Quick Guide to Calculation of Fault Along a LineRolando C. Garrido

Sr. Staff Engr.Research & Development

Introduction

This article aims to guide the Meralco Engineers in calculating the short circuit current when a shunt fault

occurs along a line. Shunt faults include Three-Phase (3Ø) fault, Single Line-to-Ground (SLG) fault,

Double Line-to-Ground (DLG) fault and Line-to-Line (L-L) fault. However, since the maximum fault will

occur either during a 3Ø fault or a SLG fault, depending on the grounding of the system, only these two

types of faults are discussed in this article. Procedure for the calculation is discussed through an actual

example. Data required for the calculation are presented as well as where the values for these data could

be obtained. This guide will particularly help the Engineers connected with Sales Offices in determining

the available fault current at the tapping point of a customer. Application of this guide, however, is for

radial distribution system only.

Data Requirements

1. Source Impedance (Thevenin’s Equivalent Impedance of the bus where the circuit is connected), Zb

a. Positive-Sequence Impedance, Zb1

b. Negative-Sequence Impedance, Zb2 (normally equivalent to the pos.-seq.)

c. Zero-Sequence Impedance, Zb0

The source impedance may be obtained from the “Bus Fault Data Report” being maintained by the

System Planning Office.

2. Type and size of wire of the circuit

Type and size of wire may be obtained from the Single Line Diagram or from the AM/FM system.

3. Circuit configuration (e.g. horizontal X-arm, horizontal Alley-arm, vertical, etc.)

Circuit configuration may be determined through field survey. However, it is normal for our circuit to

have a configuration that varies along the line. In most cases, it is sufficient to assume a single type of

configuration for a particular fault calculation. Together with the size of wire, the line configuration is

necessary to determine the per unit impedance per km of the line which is available from Research &

Development.

4. Location of the fault point (kM away from the bus), L

Distance of the fault point from the substation may be approximated through survey or through a

scaled Single Line Diagram.

Example:

Calculate the Three-Phase (3Ø) and Single Line-to-Ground(SLG) fault along a circuit connected to

the 34.5 kV bus of Sta. Mesa Bank #1. The fault is located 3 kms. away from the bus. The circuit is

utilizing 336.4 MCM ACSR. Assume that the circuit configuration is that of L10 (horizontal).

Solution:

Step 1. Draw the Single Line Diagram of the circuit at the fault point

Step 2. Compute for the required impedances

From the “Bus Fault Data Report” :

@ MVAbase = 100 MVA

KVbase = 34.5 kV

Zb1 = Zb2 = (0.000992 + j0.107236) p.u. = 0.10724 / 89.47 ° p.u.

Zb0 = (0.008190 + j0.227705) p.u. = 0.22785 / 87.94 ° p.u.

Based on the Line Impedance program from R&D, a circuit with 336.4 MCM ACSR wire and a

configuration of L10 has a per unit (p.u.) impedance, per km of the line of:

Z1 = (0.0159752 + j0.0343810) p.u./km

Z0 = (0.0438571 + j0.1229010) p.u./km

At the 3 kM fault point, the total impedance of the line from the bus is

ZL1 = (0.0159752 + j0.0343810) x 3

= (0.0479256 + j0.103143) p.u.

= 0.11373 / 65.08 ° p.u.

ZL0 = (0.0438571 + j0.1229010) x 3

= (0.1315713 + j0.368703) p.u.

= 0.39148 / 70.36 ° p.u.

L10, 336.4 MCM ACSR3Ø, SLG

3 kMZb1, Zb2, Zb0

Step 3. Construct the equivalent sequence networks as seen at the fault point

(+) Sequence Network:

(-) Sequence Network:

(0) Sequence Network:

Note:

In short circuit calculation, Ea1 is always assumed to be

Ea1 = 1.0 / 0 ° p.u.

+

Ea1

-

Zb0 ZL0

+

Ea0’

-

Ia0

Ia1

Zb2 ZL2

Ia2

+

Ea2’

-

Zeq2

Ia2

+

Ea2’

-

Zeq0

Ia0

+

Ea0’

-

Zb1 ZL1

+

Ea1

-

Ia1

+

Ea1’

-

Zeq1

Ia1

+

Ea1’

-

+

Ea1

-

Step 4. Connect the sequence network involve as required for each particular type of fault. Solve

for the sequence currents (positive-, negative- and zero-sequence)

For a Three-Phase (3Ø) fault, only the positive sequence is involved.

If = Ia1

Zeq1 = Zb1 + ZL1

= (0.000992 + j0.107236) + (0.0479256 + j0.103143)

= 0.0489176 + j0.210379 p.u.

= 0.215991 / 76.91 ° p.u.

Base Current, Ib is

Therefore, the 3Ø fault current is

If = 4.62982 x 1.673 = 7.75 kA

Zeq1

Ia1

+

Ea1’

-

For a Single Line-to-Ground (SLG) fault, all the three sequence networks are involved and are

connected in series.

If = Ia1 + Ia2 + Ia0

= 3Ia1 = 3Ia2 = 3Ia0 ( because Ia1 = Ia2 = Ia0 at the fault point )

Normally, however, the fault impedance is assumed to be zero, unless you have a given value.

If = 3Ia1 = 3(0.95735 / -76.85 °)

= 2.87206 / -76.85 ° p.u.

If = 2.87206 x 1.673 = 4.806 kA

For a more detailed study of fault calculation, you may read the references provided below.

3Zf

Zeq1

Zeq2

Zeq0

+

Ea1

-

Ia1

Ia2

Ia0

Zf - Fault Impedance

Reference:

1. Analysis of Faulted Power Systems

Paul M. Anderson

2. Elements of Power System Analysis

William D. Stevenson, Jr.

3. Electrical Transmission and Distribution Reference Book

Westinghouse Electric Corporation