Electrical TP 16

March 2017

ELECTRICAL TECHNICAL PAPER 16

IMPRESSED CURRENT ANODE MATERIAL SELECTION AND DESIGN
CONSIDERATIONS (NON-MANDATORY)

Information in this technical paper was excerpted from previous versions of DoD cathodic
protection criteria manuals. The information has been updated to reflect current practices
and technology.

1.0 ANODES FOR IMPRESSED CURRENT SYSTEMS.

Although any electrically conductive material can serve as an anode in an impressed

current system, anode materials that have a low rate of deterioration when passing
current to the environment, are mechanically durable, and are available in a form and size
suitable for application in impressed current cathodic protection systems at a low cost are
most economical. While “abandoned in place” steel such as pipelines and rails can, and
are, used as anodes, they are consumed at a rate of approximately 20 pounds per
ampere-year. The most commonly used, “purchased” materials for impressed current
anodes are high silicon chromium bearing cast iron and mixed metal oxide (ceramic).

1.1 Graphite Anodes. Although rarely used anymore on DoD CP system

installations, in the past, graphite anodes were one of the most commonly used materials
for impressed current anodes in underground applications. They are made by fusing
coke or carbon at high temperatures and are impregnated with linseed oil to reduce
porosity and increase oxidation resistance. An insulated copper cable is attached to the
anode internally for electrical connection to the rectifier. This connection must be well
sealed to prevent moisture penetration into the connection and must be strong to
withstand handling. A typical anode end connection and seal is shown in Figure 1.
However, if graphite anodes are desired, specify center connected anodes as shown in
Figure 1a to minimize failure due to “necking” and “end effect”. While more recently
developed anodes such as mixed metal oxide are now more often specified, the
information on graphite anodes is being provided in the event graphite anodes must be
specified.

1.1.1 Specifications. Table 1 lists typical specifications for commercially available

graphite anodes.

1.1.2 Available sizes. Table 1-2 provides examples of the sizes of commercially
available graphite anodes. Manufacturers can provide material catalogs with detailed
anode information and also provide the same information on their company internet
websites. The weights indicate are for the bare graphite only and do not include the
weight of the lead wire or connection, or prepackaged backfill.

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Table 1
Graphite Anode Properties
Composition
Impregnant (Linseed Oil) 6.5 wt % max
Ash 1.5 wt % max
Moisture & Volatile Matter 0.5 wt % max
Water Soluble Matter 1.0 wt % max
Graphite Remainder
Physical Properties
Density 99.84 lb/cu ft max
Resistivity 0.0011 ohm-cm max
Mechanical Requirements
Lead wire connection strength 325 lbs minimum

Figure 1
Typical anode end connection and seal

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Figure 1a
Center Connected Graphite Anode

1.1.3 Characteristics. All products from the operation or deterioration of graphite

anodes are gases. In fresh water or non-saline soil, the principal gases produces are
carbon dioxide and oxygen. In saline soils or in seawater, chlorine is also produced and
is the major gas produced in seawater applications. The gases generated, if allowed to
collect around the anode, can displace moisture around the anode which results in a local
increase is soil resistivity and an increase in circuit resistance.

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Table 1-2. Examples of Commercially Available

Graphite Anodes
Nominal Dimensions*
Nominal Weight*
(LB) (Kg) Diameter Length Surface area
(in.) (mm) (in.) (mm) (ft2) (m2)
13 (5.9) 3 (76.2) 30 (762) 2.0 (0.18)
35 (15.9) 4 (101.6) 40 (1016) 3.5 (0.32)
27 (12.3) 3 (76.2) 60 (1524) 3.9 (0.36)
70 (31.8) 4 (101.6) 80 (2032) 7.0 (0.65)

* Note for Table 1-2: Refer to manufacturers’ brochures for more specific information.
Metric sizes are soft metric conversions from U.S. standard SI units. Actual metric sizes
for materials available in foreign countries may differ.

1.1.4 Operation. Graphite anodes must be installed and operated properly in order
to insure optimum performance and life.

1.1.4.1 Current Densities. Do not exceed the current densities in Table 3 in order to
ensure optimum anode life:

Table 3
Maximum Current Densities for Graphite Anodes
Maximum Current Density (amp/sq ft)
Seawater Fresh Water Soil
3.75 0.25 1

1.1.4.2 Operating Potentials. As the potential difference between steel and graphite is
approximately 1.0 volt with the graphite being the cathode, this potential difference must
be overcome before protective current will begin to flow in the impressed current cathodic
protection system circuit. This 1.0 voltage difference must be added to the other voltage
and IR drop requirements during the selection of proper power supply driving voltage.

1.1.4.3 Consumption Rates. Assuming uniform consumption, the rate of deterioration

of graphite anodes in soil and fresh water at current densities not exceeding the values in
the table above will be approximately 2.5 pounds per ampere-year. The deterioration rate
for graphite anodes in seawater ranges from 1.6 pounds per ampere-year at current
densities below 1 ampere per square foot to 2.5 pounds per ampere-year at current
densities of 3.75 amperes per square foot.

1.1.4.4 Need for Backfill. The deterioration of any point on a graphite anode is
proportional to the current density at that point. If the resistivity of the environment at any
one point is lower than the resistivity at other points, the current density and resulting
deterioration will be higher there. This can result in uneven consumption and premature
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failure of graphite anodes, particularly if the low resistivity area is near the top of the
anode. In this case “necking” of the anode at the top occurs and the connection to the
lower portion of the anode is severed. Use carbon (coke breeze) backfill of uniform
resistivity when graphite anodes are used in soil in order to prevent uneven anode
deterioration.

1.2 High Silicon Cast Iron. Cast iron containing 14 to 15 percent silicon and one
to two percent other alloying elements such as manganese and molybdenum form a
protective film of silicon dioxide when current is passed from their surface into the
environment. This film is stable in many environments except in chloride rich
environments. The formation of this film reduces the deterioration rate of this alloy from
approximately 20 pounds per ampere-year for ordinary steel, to one pound per ampere-
year for the silicon iron. While more recently developed anodes such as mixed metal
oxide are now more often specified, high silicon cast iron anodes are still used.

1.3 High Silicon Chromium Bearing Cast Iron (HSCBCI). When using high
silicon cast iron anodes, specify the chromium-bearing alloy. The chromium-bearing alloy
of similar silicon and other alloy content was developed to help prevent premature
deterioration of high silicon cast iron in environments containing chloride.

1.3.1 Specifications. The nominal composition and typical mechanical and physical
properties of HSCBCI are as shown in Tables 4 and 5:

Table 4
HSCBCI Anode Composition
Element Percent
Silicon 14.35 min
Chromium 4.5
Carbon 0.95
Manganese 0.75
Iron Remainder

Being a metal HSCBCI has much greater mechanical strength than nonmetals such as
graphite magnetite. However, due to its low elongation under load it is brittle and should
be protected from both mechanical and thermal shock.

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Table 5
Physical Properties of HSCBCI Anodes
Tensile Strength 15,000 PSI
Compressive Strength 100,000 PSI
Hardness 520 Brinell
Specific Gravity 7
Melting Point 2,300 Degrees F
Specific Resistance 72 micro ohms/cu cm
Coefficient of Expansion per
degree F 7.33 micro inches per inch

1.3.2 Available sizes. HSCBCI anodes are available in a wide variety of standard
sizes and shapes. Tables 6 and 7 provide examples of sizes of solid rod and tubular
HSCBCI anodes. Manufacturers can provide material catalogs with detailed anode
information and also provide the same information on their company internet websites.
The weights indicated are for the bare anodes only and do not include the weight of the
lead wire or connection, or prepackaged backfill. Special configurations can be produced
at extra cost and are usually practical when standard anodes have been shown to be
unsatisfactory for a particular application and where a large number of special
configuration anodes are required. In most case, however, commercially available sizes
are specified.

The anode sizes and shapes in Tables 6 and 7 are the most commonly used anodes in
DoD applications. In addition to solid rod and tubular anode configurations, HSCBCI
anodes are also available in button, bullet, and other shapes and sizes of anodes, where
needed for particular applications. Refer to manufacturers’ brochures for more specific
information.

1.3.3 Anode Lead Wire Connection. The maximum acceptable resistance

between the cable and the anode should be 0.01 ohms. The cable to anode connection
is critical, as in the case of all impressed current anodes. Two common methods of
achieving the cable to anode connection and seal are shown in Figures 2 through 4.

The solid rod anodes shown in Table 6 had been commonly used in the past. Lead wire
connections at the end of the anode often resulted in a higher rate of consumption, or
“necking”, at that end of the anode that resulted in premature loss of lead wire connection.
To help prevent the necking of the anode at the connection point, select anodes that were
manufactured with an enlarged end if solid rods are to be specified. In addition, specify
external epoxy encapsulation as shown in Figure 3 to help reduce the necking of the
anode. The tubular anodes listed in Table 7 with the lead wire connected in the center of
the anode are now preferred over solid rod anodes as these help reduce the chance of
premature failure due to necking. Figures 4 and 5 are illustrative examples of tubular
anodes.

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Figure 2
Anode to cable connection – Teflon seal.

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Figure 3
Anode to cable connection – Epoxy seal.

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Figure 4
Anode to cable center connection

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1.3.4 Operation. HSCBCI anodes are consumed at a rate of about one (1) pound
per ampere year when used at a current not exceeding their nominal discharge rates.
The potential difference between and steel HSCBCI can be neglected in the selection of
impressed current rectifiers. HSCBCI anodes do not require backfill in most applications,
but backfill can reduce the anode to electrolyte resistance in many cases and its use
should be evaluated based upon a tradeoff between the cost of the backfill and the
savings in power associated with the lower circuit resistances and required driving
potentials.

Table 6. Examples of Commercially Available HSCBCI

Solid Rod Anodes
Nominal Dimensions*
Nominal Weight*
(LB) (Kg) Diameter Length Surface area
(in.) (mm) (in.) (mm) (ft2) (m2)
1 (0.5) 1.13 (29) 9 (229) 0.2 (0.02)
5 (2.3) 2 (51) 9 (229) 0.4 (0.04)
25 (11.4) 1.5 (38) 60 (1524) 2.0 (0.19)
44 (20.0) 2 (51) 60 (1524) 2.6 (0.24)
63 (28.6) 2.25 (57) 60 (1524) 2.8 (0.26)
110 (49.9) 3 (76) 60 (1524) 4.0 (0.37)

* Note for Table 6: Refer to manufacturers’ brochures for more specific information.
Metric sizes are soft metric conversions from U.S. standard SI units. Actual metric sizes
for materials available in foreign countries may differ.

Table 7. Examples of Commercially Available HSCBCI

Tubular Anodes
Nominal Dimensions*
Nominal Weight*
(LB) (Kg) Diameter Length Surface area
(in.) (mm) (in.) (mm) (ft2) (m2)
4.3 (1.8) 2.2 (56) 8 (203) 0.38 (0.04)
6.5 (2.9) 2.2 (56) 12 (304) 0.47 (0.05)
23 (10.4) 2.2 (56) 42 (1067) 2.0 (0.19)
31 (14.1) 2.66 (67) 42 (1067) 2.4 (0.22)
46 (20.9) 2.2 (56) 84 (2133) 4.0 (0.37)
63 (28.6) 2.66 (67) 84 (2133) 4.9 (4.6)
85 (38.6) 3.75 (95) 84 (2133) 6.9 (0.64)
110 (49.9) 4.75 (121) 84 (2133) 8.7 (0.81)
175 (79.4) 4.75 (121) 84 (2133) 8.7 (0.81)

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* Note for Table 7: Refer to manufacturers’ brochures for more specific information.
Metric sizes are soft metric conversions from U.S. standard SI units. Actual metric sizes
for materials available in foreign countries may differ.

Figure 5
Tubular HSCBCI Anodes

1.4 Aluminum. Aluminum anodes were sometimes used as impressed current CP

system anodes for the protection of the interior of water storage tanks in the past because
of their low cost, light weight and the lack of water contamination from the products of
deterioration of the anodes. They were commonly used where seasonal icing of the tank
would damage the anodes. In this case, the aluminum anodes were sized to last one
year and were replaced each spring. HSCBCI and mixed metal oxide anodes are now
more commonly used in water tanks.

1.5 Lead-Silver Alloy. Lead alloyed with silver, antimony or tin has been used in
the past as anodes for impressed current cathodic protection systems in seawater. The
chief advantage of these anodes was their low operating cost. The consumption rate for
silverized lead is 2 to 3 pounds per ampere year initially but drops off to approximately 0.2
pounds per ampere year after 2 years, as the anode surface develops a passivating film.
Alloyed lead anodes have been unreliable in many specific applications either because
they failed to passivate, their consumption rate remained in the 2 to 3 pound per ampere
year range and they were completely consumed, or they became so highly passivated
that the anode to electrolyte resistance increased substantially. When properly operating,
the current density from silverized lead anodes is typically 10 amperes per square foot.
Because of the environmental concerns with lead, these anodes are no longer used.

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1.6 Platinum. Pure platinum wire is sometimes used for impressed current
cathodic protection anodes where space is limited. Platinum is essentially immune to
deterioration in most applications. In seawater its consumption rate at current densities
as high as 500 amperes per square foot is 0.00001 pound per ampere year. Due to the
high cost of platinum, this material is more commonly used as a thin coating on other
metals.

1.7 Platinized anodes. Platinum can be bonded or deposited on other materials

for use as an impressed current cathodic protection anode. The substrate materials,
namely titanium, tantalum and niobium have the special characteristic of being covered
with a naturally formed stable oxide film that prevents current flow from their surfaces,
even when exposed to high anodic potentials. These “platinized” anodes, although high
in initial unit cost, can be used at very high current densities and have had wide
application to service in tanks and other liquid handling systems as well as in seawater.
Their use in soils has been primarily for deep well applications. Since the thin film of
platinum material is what corrodes as current flows, the platinized anodes are known as
dimensionally stable anodes since no noticeable change in the anode dimension will
occur as the platinum coating corrodes. This is contrast with other bulk anodes such as
graphite and HSCBCI where significant changes in anode dimensions occur as the
anodes corrode during operation.

1.7.1 Types. Platinized anodes are available in a wide variety of sizes and shapes
as shown below. Table 8 provides examples of platinized niobium wire anode
configurations used for impressed current anodes in water storage and processing
vessels. Manufacturers can provide material catalogs with detailed anode information
and also provide the same information on their company Internet websites. An example
of a platinized anode configuration is shown in Figure 6.

Table 8. Examples of Commercially Available Platinized Niobium

Wire Anodes
Wire Diameter Niobium Thickness Platinum Thickness
Length
(in.) (mm) (in.) (mm) (µ-in.) (µ -M)
(0.063) (1.6) As required (0.007) (0.18) (25) (0.635)
(0.063) (1.6) As required (0.007) (0.18) (50) (1.275)

* Note for Table 8: Refer to manufacturers’ brochures for more specific information.
Metric sizes are soft metric conversions from U.S. standard SI units. Actual metric sizes
for materials available in foreign countries may differ.

Platinized niobium/titanium probe type anodes are also available for use in steel vessels
such as condenser water boxes and heat exchangers where the interior is difficult to
access and its space limited. Such anodes are often mounted through the vessel shell,
with the anode part of the probe extending into the vessel, and wire connections made to
the probe on the exterior of the vessel. Table 9 provides examples of platinized niobium
wire anode configurations used for impressed current anodes in water storage and
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processing vessels. Manufacturers can provide material catalogs with detailed anode
information and also provide the same information on their company internet websites.

Figure 6
Example platinized anode configuration

Table 9. Examples of Commercially Available Platinized

Niobium/Titanium Probe Anodes
Anode Diameter Pipe Nipple NPT Platinum Thickness
(in.) (mm) (in.) (mm) (µ-in.) (µ -M)
(0.25) (6.4) (1.0) (25.4)
(0.50) (12.7) (1.0) (25.4) (50 - 300) (1.27 – 7.62)
(0.75) (19.1) (1.0) (25.4)

* Note for Table 9: Refer to manufacturers’ brochures for more specific information.
Metric sizes are soft metric conversions from U.S. standard SI units. Actual metric sizes
for materials available in foreign countries may differ.

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1.7.2 Operation. Platinized anodes can be operated at very high current densities.
100 Amperes per square foot are typical. All of the current flows from the platinum coated
portion of the anode surface. However, if this coating is damaged, current may also flow
from the substrate material resulting in its premature failure. The primary limitation of
platinized anodes is that the oxide film on the substrate can break down if excessive
anode-to-electrolyte voltages are encountered. The practical limit for platinized titanium is
12 volts. Platinized niobium can be used at potentials as high as 100 volts.

1.8 Mixed Metal Oxide and Ceramic Anodes.

1.8.1 Types. Mixed metal oxide (MMO) and ceramic anodes are available in a
variety of substrate configurations:

• Solid rod
• Tubular substrates
• Wire
• Ribbon
• Expanded mesh
• Probe

For DOD applications, Solid rod, tubular, and wire type anodes are commonly used
where bulk anodes such as HSCBCI and graphite are used. As the cost of these anodes
have decreased, these anodes are now preferred in many applications over their bulk
anode predecessors. Their smaller dimensions and lower weights can reduce and
simplify CP system installation. There generally higher current density capacities can
often result in installation of less anodes that their bulk anode counterparts. Probe
anodes are used in water storage and processing vessels. Wire, ribbon, and mesh
anodes are used in CP systems installed to protect the exterior bottoms of on-grade
storage tanks with a secondary containment liner because of limited space between the
tank bottom and the containment liner. Ribbon and mesh anodes are most often the
anode configurations of choice for CP systems installed to protect the reinforcing steel in
concrete.

Tables 10 through 13 provide examples of commercially available MMO/ceramic anode

configurations. Manufacturers can provide material catalogs with detailed anode
information and also provide the same information on their company Internet websites.
Figures 7 through 9 illustrate some examples of MMO anodes.

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Table 10. Examples of Commercially Available MMO Solid Rod

Anodes1
Nominal Dimensions1,2 Nominal Weight1,2,3
Current Rating
Diameter Length Bare Weight Packaged Wt (Amperes)4
(in.) (mm) (ft.) (mm) (oz/ft) (g/M) (LB) (Kg)
0.125 (3.18) 4 (101.6) 0.38 (35.6) 22 (10) 7.0
0.125 (3.18) 8 (203.2) 0.38 (35.6) 44 (20) 14.0
0.25 (6.35) 4 (101.6) 1.5 (43.4) 22 (10) 15.0
0.25 (6.35) 8 (203.2) 1.5 (43.4) 44 (20) 30.0
0.50 (12.7) 4 (101.6) 6.1 (173.8) 23 (10.5) 29.0
0.50 (12.7) 8 (203.2) 6.1 (173.8) 46 (21) 58.0

Notes for Table 10.

1. Refer to manufacturers’ brochures for more specific information.
2. Metric sizes are soft metric conversions from U.S. standard SI units.
Actual metric sizes for materials available in foreign countries may differ.
3. Mixed metal oxide rod anodes can be ordered either bare or pre-
packaged in coke backfill.
4. Current rating based on a 15 year design life in seawater.

Table 11. Examples of Commercially Available MMO Tubular Anodes1

Nominal Dimensions1,2 Nominal Weight1,2,3
Current Rating
Diameter Length Bare Weight Packaged Wt (Amperes)4
(in.) (mm) (ft.) (mm) (oz/ft) (g/M) (LB) (Kg)
0.75 (19.1) 2.0 (610) 3.4 (314) 23 (10.5) 23
0.75 (19.1) 4.0 (1219) 3.4 (314) 25 (11.4) 45
1.0 (25.4) 3.3 (1006) 3.8 (351) 25 (11.4) 50
1.25 (31.8) 4.0 (1219) 5.8 (538) 27 (12.3) 73

Notes for Table 12.

1. Refer to manufacturers’ brochures for more specific information.
2. Metric sizes are soft metric conversions from U.S. standard SI units.
Actual metric sizes for materials available in foreign countries may differ.
3. Mixed metal oxide tubular anodes can be ordered either bare or pre-
packaged in coke backfill.
4. Current rating based on a 15 year design life in seawater.

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Table 12. Examples of Commercially Available MMO Ribbon Anode

Components1
Nominal Dimensions1,2
Nominal Bare Weight
Component Width Thickness (LB/ft) (Kg/M)
(in.) (mm) (in.) (mm)
Ribbon Anode 0.25 (6.4) 0.025 (0.6) 6 LB/500 ft (2.7 Kg/152 M)
Ribbon Anode 0.5 (12.7) 0.025 (0.6) 10 LB/500 ft (3.0 Kg/152 M)
Conductor Bar 0.5 (12.7) 0.040 (1.0) 4 LB/100 ft (0.07 Kg/M)

Notes for Table 12.

1. Refer to manufacturers’ brochures for more specific information.
2. Metric sizes are soft metric conversions from U.S. standard SI units.
Actual metric sizes for materials available in foreign countries may differ.

Table 13. Examples of Commercially Available MMO Ribbon Anode

Components1
Nominal Dimensions1,2
Nominal Bare Weight
Component Width Profile Length (LB/ft) (Kg/M)
(in.) (mm) (in.) (mm) (in.) (mm)
Mesh Strip 2.5 (63.5) 0.125 (3.2) 24 (7.3) 6 LB/500 ft (2.7 Kg/152 M)
Conductor Bar 0.5 (12.7) 0.040 (1.0) 250 (76.2) 4 LB/100 ft (0.07 Kg/M)

Notes for Table 13.

1. Refer to manufacturers’ brochures for more specific information.
2. Metric sizes are soft metric conversions from U.S. standard SI units.
Actual metric sizes for materials available in foreign countries may differ.

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Figure 7
Roll of MMO Ribbon Anode (left). MMO Ribbon installed (right). Thinner metal
strip is the MMO anode.

Figure 8
Mixed Metal Oxide Mesh Anode installed on a concrete beam.

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Figure 9
Mixed Metal Oxide tubular anode.

1.9 Anode Backfill. For nearly all underground installations, a special carbon
backfill, referred to in the industry as coke breeze, is provided in the anode hole or well.
Some of the reasons for using this special backfill include

• Lower anode ground bed resistance, hence, resulting in lower rectifier

design voltages and reduced probability of stray current interference.
• Improved current distribution along the anode since the backfill will provide
a generally uniform environment. Helps avoid premature anode
failure.Prolonged anode life since the carbon backfill is also consumed
instead of anode.Backfill will help maintain the stability of the anode hole
or well.Backfill will provide a permeable medium for migration of gases,
thereby avoiding premature increase in anode bed resistance.

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1.9.1 Specifying the Type of Backfill. Ensure that the appropriate type of backfill
is specified and provided. Generally, calcined petroleum coke, or coke that has been
heated to remove high resistivity petroleum by-products, should be specified. The
calcined petroleum coke has a lower total bulk resistivity, and its more spherical
particles aids in the compaction of the backfill (Figure 10). Also, specify backfill that has
a carbon content greater than 92%, preferably greater than 99%, for greater anode
system life.

Figure 10
Petroleum coke backfill

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2.0 POWER SUPPLY SELECTION

2.1 Determination of Power Supply Requirements. The power supply

requirements, namely current and voltage are determined by ohms law from the required
current for protection of the structure and the calculated or measured total circuit
resistance.

2.2 Selection of Power Supply Type. Any source of direct current of appropriate
voltage and current can be used as a source of power for impressed current cathodic
protection systems. The selection of the power supply depends upon local conditions at
the site and should be evaluated based upon economics, availability of AC power or fuel,
and the availability of maintenance.

2.2.1 Transformer-Rectifiers. Transformer-rectifiers, or more simply, rectifiers, are

by far the most commonly used power supply type for impressed current cathodic
protection systems. They are available in a wide variety of types and capacities
specifically designed and constructed for use in impressed current cathodic protection
systems. The most commonly used type of rectifier has an adjustable step down
transformer, rectifying units (stacks), meters, circuit breakers, lightning arresters, current
measuring shunts, and transformer adjusting points (taps), all in one case.

2.2.2 Thermoelectric generators. These power supplies convert heat directly into
direct current electricity. This is accomplished through a series of thermocouples that is
heated at one end by burning a fuel and cooled at the other, usually by cooling fins.
Thermoelectric generators are highly reliable as they have few, if any moving parts. They
are available in sizes from 5 to 500 watts. They are very expensive and should only be
considered for remote locations where electrical power is not available and fuel is
available. They are used as a power supply for impressed current cathodic protection on
remote pipelines where the product in the pipeline can be used as a fuel.

2.2.3 Solar Power. For solar powered systems, solar cells convert sunlight directly
into direct current electricity. Their cost per watt is high but is decreasing as solar cell
technology is improved. Solar panels are used for cathodic protection power supplies at
remote sites where neither electrical power nor fuel is available. In order to supply
current continuously, solar cells are used in a system that both supplies power to the CP
system and also recharges a set of batteries when sunlight is received (Figure 11). When
sunlight is not being received, the batteries supply the required current. Security must be
considered with solar power systems as many of the solar powered systems have been
damaged by vandalism and theft. Environmental and safety considerations must include
procurement, maintenance and disposal of battery electrolytes, usually acids.

2.2.4 Batteries. When current requirements are low, storage batteries can be used
to supply power for impressed current cathodic protection systems at remote sites. They
must be periodically recharged and maintained. Again, environmental and safety
considerations must include procurement, maintenance and disposal of battery
electrolytes, usually acids.

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Figure 11
Solar powered CP system. Solar panels shown in picture on the left. Battery bank
and control center shown in picture on the right.

2.2.5 Generators. Engine or wind driven generators can also be used to supply
direct current power for impressed current cathodic protection systems at sites where AC
power is not available.

3.0 RECTIFIER SELECTION.

Since most CP systems are powered by rectifiers, the rest of this section will
concentrate on the procedures for selecting rectifiers. The rectifier selected for a specific
impressed current cathodic protection application must be matched to both the electrical
requirements and the environmental conditions at the site. Rectifiers are available in
many electrical types and specifically designed for use in impressed current cathodic
protection systems in many environments.

3.1 Rectifier Components. Figure 1-12 is a circuit diagram for a typical single
phase full wave bridge type rectifier having the components found in most standard
rectifiers of this type. The diagram also shows an external switch and circuit protection
device that is recommended for all rectifier installations to conform with the requirements
of the National Electrical Code.

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Figure 12
Single Phase, Full-Wave Bridge Rectifier

3.1.1 Transformer Component. The transformer reduces the incoming alternating

current (AC) voltage to the alternating current voltage required for the operation of the
rectifier component. The incoming alternating current is applied to one coil (primary
winding) wound on an iron core. The magnetic field produced by the flow of current
through the primary winding induces an alternating current voltage in a second coil
(secondary winding) that is also wound on the same iron core. The ratio of voltage
between the primary and secondary winding equals the ratio of turns in each coil. In most
impressed current cathodic protection rectifiers, the voltage output from the secondary
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windings can be varied by changing the effective number of secondary windings through
a system of connecting bars or “taps”. Two sets of taps are normally present, one for
coarse adjustments and one for fine adjustments. By manipulation of these taps, the
rectifier direct current (DC) output voltage can be adjusted from zero to its maximum
capacity in even steps of not more than 5 percent of the maximum output voltage.

3.1.2 Rectifying Elements. The alternating current (AC) from the secondary
windings of the transformer element is converted to direct current (DC) by the rectifying
elements or “stacks”. The stack is an assembly of plates or diodes and may be in several
configurations. The most common rectifying elements are selenium plate stacks and
silicon diodes. Each has advantages and disadvantages as discussed below. The most
common configurations of rectifying elements are the single-phase bridge, single-phase
center tap, three-phase bridge and three phase wye. These arrangements are described
in detail in section 3.2. The rectifying elements allow current to flow in one direction only
and produce a pulsating direct current. The rectifying elements do allow a small amount
of alternating current to pass. This “ripple”" is undesirable and should be held to low
levels. Rectifiers are not 100% efficient in converting alternating current to direct current.
This is due to the presence of alternating current and to inherent losses in the rectifying
elements that result in heating of the stacks. Silicon elements are more efficient than
selenium elements at high output voltages but are more susceptible to failure due to
voltage overloads or surges. The efficiency of a rectifying element is calculated by the
following equation:

Efficiency DC output power X 100

=
(%) AC input power

Typical efficiencies of single phase rectifying elements are in the order of 60 to 75 % but
can be increased by filtering the output or by using a three-phase circuit. Selection of
appropriate circuit type is discussed in section 3.3. Selection of silicon versus selenium
rectifying elements is discussed in detail in section 3.3.8.

3.1.3 Overload Protection. Overload protection in the form of circuit breakers,

fuses, or both should be used on all impressed current rectifiers. In addition to protecting
the circuits from overloads, circuit breakers provide a convenient power switch for the
unit. Circuit breakers are most commonly used on the alternating current input to the
rectifiers and fuses are most commonly used on the direct current outputs. In addition to
circuit breakers and fuses, the rectifier should be furnished with lightning arresters on
both the input and output in order to prevent damage from lighting strikes or other short
duration power surges. Due to their susceptibility to damage from voltage surges,
rectifiers using silicon elements should always be furnished with lightning arresters.

3.1.4 Meters. In order to conveniently measure the output current and potential, the
rectifier should be furnished with meters for reading these values. The meter should not
be continuously operating but should be switched into the circuit as required. This not
only protects the meter from electrical damage from surges but, when the meter is read,
its movements from zero to a value on the meter can be used to help detect defective
meters. Often, one meter and a two-position switch are used to measure both potential
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and current. Current is usually measured using a millivolt meter connected to an external
current shunt. Output voltage and current can also be conveniently measured by the use
of portable meters used across the rectifier output and the current shunt.

3.2 Standard Rectifier Types.

3.2.1 Single Phase - Bridge. The circuit for this type of rectifier is shown in Figure
12. This type of rectifier is the most commonly used type of rectifier up to an output
power of about 1,000 watts. Above 1,000 watts, the extra cost of three phase types is
often justified by the increased electrical efficiency of the three phase units. The rectifying
unit consists of four elements. If any one of the rectifying elements fails or changes
resistance, the other elements usually also fail. Current passes through pairs of the
rectifying elements through the external load (structure and anode circuit). The active
pair of elements alternates as the polarity of the alternating current reverses while the
other pair blocks the flow of current. The result is full wave rectified current as shown in
Figure 13.

3.2.2 Single Phase - Center Tap. The circuit of a single-phase center tap rectifier is
shown in Figure 14. This type of rectifier has only two rectifying elements but produces
full wave rectified output. However, as only one-half of the transformer output is applied
to the load, the transformer required is considerably heavier and more costly than in
single-phase bridge type units. This type of unit is also less sensitive to adjustment than
the single-phase bridge type, however it is electrically more efficient.

3.2.3 Three Phase - Bridge. The circuit for a three-phase bridge rectifier is shown
in Figure 15. The circuit operates like three combined single-phase bridge circuits that
share a pair of diodes with one of the other three bridges. There are three secondary
windings in the transformer that produce out of phase alternating current that is supplied
to each pair of rectifying elements. This out of phase relationship produces a direct
current output with less alternating current “ripple” than the single-phase type, typically,
only 4.5%. Due to the reduction in alternating current ripple, three-phase bridge rectifiers
are more electrically efficient than the single-phase types, and the extra initial cost of the
unit is often justified by savings in supplied power, particularly for units of over 1000 watts
capacity.

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Figure 13
Full-Wave Rectified Current

Figure 14
Single-Phase Center-Tap Circuit

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Figure 15
Three-Phase Bridge Circuit

3.2.4 Three Phase - Wye. The circuit for a three-phase wye rectifier is shown in
Figure 16. This type of rectifier supplies half-wave rectified current as shown in
Figure 17. The power to the rectifier unit is supplied by three separate windings on a
transformer but only three rectifying elements, each in series with the output, are
provided. This type of rectifier unit is practical only for systems requiring low output
voltages.

3.2.5 Special Rectifier Types. Several special types of rectifiers, specifically

designed for use in cathodic protection systems have been developed for special
applications. Some special rectifiers provide automatic control of current to maintain a
constant structure to electrolyte potential, others provide a constant current over varying
external circuit resistances, or other features desirable in specific circumstances.

3.2.5.1 Constant Current Type. A block diagram of one type of constant current
rectifier is shown in Figure 18. A DC input signal to the power amplifier is supplied from
an adjustable resistor in the output circuit. The power amplifier uses this “feedback”
signal to adjust the voltage supplied to the stack so that a constant input signal and
therefore a constant output current are supplied. The power amplifier may either be of an
electronic (silicon controlled rectifier) or saturable reactor type.

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Figure 16
Three-Phase Wye Circuit Schematic

Figure 17
Half-wave rectifier output

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Figure 18
Constant Current Rectifier Block Diagram

3.2.5.2 Automatic Potential Control Type. A block diagram for an automatic

potential control rectifier is shown in 19. This type of unit uses the potential between the
structure and a reference electrode to control the output current of the unit. As in the
constant current type of rectifier, the power amplifier can be of the electronic or saturable
reactor type. This type of rectifiers are commonly used where the current requirement or
circuit resistance varies greatly with time such as in the case of structure in an area with
high periodic tidal currents or a water storage tank where the water level changes
considerably.

Figure 19
Constant Potential Rectifier Block Diagram

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3.2.5.3 Other Special Types. Several standardized rectifiers have been developed
for commercial applications such as natural gas and electrical distribution system
protection. The use of a standardized unit allows for economy of production and
reduction in overall cost of the unit as well as the installation and maintenance of the unit.
Where a large number of similar capacity units are to be used, consider the selection of a
standardized type of rectifier.

3.3 Rectifier Selection And Specifications. Rectifiers can either be selected

from “stock” units or can be custom manufactured to meet specific electrical and site
related requirements. Many features are available either as optional “add ons” to stock
units or in custom units.

3.3.1 Available Features. Features now available on most units include:

• Constant Potential, Voltage or Current output (automatic)

• Remote Control (adjustment and/or interruption)
• Remote Monitoring
• GPS Syncronizeable interrupter
• Pulse Generator (for wave form analyzer potential surveys)
• Multiple circuits in the same enclosure
• Air cooled or oil immersed
• Any commercial input voltage
• Three Phase or Single Phase
• Center tap or bridge
• Wide range of output currents and voltages
• Efficiency filters to reduce AC ripple
• Interference noise filters
• Explosion proof enclosures
• Small arms proof enclosures
• Lightning protection on both AC input and DC output
• Surge protection on both AC input and DC output
• Silicon diodes or Selenium stacks
• Stainless Steel, Painted or galvanized cases
• Various mounting legs or brackets for floor, wall or pole mounting
• Units designed for direct burial
• External "on - off" indicators
• Variety of Price, Quality and Warranty

Factors that should be considered in selecting appropriate features for a specific

application are given below.

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3.3.2 Air Cooled vs. Oil Immersed. Rectifiers can be supplied as either entirely air-
cooled, entirely oil-immersed or with the stacks only oil-immersed (Figures 20 through
22). Air-cooled units are lowest in cost and easiest to install and repair. However, specify
oil-cooled units in marine environments, where corrosive or dirty atmospheric conditions
are encountered, or where explosive gasses may be present. Air cooled units require
more frequent maintenance to clean the air screens and other components and are also
susceptible to damage by insects and other pests. Older oil-cooled units were supplied
with oils containing Poly-Chlorinated Bisphenyls (PCBs) which have been determined to
be carcinogenic and are no longer supplied with new units. Treat units containing PCB's
according to current policy regarding PCB's.

Figure 20
Typical Air-cooled Rectifier.

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Figure 21
Typical entirely immersed oil-cooled rectifier.

Figure 22
Typical oil-cooled rectifier with only the stacks and transformer immersed.

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3.3.3 Selecting A. C. Voltage. Alternating current voltages of almost any

commercial power supply voltage. Units with 115 volts, 230 volt or 440 volt single phase
or 208, 230 or 440 volt three phase inputs are the most common. Some units are
supplied with dual input voltage selected by wiring arrangements during installation.
Choice between single phase and three phase units should be based upon a balance
between first cost and efficiency. The following Table 14 can be used to select the
combinations of rectifier capacity and input voltages that are commonly most economical
if a selection of supply voltages is available:

Table 14. Rectifier Capacity and Input Voltages

Rectifier DC Rating Single Phase Three Phase
(Watts) Voltage Voltage
Up to 2,700 115 208
2,700 to 5,400 230 230
5,400 to 7,500 440 230
Over 7,500 440 440

3.3.4 DC Voltage And Current Output. Direct current voltage outputs from 8 to 120
volts and current outputs from 4 amperes to 200 amperes are common in any
combination. Almost any current can be provided but it is generally best to select a
standard size of rectifier unit and use multiple units if very large amounts of current are
required.

3.3.5 Filters. Electrical filters are used to both increase the efficiency of the rectifier
by reducing alternating current ripple and to reduce interference with communications
equipment. Efficiency filters can increase the efficiency of single-phase bridge type
rectifiers by from 10 to 14 percent and their use should be based upon a first cost versus
operating (power) cost basis. Efficiency filters are not commonly used with three-phase
rectifiers, as the alternating current ripple in these types of units is inherently low. Use
noise interference filters when a large unit is to be installed in the vicinity of
communications lines or can be retrofitted when noise problems are encountered and are
significantly affected by turning the unit on and off.

3.3.6 Explosion Proof Rectifiers. Rectifiers and other system components such as
switches and circuit breakers are available in explosion proof enclosures conforming to
National Electrical Code (NEC) Safety Standards for Class I, Group D hazardous
conditions such as may be encountered in fuel or natural gas storage or distribution
systems. Specify such enclosures where required by the NEC or whenever explosive
hazards may exist.

3.3.7 Lightning Arresters. Lightning arresters should almost always be used on

both the AC input and DC output sides of rectifiers using silicon rectifying elements. Their
use on units using selenium elements is recommended in areas where lightning strikes
are frequent.
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3.3.8 Selenium vs. Silicon Stacks. While some old installations used copper oxide
rectifying elements, modern units use only either silicon or selenium rectifying elements.
In general, silicon units are used for larger units where their higher efficiency is more
important than their lower reliability.

3.3.8.1 Selenium Stacks. Ordinary selenium stacks deteriorate with time. This
“aging” can be reduced by variations in plate composition, and “non-aging” stacks are
available. Aging rates are determined by operating temperatures that are a function of
current flow; thus the selection of a unit using selenium rectifying elements which has a
somewhat greater capacity than required will increase stack life. The efficiency of
selenium rectifying elements is a function of operating voltage versus rated voltage as
shown in Figure 23.

3.3.8.2 Silicon Diodes. Silicon diodes are mounted in metal cases that are mounted
on either aluminum or copper plates to dissipate the heat generated during operation.
Silicon diodes do not age as do selenium stacks and, as shown in Figure 24, is more
efficient than selenium elements, particularly at higher voltage ratings. Silicon rectifying
elements are more subject to failure from voltage surges that would only cause increased
aging of selenium stacks. Surge protection should always be used on both the AC input
and DC output of rectifiers using silicon diode rectifying elements.

3.3.9 Other Options. Select other available features such as listed in Section 3.3.1
as appropriate. In remote off-base areas, small arms proof enclosures may be required
based upon local experience.

3.3.10 Rectifier Alternating Current Rating. Determine the AC current requirement

for a rectifier based upon rectifier output and efficiency using the following formulae:

3.3.10.1 Single Phase Rectifiers.

Edc x Idc
Iac =
F x Eac
where:
Iac = Alternating Current Requirement (Amps)
Edc = Direct Current Output Voltage
Idc = Direct Current Output Amperage
F = Rectifier Efficiency %
Eac = Alternating Current Voltage (per phase to ground)

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Figure 23
Efficiency versus voltage for selenium stacks

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Figure 24
Efficiency versus voltage for silicon stacks

3.3.10.2 Three Phase Rectifiers.

Edc x Idc
Iac =
3 x F x Eac
where:
Iac = Alternating Current Requirement (Amps)
Edc = Direct Current Output Voltage
Idc = Direct Current Output Amperage
F = Rectifier Efficiency %
Eac = Alternating Current Voltage (per phase to ground)

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4.0 EXAMPLES OF IMPRESSED CURRENT CP INSTALLATIONS.

4.1 Illustrative Examples of Impressed Current CP Installations. Figures 25

through 39, illustrate typical impressed current cathodic protection system installations.
Please note that these are NOT standard designs. Rather, features of these designs
may be applicable to the design of similar systems for similar applications but the design
for each specific application must be made based upon actual specific conditions and
requirements. Section 5.0 gives examples of the design of sample cathodic protection
systems.

Figure 25
CP for Building Underground Heat and Water Lines

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Figure 26
Impressed Current Type CP System for Aircraft Hydrant Refueling System

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Figure 27
Cathodic Protection of Foundation Piles

Figure 28
Impressed Current CP for Existing On-Grade Storage Tank

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Figure 29
Impressed Current CP with Horizontal Anodes for On-Grade Storage Tank

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Figure 30
Impressed Current CP for Small Water Tank

Figure 31
Deep Well Anode Impressed Current CP for Steel Sheet Pile Bulkhead

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Figure 32
Impressed Current CP with Distributed Anodes for Water Side
of a Steel Sheet Pile Bulkhead Wall

Figure 33
Suspended Anode Impressed Current CP for H-Piling in Seawater

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Figure 34
Impressed Current CP Sled Anode for H-Piling in Seawater

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Figure 35
Impressed Current CP for Cellular Earth Fill Pier Supports

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Figure 36
Impressed Current CP for Elevated Water Storage Tank Interior

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Figure 37
Typical Distributed Anode Impressed Current CP for Underground Storage
Tank Farm

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Figure 38
Impressed Current CP System for a Gasoline Service Station

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Figure 39
Impressed Current CP System for an Industrial Hot Water Storage Tank Interior,
with Separate Galvanic Anode for the Electrically Isolated Manway Cover.

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5.0 DESIGN EXAMPLES.

The following examples illustrate the application of the design principles outlined in this
Technical Paper as well as UFC 3-570-01 Chapters 3 and 4. They are intended to
illustrate the design methods to be used, but are NOT standard designs. The
examples are also not considered to be mandatory DOD policy and procedures.
The policy is described in UFC 3-570-01. Project engineers may use these examples
as guides when technically reviewing project design submissions.

In these examples, interference to or from foreign structures is not considered. In

practice, the design should be based upon field measurements whenever possible and
not on calculated estimates. In some examples, longer calculations than are actually
required are presented for illustrative purposes and shorter, more simplified calculations
would give equally applicable estimates. It is important to note that all cathodic protection
system designs make many assumptions, such as uniform environmental resistivity,
which may or may not prove to be true.

When the system is installed, it will require adjustment and possible modification in order
for effective protection to be achieved. In congested areas, interference problems are
often difficult to correct and optimum levels of protection may not be practically achieved.
In such cases, cathodic protection will reduce the incidence and degree of corrosion
damage but corrosion may not be entirely eliminated.

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5.1 Pipeline Distribution Systems. The project consists of providing cathodic

protection to a new 3 inch diameter buried fuel pipeline. For the purpose of cathodic
protection design, the pipeline is broken up into two zones:
Zone 1 Horizontal drilled segment from pump house at lower tank farm, up a
hill, and across the airfield runway.
Zone 2 Trenched pipeline from airfield to upper tank farm and then to power
plant.

5.1.1 Design Data.

A. Pipe diameter: 3” nominal with 3.4” outer diameter (OD).

B. Pipe length:
Zone 1: 3,335 feet
Zone 2: 12,276 feet
C. Pipe coating efficiency:
Zone 1: 50% (assume condition of coating may not be good after pulling
the pipe through the horizontally drilled hole)
Zone 2: 95%
D. Current Requirements: Due to the fact that this CP system is for a new
pipeline field current requirement test could not be conducted. Current
requirements for cathodic protection were assumed to be 2 milliamps per
square foot of exposed pipe surface area.
E. Soil resistivity: 3,500 ohm-cm
F. Initially assume that the entire pipeline (Zones 1 and 2) will be protected
by a single impressed current system. Power is only available at either
end of the pipeline. However, since the horizontally drilled segment starts
at the bottom of a cliff, the rectifier would be best located at that end of the
pipeline near the lower tank farm.

5.1.2 Calculations.

5.1.2.1 Find the External Surface Area (A) of the Piping.

AP = π DP LP

where
A = Surface area of piping (ft²)
DP = Diameter of pipe
LP = Length of piping

A. Zone 1:

A1 = 3.1416 x 3.4”/12 in/ft x 3,335 ft

= 2,969 ft²

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B. Zone 2:

A2 = 3.1416 x 3.4”/12 in/ft x 12,276 ft

= 10,927 ft²

C. Total Surface Area

AT = A1 + A2

AT = 2,969 + 10,927

= 13,896 ft²

5.1.2.2 Calculate the Current Requirement (I).

I = (A) (I’) (1 – CE)

where
I = Current requirement (amps)
A = Surface area of piping
I’ = Assumed current density = 2 mA/ft²
CE = Pipe coating efficiency

A. Zone 1:

I1 = 2979 x 2 x (1 – 0.5)

= 2979 mA or 2.98 amps

B. Zone 2:

I2 = 10927 x 2 x (1 – 0.95)

= 1093 mA or 1.1 amps

5.1.2.3 Attenuation Calculations.

Due to the relatively long length of this segment and the unknown quality of
the coating of the piping, attenuation calculations are performed to determine
the feasibility of utilizing a single rectifier system located at the end of the
pipeline. The end of the pipeline is chosen for the groundbed location
because that is the only spot where there is available power. Soil resistivity in
this region was measured to be an average of 3,500 ohm-cm.

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A. Determine the coating conductance (G).

Since the actual coating condition is unknown and of a concern in this
installation the coating conductance will be assumed to be 1 x 10-4
Siemens/ft2.

B. Calculate the pipe unit specific conductance (g).

gp = (G) (A)

where
gp = pipe specific conductance (Siemens)
G = coating conductance (Siemens/ft2) = 1 x 10-4
A = Surface area of piping = 13,896 ft2

gp = 1 x 10-4 x 13,896 ft2

= 1.39 Siemens

Calculate the unit specific conductance for the pipe as follows:

gp
g =
Lp/1000

1.39
g =
15,611/1000

= 0.089 per 1000 ft of pipe

C. Calculate the attenuation constant (α).

α = (r x g)1/2

where
α = attenuation constant
r = pipe resistance (ohms/1000 ft) = 0.021 ohms/1000 ft (schedule 80 pipe)
g = pipe unit specific conductance (Siemens/1000ft) = 0.089 S/1000 ft

α = [(.021ohm)(0.089 S)]1/2

= 0.043

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D. Calculate the characteristic resistance (RG).

RG = (r/g)1/2

RG = (0.021 ohm/0.089 S)1/2

= 0.486 ohm

E. Calculate resistance on opposite end of pipe from rectifier (RSO).

RSO = RG coth (αx)

where
RSO = resistance on opposite end of pipe from rectifier (ohm)
RG = pipe characteristic resistance (ohm) = 0.486 ohm
x = length of pipe (1000 ft) = 15,611/1000 ft = 15.611

RSO = (0.486) X coth[(0.043)(15.611)]

= 0.83 ohm

RSO = 0.83 ohm

F. Calculate current (IS) required to cause voltage shift of 1 volt at source

IS = VS/RSO

where
IS = current required to cause voltage shift of 1 volt at source (Amps)
VS = Source voltage (volt) = 1.0 volt
RSO = resistance on opposite end of pipe from rectifier (ohm) = 0.83 ohm

IS = (1.0 volt)/0.83 ohm

= 1.21 Amp

G. Calculate potential at end of pipe with ES = 1.0 volt

E = ES cosh (αx) – RGIS sinh (αx)

where

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E = potential at the opposite end of pipe from rectifier (volt)

ES = potential of the pipe at the rectifier (volt) = 1.0 volt
IS = current req’d to cause voltage shift of 1 volt at source = 1.21 Amp
x = length of pipe (1000 ft) = 15,611/1000 ft = 15.611

E = (1) cosh[(0.043)(15.611)] – (0.486ohm)(1.21A) sinh[(0.043)(15.611)]

= 1.234 - 0.425

= 0.809 volt

This indicates that a single power source at the lower tank farm end of the
pipeline could protect the entire line without greatly over protecting the end near the
power source. This represents a decrease of 19.1%. Generally anything less than 25%
is considered acceptable.

h. Calculate voltage shift at the mid-point along the distance of the pipe
assuming rectifier and anode groundbed are located at the bottom of the
hill near the lower tank farm.

E = ES cosh (αx) – RGIS sinh (αx)

where
E = potential at the opposite end of pipe from rectifier (volt)
ES = potential of the pipe at the rectifier (volt) = 1.0 volt
IS = current req’d to cause voltage shift of 1 volt at source = 1.21 Amp
x = length of pipe to midpoint (1000 ft) = 7,805/1000 ft = 7.81

E = (1) cosh [(0.043)(7.81)] – (0.486 ohm)(1.21A) sinh [(0.043)(7.81)]

= 1.06 - 0.20

= 0.86 volt at the midpoint

Again this indicates that a single power source at the end of the pipeline
could protect the both segments of the line without greatly over protecting the end near
the power source.

Although a single rectifier could protect this entire fuel pipeline, the
cathodic protection designer expressed concerns over this system for the following
reasons:

1) The extremely high cost of installing horizontal drilled segment which

goes underneath the airfield and the lack of economical inspection
options.

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2) The uncertainty regarding the coating condition of the horizontally

drilled segment and the fear of damaging the coating due to
overprotection.
3) The remoteness of the outpost raised the concern that proper
maintenance of the cathodic protection system may not be guaranteed.

Based on these concerns the cathodic protection system designers

decided to separate the two pipeline segments.

1) The horizontal drilled segment will be protected by a single rectifier and

anode ground bed consisting of 5 anodes.
2) The trenched segment of piping will be protected by a sacrificial anode
cathodic protection system consisting of 40 high potential magnesium
anodes with a cathodic protection test station at each anode
installation.

The two zones will be electrically isolated from each other with a dielectric insulation
flange. Provisions for bonding will be included to give versatility to the system.

End of Example

5.2 Single Underground Storage Tank (Technical Paper 17 provides example

galvanic anode system design calculations for this same tank).

5.2.1 Design data.

A. Tank Dimensions: 12' dia. X 40' long

B. Coating Efficiency: 80%
C. Design Life: 15 Years
D. Current Requirement: 0.7 Amperes (700 milliamperes)
E. Soil Resistivity: 30,000 ohm-cm
F. Other structures: None - tank is effectively electrically isolated from
pipelines and other structures

5.2.2 Calculations.

5.2.2.1 Area to be Protected.

Since we have conducted a current requirement test, there is no need to
calculate the area to be protected and current required.

5.2.2.2 Determine CP Current Requirements.

Determined by current requirement test to be 0.7 Amperes

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5.2.2.3 Calculate The Quantity Of Impressed Current Anodes.

1. Arbitrarily select a 2-21/32" dia. X 42" long, 31 Lb. tubular high silicon
chromium bearing cast iron anode in a 10" dia. X 66" long backfill column.
Try to design for a 2 ohm ground bed resistance.

Ra = [(0.0052ρ)/(N)(L)] X [ln (8L/d) - 1 + (2L/S)(ln 0.656N)]

where

Ra = Anode ground bed resistance (ohms)

N = Quantity of anode/backfill columns
ρ = Soil resistivity = 30,000 ohm-cm
L = 5.5 feet
d = 0.83 feet
I = 0.7 amperes
S = Assume 10' spacing between anode/backfill columns

Trying different quantities of anodes, we get the following:

Qty of Anodes Anode Bed Resistance

4 28. 6 ohms
6 21.2
8 17.0
80 2.6

The calculation shows that a 2-ohm ground bed is uneconomical. Choose

8 anode columns with a 17-ohm ground bed resistance.

2. Calculate current output per anode to ensure it does not exceed the
maximum recommended by the manufacturer. The maximum
recommended output for the 2-21/32" X 42" high silicon cast iron anode is
3.5 Amps

I
IA =
N

where

IA = Current output per anode

I = Total system current
N = Number of anodes

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0.7 Amp
IA =
8 Anodes

0.09 Amps/anode

This is well within the recommended maximum.

3. Calculate the life of the ground bed

NWu
L =
SI

where,

N = Quantity of anodes = 8 anodes

L = Anode life = 15 years
W = Anode weight = 31 lb
S = Anode consumption rate = 1.0 lb/ampere-year
I = Total system current required = 0.7 amperes
u = Anode utilization factor - usually 85%

(8 anodes)(31 lb/anode)(0.85)
L =
(1.0 lb/amp-yr)(0.7 amps)

= 300 years

This is well above the 15 year design life requirement.

5.2.2.4 Determine Wire Size And Resistance (Rw) For Positive And Negative
Header Cables.

Assume that the rectifier will be near the tank. Therefore, the wire resistance is
negligible.

5.2.2.5. Determine System Circuit Resistance.

RT = Ra + Rw + Rx

where,

RT = Total circuit resistance (ohms)

Ra = Anode bed resistance = 17 ohms
Rw = Wire resistance = negligible
Rx = 0 ohms (Assume no other resistance)

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RT = 17 + 0 + 0

= 17 ohms

5.2.2.6. Calculate Rectifier Voltage.

Vr = (I) (RT) (1.5)

where,

Vr = Rectifier Voltage (volts)

RT = Total circuit resistance = 17 ohms
I = Total system current = 0.7 amp (say 1 amp)

Vr = 1 amp x 17 ohms x 1.5

= 25.5 Volts

Select a commercially available rectifier with nominal DC output capacity of at

least 26 Volts DC and 1 ampere DC. A nominal size rectifier 28 volts/4 amps is
available.

5.2.2.7. Calculate AC Power Requirements.

Vr x I
PA =
e

where,

PA = AC power (VA)
Vr = Rectifier voltage = 28 volts
I = Rectifier current = 4 amps
E = Rectifier efficiency = 85%

28 Volts x 4 amps
PA =
0.85

131 VA

Field survey confirmed that 120 volts (VA) AC power is available.

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 Electrical TP 16
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PA
IAC =
VA

131 VA
IAC =
120 V

= 1.09 amps AC

One 15 or 20 ampere circuit breaker in a nearby electrical panel will be

sufficient.

End of Example

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5.3 Multiple Underground Storage Tanks. The service station shown in Figure
40 has three existing underground tanks and associated pipe. The quality of the
coating is unknown and it is not feasible to install dielectric insulation to isolate the UST
system. Because of the anticipated large current requirement, an impressed current
protection system is chosen. To distribute the current evenly around the tanks and
piping, and to minimize interference effects on other structures, a distributed anode
surface bed using vertical anodes is selected. Vertical anodes can be installed with
relative ease in holes cored through the paving around the UST system. Wiring can be
installed several inches below the paving by cutting and hand excavating narrow
slots/trenches through the paving.

5.3.1 Design Data.

A. Soil resistivity is 4500 ohm-cm.

B. Pipe is 2 in., nominal size. Total length of all buried piping is 750 ft.

C. Tanks are 8000 gal, 96 in. diameter by 21 ft-3 in. long.

D. Electrical continuity of tanks and piping has been assured.

E. It is not feasible to install dielectric insulation; system is therefore not

isolated electrically from other structures.

F. Design cathodic protection anodes for 20-year life.

G. Coating quality is unknown, assume bare.

H. The cathodic protection system circuit resistance should not exceed 2.5
ohms.

I. Electric power is available at 120 V single phase in the station building.

J. Current requirement test indicates that 8.2 amperes are needed for
cathodic protection.

K. Ceramic anodes will be specified.

5.3.2 Calculations.

5.3.2.1 Find the External Surface Area (A) of the Storage Tanks and Piping.

A. Storage Tanks

AT = 2 π rT 2 + π DT LT

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Figure 40
Cathodic Protection of Multiple Underground Storage Tanks

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where

AT = Surface area of storage tank (ft²)

rT = Radius of tank = 48 in = 4 ft
DT = Diameter of tank = 96 in = 8 ft
LT = Length of tank = 21 ft - 3 in

AT = [2 x 3.1416 x (4)2] + 3.1416 x 8 x 21.25

= 634.6 say 635 ft²

For all three tanks, the total surface area is

AT = 3 x 635 ft²

= 1905 ft²

B. Piping

AP = π DP LP

where
AP = Surface area of piping (ft²)
DP = Diameter of pipe = 2.375 in. or 0.198 ft for 2 in. nominal size pipe
LP = Length of piping = 750 ft

AP = 3.1416 x 0.198 x 750

= 467 ft²

C. Total Surface Area

AP = AT + AP

A = 1905 + 467

= 2372 ft²

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5.3.2.2 Verify the Current Requirement (I).

I = (A) (I’) (1 – CE)

where

I = Current requirement (amps)

A = Total surface area of tanks and piping = 2372 ft²
I’ = Assumed current density = 2 mA/ft²
CE = Pipe/tank coating efficiency = 0.00 for bare steel

I = 2372 x 2 x (1 – 0)

= 4744 mA or 4.7 amps

The 4.7 amp would be reasonable for the facility if it were electrically isolated
from other buried metals such as the building ground system. The actual
current requirement of 8.2 amp occurs because of current loss to these other
buried metal structures and is also reasonable in relation to that calculated for
an isolated facility.

5.3.2.3 Select An Anode and Calculate the Number of Anodes Required (NL) to
Meet the Design Life Requirements.

Calculations can be run on several size anodes, but in this case 2-in. by 60-
in. packaged ceramic rod anodes (rod size = 0.125 in x 4 ft long) are chosen
for ease of construction. Using the following equation, the number of anodes
required to meet the cathodic protection system design life can be calculated:

I
NL =
IA

where
I = Current requirement = 8.2 amps
IA = Ceramic rod anode current rating = 1.0 amps/anode
NL = Quantity of anodes to meet design life

8.2
N =
1.0

= 8.2 use 9 anodes

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5.3.2.4 Calculate the Quantity of Anodes to Meet the Maximum Anode Bed
Resistance of 2.5 Ohms (NR).

Ra = [(0.0052ρ)/(NR)(L)] X [ln (8L/d) - 1 + (2L/S)(ln 0.656NR)]

where

NR = Quantity of anode/backfill columns

Ra = Anode ground bed resistance (2.5 ohms maximum)
ρ = Soil resistivity = 4,500 ohm-cm
L = Anode backfill column length = 5 feet
d = Anode backfill column diameter = 2 in. = 0.167 feet
I = Current requirement = 8.2 amperes
S = Spacing between anodes. Assume 10' spacing between anode/backfill
columns. A distributed anode array does not lend itself to an exact
calculation using the equation above because the anodes are positioned
at various locations and are not located in a straight line. The above
equation assumes a straight-line configuration; however, to approximate
the total anode-to-earth resistance, the equation may be used.

Initially, try 9 anodes, the quantity required to meet the design life

0.0052 x 4500 8x5 2x5

RA = [ ln ( )-1 + ( ) (ln (0.656 x 9)) ]
9x5 0.167 10

= 3.26 ohms

The resistance for 9 anodes is too high. Additional calculations using an

increasing number of anodes (i.e., 11, 12, 13, 14, etc.) have to be made.
These calculations show that fourteen anodes will yield a groundbed-to-earth
resistance of 2.24 ohms.

Of the above-calculated quantities for the anodes, 9 to meet the design life
and 14 to meet the maximum anode bed resistance requirement, the larger
quantity of the two must be used to ensure all conditions are satisfied.
Therefore, use 14 each 0.125 in x 4 ft long ceramic rod anodes pre-packaged
in backfill in 2-in. by 60-in. canisters.

5.3.2.5 Calculate the Total Circuit Resistance (RT).

RT = RA + Rw + RC

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where,

RT = Total circuit resistance

RA = Anode bed resistance
Rw = Header cable/wire resistance
RC = StructurTP16-to-earth resistance

A. Ground bed resistance (RA) = 2.24 ohms from section TP16-5.3.2.4

B. Header cable/wire resistance (RW):

LW x RMFT
RW =
1000 ft

where,

Rw = Header cable/wire resistance

LW = Effective cable length. The loop circuit makes calculating effective wire
resistance complex. Since current is discharged from anodes spaced
all along the cable, one-half the total cable length may be used to
approximate the cable resistance. Total cable length = 300 ft. Effective
cable length = ½ x 300 ft = 150 ft.)

RMFT = Resistance per 1000 lineal feet of No. 4 AWG cable that has been
selected for ease of handling = 0.254 ohms/1000 LF.

150 ft x 0.254 ohm

RW =
1000 ft

= 0.038 use 0.04 ohm

C. Structure-to-earth resistance (RC):

Since the tanks and piping are essentially bare and are not electrically
isolated, structure-to-earth resistance may be considered negligible.
Therefore, RC = 0

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D. Calculate total resistance (RT):

RT = RA + Rw + RC

= 2.26 + 0.04 + 0

= 2.30 ohms

Since the design requirements call for a maximum ground bed resistance
of 2.5 ohms and RT = 2.30 ohms, the design using fourteen 2-in. by 60-in.
packaged ceramic anodes will work.

5.3.2.6 Calculate the Rectifier Voltage (VREC).

VREC = (I) (RT) (120%)

where,
VREC = Rectifier Voltage (volts)
RT = Total circuit resistance = 2.30 ohms
I = System current requirement = 8.2 amps
120% = Rectifier voltage capacity design safety factor

VREC = 8.2 amp x 2.3 ohms x 1.2

= 22.6 Volts

5.3.2.7 Select Rectifier.

Based on the design requirement of 22.6 V and 8.2 amp, a rectifier can be
chosen. A 12-amp, 24-V unit is selected because this is the nearest standard
commercial size available.

End of Example

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5.4 Elevated Steel Water Tank. This impressed current CP design is for an
elevated steel water tank that has not been built. Hence, it is not possible to determine
the current requirements, etc., by actual measurements. Calculated estimates have been
used.

5.4.1 Design Data.

A. Tank capacity - 500,000 gallons

B. Tank height (from ground to bottom of bowl) - 115 feet
C. Diameter of tank - 56 feet
D. High water level in tank - 35 feet
E. Overall depth of tank - 39 feet
F. Vertical shell height - 11 feet
G. Riser pipe diameter - 5 feet
H. Shape of tank - Ellipsoidal, both top and bottom
I. All internal surfaces are uncoated
J. Design for maximum current density - 2 mA/ft²
K. Electric power available 120/240 V ac, single-phase
L. String-type HSCBCI anodes are used
M. Design life - 10 years
N. Water resistivity - 4,000 ohm-cm
O. Tank water must not be subject to freezing
P. Assumed deterioration rate - 1.0 lbs/A yr
Q. Anode efficiency (assumed) - 50 percent

5.4.2 Calculations.

5.4.2.1 Area of Wetted Surface of Tank Bowl (see Figure 41).

A. Top section (T)

AT = 2 π r x (approximate)

where

r = 28 feet (radius of tank)

x = 10 feet
AT = 2 x 3.1416 x 28 x 10
AT = 1759 ft²

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Figure 41
Segmented Elevated Tank for Area Calculations

B. Center section (C):

Ac = 2πrh

where

r = 28 feet (radius of tank)

X = 11 feet
AC = 2 x 3.1416 x 28 x 11
AC = 1935 ft²

C. Bottom section (B):

AB = √2 π r √a2 + r2
where

r = 28 feet (radius of tank)

a = 14 feet

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AB = √2 x 3.1416 x 28 √14² + 28²
AB = 3,894 ft²

D. Total wetted area of tank bowl:

ATB = AT + AC + AB
= 1,759 + 1,935 + 3,894
= 7,588 ft²

5.4.2.2 Area of Riser Pipe.

A = 2 π rR h R

where

rR = 2.5 feet (radius of riser)

hR = 115 feet (height of riser)

AR = 2 x 3.1416 x 2.5 feet x 115

= 1,806 ft²

5.4.2.3 Maximum Design Current for Tank.

IT = 2.0 mA/ft² x 7,588 ft²

= 15,176 mA or 15.2 A

5.4.2.4 Maximum Design Current for Riser.

IR = 2.0 mA/ft² x 1,806 ft²

= 3,612 mA or 3.6 A

5.4.2.5 Minimum Weight of Tank Anode Material.

W = YSI/E

where

W = weight of anode material =

Y = design life = 10 years
S = anode deterioration rate = 1.0 lb/A-yr
I = maximum design current = 15.2 A
E = anode efficiency = 0.50

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10 x 1.0 x 15.2
W =
0.50

= 304 pounds

5.4.2.6 Minimum Weight of Riser Anode material.

W = YSI/E

where

Y = 10 years
S = 1.0 lbs/A yr
I = 3.62 A
E = 0.50

W = 10 x 1.0 x 3.62/0.50
= 72.4 pounds

5.4.2.7 Radius of Main Anode Circle.

DN
W =
2 (π + N)

where

D = 56 feet
N = Assumed number of anodes = 10

56 x 10
W =
2 (3.1416 + 10)

= 21.3 feet, use 22 feet

5.4.2.8 Spacing of Main Anodes. Generally the distance from the anode to the tank
wall and tank bottom is about equal; this distance should be about one-half
the circumferential distance between anodes.

A. Circumferential spacing:

2πr
C =
N

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where

r = radius of anode circle = 22 feet

N = assumed number of anodes = 10

2 x 3.1416 x 22
C =
10

= 13.8 feet, use 14 feet

B. Cord spacing is approximately the same as circumferential spacing: 14

feet will be used (see Figure 42).

Figure 42
Anode Spacing for Elevated Water Tank

5.4.2.9 Selection of Main Anodes.

A. Size of anode units selected is 1-1/8 inch outer diameter by ¾ inch inside
diameter by 9 inches long. This is a standard sausage type anode that
weighs one pound, and has an effective surface area of 0.25 ft2.

B. The minimum number of anode units per anode string, based on a

required weight of 304 pounds and 10 anode strings is computed as
follows:

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 Electrical TP 16
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304
Number of units =
10 x 1

= 30.4 say 31 units per string

C. Because the internal tank surfaces are uncoated, a maximum structure-to-

electrolyte potential is not a limiting factor. However, because it is desired
to limit the anode current at or below the manufacturer’s recommended
discharge current rate of 0.025 Amp for this type of anode, the minimum
number of anode units per string will be

15.2 Amps
Number of units =
10 x 0.025

= 60.8 say 61 units per string

This quantity of anode units per string is not practical for the tank bowl
since the distance between the anode hanger and the bottom of the bowl
is only 28 feet. Table 15 shows the maximum recommended discharge
current rate per anode for various types of anodes to ensure a minimum
10-year life. Using the type B anode, only three anode units per string are
required. The manufacturer does not recommend the use of more than
two type B anodes units per anode string assembly because of their
fragile nature. Therefore, the best anode unit choices for the main anode
strings are the type C or type CDD. Type CDD is recommended because
the lead wire connection is protected longer by the thicker wall of the
enlarged ends. Two type CDD anodes per string provide a current
capacity of 2 A x 10 strings = 20 A. These anodes are spaced as shown
in Figure 43.

NOTE: The anodes chosen in this example were chosen to illustrate

some of the many technical considerations during the design of cathodic
protection. For this example if HSCBCI anodes are tubular anodes, 2-
3/16 inch by 8 inch, weighing 4.3 pounds each, should be used instead of
CDD anodes.

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TABLE 15. TECHNICAL DATA - COMMONLY USED HSCBCI ANODES

ANODE ANODE MAX.

MAX. SURFACE CURRENT
DISCHARGE AREA (ft²) DENSITY
TYPE SIZE (in.) WEIGHT (lb)
(A) (A/ft²)
TA-FW 2-3/16 x 8 4.3 0.025 0.22 0.1
FW1 1-1/8 OD x 9 1 0.025 0.2 0.1
FC2 1-1/2 x 9 4 0.075 0.3 0.25
G-2 2 OD x 9 5 0.100 0.4 0.25
G-2-1/2 2-1/2 x 9 9 0.20 0.5 0.40
B3,4 1 x 60 12 0.50 1.4 0.36
C 1-1/2 x 60 25 1.00 2.0 0.50
CDD3 1-1/2 x 60 26 1.00 2.0 0.50
M3 2 x 60 60 2.5 2.8 0.9
SM 4-1/2 x 60 20 10.0 5.5 1.8
K-6 6 x 2-1/2 16 0.225 0.5 0.45
K-12 12 x 3-7/16 53 0.80 1.0 0.80
B-30 1 x 30 7 0.25 0.7 0.36
TA-2 2-3/16 x 84 46 6.4 4.0 1.6
1
For elevated fresh water tank.
2
For distributed system in ground trench.
3
Each end enlarged with cored opening for wire.
4
Not more than 2 anodes per assembly.

D. Anode current density is computed as follows:

15.2 Amps
Anode current density =
2 x 10 x 2

= 0.38 A/ft2

5.4.2.10 Resistance of Main Anodes.

0.012 ρ log (D/a)

R =
L

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where

ρ = 4,000 ohm-cm
D = 56 feet
L = 2 x 5 feet = 10 feet
a = 44 x 0.275 = 12.1 feet (0.275-equivalent diameter factor from curve - see
Figure 8-61)

0.012 x 4,000 x log (56/12.1)

R =
10

= 3.19 ohms

A. However, the L/d ratio of two 1-1/2-inch diameter by 60-inch long anodes
in tandem is less than 100 and thus the fringe factor must be used.

2 x 60
L/d =
1.5

= 80 < 100

B. The fringe factor from the curve in Figure TP16-45 corresponding to the
L/d ratio is 0.95.

R (adjusted) = 0.95 R = 0.95 x 3.19 = 3.03 ohms

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Figure 43
Anode Suspension Arrangement for Elevated Steel Water Tank

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Figure 44
Equivalent Diameter for Anodes in a Circle in a Water Tank

5.4.2.11 Stub Anodes.

A. In the design of an elevated water tank, the need for stub anodes must be
justified. The main anode radius has been calculated to b 22 feet. The
main anodes are spaced to provide approximately the same distance from
the sides and the bottom of the tank. The main anodes will protect a
length along the tank bottom equal to 1-1/2 times the spacing o the anode
from the bottom.

B. The anode suspension arrangement for the tank under consideration is

shown in Figure 43. Thus, it can be seen that stub anodes are required
for this design. Ten stub anodes are arranged equally space on a
circumference that has a radius of 8 feet in a manner illustrated in Figure
43. For smaller diameter tanks, stub anodes may not be required.

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Figure 45
Fringe Factor for Stub Anodes

5.4.2.12 Current Division Between Main and Stub Anodes.

A. Area of tank bottom protected by stub anodes (see Figure 43):

As = π (r22 – r12)

where
r = radius of protected segment (13 feet)
2
r = radius of riser (2.5 feet)
1

As = 3.1416 (169 - 6.25)

= 511.3 ft²

B. Maximum current for stub anodes:

Is = 2.0 mA/sf x 511.3 ft²

= 1022.6 mA or 1.02 A

C. Maximum current for tank bowl = 15.2 A.

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D. Maximum current for the main anodes:

Im = 15.2 - 1.02
= 14.2 A

5.4.2.13 Rectifier Voltage Rating.

A. Circuit resistance of electrical conductor to main anodes. Wire size No. 2

AWG, 0.159 ohm/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohm

B. Voltage drop in main anode feeder:

E = IR

where

I = 14.2 A
R = 0.032 ohm

E = 14.2 x 0.032
= 0.45 V

C. Voltage drop through main anodes:

E = IR

where

I = 14.2 A
R = 3.03 ohms

E = 14.2 x 3.03
= 43.0 V

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D. Total voltage drop in main anode circuit:

ET = 0.45 + 43.0
= 43.45 or 45.0 V

Use a multiplying factor of 1.5, or 67.5 V.

E. The nearest commercially available rectifier meeting the above

requirement is a single-phase, 80-V unit.

5.4.2.14 Selection of Stub Anodes. Because it is desirable to use as small an anode

as possible without exceeding the manufacturers' recommended rate, try using type FC,
HSCBCI anode measuring 1-1/2-inch by 9 inches. Use one anode per string as shown
in Figure 43. Compute anode current density as follows:

1.02
Output =
10 x 0.03

= 0.34 A/ft²

Because this exceeds the recommended maximum anode current density

(refer to Table 15), the Type B anode is the best choice.

5.4.2.15 Resistance of Stub Anodes.

0.012 ρ log (D/a)

R =
L

where

ρ = 4,000 ohm-cm
D = 56 feet
L = 5 feet
a = 16 x 0.275 = 4.4 feet (factor from Figure 44)

0.012 x 4,000 x log (56/4.4)

R =
5

= 10.6 ohms

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60
L/d =
1

= 60 < 100

The fringe factor from the curve in Figure 45 corresponding to the L/d ratio
is 0.9.

R (adjusted) = 10.6 x 0.90 R = 9.54 ohms

5.4.2.16 Voltage Drop in Stub Anode Circuit.

A. Electrical conductor to stub anodes. Wire size No. 2 AWG, 0.159

ohms/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohm

B. Voltage drop in stub anode feeder.

E = IR

where

I = 1.02 A
R = 0.032 ohm

E = 1.02 x 0.032
= 0.033 V

C. Voltage drop in anode suspension conductors. Estimated length 50 feet,

No. 2 AWG, 0.159 ohms/1,000 feet:

50
R =
1000 x 0.159

= 0.008 ohm

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 Electrical TP 16
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E = IR

where

I = 1.02/10 = 0.102 A
R = 0.008 ohm

E = negligible

D. Voltage drop through stub anodes:

E = IR

where

I = 1.02 A
R = 9.54 ohms

E = 1.02 x 9.54
= 9.73 V

E. Total voltage drop in stub anode circuit.

ET = 0.033 + 9.73
= 9.73 V

F. Since the stub anode voltage is below the 45 V calculated for the main
tank anode circuit, the necessary current adjustment can be accomplished
through a variable resistor in the stub anode circuit.

5.4.2.17 Stub Anode Circuit Variable Resistor.

A. Criteria for variable resistor. The resistor should be capable of carrying

the maximum anode circuit current and have sufficient resistance to
reduce anode current by one-half when full rectifier voltage is applied to
the anode circuit.

B. Stub anode circuit data:

Rectifier output = 80 V
Anode current = 1.02 A
Anode resistance = 9.54 ohms

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C. Variable resistor rating:

R = E/I

where

E = 80 V
I = 1.02/2 or 0.51 A

R = 80/0.51
= 156.9 ohms

Ohmic value of resistor = 156.9 - 9.54 = 147.4 ohms

Wattage rating of resistor = (1.02)² x 147.4 = 153.4 W

The nearest commercially available resistor size meeting the above

requirements is a 175-W, 200-ohm, 1-A resistor.

5.4.2.18 Resistance of Riser Anodes. In order to get the maximum desired current
in the riser (3.62 A), the resistance limit is calculated as follows:

R = E/I

where

E = 43.45 V
I = 3.62 A

R = 43.5/3.62
= 12.0 ohms

5.4.2.19 Riser Anode Design.

A. Type FW (1-1/8-inch by 9-inch) string type anodes cannot be used in the

riser because the maximum anode current discharge of 0.025 A per anode
would be exceeded. The number of type FW anodes required would be
145 and continuous throughout the riser. This is excessive. The best
choice of anode for a flexible riser string is the type G-2 (2-inch by 9-inch)
high-silicon cast iron anode.

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B. Number of units required:

(0.012 ρ log D/d)

R =
L

(0.012 ρ log D/d)

L =
R

where

ρ = 4,000 ohm-cm
D = 5 feet
D = 2 inches or 0.166 feet
R = 12 ohms

0.012 x 4000 x log (5/0.166)

L =
12

= 5.92 feet

Number of units = 5.92/0.75 = 7.9 or 8 units

In order to get proper current distribution in the riser pipe, the anode units
should not be placed too far apart. It is generally considered that each
anode unit protects a length along the riser pipe equal to 1-1/2 times the
spacing of the anode from the riser pipe wall.

Riser height = 115 feet

Spacing (center of anode to tank wall) = 2.5 feet
Length of riser protected by one anode = 1.5 x 2.5 = 3.75 feet
Number of units required = 115/3.75 = 30.7 or 31 units.

To satisfy the maximum anode discharge current for a G-2 anode:

3.62 A/0.1 amp = 36

Therefore, 36 anodes are needed instead of 31 or 8.

C. Anode resistance based on the use of 36 anode units:

R = (0.012 r log D/d)/L

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where

r = 4,000 ohm-cm
D = 5 feet
D = 2 inches or 0.166 feet
L = 36 x 9 inches = 324 inches or 27 feet

0.012 x 4000 x log (5/0.166)

L =
27

= 2.63 ohms

L/d ratio for the riser anode string is 324/2 or 162; thus no fringe factor
correction is applied.

5.4.2.20 Voltage Drop in Riser Anode Circuit.

A. Electrical conductor to riser anodes. Wire size No. 2 AWG, 0.159

ohms/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohms

B. Voltage drop in riser anode feeder:

E = IR

where

I = 3.62 A
R = 0.032 ohm

E = 3.62 x 0.032
= 0.116 V

C. Voltage drop in riser anode suspension cables. Wire size No. 2 AWG,
0.159 ohm/1,000 feet, estimated length 130 feet:

R = 130/1,000 x 0.159 = 0.02 ohm

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 Electrical TP 16
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E = IR

where

I = 3.62/2
= 1.81 A average (single current does not flow the full length of the anode
string)
R = 0.02 ohm

E = 1.81 x 0.02
= 0.04 V

D. Voltage drop through riser anodes:

E = IR

where

I = 3.62 A
R = 2.63 ohms

E = 3.62 x 2.63
= 9.52 V

E. Total voltage drop in riser anode circuit:

ET = 0.116 + 0.04 + 9.52

= 9.69 V

5.4.2.21 Riser Anode Circuit Variable Resistor.

A Criteria for the variable resistor are the same as given for the stub anode
resistor.

B. Riser anode circuit data:

Rectifier output = 80 V
Anode current = 3.62 A
Anode resistance = 2.63 + 0.032 + 0.02 = 2.68 ohms

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C. Variable resistor rating:

R = E/I

where

E = 80 V
I = 3.62/2 = 1.81 A

R = 80/1.81
= 44.2 ohms

Ohmic value of resistor = 44.2 - 2.68 = 41.5 ohms.

Wattage rating of resistor = (3.62)² x 41.5 = 543.8 W

The resistor should reduce anode current by one-half when full rectifier
voltage is applied. The nearest commercially available resistor size that
meets the above requirements is a 750-W, 50-ohm, 3.87-A resistor
(rheostat). This rheostat is 10 inches in diameter and 3 inches in depth,
and fairly expensive. This rheostat will not fit into most rectifier cases. In
addition, the power consumed by the rheostat is considerable. This power
creates substantial heat that may damage components within the rectifier
case unless adequate ventilation is provided. The problems associated
with using a large rheostat can be eliminated by using a separate rectifier
for the riser anodes. Although initial cost may be slightly high, power
savings will be substantial and damage by heat will be avoided.

5.4.2.22 Sizing Rectifier for Riser.

A. Requirements:

DC current output = 3.62 A

Anode circuit resistance = 2.68 ohms
DC voltage required = IR = 3.62 x 2.68 E = 9.70 V

B. Rectifier rating. Standard ratings for a rectifier in this size class are 18
Volts, 4 Amps.

5.4.2.23 Rectifier DC Rating for Bowl. Voltage output as previously determined, 80

V. Current rating is 15.2 A. The nearest commercially available rectifier meeting the
above requirements is 80 V, 16 A.

5.4.2.24 Wire Sizes and Types. All positive feeder and suspension cables (rectifier to
anodes) must be No. 2 AWG, HMWPE insulated copper cable. To avoid complication,
the negative rectifier cable (rectifier to structure) must be the same size and type (see
Figure 46).

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5.4.2.25 Discussion of the Design.

A. The design points out the disadvantages of achieving corrosion control

through cathodic protection without the aid of a protective coating. When
the interior of a tank is coated, the current requirement is reduced from 60
to 80 percent. On large tanks without coating, larger size and more
expensive anodes, wire, and rectifier units must be used. In addition, the
power consumed by the uncoated tank is far greater. These additional
costs usually exceed the cost of a quality coating system over 10-year
period. Corrosion above the water line of a water storage tank is usually
severe because of the corrosive nature of condensation. For this reason,
protective coatings must be used above the water line on both large and
small water storage tanks to mitigate corrosion.

B. For further assistance and guidance in the design of cathodic protection

systems for elevated water storage tanks, see Figures 46 through 48.

C. The HSCBCI anodes were selected for this particular design purely for
illustrative purposes. It does not mean that this material is superior to
other types of anode material. Other acceptable anode materials include
platinized titanium or niobium and mixed metal oxide (ceramic) anodes.
With the advent of newer tubular center connected anodes, the designer
should choose these anodes over the end connected in most cases
because of their higher current capability and longer life.

D. For this design, silicon stacks should be specified for the rectifier that
protects the bowl and selenium stacks should be specified for the rectifier
that protects the riser. Silicon stacks operate more efficiently at high DC
output voltages than selenium stacks do but require elaborate surge and
overload protection. This protection is not economical in the low power
consuming units. A guide for selection of rectifying cells is as follows:

• Use silicon stacks for single-phase rectifiers operated above 72 V dc or

three-phase rectifiers operated above 90 V dc.
• Use newer selenium stacks for single-phase rectifiers operated below
72 V dc or threTP16-phase rectifiers operated below 90 V dc.

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Figure 46
Elevated Steel Water Tank Showing Rectifier and Anode Arrangement

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Figure 47
Hand Hole and Anode Suspension Detail for Elevated Water Tank

Figure 48
Riser Anode Suspension Detail for Elevated water Tank

End of Example

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5.5 Elevated Steel Water Tank Where Ice Is Expected.

Impressed current cathodic protection is designed for an elevated steel water tank (Figure
49). The tank is already built and current requirement tests have been conducted.
Anodes cannot be suspended from the tank roof because heavy ice (up to two feet thick)
covers the water service during the winter. The anode cables could not tolerate this
weight; so another type of support must be used. An internally supported hoop shaped
wire anode system is selected.

5.5.1 Design Data.

a. The water tank is a pedestal spheroid with a ten-inch riser pipe. Only the
bowl will be protected because the riser pipe is less than 30 inches in
diameter.

b. Tank dimensions are:

Capacity - 400,000 gallons

Diameter of tank bowl – 51 ft 6 in
High water level in tank - 35 feet
Tank height (from ground to bottom of bowl) - 100 feet

c. Water resistivity - 2,000 ohm-cm

d. Anode Design life - 15 years

e. Wire type ceramic anodes to be used

f. All wetted surfaces are uncoated

g. Area above the high water level is kept well coated

h. Tank water is subject to freezing

i. The cathodic protection circuit resistance must not exceed 2 ohms

j. Electric power available is 120/240 V ac, singlTP16-phase

k. Based on structure current requirement testing on this tank, the current

required for adequate cathodic protection is 25 amps. This high current
requirement indicates that the tank internal coating is severely
deteriorated.

5.5.2 Calculations.

5.5.2.1 Calculate the Length of Wire in Feet (LB) Needed for the Current
Required.

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I
LB =
IA

where

I = Current requirement for adequate protection = 25 amps

IA = Allowable amp per foot of anode wire (varies depending on desired
anode life and diameter

Select the 0.0625 in. diameter copper cored anode wire based on the
current requirement of 25 amps and design life of 15 years.

25
LB =
0.31

= 81 feet

5.5.2.2 Calculate the Desired Diameter of the Anode Wire Ring (DR). Experience
shows that the diameter of the anode wire ring should be between 40 and 70 percent of
the bowl diameter.

A. Try 40% for the first iteration.

DR = 51.5 ft x 40% = 20.5 ft

Check to determine if the length is adequate for the desired anode life.
For an anode ring diameter of 20.5 feet the circumference (anode wire
length) is

CR = π x DR

= 64.4 feet

This length is inadequate for 0.0625 in. wire anode (which requires a
minimum of 81 feet to meet the desired anode life). Therefore, increase
the wire ring (hoop) diameter to 50 percent of the tank diameter.

DR = 51.5 ft x 50% = 25.75 ft

and

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CR = π x 25.75

= 80.9 feet

This diameter yields an anode length that is still slightly less than that
required for a 15-year anode life. Therefore, use a hoop diameter that is
about 55 percent of the tank bowl diameter.

DR = 51.5 ft x 55% = 28.3, say 29 ft

and

CR = π x 29.0

= 91 feet

5.5.2.3 Calculate the Anode-to-Water Resistance (RA) for the 0.0625 in. Diameter
Anode Wire.

0.0016 r 8 DR 2 DR
RA = ( ln + ln )
DR DA H

where,

r = Water resistivity = 2,000 ohm-cm

DR = Anode ring (hoop) diameter = 29 Ft
DA = Diameter of the anode wire anode = 0.00521 Ft (0.0625 in.)
H = Anode depth below water surface determined from the following
calculations (The anode depth below the high water line is about 60
percent of the distance between the high water line and the tank bottom).

H = 35 Ft x 60% = 21 Ft

0.0016 x 2000 8 x 29 2 x 29
RA = ( ln + ln )
29 0.00521 21

RA = 0.110 x ( ln 44,530 + ln 2.76 )

RA = 1.29 ohms

This is within the design limitation of 2.0 ohms

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5.5.2.4 Determine the Total Circuit Resistance (RT).

RT = RN + RW + RC

where,

RN = Anode-to-water resistance
RW = Wire resistance
RC = Tank-to-water resistance

A. AnodTP16-to-water resistance (RN) = 1.29 ohms from paragraph 5.5.2.3.

B. Header cable/wire resistance (RW).

LW x RMFT
RW =
1000 Ft

where,

LW = 115 Ft (Effective wire length. The positive wires from the rectifier to
each end of the anode ring will be about 115 feet long.
RMFT = 0.57 Ohm (Effective wire resistance per 1000 lineal feet. Since there
are positive wires from the rectifier to each end of the anode ring, each
wire will carry about one half of the current [12.5 amp]. The wires
selected are No. 10 AWG. Since the two wires are in parallel, the
effective resistance is one half the single wire resistance [1.02 ohms per
1000 lineal feet/2 = 0.51 ohm])

115 ft x 0.51 ohm

RW =
1000 Ft

= 0.06 ohm

C. Tank-to-water resistance (RC) and negative circuit resistance.

The negative wire is connected to the tank structure near the rectifier, so its
resistance is negligible. The tank-to-water resistance is also negligible
because the coating is very deteriorated.

D. Calculate (RT):

RT = 1.29 + 0.06 + 0.00

= 1.35 ohm

This is well below the design limitation of 2.0 ohms

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5.5.2.4 Calculate the Rectifier Voltage (VREC).

VREC = I x RT x 120%

where,

I = Current requirement = 25 amps

RT = Total circuit resistance = 1.35 ohms
120% = Rectifier voltage capacity design safety factor

VREC = 25 amps x 1.35 ohms x 120%

= 40.5 Volts

5.5.3 Select Rectifier.

Based on the design requirements of 40.5 volts and 25 amps, a commercially

available 48-volt, 28-amp unit is selected. Specify automatic potential control
to prevent over or under protection as the water level varies. The controller
maintains the tank-to-water potential through two permanent copper-copper
sulfate reference electrodes suspended beneath the anode wire ring. The
reference electrodes should have a life of at least five years. The tank-to-
water potential measured by the controller should be free of IR drop error.

5.5.4 Installation.

Figure 49 shows a typical installation while Figure 50 illustrates a typical detail

for a pressure entrance fitting for underwater power and reference electrode
wire penetrations.

End of Example

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Figure 49
Elevated Pedestal tank with ceramic anode wire ring for icing conditions.

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Figure 50
Pressure entrance fitting for underwater power and reference electrode wire penetrations in water storage tanks.

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5.6 Steel Gas Main. Design an impressed current cathodic protection system for
the 6-inch welded steel gas main shown in Figure 51. The pipeline has not yet been
constructed, so current requirement tests cannot be conducted.

Figure 51
Cathodic Protection System for Gas Main

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5.6.1 Design Data.

A. Average soil resistivity, 2,000 ohm-cm.

B. Pipe size, 6-inch outside diameter, schedule 80 pipe.

C. Pipe length, 6,800 feet.

D. Design for 15-year life.

E. Design for an estimated 2 mA/ft² of bare pipe.

F. Design for 90 percent coating efficiency, based on experience.

G. The pipeline must be isolated from the pump house with a dielectric
insulating flange on the main line inside the pump house.

H. Use HSCBCI anodes with carbonaceous backfill.

I. The pipe is coated with hot-applied coal-tar enamel and holiday checked
before installation.

J. Anode bed resistance must not exceed 2 ohms.

K. Electric power is available at 120/240 V ac, single phase, from a nearby

overhead distribution system.

5.6.2 Calculations.

5.6.2.1 Outside Area of Gas Main.

AP = π D L

where
AP = Outside surface area of the pipe
D = Outer pipe diameter = 6.625 inches for a 6-inch nominal diameter pipe
= 6.625 in/12 in/ft = 0.552 feet
L = Pipe length = 6,800 feet

AP = 3.1416 x 0.552 x 6800

= 11,792 ft²

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5.6.2.2 Area of Bare Pipe to Be Cathodically Protected Based on 90 Percent
Coating Efficiency.

A = AP x (1 – CE)

where
A = Area of bare pipe to be cathodically protected
AP = Outside surface area of the pipe = 11,792 ft²
CE = Pipe coating efficiency = 90% or 0.9

A = 11,792 ft² x (1 – 0.9)

= 1,179 ft²

5.6.2.3 Protective Current Required Based on 2 mA/ft² of Bare Metal.

I = A x CD

where
A = Area of bare pipe to be cathodically protected = 1,179 ft²
I = Current required for cathodic protection
CD = Current density = 2 mA/ft²

I = 1,179 ft² x 2 mA/ft²

= 2,358 mA or 2.36 A

5.6.2.4 Ground Bed Design.

A. Anode size: 2-inch x 60-inch (backfilled 10-inch x 84-inch), spaced 20 feet

apart.

B. Resistance of a single anode to earth:

r
RV = K
L

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where
r = Soil resistivity = 2,000 ohm-cm
K = Shape function (refer to para. 6.2.1.4a) = 0.0167
L = Backfilled anode length = 7.0 feet
L/D = 84 inches/10 inches backfill size

2,000
RV = x 0.167
7.0

= 4.77 ohms

C. Number of anodes required.

One of the design requirements is that the anode bed resistance is not to
exceed 2 ohms. Anode size used is 2-inch diameter x 60 inches long with
carbonaceous backfill having overall dimensions of 10-inch diameter x 84
inches long and spaced 20 feet apart:

1 p
Rn = RV + rs
n s

where
Rn = Anode bed resistance = 2 ohms
n = number of anodes
RV = Single anode resistance = 4.77 ohms
rs = Earth resistivity with pin spacing equal to S = 2,000 ohm-cm
p = paralleling factor (refer to para. 6.2.1.4b)
s = spacing between adjacent anodes = 20 feet

NOTE: p is a function of n as referred in para. 6.2.1.4b, and n is the number of

anodes which are determined by trial and error.

Rearranging the equation for n:

RV
n = p
Rn - ( rs )
s

4.77
n = p
2 - ( 2000 )
20

4.77
n =
2 – 100p

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Try n = 4 anodes, p = 0.00283 (refer to para. 6.2.1.4b)

4.77
4 =
2 – (100 x 0.00283)

4 = 2.78 (not very close)

Try n = 3 anodes, p = 0.00289 (refer to para. 6.2.1.4b)

4.77
3 =
2 – (100 x 0.00289)

3 = 2.79

This is the closest possible. In order to keep total resistance below 2.0
ohms, use 3 anodes.

D. Actual anode bed resistance:

1 0.00289
R3 = ( 4.77 ) + ( 2000 )
3 20

= 1.87 ohms which is less than 2.0

5.6.2.4 Next Calculate the Quantity of Anodes to Meet Recommended Maximum

Anode Current Discharge.

I
ND =
AA x IA

where:

ND = quantity of anodes to meet recommended maximum anode current

discharge limits.
I = Current required (2.36 amps)
AA = Anode surface area (ft2) = 2.6 ft2
IA = Max. recommended anode discharge current density (1 amp/ ft2)

2.36
ND =
2.6 x 1

= 0.91 say 1 anode

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Only one anode is required to stay within the maximum anode current
discharge limit. However, use the 3 anodes required to meet the 2-ohm
ground bed resistance design requirement.

5.6.2.5 Total Weight of Anodes for Ground Bed.

A. Weight of anode unit, 60 pounds (size 2 inches x 60 inches)

B. Total weight = 3 x 60 = 180 pounds

5.6.2.6 Theoretical Life of Anode Bed.

YSI
W =
E

Rearranging the equation we get:

WE
Y =
SI

where
Y = Theoretical anode bed life
W = Total anode bed weight = 180 pounds
S = Anode consumption rate = 1.0 lb/A yr
E = Anode efficiency = 0.50
I = Current required for cathodic protection = 2.36 A

180 x 0.50
Y =
1.0 x 2.36

= 38.1 years

It should be noted that the expected ground bed life greatly exceeds the
design requirement of 15 years. This is brought about by the additional
anode material required to establish a 2-ohm ground bed. The lower
ground bed resistance saves energy (power, P = I² R).

5.6.2.7 Resistance of the DC Circuit.

A. Ground bed-to-soil resistance, 2.0 ohms maximum.

B. Resistance of ground bed feeder conductor (length 500 feet, type
HMWPE, size No 2 AWG).

Conductor resistance (refer to Table 10): 0.159 ohm/1,000 feet

R = 500 ft x 0.159 ohm/1,000 feet

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= 0.080 ohms

C. Total resistance of circuit:

RT = 2.0 + 0.080

= 2.08 ohms

5.6.2.8 Rectifier Rating.

A. Minimum current requirement = 2.36 A.

B. Circuit resistance = 2.08 ohms.
C. Voltage rating:

E = IR
where
I = 2.36 A
R = 2.08 ohms

E = 2.36 A x 2.08 ohms

= 4.9 say 5.0 V

To allow for rectifier aging, film formation, and seasonal changes in the
soil resistivity, it is considered good practice to use a multiplying factor of
1.5 to establish the rectifier voltage rating.

E = 5.0 x 1.5 = 8.0 V

D. The commercial size rectifier meeting the above requirements is 115-V,

single-phase, selenium, and full-wave bridge type having a dc output of 8
A, and 8 V.

5.6.2.9 Rectifier Location.

Mount the rectifier at eye level on a separate pole adjacent to an existing

overhead electrical distribution system

End of Example

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5.7 Hot Water Storage Tank. Design impressed current cathodic protection for
the interior of the industrial hot water storage tank shown in Figure 52.

Figure 52
Cathodic Protection for an Industrial Hot Water Storage Tank

5.7.1 Design Data.

A. Tank capacity, 1,000 gallons.

B. Tank dimensions, 46 inches in diameter by 12 feet long.
C. Tank is mounted horizontally.
D. Water resistivity is 8,600 ohm-cm with a pH value of 8.7.
E. Tank interior surface is bare and water temperature is maintained at 180
degrees F (82.2 degrees C).
F. Design for maximum current density of 5 mA/ft².
G. Design life, 5 years.
H. Use HSCBCI anodes.
I. Alternating current is available at 115 V ac, single phase.

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5.7.2 Computations.

5.7.2.1 Interior Area of Tank.

AT = 2 π r² + π d L

where

r = Tank radius = 1.92 feet

d = Tank diameter = 3.83 feet
L = Tank length = 12 feet

AT = 2 x 3.1416 x (1.92)² + 3.1416 x 3.83 x 12

= 167.5 ft²

5.7.2.2 Maximum Protective Current Required.

A = AP x (1 – CE)

where
A = Area of bare pipe to be cathodically protected
AP = Outside surface area of the pipe = 11,792 ft²
CE = Pipe coating efficiency = 0% or 0

A = 167.5 ft² x (1 – 0)

= 167.5 ft²

I = A x CD

where
A = Area of bare tank surface be cathodically protected = 167.5 ft²
I = Current required for cathodic protection
CD = Current density = 5 mA/ft²

I = 167.5 ft² x mA/ft²

= 838 mA or 0.84 A

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5.7.2.3 Minimum Weight of Anode Material for 5-Year Life.

YSI
W =
E

where
Y = Anode design life = 5 years
S = Anode consumption rate = 1.0 lb/A yr
E = Anode efficiency = 0.50
I = Current required for cathodic protection = 0.84 A

5 x 1.0 x 0.84
W =
E

= 8.4 pounds

Number of anodes required. Anode size of 1-1/2 inches in diameter by 9

inches long weighing 4 pounds each is selected as the most suitable size.

W
NL =
WA

where

W = Total weight of anodes = 8.4 lbs

WA = Weight of a single anode = 4 lbs
NL = Quantity of anodes to meet design life

8.4 lbs
NI =
4 lbs

NI = 2.1 (say 3 anodes)

In order to get proper current distribution, three anodes are required.

5.7.2.3 Calculate the Resistance of a Single Anode.

0.012 r log (D/d)

R =
L

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where

r = Water resistivity = 8,600 ohm-cm

D = Tank Diameter = 3.83 feet
L = Anode length = 9 inches or 0.75 foot
d = Anode diameter = 1-1/2 inches or 0.125 foot

0.012 x 8,600 x log (3.83/0.125)

R =
0.75

= 204.5 ohms

This resistance must be corrected by the fringe factor because they are short
anodes. The fringe factor is 0.48 from the curve in Figure 45 for an L/d =
9/1.5 = 6.

R (adjusted) = 0.48 R = 0.48 x 204.5 = 98.2 ohms

5.7.2.4 Resistance of the 3-Anode Group.

1 p
RT = RV + rs
n s

where
RT = Total anode-to-electrolyte resistance
n = number of anodes
RV = resistance-to-electrolyte of a single anode = 98.2 ohms
rs = electrolyte resistivity = 8,600 ohm-cm
p = paralleling factor
s = spacing between adjacent anodes = 4 feet

1 0.00289
RT = 98.2 + 8,600
3 4

= 38.94 ohms

5.7.2.5 Rectifier Rating.

A. Calculate rectifier voltage

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E = IR

where
I = Protection current required = 0.84 A
R = Anode group resistance = 38.94 ohms

E = 0.84 A x 38.94 ohms

= 32.7 V

B. To allow for rectifier aging, film formation, it is considered good practice to

use a multiplying factor of 1.5 to establish the rectifier voltage rating.

E = 32.7 x 1.5 = 49.1 V

C. The nearest commercially available rectifier size meeting the above

requirements is a 60-V, 4-amp, single-phase unit.

5.7.2.6 Rectifier Location. Locate the rectifier adjacent to tank for the following
reasons:

A. Usually cheaper to install.

B. Easier to maintain.
C. Keeps DC voltage drop to a minimum.

5.7.2.6 DC Circuit Conductors.

A. External to tank: Use No. 2 AWG, HMWPE.

B. Interior of tank: Use No. 8 AWG, HMWPE.

No stressing or bending of the cable should be permitted.

End of Example

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5.8 Steam Heat Distribution System. Provide cathodic protection for a pre-
engineered steam conduit distribution system. Galvanic CP had been previously installed
on the outer conduit of some sections of the steam distribution lines. The system was
ineffective because of high soil resistivity, lack of adequate electrical isolation from
adjacent buried metallic structures (e.g. building H-piles, copper grounding systems,
water lines, electrical conduits, etc). The CP systems included in this design will be
impressed current type. Existing PVC condensate return lines will be replaced with steel
conduits in the near future.

5.8.1 Design Data.

A. Design life: 20 years

B. Current Density: 3 mA/ft².

C. Coating Efficiency: 85% for existing steam conduits and 95% for new
condensate lines.

D. Conventional shallow anode beds were considered, but have a high failure
rate due to third party damage. Deep well anode beds require less space
than shallow anode beds, and are not as likely to cause stray current
interference to nearby metallic structures. A number of small
rectifier/deep well systems are anticipated with each system electrically
isolated from all other systems to minimize the possibility of interference
and facilitate troubleshooting of system shorts that may occur. The deep
well will utilize mixed metal oxide tubular anodes in carbonaceous backfill.

E. Soil Resistivity: Soil resistivity measurements taken at various locations

with various pin spacing yielded a maximum resistivity measured at the
15-foot depth of 22,700 ohm-cm. Although it is not practical to measure
soil resistivity to the anticipated deep well depth, based on review of the
available geological information indicates that the resistivity is anticipated
to decline at deeper depths. However, for conservatism, the design is
based on 22,700 ohm-cm.

F. The steam lines will have insulating flanges and unions at the building tie-
ins. It is anticipated that the electrical isolation will be 90% effective.
Dielectric insulation will be provided at critical locations to electrically
segregate rectifier systems.

5.9.2 Calculations.

5.9.2.1 Outside Surface Area of Steel Conduit.

AP = π D L

where
AC = Outside surface area of the conduit

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D = Outer steam conduit diameter = 8.625 inches average = 0.72 ft
Outer steam condensate diameter = 6.625 inches average = 0.55 ft
L = Pipe length

Steam Conduit Surface Area

Location Length (ft) Diameter Area (ft²)
North Sector 50,000 0.72 113,098
South Sector 19,000 0.72 42,977
Total pipe surface area 156,075

Steam Condensate Conduit Surface Area

Location Length (ft) Diameter Area (ft²)
North Sector 50,000 0.55 86,394
South Sector 19,000 0.55 32,830
Total pipe surface area 119,224

5.9.2.2 Area of Bare Pipe to Be Cathodically Protected.

A = AC x (1 – CE)

where
A = Area of bare conduit to be cathodically protected
AC = Outside surface area of the conduit
CE = Conduit coating efficiency = 85% or 0.85
Condenstate coating efficiency = 95% or 0.95

Bare Surface Area

Structure Total Area (ft²) CE Area (ft²)
Steam Conduit 156,075 0.85 23,411
Condensate Conduit 119,224 0.95 5,961
Total pipe surface area 29,372

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5.9.2.3 Protective Current Required Based on 3 mA/ft² of Bare Metal.

I = A x CD

where
A = Area of bare pipe to be cathodically protected
I = Current required for cathodic protection
CD = Current density = 3 mA/ft²

Bare Surface Area

Structure Bare Area (ft²) CD I (ma)
Steam Conduit 23,411 3 70,233
Condensate Conduit 5,961 3 17,883
88,116
Total pipe surface area
say 88 amps

In order to keep the systems small, limit the size of each system to 15 amps.
The number of systems is

88 amps/15 amps per system = 6 systems

5.8.2.4 Calculate the Quantity of Anodes.

Use mixed metal oxide (MMO) tubular anodes in carbonaceous backfill. The
quantity of MMO anodes must be calculated to meet two different parameters:
design life based on anode maximum current discharge, and anode bed
resistance. The required quantity of anodes will be the larger of the two
calculated quantities.

1. First calculate the quantity of anodes to meet design life

Y IR
NL =
AA S

where:
IR = Current required = 8.9 amps
NL = Number of anodes to meet design life
S = Manufacturer’s MMO anode life rating (years)
AA = Manufacturer’s MMO anode current rating (amps)
Y = CP system design life = 20 years

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Example Manufacturer’s MMO Anode Ratings
Rated Output (Amperes)
Anode Size
10 years 15 years 20 years
1” x 45” 5.5 – 6.0 4.0 – 4.5 3.0 – 3.5
1” x 60” 7.0 – 8.0 5.0 – 6.0 4.0 – 5.0
1” x 90” 11.0 – 12.0 8.0 – 9.0 6.0 – 7.0

Use a 1” x 45” anode rated at an average of 3.25 amperes for a 20 year

design life.

20 x 15
NL =
3.25 x 20

= 5 anodes

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

2. Anode well resistance (RS)

The anodes will be installed in a vertical deep well. The resistance of the
anode well can be approximated by the following equation:

0.0052 r 8L
RS = [ ln ( )-1]
L d

where

RS = anode bed design resistance (2 ohm maximum desired)

r = soil resistivity (22,700 ohm-cm)
L = length of the anode backfill column (ft)
d = diameter of anode backfill column (8 in or 0.67 ft)

For five anodes try an active well depth of 100 feet.

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0.0052 x 22,700 8 x 100
RS = [ ln ( )-1]
100 0.67

= 7.2 ohms – too high

Several iterations using several different backfill column lengths yield the
following:

Anode Deep Well Resistance

Backfill Column Length (ft) Resistance (ohms)
100 7.2
150 5.1
200 4.0
250 3.3
300 2.8
400 2.2

A 2-ohm deep well ground bed is not economically feasible, therefore, select
a 200 ft backfill column with 4.0 ohms resistance. Although 5 anodes are
required to meet the design life, use 8 anodes to ensure anode current
attenuation along the backfill column is minimized. Install the anodes so that
the bottom of the anode is about ten feet above the bottom of the hole. Use
an anode spacing of twenty feet. Provide anodes with factory connected lead
wires of sufficient length to reach the anode junction box without splicing.

5.8.2.5 Rectifier Rating.

A. System current requirement = 15 A.

B. Circuit resistance = 4.0 ohms.
C. Voltage rating:

E = I R + 2 Volts
where
I = 15 A
R = 4 ohms

2 volts are added to overcome the typical rectifier back voltage.

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E = 15 A x 4.0 + 2 V

= 62 V

To allow for rectifier aging, film formation, and seasonal changes in the soil
resistivity, use a multiplying factor of 1.25 to establish the rectifier voltage and
current ratings ratings.

VR = 62 Volts x 1.25 = 78 Volts

IR = 15 Amps x 1.25 = 19 Amps

D. The commercial size rectifier meeting the above requirements is 240V,

single-phase, full-wave bridge type having a dc output of 22 Amperes, and
80 Volts.

5.8.2.4 Calculate the AC Line Load (Full Output).

1. First calculate the line current

V IR
IL =
240

where:
IL = Full output AC line current (amps).
IR = Rectifier rated output current = 22 amps
VR = Rectifier rated output current = 80 volts

80 x 22
IL =
240

IL = 8 Amps

The circuit overcurrent protection device shall not be less than 125 percent of
the continuous load. The rectifier is a continuous load, therefore,

8 amps x 125% = 10 amps

Use a 15 ampere circuit as a minimum.

End of Example

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5.9 Aircraft Multiple Hydrant Refueling System. Cathodic protection will be
provided to new underground 18-inch diameter stainless steel hydrant refueling supply
and return lines, 12-inch/10-inch diameter stainless steel lines supplying the direct
refueling stations, and 32 single point and 17 dual point hydrant outlets with 4-inch/6-inch
risers. The stainless steel lines will be coated with extruded polyethylene. Dielectric
Isolating flanges will be provided to electrically isolate the buried pipelines. Bare
galvanized steel grounding rods will be provided at the hydrant refueling pits. Also
intermittent current loads will be imposed on the CP system where copper clad tie
downs embedded in the concrete apron are connected to one or more aircraft that are
refueling. Tie downs are bonded together by a bare copper conductor encased in the
concrete apron.

5.9.1 Design Data.

A. Design life: 25 years

B. Current Density: 0.5 mA/ft² for soil, 5 mA/ft² for structures embedded in
concrete.

C. Coating Efficiency: 90% for the extruded polyethylene coating. Bare (0%)
for the hydrant outlet risers.

G. A conventional anode bed with high silicon cast iron tubular anodes
installed horizontally is planned. Due to the soft sand environment, use
ten-inch diameter pre-packaged anodes to simplify installation.

H. The anode bed design resistance should not exceed 2 ohms.

I. Soil Resistivity: Soil resistivity measurements taken at various locations

with various pin spacing yielded soil resistivities ranging from 374 ohm-cm
to 21,500 ohm-cm. The maximum resistivity measured at the 10-foot
depth was 8,600 ohm-cm. Therefore, the design is based on a resistivity
of 8,600 ohm-cm, and the anodes must be installed at this depth.

5.9.2 Calculations.

5.9.2.1 Outside Surface Area of Stainless Steel Pipe.

AP = π D L

where
AP = Outside surface area of the pipe
D = Outer pipe diameter
L = Pipe length

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Stainless Steel Distribution Pipe
Diameter (in) Length (ft) Area (ft²)
18 9,000 42,412
12 830 2,770
10 180 507
Total pipe surface area 45,689

There are 32 single point and 17 dual point hydrant outlet pits for a total of
66 hydrant outlet risers.

Stainless Steel Hydrant Riser Pipe

Diameter Length (ft) Qty Area (ft²)
4”-6” Reducer (5” Avg. Dia) 1 66* 86.4
4” Riser (4.5” OD) 2 66* 155.6
Total pipe surface area 242.0

5.9.2.2 Surface Area of Hydrant Pit Ground Rods. There are32 single point and
17 dual point hydrant outlet pits for a total of 49 pits and therefore, 49 ground rods.

AR = π D L NR

where
AR = Surface area of the ground rod
D = Ground rod diameter = ¾ inch = 0.0625 ft
L = Ground rod length = 10 ft
NR = Quantity of Ground Rods = 49

AP = 3.1416 x 0.0625 x 10 x 49

= 96.2 ft²

5.9.2.3 Surface Area of Aircraft Tiedown/Ground System.

A = πDL

where
A = Surface area
D = Ground rod/wire diameter
L = Ground rod/wire length

The tiedown/ground consists of ¾ inch diameter by 19 inches long rod plus a

¾ inch diameter by 11 inches long rod, a 2-inch diameter by 3-inch long area
receptacle, #4 bare copper wire, and a major ground bed.

Surface Area of Tiedown/Ground System

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Component Dia (in) Length (ft) Qty Area (ft²)
Ground rod (19” + 11”) 0.75 2.5 613 301
Receptacle 2 0.25 613 80
#4 Copper Wire 0.232 8490 515
Total pipe surface area 896

5.9.2.4 Area of Bare Structure to Be Cathodically Protected.

A = AS x (1 – CE)

where
A = Area of bare structure to be cathodically protected
AS = Surface area of the structure
CE = Structure coating efficiency

Area of Bare Structure to be Protected

Structure Surface Area (ft²) CE Area (ft²)
SS Distribution Pipe 45,689 0.9 4,569.0
SS Hydrant Risers 242 0 242.0
Hydrant Pit Ground Rods 96.2 0 96.2
Tiedown/Ground System 896 0 896

5.9.2.5 Protective Current Required.

I = A x CD

where
A = Area of bare pipe to be cathodically protected
I = Current required for cathodic protection
CD = Current density = 0.5 mA/ft² for pipe and ground rods in soil
= 5 mA/ft² for grounding encased in concrete

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Area of Bare Structure to be Protected
Structure Surface Area (ft²) CD (mA/ft²) I (mA)
SS Distribution Pipe 4,569.0 0.5 2,285
SS Hydrant Risers 242.0 0.5 121
Hydrant Pit Ground Rods 96.2 0.5 48
Tiedown/Ground System 896 5 4,480
6,934
Total pipe surface area
Say 6.9 Amps

In order to account for leakage current through insulating flanges and

unanticipated coating damage, add an additional 2 amperes for a total of 8.9
Amperes. To allow for rectifier aging, film formation, and seasonal changes in
the soil resistivity, select a standard size 12 ampere rectifier.

5.9.2.6 Calculate the Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes. The
quantity of anodes must be calculated to meet three different parameters,
design life, anode bed resistance, and anode maximum current discharge.
The required quantity of anodes will be the largest of the three calculated
quantities.

1. First calculate the quantity of anodes to meet design life

Y S IR
NL =
WU

where:
IR = Current required = 8.9 amps
NL = Number of anodes to meet design life
S = Anode consumption rate = 1 lb/amp-yr)
U = Anode utilization factor = 0.8
W = Weight of one anode (lb)
Y = CP system design life = 25 years

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the recommended maximum anode discharge and the ground bed resistance
requirements must still be calculated.

2. Next calculate the quantity of anodes to meet recommended maximum

anode current discharge

I
ND =
AA x IA
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where:
I = Current required = 8.9 amps)
AA = Anode surface area (SF)
IA = Max. recommended anode discharge current density (1 amp/SF)

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

3. Calculate quantity of anodes to meet 2-ohm ground bed design

The anodes will be installed in a shallow, distributed, horizontal anode

bed. The resistance of the horizontal anode bed can be approximated by
the following equation:

1.64 ρ
RS = x [ln (48L/d) + ln (L/h) - 2 + (2h/L)]
πL

where
d = diameter of anode backfill column = 10 inches
L = length of the anode backfill column (feet)
RS = anode bed design resistance = 2 ohms
ρ = soil resistivity = 86 ohm-m
h = anode depth = 10 feet

Assume one foot of backfill column beyond the ends of each anode.

4 Determine quantity of anodes. The following table summarizes the

above calculations for various common tubular anode sizes:

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CALCULATED ANODE QUANTITIES
Weight Area Backfill Backfill
Anode Size NL ND NR
(LB) (SF) Length (LF) Dia. (ft)
2.66" x 42" 31 2.4 5.42 0.833 9 4 32
2.66" x 60" 46 3.5 9.0 0.833 7 3 20
2.66" x 84" 63 4.9 9.0 0.833 5 2 20
3.75" x 84" 85 6.9 9.0 0.833 4 2 20

Of the three calculated quantities for a particular anode, the larger

quantity of the three must be used to ensure all conditions are satisfied.
The 46, 63 and 85 pound anodes utilize the same size backfill columns,
and therefore, yield the same quantity of anodes to meet the 2.0 ohm
maximum ground bed resistance requirement. For economic
considerations, use the 20 each 46 LB, 2.66 inch diameter X 60 inch
long anodes. Install the anodes so that there is one foot of backfill
column beyond each end of each anode (two feet between consecutive
anodes). Provide anode with factory connected lead wires of sufficient
length to reach the anode junction box without splicing.

5.9.2.7 Resistance of the DC Circuit.

A. Ground bed-to-soil resistance, 2.0 ohms maximum.

B. Calculate electrical conductor resistance.

Calculate Rn: #4 rectifier negative header cable resistance (90 feet):

Rn = 90 ft/1000 x 0.269 ohm/1,000 feet

= 0.024 ohms

Calculate Rp: # rectifier positive header cable resistance (1,085 feet):

Rp = 1,085 ft/1000 x 0.169 ohm/1,000 feet

= 0.208 ohms

C. Total resistance of circuit:

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RT = RS + Rn + Rp

= 2.0 + 0.024 + 0.208

= 2.23 ohms

5.9.2.8 Rectifier Rating.

A. Rectifier current rating = 12 A.

B. Circuit resistance = 2.23 ohms.
C. Voltage rating:

E = I R + 2 Volts
where
I = 12 A
R = 2.23 ohms

2 volts are added to overcome the typical rectifier back voltage.

E = 12 A x 2.23 + 2 V

= 28.8 V

D. The commercial size rectifier meeting the above requirements is 115V,

single-phase, full-wave bridge type having a dc output of 12 Amperes, and
30 V.

End of Example

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5.10 On-Grade Fuel Storage Tanks. Provide impressed current cathodic
protection for the exterior bottoms of five existing on-grade fuel storage tanks:

TANK CAPACITY DIAMETER

ID NO. (BBL) (FT)
Tank No. 1 55,000 93
Tank No. 2 55,000 93
Tank No. 3 2,300 27
Tank No. 4 10,000 45
Tank No. 5 20,000 58

Review of historical records indicates that the tanks have not had cathodic protection in
the past.

5.10.1 Design Data.

A. Design life: 25+ years

B. Current Density: Record drawings and field inspection indicates the tanks
have been constructed on an “impermeable layer” of compacted coral with
a minimum thickness of 12 inches. The compacted coral may affect the
even distribution of current and higher than normal current densities may
be necessary to ensure the entire tank bottom is adequately protected.
Usually, 1 - 1.5 ma/sf of current is sufficient to protect a tank bottom if the
current is evenly distributed. With the compacted coral layer beneath the
tanks, a current of 2 ma/sf will be used.

C. Coating Efficiency: 0% (Bare tank bottoms)

D. 120 Volt power and a spare circuit breaker is available in a panel in a

nearby pump house. Space is available in the pump house to install a
wall mount rectifier.

E. A deep well anode will be specified to: (1) improve current distribution of
current to the entire tank bottom because of the compacted coral bed;
(2) minimize excavation (historic preservation laws require an
archaeologist to monitor all excavations full time which would add
significant cost; (3) minimize disruption of on-going operations; and (4)
minimize cathodic interference on other buried metallic structures in the
tank farm.

F. The anode bed design resistance shall not exceed 1 ohm.

G. Soil Resistivity: Soil resistivity measurements taken at eight points within

the tank farm were generally in the 2,000 to 4,000 ohm-cm range. Use
2,000 ohm-cm.

5.10.2 Design Calculations.

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5.10.2.1 Calculate Tank Bottom Surface Area.

The surface area of the tank bottoms can be calculated using the following
equation:

AT = π r2

where
AT = Surface area of a tank bottom (ft2)
r = tank bottom radius (ft)

The following table summarizes the tank bottom surface areas.

Tank ID. Diameter Surface Area

NO. (FT) (ft2)
Tank No. 1 93 6,790
Tank No. 2 93 6,790
Tank No. 3 27 570
Tank No. 4 45 1,590
Tank No. 5 58 2,640

5.10.2.2 Calculate Current Requirement/Rectifier Output Current.

IR = AT x (1 - CE) x CD

where
IR = Current requirement/rectifier output current
AT = Surface area of a tank bottom = 2,640 (ft2)
CE = Coating efficiency = 0 for bare surfaces
CD = Current density = 2 ma/ ft2

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The following tables summarize the current requirements.

Tank ID Surface CE CD Current

No. Area (ft2) (ma/ft2) Req'd (ma)
Tank No. 1 6,790 0 2 13,580
Tank No. 2 6,790 0 2 13,580
Subtotal 27,160
Tank No. 3 570 0 2 1,140
Tank No. 4 1,590 0 2 3,180
Tank No. 5 2,640 0 2 5,280
Total Current 36,760
Required (IR) or 37 amperes
In order to account for current that will be drained away by the grounding systems,
use 40 amperes.

5.10.2.3 Calculate Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes.

A. Quantity of anodes to meet design life

Y S IR
NL =
WU

where:
NL = Number of anodes to meet design life
Y = CP system design life = 25 years
S = Anode consumption rate = 1 lb/amp-yr)
IR = Current required (amps) = 40 amps
W = Weight of one anode (lb)
U = Anode utilization factor (0.85)

B. Quantity of anodes to meet recommended maximum anode current

discharge

IR
ND =
AA X IA

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where:
IR = Current required (amps) = 40
AA = Anode surface area (ft2)
IA = Max. recommended anode discharge current density (1 amp/ ft2)

The following tables summarize the above calculations for various

common tubular anode sizes:

Weight Area
Anode Size NL ND
(LB) (SF)
2.19" x 84" (TA2) 46 4.0 25 10
2.66" x 84" (TA3) 63 4.9 19 9
3.75" x 84" (TA4) 85 6.9 14 6

Of the above-calculated quantities for the anodes, the larger quantity of

the two must be used to ensure all conditions are satisfied. Considering
CP current distribution in the well, use the 85 LB, 3.75 inch diameter X
84 inch long (TA-4) anode. The TA-4 appears to be a better choice to
minimize drilling.

C. Anode well resistance (RS)

The anodes will be installed in a vertical deep well. The resistance of the
anode well can be approximated by the following equation:

0.0052 r 8L
RS = [ ln ( )-1]
L d

where

RS = anode bed design resistance (1 ohm maximum)

r = soil resistivity (2,000 ohm-cm)
L = length of the anode backfill column (ft)
d = diameter of anode backfill column (10 in or 0.83 ft)

Based on the above a minimum anode well active depth of about 55 feet
minimum is required.

Considering the number of anodes required and backfill life, use two wells
of 101 feet active depth with seven anodes in each well. Install the
anodes so that the bottom of the anode is about five feet above the bottom
of the hole. Use an anode spacing of seven feet. Provide anodes with
factory connected lead wires of sufficient length to reach the anode
junction box without splicing. Calculate the anode bed circuit resistance of
each well using the appropriate numbers from the above table:

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0.0052 x 2000 8 x 101

RS = [ ln ( )-1]
101 0.83

= 0.61 ohm for each of the wells

5.10.2.4 Calculate Rectifier Circuit Output Voltage.

A. Anode bed resistance (RA) is the equivalent resistance of the two wells in
parallel. Calculate the equivalent resistance of the two wells:

1 1
RA = +
R1 R2

where,
RA = Equivalent anode well circuit resistance
R1 = Circuit resistance of the first anode well
R2 = Circuit resistance of the second anode well

1 1
RA = +
0.61 0.61

= 0.31 ohm

B. Electrical Cable Resistance

Anode Well Cable Resistance (RAW). Use No. 8 copper wire for the anode
leads. Each of the anode leads will run directly back to the anode junction
box, so the parallel cable resistance is estimated as follows:

1 1 1 1
= + +····+
RAW RA1 RA2 RAn

where,

RAW = Anode well cable resistance

RA1 = Cable resistance of the first anode in the well
RA2 = Cable resistance of the second anode in the well
RAn = Cable resistance of the nth anode in the well

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1 1 1 1
= + +····+
RAW RA1 RA2 RA7

1 1 1 1
= + +····+
RAW RA1 RA2 RAn

= 0.01 ohm for each well

(negligible)

Anode Header Cable Resistance(RHC). Use No. 2 copper wire for the
anode header cable runs. After a 100 ft run of cable to a junction box,
there are two parallel runs of 200 ft and 450 ft. The equivalent resistance
is 0.053 ohm.

Structure Header Cable Resistance. Use No. 2 copper wire for the
structure leads. After a 450 ft run of cable to anode junction box no. 1,
there are two parallel runs of 100 ft to (tanks 3, 4 and truck stand piping)
and 225 ft (tank 5 piping and tanks 1 and 2). The equivalent resistance for
all of these cable runs is 0.11 ohm.

C. Total Circuit Resistance. The total circuit resistance is

RT = RA + RAW + RHC + RSC

0.31 ohm + 0 + 0.05 + 0.11

= 0.47 ohm say 0.5 ohm

D. Calculate the rectifier output voltage using Ohm's Law:

VR = Irect x RT x 150%

40 amps x 0.5 Ω x 1.5

= 30 volts

Use the nearest nominal size rectifiers, or 30 volts/42 amps.

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5.10.2.5 Calculate Rectifier Input Power.

Available power is 120 volts. For rectifiers of the size specified above, use
single phase 120 volt AC input voltage. Rectifier input power is calculated as
follows:

VR x IR
P =
rectifier efficiency

30 volts x 42 amps
=
0.85

= 1,500 VA

AC input current is calculated to be:

P
IAC =
VAC

1,500 VA
=
120

= 12.5 amperes

5.10.2.6 Summary.

A. Anodes: 14 each high silicon chromium bearing cast iron tubular anodes,
with dimensions of 3.75 inches in diameter x 84 inches long, and weighing
85 LB. Anodes to be installed in two 130 foot deep well, seven anodes
per well, with an active depth of 101 feet. Locate the first anode so that
the bottom anode is 5 feet above the bottom of the well. Install remaining
anodes with a 14 foot spacing on center between the anodes. Anode
supplied by manufacturer with #8 AWG lead wire with HMWPE insulation.
One end is factory connected to the anode. The anode leads should be
long enough to extend to the junction box without splicing.

B. Rectifier:
Type: Air cooled
AC Input: 120 volt, 60 Hz, single phase
DC Output: 30 volts, 42 amperes
Efficiency: 85% minimum

End of Example

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5.11 On-Grade Storage Tanks with Impermeable Containment Liner. Two
each 27,000 barrel on-grade vertical fuel storage tanks are protected by an existing
impressed current CP system. However, a project proposes to install a new tank
bottom above the existing tank bottom, creating a double tank bottom. A high-density
polyethylene containment liner will be installed over the existing tank bottom. The
existing tank bottom and new polyethylene liner will prevent the existing CP system
from providing protection to the new tank bottom. Galvanic corrosion between the new
and old tank bottoms can result in rapid corrosion of the new tank bottom. Therefore, a
new CP system will be installed between the two tank bottoms to provide corrosion
protection to the new tank bottom. The space between the new and existing tank
bottoms will be 4 inches minimum; therefore, the new CP anode system must be
installable in this small space.

5.11.1 Design Data.

A. Design life: 25+ years

B. Current Density: 1 ma/ft2 for a grid system. Generally, 2 ma/ft2 is required

for protection in neutral soils. Past experience and current literature
indicates that 1 ma/ft2 is sufficient due to the even distribution of current
when using a grid system.

C. Tank bottom diameter: 70 ft

D. Tank bottom is steel

E. Slope of the new tank bottom will result in only 4 inches clearance
between the existing and new tank bottom at its lowest point. Therefore,
the proposed new CP system will be an impressed current system
consisting of mixed metal oxide coated titanium ribbon anodes.

F. Each tank will have its own independent system to allow independent
system adjustment for each tank. The rectifier will have two independent
DC output circuits, one for each tank, contained in one enclosure.

5.11.2 Calculations.

5.11.2.1 Tank Bottom Surface Area.

The tank bottom is a circular surface for which its surface area can be
calculated from the following equation:

A = π x r2

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The area of each tank bottom is

A = (3.14) (70 SF/2)2

= 3847 say 3850 SF

5.11.2.1 Current Requirement/Rectifier Output Current

The current requirement for each tank is calculated using the equation:

Ireq = A x current density

= 3850 SF x 1 ma/SF
= 3850 ma or 3.85 amps

Allow an additional 25% to account for unknown factors in the sand backfill
that will be used, for rectifier aging and other long term additional requirement:

IR = 3.85 amps x 125%

= 4.8 amps

Use the next nominal size rectifier circuit of 8 amperes for each tank.

5.11.2.2 Quantity of Anode Ribbons

Commonly used mixed metal oxide coated titanium ribbon anodes are 0.25
inch wide by 0.025 inch thick. Manufacturer's literature indicates this material to have a
current rating of 5 ma/LF for 50 year life.

The quantity of anode ribbon required for a 25+ year life for each tank is

N = 3850 ma x 5 ma/LF
= 770 LF

The ribbons will be spaced in evenly spaced parallel strips to ensure uniform
distribution of CP current. This will result in 14 strips of ribbon spaced 5 feet on center.

The anode ribbons will be connected together by titanium ribbon conductor

bars placed in a grid spaced at 30 feet on center ribbons to form a grid. The conductor
bars are 0.5 inch wide by 0.04 inch thick. The conductor bars will be fed by multiple
power feed leads. This will minimize the voltage drop through the grid, thereby allowing
an even distribution of cathodic protection current to the tank bottom.

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5.11.2.3 Circuit Resistance of Anode Ribbons

The strips are of different lengths due to the circular perimeter of the tank
bottom. The theoretical resistance of a strip of metal can be calculated using the
following equation from reference (c):

ρ 2
RS = (ln 4L + a - πab2 + ln 4L - 1)
4πL a 2(a+b) S

where

L = ½ the length of the anode strip (cm) = 335 LF or 11,735 cm

a = width of the strip (cm) = 0.25 in or 0.635 cm
b = thickness of the strip (cm) = 0.025 in or 0.0635 cm
S = twice the depth of the anode = 6 in or 15.24 cm
ρ = sand resistivity = 10,000 - 45,000 ohm-cm. Sand resistivity is
dependent on many factors dependent on the actual sand used.
Typically, the resistivity of clean sand will vary from 10,000 ohm-cm
when wet to 45,000 when dry. While the sand may be dry when first
installed, moisture will eventually intrude from rain during installation,
water used during compaction, and leakage through deteriorated
edge seals.

The resulting resistance after substituting the above numbers is

Rs = 5.7 Ω for 45,000 ohm-cm sand, and

Rs = 1.3 Ω for 10,000 ohm-cm sand

5.11.2.4 Rectifier Circuit Output Voltage

The rectifier output voltage is calculated by Ohm's Law. Use the resistance
for 45,000 ohm-cm sand to ensure proper operation when sand is dry:

VR = Ireq x Rs
= 3.85 amps x 5.7 Ω
= 21.9 volts

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Use the next highest nominal size rectifier of 24 volts to account for rectifier back
voltage, circuit resistance for the power feed leads and conductor bars, and
compensation for system aging.

5.11.2.5 Rectifier Input Power

Rectifier input power for each circuit can be calculated as follows:

VA = VR x IR
rectifier efficiency

= 24 volts x 8 amps
0.65

= 295 say 400 VA to account for 25% safety factor

The total power for the two circuits is 800 VA or 0.8 KVA. For rectifiers of this
size, use single phase 120 volt AC input voltage. AC input current is calculated to be:

IAC = VA/VAC
= 800 VA/120 volts
= 6.67 say 7 amps

5.11.3 Summary

Anode material: Mixed metal oxide coated titanium ribbon, 0.25" wide x
0.025" thick. The anode grid requires 770 LF per tank
placed in 14 parallel rows spaced at 5 FT on center.

Conductor bar: Conductor bars are uncoated titanium ribbons, 0.5" wide x
0.04" thick. Estimated quantity is 320 LF per tank.

Power feed leads: Supplied by manufacturer consisting of #8 AWG wire with

HMWPE insulation. One end is factory connected to a 5"
length of conductor bar, which will be field welded to the
anode grid conductor bars. The power feed leads should
be long enough to extend to the junction box without
splicing. 7 power feed leads per tank are required.

Rectifier:
Type: Oil cooled, enclosure suitable for class I, division 2.
AC Input: 120 volt, 7 amps, 60 Hz, single phase
DC Output: Two independent circuits, each circuit rated at 24 volts, 8
amperes
Efficiency: 65% minimum

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Reference
electrodes: Combination cell of copper copper-sulfate and zinc, each
component having a #14 AWG lead wire long enough to
reach the junction box without splicing. 5 each
combination cells per tank are required.

Due to the proximity of the tank bottom to the anode, especially at its lowest
point, dielectric mesh will be installed between the anode and new tank bottom to
prevent electrical short circuits.

End of Example

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5.12 Land Side of Steel Sheet Piling (Impressed Current).

Due to high soil resistivity, and high current requirements, an impressed

current system is to be provided for an existing sheet pile bulkhead about 660 LF in
length by 48 FT high and is constructed of PZ 27 steel sheet piles. The sheet pile was
coated between the 8 and 40 Ft depths, and the top 10 FT of the sheet pile was
encased in a reinforced concrete pile cap. Electrical continuity across each pile is
provided by existing bonding straps welded across each pile joint. The bonding straps
are electrically connected together by bonding wires. The bulkhead is anchored via
steel tie rods to an 625 FT long by eight feet high steel sheet pile deadman set back 85
FT from the bulkhead. The tie rods are coated, wrapped and enclosed in PVC sleeves
along its entire length except near the turnbuckle. Electrical continuity to and across the
turnbuckles is provided by existing bond wires between the turnbuckle and the rod
connected to each side of the turnbuckle. The deadman is constructed of coated steel
PZ 27 sheet piles. Electrical continuity across each pile is provided by existing bonding
straps welded across each pile joint. Refer to Technical Paper 17 for example
calculations for a water side galvanic anode system.

5.12.1 Design Data.

A. Seawater Resistivity - 20 ohm-centimeters

B. Design for 2 milliamperes per square foot in the soil, and 1 ma/SF for steel
in concrete.
C. High silicon chromium bearing cast iron anodes will be used.
D. The design structure to electrolyte potential for the protected structure will
be – 850 mv.
E. Design life: 15+ years
F. Coating Efficiency:
Steel sections encased in concrete: 90% (0.9)
Steel sections exposed to seawater: 80% (0.8)
Steel sections below mudline: 70% (0.7)

5.12.2 Calculations.

5.12.2.1 Calculate Total Surface Area of Sheet Pile.

1. Bulkhead. The bulkhead is constructed of PZ-27 steel sheet piles. PZ-27

sheet pile has a width of 18 inches and a surface area of 2.5 SF/LF on
face of the pile. The number of sheet piles (NSP) can be calculated as
follows:

Running Length of sheet pile wall (in inches)

NSP =
18 inches

The lengths of the bulkhead and deadman anchor wall are 660 LF and
625 LF respectively. The following table summarizes the number of piles
in each wall.

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SHEET PILE WALL RUNNING NO. OF SHEET

LENGTH (INCHES) PILES (NSP)
BULKHEAD 7,920 440
ANCHOR WALL 7,500 417

The surface area for the bulkhead can be calculated using the following
equation:

ASP = NSP X Length of sheet pile X 2.5 SF/LF

The bulkhead can be divided into the following zones:

• Land side, encased in concrete, sheet pile bare

• Land side, encased in concrete, sheet pile coated
• Land side, exposed to soil, sheet pile coated
• Land side, exposed to soil, sheet pile bare

The following table summarizes the surface areas for each zone.

No. of Pile Length Surface

Sheet Pile Zone
Piles (LF) Area (SF)
Bare 440 8 8,800
Concrete
Coated 440 2 2,200
Land side
Bare 440 8 8,800
Soil
Coated 440 30 33,000

2. Tie Rods. There are 153 tie rods, coated, wrapped, and installed in PVC
sleeves. Only the portions of the tie rod at the turnbuckle and on the
back side of the anchor wall are exposed to the soil. An estimated length
of two LF will account for the exposed tie rod and turnbuckle. The tie
rods are 2.5 inches in diameter. Since the tie rods are threaded at the
turnbuckle and anchor wall, assume that the steel is bare. Calculate the
surface area as follows.

ATR = π x d x L x Number of tie rods

= 3.14 x (2.5"/12") x 2 LF x 153
= 200 SF

3. Anchor Wall. Both sides of the anchor wall are coated and exposed to
the soil. The surface area for both sides of the pile is 5 SF/LF. The

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length of each anchor wall pile is 8 LF. The total anchor wall surface
area is:

AAW = 417 piles X 8 LF 5 SF/LF

= 16,680 SF

5.12.2.2 Calculate the Current Required for Protection of the Sheet Pile.
The current requirement is calculated using the equation:

IR = Surface area X (1 - CE) x J

where
IR = Cathodic protection current required
CE = coating efficiency
J = current density

The following table summarizes the current requirements.

LAND SIDE
Structures/Zones Surface J Current
CE
area (SF) (ma/SF) Req'd (ma)
Bare 8,800 0 1 8,800
Concrete Coated 2,200 0.9 1 220
Sheet pile
25% of above for concrete cap rebar 2,255
Bulkhead
Bare 8,800 0 2 17,600
Soil
Coated 33,000 0.7 2 19,800
Tie Rods Soil Bare 200 0 2 400
Sheet pile
Soil Coated 16,680 0.7 2 10,008
Anchor Wall
Total Land Side Current Required 59,083
or 59.1 amps

Use an 80 ampere rectifier. The extra current capacity will allow for additional
coating deterioration, system aging, and current loss to the water side of the sheet pile.

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5.12.2.3 Calculate the Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes. The
quantity of anodes must be calculated to meet three different parameters,
design life, anode bed resistance, and anode maximum current discharge.
The required quantity of anodes will be the largest of the three calculated
quantities.

1. First calculate the quantity of anodes to meet design life

Y S IR
NL =
WU

where:
IR = Current required = 59.1 amps
NL = Number of anodes to meet design life
S = Anode consumption rate = 1 lb/amp-yr)
U = Anode utilization factor = 0.8
W = Weight of one anode (lb)
Y = CP system design life = 20 years

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the recommended maximum anode discharge and the ground bed resistance
requirements must still be calculated.

2. Next calculate the quantity of anodes to meet recommended maximum

anode current discharge

I
ND =
AA x IA

where:
I = Current required (59.1 amps)
AA = Anode surface area (SF)
IA = Max. recommended anode discharge current density (1 amp/SF)

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

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3. Calculate quantity of anodes to meet 1-ohm ground bed design

The anodes will be installed in a shallow, distributed, vertical anode bed.

The resistance of the anode bed can be approximated by the following
equation:

0.0052 ρ
NR = x [ ln (8L/d) - 1 ]
RS x L

where
d = diameter of anode backfill column (1 FT)
L = length of the anode backfill column (LF)
NR = quantity of anodes to meet the anode bed design resistance
RS = anode bed design resistance (1 ohm)
ρ = soil resistivity (6,000 ohm-cm)

4 Determine quantity of anodes. The following table summarizes the

above calculations for various common tubular anode sizes:

CALCULATED ANODE QUANTITIES

Weight Area Backfill Backfill
Anode Size NL ND NR
(LB) (SF) Length (LF) Dia. (ft)
2.66" x 42" (TA1) 31 2.4 5.5 1 46 25 16
2.19" x 42" (TA2A) 23 2.0 5.5 1 62 30 16
2.19" x 60" (TACD) 32 2.8 7.0 1 45 22 14
2.66" x 60" (TAD) 45 3.5 7.0 1 32 18 14
3.75" x 60" (TAM) 60 4.9 7.0 1 23 13 14
4.75" x 60" (TAJ) 78 6.2 7.0 1 19 10 14
2.66" x 84" (TA3) 63 4.9 9.0 1 23 13 12
3.75" x 84" (TA4) 85 6.9 9.0 1 17 9 12
4.75" x 84" (TA5) 110 8.7 9.0 1 13 7 12

Of the three calculated quantities for a particular anode, the larger

quantity of the three must be used to ensure all conditions are satisfied.
Considering CP current distribution, use the 23 each 60 LB, 3.75 inch
diameter X 60 inch long anodes. Install the anodes so that the bottom
of the anode is about one foot above the bottom of the hole. Provide
anode with factory connected lead wires of sufficient length to reach the
anode junction box without splicing.

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5.12.2.4 Calculate the Total Circuit Resistance (RT).

1. Calculate the anode bed circuit resistance (Rs) using the appropriate
numbers from the above table:

0.0052 ρ
RS = x [ ln (8L/d) - 1 ]
N x L

= 0.56 ohm

2. Calculate the anode cable resistance (RC)

Each anode lead will run directly back to the anode junction box, so the
parallel cable resistance is estimated to be equivalent to a 230 LF length
of cable. Use No. 4 copper wire for the anode leads. The cable
resistance is estimated as follows:

RC = R#4 x length of cable

= 0.00025 ohm/LF x 230 LF

= 0.06 ohm

3. Calculate total circuit resistance (RT)

The total circuit resistance is 0.56 ohm + 0.06 ohm = 0.62 ohm

5.12.2.5 Calculate the Rectifier Output Voltage (VR) Using Ohm's Law.

VR = Irect x RT x 125%
= 59.1 amps x 0.62 Ω x 1.25
= 45.8 volts

Use the next highest nominal size rectifier rated for 50 volts DC.

5.12.2.6 Calculate the Rectifier Input Power

Available power is 480/277 volts. For rectifiers of the size specified above,
use three phase 480 volt AC input voltage. Three phase rectifiers are more
efficient and long term power costs outweigh the higher initial costs for the
rectifier. Rectifier input power is calculated as follows:

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VR x IR
P =
rectifier efficiency

50 volts x 80 amps
=
0.85

= 4,705 VA say 5 KVA

AC input current is calculated to be:

P
IAC =
VAC x 1.732

5,000 VA
=
480 volts x 1.732

= 6 amps

The sheet pile bulkhead is not located near a building where an electrical
panel can be used. Therefore, the rectifier will be located adjacent to a three
phase, 480 volt transformer station.

5.12.2.7 Short Circuit Calculations

A. Transformer Full Load Current (IFL)

VA
IFL =
VAC x √3

500,000 VA
=
480 volts x 1.732

= 601 amps

B. Fault Current at Transformer Secondary (ISC).

IFL
ISC =
%ZT/100

601
=
4.5/100

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= 13,356 Amps

Therefore, use 18,000 amp I.C. service breaker.

5.12.2.7 Summary

Anodes: 23 each high silicon chromium bearing cast iron tubular

anodes, with dimensions of 3.75 inches in diameter x 60
inches long, and weighing 63 LB. Locate the anodes 22 feet
from each end of the bulkhead and at 28 feet spacing in
between.

Anode Supplied by manufacturer consisting of #4 AWG wire with

HMWPE insulation. One end is factory connected to the
anode. The anode leads should be long enough to extend to
the junction box without splicing.

Rectifier:
Type: Oil cooled, enclosure suitable for outdoors use.
AC Input: 480 volt, 6 amps, 60 Hz, three phase
DC Output: 50 volts, 80 amps
Efficiency: 85% minimum

End of Example

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Figure 40
Cathodic Protection of Multiple Underground Storage Tanks

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where

AT = Surface area of storage tank (ft²)

rT = Radius of tank = 48 in = 4 ft
DT = Diameter of tank = 96 in = 8 ft
LT = Length of tank = 21 ft - 3 in

AT = [2 x 3.1416 x (4)2] + 3.1416 x 8 x 21.25

= 634.6 say 635 ft²

For all three tanks, the total surface area is

AT = 3 x 635 ft²

= 1905 ft²

B. Piping

AP = π DP LP

where
AP = Surface area of piping (ft²)
DP = Diameter of pipe = 2.375 in. or 0.198 ft for 2 in. nominal size pipe
LP = Length of piping = 750 ft

AP = 3.1416 x 0.198 x 750

= 467 ft²

C. Total Surface Area

AP = AT + AP

A = 1905 + 467

= 2372 ft²

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5.3.2.2 Verify the Current Requirement (I).

I = (A) (I’) (1 – CE)

where

I = Current requirement (amps)

A = Total surface area of tanks and piping = 2372 ft²
I’ = Assumed current density = 2 mA/ft²
CE = Pipe/tank coating efficiency = 0.00 for bare steel

I = 2372 x 2 x (1 – 0)

= 4744 mA or 4.7 amps

The 4.7 amp would be reasonable for the facility if it were electrically isolated
from other buried metals such as the building ground system. The actual
current requirement of 8.2 amp occurs because of current loss to these other
buried metal structures and is also reasonable in relation to that calculated for
an isolated facility.

5.3.2.3 Select An Anode and Calculate the Number of Anodes Required (NL) to
Meet the Design Life Requirements.

Calculations can be run on several size anodes, but in this case 2-in. by 60-
in. packaged ceramic rod anodes (rod size = 0.125 in x 4 ft long) are chosen
for ease of construction. Using the following equation, the number of anodes
required to meet the cathodic protection system design life can be calculated:

I
NL =
IA

where
I = Current requirement = 8.2 amps
IA = Ceramic rod anode current rating = 1.0 amps/anode
NL = Quantity of anodes to meet design life

8.2
N =
1.0

= 8.2 use 9 anodes

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5.3.2.4 Calculate the Quantity of Anodes to Meet the Maximum Anode Bed
Resistance of 2.5 Ohms (NR).

Ra = [(0.0052ρ)/(NR)(L)] X [ln (8L/d) - 1 + (2L/S)(ln 0.656NR)]

where

NR = Quantity of anode/backfill columns

Ra = Anode ground bed resistance (2.5 ohms maximum)
ρ = Soil resistivity = 4,500 ohm-cm
L = Anode backfill column length = 5 feet
d = Anode backfill column diameter = 2 in. = 0.167 feet
I = Current requirement = 8.2 amperes
S = Spacing between anodes. Assume 10' spacing between anode/backfill
columns. A distributed anode array does not lend itself to an exact
calculation using the equation above because the anodes are positioned
at various locations and are not located in a straight line. The above
equation assumes a straight-line configuration; however, to approximate
the total anode-to-earth resistance, the equation may be used.

Initially, try 9 anodes, the quantity required to meet the design life

0.0052 x 4500 8x5 2x5

RA = [ ln ( )-1 + ( ) (ln (0.656 x 9)) ]
9x5 0.167 10

= 3.26 ohms

The resistance for 9 anodes is too high. Additional calculations using an

increasing number of anodes (i.e., 11, 12, 13, 14, etc.) have to be made.
These calculations show that fourteen anodes will yield a groundbed-to-earth
resistance of 2.24 ohms.

Of the above-calculated quantities for the anodes, 9 to meet the design life
and 14 to meet the maximum anode bed resistance requirement, the larger
quantity of the two must be used to ensure all conditions are satisfied.
Therefore, use 14 each 0.125 in x 4 ft long ceramic rod anodes pre-packaged
in backfill in 2-in. by 60-in. canisters.

5.3.2.5 Calculate the Total Circuit Resistance (RT).

RT = RA + Rw + RC

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where,

RT = Total circuit resistance

RA = Anode bed resistance
Rw = Header cable/wire resistance
RC = StructurTP16-to-earth resistance

A. Ground bed resistance (RA) = 2.24 ohms from section TP16-5.3.2.4

B. Header cable/wire resistance (RW):

LW x RMFT
RW =
1000 ft

where,

Rw = Header cable/wire resistance

LW = Effective cable length. The loop circuit makes calculating effective wire
resistance complex. Since current is discharged from anodes spaced
all along the cable, one-half the total cable length may be used to
approximate the cable resistance. Total cable length = 300 ft. Effective
cable length = ½ x 300 ft = 150 ft.)

RMFT = Resistance per 1000 lineal feet of No. 4 AWG cable that has been
selected for ease of handling = 0.254 ohms/1000 LF.

150 ft x 0.254 ohm

RW =
1000 ft

= 0.038 use 0.04 ohm

C. Structure-to-earth resistance (RC):

Since the tanks and piping are essentially bare and are not electrically
isolated, structure-to-earth resistance may be considered negligible.
Therefore, RC = 0

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D. Calculate total resistance (RT):

RT = RA + Rw + RC

= 2.26 + 0.04 + 0

= 2.30 ohms

Since the design requirements call for a maximum ground bed resistance
of 2.5 ohms and RT = 2.30 ohms, the design using fourteen 2-in. by 60-in.
packaged ceramic anodes will work.

5.3.2.6 Calculate the Rectifier Voltage (VREC).

VREC = (I) (RT) (120%)

where,
VREC = Rectifier Voltage (volts)
RT = Total circuit resistance = 2.30 ohms
I = System current requirement = 8.2 amps
120% = Rectifier voltage capacity design safety factor

VREC = 8.2 amp x 2.3 ohms x 1.2

= 22.6 Volts

5.3.2.7 Select Rectifier.

Based on the design requirement of 22.6 V and 8.2 amp, a rectifier can be
chosen. A 12-amp, 24-V unit is selected because this is the nearest standard
commercial size available.

End of Example

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5.4 Elevated Steel Water Tank. This impressed current CP design is for an
elevated steel water tank that has not been built. Hence, it is not possible to determine
the current requirements, etc., by actual measurements. Calculated estimates have been
used.

5.4.1 Design Data.

A. Tank capacity - 500,000 gallons

B. Tank height (from ground to bottom of bowl) - 115 feet
C. Diameter of tank - 56 feet
D. High water level in tank - 35 feet
E. Overall depth of tank - 39 feet
F. Vertical shell height - 11 feet
G. Riser pipe diameter - 5 feet
H. Shape of tank - Ellipsoidal, both top and bottom
I. All internal surfaces are uncoated
J. Design for maximum current density - 2 mA/ft²
K. Electric power available 120/240 V ac, single-phase
L. String-type HSCBCI anodes are used
M. Design life - 10 years
N. Water resistivity - 4,000 ohm-cm
O. Tank water must not be subject to freezing
P. Assumed deterioration rate - 1.0 lbs/A yr
Q. Anode efficiency (assumed) - 50 percent

5.4.2 Calculations.

5.4.2.1 Area of Wetted Surface of Tank Bowl (see Figure 41).

A. Top section (T)

AT = 2 π r x (approximate)

where

r = 28 feet (radius of tank)

x = 10 feet
AT = 2 x 3.1416 x 28 x 10
AT = 1759 ft²

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Figure 41
Segmented Elevated Tank for Area Calculations

B. Center section (C):

Ac = 2πrh

where

r = 28 feet (radius of tank)

X = 11 feet
AC = 2 x 3.1416 x 28 x 11
AC = 1935 ft²

C. Bottom section (B):

AB = √2 π r √a2 + r2
where

r = 28 feet (radius of tank)

a = 14 feet

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 Electrical TP 16
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AB = √2 x 3.1416 x 28 √14² + 28²

AB = 3,894 ft²

D. Total wetted area of tank bowl:

ATB = AT + AC + AB
= 1,759 + 1,935 + 3,894
= 7,588 ft²

5.4.2.2 Area of Riser Pipe.

A = 2 π rR h R

where

rR = 2.5 feet (radius of riser)

hR = 115 feet (height of riser)

AR = 2 x 3.1416 x 2.5 feet x 115

= 1,806 ft²

5.4.2.3 Maximum Design Current for Tank.

IT = 2.0 mA/ft² x 7,588 ft²

= 15,176 mA or 15.2 A

5.4.2.4 Maximum Design Current for Riser.

IR = 2.0 mA/ft² x 1,806 ft²

= 3,612 mA or 3.6 A

5.4.2.5 Minimum Weight of Tank Anode Material.

W = YSI/E

where

W = weight of anode material =

Y = design life = 10 years
S = anode deterioration rate = 1.0 lb/A-yr
I = maximum design current = 15.2 A
E = anode efficiency = 0.50

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10 x 1.0 x 15.2
W =
0.50

= 304 pounds

5.4.2.6 Minimum Weight of Riser Anode material.

W = YSI/E

where

Y = 10 years
S = 1.0 lbs/A yr
I = 3.62 A
E = 0.50

W = 10 x 1.0 x 3.62/0.50
= 72.4 pounds

5.4.2.7 Radius of Main Anode Circle.

DN
W =
2 (π + N)

where

D = 56 feet
N = Assumed number of anodes = 10

56 x 10
W =
2 (3.1416 + 10)

= 21.3 feet, use 22 feet

5.4.2.8 Spacing of Main Anodes. Generally the distance from the anode to the tank
wall and tank bottom is about equal; this distance should be about one-half
the circumferential distance between anodes.

A. Circumferential spacing:

2πr
C =
N

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where

r = radius of anode circle = 22 feet

N = assumed number of anodes = 10

2 x 3.1416 x 22
C =
10

= 13.8 feet, use 14 feet

B. Cord spacing is approximately the same as circumferential spacing: 14

feet will be used (see Figure 42).

Figure 42
Anode Spacing for Elevated Water Tank

5.4.2.9 Selection of Main Anodes.

A. Size of anode units selected is 1-1/8 inch outer diameter by ¾ inch inside
diameter by 9 inches long. This is a standard sausage type anode that
weighs one pound, and has an effective surface area of 0.25 ft2.

B. The minimum number of anode units per anode string, based on a

required weight of 304 pounds and 10 anode strings is computed as
follows:

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 Electrical TP 16
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304
Number of units =
10 x 1

= 30.4 say 31 units per string

C. Because the internal tank surfaces are uncoated, a maximum structure-to-

electrolyte potential is not a limiting factor. However, because it is desired
to limit the anode current at or below the manufacturer’s recommended
discharge current rate of 0.025 Amp for this type of anode, the minimum
number of anode units per string will be

15.2 Amps
Number of units =
10 x 0.025

= 60.8 say 61 units per string

This quantity of anode units per string is not practical for the tank bowl
since the distance between the anode hanger and the bottom of the bowl
is only 28 feet. Table 15 shows the maximum recommended discharge
current rate per anode for various types of anodes to ensure a minimum
10-year life. Using the type B anode, only three anode units per string are
required. The manufacturer does not recommend the use of more than
two type B anodes units per anode string assembly because of their
fragile nature. Therefore, the best anode unit choices for the main anode
strings are the type C or type CDD. Type CDD is recommended because
the lead wire connection is protected longer by the thicker wall of the
enlarged ends. Two type CDD anodes per string provide a current
capacity of 2 A x 10 strings = 20 A. These anodes are spaced as shown
in Figure 43.

NOTE: The anodes chosen in this example were chosen to illustrate

some of the many technical considerations during the design of cathodic
protection. For this example if HSCBCI anodes are tubular anodes, 2-
3/16 inch by 8 inch, weighing 4.3 pounds each, should be used instead of
CDD anodes.

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TABLE 15. TECHNICAL DATA - COMMONLY USED HSCBCI ANODES

ANODE ANODE MAX.

MAX. SURFACE CURRENT
DISCHARGE AREA (ft²) DENSITY
TYPE SIZE (in.) WEIGHT (lb)
(A) (A/ft²)
TA-FW 2-3/16 x 8 4.3 0.025 0.22 0.1
FW1 1-1/8 OD x 9 1 0.025 0.2 0.1
FC2 1-1/2 x 9 4 0.075 0.3 0.25
G-2 2 OD x 9 5 0.100 0.4 0.25
G-2-1/2 2-1/2 x 9 9 0.20 0.5 0.40
B3,4 1 x 60 12 0.50 1.4 0.36
C 1-1/2 x 60 25 1.00 2.0 0.50
CDD3 1-1/2 x 60 26 1.00 2.0 0.50
M3 2 x 60 60 2.5 2.8 0.9
SM 4-1/2 x 60 20 10.0 5.5 1.8
K-6 6 x 2-1/2 16 0.225 0.5 0.45
K-12 12 x 3-7/16 53 0.80 1.0 0.80
B-30 1 x 30 7 0.25 0.7 0.36
TA-2 2-3/16 x 84 46 6.4 4.0 1.6
1
For elevated fresh water tank.
2
For distributed system in ground trench.
3
Each end enlarged with cored opening for wire.
4
Not more than 2 anodes per assembly.

D. Anode current density is computed as follows:

15.2 Amps
Anode current density =
2 x 10 x 2

= 0.38 A/ft2

5.4.2.10 Resistance of Main Anodes.

0.012 ρ log (D/a)

R =
L

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where

ρ = 4,000 ohm-cm
D = 56 feet
L = 2 x 5 feet = 10 feet
a = 44 x 0.275 = 12.1 feet (0.275-equivalent diameter factor from curve - see
Figure 8-61)

0.012 x 4,000 x log (56/12.1)

R =
10

= 3.19 ohms

A. However, the L/d ratio of two 1-1/2-inch diameter by 60-inch long anodes
in tandem is less than 100 and thus the fringe factor must be used.

2 x 60
L/d =
1.5

= 80 < 100

B. The fringe factor from the curve in Figure TP16-45 corresponding to the
L/d ratio is 0.95.

R (adjusted) = 0.95 R = 0.95 x 3.19 = 3.03 ohms

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Figure 43
Anode Suspension Arrangement for Elevated Steel Water Tank

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 Electrical TP 16
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Figure 44
Equivalent Diameter for Anodes in a Circle in a Water Tank

5.4.2.11 Stub Anodes.

A. In the design of an elevated water tank, the need for stub anodes must be
justified. The main anode radius has been calculated to b 22 feet. The
main anodes are spaced to provide approximately the same distance from
the sides and the bottom of the tank. The main anodes will protect a
length along the tank bottom equal to 1-1/2 times the spacing o the anode
from the bottom.

B. The anode suspension arrangement for the tank under consideration is

shown in Figure 43. Thus, it can be seen that stub anodes are required
for this design. Ten stub anodes are arranged equally space on a
circumference that has a radius of 8 feet in a manner illustrated in Figure
43. For smaller diameter tanks, stub anodes may not be required.

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Figure 45
Fringe Factor for Stub Anodes

5.4.2.12 Current Division Between Main and Stub Anodes.

A. Area of tank bottom protected by stub anodes (see Figure 43):

As = π (r22 – r12)

where
r = radius of protected segment (13 feet)
2
r = radius of riser (2.5 feet)
1

As = 3.1416 (169 - 6.25)

= 511.3 ft²

B. Maximum current for stub anodes:

Is = 2.0 mA/sf x 511.3 ft²

= 1022.6 mA or 1.02 A

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C. Maximum current for tank bowl = 15.2 A.

D. Maximum current for the main anodes:

Im = 15.2 - 1.02
= 14.2 A

5.4.2.13 Rectifier Voltage Rating.

A. Circuit resistance of electrical conductor to main anodes. Wire size No. 2

AWG, 0.159 ohm/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohm

B. Voltage drop in main anode feeder:

E = IR

where

I = 14.2 A
R = 0.032 ohm

E = 14.2 x 0.032
= 0.45 V

C. Voltage drop through main anodes:

E = IR

where

I = 14.2 A
R = 3.03 ohms

E = 14.2 x 3.03
= 43.0 V

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D. Total voltage drop in main anode circuit:

ET = 0.45 + 43.0
= 43.45 or 45.0 V

Use a multiplying factor of 1.5, or 67.5 V.

E. The nearest commercially available rectifier meeting the above

requirement is a single-phase, 80-V unit.

5.4.2.14 Selection of Stub Anodes. Because it is desirable to use as small an anode

as possible without exceeding the manufacturers' recommended rate, try using type FC,
HSCBCI anode measuring 1-1/2-inch by 9 inches. Use one anode per string as shown
in Figure 43. Compute anode current density as follows:

1.02
Output =
10 x 0.03

= 0.34 A/ft²

Because this exceeds the recommended maximum anode current density

(refer to Table 15), the Type B anode is the best choice.

5.4.2.15 Resistance of Stub Anodes.

0.012 ρ log (D/a)

R =
L

where

ρ = 4,000 ohm-cm
D = 56 feet
L = 5 feet
a = 16 x 0.275 = 4.4 feet (factor from Figure 44)

0.012 x 4,000 x log (56/4.4)

R =
5

= 10.6 ohms

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60
L/d =
1

= 60 < 100

The fringe factor from the curve in Figure 45 corresponding to the L/d ratio
is 0.9.

R (adjusted) = 10.6 x 0.90 R = 9.54 ohms

5.4.2.16 Voltage Drop in Stub Anode Circuit.

A. Electrical conductor to stub anodes. Wire size No. 2 AWG, 0.159

ohms/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohm

B. Voltage drop in stub anode feeder.

E = IR

where

I = 1.02 A
R = 0.032 ohm

E = 1.02 x 0.032
= 0.033 V

C. Voltage drop in anode suspension conductors. Estimated length 50 feet,

No. 2 AWG, 0.159 ohms/1,000 feet:

50
R =
1000 x 0.159

= 0.008 ohm

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E = IR

where

I = 1.02/10 = 0.102 A
R = 0.008 ohm

E = negligible

D. Voltage drop through stub anodes:

E = IR

where

I = 1.02 A
R = 9.54 ohms

E = 1.02 x 9.54
= 9.73 V

E. Total voltage drop in stub anode circuit.

ET = 0.033 + 9.73
= 9.73 V

F. Since the stub anode voltage is below the 45 V calculated for the main
tank anode circuit, the necessary current adjustment can be accomplished
through a variable resistor in the stub anode circuit.

5.4.2.17 Stub Anode Circuit Variable Resistor.

A. Criteria for variable resistor. The resistor should be capable of carrying

the maximum anode circuit current and have sufficient resistance to
reduce anode current by one-half when full rectifier voltage is applied to
the anode circuit.

B. Stub anode circuit data:

Rectifier output = 80 V
Anode current = 1.02 A
Anode resistance = 9.54 ohms

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C. Variable resistor rating:

R = E/I

where

E = 80 V
I = 1.02/2 or 0.51 A

R = 80/0.51
= 156.9 ohms

Ohmic value of resistor = 156.9 - 9.54 = 147.4 ohms

Wattage rating of resistor = (1.02)² x 147.4 = 153.4 W

The nearest commercially available resistor size meeting the above

requirements is a 175-W, 200-ohm, 1-A resistor.

5.4.2.18 Resistance of Riser Anodes. In order to get the maximum desired current
in the riser (3.62 A), the resistance limit is calculated as follows:

R = E/I

where

E = 43.45 V
I = 3.62 A

R = 43.5/3.62
= 12.0 ohms

5.4.2.19 Riser Anode Design.

A. Type FW (1-1/8-inch by 9-inch) string type anodes cannot be used in the

riser because the maximum anode current discharge of 0.025 A per anode
would be exceeded. The number of type FW anodes required would be
145 and continuous throughout the riser. This is excessive. The best
choice of anode for a flexible riser string is the type G-2 (2-inch by 9-inch)
high-silicon cast iron anode.

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B. Number of units required:

(0.012 ρ log D/d)

R =
L

(0.012 ρ log D/d)

L =
R

where

ρ = 4,000 ohm-cm
D = 5 feet
D = 2 inches or 0.166 feet
R = 12 ohms

0.012 x 4000 x log (5/0.166)

L =
12

= 5.92 feet

Number of units = 5.92/0.75 = 7.9 or 8 units

In order to get proper current distribution in the riser pipe, the anode units
should not be placed too far apart. It is generally considered that each
anode unit protects a length along the riser pipe equal to 1-1/2 times the
spacing of the anode from the riser pipe wall.

Riser height = 115 feet

Spacing (center of anode to tank wall) = 2.5 feet
Length of riser protected by one anode = 1.5 x 2.5 = 3.75 feet
Number of units required = 115/3.75 = 30.7 or 31 units.

To satisfy the maximum anode discharge current for a G-2 anode:

3.62 A/0.1 amp = 36

Therefore, 36 anodes are needed instead of 31 or 8.

C. Anode resistance based on the use of 36 anode units:

R = (0.012 r log D/d)/L

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where

r = 4,000 ohm-cm
D = 5 feet
D = 2 inches or 0.166 feet
L = 36 x 9 inches = 324 inches or 27 feet

0.012 x 4000 x log (5/0.166)

L =
27

= 2.63 ohms

L/d ratio for the riser anode string is 324/2 or 162; thus no fringe factor
correction is applied.

5.4.2.20 Voltage Drop in Riser Anode Circuit.

A. Electrical conductor to riser anodes. Wire size No. 2 AWG, 0.159

ohms/1,000 feet, estimated length 200 feet:

200
R =
1000 x 0.159

= 0.032 ohms

B. Voltage drop in riser anode feeder:

E = IR

where

I = 3.62 A
R = 0.032 ohm

E = 3.62 x 0.032
= 0.116 V

C. Voltage drop in riser anode suspension cables. Wire size No. 2 AWG,
0.159 ohm/1,000 feet, estimated length 130 feet:

R = 130/1,000 x 0.159 = 0.02 ohm

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E = IR

where

I = 3.62/2
= 1.81 A average (single current does not flow the full length of the anode
string)
R = 0.02 ohm

E = 1.81 x 0.02
= 0.04 V

D. Voltage drop through riser anodes:

E = IR

where

I = 3.62 A
R = 2.63 ohms

E = 3.62 x 2.63
= 9.52 V

E. Total voltage drop in riser anode circuit:

ET = 0.116 + 0.04 + 9.52

= 9.69 V

5.4.2.21 Riser Anode Circuit Variable Resistor.

A Criteria for the variable resistor are the same as given for the stub anode
resistor.

B. Riser anode circuit data:

Rectifier output = 80 V
Anode current = 3.62 A
Anode resistance = 2.63 + 0.032 + 0.02 = 2.68 ohms

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C. Variable resistor rating:

R = E/I

where

E = 80 V
I = 3.62/2 = 1.81 A

R = 80/1.81
= 44.2 ohms

Ohmic value of resistor = 44.2 - 2.68 = 41.5 ohms.

Wattage rating of resistor = (3.62)² x 41.5 = 543.8 W

The resistor should reduce anode current by one-half when full rectifier
voltage is applied. The nearest commercially available resistor size that
meets the above requirements is a 750-W, 50-ohm, 3.87-A resistor
(rheostat). This rheostat is 10 inches in diameter and 3 inches in depth,
and fairly expensive. This rheostat will not fit into most rectifier cases. In
addition, the power consumed by the rheostat is considerable. This power
creates substantial heat that may damage components within the rectifier
case unless adequate ventilation is provided. The problems associated
with using a large rheostat can be eliminated by using a separate rectifier
for the riser anodes. Although initial cost may be slightly high, power
savings will be substantial and damage by heat will be avoided.

5.4.2.22 Sizing Rectifier for Riser.

A. Requirements:

DC current output = 3.62 A

Anode circuit resistance = 2.68 ohms
DC voltage required = IR = 3.62 x 2.68 E = 9.70 V

B. Rectifier rating. Standard ratings for a rectifier in this size class are 18
Volts, 4 Amps.

5.4.2.23 Rectifier DC Rating for Bowl. Voltage output as previously determined, 80

V. Current rating is 15.2 A. The nearest commercially available rectifier meeting the
above requirements is 80 V, 16 A.

5.4.2.24 Wire Sizes and Types. All positive feeder and suspension cables (rectifier to
anodes) must be No. 2 AWG, HMWPE insulated copper cable. To avoid complication,
the negative rectifier cable (rectifier to structure) must be the same size and type (see
Figure 46).

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5.4.2.25 Discussion of the Design.

A. The design points out the disadvantages of achieving corrosion control

through cathodic protection without the aid of a protective coating. When
the interior of a tank is coated, the current requirement is reduced from 60
to 80 percent. On large tanks without coating, larger size and more
expensive anodes, wire, and rectifier units must be used. In addition, the
power consumed by the uncoated tank is far greater. These additional
costs usually exceed the cost of a quality coating system over 10-year
period. Corrosion above the water line of a water storage tank is usually
severe because of the corrosive nature of condensation. For this reason,
protective coatings must be used above the water line on both large and
small water storage tanks to mitigate corrosion.

B. For further assistance and guidance in the design of cathodic protection

systems for elevated water storage tanks, see Figures 46 through 48.

C. The HSCBCI anodes were selected for this particular design purely for
illustrative purposes. It does not mean that this material is superior to
other types of anode material. Other acceptable anode materials include
platinized titanium or niobium and mixed metal oxide (ceramic) anodes.
With the advent of newer tubular center connected anodes, the designer
should choose these anodes over the end connected in most cases
because of their higher current capability and longer life.

D. For this design, silicon stacks should be specified for the rectifier that
protects the bowl and selenium stacks should be specified for the rectifier
that protects the riser. Silicon stacks operate more efficiently at high DC
output voltages than selenium stacks do but require elaborate surge and
overload protection. This protection is not economical in the low power
consuming units. A guide for selection of rectifying cells is as follows:

• Use silicon stacks for single-phase rectifiers operated above 72 V dc or

three-phase rectifiers operated above 90 V dc.
• Use newer selenium stacks for single-phase rectifiers operated below
72 V dc or threTP16-phase rectifiers operated below 90 V dc.

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Figure 46
Elevated Steel Water Tank Showing Rectifier and Anode Arrangement

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Figure 47
Hand Hole and Anode Suspension Detail for Elevated Water Tank

Figure 48
Riser Anode Suspension Detail for Elevated water Tank

End of Example

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5.5 Elevated Steel Water Tank Where Ice Is Expected.

Impressed current cathodic protection is designed for an elevated steel water tank (Figure
49). The tank is already built and current requirement tests have been conducted.
Anodes cannot be suspended from the tank roof because heavy ice (up to two feet thick)
covers the water service during the winter. The anode cables could not tolerate this
weight; so another type of support must be used. An internally supported hoop shaped
wire anode system is selected.

5.5.1 Design Data.

a. The water tank is a pedestal spheroid with a ten-inch riser pipe. Only the
bowl will be protected because the riser pipe is less than 30 inches in
diameter.

b. Tank dimensions are:

Capacity - 400,000 gallons

Diameter of tank bowl – 51 ft 6 in
High water level in tank - 35 feet
Tank height (from ground to bottom of bowl) - 100 feet

c. Water resistivity - 2,000 ohm-cm

d. Anode Design life - 15 years

e. Wire type ceramic anodes to be used

f. All wetted surfaces are uncoated

g. Area above the high water level is kept well coated

h. Tank water is subject to freezing

i. The cathodic protection circuit resistance must not exceed 2 ohms

j. Electric power available is 120/240 V ac, singlTP16-phase

k. Based on structure current requirement testing on this tank, the current

required for adequate cathodic protection is 25 amps. This high current
requirement indicates that the tank internal coating is severely
deteriorated.

5.5.2 Calculations.

5.5.2.1 Calculate the Length of Wire in Feet (LB) Needed for the Current
Required.

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I
LB =
IA

where

I = Current requirement for adequate protection = 25 amps

IA = Allowable amp per foot of anode wire (varies depending on desired
anode life and diameter

Select the 0.0625 in. diameter copper cored anode wire based on the
current requirement of 25 amps and design life of 15 years.

25
LB =
0.31

= 81 feet

5.5.2.2 Calculate the Desired Diameter of the Anode Wire Ring (DR). Experience
shows that the diameter of the anode wire ring should be between 40 and 70 percent of
the bowl diameter.

A. Try 40% for the first iteration.

DR = 51.5 ft x 40% = 20.5 ft

Check to determine if the length is adequate for the desired anode life.
For an anode ring diameter of 20.5 feet the circumference (anode wire
length) is

CR = π x DR

= 64.4 feet

This length is inadequate for 0.0625 in. wire anode (which requires a
minimum of 81 feet to meet the desired anode life). Therefore, increase
the wire ring (hoop) diameter to 50 percent of the tank diameter.

DR = 51.5 ft x 50% = 25.75 ft

and

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CR = π x 25.75

= 80.9 feet

This diameter yields an anode length that is still slightly less than that
required for a 15-year anode life. Therefore, use a hoop diameter that is
about 55 percent of the tank bowl diameter.

DR = 51.5 ft x 55% = 28.3, say 29 ft

and

CR = π x 29.0

= 91 feet

5.5.2.3 Calculate the Anode-to-Water Resistance (RA) for the 0.0625 in. Diameter
Anode Wire.

0.0016 r 8 DR 2 DR
RA = ( ln + ln )
DR DA H

where,

r = Water resistivity = 2,000 ohm-cm

DR = Anode ring (hoop) diameter = 29 Ft
DA = Diameter of the anode wire anode = 0.00521 Ft (0.0625 in.)
H = Anode depth below water surface determined from the following
calculations (The anode depth below the high water line is about 60
percent of the distance between the high water line and the tank bottom).

H = 35 Ft x 60% = 21 Ft

0.0016 x 2000 8 x 29 2 x 29
RA = ( ln + ln )
29 0.00521 21

RA = 0.110 x ( ln 44,530 + ln 2.76 )

RA = 1.29 ohms

This is within the design limitation of 2.0 ohms

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5.5.2.4 Determine the Total Circuit Resistance (RT).

RT = RN + RW + RC

where,

RN = Anode-to-water resistance
RW = Wire resistance
RC = Tank-to-water resistance

A. AnodTP16-to-water resistance (RN) = 1.29 ohms from paragraph 5.5.2.3.

B. Header cable/wire resistance (RW).

LW x RMFT
RW =
1000 Ft

where,

LW = 115 Ft (Effective wire length. The positive wires from the rectifier to
each end of the anode ring will be about 115 feet long.
RMFT = 0.57 Ohm (Effective wire resistance per 1000 lineal feet. Since there
are positive wires from the rectifier to each end of the anode ring, each
wire will carry about one half of the current [12.5 amp]. The wires
selected are No. 10 AWG. Since the two wires are in parallel, the
effective resistance is one half the single wire resistance [1.02 ohms per
1000 lineal feet/2 = 0.51 ohm])

115 ft x 0.51 ohm

RW =
1000 Ft

= 0.06 ohm

C. Tank-to-water resistance (RC) and negative circuit resistance.

The negative wire is connected to the tank structure near the rectifier, so its
resistance is negligible. The tank-to-water resistance is also negligible
because the coating is very deteriorated.

D. Calculate (RT):

RT = 1.29 + 0.06 + 0.00

= 1.35 ohm

This is well below the design limitation of 2.0 ohms

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5.5.2.4 Calculate the Rectifier Voltage (VREC).

VREC = I x RT x 120%

where,

I = Current requirement = 25 amps

RT = Total circuit resistance = 1.35 ohms
120% = Rectifier voltage capacity design safety factor

VREC = 25 amps x 1.35 ohms x 120%

= 40.5 Volts

5.5.3 Select Rectifier.

Based on the design requirements of 40.5 volts and 25 amps, a commercially

available 48-volt, 28-amp unit is selected. Specify automatic potential control
to prevent over or under protection as the water level varies. The controller
maintains the tank-to-water potential through two permanent copper-copper
sulfate reference electrodes suspended beneath the anode wire ring. The
reference electrodes should have a life of at least five years. The tank-to-
water potential measured by the controller should be free of IR drop error.

5.5.4 Installation.

Figure 49 shows a typical installation while Figure 50 illustrates a typical detail

for a pressure entrance fitting for underwater power and reference electrode
wire penetrations.

End of Example

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Figure 49
Elevated Pedestal tank with ceramic anode wire ring for icing conditions.

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Figure 50
Pressure entrance fitting for underwater power and reference electrode wire penetrations in water storage tanks.

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5.6 Steel Gas Main. Design an impressed current cathodic protection system for
the 6-inch welded steel gas main shown in Figure 51. The pipeline has not yet been
constructed, so current requirement tests cannot be conducted.

Figure 51
Cathodic Protection System for Gas Main

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5.6.1 Design Data.

A. Average soil resistivity, 2,000 ohm-cm.

B. Pipe size, 6-inch outside diameter, schedule 80 pipe.

C. Pipe length, 6,800 feet.

D. Design for 15-year life.

E. Design for an estimated 2 mA/ft² of bare pipe.

F. Design for 90 percent coating efficiency, based on experience.

G. The pipeline must be isolated from the pump house with a dielectric
insulating flange on the main line inside the pump house.

H. Use HSCBCI anodes with carbonaceous backfill.

I. The pipe is coated with hot-applied coal-tar enamel and holiday checked
before installation.

J. Anode bed resistance must not exceed 2 ohms.

K. Electric power is available at 120/240 V ac, single phase, from a nearby

overhead distribution system.

5.6.2 Calculations.

5.6.2.1 Outside Area of Gas Main.

AP = π D L

where
AP = Outside surface area of the pipe
D = Outer pipe diameter = 6.625 inches for a 6-inch nominal diameter pipe
= 6.625 in/12 in/ft = 0.552 feet
L = Pipe length = 6,800 feet

AP = 3.1416 x 0.552 x 6800

= 11,792 ft²

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5.6.2.2 Area of Bare Pipe to Be Cathodically Protected Based on 90 Percent

Coating Efficiency.

A = AP x (1 – CE)

where
A = Area of bare pipe to be cathodically protected
AP = Outside surface area of the pipe = 11,792 ft²
CE = Pipe coating efficiency = 90% or 0.9

A = 11,792 ft² x (1 – 0.9)

= 1,179 ft²

5.6.2.3 Protective Current Required Based on 2 mA/ft² of Bare Metal.

I = A x CD

where
A = Area of bare pipe to be cathodically protected = 1,179 ft²
I = Current required for cathodic protection
CD = Current density = 2 mA/ft²

I = 1,179 ft² x 2 mA/ft²

= 2,358 mA or 2.36 A

5.6.2.4 Ground Bed Design.

A. Anode size: 2-inch x 60-inch (backfilled 10-inch x 84-inch), spaced 20 feet

apart.

B. Resistance of a single anode to earth:

r
RV = K
L

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where
r = Soil resistivity = 2,000 ohm-cm
K = Shape function (refer to para. 6.2.1.4a) = 0.0167
L = Backfilled anode length = 7.0 feet
L/D = 84 inches/10 inches backfill size

2,000
RV = x 0.167
7.0

= 4.77 ohms

C. Number of anodes required.

One of the design requirements is that the anode bed resistance is not to
exceed 2 ohms. Anode size used is 2-inch diameter x 60 inches long with
carbonaceous backfill having overall dimensions of 10-inch diameter x 84
inches long and spaced 20 feet apart:

1 p
Rn = RV + rs
n s

where
Rn = Anode bed resistance = 2 ohms
n = number of anodes
RV = Single anode resistance = 4.77 ohms
rs = Earth resistivity with pin spacing equal to S = 2,000 ohm-cm
p = paralleling factor (refer to para. 6.2.1.4b)
s = spacing between adjacent anodes = 20 feet

NOTE: p is a function of n as referred in para. 6.2.1.4b, and n is the number of

anodes which are determined by trial and error.

Rearranging the equation for n:

RV
n = p
Rn - ( rs )
s

4.77
n = p
2 - ( 2000 )
20

4.77
n =
2 – 100p

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Try n = 4 anodes, p = 0.00283 (refer to para. 6.2.1.4b)

4.77
4 =
2 – (100 x 0.00283)

4 = 2.78 (not very close)

Try n = 3 anodes, p = 0.00289 (refer to para. 6.2.1.4b)

4.77
3 =
2 – (100 x 0.00289)

3 = 2.79

This is the closest possible. In order to keep total resistance below 2.0
ohms, use 3 anodes.

D. Actual anode bed resistance:

1 0.00289
R3 = ( 4.77 ) + ( 2000 )
3 20

= 1.87 ohms which is less than 2.0

5.6.2.4 Next Calculate the Quantity of Anodes to Meet Recommended Maximum

Anode Current Discharge.

I
ND =
AA x IA

where:

ND = quantity of anodes to meet recommended maximum anode current

discharge limits.
I = Current required (2.36 amps)
AA = Anode surface area (ft2) = 2.6 ft2
IA = Max. recommended anode discharge current density (1 amp/ ft2)

2.36
ND =
2.6 x 1

= 0.91 say 1 anode

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Only one anode is required to stay within the maximum anode current
discharge limit. However, use the 3 anodes required to meet the 2-ohm
ground bed resistance design requirement.

5.6.2.5 Total Weight of Anodes for Ground Bed.

A. Weight of anode unit, 60 pounds (size 2 inches x 60 inches)

B. Total weight = 3 x 60 = 180 pounds

5.6.2.6 Theoretical Life of Anode Bed.

YSI
W =
E

Rearranging the equation we get:

WE
Y =
SI

where
Y = Theoretical anode bed life
W = Total anode bed weight = 180 pounds
S = Anode consumption rate = 1.0 lb/A yr
E = Anode efficiency = 0.50
I = Current required for cathodic protection = 2.36 A

180 x 0.50
Y =
1.0 x 2.36

= 38.1 years

It should be noted that the expected ground bed life greatly exceeds the
design requirement of 15 years. This is brought about by the additional
anode material required to establish a 2-ohm ground bed. The lower
ground bed resistance saves energy (power, P = I² R).

5.6.2.7 Resistance of the DC Circuit.

A. Ground bed-to-soil resistance, 2.0 ohms maximum.

B. Resistance of ground bed feeder conductor (length 500 feet, type
HMWPE, size No 2 AWG).

Conductor resistance (refer to Table 10): 0.159 ohm/1,000 feet

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R = 500 ft x 0.159 ohm/1,000 feet

= 0.080 ohms

C. Total resistance of circuit:

RT = 2.0 + 0.080

= 2.08 ohms

5.6.2.8 Rectifier Rating.

A. Minimum current requirement = 2.36 A.

B. Circuit resistance = 2.08 ohms.
C. Voltage rating:

E = IR
where
I = 2.36 A
R = 2.08 ohms

E = 2.36 A x 2.08 ohms

= 4.9 say 5.0 V

To allow for rectifier aging, film formation, and seasonal changes in the
soil resistivity, it is considered good practice to use a multiplying factor of
1.5 to establish the rectifier voltage rating.

E = 5.0 x 1.5 = 8.0 V

D. The commercial size rectifier meeting the above requirements is 115-V,

single-phase, selenium, and full-wave bridge type having a dc output of 8
A, and 8 V.

5.6.2.9 Rectifier Location.

Mount the rectifier at eye level on a separate pole adjacent to an existing

overhead electrical distribution system

End of Example

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5.7 Hot Water Storage Tank. Design impressed current cathodic protection for
the interior of the industrial hot water storage tank shown in Figure 52.

Figure 52
Cathodic Protection for an Industrial Hot Water Storage Tank

5.7.1 Design Data.

A. Tank capacity, 1,000 gallons.

B. Tank dimensions, 46 inches in diameter by 12 feet long.
C. Tank is mounted horizontally.
D. Water resistivity is 8,600 ohm-cm with a pH value of 8.7.
E. Tank interior surface is bare and water temperature is maintained at 180
degrees F (82.2 degrees C).
F. Design for maximum current density of 5 mA/ft².
G. Design life, 5 years.
H. Use HSCBCI anodes.
I. Alternating current is available at 115 V ac, single phase.

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5.7.2 Computations.

5.7.2.1 Interior Area of Tank.

AT = 2 π r² + π d L

where

r = Tank radius = 1.92 feet

d = Tank diameter = 3.83 feet
L = Tank length = 12 feet

AT = 2 x 3.1416 x (1.92)² + 3.1416 x 3.83 x 12

= 167.5 ft²

5.7.2.2 Maximum Protective Current Required.

A = AP x (1 – CE)

where
A = Area of bare pipe to be cathodically protected
AP = Outside surface area of the pipe = 11,792 ft²
CE = Pipe coating efficiency = 0% or 0

A = 167.5 ft² x (1 – 0)

= 167.5 ft²

I = A x CD

where
A = Area of bare tank surface be cathodically protected = 167.5 ft²
I = Current required for cathodic protection
CD = Current density = 5 mA/ft²

I = 167.5 ft² x mA/ft²

= 838 mA or 0.84 A

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5.7.2.3 Minimum Weight of Anode Material for 5-Year Life.

YSI
W =
E

where
Y = Anode design life = 5 years
S = Anode consumption rate = 1.0 lb/A yr
E = Anode efficiency = 0.50
I = Current required for cathodic protection = 0.84 A

5 x 1.0 x 0.84
W =
E

= 8.4 pounds

Number of anodes required. Anode size of 1-1/2 inches in diameter by 9

inches long weighing 4 pounds each is selected as the most suitable size.

W
NL =
WA

where

W = Total weight of anodes = 8.4 lbs

WA = Weight of a single anode = 4 lbs
NL = Quantity of anodes to meet design life

8.4 lbs
NI =
4 lbs

NI = 2.1 (say 3 anodes)

In order to get proper current distribution, three anodes are required.

5.7.2.3 Calculate the Resistance of a Single Anode.

0.012 r log (D/d)

R =
L

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where

r = Water resistivity = 8,600 ohm-cm

D = Tank Diameter = 3.83 feet
L = Anode length = 9 inches or 0.75 foot
d = Anode diameter = 1-1/2 inches or 0.125 foot

0.012 x 8,600 x log (3.83/0.125)

R =
0.75

= 204.5 ohms

This resistance must be corrected by the fringe factor because they are short
anodes. The fringe factor is 0.48 from the curve in Figure 45 for an L/d =
9/1.5 = 6.

R (adjusted) = 0.48 R = 0.48 x 204.5 = 98.2 ohms

5.7.2.4 Resistance of the 3-Anode Group.

1 p
RT = RV + rs
n s

where
RT = Total anode-to-electrolyte resistance
n = number of anodes
RV = resistance-to-electrolyte of a single anode = 98.2 ohms
rs = electrolyte resistivity = 8,600 ohm-cm
p = paralleling factor
s = spacing between adjacent anodes = 4 feet

1 0.00289
RT = 98.2 + 8,600
3 4

= 38.94 ohms

5.7.2.5 Rectifier Rating.

A. Calculate rectifier voltage

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E = IR

where
I = Protection current required = 0.84 A
R = Anode group resistance = 38.94 ohms

E = 0.84 A x 38.94 ohms

= 32.7 V

B. To allow for rectifier aging, film formation, it is considered good practice to

use a multiplying factor of 1.5 to establish the rectifier voltage rating.

E = 32.7 x 1.5 = 49.1 V

C. The nearest commercially available rectifier size meeting the above

requirements is a 60-V, 4-amp, single-phase unit.

5.7.2.6 Rectifier Location. Locate the rectifier adjacent to tank for the following
reasons:

A. Usually cheaper to install.

B. Easier to maintain.
C. Keeps DC voltage drop to a minimum.

5.7.2.6 DC Circuit Conductors.

A. External to tank: Use No. 2 AWG, HMWPE.

B. Interior of tank: Use No. 8 AWG, HMWPE.

No stressing or bending of the cable should be permitted.

End of Example

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5.8 Steam Heat Distribution System. Provide cathodic protection for a pre-
engineered steam conduit distribution system. Galvanic CP had been previously installed
on the outer conduit of some sections of the steam distribution lines. The system was
ineffective because of high soil resistivity, lack of adequate electrical isolation from
adjacent buried metallic structures (e.g. building H-piles, copper grounding systems,
water lines, electrical conduits, etc). The CP systems included in this design will be
impressed current type. Existing PVC condensate return lines will be replaced with steel
conduits in the near future.

5.8.1 Design Data.

A. Design life: 20 years

B. Current Density: 3 mA/ft².

C. Coating Efficiency: 85% for existing steam conduits and 95% for new
condensate lines.

D. Conventional shallow anode beds were considered, but have a high failure
rate due to third party damage. Deep well anode beds require less space
than shallow anode beds, and are not as likely to cause stray current
interference to nearby metallic structures. A number of small
rectifier/deep well systems are anticipated with each system electrically
isolated from all other systems to minimize the possibility of interference
and facilitate troubleshooting of system shorts that may occur. The deep
well will utilize mixed metal oxide tubular anodes in carbonaceous backfill.

E. Soil Resistivity: Soil resistivity measurements taken at various locations

with various pin spacing yielded a maximum resistivity measured at the
15-foot depth of 22,700 ohm-cm. Although it is not practical to measure
soil resistivity to the anticipated deep well depth, based on review of the
available geological information indicates that the resistivity is anticipated
to decline at deeper depths. However, for conservatism, the design is
based on 22,700 ohm-cm.

F. The steam lines will have insulating flanges and unions at the building tie-
ins. It is anticipated that the electrical isolation will be 90% effective.
Dielectric insulation will be provided at critical locations to electrically
segregate rectifier systems.

5.9.2 Calculations.

5.9.2.1 Outside Surface Area of Steel Conduit.

AP = π D L

where

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AC = Outside surface area of the conduit

D = Outer steam conduit diameter = 8.625 inches average = 0.72 ft
Outer steam condensate diameter = 6.625 inches average = 0.55 ft
L = Pipe length

Steam Conduit Surface Area

Location Length (ft) Diameter Area (ft²)
North Sector 50,000 0.72 113,098
South Sector 19,000 0.72 42,977
Total pipe surface area 156,075

Steam Condensate Conduit Surface Area

Location Length (ft) Diameter Area (ft²)
North Sector 50,000 0.55 86,394
South Sector 19,000 0.55 32,830
Total pipe surface area 119,224

5.9.2.2 Area of Bare Pipe to Be Cathodically Protected.

A = AC x (1 – CE)

where
A = Area of bare conduit to be cathodically protected
AC = Outside surface area of the conduit
CE = Conduit coating efficiency = 85% or 0.85
Condenstate coating efficiency = 95% or 0.95

Bare Surface Area

Structure Total Area (ft²) CE Area (ft²)
Steam Conduit 156,075 0.85 23,411
Condensate Conduit 119,224 0.95 5,961
Total pipe surface area 29,372

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5.9.2.3 Protective Current Required Based on 3 mA/ft² of Bare Metal.

I = A x CD

where
A = Area of bare pipe to be cathodically protected
I = Current required for cathodic protection
CD = Current density = 3 mA/ft²

Bare Surface Area

Structure Bare Area (ft²) CD I (ma)
Steam Conduit 23,411 3 70,233
Condensate Conduit 5,961 3 17,883
88,116
Total pipe surface area
say 88 amps

In order to keep the systems small, limit the size of each system to 15 amps.
The number of systems is

88 amps/15 amps per system = 6 systems

5.8.2.4 Calculate the Quantity of Anodes.

Use mixed metal oxide (MMO) tubular anodes in carbonaceous backfill. The
quantity of MMO anodes must be calculated to meet two different parameters:
design life based on anode maximum current discharge, and anode bed
resistance. The required quantity of anodes will be the larger of the two
calculated quantities.

1. First calculate the quantity of anodes to meet design life

Y IR
NL =
AA S

where:
IR = Current required = 8.9 amps
NL = Number of anodes to meet design life
S = Manufacturer’s MMO anode life rating (years)
AA = Manufacturer’s MMO anode current rating (amps)
Y = CP system design life = 20 years

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Example Manufacturer’s MMO Anode Ratings

Rated Output (Amperes)
Anode Size
10 years 15 years 20 years
1” x 45” 5.5 – 6.0 4.0 – 4.5 3.0 – 3.5
1” x 60” 7.0 – 8.0 5.0 – 6.0 4.0 – 5.0
1” x 90” 11.0 – 12.0 8.0 – 9.0 6.0 – 7.0

Use a 1” x 45” anode rated at an average of 3.25 amperes for a 20 year

design life.

20 x 15
NL =
3.25 x 20

= 5 anodes

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

2. Anode well resistance (RS)

The anodes will be installed in a vertical deep well. The resistance of the
anode well can be approximated by the following equation:

0.0052 r 8L
RS = [ ln ( )-1]
L d

where

RS = anode bed design resistance (2 ohm maximum desired)

r = soil resistivity (22,700 ohm-cm)
L = length of the anode backfill column (ft)
d = diameter of anode backfill column (8 in or 0.67 ft)

For five anodes try an active well depth of 100 feet.

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0.0052 x 22,700 8 x 100

RS = [ ln ( )-1]
100 0.67

= 7.2 ohms – too high

Several iterations using several different backfill column lengths yield the
following:

Anode Deep Well Resistance

Backfill Column Length (ft) Resistance (ohms)
100 7.2
150 5.1
200 4.0
250 3.3
300 2.8
400 2.2

A 2-ohm deep well ground bed is not economically feasible, therefore, select
a 200 ft backfill column with 4.0 ohms resistance. Although 5 anodes are
required to meet the design life, use 8 anodes to ensure anode current
attenuation along the backfill column is minimized. Install the anodes so that
the bottom of the anode is about ten feet above the bottom of the hole. Use
an anode spacing of twenty feet. Provide anodes with factory connected lead
wires of sufficient length to reach the anode junction box without splicing.

5.8.2.5 Rectifier Rating.

A. System current requirement = 15 A.

B. Circuit resistance = 4.0 ohms.
C. Voltage rating:

E = I R + 2 Volts
where
I = 15 A
R = 4 ohms

2 volts are added to overcome the typical rectifier back voltage.

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E = 15 A x 4.0 + 2 V

= 62 V

To allow for rectifier aging, film formation, and seasonal changes in the soil
resistivity, use a multiplying factor of 1.25 to establish the rectifier voltage and
current ratings ratings.

VR = 62 Volts x 1.25 = 78 Volts

IR = 15 Amps x 1.25 = 19 Amps

D. The commercial size rectifier meeting the above requirements is 240V,

single-phase, full-wave bridge type having a dc output of 22 Amperes, and
80 Volts.

5.8.2.4 Calculate the AC Line Load (Full Output).

1. First calculate the line current

V IR
IL =
240

where:
IL = Full output AC line current (amps).
IR = Rectifier rated output current = 22 amps
VR = Rectifier rated output current = 80 volts

80 x 22
IL =
240

IL = 8 Amps

The circuit overcurrent protection device shall not be less than 125 percent of
the continuous load. The rectifier is a continuous load, therefore,

8 amps x 125% = 10 amps

Use a 15 ampere circuit as a minimum.

End of Example

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5.9 Aircraft Multiple Hydrant Refueling System. Cathodic protection will be

provided to new underground 18-inch diameter stainless steel hydrant refueling supply
and return lines, 12-inch/10-inch diameter stainless steel lines supplying the direct
refueling stations, and 32 single point and 17 dual point hydrant outlets with 4-inch/6-inch
risers. The stainless steel lines will be coated with extruded polyethylene. Dielectric
Isolating flanges will be provided to electrically isolate the buried pipelines. Bare
galvanized steel grounding rods will be provided at the hydrant refueling pits. Also
intermittent current loads will be imposed on the CP system where copper clad tie
downs embedded in the concrete apron are connected to one or more aircraft that are
refueling. Tie downs are bonded together by a bare copper conductor encased in the
concrete apron.

5.9.1 Design Data.

A. Design life: 25 years

B. Current Density: 0.5 mA/ft² for soil, 5 mA/ft² for structures embedded in
concrete.

C. Coating Efficiency: 90% for the extruded polyethylene coating. Bare (0%)
for the hydrant outlet risers.

G. A conventional anode bed with high silicon cast iron tubular anodes
installed horizontally is planned. Due to the soft sand environment, use
ten-inch diameter pre-packaged anodes to simplify installation.

H. The anode bed design resistance should not exceed 2 ohms.

I. Soil Resistivity: Soil resistivity measurements taken at various locations

with various pin spacing yielded soil resistivities ranging from 374 ohm-cm
to 21,500 ohm-cm. The maximum resistivity measured at the 10-foot
depth was 8,600 ohm-cm. Therefore, the design is based on a resistivity
of 8,600 ohm-cm, and the anodes must be installed at this depth.

5.9.2 Calculations.

5.9.2.1 Outside Surface Area of Stainless Steel Pipe.

AP = π D L

where
AP = Outside surface area of the pipe
D = Outer pipe diameter
L = Pipe length

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Stainless Steel Distribution Pipe

Diameter (in) Length (ft) Area (ft²)
18 9,000 42,412
12 830 2,770
10 180 507
Total pipe surface area 45,689

There are 32 single point and 17 dual point hydrant outlet pits for a total of
66 hydrant outlet risers.

Stainless Steel Hydrant Riser Pipe

Diameter Length (ft) Qty Area (ft²)
4”-6” Reducer (5” Avg. Dia) 1 66* 86.4
4” Riser (4.5” OD) 2 66* 155.6
Total pipe surface area 242.0

5.9.2.2 Surface Area of Hydrant Pit Ground Rods. There are32 single point and
17 dual point hydrant outlet pits for a total of 49 pits and therefore, 49 ground rods.

AR = π D L NR

where
AR = Surface area of the ground rod
D = Ground rod diameter = ¾ inch = 0.0625 ft
L = Ground rod length = 10 ft
NR = Quantity of Ground Rods = 49

AP = 3.1416 x 0.0625 x 10 x 49

= 96.2 ft²

5.9.2.3 Surface Area of Aircraft Tiedown/Ground System.

A = πDL

where
A = Surface area
D = Ground rod/wire diameter
L = Ground rod/wire length

The tiedown/ground consists of ¾ inch diameter by 19 inches long rod plus a

¾ inch diameter by 11 inches long rod, a 2-inch diameter by 3-inch long area
receptacle, #4 bare copper wire, and a major ground bed.

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Surface Area of Tiedown/Ground System

Component Dia (in) Length (ft) Qty Area (ft²)
Ground rod (19” + 11”) 0.75 2.5 613 301
Receptacle 2 0.25 613 80
#4 Copper Wire 0.232 8490 515
Total pipe surface area 896

5.9.2.4 Area of Bare Structure to Be Cathodically Protected.

A = AS x (1 – CE)

where
A = Area of bare structure to be cathodically protected
AS = Surface area of the structure
CE = Structure coating efficiency

Area of Bare Structure to be Protected

Structure Surface Area (ft²) CE Area (ft²)
SS Distribution Pipe 45,689 0.9 4,569.0
SS Hydrant Risers 242 0 242.0
Hydrant Pit Ground Rods 96.2 0 96.2
Tiedown/Ground System 896 0 896

5.9.2.5 Protective Current Required.

I = A x CD

where
A = Area of bare pipe to be cathodically protected
I = Current required for cathodic protection
CD = Current density = 0.5 mA/ft² for pipe and ground rods in soil
= 5 mA/ft² for grounding encased in concrete

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Area of Bare Structure to be Protected

Structure Surface Area (ft²) CD (mA/ft²) I (mA)
SS Distribution Pipe 4,569.0 0.5 2,285
SS Hydrant Risers 242.0 0.5 121
Hydrant Pit Ground Rods 96.2 0.5 48
Tiedown/Ground System 896 5 4,480
6,934
Total pipe surface area
Say 6.9 Amps

In order to account for leakage current through insulating flanges and

unanticipated coating damage, add an additional 2 amperes for a total of 8.9
Amperes. To allow for rectifier aging, film formation, and seasonal changes in
the soil resistivity, select a standard size 12 ampere rectifier.

5.9.2.6 Calculate the Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes. The
quantity of anodes must be calculated to meet three different parameters,
design life, anode bed resistance, and anode maximum current discharge.
The required quantity of anodes will be the largest of the three calculated
quantities.

1. First calculate the quantity of anodes to meet design life

Y S IR
NL =
WU

where:
IR = Current required = 8.9 amps
NL = Number of anodes to meet design life
S = Anode consumption rate = 1 lb/amp-yr)
U = Anode utilization factor = 0.8
W = Weight of one anode (lb)
Y = CP system design life = 25 years

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the recommended maximum anode discharge and the ground bed resistance
requirements must still be calculated.

2. Next calculate the quantity of anodes to meet recommended maximum

anode current discharge

ND = I
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AA x IA

where:
I = Current required = 8.9 amps)
AA = Anode surface area (SF)
IA = Max. recommended anode discharge current density (1 amp/SF)

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

3. Calculate quantity of anodes to meet 2-ohm ground bed design

The anodes will be installed in a shallow, distributed, horizontal anode

bed. The resistance of the horizontal anode bed can be approximated by
the following equation:

1.64 ρ
RS = x [ln (48L/d) + ln (L/h) - 2 + (2h/L)]
πL

where
d = diameter of anode backfill column = 10 inches
L = length of the anode backfill column (feet)
RS = anode bed design resistance = 2 ohms
ρ = soil resistivity = 86 ohm-m
h = anode depth = 10 feet

Assume one foot of backfill column beyond the ends of each anode.

4 Determine quantity of anodes. The following table summarizes the

above calculations for various common tubular anode sizes:

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CALCULATED ANODE QUANTITIES

Weight Area Backfill Backfill
Anode Size NL ND NR
(LB) (SF) Length (LF) Dia. (ft)
2.66" x 42" 31 2.4 5.42 0.833 9 4 32
2.66" x 60" 46 3.5 9.0 0.833 7 3 20
2.66" x 84" 63 4.9 9.0 0.833 5 2 20
3.75" x 84" 85 6.9 9.0 0.833 4 2 20

Of the three calculated quantities for a particular anode, the larger

quantity of the three must be used to ensure all conditions are satisfied.
The 46, 63 and 85 pound anodes utilize the same size backfill columns,
and therefore, yield the same quantity of anodes to meet the 2.0 ohm
maximum ground bed resistance requirement. For economic
considerations, use the 20 each 46 LB, 2.66 inch diameter X 60 inch
long anodes. Install the anodes so that there is one foot of backfill
column beyond each end of each anode (two feet between consecutive
anodes). Provide anode with factory connected lead wires of sufficient
length to reach the anode junction box without splicing.

5.9.2.7 Resistance of the DC Circuit.

A. Ground bed-to-soil resistance, 2.0 ohms maximum.

B. Calculate electrical conductor resistance.

Calculate Rn: #4 rectifier negative header cable resistance (90 feet):

Rn = 90 ft/1000 x 0.269 ohm/1,000 feet

= 0.024 ohms

Calculate Rp: # rectifier positive header cable resistance (1,085 feet):

Rp = 1,085 ft/1000 x 0.169 ohm/1,000 feet

= 0.208 ohms

C. Total resistance of circuit:

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RT = RS + Rn + Rp

= 2.0 + 0.024 + 0.208

= 2.23 ohms

5.9.2.8 Rectifier Rating.

A. Rectifier current rating = 12 A.

B. Circuit resistance = 2.23 ohms.
C. Voltage rating:

E = I R + 2 Volts
where
I = 12 A
R = 2.23 ohms

2 volts are added to overcome the typical rectifier back voltage.

E = 12 A x 2.23 + 2 V

= 28.8 V

D. The commercial size rectifier meeting the above requirements is 115V,

single-phase, full-wave bridge type having a dc output of 12 Amperes, and
30 V.

End of Example

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5.10 On-Grade Fuel Storage Tanks. Provide impressed current cathodic

protection for the exterior bottoms of five existing on-grade fuel storage tanks:

TANK CAPACITY DIAMETER

ID NO. (BBL) (FT)
Tank No. 1 55,000 93
Tank No. 2 55,000 93
Tank No. 3 2,300 27
Tank No. 4 10,000 45
Tank No. 5 20,000 58

Review of historical records indicates that the tanks have not had cathodic protection in
the past.

5.10.1 Design Data.

A. Design life: 25+ years

B. Current Density: Record drawings and field inspection indicates the tanks
have been constructed on an “impermeable layer” of compacted coral with
a minimum thickness of 12 inches. The compacted coral may affect the
even distribution of current and higher than normal current densities may
be necessary to ensure the entire tank bottom is adequately protected.
Usually, 1 - 1.5 ma/sf of current is sufficient to protect a tank bottom if the
current is evenly distributed. With the compacted coral layer beneath the
tanks, a current of 2 ma/sf will be used.

C. Coating Efficiency: 0% (Bare tank bottoms)

D. 120 Volt power and a spare circuit breaker is available in a panel in a

nearby pump house. Space is available in the pump house to install a
wall mount rectifier.

E. A deep well anode will be specified to: (1) improve current distribution of
current to the entire tank bottom because of the compacted coral bed;
(2) minimize excavation (historic preservation laws require an
archaeologist to monitor all excavations full time which would add
significant cost; (3) minimize disruption of on-going operations; and (4)
minimize cathodic interference on other buried metallic structures in the
tank farm.

F. The anode bed design resistance shall not exceed 1 ohm.

G. Soil Resistivity: Soil resistivity measurements taken at eight points within

the tank farm were generally in the 2,000 to 4,000 ohm-cm range. Use
2,000 ohm-cm.

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5.10.2 Design Calculations.

5.10.2.1 Calculate Tank Bottom Surface Area.

The surface area of the tank bottoms can be calculated using the following
equation:

AT = π r2

where
AT = Surface area of a tank bottom (ft2)
r = tank bottom radius (ft)

The following table summarizes the tank bottom surface areas.

Tank ID. Diameter Surface Area

NO. (FT) (ft2)
Tank No. 1 93 6,790
Tank No. 2 93 6,790
Tank No. 3 27 570
Tank No. 4 45 1,590
Tank No. 5 58 2,640

5.10.2.2 Calculate Current Requirement/Rectifier Output Current.

IR = AT x (1 - CE) x CD

where
IR = Current requirement/rectifier output current
AT = Surface area of a tank bottom = 2,640 (ft2)
CE = Coating efficiency = 0 for bare surfaces
CD = Current density = 2 ma/ ft2

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The following tables summarize the current requirements.

Tank ID Surface CE CD Current

2 2
No. Area (ft ) (ma/ft ) Req'd (ma)
Tank No. 1 6,790 0 2 13,580
Tank No. 2 6,790 0 2 13,580
Subtotal 27,160
Tank No. 3 570 0 2 1,140
Tank No. 4 1,590 0 2 3,180
Tank No. 5 2,640 0 2 5,280
Total Current 36,760
Required (IR) or 37 amperes
In order to account for current that will be drained away by the grounding systems,
use 40 amperes.

5.10.2.3 Calculate Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes.

A. Quantity of anodes to meet design life

Y S IR
NL =
WU

where:
NL = Number of anodes to meet design life
Y = CP system design life = 25 years
S = Anode consumption rate = 1 lb/amp-yr)
IR = Current required (amps) = 40 amps
W = Weight of one anode (lb)
U = Anode utilization factor (0.85)

B. Quantity of anodes to meet recommended maximum anode current

discharge

IR
ND =
AA X IA

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where:
IR = Current required (amps) = 40
AA = Anode surface area (ft2)
IA = Max. recommended anode discharge current density (1 amp/ ft2)

The following tables summarize the above calculations for various

common tubular anode sizes:

Weight Area
Anode Size NL ND
(LB) (SF)
2.19" x 84" (TA2) 46 4.0 25 10
2.66" x 84" (TA3) 63 4.9 19 9
3.75" x 84" (TA4) 85 6.9 14 6

Of the above-calculated quantities for the anodes, the larger quantity of

the two must be used to ensure all conditions are satisfied. Considering
CP current distribution in the well, use the 85 LB, 3.75 inch diameter X
84 inch long (TA-4) anode. The TA-4 appears to be a better choice to
minimize drilling.

C. Anode well resistance (RS)

The anodes will be installed in a vertical deep well. The resistance of the
anode well can be approximated by the following equation:

0.0052 r 8L
RS = [ ln ( )-1]
L d

where

RS = anode bed design resistance (1 ohm maximum)

r = soil resistivity (2,000 ohm-cm)
L = length of the anode backfill column (ft)
d = diameter of anode backfill column (10 in or 0.83 ft)

Based on the above a minimum anode well active depth of about 55 feet
minimum is required.

Considering the number of anodes required and backfill life, use two wells
of 101 feet active depth with seven anodes in each well. Install the
anodes so that the bottom of the anode is about five feet above the bottom
of the hole. Use an anode spacing of seven feet. Provide anodes with
factory connected lead wires of sufficient length to reach the anode
junction box without splicing. Calculate the anode bed circuit resistance of
each well using the appropriate numbers from the above table:

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0.0052 x 2000 8 x 101

RS = [ ln ( )-1]
101 0.83

= 0.61 ohm for each of the wells

5.10.2.4 Calculate Rectifier Circuit Output Voltage.

A. Anode bed resistance (RA) is the equivalent resistance of the two wells in
parallel. Calculate the equivalent resistance of the two wells:

1 1
RA = +
R1 R2

where,
RA = Equivalent anode well circuit resistance
R1 = Circuit resistance of the first anode well
R2 = Circuit resistance of the second anode well

1 1
RA = +
0.61 0.61

= 0.31 ohm

B. Electrical Cable Resistance

Anode Well Cable Resistance (RAW). Use No. 8 copper wire for the anode
leads. Each of the anode leads will run directly back to the anode junction
box, so the parallel cable resistance is estimated as follows:

1 1 1 1
= + +····+
RAW RA1 RA2 RAn

where,

RAW = Anode well cable resistance

RA1 = Cable resistance of the first anode in the well
RA2 = Cable resistance of the second anode in the well
RAn = Cable resistance of the nth anode in the well

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1 1 1 1
= + +····+
RAW RA1 RA2 RA7

1 1 1 1
= + +····+
RAW RA1 RA2 RAn

= 0.01 ohm for each well

(negligible)

Anode Header Cable Resistance(RHC). Use No. 2 copper wire for the
anode header cable runs. After a 100 ft run of cable to a junction box,
there are two parallel runs of 200 ft and 450 ft. The equivalent resistance
is 0.053 ohm.

Structure Header Cable Resistance. Use No. 2 copper wire for the
structure leads. After a 450 ft run of cable to anode junction box no. 1,
there are two parallel runs of 100 ft to (tanks 3, 4 and truck stand piping)
and 225 ft (tank 5 piping and tanks 1 and 2). The equivalent resistance for
all of these cable runs is 0.11 ohm.

C. Total Circuit Resistance. The total circuit resistance is

RT = RA + RAW + RHC + RSC

0.31 ohm + 0 + 0.05 + 0.11

= 0.47 ohm say 0.5 ohm

D. Calculate the rectifier output voltage using Ohm's Law:

VR = Irect x RT x 150%

40 amps x 0.5 Ω x 1.5

= 30 volts

Use the nearest nominal size rectifiers, or 30 volts/42 amps.

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5.10.2.5 Calculate Rectifier Input Power.

Available power is 120 volts. For rectifiers of the size specified above, use
single phase 120 volt AC input voltage. Rectifier input power is calculated as
follows:

VR x IR
P =
rectifier efficiency

30 volts x 42 amps
=
0.85

= 1,500 VA

AC input current is calculated to be:

P
IAC =
VAC

1,500 VA
=
120

= 12.5 amperes

5.10.2.6 Summary.

A. Anodes: 14 each high silicon chromium bearing cast iron tubular anodes,
with dimensions of 3.75 inches in diameter x 84 inches long, and weighing
85 LB. Anodes to be installed in two 130 foot deep well, seven anodes
per well, with an active depth of 101 feet. Locate the first anode so that
the bottom anode is 5 feet above the bottom of the well. Install remaining
anodes with a 14 foot spacing on center between the anodes. Anode
supplied by manufacturer with #8 AWG lead wire with HMWPE insulation.
One end is factory connected to the anode. The anode leads should be
long enough to extend to the junction box without splicing.

B. Rectifier:
Type: Air cooled
AC Input: 120 volt, 60 Hz, single phase
DC Output: 30 volts, 42 amperes
Efficiency: 85% minimum

End of Example

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5.11 On-Grade Storage Tanks with Impermeable Containment Liner. Two

each 27,000 barrel on-grade vertical fuel storage tanks are protected by an existing
impressed current CP system. However, a project proposes to install a new tank
bottom above the existing tank bottom, creating a double tank bottom. A high-density
polyethylene containment liner will be installed over the existing tank bottom. The
existing tank bottom and new polyethylene liner will prevent the existing CP system
from providing protection to the new tank bottom. Galvanic corrosion between the new
and old tank bottoms can result in rapid corrosion of the new tank bottom. Therefore, a
new CP system will be installed between the two tank bottoms to provide corrosion
protection to the new tank bottom. The space between the new and existing tank
bottoms will be 4 inches minimum; therefore, the new CP anode system must be
installable in this small space.

5.11.1 Design Data.

A. Design life: 25+ years

B. Current Density: 1 ma/ft2 for a grid system. Generally, 2 ma/ft2 is required

for protection in neutral soils. Past experience and current literature
indicates that 1 ma/ft2 is sufficient due to the even distribution of current
when using a grid system.

C. Tank bottom diameter: 70 ft

D. Tank bottom is steel

E. Slope of the new tank bottom will result in only 4 inches clearance
between the existing and new tank bottom at its lowest point. Therefore,
the proposed new CP system will be an impressed current system
consisting of mixed metal oxide coated titanium ribbon anodes.

F. Each tank will have its own independent system to allow independent
system adjustment for each tank. The rectifier will have two independent
DC output circuits, one for each tank, contained in one enclosure.

5.11.2 Calculations.

5.11.2.1 Tank Bottom Surface Area.

The tank bottom is a circular surface for which its surface area can be
calculated from the following equation:

A = π x r2

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The area of each tank bottom is

A = (3.14) (70 SF/2)2

= 3847 say 3850 SF

5.11.2.1 Current Requirement/Rectifier Output Current

The current requirement for each tank is calculated using the equation:

Ireq = A x current density

= 3850 SF x 1 ma/SF
= 3850 ma or 3.85 amps

Allow an additional 25% to account for unknown factors in the sand backfill
that will be used, for rectifier aging and other long term additional requirement:

IR = 3.85 amps x 125%

= 4.8 amps

Use the next nominal size rectifier circuit of 8 amperes for each tank.

5.11.2.2 Quantity of Anode Ribbons

Commonly used mixed metal oxide coated titanium ribbon anodes are 0.25
inch wide by 0.025 inch thick. Manufacturer's literature indicates this material to have a
current rating of 5 ma/LF for 50 year life.

The quantity of anode ribbon required for a 25+ year life for each tank is

N = 3850 ma x 5 ma/LF
= 770 LF

The ribbons will be spaced in evenly spaced parallel strips to ensure uniform
distribution of CP current. This will result in 14 strips of ribbon spaced 5 feet on center.

The anode ribbons will be connected together by titanium ribbon conductor

bars placed in a grid spaced at 30 feet on center ribbons to form a grid. The conductor
bars are 0.5 inch wide by 0.04 inch thick. The conductor bars will be fed by multiple
power feed leads. This will minimize the voltage drop through the grid, thereby allowing
an even distribution of cathodic protection current to the tank bottom.

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5.11.2.3 Circuit Resistance of Anode Ribbons

The strips are of different lengths due to the circular perimeter of the tank
bottom. The theoretical resistance of a strip of metal can be calculated using the
following equation from reference (c):

ρ 2
RS = (ln 4L + a - πab2 + ln 4L - 1)
4πL a 2(a+b) S

where

L = ½ the length of the anode strip (cm) = 335 LF or 11,735 cm

a = width of the strip (cm) = 0.25 in or 0.635 cm
b = thickness of the strip (cm) = 0.025 in or 0.0635 cm
S = twice the depth of the anode = 6 in or 15.24 cm
ρ = sand resistivity = 10,000 - 45,000 ohm-cm. Sand resistivity is
dependent on many factors dependent on the actual sand used.
Typically, the resistivity of clean sand will vary from 10,000 ohm-cm
when wet to 45,000 when dry. While the sand may be dry when first
installed, moisture will eventually intrude from rain during installation,
water used during compaction, and leakage through deteriorated
edge seals.

The resulting resistance after substituting the above numbers is

Rs = 5.7 Ω for 45,000 ohm-cm sand, and

Rs = 1.3 Ω for 10,000 ohm-cm sand

5.11.2.4 Rectifier Circuit Output Voltage

The rectifier output voltage is calculated by Ohm's Law. Use the resistance
for 45,000 ohm-cm sand to ensure proper operation when sand is dry:

VR = Ireq x Rs
= 3.85 amps x 5.7 Ω
= 21.9 volts

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Use the next highest nominal size rectifier of 24 volts to account for rectifier back
voltage, circuit resistance for the power feed leads and conductor bars, and
compensation for system aging.

5.11.2.5 Rectifier Input Power

Rectifier input power for each circuit can be calculated as follows:

VA = VR x IR
rectifier efficiency

= 24 volts x 8 amps
0.65

= 295 say 400 VA to account for 25% safety factor

The total power for the two circuits is 800 VA or 0.8 KVA. For rectifiers of this
size, use single phase 120 volt AC input voltage. AC input current is calculated to be:

IAC = VA/VAC
= 800 VA/120 volts
= 6.67 say 7 amps

5.11.3 Summary

Anode material: Mixed metal oxide coated titanium ribbon, 0.25" wide x
0.025" thick. The anode grid requires 770 LF per tank
placed in 14 parallel rows spaced at 5 FT on center.

Conductor bar: Conductor bars are uncoated titanium ribbons, 0.5" wide x
0.04" thick. Estimated quantity is 320 LF per tank.

Power feed leads: Supplied by manufacturer consisting of #8 AWG wire with

HMWPE insulation. One end is factory connected to a 5"
length of conductor bar, which will be field welded to the
anode grid conductor bars. The power feed leads should
be long enough to extend to the junction box without
splicing. 7 power feed leads per tank are required.

Rectifier:
Type: Oil cooled, enclosure suitable for class I, division 2.
AC Input: 120 volt, 7 amps, 60 Hz, single phase
DC Output: Two independent circuits, each circuit rated at 24 volts, 8
amperes
Efficiency: 65% minimum

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Reference
electrodes: Combination cell of copper copper-sulfate and zinc, each
component having a #14 AWG lead wire long enough to
reach the junction box without splicing. 5 each
combination cells per tank are required.

Due to the proximity of the tank bottom to the anode, especially at its lowest
point, dielectric mesh will be installed between the anode and new tank bottom to
prevent electrical short circuits.

End of Example

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5.12 Land Side of Steel Sheet Piling (Impressed Current).

Due to high soil resistivity, and high current requirements, an impressed

current system is to be provided for an existing sheet pile bulkhead about 660 LF in
length by 48 FT high and is constructed of PZ 27 steel sheet piles. The sheet pile was
coated between the 8 and 40 Ft depths, and the top 10 FT of the sheet pile was
encased in a reinforced concrete pile cap. Electrical continuity across each pile is
provided by existing bonding straps welded across each pile joint. The bonding straps
are electrically connected together by bonding wires. The bulkhead is anchored via
steel tie rods to an 625 FT long by eight feet high steel sheet pile deadman set back 85
FT from the bulkhead. The tie rods are coated, wrapped and enclosed in PVC sleeves
along its entire length except near the turnbuckle. Electrical continuity to and across the
turnbuckles is provided by existing bond wires between the turnbuckle and the rod
connected to each side of the turnbuckle. The deadman is constructed of coated steel
PZ 27 sheet piles. Electrical continuity across each pile is provided by existing bonding
straps welded across each pile joint. Refer to Technical Paper 17 for example
calculations for a water side galvanic anode system.

5.12.1 Design Data.

A. Seawater Resistivity - 20 ohm-centimeters

B. Design for 2 milliamperes per square foot in the soil, and 1 ma/SF for steel
in concrete.
C. High silicon chromium bearing cast iron anodes will be used.
D. The design structure to electrolyte potential for the protected structure will
be – 850 mv.
E. Design life: 15+ years
F. Coating Efficiency:
Steel sections encased in concrete: 90% (0.9)
Steel sections exposed to seawater: 80% (0.8)
Steel sections below mudline: 70% (0.7)

5.12.2 Calculations.

5.12.2.1 Calculate Total Surface Area of Sheet Pile.

1. Bulkhead. The bulkhead is constructed of PZ-27 steel sheet piles. PZ-27

sheet pile has a width of 18 inches and a surface area of 2.5 SF/LF on
face of the pile. The number of sheet piles (NSP) can be calculated as
follows:

Running Length of sheet pile wall (in inches)

NSP =
18 inches

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The lengths of the bulkhead and deadman anchor wall are 660 LF and
625 LF respectively. The following table summarizes the number of piles
in each wall.

SHEET PILE WALL RUNNING NO. OF SHEET

LENGTH (INCHES) PILES (NSP)
BULKHEAD 7,920 440
ANCHOR WALL 7,500 417

The surface area for the bulkhead can be calculated using the following
equation:

ASP = NSP X Length of sheet pile X 2.5 SF/LF

The bulkhead can be divided into the following zones:

• Land side, encased in concrete, sheet pile bare

• Land side, encased in concrete, sheet pile coated
• Land side, exposed to soil, sheet pile coated
• Land side, exposed to soil, sheet pile bare

The following table summarizes the surface areas for each zone.

No. of Pile Length Surface

Sheet Pile Zone
Piles (LF) Area (SF)
Bare 440 8 8,800
Concrete
Coated 440 2 2,200
Land side
Bare 440 8 8,800
Soil
Coated 440 30 33,000

2. Tie Rods. There are 153 tie rods, coated, wrapped, and installed in PVC
sleeves. Only the portions of the tie rod at the turnbuckle and on the
back side of the anchor wall are exposed to the soil. An estimated length
of two LF will account for the exposed tie rod and turnbuckle. The tie
rods are 2.5 inches in diameter. Since the tie rods are threaded at the
turnbuckle and anchor wall, assume that the steel is bare. Calculate the
surface area as follows.

ATR = π x d x L x Number of tie rods

= 3.14 x (2.5"/12") x 2 LF x 153
= 200 SF

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3. Anchor Wall. Both sides of the anchor wall are coated and exposed to
the soil. The surface area for both sides of the pile is 5 SF/LF. The
length of each anchor wall pile is 8 LF. The total anchor wall surface
area is:

AAW = 417 piles X 8 LF 5 SF/LF

= 16,680 SF

5.12.2.2 Calculate the Current Required for Protection of the Sheet Pile.
The current requirement is calculated using the equation:

IR = Surface area X (1 - CE) x J

where
IR = Cathodic protection current required
CE = coating efficiency
J = current density

The following table summarizes the current requirements.

LAND SIDE
Structures/Zones Surface J Current
CE
area (SF) (ma/SF) Req'd (ma)
Bare 8,800 0 1 8,800
Concrete Coated 2,200 0.9 1 220
Sheet pile
25% of above for concrete cap rebar 2,255
Bulkhead
Bare 8,800 0 2 17,600
Soil
Coated 33,000 0.7 2 19,800
Tie Rods Soil Bare 200 0 2 400
Sheet pile
Soil Coated 16,680 0.7 2 10,008
Anchor Wall
Total Land Side Current Required 59,083
or 59.1 amps

Use an 80 ampere rectifier. The extra current capacity will allow for additional
coating deterioration, system aging, and current loss to the water side of the sheet pile.

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5.12.2.3 Calculate the Quantity of Anodes.

Use high silicon chromium bearing cast iron (HSCBCI) tubular anodes. The
quantity of anodes must be calculated to meet three different parameters,
design life, anode bed resistance, and anode maximum current discharge.
The required quantity of anodes will be the largest of the three calculated
quantities.

1. First calculate the quantity of anodes to meet design life

Y S IR
NL =
WU

where:
IR = Current required = 59.1 amps
NL = Number of anodes to meet design life
S = Anode consumption rate = 1 lb/amp-yr)
U = Anode utilization factor = 0.8
W = Weight of one anode (lb)
Y = CP system design life = 20 years

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the recommended maximum anode discharge and the ground bed resistance
requirements must still be calculated.

2. Next calculate the quantity of anodes to meet recommended maximum

anode current discharge

I
ND =
AA x IA

where:
I = Current required (59.1 amps)
AA = Anode surface area (SF)
IA = Max. recommended anode discharge current density (1 amp/SF)

*** IMPORTANT ***

Do not end calculations at this point. The quantity of anodes required to meet
the ground bed resistance requirements must still be calculated.

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3. Calculate quantity of anodes to meet 1-ohm ground bed design

The anodes will be installed in a shallow, distributed, vertical anode bed.

The resistance of the anode bed can be approximated by the following
equation:

0.0052 ρ
NR = x [ ln (8L/d) - 1 ]
RS x L

where
d = diameter of anode backfill column (1 FT)
L = length of the anode backfill column (LF)
NR = quantity of anodes to meet the anode bed design resistance
RS = anode bed design resistance (1 ohm)
ρ = soil resistivity (6,000 ohm-cm)

4 Determine quantity of anodes. The following table summarizes the

above calculations for various common tubular anode sizes:

CALCULATED ANODE QUANTITIES

Weight Area Backfill Backfill
Anode Size NL ND NR
(LB) (SF) Length (LF) Dia. (ft)
2.66" x 42" (TA1) 31 2.4 5.5 1 46 25 16
2.19" x 42" (TA2A) 23 2.0 5.5 1 62 30 16
2.19" x 60" (TACD) 32 2.8 7.0 1 45 22 14
2.66" x 60" (TAD) 45 3.5 7.0 1 32 18 14
3.75" x 60" (TAM) 60 4.9 7.0 1 23 13 14
4.75" x 60" (TAJ) 78 6.2 7.0 1 19 10 14
2.66" x 84" (TA3) 63 4.9 9.0 1 23 13 12
3.75" x 84" (TA4) 85 6.9 9.0 1 17 9 12
4.75" x 84" (TA5) 110 8.7 9.0 1 13 7 12

Of the three calculated quantities for a particular anode, the larger

quantity of the three must be used to ensure all conditions are satisfied.
Considering CP current distribution, use the 23 each 60 LB, 3.75 inch
diameter X 60 inch long anodes. Install the anodes so that the bottom
of the anode is about one foot above the bottom of the hole. Provide
anode with factory connected lead wires of sufficient length to reach the
anode junction box without splicing.

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5.12.2.4 Calculate the Total Circuit Resistance (RT).

1. Calculate the anode bed circuit resistance (Rs) using the appropriate
numbers from the above table:

0.0052 ρ
RS = x [ ln (8L/d) - 1 ]
N x L

= 0.56 ohm

2. Calculate the anode cable resistance (RC)

Each anode lead will run directly back to the anode junction box, so the
parallel cable resistance is estimated to be equivalent to a 230 LF length
of cable. Use No. 4 copper wire for the anode leads. The cable
resistance is estimated as follows:

RC = R#4 x length of cable

= 0.00025 ohm/LF x 230 LF

= 0.06 ohm

3. Calculate total circuit resistance (RT)

The total circuit resistance is 0.56 ohm + 0.06 ohm = 0.62 ohm

5.12.2.5 Calculate the Rectifier Output Voltage (VR) Using Ohm's Law.

VR = Irect x RT x 125%
= 59.1 amps x 0.62 Ω x 1.25
= 45.8 volts

Use the next highest nominal size rectifier rated for 50 volts DC.

5.12.2.6 Calculate the Rectifier Input Power

Available power is 480/277 volts. For rectifiers of the size specified above,
use three phase 480 volt AC input voltage. Three phase rectifiers are more
efficient and long term power costs outweigh the higher initial costs for the
rectifier. Rectifier input power is calculated as follows:

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VR x IR
P =
rectifier efficiency

50 volts x 80 amps
=
0.85

= 4,705 VA say 5 KVA

AC input current is calculated to be:

P
IAC =
VAC x 1.732

5,000 VA
=
480 volts x 1.732

= 6 amps

The sheet pile bulkhead is not located near a building where an electrical
panel can be used. Therefore, the rectifier will be located adjacent to a three
phase, 480 volt transformer station.

5.12.2.7 Short Circuit Calculations

A. Transformer Full Load Current (IFL)

VA
IFL =
VAC x √3

500,000 VA
=
480 volts x 1.732

= 601 amps

B. Fault Current at Transformer Secondary (ISC).

IFL
ISC =
%ZT/100

601
=
4.5/100

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= 13,356 Amps

Therefore, use 18,000 amp I.C. service breaker.

5.12.2.7 Summary

Anodes: 23 each high silicon chromium bearing cast iron tubular

anodes, with dimensions of 3.75 inches in diameter x 60
inches long, and weighing 63 LB. Locate the anodes 22 feet
from each end of the bulkhead and at 28 feet spacing in
between.

Anode Supplied by manufacturer consisting of #4 AWG wire with

HMWPE insulation. One end is factory connected to the
anode. The anode leads should be long enough to extend to
the junction box without splicing.

Rectifier:
Type: Oil cooled, enclosure suitable for outdoors use.
AC Input: 480 volt, 6 amps, 60 Hz, three phase
DC Output: 50 volts, 80 amps
Efficiency: 85% minimum

End of Example

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