Distributed parameters and current along a long line

Why long lines use uniformly distributed parameters

• Resistance, inductance and capacitance are not located at one point; every metre of conductor has its own series $r$, $L$ and shunt $C$ (and leakage $g$).
• In short and medium lines the total charging current is small, so lumping $C$ at one or two points gives little error.
• In a long line (above about 250 km) the charging current is large and the current changes continuously along the line: each section draws its own charging current. The voltage therefore also changes continuously.
• The line length becomes a noticeable fraction of the wavelength ($\lambda \approx 6000$ km at 50 Hz), so travelling-wave effects (phase shift, attenuation, reflection) appear. These can only be represented by distributed parameters.
• Lumped models then give wrong $V_S$, $I_S$, regulation and Ferranti rise, so an exact solution of the line's differential equations is used.

Current is different at different points

Let $z = r + j\omega L$ (Ω/km) and $y = g + j\omega C$ (S/km). Take an element $dx$ at distance $x$ from the receiving end.

 IS      x <------ measured from receiving end
 o-->--z.dx--+--z.dx--+-- ... --z.dx--+--> IR
             |        |               |
 VS        y.dx     y.dx            y.dx    VR
             |        |               |
 o-----------+--------+-- ... --------+----o
 sending                          receiving

Series drop and shunt current in the element:

$$\frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV$$

Differentiating the second and substituting the first:

$$\frac{d^2I}{dx^2} = y\frac{dV}{dx} = yz\,I = \gamma^2 I, \qquad \gamma = \sqrt{zy} = \alpha + j\beta$$

General solution:

$$I(x) = K_1e^{\gamma x} + K_2e^{-\gamma x}$$

Then $V = \frac{1}{y}\frac{dI}{dx} = Z_c\left(K_1e^{\gamma x} - K_2e^{-\gamma x}\right)$, with $Z_c = \sqrt{z/y}$.

Boundary conditions at $x = 0$: $I = I_R$, $V = V_R$:

$$K_1 + K_2 = I_R, \quad K_1 - K_2 = \frac{V_R}{Z_c} \;\Rightarrow\; K_1 = \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right),\; K_2 = \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right)$$ $$\begin{aligned} I(x) &= \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right)e^{\gamma x} + \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right)e^{-\gamma x} \\ I(x) &= I_R\cosh\gamma x + \frac{V_R}{Z_c}\sinh\gamma x \end{aligned}$$

Interpretation

• $I(x)$ depends on $x$ through $\cosh\gamma x$ and $\sinh\gamma x$, which change in magnitude and angle along the line. So the current is different at every point.
• The term $\frac{V_R}{Z_c}\sinh\gamma x$ is the accumulated charging current; it is zero at the receiving end and largest at the sending end.
• In wave form, the current is the sum of an incident wave ($e^{\gamma x}$) and a reflected wave ($e^{-\gamma x}$), whose magnitudes vary as $e^{\pm\alpha x}$ and phases as $\pm\beta x$ along the line.
• At no load ($I_R = 0$), $I(x) = \frac{V_R}{Z_c}\sinh\gamma x$, which is zero at the receiving end but not at the sending end.
• At $x = l$: $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$.