Consideration for Design of Galvanic Anode Cathodic Protection System

by Ernest Klechka, P.E, NACE International CP Instructor

N
ormally cathodic protection (CP) can be applied by a sacrificial anode or impressed current system. CP can be applied by galvanic or sacrificial system when limited amounts of current are needed, soil resistivity is low (normally less than 5,000 ohm-cm), and electric power is limited or not available. Sacrificial anode system have the added advantage of requiring minimum maintenance.
To design a sacrificial anode system, the following information is needed:
1. Current requirement (I current required)
2. Anode resistance ( R anode), calculated or based on the manufacturer's data
3. Design life

Design Current Requirements
CP current requirements can be determined by calculation based on bare or exposed surface area, assumptions about existing coatings, the coating damage factor, and the current density (CD) required for CO in environment are needed.

To calculate the current required, the total surface area (Atotal) in sq m of the structure is defined. Base on experience, the coating damage factor (fdamage) or percent bare area is determined. The total surface area to be protected by CP (Acp) in sqm is then :
ACP = Atotal fdamage

Based on the environment to which the structure is exposed, the applicable CD for CP (Icd) is estimated in mA/sqm. Using the ACP and Icd, the calculated current required (Icurrent required) is then :
Icurrent required = ACP Icd

Current Requirements Based on Field Tests

Current requirement test are the most reliable way to estimate the current requirements for existing structure and take info account the condition of the coatings and variations in soil resistivity. If the structure is coated and isolated, it is possible to directly determine the current requirements.

A temporary anode (groundbed) is established and a portable power supply (usually a battery, generator, or portable rectifier) is installed. A current is applied (I) and the change in polarized potential ( V) is determined. The ratio of the current applied dicided by the change in polarized potential (I/V) is used to calculated the current requirement.

For example, if 5A are applied to the structure and a polarized potential shift of 50 mA occur, and a 100 mV shift is needed for CP, then :

Icurrent required = shift needed x (I/V)
= 100 mV x (5A / 50mV)
= 10 A

Determining the Output of a Sacrificial Anode

Often the manufacture of an anode will provide the estimated current output in the form of table or graphs based on the shape and size of their anode and soil resistivity. The average soil (electrolyte) resistivity is needed to use these table or graphs or for calculation.

if this data is not available, the output of an anode can be estimated based on Ohm's Law :
I anode = E driving potential / R anode
Anode Resistance-to-Earth

Typically, since sacrificial anodes are buried vertically. For a single certical anode, such as a package magnesium anode, the resistance-to-earth can be calculated using Dwight's equation for a single vertical rod or pipe :
R = (0.00159 / L) [ln (8L/d)-1]
where :
R = groundbed resistance ()
= resistivity ( -m)
d = diameter of anode (m)

The resistivity value used must be representative of the volume resistivity affecting the anode.

Parallel Anode

Sunde's equation can be used to estimate the resistance of distributted parallel anodes:

RN = (0.00159 / L) [ln (8L/d)-1] - 1 + (2L/S) ln(0.656 N)
where :
RN = groundbed resistance ()
= resistivity ( -m)
d = diameter of anode (m)
L = length of anode (m)
S = spacing of anode in the groundbed (m)

If the anode are separate by 6 m or more, the parallel effect of anode is negligible.

Anode Current Output

For a single highh-[ptential (-1.75 Vcse) 7.7 kg (17 lb) magnesium anode, the current output can be calculated or determined based on the manufacturer's data.

A typical 7.7 kg anode is 1,295 mm long by roughly 51 mm square. Because Dwight's equation deals with rods, an equivalent rod with the same circumference as a 51mm square is needed; the rod will be 65 mm in diameter. The equivalent diameter d for square with sides S is given as :

3.14 d = 4 S
d = 4 S /3.14

If the structure is to be polarized to -0.850 Vcse, driving potential is then:

Enet = -1.75 -(-0.85) = -0.90 V

Anode resistance can be calculated using Dwight's equation for 5,000 -cm soil :

Ranode = (0.00159 * 5,000 / 3.14*1,295) [ln (8*1,295/65-1]
= 25

Assuming the structure and cable resistance are negligible, the expected current is then :

Ianode = Enet / Ranode
= 0.90 V / 25
= 0.036 A (36mA)

The anticipated current output is then 36mA from single 7.7 kg anode.
Anode supplier literature indicated that in 5,000 -cm soil, a high-potential magnesium anode will have an anticipated output of 0.040 A (40 mA) to a structure polarized to -0.85Vcse. The data infers that the structure has negligible resistance to earth and, therefore, no IR drop. The resistance to remote earth of single high-potential (-1.75 Vcse) magnesium anode can be calculated:

Ranode = Enet / Ianode
=(1.75 - 0.85) / 0.04
= 22.5

The two value of current output and resistance are very similar.

Number of Anode Needed Bases on Current Requirements

Once the total current required and the current output from a single anode are determined, the number of anode needed to protect the structure can be calculated. Based on current requirement :

Nanode = Icurrent required / Ianode

After the number of anode is calculated, the number of anode must be rounded up ti the next integer (no partial anode allowed).

Number of Anode Needed Based on Design Life

The number of anode needed can also be calculated based on the current requires an the design life. Total weight need for design life can be based on Faraday's Law :

Wtotal =K I T

where:
K = Consumption (kg/A-h x 24 h/day x 365 days) ( see the table I )
I = current in A (Icurrent required)
T = time in years (design life)
Then the number of anode needed is equal to the total weight needed divided by weight of single anode:

Nanode = W total / W anode

Table I


How Many Anode Are Really Needed?

The number of anode needed, therefore, is the larger of the number needed based on current requirements and the number needed based on design life.

Because the current provided can be greater than the current required, control resistor may be needed to reduce the initial output of anodes. After a period of time, the current required may change due to changes in the condition of coating or the environment. Control resistor may needed to be changed with time.

Galvanic Anode Life

Once the number of anode needed is determined, the life expectancy for the system should be checked. The following equations can be used to determine the anticipated life of the sacrificial anode system.

Magnesium years of life = 0.256 x anode weght in kg x efficiency x utilization factor Current in A
Zinc years of life = 0.0935 x anode weght in kg x efficiency x utilization factor Current in A

Post-Installation Measurements

After the systems installed, the structur-to-soil potentials sould be measured as we;; as the current output from the anode system. IF the current output needed adjustment, more anodes can be added to increase current or control resisters can be added to reduce the current output.

Anode life can be calculated based on the actual measured current of the sacrificial anode system.

Conclusions

The design of a sacrificial anode CP system can be accomplished as long as the current demand and the anode current output can be determined. Anode separated by more tham 6 m can be assumed to have minimum parallel interference.

Sacrificial of galvanic anode CP can be very effective as long as current demands are low and soil resistivities are moderate to low. Added benefits of these system are the reduce maintenance cost, no electrical power is required, and ease of installation. Low-power sacrificialsytem also cause a minimum of interference and can be used to discharge CP interference

Basic Calculation For Sacrificial System

A. Current requirement ( I ) = SA x CD

Where :
SA = Surface area (taken from Norton Corrosion calculation)
CD = Current density (taken from NACE standard RP 0176-83)
(Standard Recommended Practice Corrosion Control of Steel, fixed Offshore Platform
Associated with Petroleum Production)
Sea water = 55 mA.sq.m
Seabed = 11 mA/sq.m

B. Anode Weight = N

N = (I x Con x L) : U

Where,
I = Current requirement (Amp)
Con = Consumption rate of aluminium anode = 3.35 Kg/AY
Consumption rate of magnesium anode = 7.7 Kg/AY
L = Life time = 10 years
U = Utilization factor = 0.9

C. Example

Data:
Atmospheric Storage Tank:
- Length (L) : 5.54 m
- Diameter (D) : 9.058 m
- Water level Assume : 1 m
Current Density (CD) : 55 mA/m2
Consumption Rate of Aluminum Anode : 3.5 kg/Amp Year
Years design : 10 tahun
Utility Factor : 0.9

Calculation:

1 Surface Area (SA)

- Surface Area of Ground = pi x D2 = 3.14 x 82.047 = 257.62 m2
- Surface Area of wall = pi x D x L = 3.14 x 9.058 x 5.54 = 157.56 m2
Total Surface Area = 415.18 m2

2 Current Requirement (Ip)

Ip = SA x CD
= 415.18 m2 x 55 mA/m2
= 22,834.9 mA = 22.835 Amp

3 Weight of Anode (W)

W.Al = Ip x Y x 3.5 kg/Amp Year
U = 22.835 Amp x 10 years x 3.5 kg/Amp Year : 0.9
= 799.225 kgs

4 Number of Aluminium Anode (N)

Proposed Aluminium Anode type  500 ASOTB - Net weight = 50 kg/ea
N.Al = W.Al : 50 kg
= 799.225 kgs : 50 kg
= 15.985 pcs = 16 pcs

Total = 16 pcs Aluminum Anode type  500 ASOTB