**Pipe wall thickness calculation** is one of the important basic activities for every piping engineer. Process plants deal with the fluids which flow inside the pipe at high-pressure and temperature. So, the pipe deals with high circumferential pressure which can cause the bursting of the pipe if the pipe thickness is not enough. Hence, **The designers need to find out the required piping thickness as per section “304.1.2 of ASME B31.3” to resist the internal line pressure. **The operation must be leak-free. In this article, I have simplified the **pipe thickness calculation procedure**. A sample** pipe wall thickness calculation** problem is discussed mentioning the calculation steps. **Process Piping Code ASME B 31.3 is used as the basis for the Pipe thickness calculation**.

## Few **Important Points **required for Pipe thickness Calculation

Before starting **the piping thickness calculation**, the engineer should be aware of the following points:

- Process plants are designed for 20 years or 7200 cycles. (Considering 1 cycle per day; Total no of cycles in 20 years=20*360=7200 cycles)
- Pressure and temperature can vary from line to line and from time to time.
- Fluid could be corrosive and toxic in the system.
- Corrosion allowance for ferritic mostly taken 3 mm and zero mm for austenitic steel.
- Mill tolerance for the seamless pipe is 12.50% and 0.3 for the welded pipe.

**How do I Calculate Pipe Thickness?**

Let’s consider the following details for the **pipe thickness calculation** of a seamless Carbon Steel pipe.

- MOC (Material of Construction) of the pipe –
**A106 Gr. B** - NPS (Nominal Pipe Size) –
**4”** - Manufacturing type of the pipe (SMLS, EFW, ERW) –
**Seamless (SMLS)** - Design Pressure (PSI) –
**1200 PSIG** - Design Temperature –
**500° F** - Mechanical, corrosion and erosion allowances –
**3 mm** - Mill Tolerance –
**12.50% of the thickness**

**Pipe Wall Thickness Calculation Formula**

**As per clause 304.1.2 (a) of ASME B 31.3, the internal pressure design thickness for straight pipes with t<D/6 can be calculated using the following formula (Equation (3a):**

Here,

**P: Internal Design Gage Pressure=1200 PSIG as per problem definition**

**D: Outside Diameter of the pipe**

**The equation for the pipe wall thickness is based on the outside diameter of pipe, rather than the inside diameter. This is because the outside diameter of pipe is constant, it is independent of the wall thickness. Hence, the pipe wall thickness can directly be calculated easily using the pipe outer diameter.**

Outside diameter has to be taken from the below standards-

**ASME B36.10M:**for ferritic steel (seamless & welded wrought steel pipes).**ASME B36.19M:**for austenitic steel (stainless steel pipes)

So from Fig. 2, D= 114.3 mm

**S: Allowable Stress value of the Pipe Material** (A 106-B) at Design Temperature (**500° F**)

**Refer to Table A-1 (or Table A-1M) of the ASME B31.3** (Fig. 3) for getting the value of the allowable stress. Travel in the horizontal (x) direction for allowable stress value and vertical (y) direction for pipe material, and the match point to get the value (refer to Fig. 3). If required, use interpolation to calculate the middle value.**Note: the value** of the allowable stress in Table A-1 is given in KSI, So we need to convert the value in PSI.

As per Fig. 3, the allowable stress for ASTM A106 Gr.B is 19,000 psi at 500°F.

**E: Quality factor**

**Quality Factors are used in Pressure Design and applied at Longitudinal and Spiral Weld Joints and for Castings. The maximum Value of quality factors is 1.0.**

The value of E, Longitudinal Weld Joint Quality Factor, or Casting Quality Factor can be found in **Table A-1A or Table A-1B of the ASME B31.3**. The weld joint factor (E) is 1 for our problem case (Refer to Fig. 4).

**W: Weld Joint Strength Reduction Factor**

As per section 302.3.5(e) of ASME B31.3, **The weld joint strength reduction factor, W, is the ratio of the nominal stress to cause the failure of a weld joint to that of the corresponding base material for an elevated temperature condition of the same duration. It only applies at weld locations in longitudinal or spiral (helical seam) welded piping components.**

**Weld Joint Strength Reduction Factors are used because at elevated temperature the weld joint creep rupture strength can be lower than the base metal.**

The value of W can be found from Table 302.3.5 of ASME B 31.3 (Refer to Fig. 5) and for our problem the value of W=1

### Y: Values of Coefficient from Table 304.1.1,

**The factor “Y” depends on temperature. At elevated temperatures, the factor Y increases leading to a decrease in the calculated required pipe wall thickness.**

**Refer to Table 304.1.1 of ASME B31.3 for finding the value of Y, **It is Valid for t < D/6 and materials shown below The value of Y may be interpolated for intermediate temperatures. For material A106 Gr. B, Y is given 0.4 (refer to Fig. 6)

## Pipe Thickness Calculation Steps

**Step 1. Put the above values in the equation** **shown in Fig. 1**

**t=(1200* 114.3)/{2(19000*1*1+1200*0.4)}=3.52 mm; Hence calculated thickness (t)= 3.52 mm**

**Step 2. Add the corrosion to the calculated thickness.**

t_{c} = t + c = 3.52 + 3**t _{c }= 6.52 mm**

**Step 3. Add the mill tolerance to the thickness after adding corrosion value.**

t_{m }= t_{c} + 12.50 % of the pipe thickness**t _{m} =tc/0.875 **

**=6.52/0.875**

**= 7.45 mm**(This is required thickness)

**Step 4. Check the next Ordering thickness available in ASME B36.10M considering the required thickness.**

**So from Fig. 7, The Ordering thickness is 8.56 mm or Schedule 80.**

**Note:1. **Ordering thickness for seamless pipe will always be the next greater value available from Schedule to schedule.

2. Whereas for welded pipe any next greater value will be the ordering thickness.

3. Extra thickness can be calculated by ordering thickness minus the required thickness = (8.56 – 7.45) = 1.11 mm.

**Use of the Extra thickness available in the pipes**

- for calculating the life of a pipe after 20 years.
- For calculating maximum pressure holding capacity of the pipe.
- For checking extra thickness is sufficient to take care of thinning, if the same pipe is used for manufacturing the bend.
- The extra thickness also minimizes deflection and reduces the number of support.
- To compare with flange pressure holding capacity, to declare pipe is stronger than the flange.

Few more useful Resources for you.

Pipeline wall thickness calculation with example

Meaning of Pipe Schedule / Schedule Numbers?

Piping Layout and Design Basics

Few Job Opportunities for you

how we use the extra wall thickness available for calculating the life of a pipe after 20 years?

what is the relation between the service life (20 years) and cycles (7200 cycles)? and how it affect the original wall thickness calculations?

Hi,

Corrosion allowance is decided based on the service life of the plant, for our sample problem it is taken 3 mm for 20 years or 7200 cycles. If the project required plant life more than 20 years then the corrosion allowance will be taken more as per the number of years.

Therefore, the thickness of the pipe (Considering the same operating condition) for plant life 20 years and plant life 30 years will be different (will be greater for 30 years).

For calculating the life of the pipe after 20 years use the below formula

Life of Pipe after 20 years = (extra thickness available in the pipe)/(corrosion allowance per year)

Where,

extra thickness for our sample problem was 1.11 mm

corrosion per year = 3mm / 20 years = 0.15 mm

So,

Life of the Pipe after 20 years = (1.11)/(0.15) = 7.4 years

what is the relation between the figure 20 years and 7200 cycles?

Best regards

why not selecting 7.92 from chart

What is the basis for corrosion thickness/erosion thickness selection of 3 mm for 20 Years?

Hi,

There is no corrosion allowance exactly specified in ASME B31.3. Corrosion allowances are normally established by the end-user and are somewhat based on industry tradition or practices. 3 mm for CS piping is a common standard.

Humidity, temperature, rain, wind, impurities, metal wet times, and the types of the fluid flowing through pipe have an effect on the corrosion rate. so corrosion allowance may vary from places to places.

The relation between 20 years and 7200 cycles is simple. It assumes one temperature cycle per day (24 hours cycle) and 20 years are approximately 7200 days. Ideally, it should be 7300 cycles.