SYMMETRICAL FAULTS
CALCULATION
األستاذ الدكتور :سعدي عبد
الحميد السيد
(After decaying the DC component)
So circuit breakers have the following rating
i) Breaking capacity (rupturing capacity).
• Generally expressed in terms MVA and it's
equal to the multiplying of rated breaking
current in kilo-amperes, rated voltage in kilo-
volt and factor which depends upon the
number of phases
(breaking capacity)=
Breaking capacity = √3 V I (MVA)
Note
• British practice I is (R.M.S) but in
American practice I is max. value
ii) Making capacity
There is always a possibility that C.B
is closed under short-circuit
condition. The making capacity of the
C.B depends upon its ability to
withstand the effect of electro-
magnetic forces which are
proportional to the square of the
peak value of the making current
according to B.S.S 116.
Making capacity = 2.55 breaking
capacity
ii) Making capacity
There is always a possibility that C.B
is closed under short-circuit
condition. The making capacity of the
C.B depends upon its ability to
withstand the effect of electro-
magnetic forces which are
proportional to the square of the
peak value of the making current
according to B.S.S 116.
Making capacity = 2.55 breaking
capacity
iii) Short-time rating.
The short-time rated current is the current
can be safely applied with the C.B in its
normal condition for 3 seconds. If the ratio of
breaking current to normal current< 40 (or
0ne second other wise)
EX: 1
A circuit breaker has the following name plate 1000 amperes, 1,200 MVA,
33 K.V, 3 phases
Determine
(i) rated current (ii) breaking capacity
iii) Breaking current (iv) making current
v) short-time rating
Solution
(i) Rated currents 1000
Amperes
(ii) Rated breaking capacity =
1,200 MVA
(iii) Breaking current = 1200*106
/ √3*33*103 = 21000 A
Short-circuit currents may be several orders of
magnitude larger than normal operating currents
and, if allowed to persist, may cause thermal damage
to equipment. Windings and busbars may also suffer
mechanical damage due to high magnetic forces
during faults.
It is therefore necessary to remove faulted sections
of a power system from service as soon as possible.
Standard EHV protective equipment is designed to
clear faults within 3 cycles (50 ms at 60 Hz).
Lower voltage protective equipment operates more
slowly (for example, 5 to 20 cycles)
SERIES R–L CIRCUIT TRANSIENTS
The closing of switch SW at t = 0 represents to a first
approximation a three-phase short circuit at
the terminals of an unloaded synchronous machine.
For simplicity, assume zero fault impedance. The
current is assumed to be zero before SW closes, and
the source angle a determines the source voltage at t
= 0. Writing a KVL equation for the circuit,
The total fault current is called the asymmetrical
fault current, and has two components. The ac fault
current (also called symmetrical or steady-state
fault current), is a sinusoid. The dc offset current,
given by (decays exponentially with time constant.
The rms ac fault current is Iac = V/Z. The
magnitude of the dc offset, which depends on α,
varies from 0 when α = θ to
Note that a short circuit may occur at any instant
during a cycle of the ac
source; that is, α can have any value.
AC & DC Current Components
of Fault Current
The rms asymmetrical fault current equals the rms
ac fault current times an ‘‘asymmetry factor,’’
decreases from
Also, higher X/R ratios give higher values of Irms(τ).
The above series R–L short-circuit currents are
summarized in Table 7.1.
Momentary current
