SHORT CIRCUIT STUDIES
Dr. Nitin Dhote
Dr. Nitin Dhote, SVPCET, Nagpur
NECESSITY OF SHORT CIRCUIT STUDIES:
1. THE CURRENTS AND VOLTAGES AT DIFERENT POINTS ON THE
SYSTEM DURING FAULT CONDITION.
2. TO DESIGN POWER SYSTEM EQUIPMENT INSULATION LEVEL AND
CONDUTOR DIAMETER.
3. DESIGN EFFECTIVE AND ADEQUATE RELAYING AND SWITCHING
SYSTEM.
4. IT PROVIDES THE INFORMATION ABOUT SHORT CIRCUIT MVA
CAPACITY REQUIRED FOR CIRCUIT BREAKER AT DIFFERENT POINTS.
5. PLANNING , DESIGN AND OPERATION OF THE SYSTEM.Dr. Nitin Dhote, SVPCET, Nagpur
ASSUMPTIONS IN SHORT CIRCUIT STUDIES:
1. ALL THE SHUNT ELEMENTS INCLUDING LOADS IN THE POWER
SYSTEM ARE NEGLECTED (I L(0) = 0).
2. SYSNCHRONOUS MACHINE IS REPRESENTED BY CONSTANT
VOLTAGE BEHIND REACTANCE.
3. RESISTANCES OF ALL ELEMENTS ARE NEGLECTED.
4. POSITIVE AND NEGATIVE SOURCE REACTANCES OF ALL ROTATING
MACHINES ARE EQUAL(X1=X2)
5. TRANSFORMERES ARE REPRESENTED BY THEIR NOMINAL RATIOS
ONLY.
Dr. Nitin Dhote, SVPCET, Nagpur
G1
G2
Gn
L1 L2 Lm
machines
Loads Ep
Ei
3-Ø Transmission system
i
p
abc
abc
abc
1
E1
Dr. Nitin Dhote, SVPCET, Nagpur
machines
Ep
Ei
Transmission system with series reactances
i
p
abc
abc
Gn
G1
1
E1abc
++
+
Dr. Nitin Dhote, SVPCET, Nagpur
E1(0)
][ ,, cbaZf
Including machines reactances
machines
Ep(0)
Ei(0)
i
p
abc
abc
abc
1
Zf
[ZBUS]
Dr. Nitin Dhote, SVPCET, Nagpur
Dr. Nitin Dhote, SVPCET, Nagpur
For a power system containing m number of buses,Bus voltages during fault at bus P are given by
)()0()( abcabcabcabc IbusZbusEbusEbus
E1(+)
E2(+)
Ei(+)
EP(+)
Em(+)
=
E1(0)
E2(0)
Ei(0)
EP(0)
Em(0)
-
Z11 Z12 Z1i Z1p Z1m
Z21 Z22 Z2i Z2p Z2m
Zi1 Zi2 Zii Zip Zim
Zp1 Zp2 Zpi Zpp Zpm
Zm1 Zm2 Zmi Zmp Zmm
0
0
0
0
0
Ip(+)
0
0
Neglect prefault load currents, Fault current If =Ip(+)
abc abc abc abc
abcabcabcabcIfZppEpEp )0()( …… (a)
Voltage at fault bus is given by
Dr. Nitin Dhote, SVPCET, Nagpur
Voltage at fault bus P is also given by
abcabcabcIfZfEp .)( …… (b)
From a&b, abcabcabcabcabc IfZppEpIfZf )0(.
)1.......()0(.)(
).()0(
1 abcabcabcabc
abcabcabcabc
EpZppZfIf
IfZppZfEp
)2........(.)0()(abcabcabcabc
IfZipEiEi
Putting values of If in voltage at any other bus Ei(+) and voltage at fault bus Ep(+), we get
)3........(.)(abcabcabc
IfZfEp Dr. Nitin Dhote, SVPCET, Nagpur
If the network is expressed in sequence components 0,1,2 ,Equations 1,2&3 are modified by replacing the suffix abc by 012 asfollows:
)1.......()0(.)(0121012012012
EpZppZfIf
)2........(.)0()(012012012012
IfZipEiEi
)3........(.)(012012012
IfZfEp
Dr. Nitin Dhote, SVPCET, Nagpur
In many types of faults , Fault impedance Zf is unidentified( Infinite). Equations are obtained in terms of Fault admittance
Dr. Nitin Dhote, SVPCET, Nagpur
For a power system containing m number of buses,Bus voltages during fault at bus P are given by
)()0()( abcabcabcabc IbusZbusEbusEbus
E1(+)
E2(+)
Ei(+)
EP(+)
Em(+)
=
E1(0)
E2(0)
Ei(0)
EP(0)
Em(0)
-
Z11 Z12 Z1i Z1p Z1m
Z21 Z22 Z2i Z2p Z2m
Zi1 Zi2 Zii Zip Zim
Zp1 Zp2 Zpi Zpp Zpm
Zm1 Zm2 Zmi Zmp Zmm
0
0
0
0
0
Ip(+)
0
0
Neglect prefault load currents, Fault current If =Ip(+)
abc abc abc abc
)(..........)0()( aIfZppEpEpabcabcabcabc
Voltage at fault bus is given by
Dr. Nitin Dhote, SVPCET, Nagpur
).......()(. bEpYfIfabcabcabc
Fault current in terms of admittance is given by
From a&b, we get
)1........()0(.).()(
)0()().(
))(..()0()(
1 abcabcabcabc
abcabcabcabc
abcabcabcabcabc
EpYfZppUEp
EpEpYfZppU
EpYfZppEpEp
)3........(.)0()(abcabcabcabc
IfZipEiEi
Putting values of Ep(+) in fault Current If , we get
)2.......()(.abcabcabc
EpYfIf
Putting values of If in voltage at any other bus Ei(+) , we get
Dr. Nitin Dhote, SVPCET, Nagpur
Dr. Nitin Dhote, SVPCET, Nagpur
Dr. Nitin Dhote, SVPCET, Nagpur
Ea = Ii (Zf +Zg)
)( ZgZfIi
EaZf
aa
Dr. Nitin Dhote, SVPCET, Nagpur
Since the fault is symmetrical all transfer impedance are the
same.
Dr. Nitin Dhote, SVPCET, Nagpur
into which is given by
The fault current at the faulted bus p using equation is given by
Dr. Nitin Dhote, SVPCET, Nagpur
=Assuming pre fault bus voltage as 1p.u.
=assuming abc sequence
Dr. Nitin Dhote, SVPCET, Nagpur
• Assuming all the power system element are balanced. The Zpp 012 have only the diagonal term
Dr. Nitin Dhote, SVPCET, Nagpur
Dr. Nitin Dhote, SVPCET, Nagpur
Dr. Nitin Dhote, SVPCET, Nagpur
Simply the voltage at faulted bus is given by
Dr. Nitin Dhote, SVPCET, Nagpur
similarly the voltage at any other bus during fault is given by
for i = 1, 2,--- n--
--Ei(+) 012 =
0
√3
0
--
Dr. Nitin Dhote, SVPCET, Nagpur
=
Dr. Nitin Dhote, SVPCET, Nagpur
• Let a single line to ground fault occurs on phase a at bus ‘p’ through impedance Zf as shown in fig. below
a cb
Zf
Dr. Nitin Dhote, SVPCET, Nagpur
• THE FAULT IMPEDANCE [ZF ABC] CAN BE CALCULATED BY USING
THE DEFINITION OF DRIVING POINT AND TRANSFER IMPEDANCE
a cb
Zf
Dr. Nitin Dhote, SVPCET, Nagpur
Zfaa = Zf = Ea/Ia
Since b & c are open ckt to ground.
Zfbb = Zf
cc= ∞
And all transfer impedances are zero.
Therefore
Zfabc=
To avoid the multiplication by infinity the solution should be carried out in form of Yf
abc
Zf
∞
∞
Dr. Nitin Dhote, SVPCET, Nagpur
1/ Zf 0 0 Yf 0 0
Yfabc = 0 0 0 = 0 0 0
0 0 0 0 0 0
where Yf = 1/Zf .
Yf012 =Ts
*t . Yfabc. Ts
= 1 1 1 Yf 0 0 1 1 1
1/√3 1 α α2 0 0 0 1/√3 1 α2 α
1 α2 α 0 0 0 1 α α2
Dr. Nitin Dhote, SVPCET, Nagpur
1 1 1 Yf Yf Yf Yf Yf YfYf
012= 1/3 1 α α2 0 0 0 = 1/3 Yf Yf Yf1 α2 α 0 0 0 Yf Yf Yf
1 1 1
Yf012= Yf/3 1 1 1
1 1 1
voltage at faulted bus P during the fault is given by
Ep(+)012 = [ U + Zpp012 .Yf012 ] -1 Ep (0) 012
Dr. Nitin Dhote, SVPCET, Nagpur
1+ (Yf/3)Zpp(0) (Yf/3)Zpp(0) (Yf/3)Zpp(0) -1 0
Ep(+)012= (Yf/3)Zpp(1) 1+(Yf/3)Zpp(1) (Yf/3)Zpp(1) √3
(Yf/3)Zpp(2) (Yf/3)Zpp(2) 1+(Yf/3)Zpp(2) 0
With solving we get
-Zpp(0)
Ep(+)012 = √3/(3Zf+Zpp(0)+Zpp(1)+Zpp(2)) 3Zf+Zpp(0)+Zpp(2)
-Zpp(2)
:. Zpp(1)= Zpp(2)
-Zpp(0)
Ep(+)012= √3/(3Zf+Zpp(0)+2Zpp(1)) 3Zf+Zpp(0)+Zpp(1)
-Zpp(1)
Dr. Nitin Dhote, SVPCET, Nagpur
If012 = [Yf
012 ].Ep(+)012
1 1 1 -Zpp(0)
If012 =Yf/3 1 1 1 * 1 √3/(3Zf+Zpp(0)+2Zpp(1)) 3Zf+Zpp(0)+Zpp(1)
1 1 1 -Zpp(1)
1
If012 = √3 / (3Zf+Zpp(0)+2Zpp(1)) 1
1
Voltage at any other buses during fault
Ei (+)012 = Ei(0)012 – Zip012 .If012
Fault current
Dr. Nitin Dhote, SVPCET, Nagpur
0 Zip(0) 0 0 1
Ei(+)012= √3 - 0 Zip(1) 0 * √3/(3Zf+Zpp(0)+2Zpp(1)) 1
0 0 0 Zip(2) 1
0 Zip(0)
= √3 - √3/(3Zf+Zpp(0)+2Zpp(1)) Zip(1)
0 Zip(2)
Zip(0)
= √3/(3Zf+Zpp(0)+2Zpp(1)) 1- Zip(1)
-Zip(2)
By pre multiplying [Ts] matrix with fault current If 012, voltage of fault bus Ep(+) 012 , voltage of any other bus Ei(+) 012, current and voltages in phase quantities a bc can be obtained.
Ep(+) abc =Ts* Ep(+) 012 , If abc =Ts* If 012 & Ei(+) abc =Ts* Ei(+) 012Dr. Nitin Dhote, SVPCET, Nagpur