Analysis of Electric Power Transmission Line Resultant Wave Current in Polar Form

In this paper, a power transmission line of an electric power system [1-8] is under consideration. Its equivalent electric circuit under steady state conditions is drawn and the respective differential equations are extracted from it using as independent variable the distance x from either the rears of the line. The above mathematical model already exists in the literature and can easily be found [1-4, 8].

Solving the differential equations, the mathematical expression in polar form describing the resultant current wave is obtained (section 2). The proof that the above resultant current is a wave is the mathematical expression itself. It is the mathematical expression of a wave.

As far as I know and search in the literature, I could not find calculation and graphical representation of the current wave along an electric power transmission line. Thus, in this paper, the above mathematical expression is tested on a typical electric power transmission line and the results are presented in section 3. Furthermore, in section 3, the above results are graphed in order to have an optical image of how the resultant current wave along the line behaves. In section 4, a low voltage laboratory electric power transmission line model is used to obtain experimental results to verify the equations of section 2. Finally, in section 5, a discussion is developed, the results are studied, analyzed and in section 6, the relative conclusions are drawn.

We consider a typical electric power transmission line with the following parameters:

$\begin{array}{ll}\mathrm{R}=0.107 \Omega / \mathrm{km} & \mathrm{L}=1.362 \mathrm{mH} / \mathrm{km} \\ \mathrm{G}=0 \mathrm{S} / \mathrm{km} & \mathrm{C}=0.0085 \mu \mathrm{F} / \mathrm{km} \\ \mathrm{f}=50 \mathrm{Hz} & 1=360 \mathrm{km} \\ \mathrm{V}_{\mathrm{R}}=115470<0^{\circ} \mathrm{V} & \mathrm{I}_{\mathrm{R}}=360.844<0^{\circ} \mathrm{A}\end{array}$

Then using the list of symbols and the analysis of section 2, we can calculate the other complex parameters of the above line in polar and/or cartesian form:

$\gamma=1.085 \mathrm{x} 10^{-3}<82.98^{\circ} \mathrm{km}^{-1}=\left(0.1326 \mathrm{x} 10^{-3}+\mathrm{j} 1.07687 \mathrm{x} 10^{-3}\right) \mathrm{km}^{-1}$

$\alpha=0.1326 \times 10^{-3}$ neper $/ \mathrm{km}$

$\beta=1.07687 \times 10^{-3} \mathrm{rad} / \mathrm{km}$

$\mathrm{z}_{\mathrm{C}}=406.41<-7.02^{\circ} \Omega$

$\frac{\frac{\mathrm{v}_{\mathrm{R}}}{\mathrm{z}_{\mathrm{C}}}+\mathrm{I}_{\mathrm{R}}}{2}=321.886<3.092^{\circ}$    A

$-\frac{\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{z}_{\mathrm{C}}}-\mathrm{I}_{\mathrm{R}}}{2}=43.079<-23.767^{\circ}$  A

$\lambda=5834.674 \mathrm{km}$

$\mathrm{\upsilon}=291733.696 \mathrm{km} / \mathrm{sec}$

$\tau=1.234 \mathrm{msecs}$

$\Delta=22.212^{\circ}$

$\Delta / 1=0.0617^{\circ} / \mathrm{km}$

$\mathrm{V}_{\mathrm{S}}=132807.0<25.12620^{\circ} \mathrm{V}$

Then, Eq. (7) using the above parameters becomes:

I(x) =321.886<3.092° e^{(0.1326x10-3 + j 1.07687x10-3)x}+43.079<-23.767° e^{-(0.1326x10-3 + j 1.07687x10-3)x}     A     (9)

Table 1. Calculation results of resultant current wave

Using Eq. (8) and taking step Δx=10km, we calculate the values of resultant current wave and the results are presented in Table 1. Since the current is vector, the results are complex numbers and are given in polar form ie. in current magnitude (Amps) and current phase (^°) representation.

The graphical presentations of results obtained in Table 1 are given in Figure 2.

2.2a.png

2.2b.png

Figure 2. Current magnitude and phase (angle) from the beginning towards the end of line

The curves of graph 1 may appear common but they are not. One may look straight line or almost straight line but it is not. The above quantities have an exponential behaviour as someone can verify from the respective equation (7) in section 2. Their graphical representations depend on the values of their exponential constant factors (α and β). If their values are small and as variable x increases, the values αx and βx do not change enough in order their exponential behaviour to appear on the graphs. This is the reason they seem to be straight or almost straight lines.

The above explanation is given regarding their form. Regarding now their variation, the following reasoning is developed.

On one hand, the terms (V_R/z_C+I_R) and (V_R/z_C-I_R) are constant complex numbers since V_R, I_R and z_C are constant complex numbers. That implies that they have a constant magnitude and a constant phase.

On the other hand, the terms e^γx and e^-γx vary with distance x from the end of electric power transmission line.

 5.1 First term of resultant current

The term e^γx can be written as e^(α+jβ)x = e^αx e^jβx = e^αx[cos(βx) + j sin(βx)]

The values of α and β are real positive numbers for a typical real power transmission line. This will be understood from the following analysis.

The e^αx is the magnitude of the above term while the e^jβx is the phase (angle) of the same term.

The term e^αx increases as x increases i.e. the magnitude of current increases as we approach the beginning of line. In other words, the magnitude (intensity) of current diminishes as the wave travels from the beginning of line (where the voltage is applied and the current wave starts) to the end of line as one expects in real world (the intensity of signal diminishes as it moves away from source).

The term βx similarly increases as x increases. With similar as above reasoning, the term βx i.e. the phase of current wave diminishes as the wave travels from the beginning of line and moves to the end of line.

5.2 Second term of resultant current

Similarly, the term e^-γx can be written as e^-(α+jβ)x = e^-αx e^-jβx = e^-αx[cos(-βx) + j sin(-βx)]

With similar as above reasoning, the term e^-αx decreases as x increases. In other words, the magnitude (intensity) of current wave decreases as the wave moves from the end towards the beginning of line as one expects. It is really the part of current wave that arrives at the end of line and refracts travelling to the opposite direction of line. The opposite flow of current is indicated by the symbol minus (-) of the term.

Additionally, the term -βx decreases as x increases i.e. the phase (angle) of current wave decreases as the wave moves from the end towards the beginning of line.

5.3 Resultant current

Since from Eq. (7), the line current at point x is the algebraic summation of the above two terms and also taking into consideration the results of the above reasoning and depending on line parameters and the type of load at the end of line, we can state that in general the current magnitude and phase decrease from the beginning to the end of line. This implies having in mind the above that either or both the current magnitude and phase can also increase from the beginning to the end of line.