The following are covered: the classification of faults, sources of fault currents, sequence impedance networks, the calculation of the fault currents for different types of shunt faults, the preparation of coordination studies and the inclusion of the different current time characteristics curves, damage curves/points and inrush (energization) currents.

1. Fault calculations & protective relays co-ordination studies:
Fault classifications: faults can be classified into symmetrical short circuit currents producing faults
and unsymmetrical ones, another method of classification is whether it is of the shunt or series types.
Shunt types faults can be further classified into: single line to ground, line to line, double line to
ground, three phase grounded or ungrounded. The series types can be an open phase on one line or 2
open phases. These faults can be bolted or faulty through a resistance (impedance).
Va = Va1 + Va2 + Va0
Vb = Vb1 + Vb2 + Vb0 = a2 Va1 + a Va2 + Va0
Vc = Vc1 + Vc2 + Vc0 = a Va1 + a2 Va2 + Va0
a = 1 120o = -.5+j .866 = cos 120 + j sin 120, a2 = 1 240o = -.5-j.866 , a2 + a + 1 = 0
a3 = 1, a = ej2π/3.
Va1 = (Va + a Vb + a2 Vc)/3, Va2 = (Va + a2 Vb + Vc)/3, Va0 = (Va + Vb + Vc)/3
Where Va is the voltage of phase A, Vb of phase B & Vc of phase C (in 3-phase power systems); Va1 is
the positive sequence voltage (of the symmetrical components) of bus A, Vb1 is the positive sequence
voltage (of the symmetrical components) of bus B, Vc1 is the positive sequence voltage (of the
symmetrical components) of bus C, Va2 is the negative sequence voltage (of the symmetrical
components) of bus A, Vb2 is the negative sequence voltage (of the symmetrical components) of bus B,
Vc2 is the negative sequence voltage (of the symmetrical components) of bus C, Va0 is the zero
sequence voltage (of the symmetrical components) of bus A, Vb0 is the zero sequence voltage (of the
symmetrical components) of bus B & Vc0 is the zero sequence voltage (of the symmetrical components)
of bus C. The same is applicable phase currents and symmetrical components of phase currents.
The data required to perform fault calculations: they are the positive/negative/zero sequence
impedance values of the system under study, to achieve this requirement the following data have to be
available: size, length and methods of laying all cables in the system under study, the overhead
conductors size, type and configuration, transformers KVA/impedance (with its base)/tap changer
limits and step size/impedance per tap step/primary and secondary voltage/method of neutral grounding
and pertinent data, for the induction or synchronous motors: the rating/sub-transient reactance/transient
reactance/speed/X to R ratio and the current limiting reactors ratings (if any).
The sources of fault current: they are synchronous generators, synchronous motors & condensers,
induction machines and electric utility systems. The fault current passes through two stages before
reaching steady-state, which are the sub-transient and transient states (of current), the first lasting about
6 cycles, the second for approximately 30 cycles after which the steady state is reached. The short
circuit currents are calculated for first cycle (post fault), after 5 or 6 cycles and after 30 cycles. The
impedance of the circuit elements are established, referred to the base values (to be consistent with
each other for combination in series, parallel and to convert them from wye to delta & vice versa). In
the per unit system there are four base quantities: base apparent power (MVAb), base voltage (KVb),
base current (Ab) and base impedance (ohmb). Base MVA is taken as 100 MVA or any other suitable
MVA rating of any equipment in the system under study and stays constant through out the study. KVb
is taken at a starting point in the study (arbitrary point) and at every transformer, its value is adjusted
according to the transformer ratio.
Per unit (pu) quantity = actual value/ base value.
Base current (Ab) = base MVA x 1000 / (3)0.5 base KV.
Base impedance (ohmb) = (base KV)2 / base MVA.
Per unit impedance = actual impedance (base MVA) / (base KV)2.
Per unit impedance for transformers =
percent impedance (base KVA) / [(KVA rating of transformer) (100)]
Per unit reactance for motors = per unit reactance (base KVA) / motor KVA rating, assuming the per
unit reactance is given based on the motor KVA rating.
Shunt types fault calculations:

2. -The positive and negative sequence impedances are equal for a static device (transmission lines and
transformers), the zero sequence is different as it includes the impedance of the return path through the
ground. For rotating machines, the +ve, -ve and 0 sequence impedances are, generally, unequal. Zero
sequence impedance is much smaller than +ve & -ve sequence ones. It is, always, assumed that the +ve
sequence current flows only in the +ve sequence network impedance producing +ve sequence voltage
drop across it; the –ve sequence current flows through the –ve sequence impedance producing a –ve
sequence voltage drop across it and, as well, the zero sequence current flows through the zero sequence
impedance (network).
-Three phase faults: the fault current = Pre-fault voltage divided by equivalent +ve sequence impedance
(or reduced +ve sequence network equivalent). The prevailing conditions for this type of faults are:
Va = Vb = Vc and Ia + Ib + Ic = 0.
-Single line to ground: assuming the fault on phase A, the following relations are valid: Va = 0, Ib = Ic
= 0 (assuming no-load conditions just prior to the fault instant).
Ia1 = Ea/(Z1 + Z2 + Z0), the fault current, Ia = 3Ia1, Va1 = Ea - Ia1 Z1, Va2 = - Ia2 Z2, Va0 =
- Ia0 Z0, Vb = Vb1 + Vb2 + Vb0, Vc = Vc1 + Vc2 + Vc0, Vab = - Vb, Vac = - Vc, Vbc = Vb -
Vc.
-Line to line: assuming the fault is between phases B & C, Ia = 0 (assuming no-load conditions just
prior to the fault instant), Ib + Ic = 0, Vb = Vc, Va1 = Va2, Ia1 = Ia2 = Ea/(Z1 + Z2), Va = Va1 + Va2 +
Va0, Vb = a2 Va1 + a Va2 + 0, the fault current (Ib) = Ib1 + Ib2 + Ib0 = a2 Ia1 + a Ia2 + 0 = 2(.866)Ia1.
-Double line to ground fault: assuming the fault is between phases B & C and the ground, Ia = 0, Vb =
Vc = 0, Va0 = Va1 = Va2, Ia1 + Ia2 = - Ia0, Ia1 = Ea / [Z1 + (Z0 Z2)/(Z0 + Z2)], Ia0 = - Va0 / Z0, Va
= 3Va1, fault current (In) = Ib + Ic = a²Ia1 + aIa2 + Iao + a²Ia2 + aIa1 + Iao = - Ia1 - Ia2 + 2 Iao =
3Ia0.
-Faults with neutral impedance and fault impedance:
For line to ground: Ia1 = Ea / [Z1 + Z2 + (Z0 + 3Zn) + 3Zf]; where Zn is the impedance in the
neutral path & Zf the fault impedance.
For line to line: Ia1 = Ea / [Z1 + (Z2 + Zf)]
For double line to ground:
Ia1 = Ea / { Z1 + [ Z2 (Zo + 3Zn + 3Zf) / (Z2 + Zo + 3Zf + 3Zn)]}.
Co-ordinaion studies: for the coordination study the following data are essential: the relays current-
time characteristics curves, fuses total clearing time-current & minimum melting time-current
characteristic curves, total available 3-phase short circuit current in the system, the 58% IEEE point for
transformers connected in delta - wye, damage curves for cables & transformers, inrush currents to

3. induction motors and their durations, inrush currents for transformers and durations, relays burden data,
instrument transformers saturation & excitation curves, accuracies, ratios and taps number & ratios
data.
Coordination curves:
Current Coordination curves: they are produced for protection against over-currents/ short circuits
and over-voltages. The co-ordination between the series protective devices, relays and fuses, is
important to achieve selectivity and the operation of the device closest to the fault while leaving the
upstream devices in the normal closed condition.
Fuses are defined by two curves for each fuse size and type, the minimum melting time of the link vs.
the current flowing through it and the total clearing time vs. the current. The o/c relays depending on its
design can have settings to choose the curve shape that would give the required relation between the
current flowing in the circuit and the operating time of the relay. A safety zone, usually, exists between
the two curves of two devices connected in series (to allow for the over travel, inaccuracies, etc. in the
operating time of the devices).
Other points or curves that may appear on such composite curves are inrush currents when transformers
are energized (12 times full load current for 0.1 second), inrush current for induction motors (for 10
seconds, 4 times full load), damage curves for cables, damage curves for transformers and the ANSI
58% point for delta - wye transformers. After obtaining all the curves for the relays and fuses used to
protect the system under study, construct the curves by starting with the most downstream load going
upstream and converting each curve to the base voltage level.
For low voltage circuit breakers each of the following is an independent adjustment: long time setting,
long time delay, short time pick-up setting, short time delay and instantaneous adjustment. Add the
other curves previously mentioned as deemed applicable. Show the total short circuit available for a
three-phase fault. Construct the 1-phase ground fault coordination curves for the system under study, on
a different log-log sheet. Show on the same sheet having the set of curves the applicable portion of the
single line diagram.
The second method used (and is more suitable for studies run by computers) is the impedance matrix
method that also provide for the voltage sensitivity studies (voltages at instant of fault appearing on the

4. unfaulty buses or nodes).
Impedance network models: building the impedance network model for a power system will assist in
calculating the fault studies including voltage sensitivity for radial systems, (the voltage on the buses
upstream the faulty one at time of fault). The coverage here will cover only the representation of
branches with no mutual coupling and will cover networks and radial systems (with or without the
effect of S.C.I.M. contribution). The 4 conditions that may exist in a system are: the presentation of an
impedance connecting between the reference node (bus) which can be the neutral or ground of a system
and a new bus, the connection between an old bus and a new one, the connection between an old bus
and the reference node and finally the connection between 2 old buses. Let's assume a network with
reference node 0, then equipment and lines go out radially from the reference node or bus 0 (which is
the neutral, the 3 phases are assumed to be balanced and a single phase out of the 3 phases is
represented here) to bus 1, bus 1 is connected to bus 2, bus 2 to 3 and, finally, 1 to 3.
Step 1: Zbus = [Z01]; bus 0 to 1
Z44 = Z01+Z01+Z12+Z23 - 2 [Z01] + Z13
Step 4a:the elimination of column and row 4 using Kron reduction method. The final Zbus matrix will
have the following elements:
Z11 = Z01 - [(Z01-Z01)(Z01-Z01)/Z44]
Z12 = Z01 - [(Z01-Z01)(Z01-Z01-Z12)/Z44]
Z13 = Z01 - [(Z01-Z01)(Z01-Z01-Z12-Z23)/Z44]
Z21 = Z01 - [(Z01-Z01-Z12) (Z01-Z01)/Z44]
Z22 = Z01 + Z12 - [(Z01-Z01-Z12)(Z01-Z01-Z12)/Z44]
Z23 = Z01 + Z12 - [(Z01-Z01-Z12) (Z01-Z01-Z12-Z23)/Z44]
Z31 = Z01 - [(Z01-Z01) (Z01-Z01-Z12-Z23)/Z44]
Z32 = Z01 +Z12 - [(Z01-Z01-Z12) (Z01-Z01-Z12-Z23)/Z44]
Z33 = Z01+Z12+Z23 -[(Z01-Z01-Z12-Z23)²/Z44]
The last possibility would be the presence of an equipment ( a generator or an induction motor that its
effect has to be taken while calculating fault currents) between bus 3 - for example - and node 0
(reference). The step 4 above is replaced by the following step, and then Step 4 will be performed on
the resultant matrix and it will become Step 5.
Step 4: from Step 3 above obtain the following matrix:

5. Z44 = Z01+Z12+Z23+Z03.
Step 4a: the elimination of column and row 4 using Kron elimination method mentioned above under
4a. Then Step 5 will become the addition of the impedance between buses 3 and 1 and this will add
columns 4 and row 4 to the matrix which are then eliminated (using Kron method) to yield the final
network impedance matrix with the final impedance elements that can be used in fault studies.
Grid power systems modelling: the data required to perform fault studies on grid systems are more or
less similar to those required for radial systems. The difference is in the degree of complexity of the
grid having more than one source, feeding the connected loads at the same time. The method best
suited for grids is the impedance network. The neutral is taken as the reference node (node 0) and then
the buses are numbered sequentially, each load bus, generator bus, motor bus (if applicable). For the
synchronous machines, the subtransient reactance is used to calculate the short circuit current during
the first cycle (momentary, close and latch ratings of breaking devices), the transient reactance to
calculate the short circuit current during the 2nd or 3rd cycle up to the eighth cycle (breaking devices
interrupting rating) and finally the synchronous reactance to calculate the short circuit current for
delayed tripping. For the induction motor and to calculate its contribution for the first few cycles, the
transient reactance is used in the calculation. For stationary components like overhead conductors,
underground cables, transformers, the impedance or reactance of the component is used. Parallel
circuits, lines connecting, lets say, bus 1 to bus 3 and bus 1 to bus 2 can be presented in this model.
Types of faults calculations: the purpose of fault calculations is to get the magnitude of the short circuit
current and the voltage on the buses, other than the faulty (in radial systems will be for those upstream
the faulty bus, in grid systems, it will be for the upstream as well as the downstream faulty bus). The
voltage sensitivity calculation in this section will only be limited to the two types of faults the most
probably will happen (single phase to ground) and the most severe (3 phase faults).
•• Radial systems:
For line to ground faults: (examples: failed cable or splice, broken insulator or insulator flashover,
reduced clearances, failed single phase transformers, etc.).
The calculations here will be given for 4-bus radial system (for one feeder, as during faults the feeder
with the fault is important and is part of the study the system, with the others either supplying the fault
with motor contribution or they are not part of the study - if there are no source of short circuit current).
Step 1: construct the impedance network (that will be also used with other types of faults for +ve, -ve
and zero sequence). The fault is assumed to be on bus 4.
X111 = XX011+XX121+XX231+XX341
Y24 = XX011+XX121
Y34 = XX011+XX121+XX231

6. where XX011 is the +ve sequence reactance between node 0 and bus 1, XX121 between bus 1 and 2,
XX231 between bus 2 and 3, XX341 between bus 3 and 4 (even if the feeder had more buses, the
calculation would stop here - as the voltage on the downstream buses will be equal to that of the faulty
as long as there is no motor contribution from the downstream buses), a similar negative sequence
impedance model can be constructed.
where XX112 = XX012 + XX122 + XX232 +XX342
Y242 = XX012 + XX122, Y342 = Y242 + XX232
For the zero sequence network, it’s a little bit tricky as the connections of the transformer and neutral
grounding method (if applicable) have to be taken into consideration when building the zero sequence
impedance matrix. If the transformer connection (it is assumed in this example system that the
transformer is connected between node 0 and bus 1) is such that for a single phase to ground, the fault
current will have no path to ground eg. wye-wye (isolated ground), delta-delta, wye (grounded)-delta,
wye (grounded)-wye then the short circuit (fault) current = 0. A zig-zag transformer connection can
replace a delta and it will provide the same phase angle between the primary and secondary windings
and will provide a neutral terminal. If the transformer is connected delta - wye (grounded) or wye
(grounded) - wye (grounded) the zero sequence impedance network matrix has to be built for each case
including the source (system ahead of the transformer) of short circuit capacity or the generator (ahead
of the transformer, if applicable). The relevant elements for the first transformer connection are: X110 =
XX010 + XX120 + 3(d) + XX230 + XX340
Y240 = XX010 + XX120, Y140 = XX010 + 3(d)
Y340 = XX120 + XX230 + XX010 + 3(d)
As the example, here, is a special condition that the supply of the short circuit current can supply
infinity amperes, then the equations for the grounded wye-grounded wye transformer connection will
be identical to the above condition.
XX010: is the zero sequence reactance between node 0 and bus 1.
XX120: is the reactance between 1 and 2, and so on.
d: is the reactance in the ground (fault) current path.
Step 2: From above the short circuit current for a single phase to ground fault on bus 4 can be
calculated and is equal to 3 (Vf)/(X111 + X112 + X110). For the voltage sensitivity calculations, only
voltage at bus 1 will be given here to correlate between the elements given in the above sequence
networks and the equations:
VS10 = -Y140 (Ifault)/3, VS11 = Vf - Y14 (If/3)
VS12 = - (If) (Y142)/3, VSAf = VS10 + VS11 +VS12
VSBr = VS10 + (-.5) (VS11) + (-.5) (VS12)
VSBi = (-.866) (VS11) + (.866) (VS12)
VSBf = [VSBr)² + (VSBi)²]
VSCr = VS10 + (-.5) (VS11) + (-.5) (VS12)
VSCi = .866 (VS11) + (-.866) (VS12); where VS10: zero sequence voltage on bus 1, VS11: +ve sequence

7. voltage, VS12 is the -ve sequence voltage, VSAf is the voltage on phase A of bus 1, VSBf is the voltage
on phase B, VSCf is the voltage on phase C.
For 3 phase faults:
The S.C. fault current = Vf/X111
The voltage on bus 1 = Vf (1-Y14/X111) in p.u.
The voltage on bus 2 = Vf (1-Y24/X111) in p.u.
The voltage on bus 3 = Vf (1-Y34/X111) in p.u.
For line to line faults:
The fault current = (.866)(2)(Vf/(X111+X112))
For line to line to ground:
Int = Intermediate value = Vf/[X111 + (X112) (X110)/(X112+X110)]
The fault current = 3 (Int) (X112)/(X112 + X110)
•• Grid or network systems:
The analysis hereafter will be based on a hypothetical system having the following components: a
synchronous generator between bus 1 and the reference node 0, other sources of fault currents are
between buses 2 and reference node and another one between bus 3 node 0. The last 2 sources can be
squirrel cage induction motors and the fault current is to be calculated for the first few cycles (i.e. if
there are generators in the circuit, their reactance will be the subtransient for the calculation of the fault
current during the first few cycles and the transient value for the next period in the short circuit
duration) with motor reactance equal the transient one (afterwhich the contribution of the SCIM is
reduced to 0 - in a few cycles from fault inception). The other components are lines between the
following buses: 1& 2, 2 &3, 3 & 4, 1 & 4, 2 & 4. The full set of equations to solve this network is
given in the “Example Programs” section.
The impedance matrix is calculated based on the process given above “Impedance network models”
and will have the following form:
The same format can be used to build the negative and zero sequence networks:
For line to ground fault on bus 3:
If = 3 (Vf)/(XXC33 + XXC332 + XXC330)
The voltage sensitivity calculations can proceed following the steps given under radial systems.
For line-line-line fault on bus 3:
If = 1/XXC33
For line to line and double line to ground faults: the equations given in the radial system can be
applied here.