ELECTRICAL PROTECTION IV
Dr. Oriedi Akumu
Building 20-G01M
akumuao@tut.ac.za
Consultation Hours:
Monday: 7:00-10:00
Wednesday:7:00-12:00
Thursday:7:00-10:00
ELECTRICAL PROTECTION IV
LEARNING OUTCOMES
Understand the basic concept, definitions and be able to:
 Calculate symmetrical fault currents in power systems and
determine their distribution taking into account transient and subtransient component currents
 Calculate unsymmetrical fault currents in power systems and
determine their distribution taking into account transformer
connections
 Select current transformers, circuit breakers, fuses, over-current
relays and reclosers based on their characteristics, and evaluate
their performance
 Implement protective relaying schemes and techniques for unit and
line protection
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Text Book: Power System Analysis & Design
Authors: J. D. Glover, M. S. Sarma and T. J. Overbye
Assessment Type:
Semester Tests
Class Tests
Assignments
Design Project using Power World
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Electrical Power System
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Electrical Power System
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All plant & equipment necessary to:
Generate
Transmit
Distribute
Utilize electric power
 The equipment includes:
 Generators, transformers, transmission lines, loads,
switching devices, controls, protection, measuring
instruments and so on
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The requirements on such a system are:
Highest reliability standards
Minimum environmental impacts
Lowest operation cost
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Protection is the science, skill, and art of
applying and setting relays or fuses, or both,
to provide maximum sensitivity to faults and
undesirable conditions, but to avoid their
operation under all permissible or tolerable
conditions.
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Power System Protection
 Protection schemes are control systems that monitor the operation
of the power system and initiate appropriate corrective measures.
 Protection schemes do not prevent the occurrence of faults and
abnormal conditions.
 Protection design philosophy: Any equipment threatened with
damage by a sustained fault is to be automatically taken out of
service.
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Design Objectives
The basic design objectives of any protective scheme are to:
 Maintain dynamic stability.
 Prevent or minimize equipment damage.
 Minimize the equipment outage time.
 Minimize the system outage area.
 Minimize system voltage disturbances.
 Allow the continuous flow of power within the emergency
ratings of equipment on the system.
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Design Criteria of Relaying System
 Speed: for power system stability, reduction of damage & outage
time, and to prevent development of other faults
 Selectivity: determination of fault point & tripping only the nearest
CB
 Sensitivity: capability to reliably operate under actual conditions
that produce least operating tendency
 Reliability: trip at all times when required – dependability; not to
trip falsely – security
 Simplicity: use the minimum protective equipment necessary to
achieve the protection objectives
 Economy: about 5% cost of protected equipment
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Protection Application Concept
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Types of Faults and Abnormalities
 Short-circuit faults (3, 2, 2g, 1g)
 Open-circuit faults (open conductor)
 Complex faults (inter-circuit, cross-country, broken conductor)
 Inter-turn faults in windings
 Abnormalities
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Overload and excessive operating temperature
Real power deficit (underfrequency)
Power Swings
Power frequency overvoltage and undervoltage
Underexcitation of synchronous machines
Asynchronous operation of synchronous machines
Overfrequency
Mechanical defects (oil leaks, tap changer mechanism fault)
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Short-circuit type faults
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Short-circuit type faults with fault impedance
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Some causes of fault impedance
 Arc resistance
 Pole, tower or structure footing resistance to earth
 Resistance of things that may come into contact with a line e.g. tree,
crane etc
 Contact resistance where a conductor falls to the ground
***Fault impedance can significantly reduce the magnitude of
fault currents and should be a design consideration***
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Complex type faults
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Faults in windings
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Protective Relaying System Components
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Relaying System Design Steps
 Power flow studies – to pick all system components, including
protective devices, based on rated system quantities (voltages,
currents, power …)
 Short circuit analysis – to pick circuit breaking devices based on
maximum fault values
 Stability studies – to determine stability limits/critical clearing
times for system faults (for time setting of relays)
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Typical Power Circuit Breakers
 CBs are located such that each component of the power
system can be completely disconnected from the rest of the
system
 They must have sufficient capacity to carry momentarily the
max short-circuit current, and interrupt this current
 In cases where auto-reclosure is permitted they must be able
to withstand closing in on, and interrupting, a short circuit
Types of Circuit Breakers
Air
Oil
SF6
Vacuum
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Relay Types: Construction
 Electromagnetic attraction: An armature being attracted to the poles
of an electromagnet or a plunger being drawn into a solenoid
 Electromagnetic induction: a pivoted aluminium disc placed in two
alternating magnetic fields of the same frequency but displaced in
time and space. Torque produced by interaction of one mag. field with
current induced in the disc by the other
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 Solid state relays
 Use low power components
 Require independent power supplies
 Flexible, smaller, more accurate, settings more repeatable
 Characteristics can be shaped by adjusting logic elements
 Require less mounting space
 Not affected by vibration and dust
 Higher cost
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 Block diagram of microprocessor based relays
 Much higher precision and
more reliable and durable
 Improve the reliability and
power quality of electrical
power systems before, during
and after faults occur
 Capable of both digital and
analog I/O
 Higher cost but with more
functionality
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Relay Types: Operation Principle
 Magnitude relays: respond to the magnitude of input quantities (e.g
overcurrent relays)
 Directional relays: respond to the phase angle difference between
two quantities
 Ratio relays: respond to the ratio of two phasor quantities: ratio
can be a complex number or its magnitude (e.g impedance relays)
 Differential relays: respond to the magnitude of the algebraic sum
of two or more inputs
 Pilot relays: use communicated information from remote sites as
input signals
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MODULE 1: SYMETRICAL FAULTS
 Learning Outcome: Students will be able to explain the concepts
and methods for calculating symmetrical short circuit currents.
 Assessment Criteria:
 Effects of three-phase short circuits in series RL circuits and
unloaded synchronous generator is explained
 Calculation in p.u. and kA values of sub-transient fault currents
at the point of fault and other points of the power system is
explained
 Construction and use of bus impedance matrix are explained
 Selection procedure
demonstrated
of
circuit
breakers
and
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fuses
is
1 1: Short Circuit Conditions and Transients in Series R-L Circuits
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Example 1
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Solution
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1 2: Synchronous machine response to short-circuit conditions
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Example 2
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Solution
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Power System Three-phase Short Circuit Calculations
The following assumptions are made to compute subtransient fault
current for three-phase short circuits in power systems:
 Transformers are represented by their leakage reactances. Winding
resistances, shunt admittances, and delta/star phase shifts are
neglected.
 Transmission lines are represented by their equivalent series
reactances. Series resistance and shunt admittances are neglected.
 Synchronous machines are represented by constant voltage sources
behind subtransient reactances. Armature resistance, saliency and
saturation are neglected
 Non-rotating impedance loads are neglected
 Induction motors are either neglected (especially for small motors
rated less than 37.3kW), or represented in the same way as
synchronous machines.
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 Consider the following single-line diagram
 Assuming a 3-phase short circuit occurs at bus 1, the positive
sequence equivalent circuit becomes:
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 On closing the switch, the fault may be represented by two
opposing voltage sources;
 Using superposition, the above circuit results in:
Pre-fault condition
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 If VF is the pre-fault voltage at the fault, then the 2nd circuit represents
the system before the fault occurs, which implies I”F2 = 0. VF has no
effect and can be removed from 2nd circuit
 Hence the subtransient fault current from the 1st circuit: I”F = I”F1
 The contribution to the fault from generator and motor respectively
are: I”G = I”g1 +I”g2 =I”g1 +IL & I”M = I”m1 +I”m2 =I”m1 -IL
Where IL is the pre-fault generator current
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Example 3
The generator in the previous case is operating at rated MVA, 0.95 p.f.
lagging and at 5% above rated voltage when a bolted 3-phase short
circuit occurs at bus 1. Calculate the per unit values of:
a) subtransient fault current
b) subtransient generator and motor currents, neglecting prefault
current, and
c) subtransient generator and motor currents, including prefault
current
Solution
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1
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Bus Impedance Matrix
 The bus impedance, Zbus, is the inverse of the bus admittance, Ybus,
i.e.,
 Since we know that
then
and
 Zbus relates nodal current injections to the nodal voltages
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 In power flow studies, we work with Ybus. The reason being that the
power flow problem requires an iterative solution that can be made
very efficient when we use Ybus, due to the sparsity (lots of zeros) in
the matrix used in performing the iteration (the Jacobian matrix)
 In fault analysis, we use Zbus. The main reason is that Zbus quantities
characterize conditions when all current injections are zero except
one, corresponding to the faulted bus
 Once we have that one current injection, all bus voltages in the
network can be obtained using (4), from which all currents are then
calculated
 The Zbus is not sparse, but fault analysis does not require iterative
solutions
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Zbus Elements
 Assuming a 3 node system, we can write:
𝑉1 𝑍11
𝑉2 = 𝑍21
𝑉3 𝑍31
𝑍12
𝑍22
𝑍32
𝑍13
𝑍23
𝑍33
𝐼1
𝐼2
𝐼3
(5)
 Here the independent sources are all current sources, which
are the equivalent representation of the generator voltage
sources
 For example
𝑉2 = 𝑍21 𝐼1 + 𝑍22 𝐼2 +𝑍23 𝐼3
(6)
Hence driving point impedance becomes
𝑍22 =
𝑉2 − 𝑍21 𝐼1 − 𝑍23 𝐼3
𝐼2
(7)
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 If we set I1 and I3 to 0, that is, open any current sources at nodes
1 and 3 so that there are no sources there (but there may be
load impedances), (7) becomes:
𝑍22 =
𝑉2
𝐼2
𝐼1 =𝐼3 =0
(8)
 Z22 is the ratio of bus 2 voltage to the bus 2 current injection when
all sources are idled - this is the Thevenin impedance looking into
the network at bus 2
 Therefore, the diagonal elements of the Zbus are the Thevenin
equivalent impedances seen looking into the network at that bus
 An important attribute to building the Zbus for fault analysis is that
the generator subtransient reactances should be included
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Zbus Modifications
 It is easier to develop the Ybus. From that, one can invert it to
obtain the Zbus. This is only possible with small networks, due to
matrix inversion difficulties for large matrices
 For large networks, we must turn to the so-called Zbus building
algorithm, based on modification to an already existing Zbus
 Let’s assume that we have a 3-bus system with Ybus given by:
(9)
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 Assume the branch between buses 1 and 3 is numbered as
branch 3. Then let’s modify branch 3 by adding another circuit
between bus 1 and bus 3 having an admittance of Δy3.
 Add Δy3 to diagonal elements in positions (1,1) and (3,3)
 Subtract Δy3 from off-diagonal elements in positions (1,3) and
(3,1)
 The resulting matrix is
(10)
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 So the new Ybus is just the old Ybus with the addition of Δy3 in
four positions, as indicated by equation (10):
(11)
 Define a vector corresponding to the modification of branch 3
(connected between bus 1 and bus 3) as:
(12)
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 Then (11) becomes:
(13)
 In general, anytime we modify Ybus, then
(14)
 where ak is constructed according to:
 1 at position i and -1 at position j if we add or remove a branch
between buses i and j.
 1 at bus i if we add or remove a shunt at bus i
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 The inverse of (14) is found to be:
(15)
where bk is an n×1 vector given by:
(16)
and γ is a scalar given by
(17)
where
 μ is a scalar ( = Δy ) and
 ak is an n×1 vector.
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Example 4
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Fault calculations using Zbus
 Recall that
(18)
 Since (18) represents a set of linear equations, superposition holds,
and we may write:
(19)
 This says that the change in voltage at all buses V may be
computed if the change in injections at all buses I are known.
We can write eq. (19) in expanded form as:
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(20)
 Now consider a bolted short-circuit at bus k, where the prefault
voltage at bus k is Vf. Let the fault current be I’’f
 Since the fault is a short circuit, then ΔVk = -Vf.
 Also, since the fault current is out of bus k, then Δik = - I’’f.
Substituting these into eq. (20) results in:
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(21)
 Therefore
(22)
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 We observe from row k that:
(23)
 Solving (23) for I’’f results in:
(24)
 Substituting (24) into (22), we get:
(25)
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 Consider any bus, say bus j, with a pre-fault voltage of Vj. Then we
can compute the bus j voltage under the faulted condition as
(26)
 We know that
(27)
 Substitution of (27) into (26) results in:
(28)
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 Consider this simple system
 We can write that
 We can compute the subtransient current flowing from bus i to
bus j under the fault condition as:
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 This results in:
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Example 5
Solve problem in Example 3 using impedance matrix method
Per-unit admittance values
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𝑌𝑏𝑢𝑠 = −𝑗
9.9454
−3.2787
−3.2787
8.2787
𝑍𝑏𝑢𝑠 = +𝑗
0.11565
0.04580
0.04580
0.13893
𝑉𝑓
1.05 0°
𝐼𝑓 =
=
= −𝑗9.079 𝑝. 𝑢.
𝑍11 𝑗0.11565
"
𝑉1𝑓 = 𝑉1 −
𝑉2𝑓
𝑍11
0.11565
𝑉𝑓 = 1.05 −
∙ 1.05 = 0
𝑍11
0.11565
𝑍21
0.04580
= 𝑉2 −
𝑉 = 1.05 −
∙ 1.05 = 0.63
𝑍11 𝑓
0.11565
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