Analysis of Electric Machinery and Drive Systems Editor(s): Paul Krause, Oleg Wasynczuk, Scott Sudhoff, Steven Pekarek
Trang 19.1 INTRODUCTION
The method of symmetrical components, as developed by Fortescue [1] , has been used
to analyze unbalanced operation of symmetrical induction machines since the early 1900s This technique, which has been presented by numerous authors [2–5] , is described in its traditional form in the fi rst sections of this chapter The extension of symmetrical components to analyze unbalanced conditions, such as an open-circuited stator phase, is generally achieved by revolving fi eld theory This approach is not used here; instead, reference-frame theory is used to establish the method of symmetrical components and to apply it to various types of unbalanced conditions [6] In particular,
it is shown that unbalanced phase variables can be expressed as a series of balanced sets in the arbitrary reference frame with coeffi cients that may be constant or time varying This feature of the transformation to the arbitrary reference frame permits the theory of symmetrical components to be established analytically and it provides a straightforward means of applying the concept of symmetrical components to various types of unbalanced conditions
In this chapter, unbalanced applied stator voltages, unbalanced stator impedances, open-circuited stator phase, and unbalanced rotor resistors of the three-phase induction
Analysis of Electric Machinery and Drive Systems, Third Edition Paul Krause, Oleg Wasynczuk,
Scott Sudhoff, and Steven Pekarek.
© 2013 Institute of Electrical and Electronics Engineers, Inc Published 2013 by John Wiley & Sons, Inc.
UNBALANCED OPERATION
AND SINGLE-PHASE INDUCTION MACHINES
9
Trang 2SYMMETRICAL COMPONENT THEORY 337
machine are considered Single-phase induction motors are analyzed, and several anced and fault modes of synchronous machine operation are illustrated
9.2 SYMMETRICAL COMPONENT THEORY
In 1918, C.L Fortescue [1] set forth the method of symmetrical components for the purpose of analyzing the steady-state behavior of power system apparatus during unbal-anced operation Fortescue showed that the phasors representing an unbalanced set
of steady-state multiphase variables could be expressed in terms of balanced sets of phasors For example, the phasors representing an unbalanced three-phase set can be
expressed in terms of (1) a balanced set of phasors with an abc sequence (the positive sequence), (2) a balanced set of phasors with an acb sequence (the negative sequence),
and (3) a set of three equal phasors (the zero sequence) Although the method of metrical components is widely used in the analysis of unbalanced static networks [2] ,
sym-it is perhaps most appropriate for the analysis of symmetrical induction machines during unbalanced operations
Fortescue ’ s change of variables is a complex transformation that may be written for three-phase stationary circuits in phasor form as
F+−0s=SFabcs (9.2-1) where the symmetrical components are
(F+−0s) = ⎡⎣+ − 0 ⎤⎦
T
F F F (9.2-2) The unbalanced phasors are
(Fabcs)T
= ⎡⎣ ⎤⎦ (9.2-3) and the transformation is expressed
11
2 2
a2 e j4 3 1 j
2
32
Trang 3The inverse transformation is
is present, the method of symmetrical components can be applied to each frequency present in the system [6]
In order to compute the total instantaneous, steady-state power, it is fi rst necessary
to convert the phasors to the variables in sinusoidal form and then multiply phase voltage times phase current Thus
Uppercase letters are used to denote steady-state sinusoidal quantities We will show that the instantaneous power consists of an average value and time-varying components
9.3 SYMMETRICAL COMPONENT ANALYSIS OF
INDUCTION MACHINES
Although the method of symmetrical components can be used to analyze unbalanced conditions other than unbalanced stator voltages, this application of symmetrical com-ponents is perhaps the most common Once the unbalanced applied stator voltages are known, (9.2-1) can be used to determine V as+, V as−, and V0s It is clear that the currents due to the positive-sequence balanced set can be determined from the voltage equations given by (6.9-11)–(6.9-13) These equations are rewritten here with the + added to the subscript to denote positive-sequence phasors
V as r s j X I j X I I
e b
ls as
e b
ωω
ω
where ω is the base electrical angular velocity generally selected as rated and
Trang 4UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 339
It is customary to obtain the voltage equations for the negative-sequence quantities
by reasoning In particular, slip is the normalized difference between the speed of the rotating air-gap MMF and the rotor speed The negative-sequence currents establish
an air-gap MMF that rotates a ω e in the clockwise direction Hence, the normalized difference between the air-gap MMF and rotor speed is ( ω e + ω r )/ ω e which can be
written as (2 − s ), where s is defi ned by (9.3-3) Therefore, the voltage equations for the negative-sequence quantities can be obtained by replacing s by (2 − s ) in (9.3-1)
and (9.3-2) In particular,
V as r s j X I j X I I
e b
ls as
e b
ωω
ω
ω ( −−) (9.3-5)
If the zero-sequence quantities exist in a three-phase induction machine, the steady-state variables may be determined from the phasor equivalent of (6.5-24) and (6.5-27) The electromagnetic torque may be calculated using sequence quantities; however, the deri-vation of the torque expression is deferred until later
9.4 UNBALANCED STATOR CONDITIONS OF INDUCTION
MACHINES: REFERENCE-FRAME ANALYSIS
The theory of symmetrical components set forth in the previous sections can be used
to analyze most unbalanced steady-state operating conditions However, one tends to look for a more rigorous development of this theory and straightforward procedures for applying it to unbalanced conditions, such as an open stator phase or unbalanced rotor resistors Reference-frame theory can be useful in achieving these goals [6] Although simultaneous stator and rotor unbalanced conditions can be analyzed, the notation necessary to formulate such a generalized method of analysis becomes quite involved Therefore, we will consider stator and rotor unbalances separately
In Reference 6 , the stator variables are expressed as a series of sinusoidal functions with time-varying coeffi cients Such an analysis is notationally involved and somewhat diffi cult to follow We will not become that involved because the concept can be estab-lished by assuming a single-stator (rotor) frequency for stator (rotor) unbalances with constant coeffi cients We will discuss the restrictions imposed by these assumptions as
we go along
Trang 5Unbalanced Machine Variables in the Arbitrary Reference Frame
If we assume that the rotor is a three-wire symmetrical system and the stator applied voltages are single frequency, then unbalanced steady-state stator variables may be expressed as
F F F
as bs cs
e e
t t
F F F F
12
12
32
(9.4-4)
12
12
32
12
12
32
12
12
32
Trang 6UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 341
where ω is the electrical angular velocity of the arbitrary reference frame
It is interesting that the qs and ds variables form two, two-phase balanced sets in
the arbitrary reference frame In order to emphasize this, (9.4-2) is written with
sinu-soidal functions of ( ω e t − θ ) separated from those with the argument of ( ω e t + θ ) It is
possible to relate these balanced sets to the positive- and negative-sequence variables For this purpose, let us consider the induction machine in Figure 6.2-1 A balanced
three-phase set of currents of abc sequence will produce an air-gap MMF that rotates counterclockwise at an angular velocity of ω e By defi nition, the positive sequence is the abc sequence, and it is often referred to as the positively rotating balanced set The negative sequence, which has the time sequence of acb , produces an air-gap MMF that
rotates in the clockwise direction It is commonly referred to as the negatively rotating
balanced set Let us think of the qs and ds variables as being associated with windings
with their magnetic axes positioned relative to the magnetic axis of the stator windings
as shown in Figure 6.2-1 Now consider the series of balanced sets formed by the qs and ds variables in (9.4-2) The two balanced sets of currents formed by the fi rst and
second column of the 2 × 4 matrix produce an air-gap MMF that rotates
counterclock-wise relative to the arbitrary reference frame whenever ω < ω e and always
counter-clockwise relative to the actual stator winding Hence, the balanced sets with the
argument ( ω e t − θ ) may be considered as positive sequence or positively rotating sets
because they produce counterclockwise rotating air-gap MMFs relative to the stator windings It follows that the balanced sets formed by the third and fourth columns with
the argument ( ω e t + θ ) can be thought of as negative sequence or negatively rotating
sets Therefore, we can express (9.4-2) as
interest For example, ω = 0 for the stationary reference frame and ω = ω r for the rotor
reference frame It is also clear that the instantaneous sequence sets may be identifi ed
in these reference frames In particular, when ω is set equal to zero, the frequency
of the variables is ω ; however, the positive and negative sequence are immediately
Trang 7identifi able When ω = ω r , we see that there are two electrical angular frequencies present; ω e − ω r and ω e + ω r The latter occurs due to the air-gap MMF established by
negative-sequence stator variables Because we have assumed that the rotor circuits form a three-wire symmetrical system only one positive and one negative sequence set will be present in the rotor However, the balanced negative-sequence set of rotor vari-
ables will appear in the arbitrary reference frame with the same frequency ( ω e t + θ ) as
the negative sequence set established by the stator unbalance
It follows that for stator unbalances and the assumption that the rotor is a three-wire symmetrical system, (9.4-2)–(9.4-9) can be used to identify the positive- and negative-
sequence rotor variables We only need to (1) replace all s subscripts with r in (9.4-2)–
(9.4-7) , (2) add a prime to all quantities associated with the rotor variables and (3) set
θ in (9.4-2) equal to ω r t Recall that the rotor variables are transformed to the arbitrary reference frame by ( 6.4-1 ), wherein β is defi ned by ( 6.4-5 ) Please don ’ t confuse the β
given by ( 6.4-5 ) and the β used in the subscripts starting with (9.4-1)–(9.4-9)
The instantaneous electromagnetic torque may be expressed in arbitrary frame variables as
M b
ωω
Torque may be expressed in per unit by expressing X M in per unit and eliminating the factor (3/2)( P /2)(1/ ω b )
Phasor Relationships
The phasors representing the instantaneous sequence quantities F qs and F ds for a stator
unbalance given in (9.4-2)–(9.4-7) may be written as
Trang 8UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 343
qs ds
qs qs
j j
F F
qs qs
qs ds
j j
F F F
qs qs s
qs ds
j j
F
qs s
qs s s
s as
bs cs
Trang 912
2
321
2
12
12
j j
F F
qs s
qs s
qs s ds s
1
For stator unbalances with symmetrical three-wire rotor circuits, (9.4-29) also applies
to rotor phasors in the stationary reference frame That is
j j
F F
qr s qr s
qr s
dr s
12
1
The steady-state voltage equations in the stationary reference frame may be obtained
from (6.5-34) by setting ω = 0 and p = j ω e Thus
e b M
e b M
e b M
r b
e b rr
r b rr
r b
ωω
ωωω
ω
ωω
ωω
ωωω
ω
0
M
e b M
r b
e b rr
qs
I
ωω
ωω
ωω
dr s
I I I
Trang 10UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 345
If we now incorporate (9.4-29) and (9.4-30) into (9.4-31) , the sequence voltage tions become
e b M
b rr
s e b ss
e b M
ω
ωω
ωω
ωωω
b rr
qs s
qr
I I
ω
ωω
2
ss qs s
qr s
I I
where the asterisk denotes the conjugate
Steady-state instantaneous stator variables, F as , F bs , and F qs , may be obtained from
the phasors F as, F bs, and F cs determined from (9.4-23) With the assumption of a
sym-metrical rotor, the frequency of the stator variables is ω e ; however, as has been
men-tioned the rotor variables F ar′, F br′, and F cr′ each contain two frequencies ( ω e − ω r ) and ( ω e + ω r ) The instantaneous rotor variables can be obtained by using the rotor equiva-
lent of (9.4-15)–(9.4-18) to identify F qrA′ through F qrD′ , which can be substituted into the rotor equivalent of (9.4-2) To obtain F qr′
r
and F dr′
r
, θ in (9.4-2) must be set to ω r t ,
where-upon Kr may be used to obtain F ar′, F br′, and F cr′ Since we have assumed symmetrical
three-wire stator and rotor circuits, neither F 0 s nor F0′r exist; however, we have included the zero-sequence notation in order to show the equivalence to the symmetrical com-ponent transformation If a zero sequence were present in the rotor circuits, it would consist of two frequencies, unlike the stator zero sequence, which would contain
only ω
Trang 119.5 TYPICAL UNBALANCED STATOR CONDITIONS OF
INDUCTION MACHINES
Although it is not practical to consider all stator unbalanced conditions that might occur, the information given in this section should serve as a guide to the solution of a large class of problems Unbalanced source voltages, unbalanced phase impedances, and an open-circuited stator phase are considered, and the method of calculating the steady-state performance is set forth in each case
Unbalanced Source Voltages
Perhaps the most common unbalanced stator condition is unbalanced source voltages This can occur in a power system due to a fault or a switching malfunction that may cause unbalanced conditions to exist for a considerable period of time The stator circuit
of an induction machine for the purpose of analysis is given in Figure 9.5-1 From Figure 9.5-1 , we can write
e ga g
+
+
Trang 12TYPICAL UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 347
Substituting (9.5-4) and (9.5-5) into (9.4-29) yields
12
12
1
The steady-state phasor currents can be calculated from (9.4-32) The torque may then
be calculated using (9.4-34)
Unbalanced Stator Impedances
For this unbalanced condition, let us consider the stator circuit shown in Figure 9.5-2 ,
wherein an impedance z(p) is placed in series with the as winding The following
equa-tions may be written as
e ga =i z p as ( )+v as+v ng (9.5-8)
e gb=v bs+v ng (9.5-9)
e gc =v cs+v ng (9.5-10)
In a three-wire system, i 0 s is zero, therefore v 0 s is zero because the stator circuits are
symmetrical Let us now assume that the source voltages are balanced; hence, we can add (9.5-8)–(9.5-10) , and we have
Trang 13Equation (9.5-11) is valid as long as e ga + e gb + e gc is zero Substituting (9.5-11) into
(9.5-8)–(9.5-10) and solving for the phase voltages yields
In a problem at the end of the chapter, you are asked to write ν as , ν bs , and ν cs assuming
e ga , e ga , and e gc are not balanced Substituting the steady-state phasor equivalents of
ω
ωb M
e b M
b rr
s e b ss
e b M
ω
ω
ωω
ωω
ωω
e b M
b rr
qs s
ωω
ωω
q qr s
qs s qr s
I I
Trang 14TYPICAL UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 349
Open-Circuited Stator Phase
For the purpose of analyzing an open-circuited stator phase, which is equivalent to single-phase operation, let us consider the stator circuits of a three-phase, wye-connected
induction machine as shown in Figure 9.5-3 , where phase a is open-circuited Because the stator circuit is a three-wire symmetrical system, i 0 s and v 0 s are zero Therefore, with
M qr s
ω ′ is applied to phase a, the current
i as will be forced to remain at zero [7] From Figure 9.5-3
+
+ +
+
n
–
Trang 15v ng= 1 e gb+e gc + v as
2
12
1
The above relationships are valid for transient and steady-state conditions It is assumed that the source voltages contain only one frequency, therefore, if we substitute the steady-state phasor equivalent of (9.5-21) , (9.5-25) , and (9.5-26) into (9.4-26) , we obtain
V qs s j e X I E
b
M qr s
However, I as is zero, and since θ and I0s are both zero, then I qs
s
, which is I as, is zero Thus
I qs s I
qs s
If we substitute (9.5-30) into (9.5-27) and (9.5-28) , and then substitute the results into (9.4-32) , and if we incorporate (9.5-32) , we can write
Trang 16UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES 351
e b M
e b
12
ωω
ωω
b rr
e b M
b rr
ωω
ωωω
ω
ωω
qs s qr s
qr s
(9.5-33)
9.6 UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES
In the analysis of unbalanced rotor conditions, it will be assumed that the stator circuits are symmetrical and the stator applied voltages are balanced and have only one fre-quency Since the analysis of steady-state operation during unbalanced rotor conditions
is similar in many respects to the analysis for unbalanced stator conditions, the ships will be given without a lengthy discussion The principal difference is the refer-ence frame, in which the analysis is carried out It is convenient, in the case of rotor unbalanced conditions with symmetrical stator circuits, to conduct the analysis in the rotor reference frame since therein the variables are of one frequency
Unbalanced Machine Variables in the Arbitrary Reference Frame
Assuming only one rotor frequency is present, the rotor variables may be expressed as
ar br cr
qrA qrB qrC qrD
(9.6-2)
and
F′ = ′F αcos(ω ω− )t+ ′F βsin(ω ω− ) t (9.6-3)
Trang 1712
32
12
12
32
12
12
32
12
12
32
We have assumed that the stator is a symmetrical, three-wire system and the applied
voltages are balanced, containing only one frequency, ω e Therefore, we can use (9.6-2)– (9.6-7) to express the stator variables by (1) replacing all r subscripts with s , (2) remov- ing the primes, and (3) setting θ = 0 in (9.6-2) This gives rise to two stator frequencies;
ω e and ( ω e − 2 ω r ) It is assumed that the source voltages are of the frequency ω e , and we
will also assume a zero impedance source Thus, the stator currents and the phase
volt-ages of the stator windings will contain ω e and ( ω e − 2 ω r ) The ( ω e − 2 ω r ) frequency,
which is the stator negative sequence, is induced into the stator windings due to the
Trang 18UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES 353
rotor unbalance It is interesting to note that the stator negative sequence currents are
not present when ω r = (1/2) ω e Therefore, because I qsr− is zero at ω r = (1/2) ω e , we would
expect the negative sequence torque to also be zero
The instantaneous steady-state electromagnetic torque for a rotor unbalance may
be expressed as
M b
The pulsing component in (9.6-13) is commonly referred to as the “twice slip-frequency” torque
Phasor Relationships
It is convenient to conduct the steady-state analysis of unbalanced rotor conditions much the same as unbalanced stator conditions The phasor expressions from (9.6-2) are
Following a procedure identical to that in the case of stator unbalance, we can write
Fqr′r±0r=SFabcr′ (9.6-18) and
Fqs s± 0s =SFabcs (9.6-19)
We can write (6.5-34) in the rotor reference frame by setting ω = ω r , and then, by setting
p = j( ω e − ω r ), we can obtain the voltage equations for steady-state conditions
If we then substitute (9.4-29) and (9.4-30) for stator variables and similar relationships for rotor variables from (9.6-15) and (9.6-16) into the steady-state equations, we will obtain
Trang 19b ss
ωωω
ω
ωω
e b M
b rr
ωωω
ω
ωω
qs r
qr r qs r
qr r
(9.6-20) where
where the asterisk denotes the conjugate As we have mentioned, the negative sequence
torque is zero when ω r = (1/2) ω e , since I qs r− becomes zero; however, the pulsating torque component is still present
9.7 UNBALANCED ROTOR RESISTORS
In some applications where it is necessary to accelerate a large-inertia mechanical load,
a wound-rotor induction machine equipped with variable external rotor resistors is often used As the speed of the machine increases, the value of the external rotor resistors
is decreased proportionally so as to maintain nearly maximum electromagnetic torque during most of the acceleration period Care must be taken, however, in order not to unbalance the external rotor resistors during this process, otherwise a torque pulsation
of twice slip frequency occurs as noted in (9.6-13) and (9.6-22) , which may cause frequency oscillations in the connected mechanical system For the purpose of analyz-ing unbalanced rotor resistors, we will consider the rotor circuit given in Figure 9.7-1 The rotor phase voltages may be written as
Trang 20UNBALANCED ROTOR RESISTORS 355
v br′ =v pm− ′ ′i R br br (9.7-2)
v cr′ =v pm− ′ ′i R cr cr (9.7-3) Since the rotor is assumed to be a three-wire system, i0′ =r 0, and hence v0′ =r 0, there-
fore, if we add (9.7-1)–(9.7-3) and solve for ν pm , we obtain
R
ar br cr
ar
13
22
i i i
ar br cr
(9.7-5)
For the analysis of steady-state operation, it is convenient to express (9.7-5) as
Vabcr′ =RI′abcr (9.7-6) where the terms of (9.7-6) can be determined by comparison with (9.7-5) If we sub-stitute (9.7-6) into (9.6-18) , we obtain
0
1 0