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Nominal Π Representation of a Medium Transmission Line: I I I I I I

The document discusses transmission line parameters and modeling transmission lines using a nominal pi (Π) circuit representation. It provides the equations to calculate the A, B, C, and D parameters for a nominal pi circuit based on the series impedance (Z), shunt admittance (Y), sending voltage (VS) and receiving current (IR). It then applies these calculations to a example 100km long 11.5kV transmission line connected to a 10MW load.

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Vedaste Ndayishimiye
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568 views5 pages

Nominal Π Representation of a Medium Transmission Line: I I I I I I

The document discusses transmission line parameters and modeling transmission lines using a nominal pi (Π) circuit representation. It provides the equations to calculate the A, B, C, and D parameters for a nominal pi circuit based on the series impedance (Z), shunt admittance (Y), sending voltage (VS) and receiving current (IR). It then applies these calculations to a example 100km long 11.5kV transmission line connected to a 10MW load.

Uploaded by

Vedaste Ndayishimiye
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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INTRODUCTION

TRANSMISSION LINE PARAMETERS

An electric transmission line is modeled using series resistance, series inductance, shunt
capacitance, and shunt conductance. The line resistance and inductive reactance are important.
For some studies like for short and medium transmission line it is possible to omit the shunt
capacitance and conductance and thus simplify the equivalent circuit considerably.

Nominal Representation of a Medium Transmission Line

In case of a nominal representation, the lumped series impedance is placed at the middle of the
circuit where as the shunt admittances are at the ends. As we can see from the diagram of the
network below, the total lumped shunt admittance is divided into 2 equal halves, and each half
with value Y 2 is placed at both the sending and the receiving end while the entire circuit
impedance is between the two. The shape of the circuit so formed resembles that of a symbol ,
and for this reason it is known as the nominal representation of a medium transmission line. It
is mainly used for determining the general circuit parameters and performing load flow analysis.

As we can see here, VS and VR is the supply and receiving end voltages respectively, and Is is
the current flowing through the supply end.

IR is the current flowing through the receiving end of the circuit.


I1 and I3 are the values of currents flowing through the admittances. And
I2 is the current through the impedance Z.

Now applying KCL, at node P, we get.

I S=I 1 +I 2

I 2 =I 3 + I 3
Y
putting eq ( 2 ) into (1 ) I S=I 1 + I 2+ I R = V +I
2 S R

Y Y
2
VR (
+ I R= Z +1+ Z I R
2 )
Now by applying KVL the circuit ,V S=R R + Z I 2=V R +Z

At the end we can seethaV S= A V R + B I R I S =C V R + D I R

A= ( Y2 Z+1)
B=Z

C=Y ( Y4 +1)
D= ( Y2 Z +1)

MODEL CALCULATIONS

A three-phase 11.5-kV transmission line is connected to a 10-MW load at a 0.85


lagging power factor. The line constants of the 100Km long line are Z = 35 and
Y = 4.8e-4 S.

By using nominal- circuit representation, calculate constants A, B, C, and D of


transmission line are the parameters to be calculated.

Calculations

From section line configurations


The transmission line total series impedance and series impedance per

unit length are calculated according to the use the following notation:

z=R+ jwL , series impedence per unit length
m

S
y=G+ jwC , shunt admittance per unit length
m

Z =zl, total series impedence

Y = yl S , total shunt admittance

l=line length

z=R+ jwL=0.01273+ j 2 600.9337e-3=0.01273+ j0.351

z=0.3512

Z =zl=0.3512100=35.12

y= jwC= j 2 6012.74e-9=4.8e-6 S

Y =4.8e-6100=4.8 e4 S

The A , B, C and constant for the nominal circuit representation are calculated
from Sending Voltage, current and Receiving end Voltage, Current
V S =A V R + B I R

I S=CV R + D I R

1
A=1+ YZ
2

B=Z=95<78 0
1
C=1+ Y 2 Z
4

1
D=1+ YZ
2

SIMULATION RESULTS

To build a uniform transmission line circuit model, the process, had use the following modules:
138-kV Three-Phase AC Voltage Source Module, Pi transmission Line Module, 49-MW Three-
Phase Parallel RLC Load Module, Scope Module, Voltage Measurement Module, Current
Measurement Module and Powergui Module.

CONCLUSION

6. SUGGESTION AND FUTURE WORK

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