Page 223 - Pipeline Rules of Thumb Handbook

210 Pipeline Rules of Thumb Handbook

How to estimate the number of magnesium anodes required and their spacing for a bare line or for a
corrosion “hot spot”

,
Soil and pipeline conditions will vary widely but where 10 500 milliamperes
these given assumptions hold, an estimate can be made for Anodes required = 100 ma anode
test installations using this method. First find the area of pipe = 105 anodes distributed
to be protected, multiply by the milliamperes per sq ft nec-
essary to protect the line and divide by design output of an
,
anode in this type soil. This will give the number of anodes 10 000
Spacing = = 95 ft
required; to find spacing, divide number of anodes into the 105
length of line.
Similarly find the number of anodes required to protect ten
100-ft long “hot-spots” on a 4-in. line assuming a 100 mil-
liampere output from a 17lb magnesium anode. Estimate as
follows:

4 ¥ 314
.
Area of each “hot-spot” = = 105 sq ft
12
.
105 sq ft ¥ 1 0 ma per sq ft
.
Anodes required = = 105
100 ma per anode
One 17lb magnesium anode will produce approximately
Example. Find the number of 17lb magnesium anodes enough current in 1,000-ohm soil to protect one of these “hot-
needed and spacing used to protect 10,000ft of bare 4-in. spots.”
pipeline in a corrosive soil having a resistance of 1,000 ohm Ten anodes required for “hot-spots” on this section of
per cubic centimeter. Assume a current requirement of one line.
milliampere per sq ft, and that the anode output curve shows
an output of 100 milliamperes per anode in this type soil. Note. For these rules the protective criterion is -0.85
volts pipe-to-soil potential using a copper-copper sulfate
.
4 ¥ 3 14 reference electrode; or a negative change of potential of as
=
,
Area of pipe = ¥ 10 000 10 500 sq ft
,
12 much as -0.25 volts. Soil and pipeline conditions vary
widely, so it is well to consult an experienced corrosion
Total current requirement = 10,500 ¥ 1.0 = 10,500 engineer in the design of a complete cathodic protection
milliamperes system.

How can resistivity of fresh water be determined from chemical analysis?

Divide the total solids expressed in parts per million into 500,000 ∏ 400 = 1,250 ohm centimeters resistivity
500,000; the result is the resistivity in ohm-centimeter
units. This rule of thumb will give very accurate results for any
naturally occurring fresh waters; severely contaminated
waters may differ markedly, however, depending on the kind
Example. Total solid content is 400ppm. of contamination.