Adigrat University: College of Engineering and Technology Department of Chemical Enginnering
Adigrat University: College of Engineering and Technology Department of Chemical Enginnering
Adigrat University: College of Engineering and Technology Department of Chemical Enginnering
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SUBMISSION DATE: 12 DECEMBER 2016
ADIGRAT ETHIOPIA
Declaration
We declare that this written submission represents our ideas in our works and where other ideas
or words have been included,we have adquatly cited and referred the original sources.We also
declare that we have adherd to all principls of academic honesty and integrat and have not mis-
represented or falsified any idea(data)sources in our submission.
Acknowledgment
We are really grateful because we managed to complete our design apparatus project with in the
time given by our MscZinabu.This project can not be completed with out the effort and co-
operation from our group memmbers ,Brhane,Kibrom,Jemal,Mokenen and yemane.We also
sincerely thank our MscZinabu for the guidance and encouragement in finishing this project and
also teaching.Last but not least,we would like to express our gratitude to our librarians and
respondents for the support and willingness.
Executive summary
High pressure rise is developed in the pressure vessel and pressure vessel has to
withstand severe forces. So the selection of pressure vessel is most critical. That’s why
we can say that pressure vessel is the heart for storage of fluid. Pressure vessel
must pass series of Hydrostatic tests. These tests examine the ability of the structure to
withstand various pressures to see if protective zone around the operator station
remains intact in an overturn. The structure is to be designed, fabricated, fitted
and checked as per ASME standard. Plant safety and integrity are fundamental concern
in pressure vessel design and these of course depend on the adequacy of design
codes. The performance of a pressure vessel under pressure can be determined
by conducting a series of tests to the relevant ASME standard. Efforts are made in this
paper to design the pressure vessel using ASME codes & standards to legalize the
design.
List of figure
List of tables
material construction…………………………………………………..table 3.1.1
typical standard flange…………………………………………………table 3.2.1
column spesfication…………………………………………………….table 3.2.2
nozzle spesfication……………………………………………………...table 3.2.3
other sive plate column spesfication……………………………………table 3.2.4
Acronyms
e=minimum thickness,m
pi=internal pressure,N/𝑚𝑚2
Di=internal diameter,m
F=design stress,N/𝑚𝑚2
J=joint factor
Wv=total weight of vessel,N
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CV=factor account to weight of nozzles
Dm=mean diameter,m
Hv=height or length of the vessel,m
t=wall thickness,m
g=gravitational acceleration,9.81m/𝑠 2
𝜌=density of vessel material,kg/𝑚3
Wp=weight of the plate,KN
Wi=weight of insulation,KN
V=volume,𝑚3
Pw=wind pressre,N/𝑚2
Fw=wind load,N/m
Mx=bending moment,Nm
𝜎𝑙=longitudinal stress,N/𝑚𝑚2
𝜎h=circumsferential stress,N/𝑚𝑚2
𝜎𝑧 =resultant longitudinal stress,N/𝑚𝑚2
𝜎𝑤 =dead weight stress,N/𝑚𝑚2
𝜎𝑏 =bending stress,N/𝑚𝑚2
Iv=second moment area,𝑚𝑚4
𝜎c=elastic stability(critical buckling stress),N/𝑚𝑚2
Rc=crown radius,m
Rk=knuckle radius,m
Cs=stress concentration factor
Dn=diameter of the nozzle,m
tn=actual nozzle thickness,m
th=theoretical thickness of shell,m
C=corrosion allowance,mm
As=area of the shell,𝑚2
Av=area of compensation provide by nozzle,𝑚2
𝜎bs=bending stress in the skirt,𝑁/𝑚𝑚2
𝜎ws=dead weight stress in the skirt,𝑁/𝑚𝑚2
Ms=maximum bending moment
ts=skirt thickness,m
Ds=inside diameter of the skirt,m
fs=maximum allowable design stress for the skirt,N/𝑚𝑚2
𝜃s=base angle of conical skirt
Ab=area of bolt,𝑚𝑚2
Nb=number of bolts
fb=maximum allowable bolt stress,N/𝑚𝑚2
Db=bolt diameter,m
Fb=load on the base ring,N/m
Lb=base ring width,mm
fc=maximum allowable pressure on the concret foundation pad,N/𝑚𝑚2
tb=base ring thickness,mm
Lr=distance between the skirt and the ring,mm
f𝑐̇ =actual pressure on the base,N/𝑚𝑚2
fr=allowable design stress in the ring material,N/𝑚𝑚2
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CHAPTER ONE
1.Introduction
Sieve plate column is one of the distillation columns with cross-flow trays, which is widely
used in various mass transfer operations. Sieve plates offer high capacity and efficiency, low
pressure drop, ease of cleaning and , low capital cost When developing sieve plate column
design, It is crucial to develop and specify the basic design information for a particular vessel,
to ensure the pressure vessel work effectively. The design of such vessels requires application of
current design codes and standards to be practiced along with the stress analysis relatively. In all the
major industrialized countries, the design and fabrication of thin-walled pressure vessels is covered
by national standards and codes of practice. The primary purpose of the design codes is to establish
rules of safety relating to the pressure integrity of vessels and provide guidance on design, materials
of construction, fabrication, inspection, and testing. Improper vessel design and maintenance increase
the risk of vessel failure and may lead to serious safety hazard. Thus a chemical design of the column
is highly important in essence of keeping the safety hazard and maintenance cost at minimum.
Therefore, based on the design code drawn as well as specifications and requirements outlined, which
include column wall thickness, selection and sizing of vessel heads, reinforcement of any openings,
nozzles, flanges, column supporting skirt and base ring, the pressure vessel design is developed
accordingly. Column wall thickness , selection and sizing of vessel heads , reinforcement of openings,
nozzles and flanges , column supporting skirt and base ring as well as design loads such as internal
pressure, wind loading and dead weight of vessel and its contents are considered when developing the
intended design. Appropriate assumptions are drawn in, to execute certain conditions for the design.
To present the completed design, a dimensional sketch as well as column specifications sheet of the
design is issued at the end of the report.
1.1 Background
A column should be designed with a specific wall thickness to ensure it has sufficient rigidity to be able
to cope with the maximum amount of pressure which it is expected to be susceptible to in operation. The
minimum column wall thickness needed to withstand the vessel weight itself as well as other incidental
loads such as internal pressure is calculated and shown in the following calculation section. The design
pressure for vessels under internal pressure is conventionally taken as the pressure where the relief device
is set. This is usually about 5-10 % above the regular working pressure. This is done to avoid spurious
operation during minor process upsets. (Sinott 2008).
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1.2 Statement of the problem
The pressure vessel that not flow any standard codes can be very
dangerous.in fact many fatal accidents have occurred in the history of
their operation and development.In addition the formulas and concepts
applied in the systems are always unknown by the users.
1.3 Objective
1.3.1 General objective
The objective of this project is to design pressure vessel, to justify the pressure design and
material selection as well as the cause of the failure was investigated and finished element
analysis.
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1.3.2 Specific objective
A sieve plate column that satisfied the preliminary specifications summary given and dimensional sketch
of the design and column specification sheet is expected.
And this project is to justify the pressure vessel design and material selection as well as the cause
of the failure was investigated and finished element analysis was done by under various
simulated conditions to insure that the mechanical properties and material selection where not
the cause of the vessel failure.
CHAPTER TWO
Constraints are factors that has influence on the design of a given solutions, they may arise from
physical law, government law, regulation and standards.
2.3.1 Design pressure: A vessel must be designed to withstand the maximum pressure to
which it is likely to be subjected in operation. For vessels under internal pressure, the design
pressure is normally taken as the pressure at which relief the device is set. This will normally be
5 to 10% above the normal working pressure.
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2.3.2 Design Temperature: The strength of metals decreases with increasing temperature.so,
the maximum allowable design stress will depend on the material temperature.
2.3.3 Materials: Pressure vessels are constructed from plain carbon steels, low and high alloy
steel, otheralloys, clad plate, and reinforced plastics.
2.3.4 Design stress (nominal design strength): For design purpose it is necessary to
decide a value for the maximum allowable stress that can be accepted in the material of
construction. For materials not subject to high temperatures the design stress is based on the
yield stress or tensile strength of the material at the design temperature.
Category one: the highest class required 100% non-destructive testing of the Weld, no restriction
of the plate thickness.
Category two: requires less nondestructive testing and maximum plate thickness.
2.3.6 Corrosion allowance:is the additional thickness of metal added to allow for material
lost by corrosion and erosion. The allowance should be based on experience will the material
construction under similar service conditions to those for the proposed design. Forcarbon steel
and low alloy steel a minimum allowance of 2mm should be used, where more severe conditions
are anticipated this should be to 4mm.
2.3.7 Design loads: A structure must be designed to resist gross plastic deformation and
collapse under all the conditions of loading. Load can be classified as major load and subsidiary
loads.
2.3.8 Minimum practical wall thickness: there will be a minimum wall thickness required
to ensure that any vessel is sufficiently rigid to with stand its own weight and any incidental
loads. The wall thickness of any vessel should not be less than the given value.
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Base ring and anchor bolt design: The loads carried by the skirt are transmitted to the
foundation slab by the skirt base ring(bearing plate).
CHAPTER THREE
𝐏𝐢𝐃𝐢
e=
𝟐𝐟𝐉−𝐏𝐢
Where,
𝟏
but,Cs=𝟒 (𝟑 + √𝐑𝐜⁄𝐑𝐤)
pi=design pressure(N/mm2)
Rc=crown radius
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Rk=knuckle radius
PiDi
e= where,e=minimum thickness of ellipsoidal head(mm)
2fJ−0.2Pi
Pi=design pressure(N/mm2)
Di=internal diameter(mm)
j=joint factor
t=wall thickness(m)
weight of insulation
volume of insulation=πDHv(t)insulation
weight of insulation=ρVg
Total weight=Wvessel+Wplate+Winsulation
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Wind load
Dm=Di+2(t + tin) × 10−3
Wind load,Fw=PwDm
Analysis stress
PiDi
δl =
4t
PiDi
δh =
2t
Dead weight stress
Wv
δw =
π(Di + t)t
Bending stress
Do=Di+2t
π
Iv=64 (Do4 − Di4 )
Mx Di
δb = ± ( + t)
Iv 2
Resultant longitudinal stress
δz = δl + δw ± δb
tb
δc = 2 × 104 ( )
Do
Area of compensation
PiRcCs 1 Rc
et(torisphere)=2fJ+Pi(Cs−0.2)=th ,Cs=4 (3 + √Rk)
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PiDi
en(nozzle)=2fJ−Pi=tṅ
Ar=(DO-Di)tn,Dir=Din+tn
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4Ms
σbs=π(Ds+tbs)tbsDs
App. wt
σws(test) =
π(Ds + ts)ts
weight of vessel
σws(operating) =
π(Ds + ts)ts
Maximum ̂
σs(compressive)=σbs + σws(test)
σs
̂(tensile)< 𝐹𝑠𝑗𝑠𝑖𝑛𝜃
ts
σs
̂(compressive)< 0.125𝐸(Ds)sinθ
Properties Specification
Feed At mid point 200mm inside diameter
Vapour out At 0.7m below top of 250mm inside diameter
cylindrical section
Bottom product At center of vessel 200mm inside diameter reflux
return
At 1.0 m below top of 200mm inside diameter
cylindrical section
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ports(manholes) -at 1.5m below the top of the
column
Height of support skirt 2.5m Carbon steel, silicon killed
Thickness of insulation 50mm Mineral wool
Ladded Access ladder with platforms
Vessel Stainless steel,unestabilised(304)
Nozzle Stainless steel,unestabilised(304)
Skirt Carbon steel,silicon killed
Assumption
-no significant loading from piping and external equipment
CHAPTER FOUR
4. RESULT AND DISCUSSION
SOLUTION:
Step 1.determine the minimum plate thickness to resist the internal
pressure
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Pi = Design pressure = 1.2 N/mm2
Di = inner diameter of the column = 1500 mm
f = the design stress of the material 11of construction = 130 N/mm2
j = the weld joint efficiency = 1.0
The weld joint is double welded butt or equivalent joint and 100 percent
degree of radiography.
t= ( 1.2 (1500))/(2(130)(1)-0.2(1.2))
t= 6.93 mm
roundoff to 7 mm
t = 1.2(1500)1.77/(2(130)(1)+1.2(1.77-0.2)) = 12 mm ,
But the minimum thickness is more preferable .
Step3. So that we select & size the ellipsodial head because it has
smallest thickness and less economical than torispherical.
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Step4. For stability of the shell the column is divided into five parts
and thickness is increased from top to bottom of the shell, as 7 mm, 9
mm, 11 mm, 13 mm and 15 mm respectively.
7
9
11
13
15
For all other calculations, the average thickness is used. The average
thickness = ts =11 mm .
WV = 175.3KN
WP=150.45kN
Win=22.224KN
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W Total=347.96KN
σl=PiDi/4tb=(1.2*1500)/(4*15)=30N/𝑚𝑚2
Circumferential stress,
σh=PiDi/2tb=(1.2*1500)/(2*15)=60N/𝑚𝑚2
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step 11. Stress due to dead weight,
σw=Wv/π(Di+tb)tb=(175.3*103 )/π(1500+15)15=2.4566N/𝑚𝑚2
Up-wind
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Down-wind
The maximum resultant compressive stress will occur when the vessel is
not under pressure given by.
σz=σb+σw = 53.08+2.4566=55.54N/𝑚𝑚2
ADIGRAT ETHIOPIA
Since the maximum resultant compressive resultant stress
(55.54 N/mm2) less than that of critical buckling stress (196.07N/𝑚2 ).
So the design procedure of the vessel is correct.
Design 0f Skirt support and base ring calculation
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Figure 3.4.2: Standard ellipsoidal head
The height of each ellipsoidal head, h = D/(4 ) , 1.5/4=0.375 m
The height of vessel, Hv:
Hv = (Length of cylindrical section) + (Height of the two ellipsoidal
heads)
= 37+2(0.375) = 37.75m
Step 2: Determine the bending moment, in skirt.
x=Hs+Hv
Where,
HS= Height of skirt=2.5m
H V =Height of the vessel=37+2(0.375)=37.75m
x=37.75+2.5=40.25 m
Ms=(Fw*x^2)/2,where Fw =load per unit length is from a bove
calculation =2076.16N/m
X = Hs+Hv
Ms=bending moment at the base of the skirt
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Step 3: Determine the bending stress in the skirt.
The bending stress in the skirt is given by the following formula,
Assume as a first trial, take the skirt thickness as the same as that
of the bottom section ofthe vessel, 15 mm.
σbs= 4Ms/π(Ds+ts)Dsts
Where,
Ds =Internal column diameter, =1.5m
ts = Thickness of the skirt = Thickness of the bottom section, =15mm
Thus,
σbs=(4*1681754.48*1000)/(π*(1500+15)*15*1500)=62.85N/𝑚𝑚2
Step 4: Determine the dead weight stress for the test and operating
conditions.
For test condition,σws(test):
Formula given by, σws(test)=Wwater/π(Ds+ts)ts
Where,
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Wwater= dead weight of the vessel with contents (water) is
approximate weight
Wwater=(π*D^2*Hv*ρw*g)/4=(π*1.52 *37.75*1000*9.81)/4=653424.2
N=653.4242KN
Thus,
σws(test)=Wwater/π(Ds+ts)ts
σws(test)=653424.2/(π*(1500+15)*15)= 9.2N/mm2
For operating condition, σws(operating):
The formula given by,
σws(operating)=Wvessel/π(Ds+ts)ts
Where,
σs(compresive)=62.85+8.975=71.825 N/mm2
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For minimum:
σws(tensil)=σbs+σw(opersting)
σws(tensil)=62.85-2.45668= 60.4N/mm2
θ=90°
σs(tensil)≯fsJsinθs
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60.4≯115×1×sin90°Type equation here.
65.3N/𝑚𝑚2 ≯115 (Fulfilled)
Since all the calculations and comparison above shows that the all
design criteria were fulfilled,add 2 mm for corrosion, gives a design
thickness of new skirt thickness, ts:
New skirt thickness, ts:
ts= (Old thickness of the skirt) + (Corrosion allowance)
=15mm + 2mm= 17mm
Base ring/flange and anchor bolt design
For DS = 1.5 m
Step 1 .Determine the number of bolts required, N bolts
Since the measurements for the pitch diameter were not given, it
was assumed that the measurement of column diameter would be used to
make an assumption for the pitch diameter. An estimation of 10%
allowance was then added to the pitch circle diameter column.
Number of bolts required N bolts given by the formula:
N bolts=Dpπ/600
Where;
Ds = Internal column diameter = 1.5 m
Dp = Pitch circle diameter
= Ds + (10% Ds)
= 1.5 + 0.1 x 1.5
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= 1.65 m
= 1,650 mm
Thus, Nbolts=(1650*π)/600=8.635mm
By taking the nearest multiple of 4, 12 is the nearest value to the number
of bolts required, Nbolts.
Step 2: Determine the required bolt area, Ab
The required bolt area, A given by the formula:
Ab=1/(Nbolt*fb)(4Ms/Dp-Wv)
Where,
N bolt=Number of bolts required = 12 Bolts.
fb = Bolt design stress = 125 N/mm2
Ms =Bending moment in skirt
Wvessel= dead weight of vessel with heads
Dp = Pitch circle diameter = 1.6 5m = 1,650 mm
Ab=1/(Nbolt*fb)((4Ms )/Dp - Wvessel)
= 1/(12*125)((4*1681754.48)/1650 – 175.3*1000)
=2601.12𝑚𝑚2
Step 3: Determine the bolt root diameter, D bolt.
The bolt root diameter given by the formula:
D bolt =√(4Ab/π)=√((4*2601.12)/π)=57.56mm
ADIGRAT ETHIOPIA
With the addition of base rings, the bolt should be sufficient enough to
distribute the total compressive load to the foundation.
Step 4: Determine the total compressive load on each base ring per unit
length, Fbolt.
Given the formula:
Fbolt= 4Ms/(π*𝐷𝑠 2 ) + Wv/πDS
Where,
Wvessel=dead weight of vessel with heads = 175.3 KN
Ms=Bending moment in skirt = 1681754.48Nm
Ds=Internal column diameter = 1.5m
Fbolt=(4*1681754.48*1000)/(π*15002 ) +
(175.3*1000)/(π*1500)=989.38N/mm
Step 5: Determine minimum width of base ring, Lb.
Given by the formula:
L b = Fb/(fc*103 )
F bolt=Total compressive load on the base ring per unit length =
1,196.40N/mm
fc=Maximum allowable bearing pressure on the concrete foundation pad
bearing pressure 5 N/ 2
Lb=989.38/(5*103 )=0.1978mm
ADIGRAT ETHIOPIA
Due to the large base ring width, a flared skirt shall be used as an
alternative to the design. Therefore, the requirements and assumptions
for the new skirt are:
Flared skirt bottom diameter, Ds’ = 2.5 m = 2500 mm
Bolt circle diameter, Dolt’= Ds’+ (10% of Ds)
= 2.5 m + (1.5*0.1)= 2.5 m + 0.15 m
= 2.65 m (Assumed)
= 2,650 mm
Number of bolt, (Nbolt) ̇=(π*(Dp) ̇)/600 = but ,
D'p=D's+(10%D ṡ )
D'p=2.5+(10%2.5) = 2.75 m
= 2,750mm
(Nbolt) ̇ = (π*2750)/600=14.4mm
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Bolt specing =(Dbolt*π) ̇/(Nbolt) ̇
Where,
Dbolt'=Bolt circle diameter = 2650 mm
Nbolt'=Number of bolts =14.39
boltspecing=(π*2650)/14.4=577.8 mm (satisfactory)ss
Step 3: Determine new required bolt area, Ab
Given by the formula:
Ab=1/(Nbolt*fb) ̇ (4Ms/(Dbolt) ̇ -Wvessel)
Wvessel=l dead weight of vessel with heads = 175.3 KN
Ms=Bending moment in skirt
Nbolt'=Number of bolts required = 14.4
Dbolt'=Bolt circle diameter = 2.650m = 2,650 mm
fb=Bolt design stress = 125 N/mm2
Ab=1/(14.4*125)((4*1681754.48*1000 )/2650 -(175.3*1000))=
255.2𝑚𝑚2
Step 4: Determine the total compressive load on each base ring per unit
length, Fbolt.
Given the formula:
Fbolt=4Ms/(π*((𝐷𝑠 2 ) ̇ ) +Wv/(π(*Ds) ̇ )
Where,
W vessel= dead weight of vessel with heads =175.3 KN
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Ms= Bending moment in skirt = 1681754.48Nm
D's=Flared skirt bottom diameter=2.5m =2500mm
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Lr =Distance from the edge of the skirt to the outer edge of the ring =
178 mm (Sinnott., Coulson & Richardson's Chemical Engineering 1999)
:Using Bolt size 70 (BS 4190:1967)
Actual width = 178 + 17 +50 = 245mm
(fc) ̇=(Fbolt) ̇/(Actual Width)=365/245=1.489 N/mm2
Step 7: Determine the base ring thickness, tb
The base ring thickness given by the formula:
tb = Lr√((3*(fc) ̇)/Fr)
tb=178√((3*1.489)/140)=31.795mm
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Reinforcement of opening is given by :
A=d*tr*F+2*tn*tr*F(1-fr1)
Preliminary calculation of thickness & distance
Such as tr,trn,t,ti,tn& h
Step1.determine the required thickness under circumferential stress ,tr
for seamless shell or head is given by : tr=PiR/(fj-0.6Pi)=PiK1D/(fj-
0.6Pi)
Where tr=required thickness for a seamless sphere with radius of K1D,
R=K1D
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Step5.determine the distance nozzle projects beyond the inner surface of
the vesselwall ,h ,is given by:
h=(2.5t , 2.5ti)
h=(2.5*9,2.5*46)=(22.5,115)
Step6.determine diameter of finished opening,d, is given by :
d=D+c=600+2=602mm
Note: the opening is in a corroded condition.thus,a corrosion allowance
is added to the diameter of the opening.
Calculation of area required for reinforcement
Assuming correction factor,f=1 & d=D
A=d*tr*F+2*tn*tr*F(1-factor)
To determine whether additional reinforcement is necessary for the man
holes the actual area available for reinforcement must be calculated &
compare with the area required for reinforcement.
The following condition should be satisfied if no addition reinforcement
is required,
A<∑Ai,where ∑Ai= A1+A2+A3+A41+A43
A1(largest):
(A1 ) ̇=d(E1t-Ftr)-2tn(E1t-ftr)(1-fr1) ,where fr1=f/f=130/130=1
Since the nozzle & vessel are made-up of from the same material (i.e
from stainless steel).
Therefore the term 2tn(E1t-Ftr)(1-fr1) is equal to zero.
ADIGRAT ETHIOPIA
Then (A1 ) ̇=d(E1t-Ftr) ,but E1=F=E=1
(A1 ) ̇=d(t-tr)=602(9-5.012)=2400.77𝑚𝑚2
(A1 ) ̈=2(t+tn)(E1t-Ftr)-2tn(E1t-Ftr)(1-fr1) but f=1 , E1 =F=1,the
equation becomes
(A1 ) ̈=2(t+tn)(E1t-Ftr)=2(9+50)(9-5.012)=470.584𝑚𝑚2
Since (A1 ) ̇>(A1 ) ̈,then A=(A1 ) ̇=2400.77𝑚𝑚2
A2(smallest)
(A2 )̇ : first of all assuming Fr2 =Fr1=1
Since the both have the same formula (Sn/Sv)
( A2 ) ̇=5(tn-trn)fr2t , substituting the vessels
(A2 ) ̇=5(50-4.173)(1*9)=2062.215𝑚𝑚2
(A2 ) ̈=5(tn-trn)Fr2 tn=5(50-
4.173)50=11456.75𝑚𝑚2
Since (A2 ) ̇<(A2 ) ̈,then A2=(A2 ) ̇=2062.215𝑚𝑚2
A3(smallest)
(A3 ) ̇=5t(tiFr2)=5*9(46*1).where Fr2=1
(A3 ) ̇=2070𝑚𝑚2
(A1 ) ̈=5ti (tiFr2)=5*46*46=10580𝑚𝑚2
(A3 ) ̈=2h(tiFr2)=2*17.5(46*1)=1610𝑚𝑚2
̇ (A2 ) ̈ Then A3=(A3 ) ⃛=1610𝑚𝑚2
Since (A3 ) ⃛<(A3 ) <
ADIGRAT ETHIOPIA
For areas A41 & A43 they are not consider since the welds vary
from fillet to bolt weld type.
Thus,A41=A43=0
Then we sum up from A1 to A3
∑Ai=A1+A2+A3= (2400.77+2062.215+1610)𝑚𝑚2
=6072.985𝑚𝑚2
By comparing both the required area for reinforcement and the area
available reinforcement area equals to 3007.38𝑚𝑚2 .
A=3007.78𝑚𝑚2 and ∑Ai=6072.985𝑚𝑚2
Therefore .the condition A<∑Ai is satisfactory, no additional
reinforcement will be required .
Table 4.1 Summerization of all final results
Minimum cylindrical e or t 9mm
thickness,mm
Minimum ellipsoidal e or t 7mm
thickness,mm
Minimum torispherical e or t 12mm
thickness,mm
Average thickness,mm tave 11mm
Dead weight of Wv 175.37KN
vessel,KN
Weight of plate,Kn Wp 3.009kN
Weight of Win 22.224KN
insulation,KN
Total weight,KN wtotal 347.90KN
Wind load,N/m Fw 2076.16N/m
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Bending moment,Nm Mx 1421131.52Nm
Stress analysis,𝑁/𝑚𝑚2
Longitudinal stress 𝜎𝑙 30N/𝑚𝑚2
CHAPTER FIFE
5.1 CONCLUSION
As the process of this project is completed,it can be concluded that the
objectives of this project is successfully done.This project has lead to
several conclusions;However,the major conlusions is:The design of
apressure vessel is more selection of procdure,selection its components
to be more precise rather than designing each and every compenent.
5.2 RECOMMENDATION
In a nut shell in terms of design rules, there are many aspects to looking
from the important to have a complete and perfect pressure vessel.
The design code and standard must be appropriately revised to make
sure the safe enough.
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