Lecture note from 11-20
Lecture-11
Capacitance of 3-Phase Overhead Line
(i) Symetrical Spacing
Figure shows three con
( ii) Unsymmetrical Spacing
Fig shows 3-Phase transposed line having unsymmetrical
Lecture-12
Effect of Earth on the Capacitance of a Three Phase Line, Capacitance
Calculations for Bundled Conductors,
Effect of Earth on Capacitance
The capacitance of transmission line is influenced by the appearance of earth . As a
result of earth , electric field of a line is lowered . When we think that the earth is a perfect
conductor by means of a horizontal plane of infinite extent , we recognize that the electric field
of charged conductors above the earth is totally different from it may be if the equipotential
surface of earth were not present . The effect of earth is observed on both single phase and
three phase line .
Lecture-13
Parallel-Circuit Three Phase Lines.
If two three phase circuits that are identical in construction and operating in parallel are so
close together that coupling exists between them , the GMD method can be used to calculate
the inductive and capacitive reactance of their equivalent circuit.
Parallel circuit is shown for three phase lines on the same power.Although the line will
probably be transposed, we obtained practical values for inductive and capacitive reactance if
transposition is assumed. Conductors a and a’ are in parallel to compose phase a. Phase b and c
are similar. We assume that a and a’ take the position of band b’ and then of c and c’ as theose
conductor are rotated similarly in the transposition cycle.
To calculate Deq the GMD method requires that we use D pab , Dpbc , and Dpca ,where
the superscript indicates that these quantities are for parallel lines and where D pab means the
GMD between the conductors of phase a and those of phase b.
For inductance calculation ds is replaced by Dps , which is the geometric mean of the
GMR values of the two conductors occupying first the position of a and a’ then the positions of
b and b’ , and finally the position of c and c’.
Because of the similarity between inductance and capacitance calculation , we can
assume that the Dpsc foe capacitance and Dsp for inductance except that r is used instead of Ds
of the individual conductor.
Lecture 14
Problem solving
Prob-1
Solution
Prtob-2
Shown in figure
Solution:
Prob-3
Solution
Prob-4
plane shown in the
figure
In figure the arrangement of 3 phase conductors are shown
The equivalent equilateral spacing is
Prob-5
Solution
Lecture-15
Transmission Line Performances
Representation of Lines
Classification of Overhead Transmission Lines
Lecture-16
Short, & Medium Transmission Lines and there Performance
Performance of Single Phase Transmission lines
The effects of line capacitance are neglected in short Transmission line .Therefore
while studying the performance of such line, only the resistance and inductance of
the line are taken into account.The equivalent circuit is shown in the figure.
The phasor diagram of the line for the lagging power factor .From right angled triangle ODC,
Approximate Method
Solution in complex notation
Three Phase Transmission Line
Effect of Load power Factor on Regulation & Efficiency
2.Effect of Efficiency: Power delivered to load depends upon power factor.
VR.the load current I is inversly proportional to the load power factor cosΦR.Consequently,with
the decrease in load p.f,the load current and hence the line losses are increased.This leads to the
conclusion that transmission effinciency of a line decreases with decrease in load p.f and
viceversa.
Lecture:17
Medium Transmission line- Performance & Interpretation of Equations
End Condenser Method
Nominal T Method
Nominal Ω Method
Lecture:18
The Long Transmission Line: Hyperbolic Form of The Equations,
Analysis of Long Transmission Line (Rigorous Method)
Lecture19
The Equivalent Circuit of a Long Line, Power Flow Through Transmission Line,
The nominal 𝜋 method does not represent a transmission line exactly because it does not account
for the parameters of the line being uniformly distributed. The discrepancy between the nominal
pi and the actual line becomes larger as the length of the line increases. It is possible, however, to
find the equivalent circuit of a long transmission line and to represent the line accurately ,insofar
as measurements at the ends of the line are concerned ,by the network of lumped parameters. Let
us assume that the 𝜋 circuit similar fig shown is the equivalent circuit of the long line.Let the
series arm be called Z’ and the shunt arms be Y’/2 to distinguish them from the arms of a
nominal 𝜋 circuit.
𝑍′𝑌′
𝑉𝑠 = (
+ 1) 𝑉𝑅 + 𝑍′𝐼𝑅
2
Z’ = Zc sinhγl
Y’/2
Y’/2 = cosh γl-1/ sinhγl
Equivalent 𝜋 network
For our circuit to be equivalent to the long transmission line coefficients of VR and IR must be
identical,respectively,to the coefficient of Vr and Ir .Equating the voltage equation ,
Z’ = Zc sinhγl
Z’ = √z/y sinhγl =zl ( 𝑠𝑖𝑛ℎ𝑙 ⁄√ 𝑧𝑦 𝑙 )
Z’ = Z (sinhγl / γl )
Where Z is equal to zl ,the total series impedance of the line.The term (sinhγl / γl ) is the factor
by which the series impedance of nominal 𝜋 must be multiplied to convert the equivalent 𝜋 .For
small values of γl ,both sinγl and γl are almost identical,and this fact shows that the nominal 𝜋
represents the medium length transmission line quit accurately insofar as the series arm is
concerned.
To investigate shunt arms of the equivalent- 𝜋 circuit,we equate the coefficient of Vr and obtain
Z’Y’/2 + = coshl
Subsequently Zc sinhγl for Z’ gives
Y’Zc sinhγl/2 +1 = cosh γl
Y’/2 = cosh γl-1/ sinhγl
Another form of the expression for the shunt admittance of the expression for the shunt
admittance of the equivalent circuit can be found by
tanh γl/2 = cosh γl-1/sinh γl
By using hyperbolic function
Y’/2 = 1/Zc tanh γl/2
Y’/2 =
𝑌 𝑡𝑎𝑛ℎγ𝑙/2
γ𝑙/2
Where Y is equal to γl, the total shunt admittance of the line. In the
above equation shows correction factor used to convert the admittance of the shunt arms of the
nominal 𝜋 to that of the equivalent 𝜋.
Since 𝑡𝑎𝑛ℎγ𝑙/2 and γ𝑙/2 are very nearly equal for small values of γ𝑙 ,the nominal 𝜋 represents
the medium length transmission line quite accurately , for we have seen previously that the
correction factor for the series arm is negligible for the medium-length lines.
Power flow through a Transmission Line
If voltage, current and power factor the power can be calculated. Equation can also be derived in
terms of ABCD constants.
Vs = AVR+ BIR , IR =
𝑉𝑠−𝐴𝑉𝑟
𝐵
Letting , A=|𝐴| <α
We obtain
B=|𝐵| <β
IR =
𝑉𝑠
VR =|VR|
δ- β -
𝐵
A VR
00
Vs =|Vs|
α- β
𝐵
Then complex power VR IR* at the receiving end is
PR +j QR = -
Vs
VR
𝐵
θR
(β- α)
, (β- δ)
Real and Reactive Power at receiving end are
PR =
QR =
Vs VR
𝐵
Vs VR
𝐵
cos (β- δ) -
𝐴 𝑉𝑅2
sin (β- δ) -
𝐴 𝑉𝑅
𝐵
𝐵
cos (β- α)
sin (β- α)
β- δ
-
A VR2
𝐵
β- α
δ
Lecture 20
Reactive Compensation of Transmission Line.
By reactive compensation the performance of the transmission line can be improved. Series
compensation consists of a capacitor bank placed in series with each phase conductor line.Shunt
compensation refers to placement of inductors from each line to neutral to reduce partially or
completely the shunt suscepatance of a high voltage line,which is particularly important at light
loads when the voltage at the receiving end may otherwise become high.
Series compensation reduces the series impendence of the line , which is the principal cause of
voltage drop and most important factor determining the maximum power which the line can
transmit.
PR,max =
Vs VR
𝐵
-
𝐴
𝑉𝑅2
𝐵
cos (β- α)
The maximum power transfer is inversely proportional to B (generalized circuit constatnts)
,which for nominal pi equals to Z.
The A,B,C,D constants are function of Z, they will also change in value, but these changes will
small in comparison to the change in B
The desired reactance of the capacitor bank can be determined by compensating for a specific
amount of the total inductive reactance determined by compensating for a specific amount of the
total inductive reactance of the line .This leads to the term compensating factor.which is defined
by Xc/Xl
Xc = capacitive reactance of series capacitor bank per phase
Xl = inductive reactance of the line per phase.
When the nominal –pi circuit is used to represent the line and capacitor bank , the physical
location of the capacitor bank along the line is not taken into account.If only the sending and
receiving end condition of the line are of interest this will not create any significant error.
However when operating conditions are of interest the physical location of the capacitor bank
must be taken into account.This can be accomplished most easily by determining ABCD
constants of the portions of the line on each side of the capacitor bank and by representing the
capacitor bank by its ABCD constants. The equivalent constants combination of the line
capacitor-line can be then be determined by applying the equation.
Static capacitor Bank
Static capacitor can further be subdivided in to two categories,
(a) Shunt capacitors
(b) Series capacitor
Advantages of using shunt capacitors such as,
a) It reduces line current of the system.
b) It improves voltage level of the load.
c) It also reduces system Losses.
d) It improves power factor of the source current.
e) It reduces load of the alternator.
f) It reduces capital investment per mega watt of the Load.