Transient in Power System Electrical power system

1. TRANSIENT IN POWER SYSTEM
 Subject:- Electrical Power System – 2
 Presented By : Preet Patel (151310109032)
1

2. CONTENT
 Transients
 Voltage surge or Transient Voltage
 Types of Power System Transients
 Causes of System transients
 Overvoltage due to external causes
 Overvoltage due to internal causes
 Transients in Simple Circuit
 D. C. Source & А.С. Source
 Travelling Waves on Transmission Line
 Wave equation
 Travelling wave with open end line & Short Circuited Line
 Reflection and Refraction Coefficient
 Line Connected to a Cable
 Line terminated Through Capacitance
 Capacitor Connection at a T.L.
 Attenuation of Travelling Waves
2

3. TRANSIENTS
• The transient is a pulse which is of very short duration but of very high
intensity.
• The good example of transient is the lightening occurring in the sky in the
period of Monsoon , which is nothing but the heavy discharge of millions of
volts and millions of amperes .
• This wave travels along the length of three line at a certain velocity.
• This may cause damage to the equipments connected to the system.
• The time period when transient voltage and current occur is in the range of
micro - second to milli - seconds.
• The other examples of power system transient are, opening of the circuit
breaker , ground fault on line or in equipment etc. Insulation of the equipment
gets punctured due this high voltage.
3

4. VOLTAGE SURGE OR TRANSIENT VOLTAGE
Definition :-
 "Increase in voltage for very short time in power system is called the
voltage surge.“
 This voltage surge comes only for few micro seconds. But due to it there is
much increase in the voltage level of the system. This may cause damage to
the equipments connected to the system.
 There are different causes of occurring voltage surge. External causes
include lightning in the sky where as internal causes include switching surge
or fault.
 Flashover occurs over the insulator due to the voltage surge and the
equipments like generator, transformer may get damaged. 4

5. 5

6.  This waveform is steep fronted that means it reaches to the peak value in
very short period t1 .
 Value of t1 is 1 to 5 us. After words the voltage decreases with reduced rate.
Time period t1 is called the rise time.
 The time period during which the value of the voltage reaches to the peak
value is called the rise time.
 The value of the voltage becomes half of the peak value in time t2.
 Voltage surge is indicated by t1/t2.
 If rise time of the waveform is 3 μs and the value of peak becomes half of
the peak in 70 us then the surge is indicated by 3/70 us.
6

7. TYPES OF POWER SYSTEM TRANSIENTS
Depending upon the speed of transients, they can be broadly classified into
three groups .
 Ultra-fast transients : These types of transients are caused either by
lightning or by the abrupt .
 Medium-fast transients : These transients occur due to abrupt short-
circuits in the system causing abnormal structural changes in the system.
 Slow transient : These transients are electromechanical in nature causing
mechanical oscillations of rotors of synchronous machine.
7

8. CAUSES OF SYSTEM TRANSIENTS
 The causes of power system transients can be divided in to two categories.
(a) External causes
(b) Internal causes.
 Due to external cause
 High voltage can be produced in the line by the travelling voltage wave
produced due to the static charges in addition to the direct stroke.
 Suppose there is a cloud with negative charges over the line as shown in
figure this cloud will induce positive charges on the line.
 Suppose the strong wind blows so the cloud will be shifted away from the
line so the positive charges on the line will not remain bound and will become
free and can travel on the two sides of the line. 8

9. • The magnitude of this travelling wave is of 10 KV to 15 KV and it
has the steep wave front. Characteristic of this wave is as shown in
figure.
9

10.  Due to internal causes
 When there is sudden change in the circuit conditions of the power system,
oscillations are produced due to the inductance and the capacitance of the circuit and
there is increase in the system voltage.
 This voltage is almost double the system voltage.
 Thus the over voltage due to the internal cause is very less than that produced due to
the lightening is not necessary to provide additional protection for the over voltage
caused due to the internal causes if the insulation of the equipment is designed
properly.
 Internal causes occurred due to following reason
A. Switching Surges
B. Arcing Ground
C. Resonance
10

11. A. Switching Surges
 When switching operation is done in power system with or without
load, the voltage produced as a result of this is called the switching
surge.
B. Arcing Ground
 When neutral is not earthed in three-phase line and if line to ground
fault occurs, the phenomenon of arcing ground occurs. Due to this,
oscillations of three to four times the normal magnitude are
produced.
11

12. C. RESONANCE
 When the inductive reactance of the line becomes equal to its
capacitive reactance the net impedance of the line becomes the
minimum and it is equal to the resistance of the line.
 Series resonance occurs at this time.
 If the distortion occurs in the waveform of the e. m. f. the harmonics are
produced. So the value of Xu and Xc may become equal at the fifth or
multiple harmonics and may result in resonance.
 Over voltages are produced due to the resonance.
12

13. TRANSIENTS IN SIMPLE CIRCUIT :
D.C. SOURCE
1. Resistance only
For only resistance the relation between applied voltage
and current is given by ;
V = i*R
i = V/R
13

14. 2.) INDUCTOR ONLY
 For a inductor having value L, the relation between applied
voltage and resulting current is given by equation,
Integrating , we get
Circuit Diagram is as below
Output
14

15. 3.) CAPACITOR ONLY
 The relation between current through capacitor and voltage
across it is given by,
 Differentiating both side we get,
15

16. 4.) R-L CIRCUIT
 Consider R-L Circuit as shown below,
 When switch k is closed , the current in the circuit is given by,
Divide by R,
)(
dt
di
L+Ri(t)=V t
)(.)( t
dt
di
R
L
ti
R
V

)(.)( t
dt
di
R
L
ti
R
V

16

17. Integrating we get,
To find value of K Let take initial conditions and put t = 0 and L(t) =
0
17

18. 18

19. 5.) R-C CIRCUIT
 Circuit Diagram is as below,
The relation between current
and voltage is obtained as ;
The output response is as follows
;
19

20. 6.) R-L-C CIRCUIT
 The circuit diagram is shown below ,
 The switch S is closed , the current in the circuit is given by,
1
)(
1
1
)(
2




LCsRCs
Cs
s
V
sI
Cs
LsRs
V
sI
20

21. Let , and ; then





































LCL
R
L
R
s
LCL
R
L
R
s
L
V
sI
LC
s
L
R
s
L
s
V
sI
1
42
1
42
1
)(
1
1
.)(
2
2
2
2
2
a
L
R

2 b
LCL
R

1
4
2
  
   
   
 tbatba
ee
bL
V
ti
basbasbL
V
sI
basbass
V
sI














2
)(
11
2
)(
1
)(
21

22. WITH AC SOURCE
 With RL Circuit :-
 When switch S is closed,
the current in the circuit is given by,
=
Let
𝑅
𝐿
= a Then,
LsRs
s
s
V
Z
V
I m
s
s
s










1sincos
2222
)(
)(
)(




L
Rss
s
sL
Vm









1sincos
2222




     








 2222)(
sincos




sas
s
sas
VI ms
22

23.  Now,
And
So, L-1 I(s) =
∴
Where,
So, variation in current is shown
in figure.
   











 22222222
111
 s
s
s
a
asasas
   


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a
ss
as
asas
s
22
2
222222
1



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



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
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

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
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V
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
sincossincossincos2
 atm
t et
aL
V
i 


 )sin()sin(
22
)( 







 
R
L
 1
tan
23

24. THE RE-STRIKING VOLTAGE AFTER REMOVAL
OF S.C.
 The system consists of alternator connected to a busbar. The load is
removed after S.C. It is required to measure the voltage across C.B.
during the opening period. Some assumptions have to be done.
 Here, Fault Current= 𝐼 𝑠 =
𝑉(𝑠)
𝑍(𝑠)
=
𝑉𝑚
𝑠
×
1
𝐿𝑠
 Now the impedance between the C.B. contacts after the short circuiting
the voltage source will be the impedance of thee parallel combination of
L and C i.e.,
24

25. 25

26. DOUBLE FREQUENCY TRANSIENT
 Here, L1 and C1 are the inductance and stray capacitance on the source
side of the breaker and L2 and C2 are on the load side. Here before C.B.
operates voltage across capacitor is given by,
 Normally L2 > L1 and so capacitor voltage is less than the source voltage .
When the current passes through the zero value, the voltage is at its max.
value. When the C.B. operates, current is at zero value. So C2 will oscillate
with L2 with natural frequency of
 And C1 will oscillate with L1,
 So opening switch will result into double frequency transient.
21
2
LL
L
VVC


22
2
2
1
CL
f


11
1
2
1
CL
f


26

27. TRAVELLING WAVES ON TRANSMISSION LINE
 A transmission line is distributed parameter circuit and a distinguishing
feature of such a circuit is its ability to support travelling waves of
voltage and current.
 Transmission line is shown in figure. It is taken lossless T.L.
 A T.L. is not energized all at once this is due to presence of distributed
constants.
 Here, when switch S is closed, the inductance L1 acts as an open
circuit and C1 as short circuit instantaneously.
27

28.  At the same instant next section can not be discharged because the
voltage across capacitor C1 is zero. So unless the capacitor C1 is charged
to some value, charging of the capacitor C2 through L2 is not possible,
which will take some finite time.
 The same is applied to the third section and so on. So we see that the
voltages at the successive sections builds up gradually.
 This gradual built up of voltages over the T.L. conductors can be regarded
as a voltage wave is travelling from one end to the other end and gradual
changing of the capacitances is due to associated current waves, which
is accompanied by a voltage wave sets up a magnetic field due to which
reflection and refraction happen at terminal and junction.
 Now the electrostatic flux which is equal to the charge between the
conductors of the line up to a distance x is given by,
𝑞 = 𝑉𝐶𝑥
Now current is given by
……………..(1)
Where, v=velocity of the travelling wave
VC
dt
dx
VC
dt
dq
I 
28

29.  Now electromagnetic flux linkages created around the conductor due to
the current is,
So voltage is given by,
………………….(2)
 Dividing eq (2) by eq (1),
∴
 Now Multiplying eq (1) and (2)
ILx
ILv
dt
dx
ILV 
C
L
V
I
VCv
ILv
I
V

nZ
C
L
I
V

LC
v
LC
v
VILCvILvVCvVI
1
12
2



29

30.  Now for overhead line,
L= 2 × 10−7 𝑙𝑛
𝑑
𝑟
𝐻
𝑚𝑒𝑡𝑒𝑟
C =
2𝜋𝜖
𝑙𝑛
𝑑
𝑟
𝐹
𝑚𝑒𝑡𝑒𝑟
∴
= 3× 10-8 𝑚𝑒𝑡𝑒𝑟
𝑠𝑒𝑐𝑜𝑛𝑑
 So it means velocity of travelling wave in T.L. is equal to the velocity of
light.
2
1
7
ln
2
ln102
1















r
dr
d
v

30

31. TRAVELLING WAVE WITH OPEN END LINE
 Let us consider a T.L. open circuited at the receiving end and a steady
voltage E is suddenly applied at the sending end.
 When switch S is closed on an unenergised line of length L, the
voltage E and current I= which accompanies it travel on line at the
velocity v. At a time , they reach the far end .
 Necessarily, at the open end no current can flow and so,
 This means that voltage at the open end is raised by V volts. So total
voltage at the open end when wave reaches at the end,
V+V=2V …………(3)
Z
E
v
L
22
2
1
2
1
LdxICdxe 
EIZI
C
L
e n 
31

32.  The wave that starts travelling over the line when the switch S is
closed, can be considered as the incident wave and after the wave
reaches the open end the rise in potential V can be considered due to a
wave which is reflected at the open end and actual voltage at the open
end can be considered as refracted wave or transmitted wave and
so,
Transmitted Wave= Incident Wave + Reflected Wave
 From eq(3) it is cleared that for voltage, a travelling wave is reflected
back with +ve sign.
 Now let us see about current wave.
 Here, as incident current wave I reaches the open end the current at
the open end is zero, this could be explained by saying that a current
wave of I magnitude travels back over the T.L. So for an open end line
a current wave is reflected with –ve sign. 32

33. 33

34. SHORT CIRCUITED LINE
 Here when switch S is closed, a voltage wave of magnitude of V and
current of magnitude I starts travelling towards the S.C. end. Consider dx
element where electrostatic energy is and electromagnetic energy
is .
 Now,
=
E = iZ
i =
𝑉
𝑍
= I
 Variation of voltage and current over line is shown in fig.
2
2
1
CdxV
2
2
1
LdxI
2
2
1
CdxV 2
2
1
LdxI
34

35. 35

36.  It is seen from the figure that the voltage wave reduces to zero
periodically after it has travelled through a distance twice the length
of the line whereas after each reflection at the end the current is build
up by an amount of
.
IZ
V
n 
36

37. LINE TERMINATED THROUGH A RESISTANCE
 Consider a lossless T.L. which has a surge impedance of Z0
terminated through a resistance R. When the wave travels along the
line and absorbs any change then it is purely or totally reflected.
 But Transmitted Wave = (Incident Wave + Reflected Wave )
But & &
Since, .
So,
0Z
E
I 
𝐼′′
=
𝐸′′
0Z
0
'
'
Z
E
I 
'"
'"
EEE
III


000000
"2"'"
Z
E
Z
E
Z
EE
Z
E
Z
E
Z
E
R
E



RZ
ER
E


0
2
" 37

38.  And current
 Similarly for E” in terms of (E+E’) becomes,
And
So, coefficient of refraction for current waves=
coefficient of refraction for voltage waves =
Similarly coefficient of reflection for current waves=
coefficient of reflection for voltage waves =
0
0
0
0
0
" 222
ZR
Z
I
ZR
Z
Z
E
ZR
E
I






0
0
00
'
''
ZR
ZR
EE
Z
E
Z
E
R
EE





)(
)(
'
0
0
00 ZR
ZR
Z
E
Z
E
I



0
02
ZR
Z

)(
)(
0
0
ZR
ZR



0
2
ZR
R

)(
)(
0
0
ZR
ZR



38

39. EXTREMITIES
OPEN
CIRCUIT(R=∞)
coefficient of reflected
current wave= -1
coefficient of reflected
voltage wave= 1
coefficient of refracted
current wave= 0
coefficient of refracted
voltage wave= 2
SHORT
CIRCUIT(R=0)
coefficient of reflected current
wave= 1
coefficient of reflected voltage
wave= 0
coefficient of refracted current
wave= 2
coefficient of refracted voltage
wave= 0 39

40. LINE CONNECTED TO A CABLE
■ When a wave travels toward the cable from a line, due to difference in
impedance wave suffers from reflection.
■ Transmitted voltage wave is given by:
E′′ = 𝐸 ∗
2 ∗ 50
50 + 500
=
2
11
𝐸
■ Surge impedance of line ≅ 500Ω
Cable ≅ 50Ω
40

41. ■ Thus it can be noted that voltage transmitted to a cable is almost 20%
of the incident voltage.
■ These cable also helps in maintain the steepness of the wave.
■ Care should be taken in selecting the length of this cable, because if it
will be shorter than expected length of the wave then piling of voltage
will began and value of voltage may attain incident voltage.
41

42. ATTENUATION OF TRAVELLING WAVES
■ In practical system, there is occurrence of attenuation which
results is losses.
■ As a result there analysis becomes difficult.
■ These losses are mainly due to presence of resistance and
conductance.
So for these, Consider R,L,C & G per unit length of Overhead T.L
having V0 and I0 as there voltage and current waves at X=0.
42

43. After travelling distance-X
Power Loss in differential element:
𝑑𝑝 = 𝐼2 𝑅𝑑𝑥 + 𝑉2 𝐺𝑑𝑥
Power at distance X:
𝑉𝐼 = 𝑃 = 𝐼2 𝑍 𝑛
Differential Power:
𝑑𝑝 = −2𝐼𝑍0 𝑑𝐼
43

44. Comparing power loss and differential power, we get:
𝐼2 𝑅𝑑𝑥 + 𝑉2 𝐺𝑑𝑥 = −2𝐼𝑍0 𝑑𝐼
𝑑𝐼 = −
𝐼(𝑅 + 𝐺𝑍𝑛2)
2𝑍0
𝑑𝑥
𝑑𝐼
𝐼
= −
(𝑅+𝐺𝑍𝑛2
)
2𝑍0
𝑑𝑥
𝑙𝑛𝐼 = −
(𝑅 + 𝐺𝑍𝑛2)
2𝑍0
𝑥 + 𝐴
44

45. 𝑙𝑛𝐼 = −
(𝑅 + 𝐺𝑍𝑛2)
2𝑍0
𝑥 + 𝐴
At 𝑥 = 0,
𝐴 = 𝑙𝑛𝐼0
So,
𝑙𝑛
𝐼
𝐼 𝑂
= −𝛼𝑥 Where 𝛼 =
(𝑅+𝐺𝑍𝑛2
)
2𝑍0
Therefore,
𝐼 = 𝐼0 𝑒-αx
Hence,
𝑉 = 𝑉0 𝑒-αx
45

46. ■ Value of resistance not only depends upon the size of
conductors.
■ It also depends on the size and the shape of the wave.
■ So for this, equation of Menger and foust comes into account,
which is given by:
𝑉 =
𝑉0
1 + 𝑘𝑥𝑉0
Where,
x is distance in kms,
k is attenuation constant,
k=0.00019 (short waves)
k=0.00010 (long waves)
46

47. Arcing Ground
■ Arcing ground is the surge, which is produced if the neutral is not
connected to the earth.
■ The phenomenon of arcing ground occurs in the ungrounded three
phase systems because of the flow of the capacitance current.
■ The capacitive current is the current flow between the conductors
when the voltage is applied to it.
■ During fault,
CapacitorVoltage Reduces to Zero
47

48. Arcing Ground Phenomena
■ In a three phase line, each phase has a capacitance on earth.
■ When the fault occurs on any of the phases, then the capacitive fault
current flows into the ground.
■ If the fault current exceeds 4 – 5 amperes, then it is sufficient to
maintain the arc in the ionized path of the fault, even though the fault
has cleared itself.
48

49. Elimination of Arcing Ground
The surge voltage due to arcing ground can remove by using the
arc suppression coil or Peterson coil. The arc suppression coil has
an iron cored tapped reactor connected in neutral to ground
connection.
49

50. ■ The reactor of the arc suppression coil extinguishes the arcing ground by
neutralizing the capacitive current. The Peterson coil isolates the system, in
which the healthy phases continue supplies power and avoid the complete
shut down on the system till the fault was located and isolated.
50

51. CAPACITANCE SWITCHING
■ A hazardous condition occurs in power system when a capacitor
bank is connected to long transmission line is disconnected.
■ Consider a power system as shown in figure below:
■ Suppose switch S is interrupted at instant A and capacitor is
charged up to its max. value Em
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52. ■ At instant B the voltage across the switch reaches its peak 2Em.
■ This voltage causes reignition to generate arc and cause oscillatory
transient.
■ After reignition the current in the circuit is given by:
𝑖 𝑒 =
−2𝐸𝑚
𝐿
𝐶
sin( 𝑡 𝐿𝐶)
TransientVoltage across capacitor:
𝑉 = −2𝐸𝑚[1 − 𝑐𝑜𝑠 𝑡 𝐿𝐶 ]
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53. So total voltage across capacitor is = initial voltage + transient voltage,
which is give by:
= 𝐸 𝑚 − 2𝐸𝑚[1 − 𝑐𝑜𝑠 𝑡 𝐿𝐶 ]
This value can be maximum increased up to −3𝐸𝑚
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54. THANKYOU
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