What this page is for
These formulas estimate the current required to achieve cathodic protection on pipelines,
tank bottoms, offshore structures, and other steel assets. Depending on the information
available, current requirement may be estimated from exposed area and current density,
from voltage shift and structure resistance, from a polarization ratio test, or from
coating conductance.
Current requirement formulas
1) Surface area (pipeline)
\[
A = \pi \,\times\, d \,\times\, L
\]
\(A\) = external surface area of pipe
\(d\) = outside diameter
\(L\) = length
\(\pi\) = 3.1416
Use consistent units (ft or m).
2) Current requirement (current density method)
\[
I_{req} = A_{total} \,\times\, f_{bare} \,\times\, i_{density}
\]
\(A_{total}\) = total external surface area
\(f_{bare}\) = fraction bare (1% bare = 0.01)
3) Voltage shift / resistance method
\[
I_{req} = \frac{\Delta V_{target}}{R_p}
\]
\(I_{req}\) = required current
\(\Delta V_{target}\) = target
structure voltage shift (V)
\(R_p\) = structure (pipe-to-earth) resistance (\(\Omega\))
4) Polarization / shift ratio method
\[
I_{req} = I_{test}\left(\frac{\Delta E_{target}}{\Delta E_{test}}\right)
\]
\(I_{req}\) = required current
\(I_{test}\) = applied
test current
\(\Delta E_{test}\) = measured shift during test
\(\Delta E\) is typically based on OFF vs Native (polarization), not ON vs OFF.
5) Conductance form (Ohm’s law using conductance)
\[
I = A \,\times\, \Delta V \,\times\, G
\]
\(I\) = current (A)
\(A\) = area (ft2 or m2)
\(G\) = conductance per area (S/ft2 or S/m2)
Convert \(\mu\text{S}\) to S: \(1\ \mu\text{S} = 10^{-6}\ \text{S}\).
Calculate the total current required to protect a bare steel pipeline.
The pipe is 12 inches (1.0 ft) in diameter and 5,000 feet
long. The design current density is 1.0 mA/ft².
Reveal solution
\[
\begin{array}{c}
A = \pi \,\times\, d \,\times\, L \\[10pt]
A = 3.1416 \,\times\, 1.0 \,\times\, 5000 \\[10pt]
A = 15{,}708\ \text{ft}^2 \\[14pt]
I = A \,\times\, i_{density} \\[10pt]
I = 15{,}708\ \text{ft}^2 \,\times\, 1.0\ \text{mA/ft}^2 \\[10pt]
I = 15{,}708\ \text{mA} = 15.7\ \text{A}
\end{array}
\]
Answer: 15.7 A
Practice 2: Coated pipe (percent bare)
A 10-inch (0.833 ft) OD pipeline is 10,000 feet long.
The coating is 99% effective (1% bare). The current density required for
bare steel in this soil is 2.0 mA/ft².
Reveal solution
\[
\begin{array}{c}
A_{total} = \pi \,\times\, 0.833 \,\times\, 10{,}000 \\[10pt]
A_{total} = 26{,}170\ \text{ft}^2 \\[14pt]
A_{bare} = A_{total} \,\times\, f_{bare} \\[10pt]
A_{bare} = 26{,}170 \,\times\, 0.01 \\[10pt]
A_{bare} = 261.7\ \text{ft}^2 \\[14pt]
I = A_{bare} \,\times\, i_{density} \\[10pt]
I = 261.7 \,\times\, 2.0 \\[10pt]
I = 523.4\ \text{mA}
\end{array}
\]
Answer: 523.4 mA
Practice 3: Metric pipeline
design
Design a system for a 30 cm (0.3 m) diameter pipeline that is
6 km (6,000 m) long. The required current density is
30 µA/m². The pipe is coated, so the stated density applies to
coated surface.
Reveal solution
\[
\begin{array}{c}
A = \pi \,\times\, 0.3 \,\times\, 6000 \\[10pt]
A = 5655\ \text{m}^2 \\[14pt]
I = A \,\times\, i_{density} \\[10pt]
I = 5655 \,\times\, 30\ \mu\text{A/m}^2 \\[10pt]
I = 169{,}650\ \mu\text{A} \\[10pt]
I = 169.7\ \text{mA}
\end{array}
\]
Answer: 169.7 mA
You apply a test current of 2 A and measure a pipe-to-earth voltage
shift of 0.15 V. You require a total shift of 0.30 V
for protection. What is the total current required?
Reveal solution
\[
\begin{array}{c}
R_p = \frac{\Delta V}{I} = \frac{0.15}{2} = 0.075\ \Omega \\[14pt]
I_{req} = \frac{\Delta V_{target}}{R_p} = \frac{0.30}{0.075} = 4.0\ \text{A}
\end{array}
\]
Answer: 4.0 A
Practice 5: Polarization test
A polarization test is conducted. A current of 50 mA results in a
polarization shift of 25 mV. To satisfy the 100 mV
polarization criterion, how much total current is needed?
Reveal solution
\[
\begin{array}{c}
I_{req} = I_{test}\left(\frac{\Delta E_{target}}{\Delta E_{test}}\right) \\[10pt]
I_{req} = 50\left(\frac{100}{25}\right) \\[10pt]
I_{req} = 50 \times 4 = 200\ \text{mA}
\end{array}
\]
Answer: 200 mA
Practice 6: Tank bottom current
Calculate the current required to protect the bare bottom of an
aboveground storage tank. The tank diameter is 100 feet. The design
current density for the soil pad is 1.5 mA/ft².
Reveal solution
\[
\begin{array}{c}
A = \pi \times r^2 \\[10pt]
A = 3.1416 \times 50^2 \\[10pt]
A = 3.1416 \times 2500 = 7854\ \text{ft}^2 \\[14pt]
I = 7854 \times 1.5 = 11{,}781\ \text{mA} = 11.78\ \text{A}
\end{array}
\]
Answer: 11.78 A
Practice 7: Mixed soil resistivities
A bare steel pipe (10.75-inch OD) is 10,000 ft long.
7,500 ft is in clay and requires 2.0 mA/ft².
2,500 ft is in sand and requires 0.5 mA/ft².
Note: OD = 10.75 in = 0.896 ft.
Reveal solution
\[
\begin{array}{c}
A_{unit} = \pi \times 0.896 = 2.815\ \text{ft}^2/\text{ft} \\[14pt]
A_{clay} = 7500 \times 2.815 = 21{,}112\ \text{ft}^2 \\[10pt]
I_{clay} = 21{,}112 \times 2.0 = 42{,}224\ \text{mA} = 42.2\ \text{A} \\[14pt]
A_{sand} = 2500 \times 2.815 = 7037\ \text{ft}^2 \\[10pt]
I_{sand} = 7037 \times 0.5 = 3518\ \text{mA} = 3.5\ \text{A} \\[14pt]
I_{total} = 42.2 + 3.5 = 45.7\ \text{A}
\end{array}
\]
Answer: 45.7 A
Practice 8: Offshore zonal calculation
An offshore jacket leg has 500 m² in the water zone requiring
100 mA/m² and 200 m² in the mud zone requiring
20 mA/m². Calculate the total current required for the leg.
Reveal solution
\[
\begin{array}{c}
I_{water} = 500 \times 100 = 50{,}000\ \text{mA} = 50\ \text{A} \\[10pt]
I_{mud} = 200 \times 20 = 4000\ \text{mA} = 4\ \text{A} \\[14pt]
I_{total} = 50 + 4 = 54\ \text{A}
\end{array}
\]
Answer: 54 A
Practice 9: Using conductance
A coated pipeline has a surface area of 10,000 ft². The coating
conductance is 100 µS/ft². You want to maintain a 0.3 V
shift. Calculate current using \(I = A \times \Delta V \times G\).
Reveal solution
\[
\begin{array}{c}
100\ \mu\text{S} = 100 \times 10^{-6}\ \text{S} = 0.0001\ \text{S} \\[14pt]
I = A \times \Delta V \times G \\[10pt]
I = 10{,}000 \times 0.3 \times 0.0001 \\[10pt]
I = 0.3\ \text{A} = 300\ \text{mA}
\end{array}
\]
Answer: 0.3 A
You calculate a theoretical requirement of 10 A for a long pipeline
based on average coating quality. To account for attenuation and ensure
the midpoint receives adequate current, you apply a safety factor of 1.5.
What capacity should the rectifier be sized for?
Reveal solution
\[
\begin{array}{c}
I_{design} = I_{theoretical} \times \text{Safety Factor} \\[10pt]
I_{design} = 10 \times 1.5 = 15\ \text{A}
\end{array}
\]
Answer: 15 A
