Handbook of Electrical Engineering

592 HANDBOOK OF ELECTRICAL ENGINEERING

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

U 11 =

ρs

πLg

= 13. 263

hd=

√

(dmh) = 0. 0707

U 12 =loge( 2 Lg/hd)= 8. 13

U 13 =

K 1 Lg

A^0.^5

= 13. 8

U 21 =

ρa

2 πnlr

= 0. 531

U 22 =loge( 8 lr/dr) = 9. 903

U 23 =

2 K 1 lr

A^0.^5

= 11. 5

U 24 =(n^0.^5 − 1 )^2 = 2. 101

U 31 =

ρa

πLg

= 2. 653

U 32 =loge( 2 Lg/lr) = 1. 569

WhereLg=total length of the grid conductors
lr=average length of a buried rod, but in this example all the rods are the same length
hd=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 rodsRepbecomes:-

Rep=

R 1 R 2 −R 122

R 1 +R 2 − 2 R 12

=16.582 ohms

Find the corner mesh voltage data.