Synchronous Machine Transient Analysis and
Symmetrical Three-Phase Faults
Short circuits occur in power systems when equipment insulation fails;
due to system over-voltages caused by lightning or switching surges,
insulation contamination, and/or other mechanical causes. The resulting
short circuit current depends on the internal voltages and their
impedances to the fault. The fault currents could be several times the
normal value and may cause equipment damages if allowed to persist. If
the fault is not cleared by protective equipments, windings and bus bars
may also be damaged. The speed of the circuit breakers are given as 8, 5,
3, or 11/2 cycles, etc, which is a measure of time from occurrence of fault
to the extinction of the arc.
The selection of circuit breaker for a power system depend upon the
maximum current it may have to carry momentarily and the current it
may have to interrupt at the voltage of the line in which it is placed. In
order to understand the problem of calculating the initial current when a
synchronous generator is short circuited, consider the series R-L circuit
The applied voltage is
v(t ) Vm sin(t )
The closing of the switch at t=0 represents to a first approximation a
three-phase short circuit at the terminals of an unloaded synchronous
machine. For simplicity assume that the fault impedance is zero, i.e.,
the short circuit is a solid or ‘bolted’ fault. α determines the magnitude
of the voltage when the circuit is closed. The differential equation is
di(t )
Ri(t ) L Vm sin(t )
dt
The solution of this equation carries 2 parts – the steady state and the
transient solution and is given as
t
Vm
i (t ) sin(t ) Ae
Z
Substituting the initial condition that at t=o, i(t)=0, the complete
solution is
t
Vm
i(t ) [sin(t ) e sin( )]
Z
L L
where, Z R ( L) ; tan
2 2 1
;
R R
The expression for the current is also often written as
i(t) = iac(t)+idc(t)
The total fault current, called the asymmetrical fault current is shown
below
If the switch is closed when, 0 , there is no dc offset, and
the shape of the current plot becomes,
If the switch is closed when, / 2, there is maximum dc
offset. For / 2 , the shape of the response is
Example:
For the circuit considered, assume R=0.125Ω, L=10 mH. The
source voltage is given as,
v(t)=151sin(377t+α)
Determine current response after closing the switch for the
following cases:
a) No DC offset
b) For maximum DC offset
Three Phase Short Circuit on an Unloaded Synchronous
Machine
One way to investigate a three-phase short circuit at the terminals of a
synchronous machine is to perform a test on an actual machine. The
figure below shows such an oscillogram of the fault current in one phase
of the unloaded synchronous machine during such a test. The dc offset
has been removed.
As can be seen, the amplitude of the sinusoidal waveform decrease with
time. This can be explained in terms 3 reactances given as
Eg Eg
Direct axis subtransient reactance, xd''
I '' oc / 2
Direct axis transient reactance,
Eg Eg
x
'
d '
I ob / 2
Direct axis synchronous reactance, Eg Eg
xd
I oa / 2
Using the above reactances, the instantaneous fault current can be written
as,
t
1 1
iac (t ) 2 Eg [( '' ' )e Td''
xd xd
t
1 1 1
( ' )e Td'
]sin(t / 2)
xd xd xd
Td’’, Td’ are the direct –axis short circuit sub-transient and transient time
constants. Td’’< Td’ . Note that at t=0,
Eg
I ac ''
x d
At a later time when t is large compared to Td’’ but small compared
to Td’, the first exponent term has decayed and the second term is
predominant. When t is much larger than Td’, the rms AC current
approaches its steady state value, given by,
Eg
I ac
xd
The rms value if i(t) is of interest. Since i(t) is not strictly periodic, its
rms value is not strictly defined. We stretch the rms concept to
calculate the asymmetrical fault current with maximum dc offset as
follows:
I rms (t ) ( I ac2 ) [ I dc (t )]2
( I ac2 ) [ 2 I ac et / ]2
I ac 1 2e2t /
K (t ) I ac
Rms asymmetrical fault current= Rms ac fault current x asymmetry
factor
Example
A bolted short circuit occurs in the R-L circuit with E= 20 kV, X=8Ω,
R=0.8 Ω. Circuit breaker opens at 3 cycles after fault inception.
Determine:
(a)The rms ac fault current
(b)The rms momentary current at t=1/2 cycles which
passes through the breaker before it opens, and,
(c)The asymmetrical fault current which the breaker
interrupts
Consider maximum dc offset.
The three phase fault currents are displaced from each other by 1200
each. In addition each phase has different dc offset. The maximum dc
offset in one phase which occurs when α=0 is
2 Eg
idc max (t ) et / TA 2 I "et / TA
xd''
Example
A 50 MVA, 20 kV, 60 Hz synchronous generator with xd”=0.15, xd’=0.24
and xd=1.1 pu, and Td”=0.035, Td’=2.0 and TA=0.2 s is connected to a
circuit breaker. The generator is operating at 5% above rated voltage
and at no load when a bolted three phase short circuit occurs on the
load side of the breaker. The breaker interrupts the fault 3 cycles after
its inception. Determine:
a) Subtransient fault current in pu and in kA
b) Maximum dc offset as a function of time
c) AC current
d)Rms asymmetrical fault current the breaker
interrupts.
Assume maximum dc offset.
