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Meter Loading and Pipe-to-Soil Potentials

CP 3 IR Drop Meter Loading Pipe To Soil Potentials Polarization Reference Electrodes

What this page is for

These formulas cover several common pipe-to-soil potential calculations used in CP work, including temperature correction, meter loading, true potential correction, polarization shift, and common reference electrode conversions. The main goal is to recognize what the meter is actually reading, when a correction is needed, and how to interpret the result without mixing reference scales or mistaking IR drop for true polarization.

Pipe-to-soil potential formulas

1) Temperature correction (to 25°C)

\[ E_{25} = E_{meas} + K_t \,\times\, (T - 25) \]
\(E_{25}\) = corrected potential at 25°C
\(E_{meas}\) = measured potential at temperature \(T\)
\(K_t\) = temperature coefficient for the reference electrode
\(T\) = reference electrode temperature (°C)

\(K_t\) (CSE) ≈ +0.9 mV/°C
\(K_t\) (Ag/AgCl) ≈ −0.7 mV/°C (depends on fill solution)

2) Meter loading (voltage divider)

\[ V_m = E_{true} \,\times\, \left(\frac{R_m}{R_m + R_{ext}}\right) \]
\(V_m\) = meter reading (loaded)
\(E_{true}\) = true polarized potential (no loading error)
\(R_m\) = meter input resistance
\(R_{ext}\) = external/contact resistance in the measurement circuit

3) True potential from one reading

\[ E_{true} = V_m \,\times\, \left(\frac{R_m + R_{ext}}{R_m}\right) \]
\(E_{true}\) = corrected (true) polarized potential
\(V_m\) = measured (loaded) meter reading
\(R_m\) = meter input resistance
\(R_{ext}\) = external/contact resistance

4) Polarization shift

\[ \text{Pol} = E_{off} - E_{native} \]
Pol = polarization shift (often compared to 100 mV criterion by magnitude)
\(E_{off}\) = instant-off potential (IR-drop minimized)
\(E_{native}\) = native/free-corrosion potential

5) Two-meter true-potential method

\[ K = \frac{R_l}{R_h} \qquad\qquad E_{true} = \frac{V_h \,\times\, (1 - K)} {1 - K \,\times\, \left(\frac{V_h}{V_l}\right)} \]
\(K\) = ratio of input resistances
\(R_l\) = lower meter input resistance
\(R_h\) = higher meter input resistance
\(V_l\) = measured voltage using \(R_l\)
\(V_h\) = measured voltage using \(R_h\)
\(E_{true}\) = true polarized potential

Ag/AgCl potential depends on fill solution (for example, seawater vs saturated KCl), so always verify which reference value is being used before converting measured potentials from one scale to another.

Step-by-step method

  1. Identify the type of correction you need: temperature correction, loading correction, polarization shift, IR drop, or reference conversion.
  2. Confirm the reference electrode scale before comparing values or converting them.
  3. Keep units consistent throughout the calculation (mV vs V, MΩ vs Ω, and reference values in the same scale).
  4. Use the correct measured value for the situation, such as loaded meter reading, instant-off potential, native potential, or reference conversion value.
  5. Sanity-check the result so the direction of the correction makes physical sense.

Practice problems

Practice 1: Temperature correction (hot)

A structure potential is measured as -865 mV with respect to a Copper/Copper Sulfate (CSE) electrode. The reference electrode temperature is 45°C. Calculate the corrected potential at the standard temperature of 25°C.

Reveal solution
\[ \begin{array}{c} E_{25} = E_{meas} + K_t \,\times\, (T - 25) \\[10pt] K_t\text{ (CSE)} \approx +0.9\ \text{mV/°C} \\[10pt] T - 25 = 45 - 25 = 20^\circ C \\[10pt] \Delta E = 0.9 \,\times\, 20 = +18\ \text{mV} \\[10pt] E_{25} = -865 + 18 = -847\ \text{mV} \end{array} \]

Note: with a hot CSE, the pipe can appear more negative than the 25°C-corrected value.

Answer: -847 mV CSE
Practice 2: Temperature correction (cold)

You measure a potential of -820 mV using a CSE reference electrode in freezing conditions (-5°C). Calculate the corrected potential at 25°C. Does it meet the -850 mV criterion?

Reveal solution
\[ \begin{array}{c} E_{25} = E_{meas} + K_t \,\times\, (T - 25) \\[10pt] T - 25 = -5 - 25 = -30^\circ C \\[10pt] \Delta E = 0.9 \,\times\, (-30) = -27\ \text{mV} \\[10pt] E_{25} = -820 + (-27) = -847\ \text{mV} \end{array} \]

\(-847\ \text{mV}\) is more positive than \(-850\ \text{mV}\), so it does not meet the criterion.

Answer: -847 mV (Does NOT meet criterion)
Practice 3: Reference electrode conversion (Ag/AgCl sat KCl to CSE)

A potential of -800 mV is measured using a Silver/Silver Chloride electrode (sat. KCl). You need to report the value as CSE.

Assume: \(E_{SSC(sat\ KCl)} \approx +0.222\ \text{V vs SHE}\) and \(E_{CSE} \approx +0.316\ \text{V vs SHE}\).

Reveal solution
\[ \begin{array}{c} \Delta E = E_{CSE} - E_{SSC} = 0.316 - 0.222 = 0.094\ \text{V} = 94\ \text{mV} \\[10pt] E_{CSE} \approx E_{SSC} - 94\ \text{mV} \\[10pt] -800 - 94 = -894\ \text{mV} \end{array} \]

Note: Ag/AgCl potential depends on fill solution (seawater vs saturated KCl). This uses the saturated KCl value stated above.

Answer: ≈ -894 mV CSE
Practice 4: Meter loading (true potential)

You measure -900 mV on a structure. The meter has an input resistance \(R_m\) of 10 MΩ. The contact resistance \(R_{ext}\) is 2 MΩ. Calculate the true potential \(E_{true}\).

Reveal solution
\[ \begin{array}{c} E_{true} = V_m \,\times\, \left(\frac{R_m + R_{ext}}{R_m}\right) \\[10pt] E_{true} = -900 \,\times\, \left(\frac{10 + 2}{10}\right) \\[10pt] E_{true} = -900 \,\times\, 1.2 \\[10pt] E_{true} = -1080\ \text{mV} \end{array} \]
Answer: -1080 mV
Practice 5: Meter loading (% error)

Using Question 4, True = -1080 mV and Measured = -900 mV. Calculate the percentage error magnitude caused by loading.

Reveal solution
\[ \begin{array}{c} \%\text{Error} = \left| \frac{V_m - E_{true}}{E_{true}} \right| \,\times\, 100 \\[10pt] \%\text{Error} = \left| \frac{-900 - (-1080)}{-1080} \right| \,\times\, 100 \\[10pt] \%\text{Error} = \left| \frac{180}{-1080} \right| \,\times\, 100 \\[10pt] \%\text{Error} = 0.1667 \,\times\, 100 \\[10pt] \%\text{Error} = 16.7\% \end{array} \]
\[ \frac{R_{ext}}{R_m + R_{ext}} = \frac{2}{12} = 0.1667 \]
Answer: 16.7% error
Practice 6: Two-meter method (input impedance correction)

You measure the potential with two different input settings:

  • Low impedance: \(R_l = 1\,\text{M}\Omega\), \(V_l = -650\,\text{mV}\)
  • High impedance: \(R_h = 10\,\text{M}\Omega\), \(V_h = -800\,\text{mV}\)

Calculate the true potential \(E_{true}\).

Reveal solution
\[ \begin{array}{c} K = \frac{R_l}{R_h} \\[10pt] E_{true} = \frac{V_h \,\times\, (1 - K)} {1 - K \,\times\, \left(\frac{V_h}{V_l}\right)} \\[10pt] K = \frac{1}{10} = 0.1 \\[10pt] V_h \,\times\, (1 - K) = -800 \,\times\, 0.9 = -720 \\[10pt] \frac{V_h}{V_l} = \frac{-800}{-650} = 1.2308 \\[10pt] 1 - K \,\times\, \left(\frac{V_h}{V_l}\right) = 1 - 0.1 \,\times\, 1.2308 = 1 - 0.1231 = 0.8769 \\[10pt] E_{true} = \frac{-720}{0.8769} = -821\ \text{mV} \end{array} \]
Answer: -821 mV
Practice 7: IR drop calculation

A pipeline survey shows:

  • "On" potential: -1250 mV
  • "Instant off" potential: -950 mV

What is the magnitude of the IR drop in the "on" reading?

Reveal solution
\[ \begin{array}{c} \text{IR Drop} = \left|V_{on} - V_{off}\right| \\[10pt] \text{IR Drop} = \left|-1250 - (-950)\right| \\[10pt] \text{IR Drop} = 300\ \text{mV} \end{array} \]

The "on" potential includes 300 mV of voltage drop through the electrolyte/coating path.

Answer: 300 mV
Practice 8: Polarization shift

Determine if the 100 mV polarization criterion is met:

  • Native potential: -550 mV
  • "Instant off" potential: -700 mV

Calculate the polarization shift magnitude.

Reveal solution
\[ \begin{array}{c} \text{Pol} = E_{off} - E_{native} \\[10pt] \left|-700 - (-550)\right| = 150\ \text{mV} \end{array} \]

Since \(150\ \text{mV} > 100\ \text{mV}\), the criterion is met by polarization shift.

Answer: 150 mV (criterion met)
Practice 9: Conversion (CSE to SCE)

A potential is measured as -1.000 V vs CSE. Convert to the SCE scale.

Hint: \(E_{CSE} \approx +0.316\ \text{V vs SHE}\), \(E_{SCE} \approx +0.244\ \text{V vs SHE}\).

Reveal solution
\[ \begin{array}{c} E_{CSE} - E_{SCE} = 0.316 - 0.244 = 0.072\ \text{V} \\[10pt] V_{SCE} = V_{CSE} + (E_{CSE} - E_{SCE}) \\[10pt] V_{SCE} = -1.000 + 0.072 \\[10pt] V_{SCE} = -0.928\ \text{V} \end{array} \]

Sanity check: SCE is less positive than CSE, so the same structure appears less negative on SCE.

Answer: -0.928 V SCE
Practice 10: SHE conversion

You calculate a polarized potential of -850 mV vs CSE. Convert to the SHE scale. Calculate \(V_{SHE}\).

Reveal solution
\[ \begin{array}{c} E_{CSE} \approx +0.316\ \text{V vs SHE} \\[10pt] V_{SHE} = V_{CSE} + E_{CSE} \\[10pt] V_{SHE} = -0.850 + 0.316 = -0.534\ \text{V} \end{array} \]
Answer: -0.534 V SHE

Topics

CP 3 IR Drop Meter Loading Pipe To Soil Potentials Polarization Reference Electrodes