FAULT CALCULATION
Source Data:
Short Circuit Current (As per \data):
No. Description 33 kV BUS
1 Three phase Fault (Maximum) at 33 kV Bus in kA 20.03
3 Single phase Fault (Maximum) at 33 kV Bus in kA 3.000
Short Circuit MVA:
No. Description 33 kV BUS
1 Three phase Fault (Maximum) at 33 kV Bus in MVA 1144.0
3 Single phase Fault (Maximum) at 33 kV Bus in MVA 171.4730
1.3. 33kV Line Data:
UG-1
33 kV, 3C, 300 mm2
Positive Sequence impedance = 0.080+0.105i Ω/km
Negative Sequence impedance = 0.080+0.105i Ω/km
Zero Sequence impedance = 0.130+0.270i Ω/km
Cable Length ≈ 100m = 1.0 km
UG-2
Positive Sequence impedance = 0.080+0.105i Ω/km
Negative Sequence impedance = 0.080+0.105i Ω/km
Zero Sequence impedance = 0.130+0.270i Ω/km
Cable Length = 3.3337 km
OHL-1
Positive Sequence impedance = 0.2276+0.3378i Ω/km
Negative Sequence impedance = 0.2276+0.3378i Ω/km
Zero Sequence impedance = 0.4046+1.6290i Ω/km
Cable Length = 2.8 km
Transformers Data:
Transformer TR-1 TR-2
HV Side Voltage 33000 33000
LV Side Voltage 11500 11500
SC impedance Zsc % 12.53 12.52
X/R Ratio 24 24
Manual Calculation:
Per Unit Transformation
Ipu = ISC / Ib
Ib = MVAb/ KVb
Zb = (KVb)2 / MVAb Ω
Zpu = Zactual / Zb
Zpu new = Zpu old x [ (KVb old)2 / (KVb new)2 x (MVAb new) / (MVAb old) ]
Isc pu 1-ph = 3 x Io = 3 x E / (Z1pu + Z2 pu + Zo pu)
@ Source:
KVb = 33 kV & MVAb = 20 MVA
Zb = 332 / 20 = 54.45 Ω
Ib = 20000 / ( = 349.9 A
3-ph fault:
Z1 pu min = j 0.017472
1-ph fault:
Assume Z1 = Z2
Z1 pu = j 0.116686
ISC = 20030 A & Ib = 349.9 A
Ipu = 20030 / 349.9 = 57.22
Isc pu 3-ph = 3 x Io = 3 x E / (Z1pu + Z2pu + Zopu)
57.22 = 3 x 1 / (j 0.116686+ j 0.116686 + Zo pu)
∴ Zo pu min = j 0.18094
For Cables:
Cable-1
Z1 pu =Z2 pu = (0.08 + 0.105i) *1/ 54.45 = 0.001469 + j 0.001928
Zo pu = (0.130 + 0.270i) *1/ 54.45 = 0.0023875 + j 0.0049587
OHL -1
Z1 pu =Z2 pu = (0.2276+0.3378i)*2.8/54.45 =0.0117+0.01737i
Zo pu = (0.4046+1.6290i)*2.8/54.45 =0.0208+0.08376i
Cable-2
Z1 pu =Z2 pu = (0.08 + 0.105i) *3.337/ 54.45 = 0.004898 + j 0.006428
Zo pu = (0.130 + 0.270i)*3.337/ 54.45 = 0.007967+ j 0.016547
LINE-1 = Z C1 + Z OHL-1
Z1 pu =Z2 pu = 0.013169 + j 0.019298
Zo pu = 0.0231875 + j 0.088718
LINE-2 = Z C2
Z1 pu =Z2 pu = 0.004898 + j 0.006428
Zo pu = 0.007967 + j 0.016547
For Transformers:
TR-1&2:
Z1 pu = 0.1252 & X/R = 24
(R2+x2) = 0.12522,
R = 0.004929,
X = 0.118297
Assume Z1 pu=Z2 pu=Zo pu= 0.005212+j 0.12509
Per unit impedances summary refer to 20 MVA base:
Source
line 1& line 2
Item Max 3- Max 1-ph TR-1&2
ph Fault Fault L-1 L-2
Pos. seq.
impedance
j0.017472 j 0.116686 0.013169 + j 0.019298 0.004898 + j 0.006428 0.005212+j 0.12509
Neg. seq. Not
impedance Required
j 0.116686 0.013169 + j 0.019298 0.004898 + j 0.006428 0.005212+j 0.12509
Zero seq. Not
impedance Required
J 0.18094 0.0231875 + j 0.088718 0.007967 + j 0.016547 0.005212+j 0.12509
Max 3-ph Fault current @11.5 kV bus when both transformers working in
parallel:
For balance 3-ph fault the sequential circuit will be as in fig (1)
Z T pu = ZS1 min pu + (ZL1 1 pu + ZTR1 1 pu ) // ( ZL2 1 pu + ZTR2 1 pu )
= j 0.017472 +( 0.013169 + j 0.019298 + 0.005212+j 0.12509) //
(0.007967 + j 0.016547+ 0.005212+j 0.12509))
=0.0069408+j0.08634
= 0.08662 85.403°
IF pu = E / Z T pu (Assume the pre fault voltage equal to nominal voltage, E= 1pu)
∴IF pu = 1 / 0.07650 85.403°
= 11.5446 85.40° Lag.
@ 11.5 kV bus
Ibase = 20 / (1.732 x 11.5) = 1004 A
Isc = Ib x Ipu = 11.544x 1004 = 11590.8 -85.403° A
Isc 3-ph max = 11.59KA
Min 3-ph Fault current @11.5 kV bus when each Transformer working
separately (11 kV bus section open)
For balance 3-ph fault the sequential circuit will be as in fig (2)
Fig (2)
Z T pu = (ZS1 pu +ZLine-2 pu + ZTR2 1 pu)
= j 0.017472 +0.013169 + j 0.019298+ 0.005212+j 0.12509
= 0.018381+j0.16186
= 0.1629 83.5°
IF pu = E / Z T pu (Assume the pre fault voltage equal to nominal voltage , E= 1 pu)
∴IF pu = 1/0.1629 83.5
= 6.138 83.5° Lag.
@ 11.5 kV bus
Ibase = 20 / (1.732 x 11.5) =1004 A
Isc = Ib x Ipu = 6.138 x 1004 = 6163.2775 -85.5° A
Isc 3-ph min = 6.163 KA
Max 1-ph Fault current @11.5 kV bus when both transformers working in
parallel
For 1-ph fault the sequential circuit will be as in fig (3)
Z T pu = ZS1 min pu + (ZL1 1 pu + ZTR1 1 pu ) // ( ZL2 1 pu + ZTR2 1 pu ) +
ZS2 min pu + (ZL1 2 pu + ZTR1 2 pu ) // ( ZL2 2 pu + ZTR2 2 pu )+ ((ZTR1 o pu) // (ZTR2 o pu))
= j 0.017472 +( 0.013169 + j 0.019298 + 0.005212+j 0.12509) //
(0.007967 + j 0.016547+ 0.005212+j 0.12509)) + j 0.017472 +( 0.013169 + j 0.019298 +
0.005212+j 0.12509) // (0.007967 + j 0.016547+ 0.005212+j 0.12509)) +(0.005212+j 0.12509)//
(0.005212+j 0.12509)
= 0.01648+j0.23522
= 0.2358 86°
IF pu = E / Z T pu (Assume the pre fault voltage equal to nominal voltage, E= 1pu)
∴Io pu = 1 / 0.2358 86° = 4.24 86°Lag.
IF pu = 3 x Io pu = 3 x 4.24=12.722
= 12.722 87.965° Lag.
@ 11.5 kV bus
Ibase = 20 / (1.732 x 11.5) = 1004 A
Isc = Ib x Ipu = 12.722 x 1004 = 12773.2 - 86.0° A
Isc 1-ph max = 12.77 KA
Min 1-ph Fault current @11.5 kV bus when each Transformer working
separately (11 kV bus section open)
For 1-ph fault the sequential circuit will be as in fig (4)
Z T pu = [(j 0.017472 +0.013169 + j 0.019298+ 0.005212+j 0.12509)+ (j 0.017472 +0.013169 + j
0.019298+ 0.005212+j 0.12509)+ (0.005212+j 0.12509)]
= 0.04197+j 0.4488 = 0.4507 87.78°
IF pu = E / Z T pu (Assume the pre fault voltage equal to nominal voltage , E= 1pu)
∴Io pu = 1 / 0.2065 87.78° 2.218 84.65° Lag.
IF pu = 3 x Io pu = 3 x 2.2184 = 6.655 84.6° Lag.
@ 11.5 kV bus
Ibase = 20 / (1.732 x 11.5) = 1004 A
Isc = Ib x Ipu = 6.655 x 1004 = 6681.92 84.65° A
Isc 1-ph max = 6.68 K
