Refer to the following example calculation for Determination of the max. allowable working pressure for a given pipe wall thickness.

EXAMPLE

Given: Code : ASME B31.4

Design Temperature : 140 oF (60 oC)

Pipe Material : API 5L Grade A25, ERW (in accordance with ASME B36.10M)

Allowable Tensile Strength of Material at Design Temp., S: 18000 psi

Joint Efficiency for pipe ERW, E : 1.0

Nominal Pipe Size, NPS = Do : 48 inch (1219 mm)

Pipe Schedule (Wall Thickness): STD. (9.53 mm)

Assume a corrosion allowance, C = 3 mm (0.11811 inch)

(1 psi = 1 pound per square inch = 0.070306955 kg/cm2)

Find: Max. allowable working pressure, or what we called design pressure, P.

Data:

Nominal pipe wall thickness, tn = 9.53 mm (Schedule STD.)

Pipe Wall Thickness Tolerance = ± 12.5% tn

Remaining thickness after deducting of tolerance, t = (1-0.125) tn = 0.875 tn = 0.875 x

9.53 mm = 8.33875 mm (0.328297 inch)

Corroded Thickness = 8.33875 – C = 8.33875 – 3 = 5.33875 mm ( 0.210187 inch)

Solution:

t = PDo / 2 SE or P = 2 t S E / Do

P = 2 t S E / Do = 2 (0.210187 inch)(18000 psi)(1.0) / (48 inch) = 157.6403 psi (10.869

bar)

So, the pipe is designed to resist a design pressure P = 157 psi (11.0382 kg/cm2)

Therefore, the operating pressure shall not exceed Pop = 120 psi (8.4368 kg/cm2)

For Remaining Life (years), please refer to the API 570 Piping Inspection Code

Inspection, Repair, Alteration, and Rerating of Inservice Piping Systems.

7.1 CORROSION RATE DETERMINATION

7.1 .1 Remaining Life Calculations

The remaining life of the piping system shall be calculated from the following formula:

Remaining Life (years) = (tactual – trequired) / Corrosion Rate

Where,

tactual = the actual thickness, in inches (millimeters), measured at the time of inspection for a given location or component as specified in 5.6.,

trequired = the required thickness, in inches (millimeters), at the same location or component

as the tactualmeasurement computed by the design formulas (e.g., pressure and structural) before corrosion allowance and manufacturer's tolerance are added.

My friends, I did my best to rewright the file into DOC instead of PDF.

Refer to the following example for determination of the min. required pipe wall

thickness and calculating the remaining life of such a pipeline.

Based on API 570 for Piping Inspection Code "Inspection, Repair, Alteration, and Rerating of Inservice Piping Systems".

The remaining life of the piping system shall be calculated from the following formula:

Remaining Life (years) = (tactual – trequired) / Corrosion Rate (API 570, Para. 7.1.1)

Where, tactual = the actual thickness, in inches (millimeters), measured at the time of inspection for a given location or component as specified in 5.6.,

trequired = the required thickness, in inches (millimeters), at the same location of component

as the tactualmeasurement computed by the design formulas (e.g., pressure and structural) before corrosion allowance and manufacturer's tolerance are added.

EXAMPLE

Given: A pipeline was constructed 15 years ago, the actual thickness measured along the

full life was as follows:

Pipe Schedule:

Nominal Pipe Wall Thickness tnominal (initial thickness when the pipe was new)= 0.438"

(11.125 mm),

● After 1ST interval of 5 years, tactual5 = 10.625 mm

● After 2ND interval 5 years, tactual10 = 10.125 mm (tprevious)

● After 3RD interval 5 years, tactual15 = 9.625 mm (tactual)

Code : ASME B31.8

Pipe material: API 5L XLS, ERW

Nominal Pipe Size: NPS 30" (Outside Diameter, Do = 30")

Specified Min. Yield Strength (SMYS), S = 52000 psi at the design temp.

Max. Operating Pressure (Design pressure), P = 910 psi

Appropriate Design Factor, F = 0.72 (ASME B31.8,Para. 841.114A)

Longitudinal Weld Joint Factor, E = 1 (ASME B31.8,Para. 841.115A)

Temperature Derating Factor, T = 1 (ASME B31.8,Para. 841.116A)

Find: The Remaining Life of the Pipeline in years.

Solution:

Min. Required Thickness, trequired, (or tmin. or tr, also called designed thickness or critical

thickness) = P Do / (2 S F E T)

= (910 psi)(30 inch)/2(52000 psi x 0.72 x1 x1) = 0.364" (9.260 mm) mm/year

Long Term Corrosion Rate, LT = (tinitial – tlast)/time between last and initial = (11.125–

9.625)mm/15 years = 0.1 mm/year

Short Term Corrosion Rate for Last interval, ST = (tprevious – tlast)/time between last and previous = (10.125–9.625)mm/5 years = 0.1 mm/year

According to API 570, Section 7, the higher of these two rates LT and ST should be used to

estimate the remaining life. The interval between piping inspections shall be established and maintained depending on corrosion rate and remaining life calculation.

So, the corrosion rate to be considered = Max. (LT, ST) = 0.1 mm/year

The Remaining Life of the Pipeline = (tactual – trequired) / Corrosion Rate = (9.625 – 9.260)

mm / (0.1 mm/year) = 3 Years, 7 Months and 24 Days.

Before that interval, you have to inspect the pipeline.

Very Important Note. If there is a local corrosion or a pitting, you are requested to apply

the procedures of ASME B31.G for determining the remaining strength of corroded pipeline. Or you can use the assessment(s) provided by API 579 Fitness For Surface, FFS.

Dear inspectorjoe, please let me know your comments and any additionals you need.

(Any example derived here is formulated by myself, may be from an actual case, and not extracted from any code or manual).

♣ Sorry for typing mistake, in the following calculated equation, please delete the high

lighted distinguish:mm/year

"Min. Required Thickness, trequired, (or tmin. or tr, also called designed thickness or critical

thickness) = P Do / (2 S F E T) = (910 psi)(30 inch)/2(52000 psi x 0.72 x1 x1) = 0.364" (9.260 mm) mm/year"

♣ And, I'd like to clear that the pipeline expected remaining life is 3 years, 7 months after

its old life of 15 years, i.e. total life shall be 18 years, 7 months, if the corrosion rate still as

it is 0.1 mm/year.

You can assume a pipe material as ASTM A53, Furnace , Butt Welded & Continuous welded.

The Allowable Tensile Strength of A53, S = 0.72 SMYS = 0.72 x 25000 = 18000 psi

where, Specified Minimum Yield Strength, SMYS = 25 000 psi

Pipe Material Joint Efficiency, E = 0.6

The Min. Required Thickness, tr = P Do / (2 S E)

Operating Pressure, Pop = 150 psi

Assume a Design Pressure (more than the operating), P = 185 psi

For Pipes with NPS 6" : Outside Diameter, Do = 6.625 inch

For Pipes with NPS 8" : Outside Diameter, Do = 8.625 inch

For Pipes with NPS 10" : Outside Diameter, Do = 10.75 inch

For Pipes with NPS 12" : Outside Diameter, Do = 12.75 inch

To find the pipe nominal thickness t = tr + CA (where CA = Corrosion Allowance)

To find the pipe nominal thickness, tn = t / 0.875 (an increase represents 12.5%t)