# Three Phase pi-section model of power system transmission lines

1. Aug 18, 2011

### krishkr

I am using the "Three-Phase PI Section Line" Matlab-Simulink block from Simpowersystems blockset for modelling an electric power system transmission line.

I had obtained the data from a research paper regarding the transmission line specifications (provided as sequence values).

Since I am trying to do actual impedance calculations of each phase for distance protection & fault location purposes, I tried to compute the variation of per phase impedance using a Matlab program (given below).

Code (Text):
clear all; clc;close all;
a=exp(deg2rad(120)*1j);  %% a = -0.5000 + 0.8660j
A = [1 1 1; 1 a^2 a; 1 a a^2]; %% Matrix for Symmetrical component transformation
Fo = 60;

%% Line Specification using sequence values (taken from paper)
R10 = [0.0317   0.3192];   L10 = [0.3204   1.0083]./(2*pi*60) ;   C10 = [10.838e-9  7.6493e-9];
Distance = 1:150; %% in km

for k = 1:length(Distance)
R10 = R10*Distance(k); L10 = L10*Distance(k); C10 = C10*Distance(k);
XL10 = (2*pi*Fo)*L10;  XC10 = 1./((2*pi*Fo)*C10);
Z10 = R10 + 1j*( XL10 - XC10 );

Z0(k) = Z10(2);     Z1(k) = Z10(1);     Z2(k) = Z1(k);
Z012 = [Z0(k) 0 0; 0 Z1(k) 0; 0 0 Z2(k)];
Zabc = A*Z012*((1/3)*(A'));

Zs(k) = Zabc(1); Zm(k) = Zabc(2);

end

figure(1);subplot(2,1,1);plot(Distance,real(Z0));subplot(2,1,2);plot(Distance,imag(Z0));
figure(2);subplot(2,1,1);plot(Distance,real(Z1));subplot(2,1,2);plot(Distance,imag(Z1));
figure(3);subplot(2,1,1);plot(Distance,real(Zs));subplot(2,1,2);plot(Distance,imag(Zs));
figure(4);subplot(2,1,1);plot(Distance,real(Zm));subplot(2,1,2);plot(Distance,imag(Zm));

The program takes the system data of the transmission line (which are given in unit/km), which can be used as parameters for the Simulink 3 phase pi-section block.
The distance is multiplied to the data to get the actual values.
The impedances are computed, the sequence impedances and the per phase impedances are found out. The values of resistance and reactance are computed from the real part and imaginary part respectively. This operation is done for various distances and the plot shows that.

I find that the values plot rise to very high values (order of 1e200). How can current flow if impedance is so high ? If the system data is inappropriate, please tell me why.
Please suggest the reason for such a problem and how can I overcome it?

2. Aug 20, 2011

### Mbert

Why are you adding the capacitances to Z10? In the PI model, you should also have an admittance part, where you would have G and C (where G is usually neglected)... Check for instance "Power System Analysis and Design" from Glover.

M.

3. Mar 5, 2012

### krishkr

Thank you Mbert. It was stupid of me to add the capacitance as if it was in series with the R and L. Also, I did a programming mistake; the above code repetitively multiplies R10, L10, C10 with Distance. I unintentionally end up with a geometric progression.