EE 360 ELECTRIC ENERGY
LECTURE 31
Long Line Modeling
The material covered in this lecture will be as follows:
⇒ To develop a model for long transmission lines .
⇒ To calculate voltage and current in long lines.
⇒ To calculate voltage regulation and transmission efficiency in long lines.
At the end of this lecture you should be able to:
⇒ model long transmission lines for calculating voltages and currents .
⇒ understand the power flow between two ends of the of along transmission line.
1.
Long Line Model and Representation
The parameters of short and medium lines are assumed to be lumped.
This assumption allowed us to build models where total impedance and admittance are
used.
The voltage and current were expressed and calculated in terms of the receiving and
sending end values.
The parameters Lines longer than 250 km are considered to be distributed over the entire
line length.
This is an accurate representation and allows for the determination of voltage and current
at any point along the line length.
Figure 1 shows an equivalent representation of a long line.
···
Sending
end
Receiving
end
Figure 1: Long Line Representation
In this lesson, we will describe the mathematical model in new terms.
Let us make the following definitions
z = series impedance per unit length
y= shunt admittance per unit length
z = (r + j ω L )
y = ( g + jwC )
(1)
(2)
Figures 2 shows an equivalent circuit a long line.
Figure 2: Long Line Model
The voltage and current are shown for an elemental line length dx, the following
equations are written
dV = Izdx
dV
= zI
dx
(3)
(4)
Where x is measured from the receiving end.
A similar equation can be written for the current in terms of the admittance
dI
= yV
dx
(5)
By substituting for I and re-arranging equation3, we have
d 2V
= zyV
dx 2
(6)
This is a second order differential equation that relates the voltage to the distance.
Once the voltage expression is found, the current can be found using equation 4.
Let us make the following terms
γ = zy
γ is known as the propagation constant.
(7)
Equation 8 defines another important term , Zc, the characteristic impedance of the line:
Zc =
z
y
(8)
The solution of equation 6 will be expressed in terms of the propagation constant and
characteristic impedance.
The voltage and current at any point x along the lien length are give by equations 9-10
1
1
V x = V R (e γ x + e −γ x ) + I Z c (e γ x − e −γ x )
2
2
R
Ix =
1V R γ x
1
(e − e −γ x ) + I (e γ x + e −γ x )
2 Zc
2
R
(9)
(10)
Equations 9-10 can be written in terms of hyperbolic functions
V x =V R cosh γ x + I Z c sinh γ x
R
Ix =
VR
sinh γ x + I cosh γ x
Zc
R
(11)
(12)
At the sending end x=l, the voltage and current are found by substituting x by l in 11-12.
V S =V R cosh γ l + I Z c sinh γ l
R
IS =
VR
sinh γ l + I cosh γ l
Zc
R
(13)
(14)
Rewriting the above equations in terms of the ABCD constants, we will have
V S = AV R + BI R
(15)
I S =CV R + DI R
(16)
Where
A = cosh γ l
B = Z c sinh γ l
1
C = sinh γ l
Zc
(17)
(18)
D = cosh γ l
(20)
(19)
The solution procedure for the long line is similar to the medium line.
It will be illustrated by the following example.
Example 1
A 500 kV, 60 Hz, three phase 400 km delivers 900 MW of power at power factor of 0.8 lagging. The
line has the following parameters:
r = 0.0125 Ω per km
L = 0.35 mH per km
C = 0.0050 µ F per km
G=0.0
Find
(i)
the sending end voltage and current
(ii)
voltage regulation
(iii)
transmission efficiency
Solution
ω = 2 π f= 2x 3.14159x 60= 377
z = (r + j ω L ) =0.0125+j .13195
y= jwC= j1.885E-6
VR = 500000/
3=288683 V
3
900x 10
= 1299 A
IR =
3x 500x 0.8
The current angle θ r = − 36.870
Z c = 264.87 + j (−12.518) Ω
γ = 0.00943 + j 0.19708
A = cosh γ l
B = Z c sinh γ l
1
C = sinh γ l
Zc
== 0.98017+0.001875
= 4.9339+j52.433
D = cosh γ l
= 0.98017+0.001875
= -4.718E-7+j0.00074899
(i) sending end voltage and current
V S = AV R + BI R
The line voltage is
Vsl= 576.605 kV ∠ 8.844
I S =CV R + DI R
I s = 1156.8∠ − 28.1560
(ii) Voltage regulation
Voltage Regulation (%)= 17.654 %
Power factor angle at the sending end= Pfs= 0.7988
The sending end power= 922.705 MW
η=
PR
900
= =
*100 =97.539%
922.705
PS