In this part of the pressure vessel sizing tutorial we will discuss the minimum required thickness and maximum allowed pressure calculation of the top and bottom hemispherical shell portion of the thin walled pressure vessel according to the ASME section 8 division 1 codes.

** **

**Necessary equations as per ASME section VIII division I codes**

Minimum required thickness for hemispherical head and bottom portion of the pressure vessel (PV),

**tr1 = [P*Ri1/(2*S*E -0.2*P)]+C ………………………………..Eq.1**

** **

Maximum Allowed pressure for hemispherical head and bottom,

**Pm1= (2*S*E*nt)/(Ri1-0.4*nt)…………………………………Eq.2**

** **

Where,

P = Applied pressure inside the vessel

Ri1= Pressurevessel Inside radius after adjusting the corrosion allowance

S= Maximum allowable stress for the PV shell material

E=Efficiency of the welded joints of the shell

C=Corrosion allowance

nt=Actual thickness of the shell – corrosion allowance

The American Society of Mechanical Engineers (ASME) section eight division 1 codes suggests that the thin wall pressure vessel design is safe if:

*1. **Actual shell thickness at the hemispherical portions (t) >= tr1*

*2. **P<=pm1*

**Example Problem: **Let’s take the example discussed in Part-1 of this tutorial series.

We have the following input data:

P=**0.000075 MPa **(refer part-1)

**Ri1=1349.98 mm **(refer part-3)

S=103 MPa (refer part-1)

E=0.85 (assumed)

C=0.02 (assumed)

nt=5.98 (refer part-4)

From the **Eq.1,**

**tr1=0.020578241mm**

As, tr1 (0.02mm) is much lesser than the actual thickness (6mm) for this pressure vessel design , so the design is acceptable according to the ASME design codes.

From the **Eq.2,**

**Pm1=0.777016417 MPa**

As, Pm1 (0.77 MPa) is larger than P(0.000075 MPa) so the PV design is safe according to the Section 8 codes.

The ASME codes equations for the required minimum thickness and the maximum allowable internal pressure of the hemispherical area can be used for the thin walled pressures vessel sizing calculation.

## 4 comments

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## amit

April 3, 2012 at 3:40 pm (UTC 5.5) Link to this comment

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## amit

April 4, 2012 at 3:23 pm (UTC 5.5) Link to this comment

Fine information, many thanks to the author. It is puzzling to me now, but in general, the usefulness and importance is overwhelming. Very much thanks again and good luck!

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## Abhisar20

April 6, 2012 at 2:06 pm (UTC 5.5) Link to this comment

please provide the next details. Thanks

## MechGuru

April 10, 2012 at 8:25 pm (UTC 5.5) Link to this comment

Working on the last two parts and will be available soon