Highly Moist Soil And Stationery Water

Published Date: 02 Nov 2017

CHAPTER 4

Type of anode available :

Table 3.18 Dimension of Hi-Potential Mg Anodes

Conditions Taken into consideration ( 30" dia)

Length of the pipeline = 1km = 1000m = 3280.83ft.

Dia of the pipeline = 30" = 2.50ft.

Soil Type : Highly Moist Soil and Stationery Water

Current Density : 6mA/ft2.

Soil Resistivity: 1000 ohm cm.

Coating quality : 98%.

Selecting the 14# i.e. 14lb weight anode. With dimension 6"X46".

Hence Length of the anode 23" = 4 ft

Dia of the anode 6" = 0.5ft

Step 1

Calculation of the surface area, S

A = 3.142XDXL = 3.142 X 2.5 X 3280.83 = 25770.92 ft2.

Step 2

Calculation of Required current

IR = Ad (1-q)/1000

Where

IR = current requirement in amps

A = surface area in sq.ft.

d = current density in mA/sq.ft.

q = coating quality as a decimal

IR = 25770.92 x 6 x (1-0.98)/1000 = 3.09 Amp

Step 3

Ground bed Resistance (Dwight’s Formula)

Rgb = ( 0.00521ρ/L)(Ln (8L/d)-1)

Where,

Rgb = resistance in ohms

L = anode length (package length) in feet

d = anode diameter (package diameter) in feet

Ln = natural logarithm function

Rgb = (0.00521 X 1000/2) (ln ((8x4)/0.5416) – 1)

Rgb = 8.02 Ohm.

Step 4.

Current Output

I = (Pa – Pc)/ Rgb

Pa  Anode Potential (1.55V)

I = (1.55 – 0.85) / 8.02 = 0.087 Amp

This current output is less than the current required calculated in step 2.

Divide the current requirement by the above current output to obtain the number of anodes necessary to produce a current greater than or equal to the required current.

Number of Anodes Required N = 3.09 / 0.087 = 35

Hence the number of anodes required = 35.

Total Current in each Anode:

Ia = 3.05/35 = 0.087 Amp

Step 5. Calculation of Life of Anode

L = (WCU) / 8760 I

Where

L = years of life

W= anode weight in lbs.

C = energy capability in amp-hrs per lb.

I = current output in Amps.

U = Utilization factor as a decimal

8760 = hours in 365 days

For Mg anodes, Consumption rate is: 1230 Amp hr / Kg = 558 Amp hr / lb

L = 14 x 558 x 0.85 / 8760 x 0.087

L = 8.712 years.

Hence the life of each anode is 8.712 years.

Conditions Taken into consideration ( 30" dia)

Length of the pipeline = 1km = 1000m = 3280.83ft.

Dia of the pipeline = 30" = 2.50ft.

Soil Type : Moist Soil

Current Density : 2mA/ft2.

Soil Resistivity: 2000 ohm cm.

Coating quality : 98%.

Selecting the 12# i.e. 12lb weight anode. With dimension 6"X23".

Hence Length of the anode 23" = 2 ft

Dia of the anode 6" = 0.5ft

Step 1

Calculation of the surface area, S

A = 3.142XDXL = 3.142 X 2.33 X 3280.83 = 24015.383ft2.

Step 2

Calculation of Required current

IR = Ad (1-q)/1000

Where,

IR = current requirement in amps

A = surface area in sq.ft.

d = current density in mA/sq.ft.

q = coating quality as a decimal

IR = 24015 x 2 x (1-0.98)/1000 = 0.9606 Amp

Step 3

Ground bed Resistance (Dwight’s Formula)

Rgb = ( 0.00521ρ/L)(Ln (8L/d)-1)

Where

Rgb = resistance in ohms

L = anode length (package length) in feet

d = anode diameter (package diameter) in feet

ρ = resistivity in ohm-cm

Ln = natural logarithm function

Rgb = (0.00521 X 2000/2) (ln ((8x2)/0.5416) – 1)

Rgb = 12.43 Ohm.

Step 4.

Current Output

I = (Pa – Pc)/ Rgb

Pc  Cathode Potential (0.85V)

Pa  Anode Potential (1.55V)

I = (1.55 – 0.85) / 12.43 = 0.056 Amp

This current output is less than the current required calculated in step 2.

Divide the current requirement by the above current output to obtain the number of anodes necessary to produce a current greater than or equal to the required current.

Number of Anodes Required, N = 0.9606 / 0.056 = 17

Hence the number of anodes required = 17

Total Current in each Anode:

Ia = 0.9606/17 = 0.056 Amp

Step 5. Calculation of Life of Anode

L = (WCU) / 8760 I

Where

L = years of life

W= anode weight in lbs.

C = energy capability in amp-hrs per lb.

I = current output in Amps.

U = Utilization factor as a decimal

8760 = hours in 365 days

For Mg anodes, Consumption rate is: 1230 Amp hr / Kg = 558 Amp hr / lb

L = 12 x 558 x 0.85 / 8760 x 0.056

L = 11.6 years.

Hence the life of each anode is 11 years.

Conditions Taken into consideration ( 30" dia)

Length of the pipeline = 1km = 1000m = 3280.83ft.

Dia of the pipeline = 30" = 2.50ft.

Soil Type : Moving Freshwater containing dissolved oxygen

Current Density : 10mA/ft2.

Soil Resistivity: 1000 ohm cm.

Coating quality : 85%.

Selecting the 40# i.e. 40lb weight anode. With dimension 6.5"X66".

Length of the anode 66" = 5.5 ft

Dia of the anode 6.5" = 0.541ft

Step 1

Calculation of the surface area, S

A = 3.142XDXL = 3.142 X 2.33 X 3280.83 = 24015 ft2.

Step 2

Calculation of Required current

IR = Ad (1-q)/1000

Where

IR = current requirement in amps

A = surface area in sq.ft.

d = current density in mA/sq.ft.

q = coating quality as a decimal

IR = 24015 x 10 x (1-0.85)/1000 = 36.02 Amp

Step 3

Ground bed Resistance (Dwight’s Formula)

Rgb = ( 0.00521ρ/L)(Ln (8L/d)-1)

Where

Rgb = resistance in ohms

L = anode length (package length) in feet

d = anode diameter (package diameter) in feet

Ln = natural logarithm function

Rgb = (0.00521 X 1000/5.5) (ln ((8x5.5)/0.5416) – 1)

Rgb = 3.218 Ohm.

Step 4.

Current Output

I = (Pa – Pc)/ Rgb

Pa  Anode Potential (1.55V)

I = (1.55 – 0.85) / 3.218 = 0.217 Amp

This current output is less than the current required calculated in step 2.

Divide the current requirement by the above current output to obtain the number of anodes necessary to produce a current greater than or equal to the required current.

Number of Anodes Required, N = 36.02 / 0.217 = 165

Hence the number of anodes required = 165.

Total Current in each Anode:

Ia = 36.02/165 = 0.217 Amp

Step 5. Calculation of Life of Anode

L = (WCU) / 8760 I

Where

L = years of life

W= anode weight in lbs.

C = energy capability in amp-hrs per lb.

I = current output in Amps.

U = Utilization factor as a decimal

8760 = hours in 365 days

For Mg anodes, Consumption rate is: 1230 Amp hr / Kg = 558 Amp hr / lb

L = 40 x 558 x 0.85 / 8760 x 0.217

L = 9.95 years.

Hence the life of each anode is 9.95 years.

4.2 Impressed Current Cathodic Protection System Design

For the design of the ICCP system, the following design parameters have been considered:

Pipeline Length

120 km

Pipeline Diameter

24"

Pipeline Material

API 5L X46

Coating

3LPE

Coating resistance

10000 ohm / sqm

Safety Factor

1.3

Steel Resistivity

2.2 x 10 -7 ohm m

Design Life

25 years

Soil Resistivity

2500 ohm cm

Design Temperature

50 degrees Celsius

Maximum PSP

-1.20 volts

Minimum PSP

-0.85 volts

Anode Type

HSIC

Anode Consumption rate

0.4 kg/amp/yr

Anode Weight

23 kg

Anode Dimension

Length 1554 mm ; Dia 50 mm

Maximum Current output from anode

2.35 A

Cable type

Single core PVC of 10mm2 size

The pipeline passes through a distance of 120 km hence it passes through soils of varied soil resistance hence an average resistance has been considered. For the entire pipeline, 4 CP stations have been considered at Chainages 0, 40.00km, 80.00 km and 120.00km. Here the design is done for the 2nd CP station at chainage 40.00km.

1.Calculation of pipe surface area

Sa = π L D

Where,

Sa= area in sq.m.

L = length in m

D = diameter in m

Sa = 3.142 x 0.610 x 40000 = 76664.8 m2

2.Calculation of current requirement

It = Sa x Cd x Sf

Where,

It = CP current required Amps

Sa = Surface area sq m

Cd = Protective current density 50 μ A/ sq m

Sf = Safety Margin = 30%

It = 76664.8 x (50 x 10 ^-6) x 1.3 = 4.98 Amp

Selection of anodes

From the following table, the anode size is selected

Anode Weight

Anode Dimension

Anode Surface Size(ft2)

Package area (sq inch)

12

1X60

1.4

10X84

44

2X60

2.6

10X84

60

2X60

2.8

10X84

110

3X60

4.0

10X84

Table 3.17 Dimension of HSCI Anodes

3.Anode Current output Calculation

Saa = π d l

Where,

Saa = Anode surface area sqm

D = anode diameter m = 0.05m

L = anode length m = 1.5m

Saa = π x 0.05 x 1.5 = 0.235 sqm

Current delivery capacity of anodes

Ica = Saa x Cda

Where,

Cda = Anode Current Density (10 Amp / m2)

Ica = 0.235 x 10 = 2.35 Amp

No of anodes required considering total current

Nit = It / Ica

It = Total Protective current = 4.98 amp

Ica = Current output from one anode = 2.35 Amp

Nit = 4.98 / 2.35 = 3

4.Anode Weight requirement

Total anode Weight,Wt = Ac x It x Y / Uf

Where,

Ac = Anode consumption rate = 0.4kg / amp/yr

Y = No of years life = 25 yrs

Uf = Anode Efficiency = 0.8

It = Current required = 4.98A

Wt = 0.4x4.98x25/0.8 = 62.25kg

Number of anode required by weight

Nw = Wt / Wa

Where,

Wt = Total wt of anode = 62.25 kg

Wa = Wt of anode = 23 kg

Nw = 62.25 / 23 = 3

Though the actual number of anodes required is 3 but to keep the over all circuit resistance less than 1 ohm 10 anodes have been considered.

5.Ground bed resistance

Rs = 0.00521 Ps / N L [ 2.3 log (8L/D)-1 + 2L ( 2.3 log 0.656 N) / S)

Where,

Rs = Ground bed resistance

Ps = Soil resistivity = 2500 ohm cm

N = No of anodes = 10

D = Anode dia = 0.66ft.

L = Anode Length = 13.12 ft

S = Anode spacing = 15 ft

Rs = 0.704 ohm

6.Individual anode to backfill resistance

Rbs = 0.00521P / L [2.3 log (8L/d) -1]

Where,

P = Backfill resistivity for CPC breeze type ( 100 ohm cm)

L = Anode length = 5.08 ft, d = 5.08 ft

Rbs = 0.462 ohm

Total Anode to Backfill resistance

Rtbs = Rbs / N

Where,

Rtbs = Total Anode to backfill resistance,.

N = No of anodes.

Rtbs = 0.462 / 10 = 0.0462 ohm

7.Total Ground bed resistance

Rgb = Rs + Rtbs

Where ,

Rgb = Total ground bed resistance

Rs = Resistance of vertical anode to earth

Rgb = 0.704 + 0.0462 = 0.75 ohm

8.Total Resistance Offered by all cables

Rtot = Rta + Rac

Where,

Rac = Header cable resistance

Rta = Anode Cable resistance

Note :Anode cable resistance is very small and hence can be neglected.

Calculation of Header cable resistance

For single core copper conductor cable of 35 sq mm is selected as anode and cathode header cable with total length of 300 m

Rac = R35 (Lac + Lcc)

Where,

Rac = Resistance of header cable

Lac = Length of anode header cable = 150 m

Lcc = Length of cathode header cable = 150 m

R35 = Resistance of 35 sq mm cable = 0.56 ohm / km

Rac = 0.00056 (150 + 150) = 0.168 ohm

8.Total Circuit Resistance

Rtcr = Rgb + Rac

Where,

Rgb = Total Ground bed resistance = 0.75 ohm

Rac = Header cable resistance = 0.168 ohm

Rtcr = 0.921 ohm

9.Cross section area of the pipeline

Ap = π t (d-t)

Where,

t = thickness of pipe = 0.01 m.

d = pipe dia 0.610m

Ap = Cross section area Sq m

Ap = π x 0.01 x (0.61-0.01) = 0.0188m2

10.Linear resistance of pipe

Rs = ρs l / A

Where,

ρs = Resistivity of steel pipe ohm / sq m = 2.2 x 10-7 ohm mtr

l = Unit length of pipe = 1000m

Ap = Cross section of the pipe = 0.0188m2

Rs = 2.2 x 10-7 x 1000 / 0.0188 = 0.0117 ohm / km

11.Coating Leakage Resistance per km for PE coating

Ri = Rp / Sak , in ohm per km

Rp = Coating resistance 10000 ohm / m2

Sak = Surface area of pipe per km = 1916.62 m2

Ri = 10000 / 1916.62 = 5.217 ohm / km

12.Attenuation Constant

α = ( Rs / Ri)0.5

α = (0.0117/5.217)0.5 = 0.0473

13.Maximum Potential Shift at drain point

Edp = (PSP)max – (PSP)nat

(PSP)max = Maximum PSP at drain point -1.20 volt off(instant)

(PSP)nat = Natural PSP of pipeline – 0.45 volt

Edp = -1.20 – (-0.45) = -0.750 volts

14.Minimum Potential Shift at end point

Ed = (PSP)min – (PSP)nat

(PSP)min = Minimum PSP at end point -0.85 volts OFF (instant)

Ed= -0.85 – (-0.45) = -0.4 volts

15.Length of pipeline protected from drain point by keeping the maximum drain point potential at – 1.20 off potential

L = (1/α) cosh-1(Edp/Ed)

= ( 1/0.0473)cosh-1(0.75/0.4) = 26.24 km

It implies that 26.24 km of pipeline will be protected at both sides of CP station 2.

16.Rectifier Design

Rectifier voltage, V = I x R

Where,

I = Total current delivery capacity

R = Total Resistance

Total number of anodes provided in one CP station = 10

Current output at one anode = 2.35 A

Total current delivery capacity = 10 x 2.35 = 23.5 Amps

Total Ground bed resistance = 0.921 ohms

Voltage rating of the rectifier = 23.5 x 0.921 = 21.64 volts

Hence the rating of the CP TR is 21.64 V / 23.5 Amp (Actual Required)

Therefore a TR unit of 50V/50 A can be selected