fitness for service and api 579 - pixis.cl · fitness for service and api 579 presented by: ......

3/26/2013

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Fitness for Service and API 579

Presented by:

Ray Delaforce

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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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This another very typical example

The white flecks are from insulation that has been removed

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This is galvanic corrosion of aluminium and steel in sea water

Note the presence of red Hematite on the steel

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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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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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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-



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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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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


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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


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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


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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


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


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tam

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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


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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Then compute (trd – tam)2 for each point

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Then compute (trd – tam)2 for each point

Find the value S

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Find the value S

S = Σ (trd-ta,)2

i =1

N

= 12,9333 mm

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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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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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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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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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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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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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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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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


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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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 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



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scf = 2 + √ a

r



ε strain

σstress

elastic range

plastic range

Yield point

0,2%

This can be a source of fatigue problems

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scf = 2 + √ a

r



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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Thank you for watching this presentationAre there any questions

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