Direct lightning stroke overvoltage and tower footing

A direct stroke delivers a lightning current $I$ straight onto a line component. The voltage produced depends on the surge impedance seen from the point of strike.

1. Stroke on a phase conductor

The current divides equally into the two directions of the conductor, each of surge impedance $Z_0$:

$$V = I \cdot \frac{Z_0}{2}$$

For example, $I = 10$ kA and $Z_0 = 400\ \Omega$ give $V = 2000$ kV. This is far above the insulation level of most lines, so a direct stroke to an unshielded phase conductor almost always causes flashover. This is why ground wires are used.

2. Stroke on the earth (ground) wire at mid-span

The current divides both ways along the ground wire (surge impedance $Z_g$):

$$V_g = I\frac{Z_g}{2}$$

The phase conductor gets an induced voltage $KV_g$ through the coupling factor $K$. The voltage across the air gap between ground wire and conductor at mid-span is

$$V_{gap} = (1 - K)V_g = (1-K)\frac{I Z_g}{2}$$

The waves travel to the adjacent towers and are reflected negatively there by the low tower footing resistance. So the mid-span voltage is limited if the span is short and the clearance is adequate.

3. Stroke on the tower top

The current divides between the stroke channel $Z_s$, the ground wires $Z_g$ and the tower footing resistance $R$:

$$V_t = I\,Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

If the ground wires and stroke impedance are neglected, $V_t \approx IR$. The insulator string then sees

$$V_{ins} = (1 - K)V_t$$

     I (stroke)
       |
   ----+---- ground wire Zg  (both sides)
       |
     tower --- insulator --- phase conductor
       |
       R  footing
      gnd

Importance of tower footing resistance

• $V_t$ is roughly proportional to $R$. With $R = 40\ \Omega$ and $I = 50$ kA, $V_t$ can be as high as 2000 kV. With $R = 10\ \Omega$ it falls to 500 kV or less.
• If $(1-K)V_t$ exceeds the impulse flashover voltage of the insulator string, a back-flashover occurs: the tower, at high potential, flashes over to the phase conductor. The ground wire then fails to protect the line.
• Low footing resistance also sends a strong negative reflection back up the tower, which cuts the tower-top voltage quickly.
• Practice: keep $R$ below about 10 Ω (and preferably below 5 Ω on EHV lines). Use extra earth rods, counterpoise wires buried along the line, and chemical earthing in high-resistivity soil.

So shielding with ground wires protects the conductor only when the footing resistance is low. A high $R$ turns a shielded stroke into a back-flashover.