dc-decay, state estimation, transfer function , FE-method


70/83

Induction machine (Parameter estimation tests)

Induction machine equivalent circuit Simplified equivalent circuit Standard Specification Data of Induction machine:Rated voltage; U N ,Rated Current; I N, Rated frequency; f N. Rated output power; P N, Design class;

m x x r S =/Full load specification: Efficiency; , Power factor; cos , Slip; s

Starting specification: Current, I st, Torque, T st Maximum torque, T max , Critical slip; S cr , No-load test: 0000 ,,,, I I P P f f U U U nnn N N s ====== Locked rotor test: N SC SC N SC N S

Hanhok Book

Documents

Transcript of Hanhok Book