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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.
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