Tower footing resistance of 20 Ω electrode design

The three common footing electrodes are a hemisphere, a single driven rod and a counterpoise (buried horizontal wire). Their resistances (Begamudre, EHV AC Transmission Engineering) are:

$$\begin{aligned} \text{Hemisphere (radius } r):\quad & R = \frac{\rho}{2\pi r} \\ \text{Driven rod (length } L, \text{radius } a):\quad & R = \frac{\rho}{2\pi L}\ln\frac{2L}{a} \\ \text{Counterpoise (length } L, \text{depth } y):\quad & R = \frac{\rho}{\pi L}\left[\ln\frac{2L}{\sqrt{2ay}} - 1\right] \end{aligned}$$

Given: $R = 20\ \Omega$, $\rho = 100\ \Omega\text{-m}$, $a = 0.0125$ m, $y = 0.5$ m.

(a) Hemisphere

$$r = \frac{\rho}{2\pi R} = \frac{100}{2\pi \times 20} = 0.796\ \text{m}$$

(b) Driven rod

$$20 = \frac{100}{2\pi L}\ln\frac{2L}{0.0125}$$

This is solved by trial:

$$L \approx 5.38\ \text{m} \quad \left(\ln\frac{2 \times 5.38}{0.0125} = 6.757\right)$$

(c) Counterpoise

$$\sqrt{2ay} = \sqrt{2 \times 0.0125 \times 0.5} = 0.1118\ \text{m}$$ $$20 = \frac{100}{\pi L}\left[\ln\frac{2L}{0.1118} - 1\right]$$ $$L \approx 5.80\ \text{m} \quad \left(\ln\frac{2 \times 5.80}{0.1118} = 4.641\right)$$

Answer: hemisphere radius = 0.796 m; driven rod length = 5.38 m; counterpoise length = 5.80 m (at 0.5 m depth).

In practice the rod or counterpoise is preferred, since a 0.8 m radius hemisphere is hard to make; in high-resistivity soil several rods or longer counterpoises are used in parallel.