ee572 power system notes

Load Flow Study

Introduction Size of a power system (generation, transmission and distribution) increases day by day. Then number of interconnection increase, so that a large number power system data must be handled continuously and accurately. System variables such as voltage levels and power flow in the transmission and distribution parts are needed to be analyzed under the normal operation as well as abnormal conditions such as sudden large disturbances. Load flow becomes important in that respect. A load flow study is the determination of the voltage, convert power, power factor etc of various points in an electrical network under normal operation. Although load flow study is considered as an important section in power systems, the problem can also be solved by circuit analysis. As a simple example let’s consider a resistor (1 Ω) connected to two parallel batteries of voltage 1V. Then the current flowing to the resistor is 2A. The voltage across each battery is 1V. Now if you consider the effect of internal impedance of batteries and the impedance of the wires of the circuit, the problem becomes complicated. But it still can be analyzed using Kirchoff’s laws. Now if you replace the battery by generator transformer unit, wires by transmission lines and resistor by a load, the same problem can be considered as a three bus system. Thus the load flow problem can be analyzed using Kirchoffs laws. However, in reality, the number of buses in a power system is large, (i.e. in Sri Lankan transmission system it is about 196 buses), so that number of variables become large. So, load flow study can be easily analyzed by iterative methods rather than direct solution. The main objective of the load flow study is to calculate the bus voltages under steady state condition. Further it provides information about power flows in lines, losses in lines and transformers and any required reactive power compensation. The load flow can also provide information whether future expansion of the system will meet the load demand. In general, there are number of commercially available computer software to analyze load flow. However, this theory part covers the knowledge necessary to write your own programs to analyze the load flow study. Classification of buses The buses can be categorized in to following groups. (1) slack bus or swing bus or reference bus One bus is considered as reference bus and voltage is specified and the phase angle is set to zero. In a power system, the system losses are not known accurately before the load flow solution is calculated. Then it is not possible to specify all the injected real power in each bus. Thus the real power in the reference bus is allowed to swing and it supplies the slack between schedule real power and the sum of loads and losses. Known quantities lVl=Vspec, i.e. 1.05, δ=00

Unknown quantities P,Q (2) Generator bus or voltage controlled bus or PV bus Generators are connected to PV bus. Real power generation and voltage magnitude are specified. Known quantities lVl, and P unknown quantities δ, Q (3) Load bus or PQ pus Net real and reactive powers are known. Loads and sometimes and loads and generators are connected to this bus

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Known quantities P,Q Unknown quantities lVl, δ Usually the ratio between number of load buses to generator buses is approximately 85/15. Basic Equations Assume there is a n-bus power system. The complex power injected to the kth bus can be written as

*

1

**s

n

skskkkkkk VYVIVQPS ∑

=

==+=

If we take voltage and admittance in the polar coordinates, ,,, ksksksssskkk YYVVVV θδδ ∠=∠=∠=

Active and reactive power can be written as

)sin(

)cos(

1

1

kssk

n

sksskk

kssk

n

sksskk

YVVQ

YVVP

θδδ

θδδ

−−=

−−=

∑

∑

=

=

Alternatively if the admittance is written in rectangular coordinates as ksksks jBGY +=

[ ]

[ ]∑

∑

=

=

−−−=

−+−=

n

sskksskksskk

n

sskksskksskk

BGVVQ

BGVVP

1

1

)sin()cos(

)sin()cos(

δδδδ

δδδδ

The former set of equations for active and reactive power is used for Gauss-Seidal method whereas the latter is used for Newton-Raphson method. Limits When analyzing the load flow problem, the variations of the following parameters are checked whether they are within the limits.

The voltage magnitude of each bus must satisfy that they are between maximum and minimum limit. This limit arises because the power system components are designed to operate at a fixed voltage i.e. 5-10% of rated value.

The voltage angle difference must be less than maximum limits. This constraint limits the maximum permissible power angle of the transmission lines.

In addition active and reactive power also has limits due to physical limitations of the generators. Iterative methods The load flow problem is solved by any solution techniques used to solve non-linear algebraic solutions. The common methods are Gauss-Seidal methods and Newton Raphson method. A modified version of Newton Raphson method called fast decoupled method is also used.



3

Gauss-Seidal method Consider a n-bus system. The voltages can be written as V 1, V2, …Vk … Vn and their values need to be calculated. We need n number of different equations. Let’s rewrite each equation in the following form:

( )

( )

( )nknn

nkkk

nk

VVVfV

VVVfV

VVVfV

,......,....,.....................................

,......,....,.....................................

,......,....,

1

1

111

=

=

=

The above sets of equations help to calculate the voltages using iterative method. First initial voltage values are substituted in each equation in the right hand side, the corresponding voltages can be estimated in the left hand side after the first iteration. The estimated voltage values in the first iteration can be re-substituted so that a set of voltages can be estimated after the second iteration. This process can be repeated until satisfactory results are obtained for the voltages. After i th iteration, the set of equations looks as follows:

( )

( )

( )ni

kii

nni

ni

kii

kki

ni

kiii

VVVfV

VVVfV

VVVfV

111

1

111

1

111

111

,......,....,.....................................

,......,....,.....................................

,......,....,

−−−

−−−

−−−

=

=

=

Now we do some modification to improve the convergence of the voltage estimation. When estimating the voltage of the bus 2 from equation 2, V 1 has already been estimated from equation 1. Similarly, in equation k, when Vk is estimated, voltages up to Vk-1 have already been estimated in previous equations. So, the sets of equations can be modified as

( )

( )

( )ni

ni

kii

nni

ni

ki

kii

kki

ni

kiii

VVVVfV

VVVVfV

VVVfV

111

1111

111

111

,,......,,....,.....................................

,......,,....,.....................................

,......,....,

−−

−−−

−−−

=

=

=

In each iteration, the change of each voltage between consecutive iterations are estimated

ki

ki

k VVV 1−−=∆ If ΔVk error limit, the iteration can be stopped. Now we need to write the voltage of each bus as a function of all the other buses. Let’s consider the admittance matrix of the n-bus system.



4

=

nnnkn

knkkk

nk

YYY

YYY

YYY

Y

...............................

..............................

........

1

1

1111

The diagonal elements represent the total admittances in each bus. The off-diagonal elements represent the admittances of each line with negative sign. (i.e. -Y 12 is the admittance between bus 1 and 2). The capacitances of the lines are added to the diagonal elements if we take the nominal π network for the line model. The injected current in k th bus can be written as Ik = Yk1V1 + Yk2V2 +…….+ YkkVk + …….+ YknVn This can be simplified to

∑∑+=

−

=

++=n

ksskskkk

k

ssksk VYVYVYI

1

1

1

In addition

*k

kkk V

jQPI

−=

By simplifying

−−

−= ∑∑

+=

−

=

n

ksSks

k

ssks

k

kk

kkk VYVY

VjQP

YV

1

1

1*

1

Now we can estimate V. In the i iteration of the k th bus

−−

−= ∑∑

+=

−−

=−

n

ksS

iks

k

ss

iksi

k

kk

kk

ki VYVY

VjQP

YV

1

11

1)1*(

1

Iterate until errorVV i

ki

k <− −1

In the real situation the slack bus is not used for the iteration (V is known and fixed). For PQ buses, P and Q are known so iterations can be done directly. However, in PV buses, Q is not known and should be calculated. Q for k th bus

Qcalk = - im [V*

k (Yk1V1 + Yk2V2 + …….. + YkkVk + ……… + YknVn)] Following steps can be followed in Gauss-Seidal method. Step 1 Read data

(P for all buses excluding slack bus, Q for all PQ buses, lVl for all PV buses) Set voltage at slack bus Step 2 Calculate admittance matrix Y



5

Step 3 Assign initial conditions (for PV buses V=lVspl <00 ) (for PQ buses V=1 <00 ) Step 4 Calculations (for slack bus no calculations, for PV buses compute Q and calculate V, for PQ buses calculate V) Step 5 Test for convergence Finally calculate P and Q for slack bus, Q for all PV buses, line losses and currents. In order to accelerate the process following modification can be down when upgrading the voltages in each step.

)( 11 ik

ik

ik

ik VVVV −+= ++ α

Usually α is selected as 1.6. Newton_Raphson Method In Newton-Raphson method, we use a different iterative approach. To explain the method, let’s consider a variable x which satisfies a function y=f(x). So, if x 0 is solution, it will satisfy y=f(x). If the equation does not satisfy, now add a small x, Δx0 and check it again. Let’s take the change in y as Δy0. Now we use Tayler’s expansion and by neglecting higher order terms in the Taylar’s series, the change in y can be written as Δy0 = Δx0 (∂f/∂x)0

Now we extend the problem for n number of variables x 1….xn which corresponds to y1….yn. The relationship between Δy0 and Δx0 in matrix form

01

0

1

1

11

1

101

......

......

....................................

...................................

..........

......

......

∆

∆

∆

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∆

∆

∆

n

i

n

n

i

nn

i

i

i

ii

ni

n

i

x

x

x

xf

xf

xf

xf

xf

xf

xf

xf

xf

y

y

y

(1)

Here Δy is known values and are P and Q in load flow study. Δx are unknowns such as bus voltages. unlike in G-S method, in N-R method the bus voltages are separated into two as in complex numbers. If the bus voltages are in polar coordinates, the variables are lVl and δ. On the other hand they are e and f in rectangular coordinates where V=e+jf. Let’s assume a n-bus system with g number of PV buses. In polar coordinates, ΔP is known for all PV and PQ buses, whereas ΔQ is known only for all PQ buses. Their correspondent variables are Δδ and Δ lVl. Then Δy becomes ΔP and ΔQ whereas Δx becomes Δδ and Δ lVl Equation 1 now can be written in the following sub-matrix form.



6

∆∆

∂∂

∂∂

∂∂

∂∂

=

∆∆

////

//V

VQQVPP

QP δ

δ

δ (2)

Number of elements in Δy is now 2n-g-2 where ΔP is known for n-1 buses and ΔQ is known for n-g-1 buses. Number of elements in Δx is now 2n-g-2 where ΔlVl is known for n-g-1 buses and Δδ is known for n-1 buses. The middle matrix called Jacobian matrix has (2n-g-2)x(2n-g-2) elements. Within the J matrix ∂P/∂δ is a sub-matrix of (n-1)x(n-1) elements, ∂P/∂lVl is a sub-matrix of (n-g-1)x(n-1) elements, ∂Q/∂δ is a sub-matrix of (n-1)x(n-g-1) elements, and ∂Q/∂lVl is a sub-matrix of (n-g-1)x(n-g-1) elements. When it comes to rectangular coordinate system, the known values are ΔP, ΔQ and Δ lVl. The variables are Δe and Δf. Number of elements in Δy is now 2n-2 where ΔP is known for n-1 buses, ΔQ is known for n-g-1 buses and ΔlVl is known for g buses. Number of elements in Δx is now 2n-2 where Δe is known for n-1 buses and Δf is known for n-1 buses. The Jacobian matrix has (2n-2)x(2n--2) elements. Equation 1 now can be written in the following sub-matrix form.

∆∆

∂∂∂∂

∂∂∂∂

∂∂

∂∂

=

∆∆∆

fe

fV

fQ

eV

eQ

fP

eP

VQP

//////

where ∂P/∂e is a sub-matrix of (n-1)x(n-1) elements, ∂P/∂f is a sub-matrix of (n-1)x(n-1) elements, ∂Q/∂e is a sub-matrix of (n-1)x(n-g-1) elements, ∂Q/∂f is a sub-matrix of (n-1)x(n-g-1) elements, ∂/V//∂e is a sub-matrix of (n-1)x(g-1) elements and ∂/V//∂f is a sub-matrix of (n-1)x(g-1) elements. When comparing polar coordinates with rectangular coordinates, the size of the J matrix is large for rectangular coordinates. The elements of the J matrix in the polar coordinates (the diagonal and off-diagonal) can be written as



7

[ ]

[ ]

[ ]

[ ])cos()sin(

)sin()cos(

)sin()cos(

)cos()sin(

2

2

skksskksks

k

kkkk

k

k

k

skksskkssks

k

kkkkk

k

skksskksks

k

kkkk

k

k

k

skksskkssks

k

kkkkk

k

BGVVP

VBVQ

VQ

BGVVQ

VGPQ

BGVVP

VGVP

VP

BGVVP

VBQP

δδδδ

δδδδδ

δ

δδδδ

δδδδδ

δ

−−−=∂∂

−=∂∂

−−−−=∂∂

−=∂∂

−+−=∂∂

−=∂∂

−−−=∂∂

−−=∂∂

Following steps can be followed in Newton-Raphson method. Step 1 Read data

(P for all buses excluding slack bus, Q for all PQ buses, lVl for all PV buses) Set voltage at slack bus Step 2 Calculate admittance matrix Y Step 3 Assign initial conditions (for PV buses V=lVspl <00 ) (for PQ buses V=1 <00 ) Step 4 Calculations (for slack bus no calculations, for PQ buses compute Q and for all buses except slack calculate P). Calculate ΔP and ΔQ with assign lVl and δ. Calculate the elements in the Jaccobian matrix. Solve equation 2 and obtain ΔlVl and Δδ and update those values . Step 5 Test for convergence Finally calculate P and Q for slack bus, Q for all PV buses, line losses and currents. Decoupled Newton Method Since the Newton Raphson method is complicated, the convergence time is high. So, fast decoupled method is used for saving time for computation. In this method, it is assumed that the variation of active power w.r.t. lVl and reactive power w.r.t. δ are negligible. So, the equation (2) will simplifies to



8

∆∆

∂∂

∂∂

=

∆∆

VVQ

P

QP δδ

0

0

The matrix can be simplifies to the following two equations to obtain ΔlVl and Δδ.

[ ] [ ]

[ ] [ ]QVQV

PP

∆

∂∂

=∆

∆

∂∂

=∆

−

−

1

1

δδ

The same equations which were used in Newton Raphson method can be used to calculate the elements in partial derivatives. Now instead of finding the inverse matrix of (2n-g-2)x(2n-g-2) elements, the new sizes of matrices are (n-1)x(n-1) and (n-g-1)x(n-g-1). For example in Sri Lankan system, n=196, g=26, initially it is 364x364 matrix. Now they are 195x195 and 169x169. Fast Decoupled method The fast decoupled method was developed by B. Stott in 1974. Some assumptions were made to simplify the terms in J matrix so that convergence rate is high. The assumptions are Under normal loading conditions, the angle difference across transmission lines is negligible. In transmission line its reactance is more than the resistance so G is ignored compared to B. Based on the assumptions, the equations used in NR method simplifies to

kkss

kkkk

k

k

s

k

k

k

s

k

k

k

skkss

kkkk

k

k

VBVPVB

VQ

QQVP

VP

VVBPVBP

−=∂∂

−=∂∂

=∂∂

=∂∂

=∂∂

=∂∂

−=∂∂

−=∂∂

,

0,0,0,0

,2

δδ

δδ



9

Comparison Between Methods The relative advantages and disadvantages between Gauss-Seidal method and Newton Raphson methods are given in following table.

Gauss-Seidal Newton-Raphson Advantages Disadvantages

Simple Small computer memory required Less computational time per iteration

Solution technique is difficult Computer memory requirement is large Computing time per iteration is large

Disadvantages Advantages Slow rate of convergence Increase number of iterations with increase number of buses Effect of convergence from the selection of slack bus

More accuracy and surety of convergence No of iterations is almost independent of the system size No effect from the selection of slack bus



10

Symmetrical Components

Introduction A balanced three-phase system exists only in theory. Many systems are nearly balanced but with emergency conditions (i.e. unsymmetrical faults, unbalanced loads, open conductors and unsymmetrical condition in rotating machine etc.) system becomes unbalanced. In order to protect against the unbalanced condition, a symmetrical system is necessary. In 1918 Fortescue proposed a method for solving an unbalance set of n related phases into n set of balance phases called symmetrical components. These phases in each set are in equal magnitude and spaced 120° or 0°. We commonly use symmetrical component theory in power systems. The main objective of the symmetrical components is to analyze unsymmetrical faults. (The symmetrical components is similar when we have several vectors and how we analyze them by separating into several axes i.e. x-axis, y-axis and z-axis). Let’s consider a balance system as shown below.

Each voltage phases can be written as follows VR = VR∠0 VB = VR∠120 VY = VR∠240 Symmetrical components

Now consider an unbalanced system where we have different voltage magnitudes and phase angles spaced other than 120°. (see below)

VR VB

VY

Now we introduce the symmetrical components to solve the unbalance system. Any unbalanced 3 ∅ system shown above can be split into three-balanced system of phase sequences. 1. Positive sequence system – having the same phase sequence as the unbalance system 2. Negative sequence system – having the opposite phase sequence as the unbalance system 3. Zero sequence system – having all three phase in sequence

VB

VY

VR



11

Let’s have a look how we can split the unbalance system into three different sequences.

R Y B VB1 VR1

120

= 120 120 VY1

Positive sequence RYB

A positive sequence system is represented by a balanced system (equal magnitude & 120 phase) having some phase sequence (RYB) as the original unbalanced system. VY2 R B Y

120 120

120

VB2 VR2

negative sequence RBY A negative sequence system is represented by a balanced system having opposite phase sequence (RBY) as the original unbalanced system.

VR0 VR0 = VB0 =VY0

zero sequence A zero sequence system is represented by 3 single phases having equal magnitudes & angular displacements. The three voltage phases VR , VY , VB of the unbalanced set can be graphically represented by their symmetrical components.

R phase VR = VR0 +VR1 +VR2 (R phase voltage is the addition of all sequences of the R phase)

VR VB

VY



12

VR2 VR1 VR0 VR Y phase V Y= VY0 +VY1 +VY2 (Y phase voltage is the addition of all sequences of the Y phase)

VY0 VY

VY1 VY2 B phase VB = VB0 +VB1 +VB2 (B phase voltage is the addition of all sequences of the B phase) VB2 VB1 VB

VB0

Since we use symmetrical component theory in three-phase system, the most frequently used phase difference is 1200. Now we introduce an operator λ = 1 ∠120° = e j2π/3 = -0.5+j 0.866 for converting components.

Then the following relationships can be obtained between the operator λ and phase angles.

λ2 = λ x λ = (1 ∠120) (1 ∠120) = 1 ∠240 = 1 ∠-120 λ3 = λ2 x λ = (1 ∠240) (1 ∠120) = 1 ∠360 = 1 ∠0 λ4 = λ3 x λ = (1 ∠0) (1 ∠120) = 1 ∠120 = λ λ5 = λ4 x λ = (1 ∠120) (1 ∠120) = 1 ∠240 = λ2 λ6 = = 1

1 + λ + λ2 = 1 ∠0 + 1 ∠120 + 1 ∠240 =0 1 + λ + λ2 =0



13

In general, Parameters in Symmetrical Components Let’s obtain the relationships between parameters such as voltage, current, power and impedance of unbalanced system with their corresponding symmetrical components. Using λ, the parameters (i.e. voltage, current, impedance, power) for the Y and B phases can be written in terms of R phase. Voltages in symmetrical components The actual voltages, VR, VY and VB can be written in terms of the symmetrical components of R phase V R1, VR2, VR0 For positive sequence, VY1 = λ2 VR1 and VB1 = λ VR1

For negative sequence, VY2 = λ VR2 and VB2 = λ2 VR2 For zero sequence VR0 = VB0 = VY0 VB1= λVR1 VR1 VY2 = λVR2 VB2 = λ2VR2 VR0 = VY0 =VB0 VR2 VY1 = λ2VR1 Now consider the relationships between the actual voltages and their symmetrical components VR = VR1 + VR2 + VR0 VB = VB1 + VB2 + VB0 = VR1 + λVR2 + λ2VR0 VY = VY1 + VY2 + VY0

= λ2 VR1 + λVR2 + VR0

= = where A =

λ = λ4 = λn+1 = 1 ∠120 λ = λ5 = λn+2 = 1 ∠240 λ = λ6 = λn+3 = 1 ∠0

VR VY VB

1 1 1 1 λ2 λ 1 λ λ2

VRO VR1 VR2

VRYB A V012 1 1 1 1 λ2 λ 1 λ λ2



14

Therefore, A-1 = 1/3 Solving the matrix for symmetrical components, VR0 = 1/3( VR+ VY+ VB ) VR1 = 1/3( VR+ λVY+ λ2VB ) VR2 = 1/3( VR+ λ2VY+ λ B ) Currents in symmetrical components Similar to the voltage, the relationships between actual components and the symmetrical components of currents can be written as follows: = = Example: Determine the symmetrical components for V R = 7.3 ∠12.5°, VY = 0.4 ∠100°, VB = 4.4 ∠-154° V Ans VR1 = 3.97 ∠20.5°, VY1 = 3.97 ∠260.5°, VB1 = 3.97 ∠140.5°, VR2 = 2.52 ∠-19.7°, VY2 = 2.52 ∠100.3°, VB2 = 2.52 ∠220.5°, VR0= VY0 = VB0 = 1.47 ∠45.1°.

1 1 1 1 λ λ2

1 λ2 λ

V012 = A -1 VRYB

IR IY = IB

1 1 1 1 λ2 λ 1 λ λ2

IRO IR1 IR2 IR0

IR1 = 1/3 IR2

1 1 1 1 λ λ2 1 λ2 λ

IR IY IB

IRYB A I012

I012 A -1 IRYB



15

Power in symmetrical components The complex power S3∅ = P3∅ + jQ3∅ =SR + SY + SB = VRI*

R + VYI*Y + VBI*

B In matrix form S3∅ = = = But = and = S3∅ = = = 3 S3∅ = 3 = 3 S3∅ = 3 S3∅ = = 3 or

= 3

VR VY VB IR * IY IB

VR T

VY VB

IR *

IY IB

VRYB T IRYB *

VRYB T A T V012

T

IRYB * A * I012

*

V012 T I012 A T A *

A T A * 1 1 1 1 λ2 λ 1 λ λ2

1 1 1 1 λ λ2

1 λ2 λ

3 0 0 0 3 0 0 0 3

V012 T I012

*

VR0 VR1 VR2 IR0 *

IR1 IR2

VR0IR0 * + VR1R1

* + VR2R2*

VR IR * + VYIY

* +VBIB*

VR0IR0 * + VR1R1

* + VR2R2*



16

Impedances in symmetrical components IR VR ZR n ZB VY IY ZY VB The voltage, current and impedance relationships are

=

A

2

1

0

RN

RN

RN

VVV

=

R

R

R

ZZ

Z

000000

A

2

1

0

R

R

R

III

introducing symmetrical components

where

= =

Multiplying with A-1

2

1

0

RN

RN

RN

VVV

= 1/3 B

Y

R

ZZ

Z

000000

= 1/3 By solving the matrixes we can obtain VRN0 = 1/3 IR0 (ZR+ZY+ZB) + 1/3 IR1 (ZR+ λ2ZY+ λZB) + 1/3 IR2 (ZR+ λZY+ λ2ZB) VRN1 = 1/3 IR0 (ZR+ λZY+ λ2ZB) + 1/3 IR1 (ZR+ ZY+ ZB) + 1/3 IR2 (ZR+ λ2ZY+ λZB)

VRN VYN VBN

ZR 0 0 0 ZY 0. 0 0 ZB

IR IY IB

A

1 1 1 1 λ2 λ

1 λ λ2

1 1 1 1 λ λ2

1 λ2 λ

1 1 1 1 λ2 λ 1 λ λ2

I012

V012 1 1 1 1 λ λ2 1 λ2 λ

ZR ZR ZR

ZY λ2ZY λZY

ZB λZB λ2ZB

I012

IB



17

VRN2 = 1/3 IR0 (ZR+ λ2ZY+ λZB) + 1/3 IR1 (ZR+ λZY+ λ2ZB) + 1/3 IR2 (ZR+ZY+ ZB) For a balanced load star connected load of Z R=ZB=ZY

V RN0 = IR0ZR V RN1 = IR1ZR V RN2 = IR2ZR Similarly for a delta connected balance load, i.e. Z RY= ZYB= ZBR R

IRY IBR ZRY ZRb IYB Y ZYB B VRY0 = IRY0 ZRY, VRY1 = IRY1 ZRY, VRY2 = IRY2 ZRY Sequence networks and impedances Any unbalanced system (generators , positive lines etc.) can be divided into 3 sequence networks and analyzed separately. The networks of a particular sequence shows all the paths for flow of current at that sequence in the system. Positive, negative and zero sequence networks of loads Positive and negative sequences are simple to find out (current flows through the positive/negative impedances). In the zero sequence, zero sequence currents are in same magnitude & same phase. So 3 ∅ system operates as 1∅. But zero sequence current can occur if there is a complete path. In other cases zero sequence impedance is infinite. (open circuit) Let’s consider the most common cases: Case 1 Star connected ungrounded neutral

Z0 Z0

ref

Z0 N Z0 N0 There is no connection from neutral to ground or another neutral point. Thus the sum of current is zero (no zero sequence current).



18

Case 2 Y connected load with neutral ground IR0 R Z0 Z0 Z0 Z0 ref 3IR0 IR0 Y N0 IR0 B Neutral at a Y connected circuit is grounded through zero impedance. So zero impedance connection is inserted into then to connect the neutral point and the reference bus. Case 3 Star connected connection with ground with impedance R Z0 IR0 Z0 3ZN

Z0 3IR0 ZN Z0 ref Y B For impedance ZN, 3ZN impedance must be placed between neutral & ref bus. Case 4 Delta connected load R Z0 IR0 Z0 Z0 Z0 ref

Y

B



19

There is no return path for the current. The zero sequence network is open at the ∆ connected circuit and therefore the zero sequence current is zero. But current can circulate within the ∆ connected circuit. Sequence impedance with synchronous machines In synchronous machines the positive (Z 1), the negative (Z2) & the zero (Z0) impedances are different. The positive impedance is selected as its direct axis subtransient X” d or transient Xd reactances depending on the reaction of the time of the relays. For “fast” relays (usually for fault studies) X”d is used whereas for “slow” relays Xd is used. (Xd> X”d) Consider a star connected generator grounded through Z n impedance. Since the generator is designed to generate balance 3Ø voltages the generated voltages are positive sequence only. Positive sequence networks consist of an emf series with positive sequence impedance.

IR1 Z1 VR1 ER ref

The negative sequence reactance (Z2) is approximately equal to either Z2=(X’d+X’q)/2 or X2=(X”d+X”q)/2 For round rotor machines, X2=X1. The negative sequence network contains only the negative sequence impedances.

IR2 Z2 VR2

Zero sequence contains only the zero sequence impedances at the network. It contains the zero sequence impedance of the generator & the impedance to the ground. The zero sequence impedance is much less than positive sequence impedance. Z0 = Zgo + 3Za where Zgo = generator neutral to ground Za = neutral to ground IR0 VR0

Zgo

3Zn



20

From sequence networks, the relationship between sequence voltage, sequence current and sequence impedance can be obtained using the ohm’s law. VR0 = -IR0 Z0 VR1 = ER-IR1 Z1 VR2 = -IR2 Z2 In the matrix format, V012 = E - Z012 I012

2

1

0

R

R

R

VVV

= 0

0RE -

2

1

0

000000

ZZ

Z

2

1

0

R

R

R

III

Sequence impedance at transformers In transformers, the positive and negative impedances are same. The zero sequence impedance at the 3Ø unit is slightly different to positive and negative sequence impedances. But generally Z0= Z1= Z2= Ztrf

But for zero sequence currents, the winding arrangements in the transformer (i.e. delta star) has to considered. Here are most common types winding connections and their corresponding zero sequence networks. Case 1: Y-Y bank both neutrals and ungrounded P P Q Y Y P Q

ref There is no current path and the zero sequence current is zero.

Q Z0



21

Case 2: Y-Y bank both neutrals and grounded

Z0 P Q P Q Y Y P Q There is current path through the neutrals of the two windings. Case 3: Y-Y bank one neutral grounded P Q P Q Y Y P Q Zero sequence network is open between P and Q sothat zero sequence current cannot flow

Case 4: Δ-Y bank Y grounded

Z0 P P Q Zero sequence current can flow through the star connected ground. An induced current can circulate within Δ connected circuit. If the ground & neutral at Y has an impedance Z n The diagram looks like



22

Z0 P Q 3ZN Case 5: Y-Δ bank ungrounded Y

Z0 P P Q P Q There is no current path in the network.

Case 6: Δ -Δ bank

Δ Δ No zero sequence current can flow into Δ -Δ bank, but current can circulate through each windings.

P Q



23

Generator reactance For short circuit calculations the following three reactance are used. 1. Synchronous reactance (Xs) This reactance commonly used for performance in steady state operations and it includes reactance due to self inductance in the armature and reactance equivalent to armature reactance. The steady state short circuit current I is given by (E/Xs) where E is the emf of the machine. Xs=X+XL. Typical values of Xs in main synchronous generators at Kotmale 100% p.u., Victoria 95% p.u. and Randenigala 100% p.u. 2. Transient reactance Xd’ At a sudden short circuit across the terminator of the generator, the effective reactance is less than Xs. At the short circuit there is a sudden increase of armature current (current lags the voltage almost 90 °) and it produce a demagnetizing mmf to the main mmf. Due to this change, the magnetic flux in the air gap can change rapidly, but, in the core (in the poles) it need longer time to vary. To maintain the flux without changing at the pole base, the field current should be increased to overcome the demagnetizing effect in the armature. Increasing field current increases leakage flux and reduces the flux at base of the pole in the air gap. (The total flux equals to addition of the leakage flux and flux at pole base). Then the induced emf reduces from E to E’. The short circuit current I’ is now E’/X. For convenience we used I’=E/Xd’ where Xd’is the transient reactance. Xd’ >XL. Typical values of Xd’ in main synchronous generators at Kotmale 20% p.u., Victoria 23.9% p.u. and Randenigala 32% p.u. 3. Sub-transient reactance Xd” If the machine is a turbo-generator or fitted with damper winding the sudden air gap flux variation is not very high. This results initial short circuit current to a high value. The sub transient current I”=E/Xd”. I’=E/Xd’ where Xd’is the transient reactance. Xd”≈XL. Typical values of Xd” in main synchronous generators at Kotmale 17% p.u., Victoria 16.5% p.u. and Randenigala 21% p.u. When there is a symmetrical short circuit at the generator terminals, initially, the short circuit current I” is limited by sub-transient reactance Xd” for few cycles, then reduces to I’ limited by transient reactance and settles to steady state short circuit current I. the related time constants are Td” – sub transient and Td’- transient. Typical values Td’ and Td’’ at main synchronous generators at Kotmale power station Td’=1.6s and Td”=0.08s, Victoria power station Td’=2.08s and Td”=0.041s, and Randenigala power station Td’=1.9s and Td”=0.06s. In general, Isc = (I”-I’)e(-t/Td”) + (I’-I)e(-t/Td’) + I. Positive sequence impedance The positive impedance is selected as its sub-transient X” d or transient Xd reactances depending on the reaction of the time of the relays. For “fast” relays (usually for fault studies) X”d is used whereas for “slow” relays Xd is used. (Xd> X”d). Negative sequence impedance For turbo generators X2=(Xd”+Xq”)/2≈ Xd”. For salient pole machines it is about 20 % grater than Xd”. Typical values of X2 in main synchronous generators at Kotmale 19% p.u., Victoria 22.2% p.u. and Randenigala 23% p.u. Zero sequence impedance X0 is less than Xd” usually is 1/6 to ¾ time Xd”. Typical values of X0 in main synchronous generators at Kotmale 9% p.u., Victoria 12.1% p.u. and Randenigala 10% p.u.



24

Fault Analysis

Introduction A fault occurs in a power system due to failure of equipment or transient faults due to surges. The purpose of fault analysis is to calculate maximum and minimum fault current and voltages at different locations at the power system for different types of fault so a good protective system can be designed i.e. relay, C.B. etc… Assumptions in fault study:

1. Normal loads, line-charging capacitors, shunt elements in transformers and other shunt elements to ground are negligible.

2. All generated system voltages are equal and in phase. So pre fault bus voltages are set to 1∠0°

3. The series resistances at lines and transformers are neglected. Only reactances are contained

4. All transformers in their nominal tap positions. Fault analysis can be carried out in Pu system

5. A generator is represented by a constant voltage source and a proper reactance such as sub transient Xd″ or transient Xd’ or synchronous reactance Xd. Normally Xd″ is selected for positive sequence.

Classification All faults can be solved by using symmetrical components.

Faults

Balanced 3Φ faults (Symmetrical)

Unbalanced 3Φ faults (unsymmetrical)

Series faults

3Φ direct L-L-L 3Φ to ground L-L-L-G Single line to ground SLG Line to line –LL Double line to ground-DLG

One line open-OLO Two lines open-TLO



25

Balanced 3Φ faults 3Φ faults of no load Here 3phases are s/c in the generator.

Changing to pu notation base

pu ZZZ =

flbase I

EZ =

( )lfI

EZ

=

Ifl

EZpuZ =

( )IflEZpu

EIsc = and therefore ZpuIflIsc =

Fault MVA SF 3= VISC ( )VIflZpu

3= and therefore SF =

ZpuSbase

3Φ Fault at full load Here it is assumed that before the fault the generator supplies a load current which is significantly large (can not be neglected). Problems can be solved by using Thevenin equivalent circuit at the fault point. Unbalanced Faults Most of the 3Φ fault s are unbalanced i.e. line to ground (L-G), line to line (L-L), double line to ground (Dl-G). Accuracy percentage: LG 70% LL 15% DLG 10% LLL /LLLG 5% 3Φ faults are the severe ones. LG can be more severe than 3Φ fault under the following conditions.

1. the generators involved in the fault have solidly grounded neutrals or low impedance neutral Z.

2. on the Y side at ∆ Y transformer banks.



26

Unbalanced faults can be easily solved by using symmetrical component of an unbalanced system at currents and voltages. So the system is converted into +ve -ve and zero sequence networks. Single line to ground fault (SLG) SLG occurs when one conductor (i.e. R phase ) contacts the ground or neutral wire. Assume R phase connects the N or ground with Zf impedance. Fault current in R phase IRf Fault current in YB phases neglecting the load current Due to fault Using symmetrical components

2

1

R

R

RO

III

=31

λλλλ

2

2

11

111

B

Y

R

III

2

1

R

R

RO

III

=31

λλλλ

2

2

11

111

00RFI

So RFRRIRO IIII31

2 ===

For a terminal of a generator symmetrical components,

−

=

2

1

0

2

1

2

1

000000

0

0

R

R

RO

R

R

R

RO

III

ZZ

ZE

VVV

Then

222

111

00

RR

RRR

ROR

IZVIZEV

IZV

−=−=

−=

For voltages the symmetrical components,

Iyf=Ibf=0

VRf=Zf IRf

B Y R

R Y B

ZF



27

B

b

R

VVV

=

2

2

11

111

λλλλ

2

1

R

R

RO

VVV

B

b

eFF

VV

IZ=

2

2

11

111

λλλλ

2

1

R

R

RO

VVV

So VRO + VR1 + VR2=ZFIRF From 1,2,3 ER – Z1IR1 - Z0IR0 - Z2IR 2 = ZFIRf

ER – Z1IRF- Z1IR0 – Z1IRF = 3ZFIRf

Other calculated values

IR1 = IR2 = IR0 = F

R

ZZZZE

3210 +++

Generalized Fault Diagram

When a SLG fault occurs in a phase other than R phase Generalized Fault Diagram can be used. The +ve -ve and zero sequence networks are coupled by means of phase shifters ith complex turns ratio of 1, λ or λ2 for 0, 120 or 240 phase shifts. No magnitude change is used here.

PHASE Phase shift no n1 n2

R 1 1 1 Y 1 λ2 λ B 1 λ λ2

Z0

Z2

Z

Z1

Z

1 : n0

1: n1

1:n2

3Zf



28

Line to Line Faults(LL) LL occurs when two lines (i.e. Y, B) contact together through an impedance.

YFFBYYB

BFYF

RF

IZVVVII

I

=−=−=

= 0

−=

−

−

=

⇒

−

=

−==

−

=

⇒

=

22

11

2

1

2

1

0

2

1

2

1

0

2

1

0

2

1

21

0

2

2

2

12

2

2

1

00

000000

0

0

000000

0

0

0

0

11

111

31

11

111

31

R

RR

R

RR

R

R

RO

R

R

R

R

R

R

RO

RR

R

YF

YF

R

R

RO

B

Y

R

R

R

RO

IZIZE

II

ZZ

ZE

VVV

III

ZZ

ZE

VVV

III

II

III

III

III

λλλλ

λλλλ

F

RRR ZZZ

EII++

=−=21

21

Double line to ground fault (DLG)

DLG happens when 2 conductors (Y, B) fall and connected to ground or neutral.

R Y B

3Zf

Z1 Z2

R Y B

Zf Zf

Zg



29

(( ) ( )

( ) ( )

( ) ( ) 1220

20

1220

02

20

201

20

021

1

3

33

323

RFG

FR

RFG

GFR

R

GF

GFFF

RR

IZZZZZ

ZZI

IZZZZZ

ZZZI

ZZZZZ

E

ZZZZZZZZZZZ

EI

++++

+−=

++++

++−=

++

=

+++

+++++

=

Generalized Circuit Diagram

Three phase faults in general three phase faults are symmetrical faults which can also be solved by symmetrical components happens when phases contact together and connect the ground or neutral. Since there is no sources;

Phase Phase shift N0 N1 N2

Y-B 1 1 1 B-R 1 λ2 λ R-Y 1 λ λ2

If FZ - GZ =0

Z0

Z1

Z2

Zf

Zf

Zf +3Zg

R Y B

Zf Zf

Zg

Zf +3Zg

Z0

Zf

Z1

ER

Zf

Z2



30

F

RFBF

F

RFYF

F

RFRF

R

BF

YF

RF

F

RBF

F

RYF

F

RRF

R

BF

YF

RF

RFRRR

F

RR

RR

ZZEZV

ZZEZV

ZZEZV

VVVV

ZZEI

ZZEI

ZZEI

IIII

IZVVVZZ

EI

II

+∠

=+∠

=+

=

=

+∠

=+

∠=

+=

=

=⇒==+

=

==

1

0

1

0

1

12

2

1

0

1

0

1

12

2

1120

11

20

120,240,

0

0

11

111

120,240,

0

0

11

1110

0

λλλλ

λλλλ



31

Transient Analysis

Common Transients Transients or over-voltages are categorized into three groups depending on the time scale. The first class is very fast transients i.e. occurs within µs period. Lightning surges are examples for it. The second class is fast transients occurs in ms range. The examples are switching transients. The third class is slow transients considered as power frequency over-voltages occurs during second to minute period. Comparison of these transients are shown below.

Lightning Surges This type of transients appear in power system especially on transmission lines due to lightning surges. The surges can be due to direct or indirect hit on the transmission line. The surge travel through the line and come to the end terminals i.e. transformer at a sub-station. The usual protection methods are

(1) Using a shielding wire (2) Line insulators (3) Surge arrestors etc.

For testing purposes of the equipment in the power system, standard lightning impulses are used. It is 1.2/50 µs for voltage impulses and 8/20 µs for current impulses. The 1.2 µs is the front time with a tolerance of ± 30% (IEC 60). The front time can vary between 0.84 µs to 1.56 µs. The 50 µs is the time to half value with a tolerance of ±20%. It is the

10-6 100 10-2 10-4

2

8

6

4

Time [s]

Lighting surges

Normal voltage

Over-voltage

Switching surges

PU



32

time which the voltage reaches its half value. The time to half value can vary between 40 µs to 60 µs. The shape of the impulse is the behaviour of double exponential function as given below:

( )tt eeVtV βα −− −= 0)( The typical values for α and β are α ≈ 0.0143, β ≈ 4.87 where t is in µs. The shape of the impulse is given below.

Switching Surges The switching surges are caused by switching operation in a network. For testing purposes of the equipment in the power system, standard switching impulses are used. It is 250/2500 µs voltage impulses. The 250 µs is the time to crest value with a tolerance of ± 20% (IEC 60). The time to crest value can vary between 200 µs to 300 µs. The 2500 µs is the time to half value with a tolerance of ±60%. The time to half value can vary between 1 ms to 4 ms. Since switching transients are different in nature, a large tolerance is provided. The shape of the switching impulse is given below.

Time [µs]

Voltage

100%

50%

Thalf Tfront



33

Some of the examples for switching surges are given below. 1. Fault clearing at a short circuit point 2. Disconnecting of an unloaded transformer 3. Disconnecting of an unloaded line 4. Connecting to open unloaded line

5. Single line to ground faults – When a single line ground fault occurs a voltage equal and opposite occurs at the instant of fault inception. Traveling waves may initiate in the faulted and healthy phases due to capacitive coupling. 6. Non-simultaneous switching – For example when a transformer is connected to a network and poles of the circuit breakers do not close simultaneously.

Power Frequency Over-Voltages Following reasons may cause temporally over-voltages.

1. Line dropping – When a large load is disconnected the resistive and reactive voltage drops disappear and over-voltage occurs until the operating value is restored.

2. Ferro resonance – a high voltage can appear across the elements of a series LC circuit when it is excited near or at their natural frequency.

3. Ground faults – When a ground fault occurs with a non-earthed neutral, the healthy phases will adopt a √3 times higher voltage until the fault is cleared.

Time [µs]

Voltage

100%

50%

Thalf Tcrest



34

Wave Propagation in a Lossless Transmission Line Wave Equation Consider a small segment ∆x of a long transmission line. The series resistance and inductance of that part is given by R∆x and L∆x. The shunt admittance and capacitance of that part is given by G∆x and C∆x. The complete diagram is shown below.

Using Kirchhoff’s laws;

( )( ) ( )

( ) 0

0)(

=∆+−

∂∂

∆+∆−

=−∆++∂

∆+∂∆+∆∆+

IItVxCxGI

VVVt

IIxLxRII

By neglecting the higher order terms and considering limits of ∆x towards to zero the above two equations can be simplified to

tILRI

xV

∂∂

−−=∂∂ (1)

tVCGV

xI

∂∂

−−=∂∂ (2)

The equations can be solved when one of the variables (V or I) is removed. The voltage wave V(x,t) can be obtained by differentiating equation (1) with respect to x and equation (2) with respect to t. The modified equations are.

V+∆V I+∆I

G∆x C∆x

∆I

R∆x L∆x I

V



35

xtIL

xIR

xV

∂∂∂

−∂∂

−=∂∂ 2

2

2

(3)

2

22

tVC

tVG

txI

∂∂

−∂∂

−=∂∂

∂ (4)

By substituting equation (2) and (4) in (3)

∂∂

−∂∂

−−

∂∂

−−−=∂∂

2

2

2

2

tVC

tVGL

tVCGVR

xV

The equation can be rearranged as

( ) 2

2

2

2

tVLC

tVLGRCRGV

xV

∂∂

+∂∂

++=∂∂ (A)

Similarly an expression for I can be obtained as

( ) 2

2

2

2

tILC

tILGRCRGI

xI

∂∂

+∂∂

++=∂∂ (B)

The two equations (A) and (B) are called telegraphs equations. Now consider a lossless line. Then R=G=0. By rearranging the equations (A) and (B), the wave equations for lossless line are

2

2

2

2

tVLC

xV

∂∂

=∂∂ (C)

2

2

2

2

tILC

xI

∂∂

=∂∂ (D)

Solution The above equations are second order differential equation. Solution for the equations (C) and (D) are

( )txLCfV −= and ( )txLCV += φ . It can be proved that the above expressions are solutions of the wave equations by substituting in equations (C) and (D). Since the differential equations are linear the sum of the two solutions are also linear. The final solution is ( ) ( )txLCtxLCfV ++−= φ The wave traveling towards x direction called “incident wave” is given by

( )txLCfVI −= . The wave traveling towards negative x direction called “reflected wave” is given by

( )txLCVR += φ . The transmitted wave i.e. summation of incident wave and the reflected wave is given by

( ) ( )txLCtxLCfVVV RIT ++−=+= φ . If the transmission line is a continuous one, VR=0. So, VI=VT. The velocity of the wave v is given by

LCv 1

=



36

Typical value for overhead line is 3x108 m/s (velocity of the light in vacuum).

Typical value for a cable is r

xε

8103 m/s.

Equation for current

The equation (2), tVCGV

xI

∂∂

−−=∂∂ , simplifies to

tVC

xI

∂∂

−=∂∂ for a lossless line. By

substituting for the voltage, ( ) ( )txLCtxLCfV ++−= φ ,

[ ])()( '' txLCtxLCfCxI

+−−=∂∂

φ .

By integrating with respect to x,

[ ]

[ ]RI VVC

LI

txLCtxLCfLCCI

−=

+−−=

1

)()( φ

The CL is called the surge impedance of the line denoted by Z0. In a lossless line Z0 is

pure resistive. It is also called surge resistance denoted by R0. The transmitted current IT can be written as the summation of incident current II and reflected current IR.

[ ]

00

0

,RVI

RVI

IIR

VVI

RR

II

RIRI

T

−==

+=−

=

In a lossless line the shape and the magnitude of the wave do not change with respect to the distance x. However a delay occurs at different locations (different x) where the delay is given by x/v. Reflection, Refraction at Discontinuity Point When a surge wave goes through a transmission with surge resistance R0 and enter into a different line or impedance Z, part of the incident voltage and current wave will reflect. The rest will refract (or transmit) to the other section. The ratio between reflected wave to the incident wave is given by reflection coefficient “β”. The ratio between refracted wave to the incident wave is given by refraction coefficient “α”. Once α and β are known, the reflected and transmitted wave can be obtained.

IT

IR

VVVV

αβ

==

The relationship between α and β is



37

α = β + 1 With usual notation for incident, reflected and transmitted voltage and current the following relationships can be adapted. For currents and voltages

ZVI

RVI

RVI

TT

RR

II

=

−=

=

0

0

,

For voltage at the discontinuity point RIT VVV += (5) For current at the discontinuity point

00 RV

RV

ZV

III

RIT

RIT

−=

+= (6)

By solving equations (5) and (6) the expressions for reflected and transmitted voltage and current can be obtained as

IIT

IIR

IT

IR

IRZ

RVRZ

I

IRZRZ

VRZRZ

RI

VRZ

ZV

VRZRZV

+

=

+

=

+−

−=

+−

−=

+

=

+−

=

0

0

0

0

0

0

0

0

0

0

0

22

1

2

Then the reflection and refraction coefficients are given by

0

0

RZRZ

+−

=β and 0

2RZ

Z+

=α

The reflected and transmitted voltage and current can also be written as

IR VV β= , IT VV α= , IR II β−= , and Z

IRI IT

0α=

Reflection, Refraction at a Junction Assume that a line A terminates at a junction with another two lines B and C. The surge impedances of lines are RA, RB and RC. At the junction the voltage of line A is same as voltages in lines B or C. The voltage at A is the summation of its incident and reflected waves.

TCTBRAIA VVVV ,,,, ==+ (7)



38

The current at the junction of line A is the summation of currents in line B and C. The current at A is the summation of its incident and reflected waves.

TCTBRAIA IIII ,,,, +=+ In terms of voltages,

C

TC

B

TB

A

RA

A

IA

RV

RV

RV

RV ,,,, +=− (8)

By solving the equations (7) and (8), the reflected and transmitted waves can be obtained as

++

==

++

−−

=

CBA

AIATCTB

CBA

CBAIARA

RRR

RVVV

RRR

RRRVV

111

2

111

111

,,,

,,

If the line terminates with n number of separate lines (1 to n) the reflected and transmitted waves can be written as

+++

===

+++

−−−

=

nA

AIATnT

nA

nAIARA

RRR

RVVV

RRR

RRRVV

1.....11

2

.....

1.....11

1......11

1

,,,1

1

1,,

Terminations with Different Loads ZL When a surge wave enters to a terminal load the reflected wave to the line and the transmitted wave to the load varies depending on the type and value of the connected load ZL. The equations for voltage and currents in some of the common loads are given below.



39

Case 1: Matched load (R0=ZL) The calculated reflection and refraction coefficients are β=0 and α=1. Then the VR, VT, IR and IT are

0=RV , IT VV = 0=RI 0R

VI IT =

Case 2: Open circuit (ZL=∝) The calculated reflection and refraction coefficients are β=1 and α=2. The VR, VT, IR and IT are

IR VV = , IT VV 2= 0R

VI IR −= 0=TI

Case 3: Short circuit load (ZL=0) The calculated reflection and refraction coefficients are β=-1 and α=0. The VR, VT, IR and IT are

IR VV −= , 0=TV 0R

VI IR =

0

2RVI I

T =

Case 4: Inductor (ZL=jωL) The calculated VR, VT, IR and IT are for a step voltage function of magnitude VI=E

+−=

−L

tR

R eEV0

21 where VR varies from E (t=0) to –E (t=∝)

=

−L

tR

T eEV0

2 where VT varies from 2E (t=0) to 0 (t=∝)

−=

−L

tR

R eREI

0

210

where IR varies from –E/R0 (t=0) to E/R0 (t=∝)

−=

−L

tR

T eREI

0

12

0

where IT varies from 0 (t=0) to 2E/R0 (t=∝)

The inductor is initially acting as open circuit and later it is short circuited. Case 5: Capacitor (ZL=1/jωC) The calculated VR, VT, IR and IT are for a step voltage function of magnitude VI=E

−=

−CRt

R eEV 021 where VR varies from -E (t=0) to E (t=∝)



40

−=

−CRt

T eEV 012 where VT varies from 0 (t=0) to 2E (t=∝)

+−=

−L

tR

R eREI

0

210

where IR varies from E/R0 (t=0) to -E/R0 (t=∝)

=

−L

tR

T eREI

0

0

2 where IT varies from 2E/R0 (t=0) to 0 (t=∝)

The capacitor is initially acting as short circuit and later it is open circuited. Multiple Reflected waves – Bewley’s Zig Zag Diagram When a transmission line is terminated with mismatch loads in both ends, the incident wave reflect in both ends. These reflections repeated continuously. It is important to draw a diagram to explain how the multiple reflections effects on voltage and current at different locations and at different times. This diagram is called Bewley’s Zig Zag diagram. It provides the trace of wave front at different locations of the line and variation with respect to the time. Consider a line with surge resistance R terminates at the receiving end with different impedance ZR and at the sending end with impedance ZS. The reflection and refraction coefficients at the receiving end are βR and αR. The reflection and refraction coefficients at the sending end are βS and αS. The βR αR βS and αS are calculated as

RZZ

RZZ

RZRZRZRZ

S

SS

R

RR

S

SS

R

RR

+=

+=

+−

=

+−

=

2

2

α

α

β

β

By assuming that the length of the line is L, the time taken to travel the wave through the line is given by t=L/v. The v is the velocity of the wave. The line diagram of the selected part (line and terminal loads) and the Zig Zag diagram for the voltage waves are shown below. The diagram of the current wave is similar.



41

At the receiving end the voltage, ( ) ( ) ( )[ ].......32 +++= SRSRSRRtotal VV ββββββα

Z

ZR ZS

βR αR

αS βS

Distance [x]

Time [t]

V

αRV βRV

βRαSV βSβRV

αRβSβRV

βSβ2RV

β2Sβ2

RV

αSβSβ2RV

t

2t

4t

3t



42

Each part in the above expression appears with a time delay of 2t. In a different location the expression for the voltage is same, however, delay time varies and it should be found from the Zig Zag diagram. If the surge is a step function of amplitude E, the maximum voltage can be written as

( )SR

Rtotal

EVββ

α+

=1

Lossy Line In a lossy line R and G are not zero. Then the general wave equation can be written as

2

22

2

2

2

22

2

2

tI

xI

tV

xV

∂∂

=∂∂

∂∂

=∂∂

γ

γ

Here γ is a complex number and is given by ( )( ) βαωωγ jCjGLjR +=++=

The real part of γ, “α”, is the attenuation constant. In a lossless line, when R=G=0, α=0. The velocity of the wave is given by

βω

=v

In a lossless line LCωβ = . Now we consider of a special case of a lossy line called “distortion less line”. The condition for the distortion line is R/L=G/C. Then the γ reduces to

( )( ) ( ) LCjRGLjRLCCjGLjR ωωωωγ +=+=++=

Then the α and β are given by RG=α and LCωβ = . Both α and β are constant. The expressions for transmitted wave is given by

( ) ( )

( ) ( )[ ]txLCetxLCfeLC

I

txLCetxLCfeV

xxT

xxT

++−=

++−=

+−

+−

φ

φ

αα

αα

1

In can be clear that for a distortion less line the wave attenuate with a constant α and

travels with a constant velocity of βω

=v .

