ETL 1110-9-10(FR)

5 Jan 91

K

=

0.0213 (Shape function, from table 3-4

[where: L/d = 60 in./3 in. = 20])

L

=

5 ft (Effective anode length [canister

length))

d

=

3 in. (Anode backfill diameter [canis-

ter diameter])

RA =

2000 x 0.0213

5

RA

=

8.53 ohms

5)

Calculate the number of anodes required to meet maximum

anode groundbed-to-earth resistance requirements from

equation 1-11:

RA

RN '

%

N

Cc

Where:

RN =

Groundbed-to-earth resistance.

RA

=

8.53 ohms

(Resistance of a single

anode-to-earth

from

the

previous

calculation)

N

=

Assume 5 anodes (discussed below)

=

2000 ohm-cm (Soil resistivity)

PF

=

0.00268 (Paralleling factor from table

3-5 [discussed below])

CC

=

20 ft

(Center-to-center spacing of

anodes [discussed below])

To determine PF, a figure for N must be assumed. This

paralleling factor compensates for mutual interference

between anodes and is dependent on spacing. From the

law of parallel circuits, it appears that five anodes

might give the desired circuit resistance of 2.0 ohms

maximum, i.e.:

8.53 ohms/anode

= 1.706 ohms

5 anodes

26

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