Page 599 - Handbook of Electrical Engineering

592 HANDBOOK OF ELECTRICAL ENGINEERING

The following auxiliary equations are introduced to simplify the work involved:-

ρ s
U 11 = = 13.263
πL g
√
h d = (d m h) = 0.0707
U 12 = log (2L g /h d ) = 8.13
e
K 1 L g
U 13 = = 13.8
A 0.5
ρ a
U 21 = = 0.531
2πnl r
U 22 = log (8l r /d r ) = 9.903
e
2K 1 l r
U 23 = = 11.5
A 0.5
U 24 = (n 0.5 − 1) 2 = 2.101

ρ a
U 31 = = 2.653
πL g
U 32 = log (2L g /l r ) = 1.569
e
Where L g = total length of the grid conductors
l r = average length of a buried rod, but in this example all the rods are the same length
h d = weighted depth of the grid

Let

R 11 = resistance of the grid conductors
R 22 = resistance of all the ground rods
R 12 = mutual resistance between the whole grid and all the rods

From equations 42, 43 and 44 from IEEE80, these resistances are: -

R 11 = U 11 (U 12 + U 13 − K 2 ) = 227.86 ohms
R 22 = U 21 (U 22 − 1 + (U 23 U 24 )) = 17.541 ohms
R 12 = U 31 (U 32 + U 13 − K 2 + 1) = 30.82 ohms

From equation 41, for both the grid and the rods R ep becomes:-

2
R 1 R 2 − R 12
R ep = = 16.582 ohms
R 1 + R 2 − 2R 12
Find the corner mesh voltage data.