Corrosion Authority

Attenuation (Cathodic Protection Current Decay)

What this page is for

These formulas estimate the service life of a sacrificial anode system using either a consumption rate constant or a capacity constant. The two methods are equivalent when the constants are consistent, but references may publish one or the other. The key to correct answers is using average current, keeping units consistent (lb with lb-based constants, kg with kg-based constants), and using utilization and efficiency as fractions (for example, 0.85 and 0.50), not percents.

Anode life formulas

1) Consumption Rate Method (Cr)

\[ L = \frac{W \,\times\, U \,\times\, \eta}{I \,\times\, C_r} \]
L = anode life (years)
W = total anode weight (lb or kg — match the constant units)
U = utilization factor (dimensionless fraction, e.g., 0.85)
\(\eta\) = efficiency factor (dimensionless fraction, e.g., 0.50)
I = average current output (A)
\(C_r\) = consumption rate (lb/A·yr or kg/A·yr)

2) Capacity Method (Ca)

\[ L = \frac{W \,\times\, U \,\times\, \eta \,\times\, C_a}{I} \]
L = anode life (years)
W = total anode weight (lb or kg — match the constant units)
U = utilization factor (dimensionless fraction, e.g., 0.85)
\(\eta\) = efficiency factor (dimensionless fraction, e.g., 0.50)
I = average current output (A)
\(C_a\) = capacity (A·yr/lb or A·yr/kg)

Use the consumption rate form when your reference provides \(C_r\) in lb/A·yr or kg/A·yr, and use the capacity form when your reference provides \(C_a\) in A·yr/lb or A·yr/kg.

Example constants (verify with your reference)

Material Consumption rate \(C_r\) Capacity \(C_a\) Efficiency \(\eta\)
Magnesium 8.76 lb/A·yr (3.98 kg/A·yr) 0.114 A·yr/lb (0.250 A·yr/kg) 0.50
Zinc 23.5 lb/A·yr (10.76 kg/A·yr) 0.042 A·yr/lb (0.093 A·yr/kg) 0.90
Aluminum 6.48 lb/A·yr (2.94 kg/A·yr) 0.154 A·yr/lb (0.340 A·yr/kg) 0.85–0.95

Keep units consistent and use average current. Use \(U\) and \(\eta\) as fractions (0.85, 0.50), not percents.

Step-by-step method

  1. Pick \(C_r\) or \(C_a\) based on what your reference provides.
  2. Confirm units (lb vs kg; A·yr vs A·hr).
  3. Use average current (not peak) unless told otherwise.
  4. Solve for the unknown (life \(L\), weight \(W\), current \(I\), or efficiency \(\eta\)).
  5. Sanity-check results (bigger weight → longer life; higher current → shorter life).

Worked examples

Example 1: Magnesium life (consumption rate, \(C_r\))

A 32 lb magnesium anode outputs 0.10 A. Utilization is 0.85 and efficiency is 0.50. Use \(C_r = 8.76\ \text{lb/A·yr}\). Find life \(L\).

\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta}{I \,\times\, C_r} \\[10pt] L=\frac{32 \,\times\, 0.85 \,\times\, 0.50}{0.10 \,\times\, 8.76} \\[10pt] L=\frac{13.6}{0.876} \\[10pt] L=15.5\ \text{years} \end{array} \]
Answer: 15.5 years
Example 2: Zinc life (capacity, \(C_a\))

A 30 lb zinc anode discharges 0.05 A. Utilization is 0.85 and efficiency is 0.90. Use \(C_a = 0.042\ \text{A·yr/lb}\). Find life \(L\).

\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta \,\times\, C_a}{I} \\[10pt] L=\frac{30 \,\times\, 0.85 \,\times\, 0.90 \,\times\, 0.042}{0.05} \\[10pt] L=\frac{0.9639}{0.05} \\[10pt] L=19.28\ \text{years}\approx 19.3\ \text{years} \end{array} \]
Answer: \(\approx 19.3\) years
Example 3: Required weight (metric, \(C_r\))

Design a magnesium anode system for 20 years at 0.50 A. Use \(U=0.85\), \(\eta=0.50\), and \(C_r=3.98\ \text{kg/A·yr}\). Find required weight \(W\) (kg).

\[ \begin{array}{c} W=\frac{L \,\times\, I \,\times\, C_r}{U \,\times\, \eta} \\[10pt] W=\frac{20 \,\times\, 0.50 \,\times\, 3.98}{0.85 \,\times\, 0.50} \\[10pt] W=\frac{39.8}{0.425} \\[10pt] W=93.6\ \text{kg} \end{array} \]
Answer: 93.6 kg

Practice problems

Practice 1: Required weight (capacity, \(C_a\))

You require a magnesium system to last 15 years at 0.20 A. Use \(U=0.85\), \(\eta=0.50\), and \(C_a=0.114\ \text{A·yr/lb}\). Find required total weight \(W\) (lb).

Reveal solution
\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta \,\times\, C_a}{I}\Rightarrow W=\frac{L \,\times\, I}{U \,\times\, \eta \,\times\, C_a} \\[10pt] W=\frac{15 \,\times\, 0.20}{0.85 \,\times\, 0.50 \,\times\, 0.114} \\[10pt] W=\frac{3.0}{0.04845} \\[10pt] W=61.9\ \text{lb} \end{array} \]
Answer: 61.9 lb
Practice 2: Maximum allowable current (magnesium, \(C_r\))

A 17 lb magnesium anode (\(U=0.85\), \(\eta=0.50\)) must last 10 years. Use \(C_r=8.76\ \text{lb/A·yr}\). Find the maximum allowable current \(I\) (mA).

Reveal solution
\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta}{I \,\times\, C_r}\Rightarrow I=\frac{W \,\times\, U \,\times\, \eta}{L \,\times\, C_r} \\[10pt] I=\frac{17 \,\times\, 0.85 \,\times\, 0.50}{10 \,\times\, 8.76} \\[10pt] I=\frac{7.225}{87.6} \\[10pt] I=0.0824\ \text{A}=82.4\ \text{mA} \end{array} \]
Answer: 82.4 mA
Practice 3: Efficiency (magnesium, \(C_a\))

A 20 lb magnesium anode is consumed to 85% utilization in exactly 5 years at 0.20 A. Use \(C_a=0.114\ \text{A·yr/lb}\). Find efficiency \(\eta\).

Reveal solution
\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta \,\times\, C_a}{I}\Rightarrow \eta=\frac{L \,\times\, I}{W \,\times\, U \,\times\, C_a} \\[10pt] \eta=\frac{5 \,\times\, 0.20}{20 \,\times\, 0.85 \,\times\, 0.114} \\[10pt] \eta=\frac{1.0}{1.938} \\[10pt] \eta=0.516 \end{array} \]
Answer: \(\eta \approx 0.516\)
Practice 4: Zinc bracelet (metric, \(C_a\))

A 50 kg zinc bracelet anode discharges 0.50 A. Use \(U=0.80\), \(\eta=0.90\), and \(C_a=0.093\ \text{A·yr/kg}\). Find service life \(L\) (years).

Reveal solution
\[ \begin{array}{c} L=\frac{W \,\times\, U \,\times\, \eta \,\times\, C_a}{I} \\[10pt] L=\frac{50 \,\times\, 0.80 \,\times\, 0.90 \,\times\, 0.093}{0.50} \\[10pt] L=\frac{3.348}{0.50} \\[10pt] L=6.696\ \text{years}\approx 6.7\ \text{years} \end{array} \]
Answer: \(\approx 6.7\) years
Practice 5: Impressed-current bed (simplified consumption form)

An impressed current bed has 10 anodes at 45 lb each. Total output is 20 A. Utilization is 0.60. Use \(C_r = 1.0\ \text{lb/A·yr}\). (Efficiency is not used unless specified.) Find bed life \(L\).

Reveal solution
\[ \begin{array}{c} W_{total}=10 \,\times\, 45=450\ \text{lb} \\[10pt] L=\frac{W_{total} \,\times\, U}{I \,\times\, C_r} \\[10pt] L=\frac{450 \,\times\, 0.60}{20 \,\times\, 1.0} \\[10pt] L=\frac{270}{20} \\[10pt] L=13.5\ \text{years} \end{array} \]
Answer: 13.5 years