ETL 1110-9-10(FR)

5 Jan 91

Where:

LWB

=

291 ft (Cable

length

from

previous

calculation)

RMFT

=

0.254 ohm (Cable resistance per 1000

lin ft [same as above])

= 0.074 ohm

RWB

=

291 ft x 0.254 ohm

1000 ft

These two cables are in parallel, so that their wire

1

=

1

+

1

RT/B

RWT

RWB

1

=

1

+

1

RT/B

0.051

0.074

1

RT/B

= 19.6 + 13.5 = 33.1

RT/B = 0.030 ohm

Since current is dissipating along the portion of the

cable to which the anodes are connected, the

its total resistance as was done in example 2-2.

LW RMFT

91 ft( x 0.254 ohm/ft

1

1

RPOS '

'

x

x

1000 ft

2

1000 ft

2

*91 ft is the overall anode column length including the

interconnecting wire from the top of the top anode to the bottom

of the bottom anode (see item 4 from paragraph 2-4b.)

RPOS = 0.012 ohm

Negative circuit

wire

resistance

must

also

be

calculated:

Negative cable from rectifier to closest tank = 125 ft

of No. 4 AWG

Intertank bonds = 170 ft of No. 4 AWG**

**The two intertank bond circuits are in parallel and of about the

same length. From the law of parallel circuits, total resistance

of two parallel circuits of equal resistance is one half the

resistance of each circuit.

53

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