Friday, 8 July 2011

PRESSURE VESSELS





MECHANICS - CASE STUDY

Introduction


Cylindrical Gas Pressure Vessels
on Rail Car
A rail car needs to transport pressurized gas and due to shipping constraints, each cylinder can have only a diameter of 34 cm (inside dimension). The end caps are not considered in this initial design.
What is known:
  • Internal pressure is 10 MPa.
  • Inside diameter of a single vessel is 34 cm.
  • Yield (allowable) stress for the steel shell is 250 MPa.
  • Factor of Safety is 2.0.
  • The vessels should be designed to meet the maximum distortion energy criteria for predicting failure.

Dimensions
Question

What is the minimum wall thickness in the cylindrical section of the gas pressure vessel?
Approach

  • Determine the hoop and axial stresses as a function of the thickness.
  • Apply the factor of safety.
  • Use the maximum distortion energy criteria equation to determine the wall thickness.
         






MECHANICS - THEORY


Thin-walled Pressure Vessels


Cylindrical Pressure Vessel with
Internal Pressure
Both cylinderical and spherical pressure vessels are common structures that are used ranging from large gas storage structures to small compressed air tanks in industrial equipment. In this section, only thin-walled pressure vessels will be analyzed.
A pressure vessel is assumed to be thin-walled if the wall thickness is less than 10% of the radius (r/t < 10). This condition assumes that the pressure load will be transfered into the shell as pure tension (or compression) without any bending. Thin-walled pressure vessels are also known as shell structures and are efficient storage structures.
If the outside pressure is greater than the inside pressure, the shell could also fail due to buckling. This is an advanced topic and is not considered in this section.
Cylindrical Pressure Vessels


Cylindrical Vessels will Expierence
Both Hoop and Axial Stress in
the Mid-section
Only the middle cylindrical section of a cylinder pressure vessel is examined in this section. The joint between the end caps and the mid-section will have complex stresses that are beyond the discussion in this chapter.
In the mid-section, the pressure will cause the vessel to expand or strain in only the axial (or longitudinal) and the hoop (or circumferential) directions. There will be no twisting or shear strains. Thus, there will only be the hoop stress, σh and the axial stress, σa. as shown in the diagram at the left.

Cross Section Cut of
Cylindrical Vessel
Pressure vessels can be analyzed by cutting them into two sections, and then equating the pressure load at the cut with the stress load in the thin walls. In the axial direction, the axial pressure from the discarded sections will produce a total axial force of p(πr2) which is simply the cross section area times the internal pressure. It is generally assumed that r is the inside radius.
The axial force is resisted by the axial stress in the vessel walls which have a thickness of t. The total axial load in the walls will be σa(2πrt). Since the cross section is in equilbrium, the two axial forces must be equal, giving
     p(πr2) = σa(2πrt)
This can be simplified to
where r is the inside radius and t is the wall thickness.

Hoop Section Cut from
Cylindrical Vessel
In addition to the axial stress, there will be a hoop stress around the circumference. The hoop stress, σh, can be determined by taking a vertical hoop section that has a width of dx. The total horizontal pressure load pushing against the section will be p(2r dx) as shown in the diagram.
The top and bottom edge section will resist the pressure and exert a load of σh(t dx) (each edge). The edge loads have to equal the pressure load, or
     p(2r dx) =σh(2t dx)
This can be simplified to
where r is the inside radius and t is the wall thickness.
Spherical Pressure Vessel


Spherical Pressure Vessel
Cut in Half
A spherical pressure vessel is really just a special case of a cylinderical vessel. No matter how the a sphere is cut in half, the pressure load perpendicular to the cut must equal the shell stress load. This is the same situation with the axial direction in a cylindrical vessel. Equating the to loads give,
     p(πr2) = σh(2πrt)
This can be simplified to
Notice, the hoop and axial stress are the same due to symmetry






MECHANICS - CASE STUDY SOLUTION


Cylindrical Gas Storage Tank
A gas storage tank needs to be designed to hold pressurized gas at 10 MPa. The tank inside diameter is set at 34 cm due to tank stacking system on a rail car. For safety reasons, a factor of safety of 2.0 is required. The material is steel with a yield stress of 250 MPa. The thickness of the material needs to be determined.
To account for the stress interaction between the hoop and axial directions, the maximum distortion energy theory (von Mises' Yield Criterion) will be used to predict failure.
It is assumed that the end caps will not fail and only the cylinder middle section will be considered (end cap stresses are complex and not studied in this eBook).
Hoop and Axial Stresses

Functions for the hoop and axial stress can be determined for a cylindrical pressure vessel. These are
     σh = Pr/t = (10 MPa)(0.17 m)/t = 1.7/t
     σh = Pr/(2t) = (10 MPa)(0.17 m)/(2t) = 0.85/t
Both the stresses are functions of t.
Failure Criteria


Stress Element at Cylinder Section
The maximum distortion energy criteria takes into consideration stresses in multiple directions. The equation is
     
or for this case,
     
The yield stress is given as 250 MPa. However, to account for a factor of safety of 2.0, the actual yield stress is reduced in half. Substituting into the failure equation gives,
     
     2.890 + 0.7225 - 1.445 = 15,625 t2
     t = 0.1178 m = 11.78 mm






MECHANICS - EXAMPLE

Example


A Partial Section of a Penstock
A penstock for a hydraulic power plant has an inside diameter of 1.5 m and is composed of wooden staves bound together by steel hoops. The cross-sectional area for each steel hoop is 300 mm2. If the allowable tensile stress for the steel is 130 MPa, what is the maximum space, L, between the hoop bands under a head of water of 30 m? The mass density of water is 1,000 kg/m3. The water pressure can be assumed to be the same at all interior locations of the penstock.
Solution


Spacing of the Steel Hoops
in the Penstock
The pressure corresponding to a head of 30 m water is given by

     p = ρgh
        = (1,000) (9.81) (30)
        = 294 kPa
Circumferential stress in the steel bands is considered as the failure criteria for a safe design. If the maximum spacing between hoops is denoted as L, then each hoop must resist the water pressure over a length L of the penstock.

Direct Evaluation of Bursting Force F
The bursting force F, acting over the flat surface of the fluid equals the pressure intensity p multiplied by the area, DL, over which it acts.

     F = pDL
        = (294) (1.5) L
        = 441 L

This bursting force will be resisted by the equal forces P acting on each cut surface of the cylindrical wall.
Assuming the whole resisting force will be given by the steel hoops,

     P = Aσ
        = (300×10-6) (130×103) kN
        = 39 kN

Applying the summation of forces,

     ΣF = 0
     F - 2P = 0
     F = 2P
     441 L = 2 (39)
     L = 0.1769 m = 177 mm