Page 224 - Pipeline Rules of Thumb Handbook

Corrosion/Coatings 211

What will be the resistance to earth of a single graphite anode?

For the most common size, 3-in. by 60-in., installed in 1,800 ¥ 0.002 = 3.6 ohm
10ft of coke breeze backfill in an 8-in. hole, multiply the soil
resistivity in ohm-cm by 0.002; this will give the resistance to This assumes that the resistivity is uniform to considerable
earth in ohm. depth; if it increases with depth, the resistance will be slightly
higher.
Example. 1,800 ohm-cm soil

How to estimate the monthly power bill for a cathodic protection rectifier

Multiply the rectifier output voltage and current together, 10.2 ¥ 21 ¥ 1.5 = 321.2
1
and multiply this by 1 /2 ; the result is the power consumption
for one month, in kilowatt-hours. If the unit is on a meter of The power consumption will be about 320kwh per month; at
its own, figure the bill by the utilities rate schedule; if it rep- 3 cents this would amount to $9.60, at 1.1 cents it would be
resents an additional load, use the rate per kilowatt-hour for $3.52, etc.
the highest bracket reached. Note that the actual current and voltage values are used; if
the rating of the unit is used, the result will be the maximum
power consumption within the range of the equipment. The
Example. The terminal voltage of a rectifier is 10.2 volts, formula is approximate; it is based on an efficiency of about
and the current output is 21 amperes. 48%.

What will be the resistance to earth of a group of graphite anodes, in terms of the resistance of a
single anode?

When the spacing is 20ft (the most common value), the This rule is reasonably accurate, for soils which are uniform
resistance of the group will be equal to the resistance of a to a considerable depth; if the resistivity increases with
single anode, divided by the square root of the number of depth, the “crowding” effect will be greater, whereas if it
anodes. decreases, the individual anodes will not affect each other
so much.
1
Example. Resistance of 4 anodes = /2 resistance of single;
1
resistance of 16 anodes = /4 resistance of single, etc.

How can the current output of magnesium rod used for the cathodic protection of heat exchanger
shells be predicted?

Divide the total dissolved solids content of the water by 13; tection is obtained. If insufficient magnesium is installed, the
the answer is the current output per ft of 1.315in. (1in. pipe current flow will probably be greater, and if too much is
size) magnesium rod to be expected. installed, the current will be less.
This is an approximation to the current flow which will be
found after stabilized conditions have been reached, if pro-

What spacing for test leads to measure current on a pipeline?

Multiply the pipe weight, in lb/ft, by 4; the result is a Example. Pipe weight is 40.48lb/ft.
spacing which will give a resistance of 0.001 ohm, or one
ampere per millivolt IR drop. 40.5 ¥ 4 = 162