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Industrial and Commercial Power SystemsTopic 5

FAULT CALCULATIONS

The University of New South Wales

School of Electrical Engineeringand Telecommunications

short-circuit faults in supply system.

very high fault current causing equipment damage, fire, injury, etc.

require protection systems to detect fault currents and to take remedial action.

need to establish methods of calculating fault current in any particular location of the electrical installation.

many types of fault in three phase systems

must know impedances of all system components

fault impedance itself may be non-zero, value =?

difficult to estimate accurately earth impedance

fault current contribution from machines etc.

DC offset in initial cycles of fault current

DC system faults also include inductance effects

Difficulties:

three phase (symmetrical) faults

three phase to earth fault

single phase to earth fault

phase to phase fault

phase to phase to earth fault

Possible fault types:

We consider only symmetrical 3-phase faults. Often, this type gives maximum fault current level at any location and represents worst case situations.

important parameter that designer of an electrical installation needs to know. Can be obtained from electricity distributor

prospective short-circuit current = current which would flow as a result of a bolted 3-phase fault.

"Bolted" refers to the phases being connected via a zero impedance connection.

In reality this is not physically possible, since there will always be a resistive component with a fault occurs.

Prospective s/c current and fault level (power):

Typical value at point of supply for 230/400V NSW distribution systems

Suburban residential areas: 10 kA Commercial and industrial areas: 25 kA

Knowing fault level: impedance of upstream circuit and devices (e.g.

transformers, conductors) can be derived.

The prospective fault current varies at different points in the supply

1

PER UNIT SYSTEM

normalize electrical quantities

express voltage, current, kVA and impedance as percentages (per unit values) of their base values

simplify calculations substantially.

transformer treated as a simple impedance (no need for ratio).

Per unit values:

need to specify two base values, and the other two base values can be derived.

usually, specify VB and SB

and thus:

normally, VB is taken as rated system voltage.

SB is arbitrarily specified (say 10 or 1 MVA), or use rating of a major system component, e.g. supply transformer.

Three-phase systems

calculations done on a single-phase basis using per unit phase impedances in the one-line diagram.

VB is line voltage and SB is three-phase kVA

and note:

Example:

Example:

2

EFFECTS AND REQUIREMENTS

over-current protection requirements

peak electromagnetic forces

thermal heating effects

maximum fault current (and minimum fault current)

time discrimination requirements of protection operation

touch voltage on earthed object (personnel safety)

Knowing fault levels enables determination of:

Sources of fault currents:

electrical utility supply grid system

any in-house generation systems operating at time of the fault

any motors operating within the system at time of the fault

any electrical storage elements in the system (e.g capacitors)

Fault impedance variation:

for the utility supply, source impedance is constant (a stiff source)

for in-house generators or motors, impedance is time-varying, depending on the time after the short circuit:sub-transient reactance (Xd”) (First cycle)transient reactance (Xd’) (1.5 to 4 cycles)synchronous reactance (Xs) (30 cycles)

Synchronous motors sustain fault current longer than induction motors

DC offset:

offset can increase initial current levels substantially

magnitude of DC offset level governed primarily by X/R ratio of faulted circuit

also dependent on angle on voltage waveform at which fault occurs

The physics behind is not easy!

The system X/R determines how rapidly the DC decays.

Faults that occur at a current zero crossing don't produce a DC offset

Those that occur at a current max or min produce the largest DC offset that circuit will produce.

The actual DC offset will range between those values and will be different in each phase.

3

FAULT CALCULATION METHODS

Assumptions:

Procedure: convert all impedances to per unit values

draw single line diagram of fault circuit,

all possible sources are modeled as ideal voltage source with their source impedance

simplify circuit and reduce it to

4

FAULTS IN DC SYSTEMS

Sources: DC generators, synchronous converters, DC motors, rectifiers, battery banks, UPS

only resistance elements determine steady-state fault current.

L/R time constants usually long enough that steady state fault current will not be reached before protection operates.

arc interruption is much more difficult for DC than for AC

5

FAULT CALCULATION DATAAND EXAMPLE

Example:

Required to find fault current at location A.

Use 20 MVA base.

Example (Students try in groups)

An 11kV, 10MVA, 50Hz, three-phase generator of 0.08 per unit reactance and negligible resistance, is connected to an 11/33kV, 12MVA transformer of 0.02+j0.05 per unit impedance. A line having a series impedance of 2+j4 ohms/phase is connected to the 33kV side of the transformer.

A symmetrical fault of zero impedance develops at the far end of the line. Calculate the fault current. How much energy would be dissipated in the system if the current remained at this value for 8 cycles?

Solution

11 kV10 MVAX = 8%

11/33 kV12 MVA

Z=0.02+j0.05 pu

(2+j4) Ω11 kV10 MVAX = 8%

11/33 kV12 MVA

Z=0.02+j0.05 pu

(2+j4) Ω

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Thank you