Page 221 - Pipeline Rules of Thumb Handbook

208 Pipeline Rules of Thumb Handbook

For conventional type ground beds this chart permits a To determine the optimum size ground bed, enter the chart
rapid determination of the resistance to earth of various at the “Soil Resistivity-ohm-cm” line and extend a vertical line
ground beds and the yearly power costs based on 50% effi- from the 1,000-ohm-cm point to intersect all the anode types.
cient rectification. These curves are specifically designed for Proceed horizontally to the left for the resistance to earth of
standard 3-in. by 60-in. and 4-in. by 80-in. graphite anodes horizontal type anodes and horizontally to the right for verti-
centered in a vertical or horizontal excavation and backfilled cal type anodes. Various groups of vertical anodes with 15, 20
to the dimensions indicated with well tamped coal coke and 25-ft spacings are represented by curves.
breeze. Resistance values for horizontal anodes obtained from the
ordinate of the upper left hand section should be transferred
to the abscissa of the upper right hand section to determine
power consumption and annual costs from the lower right
Example. hand and lower left hand section. In obtaining power costs
for the two lower sections, the ordinate shows power con-
Soil resistivity = 1,000 ohm-cm sumption in AC watts based on 50% efficient rectification,
Power rate = 3 cents per kwh and the abscissa of points on the lower left hand section indi-
Rectifier rating = 20-ampere cated annual power costs in dollars.

How can output of magnesium anodes be predicted?

For a single 17lb magnesium anode, working against a fully Current output will be about 56 milliamperes.
protected pipe, divide the soil resistivity in ohm-cm units into The rule works for single anodes against fully protected
180,000; the answer is the current output in milliamperes. pipe; anode groups will have lower current output per anode,
and if the pipe is not fully protected, the current output will
Example. Soil resistivity 3,200 ohm-cm be greater. For 32lb anodes use 195,000; for 50lb anodes use
180,000/3,200 = 56.25 204,000.

How to determine the efficiency of a cathodic protection rectifier

Multiply DC voltage output times DC amperes times by the number of seconds timed. This gives AC input in watts.
100 and divide by the AC watts input. The answer is % Then go ahead and use the formula as above.
efficiency.
Example. A rectifier unit does not have a watt meter. The
Example. A rectifier has an output of 50 amperes DC at wheel of the watt hour meter revolves 11 times in one minute.
15 volts. The wattmeter gives a direct reading of an AC input Meter constant is shown as 1.8. Output of the rectifier is the
of 1155 watts. same as in the first case, 50 amperes at 15 volts DC.
¥
,
KN ¥ 3 600
( DC volts)(DC amperes )(100 ) AC watts =
Efficiency = T
( AC watts input )
()( )(100 ) where K is the meter constant and N is the number of meter
15 50
= = 65%
(1155, ) revolutions in T seconds.
(1 8 11. )( )(3 600, )
On those rectifier units that do not have a wattmeter, the AC watts = = 1 188 watts AC
,
AC input may be obtained from the watt hour meter mounted 60
on the pole. To do this, count the number of revolutions made
15 50
by the watt hour meter in so many seconds, multiply by the ()( )(100 )
Efficiency = = 63%
meter constant stamped on the meter times 3,600 and divide 1188
,