psad slides lec1 synchronous machine model
DESCRIPTION
Lec1 SynchronousTRANSCRIPT
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
48582 - POWER SYSTEM ANALYSIS ANDDESIGN
LECTURE 1 - SYNCHRONOUS MACHINEMODEL
DR. GERMANE X ATHANASIUS
School of Electrical, Mechanical and Mechatronic Systems
UNIVERSITY OF TECHNOLOGY SYDNEY
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Lecture Outline
1 Synchronous machine modelPer phase equivalent circuit
2 Two axis modelPark’s TransformationTransient modelBalanced three phase fault
3 Simplified representation for transient analysis
4 Short circuit current
5 DC components of stator currents
6 Fault on a loaded generator
Synchronous Generators
Synchronous GeneratorsSynchronous Generators
• Synchronous generators or alternators are used to convert mechanical power derived from steam, gas, or hydraulic-turbine to ac electric powerto ac electric power
• Synchronous generators are the primary source of electrical energy we consume todayenergy we consume today
• Large ac power networks rely almost exclusively on synchronous generatorsgenerators
Construction
Basic parts of a synchronous generator:
• Rotor dc excited winding• Rotor - dc excited winding • Stator - 3-phase winding in which the ac emf is generated
The manner in which the active parts of a synchronous machine are cooled determines its overall physical size and structure
Various Types
Salient-pole synchronous machine
Cylindrical or round-rotor synchronous machine
Non-uniform air-gap
Uniform air-gapUniform air gap
Salient-Pole Synchronous Generator
StatorStator
Cylindrical-Rotor Synchronous Generator
Stator
Cylindrical rotor
Operation PrincipleOperation Principle
The rotor of the generator is driven by a prime-mover
A dc current is flowing in the rotor winding which produces a rotating magnetic field within the machine
The rotating magnetic field induces a three-phase e otat g ag et c e d duces a t ee p asevoltage in the stator winding of the generator
Electrical Frequency
Electrical frequency produced is locked or synchronized toElectrical frequency produced is locked or synchronized to the mechanical speed of rotation of a synchronous generator:
120m
enP
f =
where f = electrical frequency in Hz
120
where fe = electrical frequency in HzP = number of polesn = mechanical speed of the rotor in r/minnm= mechanical speed of the rotor, in r/min
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Schematic diagram
Figure: Schematic diagram of a synchronous generator
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Flux linkages
The speed of the machine (N) in rpm is given by
N =120f
P(1)
Stator coils self inductance, Ls = Laa′ = Lbb′ = Lcc′
Mutual inductance −Ms = Lab = Lbc = LcaThe mutual inductance between field coil and stator coil, if themaximum value of mutual inductance is Msf , then mutualinductance for an angle of θd is given by,
Lfa = Msf cos θd
Lfb = Msf cos(θd − 1200
)Lfc = Msf cos
(θd − 2400
)(2)
Self inductance of the field winding Lff ′ is constant
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Flux linkages
The flux linkage of stator and rotor coil is given by,λaλbλcλf
=
Laa′ Lab Lac LafLba Lbb′ Lbc LbfLca Lcb Lcc′ LcfLfa Lfb Lfc Lff ′
iaibicif
(3)
For a balanced three phase system, ia + ib + ic = 0 andia = −(ib + ic), ib = −(ia + ic) and ic = −(ia + ib). λa
λbλc
=
Ls + MsLs + MsLs + Ms
iaibic
+
LafLbfLcf
if (4)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Flux linkages
If the field current has a constant magnitude If and the angularvelocity of the rotor is ω then we have
dθd
dt= ω and θd = ωt + θd0 (5)
Substituting in (4) and combining (2) and (4) we get, λaλbλc
=
Ls + MsLs + MsLs + Ms
iaibic
+
Msf cos (ωt + θd0)Msf cos
(ωt + θd0 − 1200)
Msf cos(ωt + θd0 − 2400)
If (6)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Induced emf
If the coil a has a resistance R, then the emf across the coil isgiven by,
ea = −Ria −dλa
dt
= −Ria − (Ls + Ms)diadt
+ ωMsf If sin (ωt + θd0) (7)
The last term in (7) represents the internal emf induced in coila by the field current and is given by,
e′a =√
2|Ea| sin (ωt + θd0) (8)
where |Ea| is the rms magnitude and is given by
|Ea| =ωMsf If√
2(9)
e′a will be the emf across the coil a when ia = 0. This voltageis known as open circuit / noload / synchronous internal /generated voltage.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Induced emf
θd0 is the angle with reference to d − axis, if δ is the angle withreference to the q − axis, then δ =
(θd0 − 900),
∴ θd = ωt + θd0 = ωt + δ + 900 (10)
Now equation (8) becomes,
e′a =√
2|Ea| sin(ωt + δ + 900
)=
√2|Ea| cos (ωt + δ) (11)
substituting in (7)
ea = −Ria − (Ls + Ms)diadt
+√
2|Ea| cos (ωt + δ) (12)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Induced emf
Equation (12) can be diagrammatically represented as inFigure 2.
Figure: Equivalent circuit of phase a
Similarly we can find λb, λc , e′b and e′c for other two stator coilsalso.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
If ia lags e′a by θl , we can write,
ia =√
2|Is| cos (ωt + δ − θl) (13)
Similarly for other phases we can write as,
ib =√
2|Is| cos(ωt + δ − θl − 1200
)ic =
√2|Is| cos
(ωt + δ − θl − 2400
)(14)
Now considering the flux linkage with the field circuit andsubstituting for Laf , Lbf and Lcf from (2) into (3) we get,
λf = Lff ′ If + Msf
[ia cos θd + ib cos
(θd − 1200
)+ ic cos
(θd − 2400
)](15)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
Now substituting for ia and θd using (10) and (13) we get,
ia cos θd =√
2|Is| cos (ωt + δ − θl) cos(ωt + δ + 900
)(16)
Using the trigonometric identity2 cos A cos B = cos(A + B) + cos(A− B) we can write,
ia cos θd =|Is|√
2{− sin θl − sin [(2ωt + δ)− θl ]} (17)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
Similarly we can write,
ib cos(θd − 1200
)=
|Is|√2
{− sin θl − sin
[(2ωt + δ)− θl − 1200
]}ic cos
(θd − 2400
)=
|Is|√2
{− sin θl − sin
[(2ωt + δ)− θl − 2400
]}(18)
In the above equations terms with 2ωt represent a balancedthree phase second harmonic components whose three phasesum will be zero. Now adding (17) and (18) we get,
ia cos θd + ib cos(θd − 1200
)+ ic cos
(θd − 2400
)=−3√
2Is sin θl (19)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
Substituting (19) in (15)
λf = Lff ′ If +
√32
Msf id (20)
where
id =
√23
[ia cos θd + ib cos
(θd − 1200
)+ ic cos
(θd − 2400
)]=
√3Is sin θl (21)
From the above discussion we conclude the field flux linkagedue to time varying currents ia, ib and ic produce a constantflux linkages and does not vary with time. We can representthis flux linkage as one coming from a fictitious coil with asteady DC current id and have an axis coinciding with daxis andthe two coils synchronously rotate and have a mutualinductance of Msf as represented in Figure 3.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
f
f
d
d
sf
Figure: Stator equivalent d − axis coil.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Field flux linkages
Now the field voltage,
Vff ′ = If Rf +dλ
dt
= If Rf sincedλ
dtdoes not vary
The current id depends on Is and θl . For lagging power factorsid will be negative causing demagnetising effect. So If has tobe increased to counteract. For leading power factors id will bepositive causing magnetising effect and If is decreased. Thiseffect is called armature reaction and this principle is used inexcitation system control.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Per phase equivalent circuit
Per phase equivalent circuit
Figure: Per phase equivalent circuit.
Figure: Phasor diagram.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
Need for the model:1 The synchronous machine model we developed so far is
based on round rotor theory. This model will be sufficientto analyse the machine under steady state conditions.
2 For transient studies we need to use two axis model.3 In salient pole machines the air gap is small above pole
faces and large in the interpolar regions. These aspectsneed to be considered.
4 Also the rotor has damper windings. These windings arerepresented as Daxis and Qaxis windings which are shortcircuited coils.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
b c
a
a
cb
d
f
f
Figure: Synchronous machine schematic diagram with damperwindings.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
We can rewrite equation (3) including the damper windings as,λaλbλcλfλDλQ
=
Laa′ Lab Lac Laf LaD LaQLba Lbb′ Lbc Lbf LbD LbQLca Lcb Lcc′ Lcf LcD LcQLfa Lab Lac Lff ′ LfD LfQLDa LDb LDc LDf LDD′ LDQLQa LQb LQc LQf LQD LQQ′
iaibicifiDiQ
(22)
The elements of the above matrix can be defined as follows:stator self inductances:
Laa′ = Ls + Lm cos 2θd
Lbb′ = Ls + Lm cos(
2θd −2π
3
)Lcc′ = Ls + Lm cos
(2θd +
2π
3
)(23)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
stator mutual inductances:
Lab = Lba = −Ms − Lm cos(
2θd +π
6
)Lbc = Lcb = −Ms − Lm cos
(2θd −
π
2
)Lca = Lac = −Ms − Lm cos
(2θd +
5π
6
)(24)
rotor mutual inductances:
LfD = LDf = Mr
LQf = LfQ = 0LDQ = LQD = 0 (25)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
stator-rotor mutual inductances:
Laf = Lfa = Msf cos θd
Lbf = Lfb = Msf cos(
θd −2π
3
)Lcf = Lfc = Msf cos
(θd −
4π
3
)(26)
stator-rotor daxis damper mutual inductances:
LaD = LDa = MsD cos θd
LbD = LDb = MsD cos(
θd −2π
3
)LcD = LDc = MsD cos
(θd −
4π
3
)(27)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Two axis model
stator-rotor qaxis damper mutual inductances:
LaQ = LQa = MsQ cos θd
LbQ = LQb = MsQ cos(
θd −2π
3
)LcQ = LQc = MsQ cos
(θd −
4π
3
)(28)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Park’s Transformation
Features of Park’s Transformation
1 The inductance elements given by (23) to (28) aredependant on the angular position of the rotor which will bevarying continuously so these inductance will be timevarying. Analysing the model with time varying inductancewill be difficult.
2 Using Park’s transformation these inductances can bemade time invariant for the purpose of analysis.
3 The stator a, b, c variables are transformed to direct axis(d), quadrature axis (q) and zero sequence (0) quantitiesusing the transformation matrix P.
4 The matrix P has the orthogonality property ie., P−1 = PT .Any stator variable can be transformed to dq0 axis bymultiplying with P.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Park’s Transformation
Park’s Transformation
For example stator currents are transformed as, idiqi0
= P
iaibic
(29)
where
P =
√23
cos θd cos(θd − 1200) cos
(θd − 2400)
sin θd sin(θd − 1200) sin
(θd − 2400)
1√2
1√2
1√2
(30)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Equation (22) can be represented in short form as,[λabcλfDQ
]=
[LSS LSRLRS LRR
] [iabcifDQ
](31)
Using Park’s transformation we can transform the stator timevarying inductance to rotor reference frame without modifyingthe rotor quantities. Now the transformation to do that is,[
λdq0λfDQ
]=
[P 00 I
] [λabcλfDQ
](32)
where I is 3 × 3 identity matrix.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
From (32) we can write,[λabcλfDQ
]=
[P−1 0
0 I
] [λdq0λfDQ
][
P−1 00 I
] [λdq0λfDQ
]=
[LSS LSRLRS LRR
] [iabcifDQ
]=
[LSS LSRLRS LRR
] [P−1 0
0 I
] [idq0ifDQ
][
λdq0λfDQ
]=
[P 00 I
] [LSS LSRLRS LRR
][
P−1 00 I
] [idq0ifDQ
](33)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Now substituting for P and P−1 in (33) we get,λdλqλ0λfλDλQ
=
Ld 0 0 kMsf kMsD 00 Lq 0 0 0 kMsQ0 0 L0 0 0 0
kMsf 0 0 Lff ′ Mr 0kMsD 0 0 Mr LDD′ 0
0 kMsQ 0 0 0 LQQ′
idiqi0ifiDiQ
(34)
where
k =
√32
Ld = Ls + Ms +32
Lm
Lq = Ls + Ms −32
Lm
L0 = Ls − 2Ms (35)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Similarly using (32) we can find the transformations for currentsand voltages without modifying the rotor quantities as,[
idq0ifDQ
]=
[P 00 I
] [iabcifDQ
](36)[
vdq0vfDQ
]=
[P 00 I
] [vabcvfDQ
](37)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
We can write the voltage equation using (7) as,
vavbvcvfvDvQ
= −
Ra 0 0 0 0 00 Ra 0 0 0 00 0 Ra 0 0 00 0 0 Rf 0 00 0 0 0 RD 00 0 0 0 0 RQ
iaibicifiDiQ
− ddt
λaλbλcλfλDλQ
(38)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Using simplified notations and transformations we get,[P−1 0
0 I
] [vabcvfDQ
]=
−[
Rabc 00 RfDQ
] [P−1 0
0 I
] [idq0ifDQ
]−d
dt
[P−1 0
0 I
] [λdq0λfDQ
][
vabcvfDQ
]= −
[P 00 I
] [Rabc 0
0 RfDQ
][
P−1 00 I
] [idq0ifDQ
]−[
P 00 I
]ddt
[P−1 0
0 I
] [λdq0λfDQ
]
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
On simplification we get,[vabcvfDQ
]= −
[Rabc 0
0 RfDQ
] [idq0ifDQ
]−[
PdP−1
dt 00 I
] [λdq0λfDQ
](39)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Now using (39),
PdP−1
dt= P
dθ
dtdP−1
dθ= ωP
dP−1
dθ
=23ω
cos θd cos(θd − 1200) cos
(θd − 2400)
sin θd sin(θd − 1200) sin
(θd − 2400)
1√2
1√2
1√2
×
cos θd sin θd1√2
cos(θd − 1200) sin
(θd − 1200) 1√
2cos
(θd − 2400) sin
(θd − 2400) 1√
2
= ω
0 1 0−1 0 00 0 0
(40)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Transient model
Transient model
Combining (34), (39) and (40) we get,
2666664vdvqv0−vf
00
3777775 =
2666664Ra ωLq 0 0 0 ωkMsQωLd Ra 0 ωkMsf ωkMsD 0
0 0 Ra 0 0 00 0 0 Rf 0 00 0 0 0 RD 00 0 0 0 0 RQ
3777775
2666664idiqi0ifiDiQ
3777775
−
2666664Ld 0 0 kMsf kMsD 00 Lq 0 0 0 kMsQ0 0 L0 0 0 0
kMsf 0 0 Lff ′ Mr 0kMsD 0 0 Mr LDD′ 0
0 kMsQ 0 0 0 LQQ′
3777775ddt
2666664idiqi0ifiDiQ
3777775(41)
The zero sequence voltage v0 is not coupled with other equations so it can be treated
separately. In (41) all the matrix coefficients are constants if ω is assumed to be
constant.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Balanced three phase fault
Balanced three phase fault
Consider a three phase fault on the terminals of a three phasegenerator, which is operating with constant speed andexcitation. Before fault we assume the phase currents ia, ib, icare zero, this will make the currents id , iq, i0 are also zero.Initial field current will be given by,
if =Vf
Rf(42)
During fault the terminal voltages will be zero, ie.,va, vb, vc = 0 so vd , vq, v0 = 0.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Balanced three phase fault
Balanced three phase fault
We shall use (41) to solve for the fault currents. Since i0 = 0, we caneliminate it from (41) and write as,
vdvq−vf00
=
Ra ωLq 0 0 ωkMsQωLd Ra ωkMsf ωkMsD 0
0 0 Rf 0 00 0 0 RD 00 0 0 0 RQ
idiqifiDiQ
−
Ld 0 kMsf KsD 00 Lq 0 0 kMsQ
kMsf 0 Lff ′ Mr 0KsD 0 Mr LDD′ 00 kMsQ 0 0 LQQ′
ddt
idiqifiDiQ
(43)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Balanced three phase fault
Balanced three phase fault
Equation (43) can be represented in short form as,
v = −Ri− Ld idt
d idt
= −L−1Ri− L−1v (44)
Equation (43) is a set of linear first order differential equationwhich can be solved for analytical solutions or by usingcomputer integration methods.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Simplified representation for transient analysis
During fault conditions, there will be a sudden rise in currentwhich will not be backed up by instantaneous changes in fluxlinkages in the rotor circuits and armature reaction effects. Itwill take a few cycles for the flux linkages to settle for steadystate values. These initial period following the fault is termed assub-transient and transient periods.Using equation (34) we get,
∆λd = Ld∆id + kMsf ∆if + kMsD∆iD (45)∆λf = kMsf ∆id + Lff ′∆if + Mr∆iD (46)∆λD = kMsD∆iD + Mr∆if + LDD′∆id (47)
The ∆ quantities indicate incremental changes.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Simplified representation for transient analysis
The field and damper fluxes cannot change instantaneously so∆λf = 0 and ∆λD = 0. Substituting in (46) and (47) we get,
∆if = −(
kMsf LDD′ − kMsDMr
Lff ′LDD′ −M2r
)∆id (48)
∆iD = −(
kMsDLff ′ − kMsf Mr
Lff ′LDD′ −M2r
)∆id (49)
Substituting (48) and (49) in (45) we get,
∆λd
∆id= L′′d = Ld − k2
(M2
sf LDD′ + M2sDLff ′ − 2Msf MsDMr
Lff ′LDD′ −M2r
)(50)
∆λd∆id
is the flux linkage change for unit current whichsub-transient d-axis inductance L′′d and sub-transient d-axisreactance X ′′
d which is less than steady state reactance Xd .
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Simplified representation for transient analysis
After the sub-transient period the effect of damper windings canbe neglected, so the flux linkage equations become,
∆λd = Ld∆id + kMsf ∆if (51)∆λf = kMsf ∆id + Lff ′∆if (52)
Equating ∆λf to zero gives,
∆if = −(
kMsf
Lff ′
)∆id (53)
substituting in (51),
∆λd
∆id= L′d = Ld −
(kMsf )2
Lff ′(54)
where L′d is known as transient d-axis inductance and transientd-axis reactance X ′
d = ωL′d which is less than steady statereactance Xd . The reactance will values will beX ′′
d < X ′d < Xd .
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Simplified representation for transient analysis
1 From this we can conclude that immediately after the faultthe reactance X ′′
d and the current decays with the timeconstant τ ′′d and has a typical value of around 0.03 s.During this period the reactance is X ′′
d .2 Once the effect of the damper becomes negligible, the
machine reactance raises to X ′d and the fault current
decays with the time constant τ ′d during this period the τ ′ddepends inversely on the field resistance Rf .
3 In steady state conditions Xd = ωLd and Xq = ωLq forsalient pole machines and Xd = ωLd for round rotormachines.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Short circuit current
1 The behavior of the short circuit current is similar to R − Lcircuit behavior when voltage is applied suddenly but in amore complex manner.
2 The short circuit currents contain dc components whichmakes the three phase currents asymmetrical.
3 If we remove the dc component of the current, we canrepresent the ac component of the fault current as,
Iac = |Ea|1
Xd+ |Ea|
(1
X ′d− 1
X ′d
)e− t
τ ′d + |Ea|(
1X ′′
d− 1
X ′d
)e− t
τ ′′d (55)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Short circuit current
From (55) it can be seen that the fault current Iac has onesteady state component and two decaying components withtime constants τ ′d and τ ′′d representing transient andsub-transient periods.
Figure: Fault current ac component
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Short circuit current
The rms value of the sub-transient current is given by,
|I′′| = oz√2
=|Ea|X ′′
d(56)
The rms value of the transient current is given by,
|I′| = oy√2
=|Ea|X ′
d(57)
The rms value of the steady state fault current is given by,
|I| = ox√2
=|Ea|Xd
(58)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
DC components of stator currents
Similar to RL circuit switching transients, during fault on thegenerator a dc component will be superimposed on the acwave. The dc component value depends on the instantaneousstator voltage Ea and the rotor angle δ at the time of fault. Thetime constant associated with dc component decay is given by
τdc =X
′′
d + X′′q
2Ra(τdc is usually around 0.05 - 0.175 s)(59)
and most of the dc decay occurs during sub-transient period.The magnitude of the dc component will be different fordifferent phases since it depends on the instantaneous voltage.The dc component for the phase a is given by,
Idc−a =√
2|Ea|X ′′
dsin δe
tτdc (60)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
DC components of stator currents
The net asymmetrical stator current is obtained bysuperimposing dc over ac wave. This asymmetrical current forphase a is given by,
iasym−a =√
2|Ea| ×[1
Xd+
(1
X ′d− 1
X ′d
)e− t
τ ′d +
(1
X ′′d− 1
X ′d
)e− t
τ ′′d +1
X ′′d
sin δet
τdc
](61)
Worst possible transient condition will occur if δ = 900. Themaximum dc component of fault current is given by,
Idc−max =√
2|Ea|X ′′
d(62)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
DC components of stator currents
The maximum rms value of the asymmetrical fault current isgiven by,
Iasym−max =√
I ′′2 + I2dc
=
√√√√(Ea
X ′′d
)2
+
(√2Ea
X ′′d
)2
=√
3Ea
X ′′d
=√
3 I′′ (63)
Iasym−max is used to determine the max asymmetrical breakingcapacity of the circuit breaker.
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Fault on a loaded generator
If the generator is loaded at the time fault and delivering a loadcurrent of Il , we have three internal voltages associated withsub-transient, transient and steady state periods viz., E ′′
a , E ′a
and Ea respectively. These voltages are given by,
E ′′a = Va + X ′′
d Il , E ′a = Va + X ′
d Il and Ea = Va + Xd Il(64)
Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator
Fault on a loaded generator
Figure: Fault on a loaded generator