Q1. (a) With the aids of suitable labelled diagrams, discuss three possible discharge paths that can cause surges on the transmission line.
(b) A lightning stroke which reachs a peak current of 35 kA in 1 µs strikes a 20 m tower on a 345 kV transmission line. The line has a ground wire joining the tops of the towers; its surge impedance is 520 Ω. The tower surge impedance os 90 Ω and the ground footing resistance is 40 Ω.
Determine whether the line insulations will flashover as a consequence of the surge, assuming that their impulse flashover strength is 1050 kV.
A coupling factor 0.3 with the phase conductor can be assumed; the impedance of the stroke channel can be ignored; a wave velocity on the tower of 2.98 x 108 m/s can be assumed. Show the surge progressions in the form of Bewley lattice diagram.
SOLUTION:-
(a) 2 leader core
earth wire space charge envelope
3
Tower conductor
Tower footing resistance
Earth place
Figure 1 – Geometry of lightning leader stroke and transmission line
In the first discharge path (1), which is from the leader core of the lightning stroke to the earth, the capacitance between the leader and earth is discharged promptly, and the capacitance from the leader head to the earth wire and the phase conductor are discharged ultimately bt travelling wave action, so that a voltage is developed across the insulator string. This is known as the induced voltage due to a lightning stroke to nearby ground. It is
not a significant factor in the lightning performance of systems above about 66 kV, but causes considerable trouble on lower voltage systems.
The second discharge path (2) is between the lightning head and the earth conductor. It discharged the capacitance between these two. The resulting travelling wave comes down the tower and, acting through its effective impedance, raises the potential of the tower top to a point where the difference in voltage across the insulation is sufficient to cause flashover from the tower back to the conductor. This is the so-called back-flashover mode.
The third mode of discharge (3) is between the leader core and the phase conductor. This discharges the capacitance between these two and injects the main discharge current into the phase condustor, so developing a surge impedance voltage across the insulator string. At relatively low current, the insulation strength is exceeded and the discharge path is completed to earth via the tower. This is the shielding failure or direct stroke to the phase conductor.
(b)
20m
I 35 kA
α3
Β2
α2 ZT = 90Ω
α1
40Ω 40Ω
1 µs t
Equivalent circuit
Find Zeq :
Zeq = 66.86 Ω (1 Mark)
Find V surge : V = Imax x Zeq
= 35 x 103 x 66.86 = 2.34 MV
= u(t) (1 Mark)
Find α and β : α1 =
=
= -0.385 (1 Mark)
α2 =
=
=
35kA ZT
Imax
Zg Zg
= 0.486 (1 Mark)
β2 = 1 α2 = 1 + 0.486
= 1.486 (1 Mark)
Time taken for wave travel from tower top to tower base
∆t =
=
= 0.067 µs = 1 T (1 Mark)
No of travel in 1 µs after the lightning strike 0.067 µs → 1T
1 µs →
= 14.9 T
V1 = u(t)
V2 = α1β2 u(t - 2T)
= (-0.385)(1.486) u(t – 2T)
= -0.572 u(t – 2T) V3 = α12α2β2 u(t – 4T)
= (-0.385)2(0.486)(1.486) u(t – 4T) = 0.107 u(t – 4T)
V4 = α13α22β2 u(t – 6T)
= (-0.385)3(0.486)2(1.486) u(t – 6T) = -0.02 u(t – 6T)
V5 = α14α23β2 u(t – 8T)
= (0.385)4(0.486)3(1.486) u(t – 8T) = 0.0037 u(t – 8T)
V6 = α15α24β2 u(t – 10T) V7 = α16α25β2 u(t – 12T)
= (0.385)5(0.486)4(1.486) u(t – 10T) = (0.385)6(0.486)5(1.486) u(t – 12T) = -0.0007 u(t – 10T) = 0.00013 u(t – 12T)
V8 = α17α26β2 u(t – 14T)
= (0.385)7(0.486)6(1.486) u(t- 14T) = -0.00002 u(t – 14T)
Vtt = V1 + V2 + V3 + V4 + V5 + V6 + V7 + V8
= u(t) – 0.572 u(t – 2T) + 0.107 u(t – 4T) – 0.02 u(t – 6T) + 0.0037 u(t – 8T) – 0.0007 u(t – 10T) + 0.00013 u(t – 12T) – 0.00002 u(t – 14T)
(4 Marks) From graph,
Vpeak = 0.56 u(t) Vphase conductor = 0.3 Vpeak
= 0.56 (2.34 x 106) = 0.3 (1.31 x 106)
= 1.31 MV (1 Mark) = 0.393 MV (1 Marks)
Vflashover = Vpeak – Vphase conductor + Vphase
= 1.31 x 106 – 0.393 x 106 +
= 1.12 MV (1 Marks)
Vflashover strength = 1050 kV = 1.05
Therefore flashover occur because flashover flashover strength
(3 Marks)
Q2. (a) Describe the lightning phenomena based on Simpson‟s theory as shown in Figure Q2a.
Figure 2a
(b) Explain what is meant by the terms „T1/T2 Impulse Wave‟ and outline the methode of lightning impulse voltage production in the laboratory.
SOLUTION:-
Q2. (a) According to Simpson‟s theory as shown in the diagram below, there are three essential regions the the cloud to be considered for charge formation.
Below region A, air currents travel about 800 cm/s and no rain drops fall through. In region A, air velocity is high enough to break the falling rain drops causing a positive charge spray in the clouds and negative charge in the air.
The spray is blown upwards but as the velocity of air decreases, the positively charge water drops recombine with the larger drops and fall again. Thus region A eventually becomes pre-dominately positively charged while region B above it becomes negatively charged by air currents. In this upper regions in the cloud the temperature is low (below freezing point) and only ice crystals exists. The impact of air on these crystals makes them negatively charged, thus the distribution of the charge within the cloud is shown as in the figure.
Lightning phenomena is based on the atmospheric process that takes place during tunderstorm where charges are accumulated in the cloud or portion of the cloud and equal and opposite charges are produced in the earth beneath.
These positive and negative charged become seperated by the heavy air currents with ice crystals in the upper part and rain in lower parts of the cloud.
As the charges increases the potential between the cloud and the earth increases and therefore the potential gradient is not uniformly distributed.
When the gradient exceeds the strength of the portion of air across which it is applied, the air breakdown and a streamer starts from the cloud towards earth. Lightning stroke may be started with potential of the order of 5 to 20 MV between the cloud and the earth.
(b) An impulse voltage wave is a unidirectional voltage which rises rapidly to a maximum voltage and the decays rather more slowly to zero as shown in the diagram below.
V
100 % 90 % 50 % 10 %
T1 t
T2
The waveshape is generally defined in terms of the time T1 and T2 in microseconds. T1 is the time taken by the voltage wave to reach its peak value, i.e from 10% to 90% of the voltage wave. T2 is the total time from the start of the wave to the instant when it has declined to one half of its peak, i.e from start of the wave to 50% of the peak during decay.
Q3 (a) Will backflashover occur on a tower that was struck by lightning of current I(t), of wave-shape with maximum value at 40kA occurring at 1.2µs and decreased linearly. The surge impedance of tower, tower ground wire and phase conductors are 200 , 250 and 300 respectively. The clearance between the tower side and the phase conductor is 2 meter. Assume that the breakdown of air is at 30kV per cm and the string insulator flashover at 1000kV.
Consider the coupling factor between ground wire and phase wire is 0.25. The lighting strike when the A.C voltage is at *
+ sin 110°kV.
Figure Q3(a) Transmission system with the top of tower was struck by lighting
(b) Figure Q3(b) shows a schematic diagram of a tilted transmission line tower with two parallel ground wire and an impulse current wave-shape, i(t).
Consider the tower top is struck by the lighting current i(t) and voltage rises to u(t).
Figure Q3(b)
(c) By referring to Fig Q3(b), discuss the validity of equation (write the detailed derivation in order to prove it):
UTT (t) = u(t) - α T (1+β)[u(t -2TT) – (αTβ)u(t -4TT) + (αTβ)2u(t -6TT)….]
Where:
UTT (t) = potential distribution on the top of tower α (Alpha) = coefficient of reflection
β (Beta) = coefficient of reflection on the tower top side
Solution:
(a) The ground potential rise at the point where lighting strikes the tower
Vsurge = [ ]
= [ ⁄
⁄ ]
= [ ⁄
⁄ ] 40kV
= *
V1 = u(t) t = 0 V2 = (α βα) u(t) t = 2TT V3 = (α2β α2β2) u(t) t = 4TT V4 = (α3β2 + 32β3) u(t) t = 6TT
VTT (t)= ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ ̅̅̅̅
= u(t) + (α βα) u (t- 2TT ) (α2β α2β2) u (t- 4TT ) (α3β2 + 32β3) u (t- 6TT ) = u(t)+α(1+β)[u(t -2TT) – (αTβ)u(t -4TT) (αTβ)2u(t -6TT)….
So the expression for VTT as given in the question is right
Q4 (a) Discuss in detail about the lighting phenomenon starting with the formation of thunderclouds.
(b) A peak lighting current of 40kA has struck a ground wire at mid-span (at the middle of two transmission towers). If the ground wire surge impedance is given as Z=500Ω, calculate the generated voltage at the point of strike. State the assumptions you made to answer this question.
Solution:
(a) During storms, charges are accumulated in clouds and equal charges of opposite polarity are formed in earth. As these charges increases, the voltage gradient in the air adjacent to the charge centre in the cloud increases.
When the gradient exceeds the insulation strength of air, a low current streamer starts downward from the cloud and continues to grow. When the streamer makes contact with the earth, it is like closing a switch between the two charges of opposite polarity, one is the earth and the other in the streamer channel and in the clouds . Thus large current flows.
In the middle stage of formation of a cloud, strong wind turbulence takes place causing the separation of charges into several layers which are:
Intra-cloud (within the clouds)
Cloud to cloud
Cloud to ground
Cloud to air
Positive stroke or “Blue sky lightning”
(b)
Assume:
1) Zs = infinity
2) No return voltage at the Earth line V at strike point,
ZL =500Ω ZL =500Ω
I = 40kA
A
Z Z
ZS
Q5 (a) Discuss two(2) of the followings i. Lightning phenomena
ii. Direct strike and indirect strike ii. Switching overvoltage
(5 marks)
(b) Lightning strike at mid-span of a transmission line ground wire at point a as shown in Figure Q5(b). This wave travels in both direction of the transmission line. Determine the transmission and reflection coefficients, β and α at points b and d.
(10 marks)
Figure Q5(b)
R1
Zg1 b Zg2 c Zg3
ZT2 ZT1
d e
R2
(c) A lightning surge of magnitude 10 kA with the voltage waveshape of 1.2/50 µs strike a ground conductor at midspan of a transmission line. If the channel surge impedance is 1500 Ω and the ground wire surge impedance is 600 Ω, determine at the point of strike;
i. The equivalent circuit and equivalent impedance, ii. The peak current, and
Benjamin Franklin has proved lightning is an electrical phenomena. An Eletrical phenomena carries the concept of charges involvement. So two types of charges are the reasons for the cloud to be considered as a „cell‟. The charge are a) positive type and b)negative type. Fig. shows the typical thundercloud structure.
Not all clouds are lightning cloud generator. It is only the cumulonimbus cloud type that can generate lightning. The Ice Splinter can be used to explain on the electrification of the cloud. The moistures and precipitation particles being is suspension in air and due to upwards action of updraft, causing supercooling to take place and resulting moistures to become ice. The ionic migration of OH- and H- in the moistures built, leaving the OH- in the front and II+ being lighter are pulled out to settle in the outer layer. The resultant
two-20km --
layer ice structure split due to different rate of ice expansion (the inner and outer layer).
The splinters are basically of positive-charged and negative-charged. The lighter splinters are pulled upwards while the lighter negatively charged splinters settle at the strike. It causes large currents to flow to the ground.
[ 2.5 Marks ]
If the damage to building and equipment due to surge propagation is because of inductive and capacitive affect. This is termed as indirect strike.
[ 2.5 Marks ]
OR
iii. Switching Voltage
With the steady increase in transmission voltages needed to fulfil the required increase in transmitted power, switching surges have become the governing factor in the design of insulation for EHV and UHV systems. In the meantime, lightning overvoltage come as a secondary factor in these networks. There is a great variety of events that would initiate a switching operations of greatest tolerance to insulation design can be classified as follows:
a. Energization of a line b. Load rejection
c. Switching on and off a equipment
d. Fault initiation and clearing [ 2.5 Marks ] (b)
Point b is to be considered:
(
)
Point d:
(c)
i. Based on the equivalent circuit:
ii Based on the circuit:
Zg1 Zg1
I peak I peak
Q6. (a) A transformer has an impulse insulation level of 1175 kV and is to be operated with an insulation margin of 15 % under lightning impulse conditions.
The transformer has a surge impedance of 400 Ω. A short length of overhead earth wire is to be used for shielding the line near the transformer from direct strikes. Beyond the shielded length, direct strokes on the phase conductor can give rise to voltage waves of the from 1000e-0.05t kV (where t is expressed in µs).
If the corona distortion in the line is represented by the expression,
= ( )
Where B = 100 m/µs and V = 200 kV, determine the minimum length of shielding wire necessary in order that the transformer insulation will not fail due to lightning surges.
For B.I.L of 1175 Kv, and an insulation margin of 15%, the maximum permissible voltage
= 998.75kV
Since the voltage is increased by b= 1.6 times at the terminal equipment (transformer), the maximum permissible incident voltage must be decreased by this factor, hence the maximum permissible incident voltage is
=998.75/1.6 =624.22 kV
Therefore the shielding wire must reduce the surge to 624.22 kV(by virtue of the corona distortion), that is
1000e-0.05t= 624.22 t = 9.425 µs Hence ΔT = 9.425 µs
The corona distortion is given by
* +
Therefore the minimum length of shielding wire required = 1.526 km.
Q7 (a) Discuss the following:
i. Backflashover of a lightning strike [2 Marks]
ii. Front time and tail time of lightning impulse voltage [2 Marks]
iii. The causes of switching surges [4 Marks]
(b) Prove that the reflection coefficient, α = (Z2 – Z1)/(Z2 + Z1) and the transmission coefficient, β = 2Z2/(Z2 + Z1) for an incident lightning surge on a transmission line where Z1 is the surge impedance of line 1 and Z2 is the surge impedance of line 2.
[4 Marks]
(c) A lightning current with the rate of rise of 25 kA/µs reaches a peak walue in 1.6 µs has struck a ground wire at mid-span (at the middle of two transmission towers). If the ground wire surge impedance is given as Zg
= 250 Ω, calculate the generated voltage at the point of strike. State and justify all assumptions made.
If the striking point is changed to the top of one tower (not the end tower) with a surge impedance of 100 Ω, calculate the generated voltage at the point of strike.
[8 Marks]
Solution (a)
i. Backflashover of a lightning strike
When a direct lightning stroke occurs on a tower, the tower has to carry huge impulse currents. If the tower footing resistance is considerable , the potential of the tower rises to a large value, steeply with respect to the line and consequently a flashover may take place along the insulator strings.
This is known as “back flashover”.
(2 Marks)
ii. Front time and tail time of lightning impulse voltage
Front – time taken for the lightning impulse waveshape to rise from start to peak value
Tail time – time taken for the lightning impulse waveshape to decay to 50% of it‟s peak value
(3 Marks)
iii. The origin of switching surges in power system;
De-energizing of transmission lines, cables, shunt capacitor, banks, etc.
Disconnection of unloaded transformers, reactors, etc.
Energization or reclosing of lines and reactive loads
Sudden switching off to loads
Shorcircuits and fault clearances 1.0
Resonance phenomenon like ferro-resonance, arcing grounds, etc.
(3 Marks)
(b)
where, E & I – incident voltage and current
ET & IT – transmitted voltage and current ER & IR – reflected voltage and current No discontinuity of potential and current at junction J, : E + ER = ET and I + IR = . . . (1)
= I(Zg/2)
Q8. (a) Fig. Q8 (a) shows a schematic diagram of a tilted transmission line tower and an impulse current waveshape, i(t). Consider the tower top is struck by the lightning current i(t) and voltage rises to u(t).
TT ZR
ZT u(t)
UR Rt
Fig. Q8(a) Inverted tower for analysis
Also Zg is the surge impedance of the ground wire Zt is the surge impedance of the tower
u(t) is the impulse surge function i(t) is the current wave function
TT is the time of surge propagation from tower top to the tower footing Rt is tower footing resistance
UTT(t) is the potential distribution on the top of tower α is the coefficient of reflection on the tower bottom side β is the coefficient of reflection on the tower top side
α β i(t)
UTT(t)
Current (kA)
Time (microsec) i(t)
(i) Show that
u(t)= Zg Zt / (Zg + 2Zt ). i(t) (3 marks)
(ii) Determine whether the following equation is right or wrong (write the detailed derivation in order to prove it)
UTT(t)=u(t)+αT(1+β)[u(t-2TT)+(αTβ)u(t-4TT)+(αTβ)2u(t-6TT)+……..]
V1=u(t) V2=(α+ βα)
V3=(α2β+α2β2) u(t) V4=(α3β2+α3β3) u(t)
t=0 t=2TT
t=4TT
t=6TT
TT
2TT
3TT
4TT
UTT
Ut Ut
=αUt
=α β Ut
=α2 β Ut
=α2 β2Ut
=α3β 2Ut
=α3β 3Ut
5TT
(b) A lightning current surge with the wave shape as shown in Fig Q8 (b), strike a tower, which has a single ground wire in both directions. The characteristic are as follows :
Surge impedance of lightning channel, Zl = infinity Surge impedance of the tower, Zt = 150 Ω Surge impedance of ground wire, Zg = 340 Ω Velocity of wave propagation on lines = 298 m/ µs Velocity of wave propagation on tower = 240 m/ µs
Height of tower = 30m
Effective tower footing resistance = 40 Ω Lightning current peak magnitude = 40 kA
Based on Fig.8 (b), determine the maximum tower top potential for a duration 5 times the time of surge propagation from the tower top to the tower base after the lightning strike the tower.
(15 marks)
Solution:
(a) i. u(t)= [Zg // Zt // Zg ] x i(t) = // Zt x i(t)
ID Orientation of propagation
20 µs 1 µs
Fig.Q8 (b) The simplified lightning current wave shape
ii. UTT=V1+V2+V3+V4
=u(t)- (α βα ) u(t-2TT) - (α2β α2β2) u(t-4TT) – (α3β2 α3β3) u(t-6TT) =u(t)- α(1 β) [u(t-2TT) - αβu(t-4TT) – α2β2u(t-6TT) …..
Replace α = - αt
=u(t)- αt (1 β) [u(t-2TT) - αtβu(t-4TT) – αt2β2u(t-6TT) …..
The equation proves that UTT is right.
UTT=Ut(t) β1α1ut (t-2 T) β1α12 α2ut (t-4 T) β1α13α22ut (t-6 T) …..
=Ut(t) + (1 - β) α1ut (t-2 T) β (1 - β) α12 α2ut (t-4 T) + (1 - β) α13α22ut (t-6 T) …..
=Ut(t) α1 (1 - β) [ut (t-2 T) - α1α2ut (t-4 T) α12α22ut (t-6 T) …..
=Ut(t) αT (1 - β) [ut (t-2 T) – (αTβ) ut (t-4 T) + (utβ)2 ut(t-6 T) …..
=Ut(t) + 0.579 (1 – 0.0625) [ut (t-2 T) – (0.579 x 0.0625) ut (t-4 T)]
=Ut(t) + 1.036 [ut (t-2 T) – (0.036)ut (t-4 T)]
=Ut(t) + 1.036ut (t-2 T) – 0.036ut (t-4 T)]
(b) The lightning current waveshape 1/20 µs Comparison:
1 µs = 1/0.125 = 8 T 20 µs = 20/0.125 = 160
The Vsurge=
8 T 3.3 MV
=1 pu
16 T
=Ut(t)
The peak voltage is 0.91 MV
Q9. (a) Discuss two (2) of the followings:
i. Lightning Phenomena (3 marks)
ii. Direct Strike and Indirect Strike (3 marks)
iii. Switching Overvoltage (3 marks)
(b) Lightning strike at mid-span of a transmission line ground wire at point a as shown in figure Q9(b). This wave travels in both direction of the transmission line. Determine the transmission and reflection coefficients, α and β at points b and d.
R1
Zg1 b Zg2 c Zg3
ZT2 ZT1
d e
R2
Figure Q9 (b)
(6 marks) (c) A lightning surge of magnitude 10kA with the voltage waveshape of 1.2/50 µs strike a ground conductor at midspan of a transmission line. If the channel surge impedance is 1500Ω and the ground wire surge impedance is 600Ω, determine at the point of strike :
i. The equivalent circuit and equivalent impedance (2 marks)
i) The peak current (3 marks)
ii) The peak voltage (3 marks)
(d) A lightning current surge with the wave shape of figure 7 strikes a tower which has a single ground wire in both directions. The characteristics are as follows;
Surge impedance of lightning channel = infinity Surge impedance of tower = 150Ω Surge impedance of ground wire = 340Ω Velocity of the wave propagation on lines = 298 m/µs Velocity of the wave propagation on tower = 240 m/µs Coupling factor of phase conductors = 0.25
Height of tower = 30m Effective tower footing resistance = 40Ω
Determine the maximum tower top potential, after 0.4 µs the tower has been struck by the lightning. Please show clearly all the calculations involving coefficients of the reflection and refraction. Show the surge progressions in the form of the Bewley Lattice Diagram.
a) What will happen if the tower footing resistance increases in the value?
b) Provide one reason for the tower footing resistance increase in the value.
c) Why the speed of surge is higher in the conductor than in the tower structure?
(17 marks) 25 kA
20 µs t 1 µs Figure 9(d) the wave shape of the lightning current
Solution:
(a) i. Benjamin Franklin has prove lightning is an electrical phenomena. An electrical phenomenon carries the concept of charges involvement. Two types of charges are the reasons for the cloud to be considered as a „cell‟.
The charges are:
a) Positive Type b) Negative Type
Figure above show the typical thundercloud structure. Not all clouds are lightning cloud generator. It is only the cumulonimbus cloud type that can generate lightning. The Ice Splinter can be used to explain on the electrification of the cloud. The moistures and precipitation particles being is suspension in air and due to upwards action of updraft, causing super cooling to take place and resulting moistures to become ice.