Pressure vessel project.docx

PRESSURE VESSEL DESIGN
PREPAREDBY YARED BERIHUN Page 1
DIRE DAWA UNIVERSITY
Institute of technology
School of mechanical and industrial engineering
Department of mechanical engineering
Machine design project
Student project: Pressure vessel with acidic solution medium
Name: YARED BERIHUN
ID: DDU11004374
 INSTRUCTOR:BALASUADHAKAR

CONTENTS………………………………………………………….PAGES
CHAPTER ONE……………………………………………………………………………………….……………4
1.1 INTRODUCTION OF PRESSURE VESSEL…………………………………………………………………………………………………4
1.2 typesof pressure vessel…………………………………………………………………..………………………………5
1.3 Industrial application’s………………………………………………………………………………………….…………7
1.4. Componentof pressure vessel……………………………………………………………………………….…………7
1.5 . Determiningthe diameterandlengthof a pressure vessel………………………………………………12
CHAPTER TWO………………………………………………………………………………………………..………………….14
2.1.MATERIAL SELECTION………………………………………………………………………………………………………14
2.2.weldingtype…………………………………………………………………………………………………….……………...18
CHAPTER THREE ……………………………………………………………………………………………………………………22
3.1CALCULATE THE SELL THICKNESS………………………………………………………………………………………22
3.2 Review different type of head ………………………………………………………………………………………….…………..22
3.3 calculaterequired head thickness………………………………………………………………………………………………….24
3.3 calculaterequired head thickness……………………………………………………………………………………..…………..25
3.4. Head to shell transition………………………………………………………………………………………………….……………..26
CHAPTER F0UR
4.1 Selection of flange based on the temperature and pressure rating……………….………………………..28
4.2flange and desgin…………………..……………………………………………………………………………..30
4.3Gasket design………………………………………………………………………………………………38
CHAPTER FIVE
5.1 calculating wall thickness of the nozzle…………………………………………………… 39
5.2 reinforcment of opening…………………………………………..………………………………39
5.3 design of manhole or inspection…………………………………………………………………… 41
5.4 suporetdesignandcheckof pressure vessel……………………………………………………………… 42

5.5Fabric ability and Welded joint design……………………………………………………50
summery………………………………………………………………………………………………………54
Reference……………………………………………………………………………………………………..55
CHAPTER ONE
WHAT IS PRESSURE VESSEL
A pressure vessels is defined as a Vessels, tanks, and pipelines that carrying, storing, or receive fluid(liquid
and gas) such as water, oil, and gas at a pressure of above atmospheric pressure, such a pressure vessel
are designed according to national(BIS) and international codes (ASME,BE-D0,AD-2000).It is defined as a
container with a pressure differential between inside(internal) and outside(external). The inside pressure
is usually higher than the outside, except for some isolated situations. The fluid inside the vessel may
undergo a change in state as in the case of steam boilers, or may combine with other reagents as in the
case of a chemical reactor.

Pressure vessels often have a combination of high pressures together with high temperatures, and in some
cases flammable fluids or highly radioactive materials. Because of such hazards it is imperative that the
design be such that no leakage can occur. In addition these vessels have to be designed carefully to cope
with the operating temperature and pressure. It should be borne in mind that the rupture of a pressure vessel
has a potential to cause extensive physical injury and property damage. Plant safety and integrity are of
fundamental concern in pressure vessel design and these of course depend on the adequacy of design codes.
When discussing pressure vessels we must also consider tanks. Pressure vessels and tanks are
significantly different in both design and construction: tanks, unlike pressure vessels, are limited to
atmospheric pressure; and pressure vessels often have internals while most tanks do not (and those that do
are limited to heating coils or space. Boiler drums, heat exchangers, chemical VE
CLASSIFICATION OF PRESSURE VESSEL
❖ According to the position of the vessel: the pressure vessel divides in to two according to
their position.
 HORIZONTAL PRESSURE VESSEL: the pressure vessel lay horizontally.
 VERTICAL PRESSURE VESSEL: the pressure vessellay vertically.

❖ ACCORDING TO FUNCTION
 Storage
 Reactor
 Boiler
 Mixer
 Heat exchanger
❖ ACCORDING TO MATERIAL
 Mild steel
 Cast iron
 Copper
 plastic
 etc…
❖ ACCORDING TO FABRICATION
 Welded
 Casted
 flanged
 Bras
❖ ACCORDING TO LOADING
 Internal pressure
 External pressure
❖ ACCORDING TO WALL TEMPRETURE
 Heated
 Unheated
❖ ACCORDING TO CORROSION ACTION
 High corrosion effect
 Moderate corrosion effect
❖ ACCORDING TO ASSAMBLEY
 Detachable
 No detachable
❖ ACCORDING TO WALL THICKNESS
 thin shell: if the wall thickness of the shell is less than 1/10 used in boilers, tanks and
pipes.
 thick shell: If the wall thickness of the shell is greater than 1/10used in high pressure
cylinders, tanks, gun barrels.
❖ ACCORDING TO END OF CONSTRACTION
 Open end
 Closed end

❖ ACCORDING TO MATERIAL
 Brittle
 Ductile
❖ ACCORDING TO DIRECTION OF FORCE
 Internal pressure
 External pressure
Component of pressure vessel
Pressure vessel have five main components these are
 Shell
 Head
 Nozzle
 Support
 Flanges
 Manhole
 Drain
 Inlet and outlet
SHELL
The shell is the main component of any vessels that contains the pressure. Material of shell
normally come in plate or rolled steel. Commonly, some pressure vessel shells have a
rotational axis and be welded together to form a structure. Most pressure vessel shells are
cylindrical, spherical, or conical in shape.
 It Pressure vessel shells in the form of different plates are welded together to form a
structure that has a common rotational axis.
 Shells are either cylindrical, spherical or conical in shape.
 It is the primary component that contains the pressure.
HEAD

All pressure vessels shells must be closed at the ends by heads (or another shell section).
Heads are typically curved rather than flat. Curved configurations are stronger and allow the
heads to bethinner, lighter, and less expensive than flat heads. Heads can also be used inside a
vessel. These “intermediate heads” separate sections of the pressure vessel to permit different
design condition in each section. Head are usually categorized by their shapes.
 Ellipsoidal,
 hemispherical,
 torispherical,
 conical
 flat are the common types of heads.
Ellipsoidal head⇒ this is also called 2:1 elliptical head. The shape of this head is more
economically, because the height of the head is just a quarter of the diameter. Its radius
varies between major and minor axis.
- All the pressure vessels must be closed at the ends by heads (or another shell
section).
- Heads are typically curved rather than flat.
- The reason is that curved configurations are stronger and allow the heads to be
thinner, lighter and less expensive than flat heads.

NOZZLE
A nozzle is a cylindrical component that penetrates and mounts whether at the shell or heads of a
pressure vessel surface. The nozzle ends are generally flanged. Flanges function is to allow the
necessary connections. Flanges also use to permit easy disassembly for routines maintenance or easy
access.
- A nozzle is a cylindrical component that penetrates into the shell or head of pressure vessel.
- Attach instrument connection (level gauges, thermos wells, pressure gauges
- Attach piping for flow into or out of the vessel.
- Provide access to the vessel interior at man way.
- Provide for direct attachment of other equipment item (e.g. heat exchanger)
SUPPORT
Orientation of the pressure vessel whether horizontally or vertically. In any situation, the
pressure vessel support must be the type of support that is designed and used depends on the
enough to support the applied weight and other loads. Design pressure of the vessel is not
being considered in the design of its support because the support is not being pressurized, but
design temperature should be considered for support design. It should be considered from the
perspective of material selection and provision for differential thermal expansion.
Supports can be classified as follow
Saddles
Legs

Skirt
Bracket supports
SADDLE SUPPORT
Horizontal drums are normally supported by saddle. This type of support divides the weight load
over a large area of the shell to avoid an unnecessary stress in the shell at two different locations.
The width of the saddle is considered by the specific size and design conditions of the pressure
vessel. One saddle support is normally fixed or anchored to its foundation
.
LEGS SUPPORT
Small vertical drums are normally supported by legs that are welded to the bottom of the
pressure shell. The maximum ratio of support provides for leg length to drum diameter is
typically 2:1. The number of legs is designed depends on the drum size and the loads to be
carried. Support legs are also usually designed for spherical pressure vessels. The support
legs for small vertical. vessels and spherical storage vessels normally made from high
carbon material such as structural steel columns or pipe sections, which provides a more
efficient and perfect design.

SKIRT SUPPORT
This type of support generally been used for tall, vertical, cylindrical pressure vessels. This
type of support is a cylindrical shell section which is be weld either to the bottom of the
vessel shell or to the bottom head for the cylindrical vessels. Skirt support for spherical
vessel is welded to the vessel near the mid plane of the shell. The skirt is normally design
long enough to provide flexibility so that radial thermal expansion of the shell does not
cause high thermal stresses at its junction with the skirt.
LUGS SUPPORT/ Bracket supports
support Brackets, or lugs, can be used to support vertical vessels. The bracket may rest on the
building structural steel work, or the vessel may be supported on legs. The main load carried by
the brackets will be the weight of the vessel and contents; in addition the bracket must be
designed to resist the load due to any bending moment due to wind, or other loads. If the bending
moment is likely to be signiﬁcant skirt supports should be considered in preference to bracket
supports. As the reaction on the bracket is eccentric, the bracket will impose a bending moment
on the vessel wall. The point of support, at which the reaction acts, should be made as close to
the vessel wall as possible; allowing for the thickness of any insulation.

FLANGE
Used for coupling the pipe which the fluid enter or live the pipe to the pressure vessel .Flanged joints are
used for connecting pipes and instruments to vessels, for manhole covers, and for removable vessel heads when ease
of access is required.
circular gasket should have circular tool marks in the flange face. ellipsoidal, tori-spherical, hemispherical, conical
and tori-conica .
MANHOLE
Identical to a nozzle except it not bolted to piping and it has a cover plate (or blind flange), which is bolted to the
flange. When unbolted it allows access to the inside of the vessel. Manholes are made for vessel diameter greater than
90cm.Generally larger in size.
DRAIN
Drain is one part or component of pressure vessels that used to removed wasted material
fromthe vessels.
INLET AND OUTLET
Inlet and exit (outlet) are also other parts of pressure vessels that used to add up or to exhaust
anytypes of fluid to the pressure vessels. Also used to filter the fluid from the pressure vessels.
Application area of pressure vessels

Pressure vessels are used in a variety of application in both industry and the private sector. Theyappear in
these sector as in industrial compressed air receives and domestic hot water storage tanks. Pressure
vessels are also used in a number of industries. For example, the power generation for fossil and nuclear
power, the photochemical industry for storing and processing cruel petroleum oil in a tank etc. Pressure
vessel industries sector uses as storage vessels like:
- Petroleum refining
- Chemical reactor
- Nuclear power
- Distillation tower
- Heat exchanger mechanism
- For mixer case
- For separation means.
□ Pulp, paper and food. Pressure vessels are also used in so many aspect or area like diving cylinders,
pressure reactor,autoclave, in mining operation, oil refining, photochemical plants, nuclear reactor
vessels, submarine and space ship habitats, pneumatic plants or reservoir, hydraulic reservoirs
under pressure and rail vehicle air brake reservoirs.
Pressure Vessel with Acidic solution medium
Acid solution: Acid is one of the most corrosive and dangerous materials chemical processing
workers handle — and sulfuric acid is one of the most common agents within this group. In fact,
global sulfuric acid production stands at approximately 180 million tons per year. Sulfuric acid is
a clear liquid solution, soluble in water with no smell that packs a punch. Known as the “king of
chemicals” due to its wide array of uses, sulfuric acid is common in operations such as chemical
processing, mineral processing, petroleum refining, water treatment, and more. If you utilize
sulfuric acid in chemical or manufacturing processes at your plant, it’s incredibly important to
store and handle the acid properly. This includes careful storage tank design, use, and upkeep.
When determining the best fabrication options for your acid storage tanks, consider the
following:
Materialsof construction
Ventilation systems
Secondary containment liners

CHAPTER TW0
CHAPTER 2
2.1. Designing procedure
Determine the length and diameter of the vessel
Designspecification
1. Main parameter
1.Medium Acid solution
ii. Inner service pressure: [MPa] 14
iii. Nominal Volume: [m3] 3.4
iv. Service temperature: [℃] 405
V .Position VERICAL
2. Supporting Element
Supporting element LEG
3. Connecting Pipe
a. Two longitudinal, one lateral, with
Nominal Diameter, [mm] 70
b. One Bottom discharge pipe, if necessary with flange
Nominal Diameter, [mm] 30

CHAPTER THREE
Materials of construction
Most tanksholdingsulfuricacidare made of carbon steel or stainlesssteel due totheirabilitiestoresistthe acid’s
corrosive nature.Otherfactorsthat playa role inmaterial selection include tanksize,costs,desiredacidquality,
acid concentration,and storage temperature.Toensure you’re optingforthe best — andsafest— materialsof
construction,partnerwithatank fabricatorwhounderstandswhichmaterialscanbestwithstandhighlycorrosive
acidicproperties.
Ventilation systems
Ensuring proper ventilation systems is another crucial factor in tank design and safety. Tank
ventilation not only allows the tanks to breath but also mitigates “harmful and potentially
dangerous gases” from accumulating in the tanks. Ventilation begins with expert design but
also requires employee inspection and upkeep such as looking for signs ofdamage or clearing
debris that could block vents.
Secondary containment liners
Acid containment is of the utmost importance. Erosion is “especially common” in sulfuric acid
storage tanks. Secondary containment liners help seal andprotect the metal, creating an
additional barrier to fight back against such highly abrasive materials. This can help reduce
wear, improve tank integrity, and safeguard your facility from detrimental tank failures.

DETERMINING THE LENGTH AND DIAMETER OF PRESSURE
VESSLE
I knowthat the volume of a pressure vessel is3.4m3
and the volume isequal tothe sumof the volume of the
head/bottom(hemispherical) andthe volume vesselshell (whichisincylindricalinshape. Accordingtoprocess
equipmentdesign,brownell andyoungsuggestthatthe ratioof L/D canbe determined byaccordingtotheir
presservalue. Andthe ratioof L/D isgivenbelow.
OPERATING PRESSURE is a pressure which is required for the process, served by
the vessel, at which this vessel normally operated.(14MPA)
DESIGN PRESSURE is the pressure used in the design of a vessel. It recommended to
design a vessel and its parts for higher pressure than the operating pressure.
Design pressure = operating pressure ie 5 to 10 percent operating pressure (Coulson &
Richardson's 2005, #)
I select 10 percent operating pressure for more safety purposes for my design pressure
DP=OP+OP*(10/100)
DP=(14MPA)+14*(10/100)MPA
DP=14MPA+1.4MPA
DP=15.4MPA
After finding the design pressure I find the length to diameter ratio by comparing the design
pressure value
In below table
Pressure
L/D ratio
Psi MPa
3 0-250 0.000-1.724
4 250-500 1.724-3.448
5 >500 >3.448
So my pressure vessel DP>3.448MPA
L/Di= 5 OR L=5Di.
V=Vh+Vs
Vh=(4/3)*π*Ri3 and Vs=π*Ri2*L

Vh =(4/3)*π*(Di/2)3 Vs =π*(Di/2)2*L
Vh =(1/6)*π*Di3 Vs =(1/4)*π*Di2*5Di
Vs=(5/4)* π*Di3
V=(1/6)*π*Di3 + (5/4)* π*Di3
3.4 =17/12* ΠDi3
Di=0.914m OR Di=914mm
SO the length l=5Di
L=5*0.914m
L=4.571m
MATERIAL SELECTION
Pressure vessels are constructed from plain carbon steels, low and high alloy steels, other alloys,
clad plates, and reinforced plastics. Selection of a suitable material must take into account the
suitability of the material for fabrication (particularly welding) as well as the compatibility of the
material with the process environment. The pressure vessel design codes and standards include
lists of acceptable materials; in accordance with the appropriate material standards.
According to different kind of standard such as ASMS the following given below are used for
the selection of materials
Strength
Corrosion resistance
Resistance to hydraulic attack
Fracture toughness
Fabric ability/predictability
STRENGTH
The strength of the martial are based on the mechanical propriety of the material that used

for themost easy, fast, strong and low cost of production .some of the mechanical
properties are
 yield strength
 ultimate tensile strength
 creep strength
 rupture strength
Yieldstrength
Yield strength is the region which a material changes from plastic to elastic deformation.
Someof the material I selectfor my presser vessel are shown below and there yield strength
.The stress level at which the plastic deformation begins.
Table :Yieldstrengthof alloys
Metal alloys Aluminum Copper Nickel Cast
iron
Stain less
steel18Cr/8Ni(30
4)
Yield strength(MPa) 35 69 138 197 205
Ultimatetensilestrength (UTS)
The ultimate tensile strength (tensile stress) i a measure of the basic strength of the material .it
isthe maximum stress that the material will stand and measure by a standard tensile test.
Table -Ultimate tensile strength
Material Aluminum stainless steel
18Cr/8Ni(304)
Cast iron Copper Nickel
UTS(MPa) 90 510 414 200 520

Creep strength
Material are often placed in a service at elevated temperature and exposed to static
mechanical stress the deformation under such circumstance is termed creep .The time
dependent and permanent deformation of material when subjected to a constant lode
andstress.
Rupturestrength
Is the martial resistance to the fracture caused by the internal pressure of the fluid
that was present inside the tank and it is proportional to tensile strength(UTS) which means the
martial have high UTS and it have good Rupture strength.
Property Material
Carbon
carbon-manganese
low alloy steel
Austenitic stainless
steels
Non-ferrous metals
Minimum yield stress or
0.2 percent proof stress,
at the design temperature
15 15 15
Minimum tensile
strength, at room
temperature
2.35 2.5 4.0
Mean stress to produce
rupture at 105 h at the
design temperature
1.5 1.5 1.0
Table strength of somematerial
Corrosion resistance

The “corrosion allowance” is the additional thickness of metal added to allow for material lost bycorrosion
and erosion, or scaling. The allowance to be used should be agreed between the customer and
manufacturer. Corrosion is a complex phenomenon, and it is not possible to give specific rules for the
estimation of the corrosion allowance required for all circumstances. The allowance should be based on
experience with the material of construction under similar service conditions to those for the proposed
design. For carbon and low-alloy steels, where severe corrosion is not expected, a minimum allowance of
2.0 mm should be used; where more severe conditions are anticipated this should be increased to 4.0
mm. Most design codes and standards specify a minimum allowance of 1.0 mm.
Table corrosion resistance of some material
Chemical Cast
iron
Aluminum Nickel Copper Steen less
steel18Cr/8Ni(304)
Oil G G C C G
G-good
C-caution-depending on the martial
So by use of the following corrosion resistance method, we minimize the destruction done by
corrosion
by using of cathodic protection
by using of protective coating or surface coating
galvanization
hot dipping
tinning
Resistance to hydraulic attack
The tensile strength (UTS) and elastic modulus of metals decrease with increasing temperature.
The temperature that gives for my pressure vessel is 4050C(757F) above 600F, hydrogen attack
causes irreparable damage through the component thickness& can damage carbon and low alloy
steel.
Table temperature effect on UTS

Material Cast
iron
Aluminum Nickel copper Stainless steel
18Cr/8Ni(304)
Tensile
strength(ULT)MPa
414 90 520 200 510
Design stress at
temperature
(4000C)MPa
-- -- 220 -- 107.6
Fracture toughness
The ability of the material that absorbs energy up to fracture. Brittle
fracture without applicable deformation and by rapid crack propagation .The direction the crack motion
is very nearly perpendicular to the direction the applied tensile stress and yield stress related with
fracture surface. Brittle fracture is dependent on the stressconsternation (K). And the stress consternation
facer is shown below.
Table fracturetoughness of somematerial
Material Steel less
steel 18 8/8Cr
Aluminum Copper Nickel Cast
iron
Fracture
toughness(K),MPa
76 44 Low Low Low
Fabric ability
Based on how it made and cost of production. A guide to the fabrication properties of common
metals and alloys are shown blow.
S -satisfactory
D -Difficult, special techniques needed.
U – Unsatisfactory

Table fabric ability of somematerial
Martial Mashing Cold
work
Hot
work
Castin
g
Welding Annealing
temperate(co.
)
Cast iron S U U S U/D _
Stainless
steel(18Cr,8Ni)30
4
S S S D S 1050
Nickel S S S S S 1150
Aluminum S S D S S 550
Copper(dioxides) D S S S D 700
Conclusion about the Material Selection
Pressure vessel material is selected based on the medium it hold and since my medium is acid
solution and it’s a very high corrosive element I select a material with good corrosive resistance and
high strength which can withstand high temperature and I select for this material to be
stainless steel 18Cr/8Ni unstabilized (304) with specific mechanical and physical
properties. But the stainless steel occurs when the concentration of chromium exceeds
about 12 wt. %. However, even this is not adequate to resist corrosion in acids such as HCl or
H2SO4; higher chromium concentrations and the judicious use of other solutes such as
molybdenum, nickel and nitrogen can be used as insulating and regenerating protective film on
the surface of stainless steel.
For materials not subject to high temperatures the design stress is based on the yield stress (or
proof stress), or the tensile strength (ultimate tensile stress) of the material at the design
temperature. For materials subject to conditions at which the creep is likely to be a consideration,
the design stress is based on the creep characteristics of the material:

Material Tensile(
UTS)strengt
h, MPa
Modulus
of
elasticity,
GPa
Hardness
Brinell
Specific
gravity
corrosion
property
Fracture
toughness(K),
MPa
Stainless
steel
18Cr/8NI
(304)
510 210 160 8.0 G 76
Table stainless steel material property
WELDING TYPE
Welding joint is the permanent joint which is obtained by the fusion of the edges of the
two parts to be joined to gather, which or without the application of pressure and a filler
material. The heat required for a fusion of the material may be obtained by burning of gas (in
case of gas welding) or by an electric arc (in case of electric arc welding).
The letter method is extensively used because of greater speed of welding.
Generally welding, in engineering, any process in which two or more pieces of metal are
joined together by the application of heat, pressure, or a combination of both.
There are several methods makewelded joints. In a particular case the choice of a type from the
numerous alternatives depend on:
 The circumstances of welding.
 The requirements of the code.
 The aspect of economy.
THE CIRCUMSTANCES OF WELDING
In many cases the accessibility of the joint determines the type of welding. In a small diameter
vessel (under 18-24 inches) from the inside, no manual welding can be applied. Using backing
strip it must remain in plate. In larger diameter vessels if a man way is not used, the last
(closing) joint can be welded from outside only. The type of welding may be determined also
by the equipment of the manufacturer.
CODE REQUIREMENTS
Regarding the type of joint the Code establishes requirements based on service, material and
location of the welding. The welding processes that may be used in the construction of vessels are

also restricted by the Code. The Code-regulations are tabulated on the following pages under the
titles:
TYPES OF WELDED JOINTS
Joints permitted by the code, their efficiency and limitations of their
application. Table UW-12
 Type-1: -butt joint by double welding to obtain the same quality of deposited weld
metalon the inside and outside of weld surface. full radiographic examination, joint
efficiencyis 1.
 Type 2: - single welded butt joint with backing strip, joint efficiency is 0.90
 Type 3: - single welded butt joint without the use of backing strip.
 Type 4: -double full fillet lap joint
 Type 5: -single full fillet lap joint with plug welds
 Type 6: -single full fillet lap joint without plug welds

DESIGN OF WELDED JOINTS: Types of joints to be used for vesselsin various
services andunder certain design conditions.) UW-2, UW-3
JOINT EFICIENCIES AND STRESS REDUCTIONS: Efficiencies of joints at
certainlocations and reduced allowable stress to be used in calculations
of vessel components.
The data of the table are based on the following Code regulations: Full, spot, partial
radiographic examination or no radiography of A, B, and C joints. UW-11
For longitudinal stress calculation the efficiency of partially radio graphed joints is the same
as for spot radio graphed joints. Seamless vessel sections and heads with Category B,C or D
butt joints that are spot radio graphed shall be designed for circumferential stress using a

stress value equal to 85% of the allowable stress value of the material; UW-12(b)
Whenthe jointsare not radiographedand for jointefficiency,Ethe value incolumnof table “Types of welded
joints”are used, inall otherdesigncalculation, astressvalue equal to80% ofthe allowable stressvalue of
material shall be usedexceptforunstainedflatheads,etc.UW- 12(c)
THE ECONOMY OF WELDING:
If the two preceding factors allow free choice, then the aspect of economy must be the
deciding factor.
Some consideration concerning the economy of welding-edge preparation, which can be
madeby torch cutting, is always more economical than the use of J or U preparation.
Double V preparation requires only half the deposited weld metal required for single
Vpreparation.
Increasing the size of a fillet weld, its strength increases in direct proportion, while the
deposited weld metal increases with the square of its size. Lower quality welding makes
necessary the use of thicker plate for the vessel. Whether using stronger welding and thinner
plate or the opposite is more economical, depends on the size of vessel, welding equipment,
etc. this must be decidedin each particular case.
There are different kind of weldingS based on IS-28-25 it categorized in to 4 select
 category A: longitudinal welded joints within the main sheet, communicating,
Chambers,nozzles and any welded joints within a formed or flat Head.
 Category B: circumferential welded joints within the main shell communicating
chambers, nozzles and transitions in diameter including joints between the translations
and a cylinder at either the large of small end, circumferential welded joints connecting
from heads to main shellsto nozzles and to communicating hampers.
 Category c: welded joints connecting flanges, tubes sheets and flat heads to main
shells, toformed heads, to nozzles or to communicating chambers and any welded joints
connecting oneside plate to another side plate of a flat sided vessel.
 Category d: welded joints connecting communicating chambers or nozzles to main

shell, to heads and to flat sided vessels and those joints connecting nozzles to
comm
unicati
ng
chamb
ers
CONCLUSION ON THE TYPE OF WELDING TYPE
I select Category A (type 1 )andbutt jointforthe shell andhead. And itsradiographicexamination (full),joint
efficiency is(E=1).
CALCULATION OF SHELL THICKNESS CORROSION ALLOWANCE
The “corrosion allowance” is the additional thickness of metaladded to allow for material lost by
corrosion and erosion, or scaling. The allowance to be used should be agreed between the
customer and manufacturer. For carbon and low-alloy steels, where severe corrosion is not
expected, a minimum allowance of 2.0 mm should be used; where more severe conditions are
anticipated this should be increased to 4.0 mm. Most design codes and standards specify a
minimum allowance of 1.0 mm. (Coulson & Richardson's 2005, #)
To find the thickness of the pressure vessel which is subjected to internal design pressure of
P=1.6MPa there are different equations and its corrosion allowance for stainless steel
18Cr/8Ni is (2mm) from the table. There will be a minimum wall thickness required to ensure

that any vessel is sufficiently rigid to withstand its own weight, and any incidental loads.
Where t = min. required thickness of shell, mm
P= internal design pressure, pa
R = inside radius of shell, mm
δ = max. Allowable stress, Pa
E = joint efficiency (min)
C.A=corrosion allowance, mm
CIRCUMFERENTIAL STRESS (LONGITUDINAL JOINTS)
It means that the governing stress will be the circumferential stress (hoop stress) in the
long seam. For this it has to satisfy that P does not exceed 0.385δE .In which case we
shall use the following formulae for thickness of shell.
t = PR/ (2δE +0.4P)+ C.A
LONGITUDINAL STRESS(CIRCUMFERENTIAL JOINTS
It means that the governing stress will be the longitudinal stress in the circumferential
joint. For this it has to satisfy that P does not exceed 1.25δE. Or if the circumferential
joint efficiency is less than ½ the longitudinal joint efficiency. In which case we use
the formula for thickness is
t = PR/ (δE -0.6P) + C.A
Given
P=14MPa
R=0.4571m
δ= 103MPa
E=1
C.A=2mm=0.002m
Let's use these two equations to find the thickness of the pressure vessel and select the
smallest value of them for safety purposes. The weight of the pressure vessel can
affect the support leg.

Case1 using
Circumferential stress (longitudinal joint)
P ≤ 0.385δE
P ≤ 0.385*103MPa*1
14MPa ≤ 39.655MPa …………………. (satisfied)
Therefore
t=PR/ (δE-0.6P) + C.A
t= (14MPa*0.4571)/ (( 103MPa*1) - (0.6*14Mpa) ) + 0.002m
t=0.067646m+0.002m
t=0.06964m=69.6469mm≈70mm
ts=70mm (standard)
case 2 using
Longitudinal stress (circumferential stress)
P ≤ 1.25δE
P ≤ 1.25*103MPa*1
14MPa ≤ 128.75MPa …………………… (satisfied)
Therefore
t=PR/(2δE+0.4P) + C.A
t= (14MPa*0.4571)/ ((2*103MPa*1) + (0.4*14MPa)) + 0.002m
t=0.0310048m+0.002m
t= 0.0330048m = 33.004845mm
t=34mm ……………………..(standard)
And the largest value is 70mm
Finally let’s find the external radius and diameter
Ro =Ri+t
Ro=0.4571m+0.07m
Ro=0.5271m
Ro =0.5271m
And
Do=2*Ro

Do=2*0.5271m
Do=1.0542m (ans.)
Design of HEAD
My pressure vessel is vertical with hemispherical head in shape and 1.76MPa internal
design pressure is applied on it and it’s thickness is given blow
According on the two theory
1) thin-shell theory
δ= (PR)/ (2t)
2) ‘’exact’’ theory
δ= [PRi
3/Ro
3 -Ri
3 ] [1+Ro
3 /2Ri
3 ]
Given values
Ri=0.4571m
Ro=0.5271m
P=14MPa
t=70mm=0.07m
E=1
C.A=2mm=0.002m
thin-shell theory
δ= (PRi)/ (2t)
δ= (14MPa*0.4571)/ (2*0.07m)
δ=45.71MPa …………………….. (ans.)
‘’exact’’ theory
δ= [PRi
3/ (R0
3-Ri
3)] [1+R0
3/2Ri
3]
δ= [(14MPa*(0.4571) 3)/ (0.52713-0.45713) m3] *[1 + (0.52713/2*0.45713)]
δ=46.372MPa …………………….(ans.)
Therefore let's find the head thickness and take the largest value for δ=46.372MPa.

th= PRi/ (2δE-0.2P) + C.A
th= (14MPa*0.4571m)/ (2*46.372MPa*1 – 0.2*14MPa) + 2mm
th =0.071148m+0.002m
th = 0.073148m
th=73.148mm
The standard head thickness is 74mm
Head to shell transition
To join the head and shell I use butt walledjoint according to the (category) and the
head shell has different thickness according to CODE UW-9(c),UW-13. Joining a
plate unequal with butt weld, the thickness of the plate shall be tapered is more than
3.125mm (1/8in).Thickness of head (th=74mm) and shell (ts=70mm), their deference
(2mm) is less than 3.125mm (1/8in) or 3.175mm.
l ≥ 3*y and y=2mm
l ≥ 6mm.

Closure Head
Types of closure head
The ends of a cylindrical vessel are closed by heads of various shapes. The principal types used are:
Flanged head
Hemispherical heads
Ellipsoidal heads
Torispherical heads
Conical head
Hemispherical, ellipsoidal and torispherical heads are collectively referred to as domed heads. They are
formed by pressing or spinning; large diameters are fabricated from formed sections. Torispherical heads
are often referred to as dished ends. The preferred proportions of domed heads are given in the standards
and codes.
Where
P=internal design pressure
E=joint efficiency
R0= external radius
Ri=internal radius
S= allowable shear
Flangedhead
Formed domed heads are made with a short straight cylindrical section, called a flange or skirt. This ensures
that the weld line is away from the point of discontinuity between the head and the cylindrical section of the
vessel.
Hemisphericalhead
Equal stress in the cylindrical section and hemispherical head of a vessel the thickness of the head need
only be half that of the cylinder. However, as the dilation of the two parts would then be different,
discontinuity stresses would be set up at the head and cylinder junction. For no difference in dilation
between the two parts (equal diametral strain) it can be shown that for steels (Poisson’s ratio D 0.3) the ratio
of the hemispherical head thickness to
cylinder thickness should be 7/17. However, the stress in the head would then be greater than that in the
cylindrical section; and the optimum thickness ratio is normally taken as 0.6.
The design thickness of hemispherical head is given by

t=PRi/ (2SE-0.2P)
Where
E=joint efficiency
R0= external radius
Ri=internal radius
S= allowable shear
Ellipsoidalhead
Most standard ellipsoidal heads are manufactured with a major and minor axis ratio of 2 : 1. For this ratio,
the following equation can be used to calculate the minimum thickness required:
The design thickness of elliptical head is given by
t=PDK/ (2SE-0.2P)
where D-shell diameter
K-stress intensity factor
K= [1/6 + (a/b)2]
a and b semi-major, semi-minor axes of ellipse
E=joint efficiency
S= allowable shear
Torispherical head
There are two junctions in a torispherical end closure: that between the cylindrical section and the head,
and that at the junction of the crown and the knuckle radii. The bending and shear stresses caused by the
differential dilation that will occur at these points must be taken into account in the design of the heads. One
approach taken is to use the basic equation for a hemisphere and to introduce a stress concentration, or
shape, factor to allow for the increased stress due to the discontinuity. The stress concentration factor is a
function of the knuckle and crown radii.
The design thickness of Tori spherical head is given by
t=PLM/ (2SE-0.2P)
where L-spherical cross radius
M=1/4[1/3 + (L/r) 1/2]
M-shear intensity factor
r-knuckle radiuses
Conical head
Conical sections (reducers) are used to make a gradual reduction in diameter from one cylindrical section
to another of smaller diameter. Conical ends are used to facilitate the smooth flow and removal of solids
from process equipment; such as hoppers, spray-dryers and crystallisers.

The design thickness conical head is given by
t=PD/2cosα (SE-0.6P)
where α-seim-apex angel
Gasket
Gaskets are used to make a leak-tight joint between two surfaces. It is impractical to machine flanges to the
degree of surface finish that would be required to make a satisfactory seal under pressure without a gasket.
Gaskets are made from “semi-plastic” materials; which will deform and flow under load to fill the surface
irregularities between the flange faces, yet retain sufficient elasticity to take up the changes in the flange
alignment that occur under load.
Gasket selection
Gaskets are used to make a leak-tight joint between two surfaces. It is impractical to machine flanges to the
degree of surface finish that would be required to make a satisfactory seal under pressure without a gasket.
Gaskets are made from “semi-plastic” materials; which will deform and flow under load to fill the surface
irregularities between the flange faces, yet retain sufficient elasticity to take up the changes in the flange
alignment that occur under load. The minimum seating stress “y” is the force per unit area (pressure) on the
gasket that is required to cause the material to flow and fill the surface irregularities in the gasket face. The
gasket factor “m” is the ratio of the gasket stress (pressure) under the operating conditions to the internal
pressure in the vessel or pipe. The internal pressure will force the flanges’ faces apart, so the pressure on the
gasket under operating conditions will be lower than the initial tightening-up pressure. The gasket factor
gives the minimum pressure that must be maintained on the gasket to ensure a satisfactory seal. The
following factors must be considered when selecting a gasket material:
1. The process conditions: pressure, temperature, corrosive nature of the process fluid.
2. Whether repeated assembly and disassembly of the joint is required.
3. The type of flange and flange face.
Up to pressures of 20 bar, the operating temperature and corrosiveness of the process fluid will be the
controlling factor in gasket selection. Vegetable fiber and synthetic rubber gaskets can be used at
temperatures of up to 100C. Solid perfluorocarbon (Teflon) and compressed asbestos gaskets can be used to
a maximum temperature of about 260C. Metal-reinforced gaskets can be used up to around 450C. Plain soft
metal gaskets are normally used for higher temperatures.
Based on this criteria I select

CHAPTER FOUR
Flange selection
Flange
joints are used for connecting pipes and instruments to vessels, for manhole covers, and for removable
vessel heads when ease of access is required. Flanges may also be used on the vessel body, when it is
necessary to divide the vessel into sections for transport or maintenance.
Flanged joints are also used to connect pipes to other equipment, such as pumps and valves. Screwed joints
are often used for small-diameter pipe connections, below 40 mm. Flanged joints are also used for
connecting pipe sections where ease of assembly and dismantling is required for maintenance, but pipework
will normally be welded to reduce costs. Flanges range in size from a few millimetre diameter for small
pipes, to several metres diameter for those used as body or head flanges on vessel.
Types of flangeand selection
Several different types of flange are used for various applications. The principal types used in the process
industries are:
Welding-neck flanges.
Slip-on flanges, hub and plate types.
Lap-joint flanges.
Screwed flanges.
Blank, or blind, flanges.
Welding-neck flanges
have a long tapered hub between the flange ring and the welded joint. This gradual transition of the section
reduces the discontinuity stresses between the flange and branch, and increases the strength of the flange
assembly. Welding-neck flanges are suitable for
extreme service conditions; where the flange is likely to be subjected to temperature, shear and vibration
loads. They will normally be specified for the connections and nozzles on process vessels and process
equipment.
Slip-on flanges
slip over the pipe or nozzle and are welded externally, and usually also internally. The end of the pipe is set
back from 0 to 2.0 mm. The strength of a slip-on flange is from one-third to two-thirds that of the
corresponding standard welding-neck flange. Slip-on flanges are cheaper than welding-neck flanges and are

easier to align, but have poor resistance to shock and vibration loads. Slip-on flanges are generally used for
pipe work. For light duties slip-on flanges can be cut from plates.
Lap-jointflanges
are used for piped work. They are economical when used with expensive alloy pipes, such as stainless
steel, as the flange can be made from inexpensive carbon steel. Usually a short lapped nozzle is welded to
the pipe, but with some schedules of pipe the lap can be formed on the pipe itself, and this will give a cheap
method of pipe assembly. Lap-joint flanges are sometimes known as “Van-stone flanges”.
Screwed flanges
are used to connect screwed fittings to flanges. They are also sometimes used for alloy pipe which is
difficult to weld satisfactorily.
Blind flanges(blank flanges)
are flat plates, used to blank off flange connections, and as covers for manholes and inspection ports.
❖ For selecting the standard dimension and material for the flange by applying the American
National Standard ANSI B16.5-1981 of temperature and pressure rating.
Flangedesign

For selecting the standard dimension and material for the flange by applying the American National Standard ANSI
B16.5-1981 of temperature and pressure rating.
The given temperature is 405O
C (761o
F) and pressure is 14MPa (2030.53PSi). And I can’t find the Machu value
therefore design pressure lets change into standard value which is 15.3408 (2225Psi).
Class 600lb.
Hydrostatic test/design pressure MPa(PSi) 15.3408(2225)
Temperature ,o
C(o
F) Maximum allowable non-shock pressure,
MPa(PSi)
405(761) 5.68816(825)
FROM MAXIMUM PRESSURE AND TEMPRETURE RATING TABLE

I know that the flange is class 600lb and the material is `Forged steel SA105 . I select Welding-neck flanges,
because they have a long tapered hub between the flange ring and the welded joint. This gradual transition of the
section reduces the discontinuity stresses between theflange and branch, and increases the strength of the flange
assembly. Welding-neck flanges are suitable for extreme service conditions; where the flange is likely to be
subjected to temperature, shear
and vibration loads. They will normally be specified for the connections and nozzles on process vessels and
process equipment.
There are 7 flanges that we used for our pressure vessel; their value is shown below.
Two longitudinal pipes with D=70mm (2.75576in).
One lateral pipe with D=70mm (2.75576in).
One bottom discharge D=30mm (1.16916in)
One sample opening D=30mm(1.16916in)
Temperature gage D=25mm(1in)
Safety valve D=25mm(1in)
Nom.
Pipe
Size
O T1 R X
No.2& Bolt
L21 H B2 L B R L3 B3 D C THr
Dia. of Circle
Holes Dia.
1/2 3.75 0.56 1.38 1.50 4-0.62 2.62 2.06 0.84
To
be
specified
by
purchaser.
0.88 0.88 0.12 0.88 0.90 0.38 0.93 0.62
3/4 4.62 0.62 1.69 1.88 4-0.75 3.25 2.25 1.05 1.00 1.09 0.12 1.00 1.11 0.44 1.14 0.62
1 4.88 0.69 2.00 2.12 4-.075 3.50 2.44 1.32 1.06 1.36 0.12 1.06 1.38 0.50 1.41 0.69
1-1/4 5.25 0.81 2.50 2.50 4-0.75 3.88 2.62 1.66 1.12 1.70 0.19 1.12 1.72 0.56 1.75 0.81
1-1/2 6.12 0.88 2.88 2.75 4-0.88 4.50 2.75 1.90 1.25 1.95 0.25 1.25 1.97 0.62 1.99 0.88
2 6.50 1.00 3.62 3.31 8-0.75 5.00 2.88 2.38 1.44 2.44 0.31 1.44 2.46 0.69 2.50 1.12
2-1/2 7.50 1.12 4.12 3.94 8-0.88 5.88 3.12 2.88 1.62 2.94 0.31 1.62 2.97 0.75 3.00 1.25
3 8.25 1.25 5.00 4.62 8-0.88 6.62 3.25 3.50 1.81 3.57 0.38 1.81 3.60 0.81 3.63 1.38
3-1/2 9.00 1.38 5.50 5.25 8-1.00 7.25 3.38 4.00 1.94 4.07 0.38 1.94 4.10 4.13 1.56
4 10.75 1.50 6.19 6.00 8-1.00 8.50 4.00 4.50 2.12 4.57 0.44 2.12 4.60 4.63 1.62
5 13.00 1.75 7.31 7.44 8-1.12 10.50 4.50 5.56 2.38 5.66 0.44 2.38 5.69 5.69 1.88
6 14.00 1.88 8.50 8.75 12-1.12 11.50 4.62 6.63 2.62 6.72 0.50 2.62 6.75 6.75 2.00
8 16.50 2.19 10.62 10.75 12-1.25 13.75 5.25 8.63 3.00 8.72 0.50 3.00 8.75 8.75 2.25
10 20.00 2.50 12.75 13.50 16-1.38 17.00 6.00 10.75 3.38 10.88 0.50 4.38 10.92 10.88 2.56
12 22.00 2.62 15.00 15.75 20-1.38 19.25 6.12 12.75 3.62 12.88 0.50 4.62 12.92 12.94 2.75

14 23.75 2.75 16.25 17.00 20-1.50 20.75 6.50 14.00 3.69 14.14 0.50 5.00 14.18 14.19 2.88
16 27.00 3.00 18.50 19.50 20-1.62 23.75 7.00 16.00 4.19 16.16 0.50 5.50 16.19 18.19 3.06
18 29.25 3.25 21.00 21.50 20-1.75 25.75 7.25 18.00 4.62 18.18 0.50 6.00 18.20 18.10 3.12
20 32.00 3.50 23.00 24.00 24-1.75 28.50 7.50 20.00 5.00 20.20 0.50 6.50 20.25 20.19 3.25
22 34.25 3.75 25.25 26.25 24-1.75 30.63 7.75 22.00 5.25 22.22 0.50 6.88 22.25 -- --
24 37.00 4.00 27.25 28.25 24-2.00 33.00 8.00 24.00 5.50 24.25 0.50 7.25 24.25 24.19 3.62
Nominal
pipe size(m
m)
Diameter of
the bore(mm)
[A]
Length
through the
hub(mm
)[C]
Diameter of
the hubat
the point of
welding[E]
Diameter of
the hubatthe
base(mm)[G]
Outside
diameterof
flange(m
m)[H]
Thickness of
flange(mm
)[J]
Outside
diamete r of
raiseface(m
m)[K]
Bolt
ing
70mm(2
…
8
in)
25mm(1in) 81.25mm
(3.25in)
87.5mm(3.5in
)
115.5mm(4.62in
)
206.25mm(8.25i
n)
31.25mm(1.25in) 125mm
(5in)
M16
30mm(1.2IN
)
25(1in) 65.5mm
(2.62in)
41.5mm(1.66i
n)
62.5mm(2.5in) 131.25mm(5.25i
n)
20.25mm(0.81in) 62.5mm(2.5in
)
M12
25mm(1
in)
25(in) 97mm(2.44in) 33mm(
1.32in)
53mm(
2.12in)
122mm(
4.88in)
17.25mm(0.69
in)
50mm(2in) M10
FLANGE APPLIEDLODE AND FALNGE MOMEANT
The lode on the flange can be given as fallow
m=gasket factor
Pi=internal pressure of flange
Pi=14MPa
B=in side diameter of a flange for each pipe
B=A/2
b=effective gasket selling width
2b=effective gasket pressure width
tf=thickness of flange
tf=J
hd= (G + H – 2E)/4
hg= (H - G)/4
ht= (G + H)/4
G’=mane diameter of gasket
=B + (ht-hg)

Hg=gasket reaction forces (pressure forces)
=πG’ (2b) mPi
H=total pressure
= (π/4)G’2
Pi
Hd=pressure force of area inside the flange
= (π/4) B2
Pi
Ht=pressure for a flange face
= H - Hd
The moment on the flange is
Mop=Hd*hd + Ht*ht + Hg*hg
Let’s calculate the lode and moment
 There gasket factor width and internal pressure for the flange are the same
m=1.75, b=10mm and Pi=14MPa
hd= (G + H – 2E)/4
For 70mm= (115.5+206.25-2*87.5)mm/4=36.682mm
For 30mm = (62.5+131.25-2*41.5)mm/4=27.688mm
For 25mm = (53+122-2*33)mm/4=27.25mm
hg= (H - G)/4
For 70mm= (206.25-115.5)mm/4=22.6873mm
For 30mm= (131.25-62.5)mm/4 =17.1875mm
For 25mm= (122-53)mm/4=17.25mm
ht=(G+H)/4
For 70mm= (206.25+115.5)mm/4=80.4375mm
For 30mm= (131.25+62.5)mm/4=48.4375mm
For 25mm= (122+53)mm/4=43.75mm
tf=J
For 70mm=31.25mm
For 30mm= 20.25mm
For 25mm=17.25mm

B=A/2
For 70mm=25mm/2=12.5mm
G’=B + (ht - hg)
For 70mm=12.5mm + (80.4375-22.6875) mm=70.25mm
For 30mm=12.5mm + (48.4375-17.1875) mm=43.75mm
For 25mm=12.5mm + (43.75-17.25) mm= 39mm
H= (π/4) G’2Pi
For 70mm= (π/4)(70.25mm)2
*14MPa=217,055N
For 30mm= (π/4)(43.75mm)2
*14MPa=21,046.21N
For 25mm = (π/4)(39mm)2
*14MPa=428.827N
Hd= (π/4) B2Pi
For 70mm= (π/4)(12.5mm)2
*14MPa=6872.233N
For 30mm= (π/4)(12.5mm)2
*14MPa=6872.233N
For 25mm = (π/4)(12.5mm)2
*14MPa=6872.233N
Ht =H-Hd
For 70mm= (217,055-6872.23)N=210,182.77N
For 30mm= (21,046.21-6872.23)N=14,173.98N
For 25mm= (428.872-6872.23)N=-6443.335N
Hg=πG’ (2b)mPi
For 70mm=π (40.3725mm)(2*10mm)1.5*3.48MPa=12,234.75N
For 30mm=π(65.24mm)(2*10mm)1.5*3.48MPa=21,386.72N
For 25mm=π (117.89mm)(2*10mm)1.5*3.48MPa=38,646.23N
The moment is
Mop=Hd*hd + Ht*ht + Hg*hg
For 70mm pipe
Mop=(6872.233N)(36.682mm)+(210182N)(22.6873mm)+(12234.75N)(80.4375mm)
=6,005,689.9165Nmm=6005.689Nm

For 30mm pipe
Mop=(6872.233N)(27.688mm)+(14173.98N)(17.1875mm)+(21396.72N)(48.4376mm)
=6,314,047.2925Nmm=6314.047Nm
For 25mm pipe
Mop=(6872.233N)(27.25mm)+(6443.355N)(17.25mm)+(38646.23N)(43.75mm)
=1,989,188.7855Nmm=1989.1887Nm
The minimum load required bolt load under the operating condition given by
Wm1=H + Hg
Wm1, 70mm=217,055N + 12,234.75N=229,289.75N=229.2897KN
Wm1, 30mm=21,046N + 21,386.72N=42432.72N=42.4327KN
Wm1, 25mm=428.827N + 38,646.23N=39,075.05N=39.075KN
The force and the momentum must be checked under the bolting up conditions. The moment is given by
Where Wm2 is the bolt load required to sat the gasket,given by
Wm2=yπG’b
y=gasket sating pressure (stress) =2.28MPa
Wm2, 70mm=2.28MPa*π*70.25mm*10mm=5,029.94N=5.029KN
Wm2, 30mm=2.28MPa*π*43.75mm*10mm =3132.19N=3.132KN
Wm2, 25mm=2.28MPa*π*39mm*10mm=2792.51N=2.792KN
Matm=Wm2*hg
Matm,,70mm=5029.94N*22.6873mm=114,115.7577Nmm=114.115Nm
Matm, 30mm =3132.19N*17.1875mm =53,834.51Nmm=53.834Nm
Matm,25mm=2792.51N*17.25mm=48,170.7975Nmm=48.1707Nm
Flange stress

Flange stress are given by
Longitudinal hub stress
σhb=F1M
Radial flange stress
σrd=F2M
Tangential flange stress
σtg=F3M - F4σrd
where M is taken as Mop/B’ or Matm/B’, whichever is the greater; and the factors F1,F2, F3 and F4 are functions of
the flange type and dimensions, and are obtained as followed
F1= 1/λ gt
2
gt= (G-A)/2
λ=δ + γδ=t3/d and
δ=t3/d, γ=α/t, α=tf* e+1
d= (u/v)hogo
2 , go=J and
ho= (B’go)1/2
ho =(B’J)1/2
ho =(u/v)(B’J)1/2J2, B’=(G-A)/2
F 2=β/λtf2
tf=thickness of the flange=J
β=1.333tfe+1
F3=y/t2
y = (1-v2) u
F4=z
z=k-factor
K=A’/B’
A’= (H-G)/2 and B’= (G-A)/2
K = (H-G) / (G-A)
K70mm= (206.25-115.15) mm/ (115.15-25) mm =1.0105
K30mm=(131.25-62.5)mm/(62.5-25)mm =1.833
K25mm=(122-53)mm/(53-25)mm =2.464

Nominal pipe
Dia.(mm)
K T Z Y, MPa(Psi) U
K70mm 1.0105 1.85 6.04 1.170 1.286
K30mm 1.833 1.36 1.43 2.36 2.59
k25mm 2.464 1.35 1.40 2.29 2.52
Table value of (K, T, Z, Y and U)

Y-gasket seating design stress
Let’s find the value of (e)
e=F/ho =F/ (B’J) ½
B’=(G-A)/2
B’70mm=(115.5-25)mm/2
=45.25mm
B’30mm =(62.5-25)mm/2
=18.75mm
B’25mm =(53-25)mm/2
=14mm
Let’s find the value of F (in y-axis) by using ASME code, section viii,
Dev. 1.in x-axis

g1/go (or B’/J) and
H / ho (or (C-J)/ (B’-J) 1/2)
After calculating of this value I fined F for each flange
And the values are shown blow.
F70mm=0.903
F30mm=0.823
F25mm=0.801
e=F/(B’J)1/2
e70mm=0.903/(45.25mm*31.25mm)1/2
e70mm =0.0240/mm
e30mm=0.823/(18.75mm*20.25mm)1/2
e30mm =0.0422/mm
e25mm=0.801/(14mm*17.25mm)1/2
e25mm =0.0515/mm
and let’s find (α and β)
α= TFe+1
α =Je+1
α70mm=31.25mm (0.024/mm)+1
α70mm =1.75
α30mm=20.25mm (0.0422/mm)+1
α30mm =1.854
α25mm=17.25mm (0.0515/mm)+1
α25mm =1.888
β = 1.333Je + 1
β 70mm=1.333*31.25mm (0.0240/mm)+1
β 70mm =1.999
β 30mm=1.333*20.25mm (0.0422/mm)+1
β 30mm =1.139
β 25mm=1.333*17.25mm (0.0515/mm)+1
β 25mm =2.184
and γ will be
γ=α/T
γ70mm=1.75/18.5
γ70mm =0.0945
γ30mm=1.854/1.36
γ30mm =1.3632

γ25mm=1.888/1.35
γ25mm =1.39
lets find the value of v from ASME code Viii Div. 1 using by the value of (**) and it will be
v25mm=0.36
v30mm=0.173
v70mm=0.12
d will be
d =(u/v) (B’J)1/2J2
d70mm=(1.286/0.12)(45.24mm*31.25mm)1/2
(31.25mm)2
d70mm =393,501.21mm3
d30mm =(2.59/0.173)(18.75mm*20.75mm)1/2
(20.75mm)2
d30mm =127,145.077mm3
d25mm=(2.52/0.36)(14mm*17.15mm)1/2
(17.15mm)2
d25mm =31,902.3274mm3
The value of δ will be
δ = t3
/d
δ70mm=(31.25mm)3
/393,501.21mm3
δ70mm =0.07755
δ30mm=(20.25mm)3
/127,145.077mm3
δ30mm =0.0653
δ 25mm=(17.25mm)3
/31,902.3274mm3
δ 25mm =0.16089
Let’s find the value of λ
λ = δ + γ
λ70 mm=0.07755+0.0945
λ70 mm= 0.17205
λ30mm=0.065+1.3632
λ30mm=1.4282
λ25mm=0.16089+1.39
λ25mm=1.55089
y will be
𝑦 = (1 − 𝑣2)𝑢
Y70mm= (1-0.362
)1.286
Y70mm =1.1193

Y30mm= (1-0.1732
)2.59
Y30mm =2.5124
Y25mm= (1-0.122
)2.52
Y25mm =2.4837
gt will be
𝑔𝑡 = (𝐺 − 𝐴)/2
gt70mm=(115.5-25)mm/2
gt70mm =45.25mm
gt30mm =(62.5-25)mm/2
gt30mm =18.75mm
gt25mm =(53-25)mm/2
gt25mm =14mm
Table 11.value of the constant
Nominal Dia.
(mm)
F1(1/mm2)
F1= 1/λ gt2
F2 (1/mm2)
F2=β/λt2
F3(1/mm2)
F3=y/t2
F4
F4=z
70mm 0.000341 0.00661 0.0020469 6.09
30mm 0.00199 0.004289 0.00165 1.43
25mm 0.003289 0.005381 0.0733 1.40
To find the stress on the flange we must the value of the moment M and it is the largest of them (Mop/B’ or Matm/B’).
And it is given blow
Nominal diameter(mm) Mop, Nm(KNmm) Matm(Nm)
70 6005.689(6005.689) 114.115
30 6314.047(6314.047) 53.834
25 1989.1889(1989.1889) 48.1707
Table 12.
There for M=Mop/B’ the stress is
M=Mop/B’
M70mm=6005.689KNmm/45.25mm
M70mm=132.722KN
M30mm=6314.047KNmm/18.75mm
M30mm=336.7491KN
M25mm=1989.1889KNmm/14mm
M25mm=142.084KN
σhb=F1M
σhb 70mm=0.000341(1/mm2
)(132.722KN)
σhb 70mm =45.258MPa

σhb 30mm =0.00199(1/mm2
)(336.749KN)
σhb 25mm =0.003289(1/mm2
)(142.084KN)
σrd=F2M
σrd70mm =0.00661(1/mm2
)(132.722KN)
σrd70mm =877.2924MPa
σrd30mm =0.004284(1/mm2
)(336.749KN)
σrd30mm =1442.643MPa
σrd25mm =0.005381(1/mm2
)(142.084KN)
σrd25mm =764.41MPa
σws=F3M
σws70mm =0.002064(1/mm2
)(132.722KN)
σws70mm =273.9382MPa
σws30mm =0.00165(1/mm2
)(336.749KN)
σws30mm =555.6358MPa
σws25mm =0.00733(1/mm2
)(142.084KN)
σws25mm =1041.4757MPa
σtg = σws -F4σrd
σtg70mm=2739.39MPa – 6.09*877.2924MPa
σtg70mm=-2603..14
σtg 30mm=555.635MPa-1.43*1442.643MPa
σtg 30mm =-206.664MPa
σtg 25mm=1041.4757MPa-1.49*764.41MPa
σtg 25mm=-97.4953MPa
The flange must be sized so that the stresses given by equations satisfy the following criteria: when
ffo=102.04MPa(14.8KPSi) is the maximum allowable design stress for the flange material at the operating conditions.
R[1] and [2]
σhb > 1.5ffo
For 70mm 45.258MPa<153.06MPa -----(unsatisfied)
For 30mm 670.164MPa > 153.06MPa ---- (satisfied)
For 25mm 467.31MPa > 153.06MPa ---- (satisfied)
σrd > ffo
For 70mm 877.2924MPa >102.02MPa ------- (satisfied)
For 30mm 1442.643MPa > 102.04MPa ------ (satisfied)
For 25mm 764.41MPa > 102.04MPa ------ (satisfied)
0.5(σhb+σrd) > ffo
For 70mm 461.27MPa > 102.04MPa --- (satisfied)

0.5(σhb+σtg) > ffo
For 70mm -1278.94MPa<102.34 ------(unsatisfied)
The bolt spacing must be selected to give a uniform compression of the gasket. It will not normally be less
than 2.5 times the bolt diameter, to give sufficient clearance for tightening with a wrench or spanner. The following
formula can be used to determine the maximum bolt spacing:
From the table R [1] and R [2] the material [SA-193-B7] and maximum allowable stress (fb) for the bolt is 172.37MPa
(25KPSi).
The minimum bolt area is given by
Abf=Wm/fb
Where Wm greatest of Wm1or Wm2
Nominal Dia.(mm) Wm1(KN) Wm2(KN)
70 229.2897 5.029
30 42.4327 3.132
25 39.075 2.792
Abf=Wm/fb
For 70mm nominal diameter greatest number is Wm1 =229.257KN
Abf70mm=229.2897KN/172.37MPa
Abf70mm =1330.21mm2
…………………Abf30mm =42.4327KN/172.37MPa
Abf30mm =246.1721mm2
Abf25mm =39.075KN/172.37MPa
Abf25mm =226.6925smm2
The bolt spacing must be selected to give a uniform compression of the gasket. It will not normally be less than
2.5 times the bolt diameter, to give sufficient clearance for tightening with a wrench or spanner. The following formula
can be used to determine the maximum bolt spacing:
Pb=2db+6tf/ (m+0.5)
Where Pb-bolt pitch (spacing), mm
db- bolt diameter, mm
tf- flange thickness, mm
m- Gasket factor,1.75
Pb70mm =2*16mm+6*31.25mm/(1.75+0.5)
Pb70mm =115.33mm

Pb30mm =2*12mm+6*20.25mm/(1.75+0.5)
Pb30mm =78mm
Pb25mm =2*10+6*17.25/(1.75+0.5)
Pb25mm =66mm
Gaskets
Gaskets are used to make a leak-tight joint between two surfaces. It is impractical to machine
flanges to the degree of surface finish that would be required to make a satisfactory seal under
pressure without a gasket. Gaskets are made from “semi-plastic” materials; which will deform
and flow under load to fill the surface irregularities between the flange faces, yet retain sufficient
elasticity to take up the changes in the flange alignment that occur under load
GASKETDESIGN
Gaskets are used to make a leak-tight joint between two surfaces. It is
impractical to machine flanges to the degree of surface finish that would be
required to make a satisfactory seal under pressure without a gasket. Gaskets
are made from “semi-plastic” materials; which will deform and flow under
load to fill the surface irregularities betweenthe flange faces, yet retain
sufficient elasticity to take up the changes in the flange alignment that occur
under load
The bolt spacing must be selected to give a uniform compression of the
gasket. It will not normally be less than 2.5 times the bolt diameter, to give
sufficient clearance for

CHAPTER FIVE
calculating wall thickness of the nozzle
To find the thickness of the nozzle
tn=PR/ (SE-0.6P)+ C.A
Where tn-thickness of the nozzle
P-maximum allowable
E-joint efficiency (E=1)
S-maximum allowablestress
R-internal radius of the pipe
C.A-corrosion allowance
For my nozzle I selects X-STGfrom the table and S=103.42MPa (15.0KPSi). R [2]
Given Table 5.1
Nominal Dia. (mm) P, MPa(PSi) C.A,mm(in) Radius(mm)
25 27.2 (3946) 1.6 (1/16) 12.5
30 16.159 (2348) 3.12 (1/8) 15
70 8.0811(1175) 6.35(1/4) 35
E=1
S=15000Psi =103.421MPa
tn will be
tn=PR/ (SE-0.6P) + C.A
tn,25mm=[(27.2MPa*12.5mm)/(103.421MPa*1-0.6*27.2MPa)]+1.6mm
tn,25mm =5.5035mm
tn,30mm=[(16.159MPa*15mm)/(103.421MPa*1-0.6*16.159MPa)]+3.12mm
tn,30mm =5.7061mm
tn,70mm=[(8.0811MPa*35mm)/(103.421MPa*1-0.6*8.0811MPa)]+6.35mm
tn,70mm =9.219mm
reinforcment of opening

The “equal area method” is the simplest method used for calculating the amount of reinforcement required, and
is allowedin most design codes and standards. The principle used is to provide reinforcement localto the
opening, equal in cross-sectional area to the area removed in forming the opening. If the actual thickness of the
vessel wallis greater than the minimum required to resist the loading, the excess thickness can be taken into
accountwhen estimating the area of reinforcement required. Similarly with a branch connection,if the wall
thickness of the branch or nozzleis greater than the minimum required, the excess material in the branch can be
taken into account.Any corrosion allowancemust be deducted when determining the excess thickness available
as compensation. The standards and codes differ in the areas of the branch and shell considered to be effective
for reinforcement, and should be consulted to determine the actual area allowedand the disposition of the
various types of reinforcement. For branch connections of small diameter the reinforcement area can usually be
provided by increasing the wall thickness of the branch pipe. Some design codes and standards do not require
compensation forconnections below 89 mm (3 in.) diameter. If anything, the equal area method tends to over-
estimate the compensation required and in some instances the additional material can reduce the fatigue life of
the vessel. More sophisticated methods for determining the compensation required have been introduced into
the latest editions of the codes and standards. The equal-area method is generally used forestimating the
increase in thickness required to compensate formultiple openings. R [1]
First calculatethe reinforcement area R [3]
A=D*t*F
where tr-shell thickness=0.64((D+tn) tn) 1/2 (,)
F-correctionfactor=1
D-diameter of the pipe
tn-nozzle material thickness
E-jointefficiency=1
T-Shell material thickness= (forshell 70mm or forhead 74mm)
A=D*t*F
A30mm=30mm*74mm*1=2220mm2
A70mm=70mm*74mm*1=5180mm2
tr =0.64((D+tn) tn) ½
tr30mm=0.64((30mm+5.7061)*5.7061)1/2
tr30mm =9.135mm
tr70mm=0.64((70mm+9.219)*9.219)1/2
tr70mm =49.3642mm
By taking A1 as the largest of the A11 or A12 calculating the reinforcement area of the vessel
A11= (Et-F*tr)D and A12=2(E*t-F*tr)*(t+ tn)
For 30mm tr=9.135mm

For 70mm tr=49.3624mm
A11= (Et-F*tr) D
A12=2(E*t-F*tr)*(t + tn)
A11, 30mm = (1*74mm-1*9.135mm)*30mm
A11, 30mm =1945.95mm2
A12, 30mm =2(1*74mm-1*9.135mm) (74mm+5.7061mm)
A12, 30mm =10,340.27mm2
A11, 70mm = (1*74mm-1*49.363mm)*70mm
A11, 70mm =1724.59mm2
A12, 70mm =2(1*74mm-1*49.356mm) (74mm+9.219mm)
A12, 70mm =4101.6980mm2
There for A1 (reinforcement in shell) will be the largest value of (A11 or A12)
A1, 30mm=10340.27mm2
A1, 70mm=4101.6980mm2
By taking A2 as the smaller of A21 or A22 and calculate the available nozzlewall.
For 30mm tr n=2.591
For 70mm tr n =2.8746
A21= (tn-trn) 5t
A22=2(tn-trn) (2.5tn-te) , te=0(nopad)
A21, 30mm= (5.7061mm-2.5911mm) 5*74mm
A21, 30mm =1168.124mm2
A22, 30mm=2(5.7061mm-2.5911mm) (2.5*5.7061mm-0)
A21, 30mm =88.8725mm2
A21, 70mm= (9.219mm-2.8746mm) 5*74mm
A21, 70mm =2347.428mm2
A22, 70mm=2(9.219mm-2.8746mm) (2.5*9.219mm-0)
A22, 70mm =292.4451mm2
A2 will be
A2, 30mm=1168.124mm2

A2, 70mm=2347.428mm2
Atotal=A1+A2
A30mm, total = 10340.27mm2 +1168.124mm2=11508.394mm2
A70mm, total = 4101.6480 mm2+2347.428mm2=6449.076mm2
Nominal Dia.(mm) Atotal(mm2) A(mm2)
30 11508.394 2220
70 6449.076 5180
Atotal>A it is adequately reinforced.
DESIGN OF MANHOLEOR INSPECTION
All pressure vessels for use with compressed air and those subject to internal corrosion, erosion or mechanical
abrasion, shall be provided with suitable manhole, hand hole, or other inspection openings for examination and
cleaning.The requiredinspectionopeningsshowninthe table below are selectedfromthe alternativesallowedbythe
Code, UG46, as they are considered to be the most economical. The inside diameter of my pressure vessel is 1m
(39.37in).
Accordingto the diameteritsmanhole recommendedis381mm(15in).
Di mm(in) Inspectionopeningrequired,mm(in)
1000(39.37) 381(15)
SUPORETDESGEN AND CHECK OF PRESSUREVESSLE
STRESSES IN RESPONSE TO DIFFERENT LOADS

 DUE TO INTERNAL PRESSURE
As we are treatingmediatorasa thincylindersothe valuesof hoopstress&longitudinal stressare calculatedas
under
Therefore radial stressesare ignored(verysmall) sowe considerthe followingprimarymembrane stresses
(a)HoopStressesand(b) Longitudinal Stresses
 HOOPSTRESSES (H1)
H1= PIDI /2t
H1= (14MPa) (914mm) / 2(70mm)
H1= 91.4MPa
 LONGITUDINAL STRESS (L1)
L2= PIDI/ 4t
L2= (14MPa) (914mm) / 4(70mm)
L2= 45.7MPa
As hoop stress is greater so design is based on hoop stress.
 STRESSDUE TO WEIGHT OF VESSEL & ATTACHMENT
It is assumed that weight of the vessel and its attachments results in compressive stress only & eccentricity
doesn’t exist and the resulting force coincides with the axis of the vessel.
The weightshall be calculatedforthe variousconditionsof the towerasfollows.
A. Erectionweight
B. Operatingweight
C. Testweight
The compressive stressdue tothe weightisgivenby
S = W / Ct
Where S = unitstress,MPA
W = weightof vessel above the sectionunderconsideration,N
c = circumference of shell orskirtonthe meandiameter,mm
t = thicknessof shell orskirt,mm
The weightsof differentvessel elementsare giveninthe tablesattached.
WEIGHT

ERECTION WEIGHT
SHELL=ρ shell* V shell *g
=8,000Kg/m3*2.75m3*9.82m/s2
=216.04KN
HEADS = ρ head* V head *g
=8,000Kg/m3*0.75m3*9.82m/s2
=33.1087KN
FLANGES (6) = n* (SIZE)
= 4* m flange*g
=4*4350Kg*9.82m/s2
=170.86KN
ERECTION WEIGHT=216.04KN+33.1087KN+170.86KN=420KN
OPERATING WEIGHT
Which include the weight of the erection + operating liquid
ERECTION WEIGHT =420 KN
WEIGHT FOR OPERATING LIQUID = ρliquid*vpressurevessel*g
= 800Kg/m3*3.4m3*9.81m/s2
=26.6832KN
Total weight=26.6832KN + 420KN=446.6832KN------ (ans.)
 STRESS DUE TO WEGHT
Putting values in the formula (a)
Where,
c = π * D mean
C = 3.14 *944mm =2964.16mm
t = 70mm
SW = 446.6832KN/ (2964.16mm*70mm) =2.1527MPa (compressive) ---------- (1)
STRESS DUE TO WIND LOAD:
Towers under wind pressure are considered as uniformly loaded cantilever beams. The
computation of wind is based on standard ANSI A58.1-1982. Where terrain features and local records

indicate that 50 years at standard height are higher than those shown in the map, those higher values shall
be the minimum basic wind speed.
The minimum basic wind speed for determining design wind pressure shall be taken from the map
of wind speed. Design wind pressure shall be determined by the following formula:-
F=wind load F=q*G*CF* As
P = QS*G* CFA/A
=q* G *CF
Where, P= Design wind pressure, MPA
q = Wind stagnation pressure at the standard height of 9.144m (30 feet) as tabulated:
Basic wind speed,
mph, V
70 80 90 100 110 120 130
Pressure PSF q 13 17 21 26 31 37 44
Table 1 wind pressure
CF = Pressure coefficient (shape factor):
Round or elliptical towers----------------------------0.8
G = Combined height, exposure and gust factor coefficient as tabulated:
Height above ground, ft.
Coefficient G
Exposure C Exposure B
0-20 1.2 0.7
20-40 1.3 0.8
40-60 1.5 1.0
60-100 1.6 1.1
100-150 1.8 1.3
150-200 1.9 1.4
Table 2 Coefficient of C and B

Exposure C---------------------The most severe exposure
Exposure B ---------------------Intermediate exposure
For the metanator we will take a wind speed of 112.7Km/HR (70mph), so the value of
QS =0.089MPa (13psf)
CF= 0.8-------------------------For circular vessel
G = 1.3 ---------------Intermediate exposure & vessel height of 5.524m (18.1ft)
There for the value of wind pressure using the above formula will be;
P = 14MPa
We will take the wind pressure 0.09MPa.
QUANTITIES FORMULAS
Shear V= Pw*D1*H
Moment at base M=Pw*D1*H*h1
Moment at height h(t) Mt = M- HT{V-0.5PwD1ht}
Stress S= 12M / R2*π*t
Where, D1= width of the vessel with insulation, 0.915m=915mm
E = Efficiency of the welded joints = 1.0
h1= lever arm, ft = H / 2 = 3.26m=3260mm

ht = distance from base to section under consideration,
3.9m=3903mm
H = length of vessel section, 5.524m=5524mm
M = Maximum moment (at the base),
Nm
Mt= Moment at height h t, Nm
Pw= Wind pressure, 0.089MPa
R = Mean radius of vessel, 0.467m=467mm
S = Stress due to wind, MPa =?
V = Total shear, N
t = Thickness of shell excluding corrosion, 30mm
The values of shear, moment at base & moment at skirt joint are calculated as under and then the
stress developed in response to the moment M(t) using the formulae listed in the table above. By
putting the values of the parameters listed above for methanator.
Shear
V = 0.089MPa*0.914m*5.524m
V = 0.4493MN=44.935KN
Moment (at base)
M = 0.089MPa*1.06m*5.524m*3.26m
=0.96136MNm=96.136KNm
Moment at head to skirt joint
Mt = M – 1.52m*V
Mt=96.136KNm-1.52m*44.935KN
Mt=27.8348KNm
STRESS DUE TO SEISMIC LOAD
PERIOD OF VIBRATION
As a result of wind tall towers develop vibration. The period of vibration should be limited, since
large natural periods of vibration can lead to fatigue failure. The allowable period has been computed
from the maximum permissible deflection.

Table 5.4
QUANTITIES FORMULAS
Period of vibration, T sec T=0.0000265(H / D)2
*(w*D /t)½
Maximum allowable period of vibration, Ta
(sec)
Ta=0.80(WH /Vg) ½
s
Formula for tim
e of vibration (PRESSURE VESSEL)
Where, D = Outside diameter of vessel, 1.0542m=1054.20mm
H = Length of vesselincluding skirt, 4.571m=4571mm
g = 9.8m / sec2
gravitational acceleration
t = Thickness of skirt at the base, 70mm
V = Total shear = ZICW/Rw (calculated ahead)
=1038.6NW= Weight of tower, =394.85KN
w= weight of tower per mater of height, = 60.52KN
Putting values to get period of vibration for methanator
T = 0.0000265(6.524/1.054) 2
*(60.52*1.054*1/0.007) ½
T = 0.09691sec
Now allowable period of vibration
Ta = 0.80{W*H / V*g} ½
=0.8{(60.52*6.524)/(9.81*1038.6)}1/2
Ta = 0.1575sec
As ‘T’ is less than ‘Ta’ hence the condition is satisfied
STRESS DUE TO EARTHQUAKE
The loading condition of the tower under seismic forces is similar to that of the cantileverbeam
when the load increases uniformly towards the free end
Table 5.5
FORMULAS
Shear Moment
V=ZICW/Rw M=[FtH+(V-Ft)(2H /3)]
Mx= M(x/H)

Where
C= Numerical coefficient=1.25S/T2/3

= 1.25*1/(0.076644)2/3 =2.71
=2.71(should not be more than 2.75)
Rw=Numerical coefficient (use 2.9 for vessels)
E = Efficiency of welded joints = 1.0
Ft = Total horizontal seismic force at the top of the vessel, 0N(because T < 0.7)
= 0.07TV (Ft shall not exceed 0.25V)
= 0, for T < 0.7
H =Length of vessel including skirt, 4.571m=4.571mm
I = Occupancy importance coefficient (use 1.0 for
vessels)
K = Horizontal force factor (use 2.0 for vessels)
M = Maximum moment at the base,
Nm
Mx= Moment at distance x, Nm
S = Numerical coefficient for site structure resonance
= 1.0
The product CS shall not exceed 0.14
W = Weight of the vessel, 394.85KNZ =
Seismic factor
= 0.15
Shear = (0.15*1*0.0299*394.85KN)/2.9 V =
1038.6N
Ft = 0.07*T*V =0.07*0.08644*1038.6N=6.284N
0.25V= 259.65N
As condition is that Ft should not exceed 0.25V so it is satisfied for methanator
Therefore Moment
M = [6.284*4.5124 + (1038.6-6.284)*(2*4.524/3)] M
=3140.089Nm
Moment at skirt to head joint
Mt = M(x/H) where x=H/3=4.57m/3=1.5233m
=3140.089Nm*(1.5233m/4.57m)
=1046.67Nm

Therefore stress due to earthquake
Seq = 12* Mt / R2
*π* t
= 12*1046.6/ (0.457)2
*π*0.07
=273.452MPa
FABRICATION AND WELDING METHOD
Fabrication and welding method of the pressure vessel is based on the economic aspects of the shell and
the head. I used for the fabrication of the head and the shell by using but joint weldingfor the contacting
of the shell and head. We know that the martial for the pressure vesselis stainless steel 18Cr/8Ni type
304 and its length is 3m by 1m and this stainless steel are join by but welding joint.
Shell fabrication
Shell is fabricated by rolling the sheet of stainless steel and welds it together to get the
cylindrical shape of the vessel by using but weld joint.
Head fabrication
Head shall is fabricated by making it in to different parts and by using but weld joint.And
finally support joining with the shell by using but joint welding. And different parts of welding is
shown blow in the figure.

CHAPTERSIX
CAD DRAWING
SHELL

HEMISPHERICAL HEAD
LEG

ASSAMBLEY DRAWING
Summary
A pressure vessel is a closed container to hold (fluid) such as, gas, liquids
at a pressuresubstantially different from the ambient pressure.
Pressure vessels are used in a variety of application in both industry & the private
sector. They appear in these sectors as industrial compressed air receivers and
domestic hot water storage tanks. Pressure vessels may theoretically be almost any
shape, but shapes made of sections of spheres, cylinders, heads, & cones are
usually employed. A common design is a cylinder with end caps called heads,
heads shapes are frequently either hemispherical or dished (torispherical). We can
calculate length & diameter of vessels by using nominal vessel volume for typical
ratio of length to diameter by using standard table used to find L/D ratio depending
on the inner service pressure given.
While selecting suitable materials, the requirements of the relevant part relating to
the function, stress conditions, and service life have first of all to be considered.
The ends of a cylindrical vessel are closed by heads of various shapes. The
corrosion allowance is the additional thickness of metal added to allow for material
lost by corrosion & erosion, or scale. The strength of metals decreases with
increasing temperature. Pressure vessels are built up from preformed parts;
cylinders, heads,& fitting. Joined by fusion welding. Heads to shells attachment by
butt welded joints of plates of unequal thickness. The method used to support a
vessel will depend on the size, shape & weight of the vessels, the design
temperature & pressure the vessel location & arrangement & the internal &
external fittings attachments.
Generally when we design pressure vessel we have to used gasket, flanges,
consider reinforcement of openings, manhole ,earthquake loading, wind load,
external load ,dead weight of vessel , pressure,…etc.

REFERENCE

1. “Acid storage tank fabrication.” n.d. moontanks.com.
2. Coulson & Richardson's. 2005. chemical engineering design. fourth ed. Vol. six. six
vols.
London: elsevier butterworth-Heinemann.
3. Machine Design databook. n.d.
4. Megyesy, Eugene F. n.d. pressure vessel handbook. tenth ed. Tulsa,
Oklahoma, USA:pressure vessel publishing INC.
5. DDU student project

Pressure vessel project.docx

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