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3/26/2013
1
Fitness for Service and API 579
Presented by:
Ray Delaforce
3/26/2013 1
First consider Unfitness for Service
First consider Unfitness for Service
Here are some of the problems – first we consider corrosion
This is generalised corrosion of steel
The red areas are Hematite
The black spots are Magnetite
Red Hematite is flaky and porous. Hematite is the main problem
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First consider Unfitness for Service
First consider Unfitness for Service
This another very typical example
The white flecks are from insulation that has been removed
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First consider Unfitness for Service
First consider Unfitness for Service
This is galvanic corrosion of aluminium and steel in sea water
Note the presence of red Hematite on the steel
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First consider Unfitness for Service
The various mechanisms of Corrosion
� General corrosion� Occurs over large surfaces� Generally in the form of Hematite
� Crevice corrosion� Where crevices give limited access to contained fluid� Often where parts are fitted together but not welded
� Microbial corrosion� Caused by microorganisms� Often producing Hydrogen Sulfide� Giving rise to accelerated corrosion
� High temperature corrosion� Causes chemical deterioration� Causing products that migrate to the grain boundaries
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First consider Unfitness for Service
Consider for a moment the chemistry of corrosion
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Fill then as follows: 1 tap water, 2 boiled water with at film of oil as a seal to keep air out, 3 dessicant
Take 3 test tubes and set them up in a stand each with a nail
1 2 3
The nails in tubes 2 and 3 are the only ones that do not rust
Corrosion requires both Oxygen and Water to produce rust
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First consider Unfitness for Service
Consider for a moment the chemistry of corrosion
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Here we have a piece of steel immersed in water
The OH- ions and are distributed in the water
O2 + 2H2O + 4e-
4HO-(OH- is a hydroxyl ion)
By oxidation, the iron loses 2 electrons
Fe Fe2+ (Iron ion) + 2e-
At a site in the metal, an Anode is formed with the Iron ions
Fe2+
anode
In the presence of O2, hydroxyl ions are formed
OH-
OH-OH-
OH-OH-
First consider Unfitness for Service
Consider for a moment the chemistry of corrosion
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Negative OH- ions combine with positive Fe2+ ions, to form Iron Hydroxide
Fe2+
anodecathode
OH-
OH-OH-
OH-OH-
Fe2+ + 2OH- Fe(OH)2
The Iron Hydroxide is deposited on the plate at another site
Fe(OH)2
Forming a Cathode
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First consider Unfitness for Service
Consider for a moment the chemistry of corrosion
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Negative OH- ions combine with positive Fe2+ ions, to form Iron Hydroxide
Fe2+
anodecathode
Fe2+ + 2OH- Fe(OH)2
The Iron Hydroxide is deposited on the plate at another site
Fe(OH)2
Forming a Cathode
The electrons lost by the Iron now migrate through the metal
e-
In the presence of O2 the Iron Hydroxide Oxidises further
4Fe(OH)2 + O2 Fe2O3.H2O + 2H2O Iron Oxide – Red Rust
This rust is known as Hematite
OH-
We now move on to API 579
What to do when a vessel suffers wear and tear
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We now move on to API 579
After some time in service, a vessel can suffer damage
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� General corrosion� Pitting corrosion� Grooves and gouging� Surface cracks� Susceptibility to brittle facture� Welding misalignment� Dents� Fire damage
In this presentation we are going to consider just a few of the effects of wear and tear
Consider a new cylinder
Look at the requirement per ASME VIII, Division 1
P = 1,5 MPa Design pressureD = 1500 mm Inside diameterc = 3 mm Corrosion allowanceS = 138 Mpa Allowable (design) stressE = 0,85 Joint Efficiency
According to the code the required thickness is – by PV Elite:
If the thickness is below 12,7038 mm there is a code violation
But, that is for new construction – so select a 14 mm plate.
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Now consider the same cylinder after several years service
Some local general corrosion is detected like this
In some locations the thickness is reduced to 11 mm (was 14 mm)
According to the original (new) calculation:
� The corroded required thickness was 12,7038 –3 = 9,7038 mm� The chosen plate thickness was 14 mm� The corroded thickness is still thick enough at 11 mm� Corrosion allowance is reduced to 11– 9.7038 = 1,296 mm� BUT: we still need a corrosion allowance of 1,5 mm for future . service !
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Now consider the same cylinder after several years service
We need 1,5 mm corrosion allowance for the remaining life
� The corroded required thickness was 12,7038 –3 = 9,7038 mm� The chosen plate thickness was 14 mm� The corroded thickness is still thick enough at 11 mm� Corrosion allowance is reduced to 11– 9.7038 = 1,296 mm� BUT: we still need a corrosion allowance of 1,5 mm for future . service !
The remaining metal available for corrosion is only 1,296 mm
According the original code, we would be in violation !
Remember: The Code is for new construction
Clearly we need additional technical assistance
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API 579 is the help we need
This is Fitness For Service, and modifies the code requirement
It allows us to use thinner plate, but certain criteria must be met
It provided a procedure to assess the corrosion damage
These are the steps that initially must be followed:
� First a grid must be set over the corroded area� The at the node points, thicknesses have to be surveyed� An analysis must be carried out for future service
Then we must consider the future possibilities:
� Can vessel be put back into service, or,� Must the operating pressure be reduced� Must the future service life be reduced� Must a repair be carried out� Must the vessel be scrapped !
API 579 can answer these questions
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API 579 – Level 1 Assessment
The first step: Survey the corroded area for thickness at the nodes
C1 C2 C3 C4 C5 C6 C7 C8
M1
M2
M3
M4
M5
M6
M7
1 Overlay the corroded area with a grid
2 Label the circumferential and longitudinal lines in the grid
3 Measure the thicknesses at the node points
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Next: Make a table of the thickness measurements inches
0.48
0.75
0.55
0.36
0.48
049
0.75
0.360.75 0.48 0.47 0.55 0.48 0.49 0.75
1 Find the smallest thickness in the circumferential direction
2 Find the smallest thickness in the longitudinal direction
2 Complete finding the minimum thicknesses in each direction
API 579 – Level 1 Assessment
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Next: Find the absolute smallest thickness tmm
0.48
0.75
0.55
0.36
0.48
049
0.75
0.360.75 0.48 0.47 0.55 0.48 0.49 0.75
Here is the Critical Thickness Profile (CTP) in the longitudinal direction
We can plot the CTP to find the average thickness
API 579 – Level 1 Assessment
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Here is the plot of the CTP (critical thickness profile)
0.470.55
0.36
0.48 0.49tmm
1,5” 1,5” 1,5” 1,5”
This is the grid distance.
Now, we need the Critical Length LQ = 0,4581 (read from Table 4.5)
Thickness to be used in the assessment tc
tc = Original thickness – Corrosion allowance = 0,75-0,1 = 0,65 in
L
L = Q(D.tc)1/2 = 0,4581.(48,2.0.65) =2,564 in (D corroded)
2,564 in
API 579 – Level 1 Assessment
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Now we need the thicknesses t1 and t2
0.470.55
0.36
0.48 0.49tmm
1,5” 1,5” 1,5” 1,5”
L
2,564 in
API 579 – Level 1 Assessment
t1 t2
From the geometry t1 = 0,522 in and t2 = 0,463 in
Find the average thickness tam
tam = this shaded area divided by L tam = 0,426 in
tcmin From the code = = = 0,430 inP.R
S.E-0,6.P
300.24,1
20000.0,85-0,6.300
tam Corroded = tam - FCA = 0,426 – 0,1 = 0,326 in
(FCA = Future corrosion allowance)
Conclusion: Fails Level 1 assessment - try Level 2 assessment
FCA
API 579 – Level 2 Assessment
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tam
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API 579 – Level 2 Assessment
tam Corroded = tam - FCA = 0,426 – 0,1 = 0,326 in
We have already computed the corroded average thickness:
We need the Remaining Allowable Strength Factor RSFa
The calculated required thickness (code formula) tcmin= 0,430 in
This is normally taken as RSFa= 0,9
Now: RFSa x tcmin = 0,9 x 0,430 = 0,387 in
The remaining thickness Fails Level 2
We can drop the MAWP of the component
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API 579 – Reduced MAWP
tam Corroded = tam - FCA = 0,426 – 0,1 = 0,326 in
We have already computed the corroded average thickness:
MAWP = = = 228,1 psi t.S.E
R+0,6.t
0,326.20000.0,85
24,1+0,6.0,326
Thus the MAWP has to be reduced from 300 psi to 228,1 psi
This method used the formal grid, but we can take random points
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API 579 – Consider Random Points
First, we make a table of the random point thicknesses
Next we compute the average thickness
tam = Σ (trd)1
N i =1
N
= 12,0667 mm
Next we compute (trd – tam) for each point
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API 579 – Consider Random Points
First, we make a table of the random point thicknesses
Next we compute (trd – tam) for each point
Then compute (trd – tam)2 for each point
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API 579 – Consider Random Points
First, we make a table of the random point thicknesses
Next we compute (trd – tam) for each point
Then compute (trd – tam)2 for each point
Find the value S
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API 579 – Consider Random Points
First, we make a table of the random point thicknesses
Find the value S
S = Σ (trd-ta,)2
i =1
N
= 12,9333 mm
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API 579 – Consider Random Points
Now compute the Coefficient Of Variance COV
COV = =1
tam
S
N-1[ ]
0,5 1
12,0667
12,9333
14-1[ ]
0,5
= 0,080 or 8%
Because COV is less than 10% we can use the average thick. tam
Now the original nominal thickness tnom = 16 mm
LOSS is the amount of corrosion that has already taken place
LOSS tnom - tam= 16 – 12,0667 = 3,9333 mm
We now have to use code formula to compute required thickness
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API 579 – Consider Random Points
This is the cylinder we with which are dealing
P = 3,85 MPa Current MAWPD = 484 mm Inside diameterS = 96,196 MPa Allowable design stress E = 1,0 Joint Efficiencyc = 2 mm Corrosion allowance – Also FCAtnom = 16 mm Actual new thickness of the cylinder
From the code formula:
tcmin = = = 10,1670 mmP.(R+c+LOSS)
S.E-0,6.P
3,85.(242+2+3,9333)
96,196.1,0-0,6.3,85
tam – FCA = 12,0667 – 2,0 = 10,0667mm
The cylinder is not thick enough for future service per Level 1
We can:
� Reduce the MAWP, or,� Reduce the future corrosion allowance (FCA)� Try Level 2 assessment
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API 579 – Consider Random Points – Level 2
We already know the tam – FCA = 10,0667 mm
The minimum measured thickness tmm- FCA = 8 mm
tlim = max(0,2.tnom; 2,5) = max(0,2.16; 2,5) = 3,2 mm
Compute max(0,5.tcmin ; tlim) = max(0,5.10,167; 3,2)
= 5,0835 mm
Level 2 assessment passes
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Simple calculation – Remaining Life
Suppose we have the following data over a period of 5 years
� The total uniform corrosion is 2,5 mm for the 5 years� We need to know the remaining life of a head
This is the head whose remaining life we require:
P = 1,5 MPa MAWP if the headD = 2500 mm Original Diameter (New)c = 3 mm Original corrosion allowanceS = 120 Mpa Design stress of the materialE = 1,0 Joint efficiency
Uniform corrosion rate = 2,5 / 5 = cRate = 0,5 mm per year
tFinal = = = 15,645 mmP.D
2SE–0,2P
1,5.2500
2.120.1,0–0,2.1,5
t = Current corroded thickness = 20 mm
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Simple calculation – Remaining Life
tFinal= 15,645 mm, and t = 20 mm and cRate = 0,5 per year
Remaining corrosion = 20 - 15,645 = 4,355 mm
Remaining life = 4,355 / corrosion rate = 4,355 / 0,5 = 8,71 years
This assumes the corrosion continues at a uniform rate
We move onto something else now
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Simple Example – Pitting Corrosion
This is a typical example of pitting corrosion
The Level 1 assessment is very simple
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Simple Example – Pitting Corrosion
First, we get the data for the cylinder – original design
D = 60 in Inside diametertnom = 0,75 in Original thicknessLOSS = 0,05 in Uniform metal loss so farFCA = 0,07 in Future corrosion allowanceS = 17500 psi Design stress E = 0,85 Joint Efficiency
Step 1: Find the worst the area with the highest number of pits
Step 2: Measure maximum pit depth wmax= 0,3 in
Step 3: Get the value tc from this equation
tc = tnom – LOSS – FCA = 0,75 – 0,05 – 0.07 = 0,63 in
Step 4: Get the remaining thickness ratio Rwt from this equation
tc
tc + FCA-wmax
0,63
0,63 + 0,07-0,3Rwt = = = 0,6349
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Simple Example – Pitting Corrosion
tc = 0,63 in Rwt = 0,6349 LOSS = 0,05 in FCA = 0,07 in
Step 5: Get the effective inside radius of the shell Rc
Rc = D/2 + LOSS + FCA = 30+0,05+0,07 = 30,12 in
Step 6: Compute the MAWP of the cylinder
MAWP = = = 307 psiS.E.tc
Rc+0,6.tc
17500.0,85.0,63
30,12+0,6.0,63
Step 7: Check to see if the MAWP has to be reduced
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Simple Example – Pitting Corrosion
Here is an example of a pit measuring guage
We move on to a new subject
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Out of Roundness - Peaking
This is what is meant by peaking:
The ends of the cylinder at straight instead of cylindrical
Measure the peak height δ = 0,31 in
δ
We perform a Level 2 assessment
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Out of Roundness - Peaking
Get all the data for this cylinder
δ
36 in
0,31 in
Do = 36 in cylinder outside diametertnom = 0,5 in wall thicknessP = 315 psi design pressureE = 1,0 joint efficiencyFCA = 0,05 in future corrosion allowanceEY = 25,2.106 psi Elastic ModulusSa = 16 800 psi design stressHf = 3,0 factor for secondary stress – from API 579
0,5 in
LOSS= 0 in metal lost so far
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Out of Roundness - Peaking
Get all the data for this cylinder
Do = 36 in cylinder outside diametertnom = 0,5 in wall thicknessP = 315 psi design pressureE = 1,0 joint efficiencyFCA = 0,05 in future corrosion allowanceEY = 25,2.106 psi Elastic ModulusSa = 16 800 psi design stressHf = 3,0 factor for secondary stress – from API 579
LOSS= 0 in metal lost so far
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Wall thickness to be used in the assessment tc
tc = tnom – LOSS – FCA = 0,5 – 0,0 – 0,05 = 0,45 in
Get the current membrane stress from the Code σm
σm = = = 12 474 psiP.(Ro- 0,4.tc)
tc.E
315(18 - 0,4.0,45)
0,45.1,0
R internal radius = Ro – tnom + FCA + tc/2 = 17,775 in
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Out of Roundness - Peaking
Induced extra bending stress ratio SP
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ν is Poisson’s Ratio = 0,3
SP = ( )0,5 = ( )0,5
12(1- ν 2)PR3
EY.tc3
12(1- 0,3 2)315.17,7753
25,5.106.0,453
= 2,88
From SP and δ/R and Figure 8.13 to get Cf = 0,83
Out of Roundness - Peaking
Induced extra bending stress ratio SP
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ν is Poisson’s Ratio = 0,3
SP = ( )0,5 = ( )0,5
12(1- ν 2)PR3
EY.tc3
12(1- 0.3 2)315.17,7753
25,5.106.0,453
= 2,88
From SP and δ/R and Figure 8.13 to get Cf = 0,83
We need 3 other values: Rb1, Rb2 and Rbs the calculation is long
Rb1 = 3,43, Rb2 =3,43 and Rbs= -1,0 (no calculation shown)
Finally, compute the Remaining Strength Factor RSF
Hf.Sa
σm.(1+Rb1)+LOSS(1+Rbs)RSF = min( ; 1,0)
3,0.16 800
12,474.(1+3,43)+0= min( ; 1,0 ) = 0,91
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Out of Roundness - Peaking
RSF = 0,91 from the previous slide
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RSF allowed = 0,90 API 579 requirement
Conclusion: passed, cylinder can be put into service
A word or two about grooves
A groove can be thought of as a blunt crack in the vessel
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σ σ
This can be analysed mathematically (modeled as) as a half ellipse
The two important dimension are: crack length a, and tip radius r
ar
Adjacent to the edge of the crack, the stresses increase
Sa
Sc
Sa is the average stress, and Sc is the increased stress
Sa / Sc is known as the Stress Concentration Factor (scf)
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A word or two about grooves
A Cambridge researcher, Inglis did the mathematical work
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σ σ
ar
Sa
Sc
This is the equation he devised for the scf
scf = 2 + √ a
r
If the crack tip has a very sharp radius, the scf if very high
The stress can be in the Plastic range of the stress-strain diagram
A word or two about grooves
A Cambridge researcher, Inglis did the mathematical work
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This is the equation he devised for the scf
scf = 2 + √ a
r
If the crack tip has a very sharp radius, the scf if very high
The stress can be in the Plastic range of the stress-strain diagram
ε strain
σstress
elastic range
plastic range
Yield point
0,2%
This can be a source of fatigue problems
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A word or two about grooves
A Cambridge researcher, Inglis did the mathematical work
3/26/2013 45
This is the equation he devised for the scf
scf = 2 + √ a
r
If the crack tip has a very sharp radius, the scf if very high
The stress can be in the Plastic range of the stress-strain diagram
This can be a source of fatigue problems
We have not analysed grooves and pitting
This is the end of the presentation
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