j. mccalley
DESCRIPTION
J. McCalley. Machine Transformations. Consider a 2-pole induction machine with the a-phase stator winding, having N a turns, is distributed about its axis as shown below. Space vectors. The conductor density of the phase a winding is. - PowerPoint PPT PresentationTRANSCRIPT
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J. McCalley
Machine Transformations
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Space vectors
2
sin2a
aNn
Consider a 2-pole induction machine with the a-phase stator winding, having Na turns, is distributed about its axis as shown below.
The conductor density of the phase a winding is
where a negative density is interpreted as the return path of the turns.
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Space vectors
3
If the coil carries a current ia, then the magneto-motive force (MMF, the magnetic circuit analogue to voltage in an electric circuit) is given by an application of Ampere’s circuital law, which relates the intensity of a magnetic field to the current that produces it.
A
dAJdlHF
Apply to the closed path shown in the figure:
cos
sin2aa
aa
aaa
iN
diN
dinF
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Space vectors
4
Symmetry allows us to associate half of this path mmf with each side of the machine. Thus, the mmf per airgap is
Because this and other variables are distributed around the air-gap periphery, it is convenient to represent them by space vectors. A space vector is represented by a complex number having•a magnitude equal to the peak amplitude of the variable and •an angle equal to the angular position of this positive peak value. Each space vector is considered to have a cosinusoidal distribution around the machine periphery.
cos2as
aiNF
The MMF space vector is denoted aFWe may also denote the current as a space vector: ai
Thus, aa
a iNF2
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Space vectors
5
Symmetry allows us to associate half of this path mmf with each side of the machine. Thus, the mmf per airgap is
Because this and other variables are distributed around the air-gap periphery, it is convenient to represent them by space vectors. A space vector is represented by a complex number having•a magnitude equal to the peak amplitude of the variable and •an angle equal to the angular position of this positive peak value. Each space vector is considered to have a cosinusoidal distribution around the machine periphery.
cos2as
aiNF
The MMF space vector is denoted aFWe may also denote the current as a space vector: ai
Thus, aa
a iNF2
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Space vectors
6
The two space vectors corresponding to F and i are illustrated below.
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Space vectors
7
We may go through the same procedure for the phase currents b and c, in whichcase we represent them with space vectors
bi ciDefine a positive rotation of 2π/3 as:
23
21)3/2( jea j
Because the b and c phase windings are spatially displaced from the a phase winding by 120 and 240 degrees, respectively, then:
cc
bb
iai
aii2
The total effect of a set of instantaneous currents in all three phase windings is then found by vector addition according to:
cba iiii This is not the same as the sum of the currents: ia+ib+ic, which is zero!
Then a positive rotation of 4π/3 is: 23
21)3/4(2 jea j
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Space vectors
8
Time-domain waveforms Space vectors and their summation for the “instant” indicated on the other plot
cbacba iaaiiiiii 2
Observe: at the “instant,” ia is positive and just after its peak, ib and ic are negative.
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Space vectors
9
cbacba iaaiiiiii 2Let’s add them mathematically:
23
21)3/2( jea j
23
21)3/4(2 jea j
Now use:
And we obtain:
cbcba
cba
iijiii
ijijii
23
21
23
21
23
21
But ia+ib+ic=0ia=-(ib+ic). Making this substitution in the above results in:
cbacbaa iijiiijiii 23
23
23
21
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Space vectors
10
It will be convenient for us later on to scale this vector by 2/3 resulting in :
cba iijii 23
23
cba
cbacbas
iiji
iijiiijii
31
33
23
23
32
The above relation concentrates the effects of the three phase currents into a single complex variable. Very nice!
Now, let’s consider the converse process….
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Space vectors
11
The converse process….
Consider that we know the space vector and that we want to find the instantaneous values of the individual phase currents. How to do this?
si
We will project the vector onto the respective a, b, and c axes, as shown below. si
The analytic equivalent of this projection is:
sc
sb
sa
iai
iai
ii
Re
Re
Re
2
Just take the real part of si
Rotate the is vector forward by 240°. This places +b-axis where the +a-axis was. Then take real part. Rotate the is vector forward by 120°. This places +c-axis where the +a-axis was. Then take real part.
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Space vectors
12
We have represented the phase currents as space vectors. In doing so, however, the only thing required was they were balanced three-phase quantities. Any other variables can be similarly represented as long as they are balanced three-phase quantities, e.g., currents, voltages, and fluxes.
Let’s generically refer to any such variables as xa, xb, and xc and the corresponding space vector as sx
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α-β transformation
13
The space vector can also be represented by two “phase magnitudes,” called xα and xβ in the real-imaginary complex plane, as illustrated below.
sx
We can express this relationship mathematically according to:
cbas xaaxxjxxx 2
32
The α-β components of the space vector can be calculated from the abc magnitudes according to:
cbas xxxxx
21
21
32Re
cbs xxxx23
23
32Im
Real partof axb
Real partof a2xc
Im partof axb
Im partof a2xc
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α-β transformation
14
These two relations can be represented in matrix form as follows:
We define the matrix as the T-matrix:
cbas xxxxx
21
21
32Re
cbs xxxx23
23
32Im
c
b
a
xxx
xx
23
230
21
211
32
23
230
21
211
32T
This transformation is also called the “Clarke transformation” for the person who developed it, Edith Clarke.
February 10, 1883 - October 29, 1959
AT&T, MIT, GE, U-Texas
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α-β transformation
15
We will often represent the Clarke transformation in one of the two equivalent ways:
cbas xxxxx
21
21
32Re
cbs xxxx23
23
32Im
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α-β transformation
16
It is interesting to observe, in a balanced system, so that xa+xb+xc=0,
cbas xxxxx
21
21
32Re
cbs xxxx23
23
32Im
aaaa
cbas
xxxx
xxxxx
23
32
21
32
21
32Re
We can solve for xb and xc in terms of xα and xβ, in which case we obtain the inverse Clarke transformation as
xx
xxx
c
b
a
23
21
23
21
01
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α-β transformation
17
We often represent the inverse Clark transformation in two equivalent ways:
xx
xxx
c
b
a
23
21
23
21
01
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Other transformations
18
When we transform variables to the α-β transformation we have just established, we are said to be working in the Stator Reference Frame.
This reference frame is aligned with the stator, and the rotational speed of this reference frame, since it is aligned with the stator, is 0.
The space vector referred to it rotates at the synchronous speed ωs. We denote the corresponding space vector with a superscript “s” (stationary) according to:
jxxxs
We can also define a space vector aligned with the rotor. The reference frame in this case is called the D-Q reference frame and it rotates with angular speed of ωm. Therefore the space vector referred to it rotates at the slip speed of ωr.
Finally, we can also define a space vector aligned with the synchronous reference frame, at a speed of ωs. The space vector referred to it does not rotate, that is, it presents constant real and imaginary parts. This is called the d-q frame.
QD
rjxxx
qd
ajxxx
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Other transformations
19
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Reference frame transformations
20
Our objective now is to represent balanced, but time-varying (dynamic) behavior of a DFIG. There will be three types of equations that we will need:•Voltage equations•Flux linkage equations•Motion equations
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Revolving rotor with constant flux
21
Consider the rotating magnetic field (or a sinusoidal traveling wave) of an air gap, i.e., the plot on the left of the below figure and how it “moves” with time for a revolving rotor with constant flux (like that of a synchronous machine rotor). We see that, for fixed time (just one of the plots), there is sinusoidal variation of flux density with space. Also, if we stand on a single point on the stator (e.g., θ=90°) and measure flux (or flux density) as a function of time, we see that for fixed space (the vertical dotted line at 90°, and the red eye on the pictures to the right), there is sinusoidal variation of flux density w/time.
0θ
N
N
N
θ
θ
θ
θ
θ
θ=90°
We can show that 3-phase, balanced currents of 3 stator windings separated by 120 degrees achieve the same thing. Let’s look at that….
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Rotating magnetic field
22
Consider the 3-phase currents developed in the stator windings concentrated at points along the stator circumference 120 degrees apart. The currents are given by:
)240cos()120cos(
cos
tIitIitIi
sc
sb
sa
Whenever we have a current carrying coil, it will produce a magnetomotive force (MMF) equal to Ni. (MMF is the magnetic circuit analogue to voltage in an electric circuit.) And so each of the above three currents produce a time varying MMF around the stator. Each MMF has a maximum in space, occurring on the axis of the phase, of Fam, Fbm, Fcm, expressed as
)240cos()240cos()()120cos()120cos()(
coscos)(
tNItFtFtNItFtF
tNItFtF
ssmcm
ssmbm
ssmam
Define the angle θ as measured from the a-phase axis, and consider points in the airgap. At any time t, the spatial maximums expressed above occur on the axes of the corresponding phases and vary sinusoidally with θ around the air gap. We can combine the time variation with the spatial variation in the following way:
)240cos()(),()120cos()(),(
cos)(),(
tFtFtFtFtFtF
cmc
bmb
ama
Now substitute (*) into these equations….
(*)
Each individual phase MMF here varies with θ around the air gap and has an amplitude that varies with time.
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Rotating magnetic field
23
Now add the three MMFs above:
)240cos()240cos(),()120cos()120cos(),(
coscos),(
tFtFtFtFtFtF
smc
smb
sma
)240cos()240cos()120cos()120cos(
coscos),(),(),(),(
tFtFtF
tFtFtFtF
sm
sm
sm
cba
Use cosαcosβ=0.5[cos(α-β)+cos(α+β)] and then simplify, and you will obtain:
)cos(23),( tFtF sm
If we plot the above along the airgap, we will see exactly the same MMF distribution that was created by the revolving rotor with constant flux.
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Space vectors
24
)cos(23),( tFtF sm
0θ
θ
θ
θ=90°