Learn A Real World Cathodic Protection Calculation

Cathodic protection is an effective method to mitigate corrosion. To illustrate, I’ll go over a solar farm cathodic protection calculation.

We’ll calculate the number of required anodes per stanchion, on which the solar panels will mount.

Before we begin though, we’ll discuss the details of cathodic protection. This will give you a better understanding of the calculation.

What is cathodic protection?

corrosion cell schematic

It’s a technique used to control the corrosion of a metal surface. The goal is to turn the metal you want to protect into a cathode.

When the metal becomes a cathode, you prevent metal oxidization, as oxidation only happens to an anode.

It’s also important to know, rust is the result of corroding metal. More specifically, when you expose the metal to any one of the following:

  • Water
  • Moist air
  • Moist soil and even moist concrete

Important Note: The deterioration of metals is caused by electron transfer, known as corrosion. There are two types of corrosion processes: oxidation and reduction.

Oxidation occurs when a metal atom loses electrons in a chemical reaction with oxygen, resulting in the formation of a metal oxide. On the other hand, reduction is the transfer of electrons from a metal to another material.

We define a cathode and anode as follows:

  • Anode (active site): the metal, which loses electrons during a chemical reaction
  • Cathode (less active site): the metal, liquid, or gas, which gains electrons during a chemical reaction

Next, for corrosion to occur, the following three things must be present:

  • Two different types of metal, such as steel and aluminum
  • A medium (electrolyte) such as seawater and the Earth, which allows ions to flow and transport electric charges
  • An electrical connection between the cathode and anode, enabling current flow between the metals

Important Note: the two different metals can be separate metals. But also, a single piece of metal with metallurgical differences on the surface will work. These differences can be macroscopic, resulting in non-uniformity on the metal surface. For example, a single pipe can have an anode part and a cathode part.

Imagine a pipe half dipped in water with low oxygen concentration. The dipped half becomes the anode, while the other half in high oxygen concentration becomes the cathode. 

single pipe corrosion schematic

How does cathodic protection prevent corrosion?

Cathodic protection prevents corrosion by converting all anodic (active) sites on metal to cathodic (passive) sites. This is done by pumping electrons onto the metal, which needs protection. It can be a large steel pipe or the hull of a ship, it doesn’t matter.

Additionally, there are two different cathodic protection methods. Each has its own advantages and disadvantages in real-world applications.

Method #1: sacrificial anode

sacrificial anode cathodic protection method

The anode is a metal, which is more reactive than the protected metal. When a metal is more reactive, it loses electrons more easily and forms ions.

For example, when protecting iron, you’ll use a more reactive metal than iron such as zinc or magnesium.

Important Note: the additional metal source is also called a sacrificial anode. Because the galvanic anode sacrifices itself to protect another metal from corrosion. 

What happens next is the sacrificial anode oxidizes in the electrolyte, which can be soil or the ocean. The oxidation generates electrons.

These electrons then flow to the metal you want to protect, forcing it to become a cathode. Additionally, the incoming electrons heal the protected metal by causing any oxidized parts of the cathodic metal to reduce and return to their original state.

The presence of the electrons will cause the ferrous ions to turn back into iron solid.

Important Note: the sacrificial material needs to oxidize before the protected material. Otherwise, cathodic protection won’t work. 

Method #2: impressed current protection

impressed current cathodic protection method

This method is similar to the sacrificial anode method. Except, we use an external power source to generate the electric current.

We pump electrons onto the protected metal through an alternate power source. For example, using a DC power supply, connect the negative end to the protected metal. This makes the protected metal a cathode, thus protecting it from oxidization.

The cool thing is, the anode material can be anything as long as it’s electrically conductive. For example, iron would work, while plastic wouldn’t.

An important benefit of this approach is you don’t need to replace the anode. Whereas with the sacrificial anode method, you need to replace the anode when it corrodes.

Important Note: a big difference between the two methods is oxidation. In the anode of a sacrificial system, the metal oxidizes. Whereas in an impressed current system, the water oxidizes.

Example of cathodic protection with a ship

Imagine a large ship traveling through the ocean. The ship sits in an electrolyte solution of saltwater, the ocean. The goal is to protect the ship’s steel from corrosion.

To do this, a zinc anode is dropped into the water and connected electrically to the hull of the ship. The reactive metal zinc oxidizes and produces electrons, which are then pumped onto the hull of the ship, making it a cathode.

Now assume the ship’s hull’s material is iron (Fe) and it has already started to oxidize. The anode’s electrons will reduce the Fe ions, by forming Fe atoms again. This prevents corrosion, as only Fe bonds can bond to other atoms and cause corrosion.

Important Note: Zinc anode oxidation: Zn \rightarrow Zn^{2+}(aq) + 2e^{-}

Assume some iron has oxidized on the ship’s hull. The below chemical reaction shows the Fe ions capturing the two electrons from zinc. In return, we again have solid iron, which is what we want!

Reduction: Fe^{2+} + 2e^{-} \rightarrow Fe(s)

Now, assume iron hasn’t oxidized on the ship’s hull.  Instead of reducing Fe ions back to Fe solid, a different reaction occurs. We reduce water and oxygen into OH^{-} with the addition of the electrons. The electrons need to go somewhere. 

Reduction: O_{2}(aq) + 2H_{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq)

Your water heater and submarines

Your water heater works using cathodic protection. If you replace your water heater’s sacrificial anode every couple of years, your unit will last longer.

What’s more, submarines use cathodic protection. It’s how these large engineering marvels can safely stay underwater for so long.

In short, you use cathodic protection when the environment around a metal acts as an electrical conductor.

Cathodic protection calculation for a solar farm

We want to install 378 steel stanchions for our solar farm, and they need corrosion protection.

First, we calculate how much current we need to output from each anode. Or as we learned, the number of electrons, which will flow out from an anode.

The required current output for the protection of embedded steel structures we define as:

i_{m} = \dfrac{(S_{cm})(f)(Y)}{\rho}

Where,
S_{cm}: factor for uncoated steel is 150,000
f: anode factor is 1.90 for a 42lb long shape 3″ x 3″ x 72″ magnesium anode
Y: structure to soil potential correction factor for standard -0.85 volt differential is 1.0
\rho: soil resistivity in ohm-centimeters = 1951 ohm-centimeters for our location per test report
i_{m}: current output in milliamperes

i_{m} = \dfrac{(150,000)(1.0)(1.90)}{1951} = 146.08 milliamps output per anode

Our install location has low-resistivity (high corrosion) soil. And in our application, we’ll require a current of 15 milliamperes/ft^{2} of surface.

The total embedded surface area for 378 stanchions is 1,669,920 in^{2} or 11,597 ft^{2}

So, (11,597 ft^{2}) x (15 x 10^{-3} Amps/ft^{2}) = 173.95 Amps

I = 173.95 Amps (total current requirement) = 173,950 milliamps

Based on a 10-year life, the total weight of anodes is given by the following equation:

W = \dfrac{L_{m} \times I }{42.81}

Where,
W: weight of anodes
L_{m}: projected life in years
I: total current required in milliamps

W = \dfrac{10 \times 173,950 }{42.81} = 40,633 pounds of magnesium anodes

Next, we calculate the number of anodes required:

\dfrac{W}{weight \: per \: anode} = \dfrac{40,633}{42} = 967.5 anodes

The number of anodes per stanchion calculates to:

\dfrac{967.5}{378} = 2.56

With 3 anodes per stanchion = 3 x 378 = 1134 total anodes

L_{m} = \dfrac{42.81(1134 \times 42)}{173,950} = 11.72 years anticipated life

Important Note: anodes are 3-inch x 3-inch x 72-inches, installed in 6-inch diameter holes, which are by 72-inches in length. Additionally, you install the anodes approximately 10-feet from the stanchion they’re protecting. 

General rules of thumb with sacrificial anode system designs

I = \dfrac{V}{R} drives the amount of electric current an anode can discharge. This is Ohms’s Law!

Where,
I: the flow of current in amps
R: circuit resistance
V: voltage difference between the anode and cathode

In application, the current is high at first due to the large voltage difference between the anode and cathode. Over time, however, the potential difference drops as more current travels from the anode to the cathode. Eventually, the current decreases with the polarization of the cathode.

Finally, regarding circuit resistance, we’ll go back to our ship example. The path through the saltwater and metal, and any connected cables, make up the resistance.

Important Note: the following are general rules of thumb with anodes:

  • The length of an anode determines the amount of generated current. This in turn determines the area of metal, which you can protect.
  • The cross-sectional area of an anode, determined by its weight, determines how long you can protect a metal.

Cathodic protection calculation wrap up

Without cathodic protection, the world we live in today would be much different. Because we constantly install metals in corrosive environments, such as ships traveling across the oceans.

It’s always surprising how much goes into every installation, which isn’t visible to the naked eye.

What are your thoughts on the cathodic protection calculation? Which type of cathodic protection do you most commonly see installed? 

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5 thoughts on “Learn A Real World Cathodic Protection Calculation”

  1. The soil resistivity will change with the weather (rainfall). Is the goal to keep a constant current as the resistivity changes, or a constant voltage (increasing the current when the soil gets wet)? How much total power is generally needed per square foot of buried metal?

    Reply
    • Great questions!

      Correct; the upper layer resistivity greatly varies with the season. Low temperature and low moisture both increase the soil resistivity.

      There are many applications to consider – this isn’t a black and white question; generally speaking though, say the cathodic protection circuit resistance changes greatly at a field location due to a seasonal change (dry to wet) – the anode to earth resistance alters – an automatic rectifier, which is current controlled, would be used to maintain the correct applied current to the protected structure.

      Now, if you have an above grade water tank, with varying water elevations inside, you’d typically find a potential controlled unit installed. This will allow you to maintain a fixed potential on a set reference cell in said tank, as the surface area requiring protection constantly changes with the fluctuating water level (i.e. the current requirement constantly changes).

      To your last question, this is a highly subjective question. Too many variables at play.

      Reply
  2. Interesting. So the idea is to maintain a constant current per unit surface area. If the surface area is fixed, e.g. for a buried pipe, then the applied voltage needs to change dynamically as the soil resistivity changes to maintain the fixed current. If several variables are changing simultaneously, as in your tank example if the water level, temperature, and perhaps salinity or mineral content are all changing at once, then the system needs to be able to sense the water level to determine the surface area involved, then, like the first example, adjust the voltage to maintain a fixed current appropriate for the surface area involved at that moment.

    In the case of something like a pipe network that is essentially infinite in size, does the pipe need to pass through electrical isolators periodically so the current and voltage can be kept manageable and safe for those living or working around the equipment?

    I am trying to get a sense of the cost tradeoffs of cathodic protection versus material cost, which is why I needed a wattage figure. For your solar farm example, where the needed current is 174A, I would assume that it would take a lot of voltage to drive that much current through the soil so there are substantial ongoing operating costs for cathodic protection, which is hopefully justified by the lower up-front cost of using uncoated steel versus steel that is hot-dip-galvanized (HDG), plastic-coated, or stainless to block the galvanic corrosion process.

    Reply
    • Sorry, missed your response – I never received a notification. 

      Your overview is correct – maintaining the current in the fashion you expounded over would be optimal when possible.

      You can utilize one cathodic protection system for say a single relatively large facility if you properly calculate the current demand, and rectifier and anode sizes. But for a pipeline stretching miles, you’d probably need to utilize separate cathodic protection systems. The drive voltage would become compromised (i.e. voltage drop) as you alluded. Also, impressed current anodes are supplied with recommended current densities – higher current densities would reduce the anode’s life. 

      Overall though, there are many variables you need to consider, which a cathodic protection engineer would hopefully be able to detail/design. Then you need to do a cost-benefit analysis. 

      Reply

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