Chapter 5. Line Model and Performance
5.1 Introduction
We shall represent lines as a "pi" section with
lumped parameters. There are three cases: short,
medium and long lines.
Z = R+jX
IS
+
VS
_
IR
+
VR
_
IL
In the short line model (under 50 miles), we
Y/2
Y/2
assume the shunt capacitance (the legs of the π )
are so small that they are open circuit (i.e. neglected). This leaves only the series RL
branch.
In the medium line model (50 to 250 miles), we assume that the capacitance may be
represented as two capacitors (legs of the π ) each equal to half the line capacitance. This
is known as the nominal π model.
In the long line model (over 250 miles), we assume the line has distributed parameters
instead of lumped parameters. This yields exact results. After finding the exact model,
an equivalent π is found. This is called the equivalent π model because it has lumped
parameters which are adjusted so that they are equivalent to the exact distributed
parameters model.
5.2 Short Line Model
A short line with a load is shown (per phase) in the
figure to the right. Using circuit theory we
S*
have: I R = R* . We also know VS = VR + ZI R . Note that
VR
I S = I R . Now we will find the transmission parameters
(or ABCD parameters) of the short line:
IS
Z = R+jX
+
+
VS
VR
_
_
For a 2-port shown as "ABCD" the equations are as
IS
V   A B  VR 
follows:  S  = 
.
Comparing
with
the
 
+
 I S  C D   I R 
ABCD
VS
_
equations for the short line above, we see that: A = 1 ,
B = Z , C = 0 and D = 1 . This completes the equations
needed to represent the short line. The voltage regulation in percent is given by:
VR ( NL) VR ( FL )
Percent VR =
× 100
VR ( FL )
IR
SR
IR
+
VR
_
At no-load I R = 0 and thus VS = AVR ( NL ) , and since A = 1 , VR ( NL) = VS . This conclusion is
easily made by observation of the circuit above.
Example 5.1
A 220-kV, three phase transmission line is 40 km long. The resistance per phase is
0.15Ω per km and the inductance per phase is 1.3263 mH. The shunt capacitance is
negligible. Use the short line model to find the voltage and power at the sending end and
the voltage regulation and efficiency when the line is supplying three phase load of
(a) 381 MVA at 0.8 power factor lagging at 220 kV.
(b) 381 MVA at 0.8 power factor leading at 220 kV.
The Matlab program follows:
VRLL=220; VR = VRLL/sqrt(3);
Z = (0.15+j*2*pi*60*1.3263e-3)*40;
disp('(a)')
SR=304.8+j*228.6;
IR = conj(SR)/(3*conj(VR)); IS = IR;
VS = VR + Z*IR;
VSLL = sqrt(3)*abs(VS)
SS = 3*VS*conj(IS)
REG = (VSLL - VRLL)/VRLL*100
Eff = real(SR)/real(SS)*100
disp('(b)')
SR=304.8-j*228.6;
IR = conj(SR)/(3*conj(VR)); IS = IR;
VS = VR + Z*IR;
VSLL = sqrt(3)*abs(VS)
SS = 3*VS*conj(IS)
REG = (VSLL - VRLL)/VRLL*100
Eff = real(SR)/real(SS)*100
(a)
VSLL =
250.0186
SS =
3.2280e+002 +2.8858e+002i
REG =
13.6448
Eff =
94.4252
(b)
VSLL =
210.2884
SS =
3.2280e+002 -1.6862e+002i
REG =
-4.4144
Eff =
94.4252
2
5.3 Medium Line Model
As the length of the line is increased to over 50 miles (but less than 150 miles) the
medium line model is used. This is also known as the nominal π model as shown below.
The admittance Y and impedance Z shown are shown for the whole line, per-phase (and
not per mile). For the "pi" model shown we can
Z = R+jX
IS
IR
Y
write the following equations: I L = I R + VR and
2
+
+
IL
VR
VS
VS = VR + ZI L . Using the first equation in the
_
_
Y/2
Y/2
 ZY 
second we have: VS =  1 +
V
+
ZI
.
We
also
R
2  R

Y
Y
Y  ZY 

see that I S = I L + VS and thus we have: I S = I R + VR + 1 +
 VR + ZI R  .
2
2
2 
2 

 ZY 
 ZY 
= Y 1 +
VR + 1 +
 IR
4 
2 


Thus we now have the ABCD parameters:
 ZY 
A = 1 +
B=Z

2 

 ZY 
 ZY 
C = Y 1 +
D = 1 +


4 
2 


In general for a symmetrical 2-port A = D and AD − BC = 1 (the determinant of the
ABCD matrix in this case is unity). Thus it is easy to find the inverse relationship:
VR   D − B  VS 
 =
 
 I R   −C A   I S 
Note the two Matlab functions for computing the ABCD and ZY parameters:
[ Z , Y , ABCD ] = rlc 2abcd (r , L , C, g, f , Length )
[ Z , Y , ABCD ] = zy 2 abcd ( z, y , Length )
Note also that f is frequency in Hz and g is conductance to ground per unit length (zero
in the specific case of the medium line model). Note also that r , g, z, and y are all per
unit length of line. The series impedance is Z = ( r + jω L ) l = R + jX and the shunt
admittance is Y = ( g + jωC ) l .
Example 5.2
A 345-kV three-phase line is 130 km long. The resistance per phase is 0.036Ω per km
and the inductance per phase is 0.8 mH per km. The shunt capacitance is 0.0112 µ F per
km. The receiving end load is 270 MVA with 0.8 power factor lagging at 325 kV. Use
the medium line model to find the voltage and power at the sending end and the voltage
regulation.
3
Note that we cannot run this program from within MS-Word since it has interactive
statements. A listing of the program for example 5.2 is shown below. Run example 5.2
in the Matlab window (chp5ex2). DO NOT EXECUTE IN THIS NOTEBOOK!
(Reason: edit the function "rlc2abcd.m", it has interactive code!)
r = .036; g = 0; f = 60;
L = 0.8;
% milli-Henry
C = 0.0112;
% micro-Farad
Length = 130; VR3ph = 325;
VR = VR3ph/sqrt(3) + j*0;
% kV (receiving end phase voltage)
[Z, Y, ABCD] = rlc2abcd(r, L, C, g, f, Length);
AR = acos(0.8);
SR = 270*(cos(AR) + j*sin(AR));
%
MVA (receiving end power)
IR = conj(SR)/(3*conj(VR));
% kA (receiving end current)
VsIs = ABCD* [VR; IR];
%
column vector [Vs; Is]
Vs = VsIs(1);
Vs3ph = sqrt(3)*abs(Vs);
% kV(sending end L-L voltage)
Is = VsIs(2); Ism = 1000*abs(Is);
%
A (sending end current)
pfs= cos(angle(Vs)- angle(Is));
% (sending end power factor)
Ss = 3*Vs*conj(Is);
%
MVA (sending end power)
REG = (Vs3ph/abs(ABCD(1,1)) - VR3ph)/VR3ph *100;
fprintf(' Is = %g A', Ism), fprintf(' pf = %g\n', pfs)
fprintf(' Vs = %g L-L kV\n', Vs3ph)
fprintf(' Ps = %g MW', real(Ss)),
fprintf(' Qs = %g Mvar\n', imag(Ss))
fprintf(' Percent voltage Reg. = %g\n', REG)
Example 5.3
A 345-kV, three-phase transmission line is 130 km long. The series impedance is
z = 0.036 + j 0.3 Ω per phase per km. The shunt admittance is y = j 4.22 × 10−6 siemens
per phase per km. The sending end voltage is 345-kV and the sending end current is 400A at 0.95 power factor lagging. Use the medium line model to find the voltage, current
and power at the receiving end and the voltage regulation.
Note that we cannot run this program from within MS-Word since it has interactive
statements. A listing of the program for example 5.3 is shown below. Run example 5.3
in the Matlab window (chp5ex3). DO NOT EXECUTE IN THIS NOTEBOOK! The
culprit is the command "zy2abcd" which has interactive code.
z = .036 + j* 0.3; y = j*4.22/1000000; Length = 130;
Vs3ph = 345; Ism = 0.4; %KA;
As = -acos(0.95);
Vs = Vs3ph/sqrt(3) + j*0;
% kV (sending end phase voltage)
Is = Ism*(cos(As) + j*sin(As));
[Z,Y, ABCD] = zy2abcd(z, y, Length);
VrIr = inv(ABCD)* [Vs; Is];
%
column vector [Vr; Ir]
Vr = VrIr(1);
Vr3ph = sqrt(3)*abs(Vr);
% kV(receiving end L-L voltage)
Ir = VrIr(2); Irm = 1000*abs(Ir);%
A (receiving end current)
pfr= cos(angle(Vr)- angle(Ir)); % (receiving end power factor)
4
Sr = 3*Vr*conj(Ir);
%
MVA (receiving end power)
REG = (Vs3ph/abs(ABCD(1,1)) - Vr3ph)/Vr3ph *100;
fprintf(' Ir = %g A', Irm), fprintf(' pf = %g\n', pfr)
fprintf(' Vr = %g L-L kV\n', Vr3ph)
fprintf(' Pr = %g MW', real(Sr))
fprintf(' Qr = %g Mvar\n', imag(Sr))
fprintf(' Percent voltage Reg. = %g\n', REG)
5.4 Long Line Model
This is a distributed parameters model of the line. It is very exact and gives the voltage
and current anywhere along the line. The distance x is measured from the receiving end
towards the sending end. If we let the x = l (the length of the line) the equations for the
long line model have the following ABCD parameters:
A = cosh γ l
B = Z c sinh γ l
C=
1
sinh γ l
Zc
D = cosh γ l
z
and γ = α + j β = zy .
y
It is now possible to find an accurate π model whose ABCD parameters are equal in
value to the ones shown above. Without proof the results are outlined below:
sinh γ l
Z ′ = Z c sinh γ l = Z
γl
Y′ 1
Y tanh γ l / 2
=
tanh γ l =
2 Zc
2 γl/2
where the primed quantities are for the exact π or equivalent π model, and the unprimed
quantities are those for the nominal π . This way we can transform the nominal π to an
exact or equivalent π . The Matlab functions mentioned earlier (rlc2abcd and zy2abcd)
with option 2 will perform the equivalent π model for a long line.
where Z c is the line characteristic impedance given by Z c =
Example 5.4
A 500-kV, three phase transmission line is 250 km long. The series impedance is
z = 0.045 + j0.4 Ω per phase per km and the shunt admittance is y = j 4 × 10−6 siemens
per phase per km. Evaluate the equivalent π model and the transmission matrix.
Below is the program for example 5.4:
z = 0.045 + j*.4;
y = j*4.0/1000000; Length = 250;
gamma = sqrt(z*y); Zc = sqrt(z/y);
A = cosh(gamma*Length); B = Zc*sinh(gamma*Length);
C = 1/Zc * sinh(gamma*Length); D = A;
ABCD = [A B; C D]
Z = Zc * sinh(gamma*Length)
Y = 2/Zc * tanh(gamma*Length/2)
5
ABCD =
0.9504
-0.0000
Z =
10.8778
Y =
0.0000
+ 0.0055i
+ 0.0010i
10.8778 +98.3624i
0.9504 + 0.0055i
+98.3624i
+ 0.0010i
6