Surges  in  Power  System  

§  High voltage surges induced on the transmission line due to; 1. Direct stroke 2. Indirect stroke 3. Switching operations ■ The surge travels along the line at the speed of light ■ As they reach the end (termination) or junction of the line, partly are reflected and transmitted.

A wave may be seen traveling down the rope's length until it dissipates entirely due to friction

This is analogous to a long transmission line with internal loss: the signal steadily grows weaker as it propagates down the line's length, never reflecting back to the source

If the far end of the rope is secured to a solid object at a point prior to the incident wave's total dissipation, a second wave will be reflected back to your hand.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

The incident wave, the reflected wave and the transmitted wave are formed in accordance with Kirchhoff's laws. They must also satisfy the

differential equation of the line.

Figure 1

Z1 Z2

Consider a step-voltage wave of magnitude E incident at junction J between two lines of surge impedances Z1 and Z2. A portion ET of this surge would be transmitted and a portion ER would be reflected.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

No discontinuity of potential at the junction J. Therefore

TR EEE =+No discontinuity of current at the junction J. Therefore

I + IR = IT

The incident surge voltage E is related to the incident surge current I by the surge impedance of the line Z1. Similarly the transmitted surge voltage ET is related to the transmitted surge current IT by the surge impedance of the line Z2 and the reflected surge voltage ER is related to the reflected surge current IR by the surge impedance of the line Z1.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

Knowing that I + IR = IT

Substituting these values gives

E/Z1 - ER/Z1 = ET/Z2 = (E + ER)/Z2

However it is to be noted that the reflected wave is a reverse wave. Thus we can write,

E = Z1 I , ET = Z2 IT , and ER = - Z1 IR

After simplification; reflected wave can be written as : Similarly, the transmitted surge may be written as:

Reflec%on  of  Traveling  waves  at  a  Junc%on  

Since both these surges are a definite fraction of the incident surge, a transmission factor β and a reflection factor α are defined as follows.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

Note : When the junction is matched (i.e. Z1 = Z2), then there is no reflection and the reflection factor

can be seen to be zero.

12

2

12

12

21

;

ZZZ

ZZZZ

+=+=

+

−=

αβ

α

Example 1

An overhead line A with a surge impedance 450 Ω is connected to three other (overhead lines B and C with surge impedances of 600Ω each, and a cable D with surge impedance 60Ω) at the junction J. A traveling wave of vertical front of magnitude 25 kV and very long tail travels on A towards the junction J. Calculate the magnitude of the voltage and current waves which are transmitted and reflected when the surge reaches the junction J. Attenuation in the lines can be neglected.

Solution

When the line Z1 is open circuited at the far end (i.e. Z2 = ∞), then the full wave is reflected back and the reflection factor becomes 1.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

Case (i): Open ended transmission line of surge impedance Z

α = 1 and no wave is transmitted (β does not exist)

Since

Thus ;

Extreme cases :

12

2

12

12 21;ZZZ

ZZZZ

+=+=

+

−= αβα

When the line Z1 is short circuited at the far end (i.e. Z2 = 0), then no voltage can appear and the full wave is reflected back negated so as to

cancel the incident wave and the reflection factor becomes - 1.

Reflec%on  of  Traveling  waves  at  a  Junc%on  

Case (ii): Short –circuit ended transmission line of surge impedance Z

α = -1 ; β = 0

Since

Thus ; 12

2

12

12 21;ZZZ

ZZZZ

+=+=

+

−= αβα

Bewley Lattice Diagram Shows at a glance the position and direction of motion of every incident, reflected, and transmitted wave on the system at every

instant of time.

■ All waves travel downhill because time increases ■ Position of any wave at any time can be deduced directly from the diagram ■ Total potential at any point is the superposition of all the waves arrived at that point ■ History of wave can easily be traced ■ Attenuation is included

Bewley Lattice Diagram – Open end line

Reflected without loss or change of sign

K = Attenuation factor

Bewley Lattice Diagram – Open end line

A step voltage wave of amplitude unity starting from the generator end at time t = 0.

Along the line the wave is attenuated and a wave of amplitude 0.7536 reaches the open end at time ζ

At the open end, this wave is reflected without a loss of magnitude or a change of sign. The wave is again attenuated and at time 2ζ reaches the generator end with amplitude 0.5679

t=0   0  

t=ζ   ζ  

t=2ζ          2ζ  

+1  

+0.7536  

+0.7536  

+0.5679  

Receiving  end  α =1  Sending  end  (α =-1)  

In order to keep the generator voltage unchanged, the surge is reflected with a change of sign (-0.5679), and after a time 3ζ reaches the open end being attenuated to -0.428.

Bewley Lattice Diagram – Open end line

It is then reflected without a change of sign and reaches the generator end with amplitude -0.3225 and reflected with amplitude +0.3225.

t=2ζ  

t=3ζ   3ζ  

t=4ζ          4ζ  

-‐0.428  

-‐0.428  

-‐0.3225  

     2ζ  

       5ζ        t=  5ζ  

+0.3225   +0.2431  

Sending  end  (α = -1)   Receiving  end  (α = 1)  

-‐0.5679  

Assume  that  R  =  3  Z1  The  reflec?on  factor  at  receiving  end,    αreceiving = (3Z1- Z1)/(3Z1 + Z1) = 0.5. At the sending end, the reflection operator αsending = (0- Z1)/(0 + Z1) = -1 Assume the line is short and no attenuation involved, draw the

lattice diagram from t = 0 to t = 9ζ

 

Bewley Lattice Diagram – Line terminated with resistance Z2 =

R

Bewley Lattice Diagram – Line terminated with resistance Z2 =

R

t=0   0  

t=ζ   ζ  

t=2ζ          2ζ  

+1  

+1  

+0.5  

+0.5  

Receiving  end  α =0.5  Sending  end  (α =-1)  

Bewley Lattice Diagram – Line terminated with resistance Z2 =

R

t=2ζ  

t=3ζ   3ζ  

t=4ζ          4ζ  

-‐0.5  

-‐0.25  

-‐0.25  

     2ζ  

       5ζ        t=  5ζ  

+0.25  +0.25  

Sending  end  (α = -1)   Receiving  end  (α = 1)  

-‐0.5  

t=5ζ  

t=6ζ   6ζ  

t=7ζ          7ζ  

+0.125  

-‐0.125  

-‐0.125  

     5ζ  

       8ζ        t=  8ζ  -‐0.0625   -‐0.0625

 

Sending  end  (α = -1)   Receiving  end  (α = 1)  

+0.125  

     t=  9ζ        t=  9ζ  

+0.0625  

+0.0625  

Bewley Lattice Diagram – Line terminated with resistance Z2 =

R

After time t = 9ζ , voltage at the receiving end is given by

Vr= 1 + 0.5 – 0.5 +0.25 - 0.25 + 0.125 – 0.125 + -0.0625 +0.0625 = 1

It can be seen that the voltage and current oscillate

around the value 1 and finally settle

down to this value.