Transcript Pipline Cathodic Corrosion

Corrosion Management on
Pipeline with Cathodic
Protection
Ilmu Bahan dan Korosi
Outline
• Pipeline Risks and Corrosion
• Cp Technique : Galvanic Sacrificial
Anode (Principle, Installation)
• Cp Technique : Impressed Current
(Principle, Installation)
• Case Study
: Sacrificial Anode
• Case Study
: Impressed Current
Risks on Pipeline
The product of the frequency with which
an event is anticipated to occur and the
consequence of the event outcome.
RISK = PROBABILITY x SEVERITY
Risks on Pipeline
PRESSURE
DROP
LEAKAGE
CORROSION
Corrosion Control
• Coating : insulation
• Cathodic Protection: DC current
injection towards pipeline, result:
pipeline is “shifted” as cathode.
 Sacrificial anode
 Impressed current
Galvanic Sacrificial Anode Principle
• Metal
Driving voltage (Vd)
• Negative Vd = active metal =tendency
to corrode
• Pipeline: connected with active metal
• Anode: Mg, Zn
Cathode: pipeline
• Corrosion on Mg, Zn hence “sacrificial”
Galvanic Sacrificial AnodePrinciple
Material
Driving Voltage (V)
Silver (Ag/Ag+)
+0.8
Copper (Cu/Cu2+)
+0.34
Water (O2+H20+4e- = 4OH-)
+0.401
Hydrogen (H2)
0
Iron (Fe/Fe2+)
-0.44
Zinc (Zn/Zn2+)
-0.76
Magnesium (Mg2+)
-2.36
(cathodic)
(reference)
(anodic)
Galvanic Sacrificial AnodeInstallation
(Source: Peabody’s Control of Pipeline Corrosion)
Galvanic Sacrificial AnodeInstallation
(Source: Peabody’s Control of Pipeline Corrosion)
Impressed Current
(Source: Peabody’s Control of Pipeline Corrosion)
• Minimum potential: -0.85V
• Reference electrode: Copper Sulfate
Galvanic Sacrificial AnodeDesign
•
•
•
•
•
•
Calculate total area to be protected (Ap)
Determine the current density (ρ)
Calculate total protection current (Itot = ρ .Ap)
Calculate R per anode : R = f(d,l,ρ)
Calculate Ia per anode : (Vd-0.85/R)
Calculate total anode needed :
Initial : N=Itot/In
Lifetime: N = f(mass,lifetime,A/poundyear)
Impressed Current-Principle
• Rectifier
• Variable voltage, variable current
• Higher current density (> 1 A)
• Higher soil resistivity (> 104 Ωcm)
Impressed Current-Installation
• Types of anode:
 Graphite : big CR (pound/A/year)
 High Silicon Cast Iron : medium CR
 Platinum & niobium : small CR
• Rectifier rating:
 Voltage, Amperes, Watt, Freq.
 Cooling System : Air, Oil
 Efficiency
CR: Consumption Rate
Impressed Current - Installation
(Source: Peabody’s Control of Pipeline Corrosion)
Impressed Current - Design
•
•
•
•
Calculate total area to be protected (Ap)
Determine curent density (ρ)
Calculate total protection current (Ip)
Calculate total anode needed (N)
 Initial
 Lifetime
• Calculate total anode resistance (Rtot =
f(N,ρ,d,L,spacing))
• Calculate rectifier specification (Vdc, Idc, Pdc =
f(Rtot,Ip)
Pipeline Cathodic Protection–
Case Study
Galvanic Sacrificial Anode –
Case Study
Variabel
Nilai
Pipe Diameter (d)
a. 2 inch = 0.0508m
b. 1 inch = 0.0254m
Pipe Length (l)
a. 2 inch pipe: 300m
b. 1 inch pipe: 1300m
Soil Resistivity (average)
4500 Ωcm
Current Density (ρ)
9mA/sqmm
Type of Anode
Magnesium, weight 9 pon, r = 8.89cm, l = 53.54cm
Coating efficiency (n)
50%
Lifetime
20 years
Galvanic Sacrificial Anode –
Case Study
• Step 1: surface area of pipeline
• Step 2: total protection area
Ac  n  Ap  50% 151,613m 2  75,806m 2
• Step 3: total protection current
Ip    Ac  9mA / sqm  75,806m 2  682,254mA  0,682 A
Galvanic Sacrificial Anode –
Case Study
• Step 4: Anode resistance (Dwight’s Equation)
0,159  4 L 
0,159  4  53,54 
R
ln
 1  4500 
ln
 1  29,15


L  r
53,54 
8,89


• Step 5: Current per Anode
Ia 
DrivingVol tage  Polariseds teelVoltage E 1,55  0,85


 0,024mA
Re sis tan ceAnodeElectrolyte
Rae
29,15
Galvanic Sacrificial Anode –
Case Study
• Step 6: Total Number of Anode
Initial
Lifetime
N
Ip 0,682 A

 28.41anodes
Ia 0,024mA
N
Lifetime  Ip
20  682,254

 30.75anodes
49,3  AnodeWeight
49,3  9
Selected number of anodes: 31.
Galvanic Sacrificial Anode –
Case Study
• Step 7: Distribution and Layout
Distribution
Ac 75.806
A

 2.445m 2
N
31
Layout
Type of Pipe
Effective Surface (Ac)
Number of Anode
(Ac/A)
Diameter 2inch
(0.0508m)
0.5    0,0508  300  23.935m 2 23,93 : 2,445  10
Diameter 1inch
(0.0254m)
0,5    0,0254 1300  51.87m 2
51.87 : 2,445  21
Total
31
Impressed Current
Case Study
Variabel
Value
Pipe Diameter (d)
6 inch = 0,1524 m
Pipe Lengyth (l)
6800 feet = 2072 m = 2,072 km
Soil Resistivity (average)
2000 Ωcm
Current Density (ρ)
30mA/sqmm
Type of Anode
Stated in next section
Coating Efficiency (n)
50%
Coating Impedance
25000Ω/m2
Maximum Groundbed
Resistance
2Ω
Lifetime
15years
Impressed Current
Case Study
• Step 1: surface area of pipeline
Ap  dl    0,1524  2072  992,03m
2
• Step 2: total protection area
Ac  n  Ap  50%  992,03m 2  496,015m 2
• Step 3: total protection current
Ip    Ac  30mA / sqm  496,015m 2  14880,45mA  14,88 A
• Step 4: determine type of anodes
Iron Silicon Chromium. Weight: 110 pon, cross section 4 sqft (0,371m2),
diameter 10 inch (0,833 feet = 2,54cm), length 84 inchi (7 feet = 20,32cm)
Impressed Current
Case Study
• Step 5: total number of anodes
• Current limit (silicon: 10.76 A/m2)
N
Ip
14,88 A

Ai  Ic 0,371m 2 10,76 A
• Lifetime
N
 3,72anode
m2
Lifetime  Ip
15 14880mA

 2,02anode
1000  AnodeWeight 1000 110 pon
Total anodes: 4.
Impressed Current –Case
Study
• Step 6: groundbed resistance calculation
Resistance limit: 2Ω (to limit rectifier’s spec)
Spacing: S = 30 feet , Anode Length: L = 7 feet
Anode Diameter: D = 0.8333 feet
0.00521
8L
2L
(ln
1 
ln 0.656 N )
L N
d
S
0.00521 2000  8  7
27


ln

1

ln
0
.
656

4


74
30
 0.833

 1.36
Rg 
Groundbed resistance is below limit.
If not, choose longer spacing.
Impressed Current Principle
• Step 7: rectifier specification
Vrec  Ip  Rg  14,88  1.36  20,237V .
Srec  Vrec  Ip  301,123W
Specification: 21VDC, 302W .
Final Results:
• Anodes : Iron Silicon Chromium, 4 anodes.
• Rectifier : 21VDC, 302W.