Making communications land - Are they received and understood as intended? we...
See4423 chapter1 introduction[1]
1. OFFICE: CENTER OF ELECTRICAL ENERGY SYSTEM (CEES)
FACULTY ELECTRICAL ENGINEERING
UNIVERSITI TEKNOLOGI MALAYSIA
81310 UTM JOHOR BAHRU
JOHOR DARUL TAKZIM
ROOM NO: P07-412
Tel: 07-5536262
2. WELCOME
• POWER SYSTEM ENGINEERING
• SEE 4423
• LECTURER: ASSOC PROF MD SHAH MAJID
• ROOM NO: P08-409/P07-417(CEES)
• TEL: 07 -5535295/5536262 (CEES)
• E-mail: mdshah@fke.utm.my, mdshah@ieee.org
3. SEE 4423
POWER SYSTEM ENGINEERING
SEMESTER II SESSION 2009/2010
Assoc Prof MD SHAH MAJID
4. SEE 4423
POWER SYSTEM ENGINEERING
SEMESTER 1 SESSION 2009/2010
4
6. Introduction:Need for Protective Systems
• Power system consists of; generators, transformers, transmission and
distribution lines, etc…..etc.
– Sistem kuasa mengandungi: penjana, pengubah, talian penghantaran dan pengagihan dsbnya
• Short circuits and other abnormal conditions often occur on a power system
– Litar pintas dan keadaan tidak normal kerap berlaku pada sistem kuasa
• Heavy current associated with short circuits is likely to cause damage to
equipments
– Arus yg tinggi ketika litar pintas mungkin menyebabkan kerosakan pada peralatan jika geganti
perlindungan dan pemutus litar tidak di bekalkan untuk perlindungan bagi setiap seksyen
sistem kuasa
7. Need for Protective Systems
• Short circuit is a faults (power engineer)
• Failure of conducting path due to a break in a conductor is a type of fault
• Fault occurs – automatic protective device is needed to isolate the faulty
element as quickly as possible to keep the healthy section of the system in
normal operation
• Fault must be cleared within fraction of a second
• Uncleared short circuits may cause total failure of the system
8. • A protective scheme includes transducers, protective relays and circuit
breakers to isolate the faulty section of the system from the healthy section
• Protection is needed not only against short circuit but also against any
abnormal condition
– Overspeed of generators and motors
– Overvoltage
– Underfrequency
– Loss of excitation
– Overheating of stator or rotor of an alternator
– Etc…etc…etc
• Protection is a pre-requisite for an effective and reliable system
• A protective relay does not anticipate or prevent the occurrence of fault,
rather it takes action only after a fault has occurred – except Buchholz relay; a
gas actuated relay
9. Nature and causes of Faults
• Faults are caused either by;
– Insulation failure –
– Conducting path failure
• results in short circuit
• Faults on transmission and distribution lines are caused by
overvoltage
– Lightning and switching surges
– External conducting objects falling on overhead line.
• Birds also may cause faults on overhead line if their bodies touch one
of the phase and earth wire
10. • If conductors are broken – failure of conducting path and the conductor
becomes open-circuited
– If broken conductor fall to the ground, results in short circuit
• Joint failure on cables and overhead lines also cause a failure of the
conducting path
• Opening of one or two of the three phases makes the system unbalanced
– Set-up harmonics
11. • Other causes of faults in o/h line
– Direct lightning strokes
– Aircraft
– Snakes
– Ice and snow loading
– Abnormal loading
– Storm
– Earthquakes
– Creepers
– Etc…..etc……etc……etc
12. • Cables, transformer generators and other equipment:
– Failure of solid insulation due to aging
– Heat
– Moisture
– Overvoltage
– Mechanical damage
– Accidental with earth
– Flashover due to overvoltages
– Etc…etc…..etc
13. Types of faults
• Symmetrical faults
• Unsymmetrical faults
• Symmetrical faults
– Kerosakan tiga fasa atau tiga fasa ke bumi
• Unsymmetrical faults
– Single line to earth, line-to-line, double line to
earth, open-circuited phases
14. • Faults can interrupt power system in several ways:
– Heavy current to flow
• Effect : overheating of power system component
– Fault is a short circuit and exist as electrical arc and liquid e.g air
• Effect: equipment faulty and fire
15. • Fault
– can increase/decrease voltage system outside its acceptable range
– Can cause unstable three phase system, improper operation of three phase
equipments
– Prevent power flow
– Can cause system to be unstable and collapse
16. • Faults incur MONEY.
• If fault is isolated as quickly as possible and
accurate –less money required
• Cost of protective equipment is 5% of the total
cost of the system
17. • PROTECTION DOES NOT MEAN
PREVENTION
– A protective relay does not anticipate or prevent the occurrence of fault,
rather it takes action only after a fault has occurred – except Buchholz relay; a
gas actuated relay
18. Functions of Protective
System
• Fast and automatically opens the faulty section in the power system
• Increase system reliability and security – only affected area will be
isolated and maintain the healthy line
– Ensure consumers receive continuity of supply
• Delay in isolating the fault
– System unstable, loss of synchronism, total failure of the system
– Fire
19. Requirements of Protective
System
• The basic requirements of a protective
system are as follows:
– Discrimination/selectivity
– Sensitivity
– Reliable
– Stability
– Speed
20. (Discrimination/Selectivity)
• Ability to select either to operate or not
– Keupayaan untuk memilih sama ada bekerja atau tidak
• Select to isolate the faulty section only (the
rest normal condition)
– Memilih untuk mengasingkan bahagian rosak sahaja (yang lain berkeadaan
normal)
• The nearest circuit breaker will trip
– Pemutus litar yang hampir sahaja terbelantik (trip)
22. Sensitivity
• Relay should operate when the magnitude of the current exceeds the
preset value
• This value is called pick up current
• Should not operate when current is below its pick-up value
• Should be sufficiently sensitive to operate when the operatinfg current
just exceeds the pick-up value – relate with minimum operating current
23. Reliability
• A protective system must operate reliably
when a fault occurs in its zone of protection
• Failure may due to its protective element
system; CT, PT, CB, relay etc….
• Reliability of protective system 95%
24. Stability
• A protective system should remain stable even
when a large current is flowing through its
protective zone due to an external fault
• Concerned circuit breaker is supposed to clear
the fault
26. Speed-fast operation
• Isolate fault as quickly as possible (at shortest time possible )
• Isolate disturbances before loss of synchronism and plant stop operation – time
should not exceed critical clearing time
• Speed balanced with economy
– Cost of protective equipment should be relevant to the cost of the protected
zone
• Avoid fire to equipments, interruption of supply to consumer, voltage drop
operating time; 1 cycle, ½ cycle are also available; distribution system > 1 cycle
27. Operating time
• Total time to accomplish the isolation
• Calculated at the instance of fault until the
given trip signal
• Must be low for the sake of plant and
equipment safety
• Delay time – for discrimination
28. Protection economy
• Protection system can be designed as:
– Simple and cheap
– Complex and expensive
• Baesd on:
– The cost of fault
– Safety level requirement
29. Protection economics…
• Cost of fault
– Cost of damage toward the plant
– Cost loss of revenue due to cut-out supply
– Cost of customers confidence
• Higher the cost of fault, more expensive the protection system
• Higher the plant kVA, more complex protection system required
30. Protection system at customer level
• Simple fuse – protect equipment and certain
circuits
• Fius ringkas melindungi alat-alat dan litar
tertentu
• Miniature circuit breaker- MCB
31. Protection system at distribution
level
• Fuse and switch fuse
• Automatic reclosing circuit breaker
– For rural area
– Transient fault is corrected by self clearing
32. Protection system at transmission
level
• Technical consideration outweighs the economy
• 275kV dan 400kV System require a protection system which is:
– very reliable
– Full discrimation
– High speed
• Expensive and complex
• Provide back-up protection
33. Zone of Protection
GENERATOR
PROTECTION
CIRCUIT BREAKER
• Power system is divided into HV SWITCHGEAR
several zones PROTECTION
• Each zone of protection is TRANSFORMER
provided with 2 types of
PROTECTION
protection:
EHV SWITCHGEAR
– Primary protection PROTECTION
– Backup protection
TRANSMISSION LINE
PROTECTION
EHV SWITCHGEAR
PROTECTION
ZONE OF
PROTECTION
35. Protection zone
CB3
CB3
CB1
CB2
SD1 SD2 SD3
Radial protection system
36. Main/primary protection
• In general primary protection is provided for
each transmission line segment, major piece
of equipment and switchgear
• If fault occurs, it is the duty of the primary
protective scheme to clear the fault
• First line of defence - Responsible to isolate
fault as quickly as possible
• Fails, back-up protection clears the fault
37. Backup Protection
• Act when primary protection fails – second
line of defence
• Usually several back up protection scheme will
act to control the power system
• Longer time delay
40. Fault detection
• Current magnitude
• Current in abnormal path (earth)
• Current balance (current out#current in)
• Voltage balance
• Changes in impedance
• Protective system damage(Buchholz), power flow direction, temeprature
& pressure
41. Protective system components
1. Circuit breaker
2. Transducer – current transformer (CT), voltage transformer (VT),
potential transformer (PT)
3. Communication links
– Pilot wire
– Radio link
– Overlapping signal, power line carrier (PLC)
42. Protective system components
4. Relay
– Electromagnetic
– Static – improve reliability,versatile, fast response (1/4 cycle)
– Microprocessor- VLSI teknologi – current interest to power engineers;
adv:attractive flexibility due to programmable approach, can provide
protection at low cost and compete with conventional relays
5. Fuse
43. Transducer
Serves as a sensor to detect
abnormal system conditions and to
transform the high values of short
circuit current and voltage to a lower
values
43
44. Protective relays
• Process the signals provided by the
transducers which may be in the form of
current, voltage or a combination of current
and voltage
44
47. Classifications of relay based on its
function
• Overcurrent relay
• Undervoltage relay
• Impedance relay
• Under frequency relay
• Directional relay etc….etc….etc
48. Classification of protective scheme
• Overcurrent protection
• Distance protection
• Current carrier protection
• Differential protection
49. Transformer in a protective system
• CT assume ideal
Feeder Ip • Normal rating:
conductor – 1:1, 2:1. 2.5:1, 4:1, 5:1
– 20:1, 40:1, 100:1, 200:1, 300:1,
600:1
Is CT – 1000:1, 2000:1, 4500:1
• Secondary coil is connected to a
Vp “sensitive device” eg voltmeter
Vs
• Low Stray impedance so that voltage
drop small
VT
50. Current transformer (CT)
• Normal rating
– 1A (Europe)
– 5A (USA)
• Example :
– 50:5, 100:5, 150:5, 200:5, 250:5
– 450:5, 500:5,……1000:5, ……, 6000:
• Secondary coil is connected to a “current sensing device” of zero
impedance
• Shunt impedance high so that I0 low.
51. (current transformer CT)
• Working principle similar to voltage transformer
• Supply by current source
• Primary winding connected in series with power circuit
– Carries full load current
• Transformer impedance (referred to secondary and can be neglected)
• Secondary coil connected to load (burden)
52. (current transformer CT)
• Secondary winding feeds the protective system
• Current is reduced but must be almost the same with the power system
current
• Secondary rated current 5 A or 1 A
• Fault current 10 – 20 X rated current. Transformer must be capable to
operate reliably/accurately at this value
• Burden usually usually small – flux do not saturate
53. Current transformer construction
Bar Type
Core
Power system
current
Primary winding
(1 turn)- from power system
Secondary winding wound
on the core
55. design
• CT – similar to a normal transformer emf
equation
• Average induced voltage = product of No. of
turn and rate of change of flux magnet; eaverage
= N(dΦ/dt)
• Knee point voltage(rms) = 4.44 BAfN
56. Burden
• Defined as the load connected across its
secondary CT
– express in VA (VA taken as nominal secondary
current of CT) or
– Impedance (at the rated secondary current at a
given power factor usually 07. lagging
• Increase in impedance – increase burden
• CT unloaded if secondary winding is short
circuited
57. Example
• 5VA burden at 1A transformer, gives 5 Ohms
impedance.
• 5VA/1A = 5V
• impedance = 5V/1A = 5Ω
• or
• At 5A CT
• 5VA/5A
• impedance = 1V/5A = 0.2 Ω
58. Example
• CT ratio 300/1, core area 40x30 mm
• Voltage at knee point?
• V = 4.44 x 0.0018 x 300 x 50 = 120 V
• 1.5 x 40 x 30 x10-6 = 0.0018 Wb ( B for sheet
steel, 1.5 tesla at knee point)
59. CT open circuit
• If burden high, Es high– exceed Vkp (knee point voltage)
• Io high ; I2 less
• Limited value when the secondary CT open circuit; I2 = 0;
• Then N1I1 = N2 (I2+Io) = N2Io
• This will drive the CT to saturation level
• dΦ/dt = 100 x Vkp (induced = 100 Vkp )
• Cause insulation failure and overheated
60. Current transformer equivalent circuit
Ip’ x r Is
I0’
Ic ’ Load (burden)
Es Im’ Vs
R
Ip = primary current
(nominal ratio) Kn
I0 = primary excitation
current (no load)
I 'p = I p / Kn '
I0 = I0 / Kn
61. Current transformer equivalent circuit
Ip’ x r Is
I0’
Im’ Ic ’ Load (burden)
Es Vs
R
Es
Ip’
I 'p = Is + I0
'
Is
θ
Is = I 'p − I 0
'
I0’
Ic ’ φ
Im’
62. relationship Es and φ (flux)
N=no of turn in secondary
dφ
es = N
dt π
es = Nωφm sin(ωt + )
2
φ = φ m sin ωt
Nωφ m
es = Nωφm cos ωt Es = ∠90°
2
63. (Nominal Turn Ratio)
• Determine from the given current ratio
– Eg.1000/5 (bar type)
Ip
200
K n = nominal ratio = =
Is 1
64. Actual Ratio
• Actual ratio value Ip and Is
′
Ip
actual ratio = winding ratio ×
Is
′
Ip
= Kt ×
Is
65. • Winding ratio= secondary coil turn if transformer is of bar type l
– Kt = N2/1
• Ideally ′
Ip
=1
Is
• Practically
′
Ip
〉1
Is
Actual turn≥ nominal ratio
67. Ratio error or current error
nominal ratio -actual ration
ratio error = × 100%
actual ratio
Nominal ratio= Kn ′
Ip
actual ratio = K t ×
Turn ratio= Kt Is
No compensation Kn=Kt
68. Cont’d
'
Ip
K n − Kt
Is
ratio error = '
Ip Kn=Kt
Kt
Is
(no compensation)
K n I s − Kt I '
=
p
'
Kt I p
difference in
I s − I 'p magnitude between
ratio error = × 100% Ip’ and Is
I 'p
70. Compensating winding
• Secondary coil winding is reduced (1 or 2 turn)
to compensate current error due by I0’.
• Secondary coil current will be high, but the
value will be minimised by the excitation
current component
• This can minimised the current error
71. Example T1
• A current transformer 50 Hz has a secondary current of 3A,Sebuah
• Secondary impedance circuit (0.6+j0.45),
• Maximum flux 0.253 weber
• Magnetisation current and core loss current (primary circuit) at this load
are 0.2 and 0.15 A respectively
• Turn ratio1:10, assume no compensation
• Determine no of turn, ratio error and phase error of transformer
72. CT equivalent circuit
r
Ip’ Is=3 ∠ αº
I0’
Im’ Ic ’ Load (burden)
Es Vs
= 0.6+j0.45
Es
'
Ip’ I 'p = I p / Kn I0 = I0 / Kn
Is
θ I 'p = Is + I0
'
I0’
Ic ’ φ
Is = I 'p − I 0
'
Im’
α
73. N1:N2 = 1:10 Kt = 10 = Kn
Is=3 ∠ αº Im = 0.2 A
Es = I s Z
Ic = 0.15 A
= 3∠α ° × 0.75∠36.87°
N s × φm × ω = 2.25∠(α + 36.87)°
Es = ∠90°
2
Es
(α + 36.87)° = 90°
Ip’ α = 90 − 36.87
α Is = 53.13°
θ
I0’
Ic’
Im’ φ
75. I 0 = 0.20 + j 0.15
Is=3 ∠ αº
′ 1
I0 = (0.20 + j 0.15) ×
Im = 0.2 A 10
Ic = 0.15 A = 0.02 + j 0.015
′ ′
I p = I0 + Is
Es
= 0.020 + j 0.015 + 3∠53.13°
Ip’
= 3.024∠52.9975°
Is
θ
I0’
Ic’ φ
Im’
76. ′
Is + I p 3 − 3.024
ratio error = = × 100%
′ 3.024
Ip
= −0.794 %
phase error = 53.13 − 52.9975
= 0.1325
77. Ex 1
• A current transformer has 15 turn on the primary and 75 turn on the
secondary. The relay burden is (1.2+j2.6)Ω and secondary winding
impedance of the transformer is (3.6+j0.4)Ω. If the magnitude of the
secondary voltage is 26 volt and magnetization current referred to
secondary is 0.24∠-45o amp. Determine the ratio error and phase error.
78. Ex 2
• A new non compensated CT with a cross sectional area of 40 cm2 has a nominal
ratio of 2000/5. When 5A is supplied to a relay, the combination of load relay and
secondary winding impedance is (10+j5.77) Ω. If the magnetisation amp-turn is
100 and core loss amp-turn is 50, determine the ratio error and phase error. Using
the same data, calculate also the compensating turn to reduce the ratio error to a
minimum.
79. Ex. 3
• A CT, 50 Hz has a primary winding of 10 turn and its cross sectional area is 8 cm2.
Secondary current is 5A. The total load and secondary impedance is 0.8+j0.5 Ω.
The excitation current required at the primary winding to establish a working flux
with the secondary open is 0.6∠45o amp. If the flux density is 0.1878
wb/m2,determine the nominal ratio of the transformer. Determine also the ratio
error and phase error of the transformer.
80. Composite error
• BS 3938:1973
– (r.m.s.) value; difference between ideal secondary
current (i.e CT is ideal ; no excitation component)
with actual secondary current.
• Encompass (current error, phase error and
harmonic effect)
81. Accuracy limit current of a
protection system CT
• Accuracy limit current (saturation current)
– The maximum current a CT can sustain before
exceed its accuracy
– Specify either in term of primary current or
secondary current
• Accuracy limit factor (saturation factor)
– Ratio of accuracy limit primary current and
rated primary current
– Standard values are 5, 10, 15, 20 and 30
82. CT specification
• Expressed in
– VA at rated current /class accuracy/accuracy limit factor
– Burden standard rated values
• 2.5,5,7.5,10,15 and 30 VA
– 2 accuracy class: 5P and 10P – gives composite error at rated accuracy limit 5%
and 10% respectively
– Standard accuracy limit factor:
• 5, 10, 15, 20 and 30
83. Rating of CT
• Method to explain CT
– 10VA/5P/25
Accuracy limit current is 25 x rated
current
Transformer works in good
condition/reliably until 25 x its
Rated burden Class – error rated curent
do not exceed
5%
84. Ex 1
• CT 10VA/5P/15
• Calculate the accuracy limit factor when the
burden is half of its initial value.
• Secondary resistance of CT 0.15 ohm
86. Solution
Total impedance referred to r
secondary winding
= 0.15 + 0.4 = 0.55 ohm
Es
V
Burden (R)
Accuracy limit factor= ratio accuracy
limit current and rated current
Accuracy limit current= Accuracy limit factor × Rated current
accuracy limit current = 15 × 5
= 75 A
87. secondary emf = 0.55 × 75 V
= 41.25 V
r
Es
Burden
(R=0.4)
When burden is half
R = 0.2 ohm
Total secondary impedance= 0.15 + 0.2 = 0.35 ohm
88. new accuracy limit current = 41.25 / 0.35
= 117.8 A
new accuracy limit factor = 117.8 / 5
= 23.56
Smaller the burden, the higher the accuracy
Therefore :
burden has to be low possible
89. CT Class X
• BS 3938 – CT for special application is known as Class “X’
• Expressed in turn ratio, knee point voltage, excitation current at the specified voltage and
secondary winding resistance
– Application – scheme where the phase fault stability and/or accurate time grading required
• Transformer rating specify in term of e.m.f maximum that can be use by the transformer
• knee point
– Point where an increase of 10% in secondary e.m.f require an addition of 50% excitation current
90. Definition of knee point
Excitation
voltage
+10%
Knee
point
+50%
Excitation current
91. Voltage transformer/potential transformer(VT/PT)
• Supply voltage lower than the sistem voltage
• Nominal sekunder voltage110V
• 2 types
– Wound ( electromagnetic) type(> 132 kV non economical)
– Capacitor type- CVT
• Element X and C consist of tuning circuit – to reduce the ratio and phase
angle error secondary voltage
92. 1/ωC2 = ωL pd 50 Hz; V2 = V’
Talian
C1
L
Vp
Rb
V’
Vs
C2 V2
Xb
V’ = 12 kV, C1 = 2000pF
L
C2
V2 110 V
V’
93. Protection techniques
• Protection scheme:
– Arrangements of CT, VT, pilot wire and relay so
that the required operating characteristics can be
obtained
• Each section of the power system network is monitored by several
protection scheme.
96. Unit protection scheme
CT CT
Protected zone
P
Pilot wire
R
relay
Q
Difference current scheme: circulating current
FAULT:
Non similar current flow in the relay
CB will trip
R
97. CT
Healthy and fault
e1 Protected zon e2 external zone (through
fault):
No current in relay,
e1=e2
R R
FAULT:
Current flow in circuit, result e1≠e2
Differencec current scheme: voltage balance
99. CB1 CB2
A
B
R1 R2
Trip
Trip
Restrain
Restrain
Power flow entering the feeder (from point A) will cause relay (R1)
close trip contact CB1
Power flow exit fron the feeder (at point B) will cause relay R2 rotate
and close restrain contact of the further circuit breaker (remote) (CB1)
(relay R2 send restrain signal CB1)
100. CB1 CB2
A
B
R1 R2
Trip
Trip
Restrain
Restrain
Power enter at A and exit at B, block trip will prevent the CB from
opening.
101. CB1 CB2
A
B
R1 R2
Trip
Trip
Restrain
Restrain
Power enter at A and at B (fault occurs), both CB open
102. CB1 CB2
A B
R1 R2
Trip
Trip
Restrain
Restrain
Power enter at A and no power at B, circuit breaker A, CB1 will open
103. NON UNIT SCHEME
supervise/monitor one or several points in the power system
signal will be detected at the points
this signal is the input to the relay
section monitored/supervised is not clearly defined
ZONE OF PROTECTION IS NOT WELL DEFINED
104. 2 examples of non unit scheme
(1.) Over Current Protection
(2.) Distance Protection
