UNSYMMETRICAL FAULT CURRENT
This page will demonstrate calculations of two unbalanced fault currents , ie line-to-ground and line-to-line faults.
Line-to-ground fault is influenced by positive equence impedance Z1, negative sequence impedance Z2 and zero sequence impedance Z0 while that of Line-to-line fault only involves Z1 and Z2. For this example problem, Z1 is the impedance as calculated in previous page on symmetrical fault. All impedance values are in % on chosen 100,000kVA base.
For static plants (eg cable, transformer, source network), Z2 = Z1. In the case of rotating equipment (eg generator), Z2 = 0.73 Z1.
Z0 exists only at the point of fault.
| Plants | Positive sequence impedance, Z1 | Negative sequence impedance, Z2 | Zero sequence impedance, Z0 |
| Source impedance,Zs | 0 + j40 | 0 + j40 | - |
| 11kV cable impedance,Zcab11kv | 0.21 + j 0.24 | 0.21 + j 0.24 | - |
| Transformer Impedance, Ztx | 0 + j475 | 0 + j475 | - |
| Low Voltage cable Impedance,Zcab415v | 146.3 + j 153.3 | 146.3 + j 153.3 | 146.3 + j 153.3 |
| Total impedances | 146.51 + j668.54 | 146.51 + j668.54 | 146.3 + j 153.3 |
| Z1 + Z2 + Z0= | 439.32 + j2158.92 = 2203 /_78.5° % |
| Z1 + Z2 = | (146.51 + j668.54 ) + (146.51 + j668.54) = 293.02 + j 1336.8 =1368.5 /_77.6 ° % |
| Z1 = | 146.51 + j668.54 = 684.4 /_77.6 ° % |
I, Current due to 100,000kVA base at fault point (ie 415V) = 100,000 kVA/(1.732 x 0.415 V) = 139,120.5 A
If (line to earth) = 3 I x 100/ (Z1 + Z2 + Z3) = 3 x 139,120.5 A x 100/ 2203 % = 18.9 kA
If (line to line) = Root (3) x I x 100/ (Z1 + Z2) = Root (3) x 139,120.5 A x 100/ 1368.5 % = 17.6 kA
If (symmetrical) = I x 100/ Z1 = 139,120.5 A x 100/ 684.4 % = 20.3 kA (exactly similar to that calculated in symmetrical fault calculation's page)
(1) Distribution of currents throughout the complete network (case: Line-to-ground fault).
LINE CURRENTS IN SECONDARY OF TRANSFORMER
Line-to-ground fault on Line R, IR = 18,900 A
Sequence fault current of line R:
Positive sequence current, I1 = 1/3 ( 18,900) = 6300A
Negative sequence current, I2 = 1/3 ( 18,900) = 6300A
Zero sequence current, I0 = 1/3 (18,900) = 6300A
Fault currents on Line R
IR = I1 + I2 + I0 = 18,900 A (proved !)
Fault currents on Line Y
IY = I1/_240 + I2/_120 + I0 =6300 /_240 + 6300/_120 + 6300 = 0 A (proved !)
Fault currents on Line B
IB = I1/_120 + I2/_240 + I0 =6300 /_120 + 6300/_240 + 6300 = 0 A (proved !)
LINE CURRENTS IN PRIMARY OF TRANSFORMER
turns per phase ratio brought about by delta/star transformation = 240 V/(1.732 x 11000 V)= 0.0126
Irr = IY - IB = -0.0126 (Iy - Ib) = - 0.0126(0 - 0) =0 A
Iyy = IB - IR = - 0.0126 (Ib - Ir) = - 0.0126 (0 - 18,900 A) = 238 A
Ibb =IR - IY = - 0.0126 (Ir - Iy) = - 0.0126 (18,900 A - 0) = - 238 A
negative sign before turn per phase ratio means reversal of line current brought about by cascade delta/star transformation
(2) Distribution of currents throughout the complete network (case: Line-to-line fault).
LINE CURRENTS IN SECONDARY OF TRANSFORMER
Line-to-line fault on Line R, IR = 17,600 A
Sequence fault current of line R:
Positive sequence current, I1 = 1/3 (IR + IY /_120° + IB /_240°) = 1/3 (17,600 - 17,600 /_120° + 0 ) = 8800 - j15242
Negative sequence current, I2 = 1/3 (IR + IY /_240° + IB /_120°) = 1/3 (17,600 - 17,600 /_240° + 0 ) = 8800 + j5080.7
Note that line-line fault is influenced by positive and negative sequence only.
Therefore, total line-line fault current = I1 + I2 = (8800 - j15242) + (8800 + j5080.7) =17,600 A (proved ! )
Fault currents on Line R
IR = 17,600 A
Fault currents on Line Y
IY = - 17,600 A
Fault currents on Line B
IB =0 A
LINE CURRENTS IN PRIMARY OF TRANSFORMER
turns per phase ratio brought about by delta/star transformation = 240 V/(1.732 x 11000 V)= 0.0126
Irr = IY - IB = -0.0126 (Iy - Ib) = - 0.0126(-17,600 A - 0) = 221.8 A
Iyy = IB - IR = - 0.0126 (Ib - Ir) = - 0.0126 (0 - 17,600 A) = 221.8 A
Ibb =IR - IY = - 0.0126 (Ir - Iy) = - 0.0126 (17,600 A- 0) = - 221.8 A
