Coating conductance and coating resistance calculations are used to evaluate the
electrical quality of pipeline coatings using interruption test data. By applying
a known test current and observing potential shifts along the pipeline, engineers
can estimate pipe-to-earth resistance, specific coating resistance, coating conductance,
and the current required to meet cathodic protection criteria.
These calculations are commonly used for HDD crossings, isolated pipeline sections,
casing investigations, and general coating condition evaluation.
Reference 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
2) Average voltage shift (IR/Ohmic drop during test)
\[
\Delta V_{avg} = \left\langle \left|V_{on} - V_{off}\right| \right\rangle
\]
\(\Delta V_{avg}\) = average shift magnitude used for \(R_p\)
Average the shift magnitude across all
test locations.
3) Pipe-to-earth resistance during test
\[
R_p = \frac{\Delta V_{avg}}{I_{test}}
\]
\(R_p\) = pipe-to-earth resistance (\(\Omega\))
\(\Delta V_{avg}\) = average voltage shift (V)
\(I_{test}\) = applied test
current (A)
4) Specific coating resistance
\[
R_c = R_p \times A
\]
\(R_c\) = specific coating resistance (\(\Omega\cdot\text{ft}^2\) or \(\Omega\cdot\text{m}^2\))
\(R_p\) = pipe-to-earth resistance (\(\Omega\))
\(A\) = pipe surface area (\(\text{ft}^2\) or \(\text{m}^2\))
5) Coating conductance (per area)
\[
G = \frac{1}{R_c}
\]
\(G\) = coating conductance per area (S/\(\text{ft}^2\) or S/\(\text{m}^2\))
\(R_c\) = specific coating resistance (\(\Omega\cdot\text{ft}^2\) or \(\Omega\cdot\text{m}^2\))
Convert \(\mu\text{S}\) to S: \(1\ \mu\text{S} = 10^{-6}\ \text{S}\).
6) Current requirement from polarization/shift ratio
\[
I_{req} = I_{test}\left(\frac{\Delta E_{target}}{\Delta E_{test}}\right)
\]
\(I_{test}\) = test current applied (A)
\(\Delta E_{target}\) = required shift (V)
\(\Delta E_{test}\) = shift achieved during test (V)
\(\Delta E\) is typically based on OFF vs Native (
polarization), not ON vs OFF.
Practice 1: The Standard HDD Crossing
Scenario: A 12-inch (1.0 ft) OD pipeline is installed via HDD for 3,000 feet. You perform a current interruption test.
Data: Test Current (\(I_{test}\)) = 20 mA (0.020 A). Target Potential = -0.850 VCSE.
| Location | Native | ON | OFF |
|---|
| V1 (Entry) | -0.600 V | -1.500 V | -1.100 V |
| V2 (Exit) | -0.550 V | -1.400 V | -1.050 V |
Objectives: Calculate \(R_p\), \(R_c\), and the total current required.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Calculate voltage shifts } (|On-Off|): \\[6pt]
\quad V1\ \text{shift} = |(-1.500)-(-1.100)| = 0.400\ \text{V} \\[6pt]
\quad V2\ \text{shift} = |(-1.400)-(-1.050)| = 0.350\ \text{V} \\[6pt]
\quad \Delta V_{avg} = \dfrac{0.400+0.350}{2} = 0.375\ \text{V} \\[12pt]
2.\ R_p = \dfrac{\Delta V_{avg}}{I_{test}} = \dfrac{0.375}{0.020} = 18.75\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 1.0 \times 3000 = 9424.8\ \text{ft}^2 \\[12pt]
4.\ R_c = R_p \times A = 18.75 \times 9424.8 = 176{,}715\ \Omega\cdot\text{ft}^2 \\[12pt]
5.\ \text{Current requirement (OFF vs Native):} \\[6pt]
\quad \Delta E_{target} = |-0.850-(-0.550)| = 0.300\ \text{V} \\[6pt]
\quad \Delta E_{test} = |-1.050-(-0.550)| = 0.500\ \text{V} \\[6pt]
\quad I_{req} = 20\ \text{mA}\left(\dfrac{0.300}{0.500}\right) = 12\ \text{mA}
\end{array}
\]
Practice 2: Metric Large Diameter
Scenario: A 762 mm (0.762 m) OD pipeline section, 500 meters long, is isolated for testing.
Data: Test Current = 100 mA (0.100 A). Target Potential = -0.850 VCSE.
| Location | Native | ON | OFF |
|---|
| V1 | -0.500 V | -1.200 V | -0.900 V |
| V2 | -0.500 V | -1.150 V | -0.880 V |
Objectives: Calculate \(R_p\), \(R_c\) (\(\Omega\cdot\text{m}^2\)), and current required.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Voltage shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-1.200)-(-0.900)| = 0.300\ \text{V} \\[6pt]
\quad V2 = |(-1.150)-(-0.880)| = 0.270\ \text{V} \\[6pt]
\quad \Delta V_{avg} = \dfrac{0.300+0.270}{2} = 0.285\ \text{V} \\[12pt]
2.\ R_p = \dfrac{0.285}{0.100} = 2.85\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 0.762 \times 500 = 1197\ \text{m}^2 \\[12pt]
4.\ R_c = R_p \times A = 2.85 \times 1197 = 3411\ \Omega\cdot\text{m}^2 \\[12pt]
5.\ \text{Current requirement (OFF vs Native):} \\[6pt]
\quad \Delta E_{test} = |-0.880-(-0.500)| = 0.380\ \text{V} \\[6pt]
\quad \Delta E_{target} = |-0.850-(-0.500)| = 0.350\ \text{V} \\[6pt]
\quad I_{req} = 100\ \text{mA}\left(\dfrac{0.350}{0.380}\right) = 92.1\ \text{mA}
\end{array}
\]
Practice 3: The Excellent Coating
Scenario: A 10-inch (0.833 ft) OD pipeline is 1 mile (5,280 ft) long.
Data: Test Current = 10 mA (0.010 A). Target = -0.850 VCSE.
| Location | Native | ON | OFF |
|---|
| V1 | -0.600 V | -2.500 V | -1.200 V |
| V2 | -0.600 V | -2.400 V | -1.150 V |
Objectives: Calculate \(R_c\) and current requirement.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Voltage shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-2.500)-(-1.200)| = 1.300\ \text{V} \\[6pt]
\quad V2 = |(-2.400)-(-1.150)| = 1.250\ \text{V} \\[6pt]
\quad \Delta V_{avg} = \dfrac{1.300+1.250}{2} = 1.275\ \text{V} \\[12pt]
2.\ R_p = \dfrac{1.275}{0.010} = 127.5\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 0.833 \times 5280 = 13{,}823\ \text{ft}^2 \\[12pt]
4.\ R_c = R_p \times A = 127.5 \times 13{,}823 = 1{,}762{,}433\ \Omega\cdot\text{ft}^2 \\[12pt]
5.\ \text{Current requirement (OFF vs Native):} \\[6pt]
\quad \Delta E_{test} = |-1.150-(-0.600)| = 0.550\ \text{V} \\[6pt]
\quad \Delta E_{target} = |-0.850-(-0.600)| = 0.250\ \text{V} \\[6pt]
\quad I_{req} = 10\ \text{mA}\left(\dfrac{0.250}{0.550}\right) = 4.54\ \text{mA}
\end{array}
\]
Practice 4: Damaged River Crossing
Scenario: 20-inch (1.667 ft) OD pipeline, 2,000 ft long. Suspected coating damage.
Data: Test Current = 500 mA (0.500 A). Target = -0.850 VCSE.
| Location | Native | ON | OFF |
|---|
| V1 | -0.600 V | -0.900 V | -0.750 V |
| V2 | -0.600 V | -0.850 V | -0.720 V |
Objectives: Calculate \(R_c\) and current requirement.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Voltage shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-0.900)-(-0.750)| = 0.150\ \text{V} \\[6pt]
\quad V2 = |(-0.850)-(-0.720)| = 0.130\ \text{V} \\[6pt]
\quad \Delta V_{avg} = \dfrac{0.150+0.130}{2} = 0.140\ \text{V} \\[12pt]
2.\ R_p = \dfrac{0.140}{0.500} = 0.28\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 1.667 \times 2000 = 10{,}474\ \text{ft}^2 \\[12pt]
4.\ R_c = R_p \times A = 0.28 \times 10{,}474 = 2{,}933\ \Omega\cdot\text{ft}^2 \\[12pt]
5.\ \text{Current requirement (OFF vs Native):} \\[6pt]
\quad \Delta E_{test} = |-0.720-(-0.600)| = 0.120\ \text{V} \\[6pt]
\quad \Delta E_{target} = |-0.850-(-0.600)| = 0.250\ \text{V} \\[6pt]
\quad I_{req} = 500\ \text{mA}\left(\dfrac{0.250}{0.120}\right) = 1042\ \text{mA}\ (1.04\ \text{A})
\end{array}
\]
Practice 5: Mixed Unit Challenge
Scenario: 508 mm (0.508 m) OD pipeline, 1.5 km (1500 m) long.
Data: Test Current = 50 mA. Target = -0.850 VCSE.
| Location | Native | ON | OFF |
|---|
| V1 | -0.600 V | -1.350 V | -0.950 V |
| V2 | -0.600 V | -1.300 V | -0.920 V |
Objectives: Calculate \(R_c\) (metric) and current requirement.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Voltage shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-1.350)-(-0.950)| = 0.400\ \text{V} \\[6pt]
\quad V2 = |(-1.300)-(-0.920)| = 0.380\ \text{V} \\[6pt]
\quad \Delta V_{avg} = \dfrac{0.400+0.380}{2} = 0.390\ \text{V} \\[12pt]
2.\ R_p = \dfrac{0.390}{0.050} = 7.8\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 0.508 \times 1500 = 2394\ \text{m}^2 \\[12pt]
4.\ R_c = R_p \times A = 7.8 \times 2394 = 18{,}673\ \Omega\cdot\text{m}^2 \\[12pt]
5.\ \text{Current requirement (OFF vs Native):} \\[6pt]
\quad \Delta E_{test} = |-0.920-(-0.600)| = 0.320\ \text{V} \\[6pt]
\quad \Delta E_{target} = |-0.850-(-0.600)| = 0.250\ \text{V} \\[6pt]
\quad I_{req} = 50\ \text{mA}\left(\dfrac{0.250}{0.320}\right) = 39.1\ \text{mA}
\end{array}
\]
Practice 6: Large Diameter Short Section
Scenario: A 30-inch (2.5 ft) OD pipeline section is 1,000 ft long. You apply a test current of 15 mA.
| Location | Native | ON | OFF |
|---|
| V1 | -0.600 V | -1.000 V | -0.900 V |
| V2 | -0.620 V | -0.980 V | -0.880 V |
Objectives: Calculate \(R_c\) and identify the coating quality rating.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-1.000)-(-0.900)| = 0.100\ \text{V} \\[6pt]
\quad V2 = |(-0.980)-(-0.880)| = 0.100\ \text{V} \\[6pt]
\quad \Delta V_{avg} = 0.100\ \text{V} \\[12pt]
2.\ R_p = \dfrac{0.100}{0.015} = 6.67\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 2.5 \times 1000 = 7854\ \text{ft}^2 \\[12pt]
4.\ R_c = R_p \times A = 6.67 \times 7854 = 52{,}386\ \Omega\cdot\text{ft}^2 \\[12pt]
5.\ \text{Rating: approximately } 50k\ \Omega\cdot\text{ft}^2 \\
\quad \text{commonly classified as Fair to Good coating quality.}
\end{array}
\]
Practice 7: High Resistance Calculation
Scenario: A 6-inch (0.15 m) OD pipe is 100 m long. Test current is 1 mA.
| Location | Native | ON | OFF |
|---|
| V1 | -0.500 V | -1.500 V | -1.100 V |
| V2 | -0.500 V | -1.400 V | -1.050 V |
Objectives: Calculate \(R_c\) (metric) and convert to Imperial units.
Reveal solution
\[
\begin{array}{l}
1.\ \text{Shifts } (|On-Off|): \\[6pt]
\quad V1 = |(-1.500)-(-1.100)| = 0.400\ \text{V} \\[6pt]
\quad V2 = |(-1.400)-(-1.050)| = 0.350\ \text{V} \\[6pt]
\quad \Delta V_{avg} = 0.375\ \text{V} \\[12pt]
2.\ R_p = \dfrac{0.375}{0.001} = 375\ \Omega \\[12pt]
3.\ A = \pi \times d \times L = 3.1416 \times 0.15 \times 100 = 47.12\ \text{m}^2 \\[12pt]
4.\ R_c = 375 \times 47.12 = 17{,}670\ \Omega\cdot\text{m}^2 \\[12pt]
5.\ \text{Convert to } \Omega\cdot\text{ft}^2: \\[6pt]
\quad 1\ \text{m}^2 = 10.7639\ \text{ft}^2 \\[6pt]
\quad R_c = 17{,}670 \times 10.7639 = 190{,}200\ \Omega\cdot\text{ft}^2\ (\text{approx.})
\end{array}
\]
Practice 8: Shorted Casing Check
Scenario: A 48-inch (4 ft) OD pipe inside a casing is 500 ft long. You suspect a dead short. You apply 1.0 Amp.
Result: The average voltage shift (\(\Delta V\)) observed is only 0.05 V.
Objectives: Calculate the effective specific coating resistance \(R_c\).
Reveal solution
\[
\begin{array}{l}
1.\ R_p = \dfrac{\Delta V}{I} = \dfrac{0.05}{1.0} = 0.05\ \Omega \\[12pt]
2.\ A = \pi \times d \times L = 3.1416 \times 4 \times 500 = 6283\ \text{ft}^2 \\[12pt]
3.\ R_c = R_p \times A = 0.05 \times 6283 = 314\ \Omega\cdot\text{ft}^2 \\[12pt]
4.\ \text{Conclusion: this extremely low } R_c \\
\quad \text{suggests a metallic short or bare contact} \\
\quad \text{such as a shorted casing.}
\end{array}
\]
Practice 9: Current Requirement Extrapolation
Scenario: You perform a linear polarization-style test. You need to reach -0.850 V polarized.
Data: Native = -0.650 V. Apply 0.5 A. Instant-off becomes -0.750 V.
Objective: Calculate total current required.
Reveal solution
\[
\begin{array}{l}
1.\ \Delta E_{test} = |-0.750-(-0.650)| = 0.100\ \text{V} \\[12pt]
2.\ \Delta E_{target} = |-0.850-(-0.650)| = 0.200\ \text{V} \\[12pt]
3.\ \dfrac{0.200}{0.100} = 2.0 \\[12pt]
4.\ I_{req} = 0.5\ \text{A} \times 2.0 = 1.0\ \text{A}
\end{array}
\]
Practice 10: Coating Conductance Conversion
Scenario: A historical report lists the coating conductance as 200 \(\mu\text{S}/\text{ft}^2\) (microSiemens per square foot).
Objective: Convert this to specific coating resistance \(R_c\) in \(\Omega\cdot\text{ft}^2\).
Reveal solution
\[
\begin{array}{l}
1.\ 200\ \mu\text{S}/\text{ft}^2 = 200 \times 10^{-6}\ \text{S}/\text{ft}^2 = 0.000200\ \text{S}/\text{ft}^2 \\[12pt]
2.\ G = \dfrac{1}{R_c}\ \Rightarrow\ R_c = \dfrac{1}{G} \\[12pt]
3.\ R_c = \dfrac{1}{0.000200} = 5000\ \Omega\cdot\text{ft}^2
\end{array}
\]
