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 decimals (for example, 0.85 and 0.50), not percents.
1) Consumption Rate Method (Cr)
\[
L = \frac{W \times U \times \eta}{I \times C_r}
\]
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)
\(C_r\) = 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)
\(C_a\) = 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.
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
\[
\begin{array}{c}
L=\frac{W \times U \times \eta \times C_a}{I}\Rightarrow
W=\frac{L \times I}{U \times \eta \times C_a} \\[10pt]
W=\frac{15(0.20)}{(0.85)(0.50)(0.114)} \\[10pt]
W=\frac{3.0}{0.04845} \\[10pt]
W=61.9\ \text{lb}
\end{array}
\]
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
\[
\begin{array}{c}
L=\frac{W \times U \times \eta}{I \times C_r}\Rightarrow
I=\frac{W \times U \times \eta}{L \times C_r} \\[10pt]
I=\frac{17(0.85)(0.50)}{10(8.76)} \\[10pt]
I=\frac{7.225}{87.6} \\[10pt]
I=0.0824\ \text{A}=82.4\ \text{mA}
\end{array}
\]
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
\[
\begin{array}{c}
L=\frac{W \times U \times \eta \times C_a}{I}\Rightarrow
\eta=\frac{L \times I}{W \times U \times C_a} \\[10pt]
\eta=\frac{5(0.20)}{20(0.85)(0.114)} \\[10pt]
\eta=\frac{1.0}{1.938} \\[10pt]
\eta=0.516
\end{array}
\]
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
\[
\begin{array}{c}
L=\frac{W \times U \times \eta \times C_a}{I} \\[10pt]
L=\frac{50(0.80)(0.90)(0.093)}{0.50} \\[10pt]
L=\frac{3.348}{0.50} \\[10pt]
L=6.696\ \text{years}\approx 6.7\ \text{years}
\end{array}
\]
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
\[
\begin{array}{c}
W_{total}=10\times 45=450\ \text{lb} \\[10pt]
L=\frac{W_{total} \times U}{I \times C_r} \\[10pt]
L=\frac{450(0.60)}{20(1.0)} \\[10pt]
L=\frac{270}{20} \\[10pt]
L=13.5\ \text{years}
\end{array}
\]
Answer: 13.5 years
