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Tower footing resistance of 16 Ω electrode design

Short answer

Using R=ρ/(2πr) for hemisphere, R=(ρ/(2πL))ln(2L/a) for driven rod, and the counterpoise formula with R=16 Ω, ρ=100 Ω-m, a=1.25 cm and y=0.6 m gives: hemisphere radius 0.995 m, driven rod length 6.98 m, and counterpoise length 7.60 m buried 0.6 m deep.

It is necessary to obtain a tower footing resistance of 16 Ω in a soil of resistivity ρs = 100 Ω-m using the three common types of electrodes. Calculate the required dimensions if the radius of rod and counterpoise is 1.25 cm and depth of counterpoise is 0.6 m.

The three common electrodes are a hemisphere, a driven rod and a counterpoise. Using the standard relations (Begamudre, EHV AC Transmission Engineering):

Hemisphere:R=ρ2πrDriven rod:R=ρ2πLln⁡2LaCounterpoise:R=ρπL[ln⁡2L2ay−1]\begin{aligned} \text{Hemisphere:}\quad & R = \frac{\rho}{2\pi r} \\ \text{Driven rod:}\quad & R = \frac{\rho}{2\pi L}\ln\frac{2L}{a} \\ \text{Counterpoise:}\quad & R = \frac{\rho}{\pi L}\left[\ln\frac{2L}{\sqrt{2ay}} - 1\right] \end{aligned}

Given: R=16 ΩR = 16\ \Omega, ρ=100 Ω-m\rho = 100\ \Omega\text{-m}, a=0.0125a = 0.0125 m, y=0.6y = 0.6 m.

(a) Hemisphere

r=ρ2πR=1002π×16=0.995 mr = \frac{\rho}{2\pi R} = \frac{100}{2\pi \times 16} = 0.995\ \text{m}

(b) Driven rod

16=1002πLln⁡2L0.012516 = \frac{100}{2\pi L}\ln\frac{2L}{0.0125}
LL (m)RR (Ω\Omega)
5.021.28
6.018.22
7.015.96
6.9816.00
L≈6.98 m(ln⁡2×6.980.0125=7.018)L \approx 6.98\ \text{m} \quad \left(\ln\frac{2 \times 6.98}{0.0125} = 7.018\right)

(c) Counterpoise

2ay=2×0.0125×0.6=0.1225 m\sqrt{2ay} = \sqrt{2 \times 0.0125 \times 0.6} = 0.1225\ \text{m} 16=100πL[ln⁡2L0.1225−1]16 = \frac{100}{\pi L}\left[\ln\frac{2L}{0.1225} - 1\right]
LL (m)RR (Ω\Omega)
7.017.00
8.015.41
7.6016.00
L≈7.60 m(ln⁡2×7.600.1225=4.821)L \approx 7.60\ \text{m} \quad \left(\ln\frac{2 \times 7.60}{0.1225} = 4.821\right)

Answer: hemisphere radius = 0.995 m; driven rod length = 6.98 m; counterpoise length = 7.60 m (buried 0.6 m deep).

✓ Answer checked by Claude Opus 5.5 (high effort)

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