1. POWER SYSTEM ENGINEERING
EET 301/4
EET301 POWER SYSTEM ENGINEERING

2. PROTECTION IN POWER SYSTEM
CHAPTER 7
PROTECTION IN POWER
SYSTEM
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3. On completion of this lesson, a student should be able to:
Ability to explain and evaluate symmetrical fault and protection in power system
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4. TOPIC OUTLINE 7.1 Introduction
7.2 Basic Components of Protection System
7.3 The Concept of Protection System
7.4 Fuse
7.5 Computer Relaying
7.6 Zone of Protection
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5. INTRODUCTION
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6. 7.1 INTRODUCTION
The primary function of a protection system in an electrical power network is to ensure the continuity of the electricity supply.
To achieve this, the protection system must be able to:
Detect the abnormal condition in an electrical circuit or piece of equipment
Identify the location of the fault
Isolate the faulty section of the circuit or the faulty equipment only without the electricity supply to the rest of the network.
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7. INTRODUCTION
Requirements To Protective System Provide For Basis To Design Criteria
Reliability: Operate dependably when fault conditions occur, even after remaining idle for months or years. Failure to do so may result in costly damages.
Speed: Operate rapidly to minimize fault duration and equipment damage. Any intentional time delays should be precise.
Selectivity: A relay system should provide maximum possible service continuity with minimum system disconnection.
Simplicity: Minimize protection equipment and circuitry.
Economy: Provide maximum protection at minimum cost.
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8. BASIC COMPONENTS OF PROTECTION SYSTEM
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9. 7.2 BASIC COMPONENTS Protection systems have 3 basic components:
Instrument transformers
Protective Relays
Circuit breakers
A simple overcurrent protection schematic with:
(1) one type of instrument transformer—the current transformer (CT),
(2) An overcurrent relay (OC), and
(3) a circuit breaker (CB) for a single-phase line.
Figure 1: Overrcurrent Protection Schematic
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10. EET 102 ELECTRIC CIRCUIT II

11. BASIC COMPONENTS 7.2.1 Instruments Transformer
In general, there are 2 basic types of instruments transformers:
Current Transformer (CT)
Voltage Transformer (Potential Transformer) (VT)
Figure 2 shows a schematic representation for the VT and CT. The transformer primary is connected to or into the power system and is insulated for the power system voltage.
The VT reduces the primary voltage and the CT reduces the primary current to much lower, standardized levels suitable for operation of relays.
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12. BASIC COMPONENTS Figure 2: VT and CT Schematic
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13. BASIC COMPONENTS Currents Transformer (CT)
For small currents at low voltage, trip coils and ammeter can be connected directly into the current, for example in series with the load.
However, series connection is not practical for HV systems and for larger currents, for example 100A and above.
Current transformers are used as a link between the main circuit and the protection or measuring equipment.
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14. BASIC COMPONENTS
The function of the CT is to reproduce in its secondary winding a current I’ that is proportional to the primary current I.
The CT converts primary currents in the kilo ampere range to secondary currents in the 0-5 ampere range for convenience of measurement, with the following advantages.
Safety: Instrument transformers provide electrical isolation from the power system so that personnel working with relays will work in a safer environment.
Economy: Lower-level relay inputs enable relays to be smaller, simpler and less expensive.
Accuracy: Instrument transformers accurately reproduce power system currents and voltages over wide operating ranges.
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15. BASIC COMPONENTS Currents transformers are divided into 2 main types:-
Protection currents transformer
Measuring current transformer
The main parameters that define a current transformer are:-
Rated Primary Current: The maximum primary current that the transformer has been designed, for example the maximum current at which it can still comply with the relevant standard specifications.
Rated Secondary Current: The maximum secondary current that transformer has been designed for. The values are usually 1 A or 5 A .
Burden (in VA):The maximum load which can be connected to secondary of the transformer.
Accuracy class: The designation given to a current transformer that complies with the error limits specified in the standards specification.
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16. BASIC COMPONENTS Accuracy class
In BS 3938: 1973, measuring CTs are divided into 5P of 10P. A CT marked “5P20” indicates that this is a protection CT with composite error of 5% which is applicable for currents up to 20 times the rated current.
Special purpose CT are designated as Class X.
Example:
A CT with a specification of 800/5A, 15 VA, Class 0.5, means that the rated primary current is 800A, the secondary current is 5A (when primary is 800A), the burden is 15 VA, and it is a measuring CTs with a current ratio error of 5% at the rated current.
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17. BASIC COMPONENTS
Currents transformer used in distribution system have a single-turn primary windings and multi-turn secondary windings
The 2 most common types of current transformer are:
The primary winding is a straight copper bar (round or rectangular) the secondary windings is wound round this bar.
A round iron core which the secondary winding is wound. Also known as a ring CTs. This types usually used extensively in the low voltage systems.
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18. CURRENT TRANSFORMER
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19. BASIC COMPONENTS Current Ratios Table 1: Standard CT Ratios 50:5 100:5
150:5
200:5
250:5
300:5
400:5
450:5
500:5
600:5
800:5
900:5
1000:5
1200:5
1500:5
1600:5
2000:5
2400:5
2500:5
3000:5
3200:5
4000:5
5000:5
6000:5
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20. BASIC COMPONENTS
Ideally, the CT secondary is connected to a current- sensing device with zero impedance, such that the entire CTs secondary current flows through the sensing device.
In practice, the secondary current divides, with most flowing through the low-impedance sensing device and some flowing through the CT shunt excitation impedance.
CT excitation impedance is kept high in order to minimize excitation current.
An approximate equivalent circuit of a CT is shown in Figure 3, where:-
Z' = CT secondary leakage impedance
Xe = (Saturable) CT excitation reactance
ZB = Impedance of terminating device (relay, including leads)
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21. BASIC COMPONENTS Figure 3: Current Transformer Equivalent Circuits
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22. BASIC COMPONENTS Determine Current Transformer Performance
The following procedure can be used to determine CT performance.
Step 1 : Assume a CT secondary output current I’
Step 2: Compute E’=(Z’+ZB) I’
Step 3: Using E’, Find Ie from the excitation curve
Step 4: Compute I = n(I’+ Ie) , note: n=CTs ratio
Step 5: Repeat step 1-4 for different value of I’, then plot I’ versus I
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23. BASIC COMPONENTS CT Error
CT error is the percentage difference between
(I’+ Ie) and I’.
Where,
Ie=CT Secondary excitation current
I’=CT Secondary output current
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24. EXAMPLE 7.1
Evaluate the performance of the multi-ratio CT in Figure 3 with a 100:5 CT ratio, for the following secondary output currents and burdens:
I’ = 5 A and ZB = 0.5 ohm.
I’ = 8 A and ZB = 0.8 ohm.
I’ = 15 A and ZB = 1.5 ohm.
Figure 3: Current Transformer Equivalent Circuits
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25. EXAMPLE 7.1 Figure 4: Excitation curve for a multi-ratio bushing CT
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26. SOLUTION (a) STEP 1: I’ = 5 A STEP 2: From Figure 4,
E’ = (Z’ + ZB)I’ = ( )(5) = 2.91 V
STEP 3: From Figure 4,
Ie = 0.25 A
STEP 4: I = n (I’ + Ie) = (100/5) (5+0.25) = 105 A
CT error = 0.25/5.25 x 100% = 4.8%.
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27. SOLUTION Figure 4: Excitation curve for a multi-ratio bushing CT
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28. SOLUTION (b) STEP 1: I’ = 8 A STEP 2: From Figure 4,
E’ = (Z’ + ZB)I’ = ( )(8) = 7.06 V
STEP 3: From Figure 4,
Ie = 0.4 A
STEP 4: I = n (I’ + Ie) = (100/5) (8+0.4) = 168 A
CT error = 0.4/8.4 x 100% = 4.8%.
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29. SOLUTION Figure 4: Excitation curve for a multi-ratio bushing CT
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30. SOLUTION (c) STEP 1: I’ = 15 A STEP 2: From Figure 4,
E’ = (Z’ + ZB)I’ = ( )(15) = V
STEP 3: From Figure 4,
Ie = 5 A
STEP 4: I = n (I’ + Ie) = (100/5) (15+5) = 400 A
CT error = 20/35 x 100% = 57.1%.
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31. SOLUTION Figure 4: Excitation curve for a multi-ratio bushing CT
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32. SOLUTION
Note that for the 15 A secondary current in (c), high CT saturation causes a large CT error of 57.1%.
Standard practice is to select a CT ratio to give little less than 5-A secondary output current at maximum normal load.
From (a), the 100:5 CT ratio and 0.5 ohm burden are suitable for a maximum primary load current of about 100 A.
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33. EXAMPLE 7.2
An overcurrent relay set to operate at 8 A is connected to the multiratio CT with a 100:5 CT ratio. Will the relay detect a 200 A primary fault current if the burden ZB is
0.8 ohm
3 ohm 
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34. BASIC COMPONENTS Voltage Transformer (VT)
It is not practical to manufacture meters and indicating lamps for direct operation on 3.3 kV, 6.6 kV, 22 kV or higher voltages.
Voltage transformer that steps down the high voltages to working voltages for example 120 V, 240 V and 415 V are used.
The are 2 types of voltage transformer :-
Protection transformer
Measuring transformer
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35. BASIC COMPONENTS
The variation of voltage is a electricity network usually very small (5%) so the operating range of a measuring voltage transformer is small compare with that of a measuring.
However, if a fault occurs, the voltage drops to a very low value. Protection VT must be able to operate over a wide range from as low as 5% to 190% of the rated voltage.
(Alternatively, a separate battery power supply unit is used for supply power to the protection system/gear – the voltage is normally 30V d.c)
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36. BASIC COMPONENTS
There are 2 types of construction for voltage transformer:-
Electromagnetic transformers with primary and secondary windings. It is impregnated with synthetic resin. Normally used up to 66 kV.
Capacitor VT which uses two capacitors in a voltage divider arrangement. The voltage across one of the capacitor connected to the primary of an auxiliary of an auxiliary wound-type transformer.
The required voltage for the protection or measuring device is taken from the secondary of this auxiliary transformer.
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37. VOLTAGE TRANSFORMER
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38. BASIC COMPONENTS 7.2.2 Protective Relays
The function of the relay is to discriminate between normal operation and fault conditions.
The OC relay in Figure 1 has an operating coil which is connected to the CT secondary winding, and a set of contacts.
When |I’| exceeds a specified "pickup" value, the operating coil causes the normally open contacts to close.
When the relay contacts close, the trip coil of the circuit breaker is energized, which then causes the circuit breaker to open.
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39. BASIC COMPONENTS Class of Measuring Relays
Current relay: Operate at predetermined threshold value of current.
Voltage relay: Operate at predetermined threshold value of voltage.
Power relay: Operate at predetermined threshold value of power.
Directional relay:
Alternating current: Operate according to the phase relationship between alternating quantities.
Direct current: Operate according to the direction of the current and are usually of the permanent-magnetic, moving-coil pattern
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40. BASIC COMPONENTS
Frequency relays: Operate at a predetermined frequency. These include over frequency and under frequency relays.
Temperature relays: Operate at a predetermined temperature in the protected component.
Differential relays: Operate according to the scalar or vectorial difference between two quantities such as current, voltage, etc.
Distance relays: Operate according to the "distance" between the relay's current transformer and the fault. The "distance" is measured in terms of resistance, reactance, or impedance.
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41. BASIC COMPONENTS Electromechanical and Induction Disc Relays
Two types of electromagnetic attraction type of relay are:
The plunger unit.
It has cylindrical coils with an external magnetic structure and a center plunger. The plunger moves upward to operate a set of contacts when the current or voltage applied to the coil exceeds a certain value.
The clapper unit.
It has a U-shaped magnetic frame with a movable armature across the open end. The armature is hinged at one side and spring-restrained at the other.
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42. BASIC COMPONENTS
When the electrical coil is energized, the armature moves toward the magnetic core, opening or closing a set of contacts with a torque proportional to the square of the coil current.
Induction disc relay
Induction disc units are based on the watt-hour meter design and use the same operating principles. They operate by torque resulting from the interaction of fluxes produced by an electromagnet with those from induced currents in the plane of a rotatable aluminum disc.
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43. Figure 7: Induction Disc Type Relay Unit.
BASIC COMPONENTS IS I
Magnet plugs
Keeper
Main coil
Disc
Electromagnet ф
.фL
.фR
Figure 5: Electromagnet Plunger type 1 2 3 4 5 6
magnetic core
armature
relay coil
spring
moving contact Bridge
normally open relay contact
Figure 7: Induction Disc Type Relay Unit.
Figure 6: Electromagnet clapper type
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44. Induction Disc Relay This relay Inverse Definite Minimum Time (IDMT).
The operating characteristic of a standard IDMT relay is defined as:
TMS – time multiplier setting
PSM – plug setting multiplier
I – measured current
Is – relay setting current
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45. Induction Disc Relay Example:
Determine the time of operation of a 1 A, 3 s overcurrent relay having a Plug Setting of 125% and a Time Multiplier of 0.6. The supplying CT is rated 400:1 A and the fault current is 4000 A.
Solution
the relay coil current for the fault
The nominal relay coil current
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46. Induction Disc Relay The plug setting multiplier the time of operation
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47. BASIC COMPONENTS 7.2.3 Circuit Breaker (CB)
CB is a reusable device that may be rest after operation.
It set to protect circuit and equipment from over current resulting from overloads and short circuits.
Electrically, CB connected in series with the load and carry load current under normal condition.
Have three operation condition ON (Contact closed) allowing, load current to flow through breaker, OFF (Contact open) breaker contact open , circuit is deernergized and TRIPPED ( Contact open need reset) when overcurrent and overload occur or by the operation of the shunt trip on under voltage.
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48. Figure 8: Circuit breaker trip mechanisms.
BASIC COMPONENTS
Bimetal
Current in
Latch
Spring
Current out
Normal
Tripped
Circuit breaker contacts
A. Thermal
Normal
Current out
Current in
Latch
Spring
Armature
Tripped
Circuit breaker contacts
Electromagnetic coil
Electromagnetic coil
B. Magnetic
Figure 8: Circuit breaker trip mechanisms.
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49. THE CONCEPT OF PROTECTION SYSTEM
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50. 7.3.1 TRANSFORMER PROTECTION
A number of fault conditions can arise within a power transformer.
Earth fault
Core faults
Inter turn faults
Phase to phase faults
Tank fault
In addition to the fault condition cause stresses on the transformer, such as overloading, system faults , over voltages and under frequency operation
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51. TRANSFORMER PROTECTION
High impedance relay I
im1
im2
im2 – im1
PROTECTED APPARATUS
Relay
Basic Differential Connections.
Restricted Ground Fault Protection for a Y winding.
Main scheme use for transformer. The main principle , the current at each side of protected apparatus for each phase are compared in differential relay.
Any difference current will operate the relay.
The sum of the phase currents is balanced against the neutral current, hence the relay will not respond to fault outside the winding.
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52. TRANSFORMER PROTECTION
Number of considerations should be dealt with in applying differential protection including:- I
im1
im2
im2 – im1
PROTECTED APPARATUS
Relay
Transformer ratio
The CT should have rating to match the rated currents of the transformer winding which they are applied.
The current transformer should be connected in Y for ∆ winding and ∆ for Y winding ( see next figure)
Allowance should be made for taping changing by providing restraining coil.
Figure 9: Basic Differential Connections.
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53. TRANSFORMER PROTECTION
C. T.’s
Circuit Breaker
Power transformer
Phase A
Phase B
Phase C R o A B C
Relay
Example of Differential Protection of a Δ/Y transformer.
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54. EXAMPLE 7.3
Consider a Δ/Y connected, 20 MVA, 33/11 kV transformer with differential protection applied, for the current transformer ratios shown in Figure 10. Calculate the relay currents on full load. Find the minimum relay current setting to allow 125 percent overload.
Figure 10
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55. SOLUTION The primary line current is given by
The secondary line current is
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56. SOLUTION
The C.T. current on the primary side is thus The C.T. current in the secondary side is Note that we multiply by to obtain the values on the line side of the Δ connected C.T.’s. The relay current on normal load is therefore With 1.25 overload ratio, the relay setting should be
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57. TRANSFORMER PROTECTION
Buchholz Protection
it is common practice in transformer protection to employ gas-actuated relays for alarm and tripping.
One such a relay is the Buchholz relay. Faults within a transformer will result in heating and decomposing of the oil in the transformer tank.
The decomposition produces gases such as hydrogen, carbon monoxide, and light hydrocarbons, which are released slowly for minor faults and rapidly for severe arcing faults.
In the gas activated relay named after its inventor, this phenomenon is utilized.
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58. TRANSFORMER PROTECTION
The relay is connected into the pipe leading to the conservator tank.
As the gas accumulates, the oil level falls and a float F is lowered and operates a mercury switch to sound an alarm.
Sampling the gas and performing a chemical analysis provide a means for classifying the types of fault.
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59. TRANSFORMER PROTECTION
Buchholz protection provides an alarm for a number of fault conditions including:
Inter turn faults or winding faults involving only lower power levels.
Core hot spots due to short circuits on the lamination insulation.
Faulty joints.
Core bolt insulation failure.
Major faults resulting in tripping include severe earth or inter phase winding faults and loss of oil through leakage.
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60. 7.3.2 BUSBAR PROTECTION
Bus bars are an essential link in the electric power system, and short circuits in their zone have to be interrupted in the shortest possible time.
In distribution systems (6-20 kV) with supply through transformers, overcurrent time-relays provide an easy protection mechanism.
The relays interrupt the supply to the bus bars if one or more supplies conducting a fault current.
Power direction relays are used in the transformer supply side to respond to a fault at the bus bars.
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61. BUSBAR PROTECTION
Differential protection is the most reliable method of protection for bus bars.
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62. 7.3.3 GENERATOR PROTECTION
There are a number of abnormal conditions that may occur with rotating equipment, including:
Faults in the windings.
Loss of excitation.
Motoring of generators.
Overload.
Overheating.
Overspeed.
Unbalanced operation.
Out-of-step operation.
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63. GENERATOR PROTECTION
Various types of faults that may occur in the insulation system of a generator's windings.
The faults shown are identified as:
1. Interphase short circuit.
2. Interturn fault.
3. Stator earth fault.
4. Rotor earth fault.
5. Interturn fault in rotor.
Note that A denotes the insulation of individual windings and B denotes the stator core.
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64. GENERATOR PROTECTION
Allocation of Protective Devices for the Stator, Rotor and Prime Mover.
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65. 7.3.4 TRANSMISSION LINE PROTECTION
It is necessary to provide the desired selectivity such that relay operation result in the least service interruption while isolating the fault.
The excessive currents accompanying a fault are the basis of over current protection scheme.
In transmission line protection system, it is necessary to provide the desired selectivity such that relay operation result in the least service interruption while isolating the fault.
In transmission line protection scheme, relay coordination is MUST.
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66. TRANSMISSION LINE PROTECTION
Three methods of relay grading:
Current Grading.
Time Grading
Inverse Time Over Current Relaying
CURRENT GRADING
Operates at suitable graded current
Fault current are higher near the source.
Relays are set that decreases as distance from the source is increased.
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67. TRANSMISSION LINE PROTECTION
Time Grading
Ensure the breaker nearest to the fault open first, by choosing an appropriate time setting for each of the relay.
Time setting increase as the relay get closer to the source.
Inverse Time Over Current Relaying
Evolved because of the limitation imposed by the use either current or time alone.
Time of operation is inversely proportional to the fault current level. Relays type CO-7 is in common use.
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68. TRANSMISSION LINE PROTECTION
Inverse Time Over Current Relaying
The operating time over current relay varies with the current magnitude.
There are two setting for this type of relay:
Pick-up current is determined by adjusted current coil tap setting (C.T.S).
The pick-up current is the current that causes the relay to operate and close the contacts.
Time dial setting (or time multiplier setting) refers to the reset position of moving contact, and it is varies the time of operation at a given tap setting and current magnitude.
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69. TRANSMISSION LINE PROTECTION
Radial System Protection
Many radial systems are protected by time delay overcurrent relays.
Adjustable time delays can be selected such that the breaker closest to the fault opens, while other upstream breakers with larger time delays remain closed.
That is, relays can be coordinated to operated in sequence so as to interrupt minimum load during faults.
Successful relay coordination is obtained when fault currents are much larger than normal load currents
Also, coordination of overcurrent relays usually limits the maximum number of breakers in a radial system to five or less, otherwise the relay closet to the source may have an excessive time delay
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70. TRANSMISSION LINE PROTECTION
Consider a fault at P1 to the right of breaker B3 for the radial system of Figure 10.
We want breaker B3 to open while B2 and B1 remains closed.
Under these condition, only load L3 is interrupted
We could select a longer time delay for the relay at B2 so that B3 operates first
Thus, for any fault to the right of B3, B3 provides primary protection
If B3 fails to open, B2 will open after time delay, thus providing backup protection
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Figure 10

71. TRANSMISSION LINE PROTECTION
Similarly, consider a fault at P2, we want B2 to open while B1 remains closed
Under these conditions, loads L2 and L3 are interrupted
Since the fault is closer to the source, the fault current will be larger than for the previous fault considered
We also select the B1 relay with a longer time delay than B2, so that B2 opens first.
Thus, B2 provides primary protection for faults between B2 and B3, as well as back up protection for faults to the right of B3.
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Figure 10

72. TRANSMISSION LINE PROTECTION
Similarly, B1 provides primary protection for faults between B1 and B2, as well as back up protection for further downstream faults
The coordination time interval is the time interval between the primary and remote backup protective devices
It is the different between the time that the backup relaying operates and the time that circuit breakers clear the fault under primary relaying
Typical time interval between 0.2 to 0.5 seconds are selected to account in most practical applications
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Figure 10

73. EXAMPLE 7.4
The CO-8 relay with a current tap setting of 6 amperes and time dial setting of 1 is used with the 100:5 CT in Example Determine the relay operating time for each case.
Solution:
From example 7.1(a)
The relay does not operate. It remains in the blocking position.
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74. SOLUTION b) Using curve 1 in Figure 11, toperating = 6 seconds c)
From curve 1, toperating = 1.2 seconds
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75. Figure 11: CO-8 time delay overcurrent relay characteristic
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76. EXAMPLE 7.5
An 11 kV, 50 Hz radial system is shown in Figure 12. Assuming a CO-7 relay with relay characteristic given in Figure 13 and the same power factor for all loads, select relay settings to protect the system with coordination time interval 0.3 seconds.
Figure 12
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77. EET301 POWER SYSTEM ENGINEERING
Figure 13

78. SOLUTION The load currents are calculated as
The normal current through the sections are then given by
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79. SOLUTION With the given CT ratios, the normal relay currents are
Now obtain C.T.S. (currents tap setting) or pick up current in such a way that the relay does not trip under normal currents. 
For this type of relay, C.T.S. available are 4, 5, 6, 7, 8, 10 and 12 A.
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80. SOLUTION Choosing the nearest setting higher than the normal current
For position 1, the normal current in the relay is A, so choose C.T.S.1 = 6 A
For position 2, the normal current in the relay is A, so choose C.T.S.2 = 10 A
For position 3, the normal current in the relay is A, so choose C.T.S.3 = 10 A
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81. SOLUTION
Next, select the intentional delay indicated by TDS (Time Dial Setting). Utilize the short circuit currents to coordinate the relays
The current in the relay at 1 on short circuit is
Expressed as a pick up value
Choose the lowest TDS for this relay for fastest action. Thus,
TDS1 = ½
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82. SOLUTION
Referring to the relay characteristic, for TDS1 = ½ and ratio, the operating time for the relay 1 for a fault at 1 is obtained as T11 = 0.15 s.
To set the relay at 2 responding to a fault at 1, allow breaker operating time,
Tbreaker = 5cycle x 1/50Hz = 0.1 s.
With coordination time 0.3 seconds, thus primary protection to clear this fault
T21 = T11+Tbreaker+Tcoordination = = 0.55 s
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83. SOLUTION
Short circuit for a fault at 1 as a multiple of the CTs at 2 is
From the characteristics for 0.55 s operating time and 6.25 ratio,
TDS2 = 2
Now setting the relay at 3
For a fault at bus 2, the short circuit current is 3000 A, for which relay 2 responds in a time T22 calculated as
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84. T32 = T22 + Tbreaker + Tcoordination = 0.5 + 0.1 + 0.3 = 0.9 s
SOLUTION
For TDS2 = 2 and 7.5 ratio, from relay characteristic,
T22 = 0.5 s
Allowing the same margin for relay 3 to respond for a fault at 2, as for relay 2 responding to a fault at 1,
T32 = T22 + Tbreaker + Tcoordination = = 0.9 s
The current in the relay expressed as a multiple of pickup is
Thus, for T3 = 0.9 s and 3.75 ratio, from the relay characteristic
TDS3 = 2.5
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85. Jumaat
EET 102 ELECTRIC CIRCUIT II

86. FUSE
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87. 7.4 FUSE
Fuse is a device connected in series with the conductors that caries the full load current supplied to the load.
Under the normal condition, the load current is less than the ampere rating of the fuse, and the fuse remain intact to connect the load to the source.
When overload or short circuit occurs, the current through link will also increase. Once the link totally melted, an open circuit exist in the fuse, an open circuit exist and the circuit will remain open.
Then it must be replaced. Fuse are “one time” device.
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88. FUSE Reduced area of link Filler material
Short circuit element
Trigger spring
Connector
C. Current Limiting Time Delay Fuse
A. Non-Current Limiting , Non-Time Delay Fuse
Filler material
B. Current Limiting Non-Time Delay Fuse
EET301 POWER SYSTEM ENGINEERING

89. 7.4 FUSE
During normal operation, when the fuse is operating below its continuous current rating, the resistance of the link is so low
If an overload current occurs and persists for more than a short interval of time, the temperature of the link eventually reaches a level that causes a restricted segment of the link to melt.
A gap is then formed and an electric arc is established.
As the arc causes the link metal to burn back, the gap width increases.
The resistance of the arc eventually reaches such a high level that the arc cannot be sustained and it is extinguished.
The current flow within the fuse is then completely cut off.
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90. 7.4 FUSE
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91. 7.4 FUSE
Fuse specification is normally based on the following four factors.
Voltage rating: this rms voltage determines the ability of a fuse to suppress the internal arc that occurs after the fuse link melts. A blown fuse should be able to withstand its voltage rating. Most low voltage fuse have 250 – 600 V rating and medium voltage rating fuse range from 2.4 – 34.5 kV.
Continuous current rating: the fuse should carry this rms current indefinitely, without melting and clearing.
Interrupting current rating: Is the largest rms asymmetrical current that the fuse can safely interrupt. Standard interrupting rating for medium voltage include 65, 80 and 100 kA.
Time response: the melting and clearing time of a fuse depends on the magnitude of the overcurrent and specified by a ‘time-current’ curve.
EET301 POWER SYSTEM ENGINEERING

92. Example
A 415 V distribution system is shown in Figure 14. Select the suitable rating of fuse for each load and incoming circuit using fuse time-current characteristic in Figure 15.
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Figure 14

93. Example
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Figure 15

94. EET301 POWER SYSTEM ENGINEERING
Figure 15

95. Solution
Fuse selection for each circuit Lighting load: select 32 A fuse Heating load: select 50 A fuse
EET301 POWER SYSTEM ENGINEERING

96. Solution
Motor load: The motor full load current is The starting current for 10 seconds is 7 time than the full load current, therefore
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97. Solution
From the time-current curve, an 80 A fuse would withstand 383 A for only 6 seconds.
Therefore, for a 100 A fuse, which would withstand 383 A for longer than 10 seconds, would be necessary.
To provide discrimination the fuse at incoming circuit must meet the following requirements:
It must carry the normal load current:
It must carry the load + the starting current of the motor
From time-current curve, a 125 A fuse would withstand A for more than 10 s.
EET301 POWER SYSTEM ENGINEERING

98. COMPUTER RELAYING
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99. 7.5 COMPUTER RELAYING
The electric power industry has been one of the earliest users of the digital computer as a fundamental aid in the various design and analysis aspects of its activity.
There are many perceived benefits of a digital relaying system:
Economics: cost effective
Reliability: continuously active providing a high level of self-diagnosis.
Flexibility: Revisions or modifications made necessary by changing operational conditions
System interaction: The availability of digital hardware that monitors continuously the system performance at remote substations.
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100. COMPUTER RELAYING Example of Functional Block of Digital Relay
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101. ZONE OF PROTECTION
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102. 7.6 ZONE OF PROTECTION
Power system configurations, a fundamental concept is the division of a system into protective zones.
If a fault occurs anywhere within a zone, action will be taken to isolate that zone from the rest of the system. Zones are defined for:
Generators
Transformers
Buses
Transmission and distribution lines
Motors
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103. ZONE OF PROTECTION
Each zone is defined by a closed, dashed line. As shown in Figure 16, Zone 1 for example, contains a generator and connecting leads to a transformer ,.
In some cases a zone may contain more than one component.
For example, Zone 3 contains a generator-transformer unit and connecting leads to a bus, and Zone 10 contains a transformer and a line.
Protective zones have the following characteristics:
Zones are overlapped.
Circuit breakers are located in the overlap regions.
For a fault anywhere in a zone, all circuit breakers in that zone open to isolate the fault.
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104. ZONE OF PROTECTION Figure 16 : Power System Protective Zones
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105. END OF CHAPTER 7
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106. T H A N K
YOU
EET305 POWER SYSTEM FUNDAMENTAL