Design and Analysis of Pressure Vessel using ASME Code
Mukesh S. Kumavat
1Valmik G. Garje
2Basanagouda Patil
31,2
BE Student
3Assistant Professor
1,2,3Department of Mechanical Engineering
1,2,3MGMCET, Navi Mumbai-410209, India
Abstract—Pressure vessels are widely used for the storage of high pressure fluids in industries. This technical paper represents the design and analysis of pressure vessel which is a vertical container used as activated carbon absorber in helium gas purification system. Shell, head, flange, nozzle of pressure vessel is designed by using ASME section -8 division-1 and supporting legs and base plate design is completed using Pressure Vessel Design Manual by Dennis Moss, third edition. Efforts are made in this paper to design the pressure vessel using ASME codes & standards to legalize the design. The objective of present work is to design a pressure vessel whose sole purpose is to withstand the pressure of helium gas flowing through it. A detailed study of various parts of pressure vessel like shell, head, support, flange, nozzle, etc. is carried out and accordingly the vessel is designed.
Key words: Pressure Vessel, ASME Code, Design, Maximum
Allowable Stress, Shell, Nozzle, Flange I. INTRODUCTION
Tanks and vessel that carry, store or receive high pressure fluids are called pressure vessel. A pressure vessel is defined as a container with a pressure differential between inside and outside. It is designed to hold gases or liquids at a pressure substantially different from the ambient pressure. Pressure vessel often has a combination of high pressure together with high temperature and in some cases flammable fluids or highly radioactive material. Because of such hazards it is imperative that the design be such that no leakage can occur. In addition vessel has to be design carefully to cope with the operating temperature and pressure.
The pressure differential is dangerous, and fatal accidents have occurred in the history of pressure vessel development and operation. Consequently, pressure vessel design, manufacture, and operation are regulated by engineering authorities backed by legislation. Therefore in this presented paper, the design of pressure vessel is governed by ASME codes which specify requirements of design, fabrication, inspection and testing of pressure vessel. While designing a pressure vessel primary consideration is safety to avoid damages due to various untoward accidents because due to high pressure stresses are induced in pressure vessel if this stresses are more than permissible stresses than the failure of pressure vessel occurs. The size and geometry from of pressure vessel are differs as per application. The large cylinder-shaped vessel used for high pressure gas loading to the minor size used as hydraulic components for air craft.
II. DESIGN PARAMETERS Design pressure - 25 bar
Operating pressure & temperature:
Maximum Minimum
Pressure 20 bar 20 bar
Temperature 150 ᴼC -28 ᴼC Table 1: Operating pressure & temperature
Design Code - ASME Section: VIII (Division-I), 2015
Material selection code – ASTM standards
Outside Diameter of vessel Do – 20.18 in.
Height of vessel – 1850 mm
Maximum allowable stress (S) – 20015.21 Psi
Joint efficiency- 1.00
III. MATERIAL SELECTION
From ASME BPVC Section-II Part-D through Section-VIII Division-1, UG-4 following materials are selected for different components: Componen t Materia l Yield strength (Psi) Tensile strength (Psi) Allowabl e stress (Psi) Shell and head SA516 Grade 70 (Plate) 37709.8 5 70343.3 7 20015.21 Nozzle SA106 Type-B (Pipe) 34809.1 0 60190.7 2 17114.45
Table 2: Materials and their mechanical properties IV. DESIGN OF CYLINDRICAL SHELL
A. Notations
A = Factor obtained from ASME BPVC Section-II Part-D
B = Maximum allowable compressive stress obtained from ASME BPVC Section-II Part-D
D0 = Outer Shell diameter = 20.18 in. E = Joint efficiency = 1
L = Shell height = 1250 mm = 49.21 in.
PEXETERNAL = Atmospheric Pressure = 1.013 bar = 14.7
Psi
P = Internal pressure = 25 bar = 362.6 Psi
𝑅0 = Outer Shell radius = 10.09 in. 𝑅 = Inner radius of shell = 9.69 in.
S = Maximum allowable stress from ASME BPVC Section-II part-D = 138 MPa = 20015.21 Psi
t = Minimum required thickness for shell
B. Design under Internal Pressure
Thickness of cylindrical and spherical shells t = PR0
SE+0.4P =
362.6∗10.09
20015.21∗1+0.4∗362.6= 0.18 in. Considering availability and corrosion allowances, Assume, t = 0.4 in.
Checking for the maximum pressure for t = 0.4 in.:
1) Circumferential Stress (Longitudinal Joints):
P = SEt R+0.6t =
20015.21∗1∗0.4
2) Longitudinal Stress (Circumferential Joints):
P = 2SEt R−0.4t =
2∗20015.21∗1∗0.4
9.69−0.4∗0.4 =1680 Psi
The maximum allowable working pressure of cylindrical shells is less than above two pressures.
Therefore, design is safe.
C. Design under External Pressure
Assume t = 0.4 in. Now, L D0 = 2.44 & D0 t = 50.45
From ASME BPVC Section- II, Part-D, Subpart-3, we get
A = 0.0005
B = 52 MPa = 7541.97 Psi
Maximum allowable pressure in the vessel considering external pressure,
PALLOWABLE = 4B
3(D0/t) = 100.11 psi
Thus PALLOWABLE (100.11 Psi) > PEXETERNAL (14.7
Psi).
Since the maximum allowable pressure in the shell for thickness, t = 0.4 in. is greater than the actual external pressure acting on the shell, therefore it can be stated that design of the shell is safe for the specified conditions.
V. DESIGN OF HEAD
Fig. 1: Torispherical head design
A. Notations
A = Factor obtained from ASME BPVC Section-II Part-D
B = Maximum allowable compressive stress obtained from ASME BPVC Section-II Part-D
D0 = Outer head diameter = 20.18 in. E = Joint efficiency = 1.0
L = Crown radius = D =20.18 in.
Le = Effective length of head. M = Factor = 1 4 (3+√ L r ) = 1 4 (3+√ 20.18 1.21 ) = 1.77 P = Internal design pressure = 362.6 Psi.
PEXETERNAL = Atmospheric Pressure = 1.013 bar = 14.7
Psi
r = Knuckle radius =6 % of L = 1.21 in.
S = Allowable stress for the material = 20015.21 Psi.
ts = Thickness of shell = 0.4 in.
t = Minimum thickness required for the head.
B. Heads under internal pressure
For torispherical Heads with tS / L ≥ 0.002:
Minimum thickness required,
t = 0.885PL SE−0.1P =
0.885× 362.6× 20.18
20015.21 × 1−0.1× 362.6 = 0.32 in.
Minimum required thickness of head can also be find using following method:
t = PLM 2SE−0.2P =
362.6 × 20.18 × 1.77
2× 20015.21× 1 −0.2× 362.6 = 0.32 in. Considering availability and corrosion allowances, Assume, tHEAD = tS = 0.4 in.
C. Heads under external pressure
Le = h/3 = 200/3 = 66.67 mm = 2.62 in.
Now,
Le / D0 = 0.13 & D0 / t = 50.45
From ASME Section- II, Part-D, Subpart-3, we get A = 0.04
B = 120 MPa = 17404.52 Psi PALLOWABLE =
4B
3 (D0/t)= 459.98 Psi
Thus PALLOWABLE (369.67 Psi) > PEXTERNAL (14.7 Psi)
Since the maximum allowable external pressure for head is greater than the actual external pressure acting on the shell i.e., atmospheric pressure, therefore it can be stated that design of the head is safe.
VI. DESIGN OF NOZZLE
A. Notations
A = Factor obtained from ASME BPVC Section-II Part-D
B = Maximum allowable compressive stress obtained from ASME BPVC Section-II Part-D
D0 = Outer Diameter of nozzle = 168.3 mm = 6.625 in. E = Joint efficiency = 1.0
L = Length of nozzle = 64.77 mm = 2.56 in.
P = Internal design pressure = 362.6 Psi
PEXTERNAL = Atmospheric pressure = 1.013 bar = 14.7
Psi.
S = Maximum allowable stress = 118 MPa = 17114.47 Psi
R = inner radius of nozzle = 𝐷0
2 - 0.28 = 3.033 in. R0 = outer radius of nozzle = 3.3125 in.
t= Minimum required thickness of nozzle.
B. Nozzles under internal pressure
Minimum required thickness for the nozzle is given as below, t =
P𝑅0 SE+0.4P =
362.6 × 3.3125
17114.47 × 1+0.4 × 362.6 = 0.070 =1.8 mm From standard size tables of ASTM,
Selecting, t= 0.28 in.
Checking for various stress induced in the nozzle for the specified pressure and considering selected thickness:
1) Considering circumferential stress:
P = SEt R+0.6t=
17114.47 × 1 × 0.28
3.033+0.6 × 0.28 =1497.05 > 362.6 Psi
2) Considering longitudinal stress:
P = 2SEt R−O.4t=
2× 17114.47 × 1 × 0.28
3.033−0.4 × 0.28 = 3281.10 > 362.6 Psi The maximum working pressure of cylindrical shells shall is less than above two allowable pressures.
Therefore, design is safe.
C. Nozzle under external pressure
Assume, t = 0.28 in.
From ASME Section- II, Part-D, Subpart-3, A = 0.025
B = 130 MPa = 18854.9 Psi
Maximum allowable pressure in the nozzle considering external pressure,
PALLOWABLE = 4B
3(D0/t) = 1062.52 psi
Thus PALLOWABLE (1062.52 Psi) > PEXETERNAL (14.7
Psi)
Since the maximum allowable pressure on the nozzle for thickness, t = 0.28 in. is greater than the actual external pressure acting on the shell i.e., atmospheric pressure, therefore it can be stated that design of the nozzle is safe for the specified conditions.
VII. DESIGN OF FLANGE AND GASKET Flange Dimensional Specifications
NPS Outer Diameter (in.) Dia. of Bolt circle (in.) Dia of Bolt Holes (in.) Number of Bolts Dia. of Bores (in.) 6 12.6 10.62 7⁄ 8 12 3⁄ 4
Gasket Dimensional Specifications
NPS Internal Dia. (in.) Outer Diameter (in.) Thickness (in.) 6 6.61 9.88 0.122
Stud Bolt Dimensional Specifications
Nominal Size (in.) Diameter (in.)
3 4
⁄ O.75
Table 3: Flange dimensions
Fig. 2: Flange dimensions (All dimensions are in inches).
A. Notations
Ab = cross sectional area of the bolts using the root diameter of the thread = (π
4 × d2
2 ) × 12 = 3.62 in2. a = nominal bolt diameter = 0.75 in.
a1 = Zero (except where the small female face is on the
end pipe, a = 0.19 in.)
BSC = bolt spacing factor Bs = bolt spacing. = 2.78 in.
BS max = maximum bolt spacing
b0 = N 2= 1.62 2 = 0.82 in. b = 0.5√b0= 0.5√0.82 = 0.45 in. B = Inside diameter of flange = 6.065 in.
C = bolt circle diameter = 10.62 in. d = factor = U
V× h0× g0 2 = U
0.550103× 1.36 × 0.28
2 = 0.679
d1 = heavy nut thickness (equals nominal bolt diameter
see ASME 18.2.2) = 0.75 in
d2 = Bolt root diameter = 0.38 in.
E = Modulus of elasticity for flange = 28.8 × 106 Psi. e = factor for integral type = 𝐹
ℎ0 = 1
1.36 = 0.67 F = Factor for integral type flanges = 0.908920
𝑓 = Hub stresses correction factor = 1
F1 = total height of facing or depth of ring joint groove
for both flanges = 0.12 in.
G1 = gasket thickness for raised face = 0.12 in G = OD of gasket contact face less of 2b = 8.97 in.
g0 = thickness of hub at small end = 0.28 in. g1 = Thickness of hub at back of flange = 0.28 in. h0 = factor = √𝐵𝑔0 = 1.36
hD = radial distance from the bolt circle, to the circle on
which HD acts = R+0.5g1 =1.72+0.5(0.28) = 1.85in. hG = radial distance from gasket load reaction to the bolt
circle = C − G 2 =
10.62−8.97
2 = 0.825 in.
hT = radial distance from the bolt circle to the circle on
which HT acts = R +g1+hG
2 =
1.72+0.28+0.825
2 = 1.412in. HD = hydrostatic end force on area inside of flange
considering internal pressure = 0.785B2P = 12493.07 lb. HG = gasket load = W – H where, W = Wm1 = 18383.23
lb.
H = total hydrostatic end force considering internal pressure = 0.785G2P = 22902.5 lb.
HT = H – HD = 10409.43 lb.
Hp = Total joint-contact surface compression load =
2b×3.14GmP = 2×0.45×3.145×8.97×2×362.6 = 18383.25 lb.
KI = Rigidity factor for integral flange = 0.3
L = factor = te+1 T + t3 d = 5.114 m = Gasket factor = 2 MD = Component of moment = HD ×hD =23112.18 lb-in.
MT = Component of moment due to HT = HT ×hT =
14573.202 lb-in.
MG = Component of moment due to HG = HG ×hG =
15166.16 lb-in.
M0 = Total moment acting upon the flange for the
operating condition or gasket seating
n = Negative tolerance on bolt length = 0.25 in N = (𝑂𝑢𝑡𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑔𝑎𝑠𝑘𝑒𝑡 −𝐼𝑛𝑛𝑒𝑟 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑔𝑎𝑠𝑘𝑒𝑡)
2 =
( 251−168
2 ) = 1.62 in.
P = Internal design pressure = 25 bar = 362.6 Psi
Pe = External design pressure = 14.7 Psi
Sa = Allowable design stress for stud = 14500 Psi. Sf= Allowable design stress for flange = 20000 Psi. Sn= Allowable design stress for nozzle = 17114.45 Psi. t = flange thickness = 1.38 in.
T = factor involving K = 𝐾
2(1+8.55246 log 10𝐾)−1
(1.04720+1.9448𝐾2)(𝐾−1) = 1.602 t1 = bolt length = 0.12 in.
R = radial distance from bolt circle to point of intersection of hub and back of flange = C − 𝐁
2 − 𝑔1 = 1.71 in.
U = 𝐾
2(1+8.55246 log 10𝐾)−1
1.36136(𝐾2−1)(𝐾−1) = 3.507 V = Factor for integral type flanges = 0.550103
W= flange design bolt load, for the operating conditions or gasket seating
y = Minimum design seating stress = 1200 Psi.
Y = Factor involving K (where k = 𝐴
𝐵 = 1.9) = 1 𝐾−1 [0.66845 + 5.71690 × 𝐾2𝑙𝑜𝑔10 𝑘 𝑘2−1 ] = 3.19 Z = Factor = 𝐾 2+1 𝐾2−1 = 1.92+1 1.92−1 = 1.766
Fig. 3: Integral type flange
1) Flange and gasket material selection
Flange material: A516 GRADE 70 (PLATE) Allowable stress Sf = 20000 Psi
From ASME B16.5, according to pressure temperature rating for 150O C and pressure range 15.8 to 45.1
bar, flange type selected is class 300. Gasket Material: PTFE (TEFLON).
2) Bolt loads
The minimum required bolt load for the operating conditions: Wm1= H + Hp =0.785G2P + (2b×3.14GmP = 41285.73 lb
The minimum required bolt load for gasket seating, Wm2 = 3.14bGy= 3.14×0.45×8.97×1200 = 15209.653 lb
Am = total required cross sectional area of bolts, taken as the greater of Am1 and Am2 = 2.85 in2.
Where, Am1 = area required from operating
conditions considerations =Wm1
𝑆𝑎 = 2.85 in 2
Am2 = area required from gasket seating conditions
considerations = Wm2
𝑆𝑎 = 1.04 in 2
a) Bolt load for flange Design
The bolt loads used in the design of the flange shall be the values obtained from,
For operating conditions: W = W1
For gasket seating conditions: W= (Am+ Ab)Sa
2 =
(2.85 + 3.62)14500
2 = 46907.5 lb
The maximum bolt spacing shall not exceed the value calculated as below:
BS max = 2a + 6t m+ 0.5 = 2 × 0.75 + 6 × 1.38 2+0.5 = 4.812 in. 3) Flange moments
a) For operating conditions,
M0 = MD + MT + MG = 23112.18 + 14573.202 + 15166.16 =
52851.54 lb-in. For gasket seating conditions, M0 = W×
C − G
2 = 46907.5 × 0.825 =38698.69 lb-in.
∴ M0 = 46907.5 × 0.825 = 38698.69 lb-in.
b) For bolt spacing,
When the bolt spacing exceeds 2a + t, multiply M0 by the bolt
spacing correction factor BSC for calculating flange stress.
Here, 2a + t = 2 × 0.75 +1.38 = 2.88 in.
Therefore bolt spacing exceeds 2a+t, hence for operating conditions M0 = M0 × BSC ∴ BSC = √ BS 2a+t = √ 2.78 (2 × 0.75)+1.38 = 0.98
And for flange moment for operating conditions is given as, M0 = 52851.54 × 0.98 = 51794.50 in-lb.
4) Calculation of flange stresses
a) For operating conditions:
Longitudinal hub stress: SH =
𝑓 M0 Lg12B=
1 × 51794.5
5.114 × 0.28 × 0.28 × 6.625 =19499.38 Psi. Radial flange stress:
SR=
(1.33te + 1)M0 Lt2B =
(1.33 × 1.38 × 0.67+1) × 51794.50
5.114 × 1.38 × 1.38 × 6.625 =1789.10 Psi Tangential flange stress:
ST = YM0
t2B − ZSR =9927.62 Psi. b) For gasket seating conditions:
Longitudinal hub stress: SH =
𝑓 M0 Lg12B =
1 × 38698.69
5.114 × 0.28 × 0.28× 6.625 =14569.12 Psi. Radial flange stress:
SR =
(1.33te+1)M0 Lt2B =
(1.33 × 1.38 × 0.67+1) × 38698.69
5.114 × 1.38 × 1.38 × 6.625 =1337.34 Psi Tangential flange stress:
ST = YM0
t2B − ZSR = 8445.48 Psi.
5) Checking for allowable flange design stresses
The flange stresses calculated by the above equations, shall not exceed the following values:
Longitudinal stress SH not greater than the smaller
of 1.5Sf or 2.5Sn
1.5Sf = 30000 Psi or 2.5Sn = 42786.12 Psi 19499.38 Psi < 30000 Psi
Radial flange stress SR not greater than Sf ; 1789.90 Psi < 20000 Psi
Tangential flange stress ST not greater than Sf 9927.62 Psi < 20000 Psi
(SH + SR)/2 not greater than Sf 10644.64 Psi < 20000 Psi
(SH + ST)/2 not greater than Sf. 14713.5 Psi < 20000 Psi
Hence, the various stresses induced in flanges are less than the required permissible value.
Therefore, it can be stated that the design of flange is safe for the specified operating and design conditions.
6) Flanges subject to external (atmospheric) pressures
For external atmospheric pressure acting on flange hydrostatic loads are given as below:
HD = 0.785B2Pe = 0.785 ×6.625×6.625×14.7 = 506.47 lb. HT = H - HD = 928.47 – 506.47 = 422.00 lb.
H = 0.785G2P
e =0.785×8.97×8.97×14.7 = 928.47 lb.
The design of flanges for external pressure only are based on the equations given below:
a) For operating conditions,
M0 = HD (hD - hG) + HT ( hT - hG ) =761.78 lb-in.
b) For gasket seating,
M0 = W hG =27872.63 lb-in.
Where,
7) Flange Rigidity
Here, the flange moment M0 without correction for bolt
spacing is used for the calculation of the rigidity index. Flange rigidity is given by following formula,
J= 52.14VM0 LEg02KIh0 =
52.14 × 0.55 × 52851.54
5.114 × 28.8 × 106 × 0.28 × 0.28 × 0.3 × 1.36= 0.32
J < 1.00, therefore theoretically design is safe from leakages.
8) Stud Bolt Design
Standard parameters for stud & Hex Nut:
M20 Coarse series; pitch = 2.5 (from ASME B18.31.2 2008, ASME B18.31.4M 2009)
From ASME B16.5, 2013
Stud bolt length exclusive of negative length tolerance is given as following,
A = 2 (t+ t1 + d) + G1 + F1 – a1 = 4.74 in.
Length of stud bolt:
LCSB = calculated stud bolt length = A + n = 4.99 in.
VIII. SUPPORTS DESIGN – LEGS
Leg support selected: ISA 75 × 75
A = t (2a-t) =725 𝑚𝑚2 I = 398277.67 𝑚𝑚4 r = 23.40 mm
y = 54.40 mm
Fig. 4: Legs support design.
A. Various parameters to be considered for the design of supporting legs are calculated as follows
1) Deflection in legs: y = 2W𝐼 3 3nE∑𝐼2 = 2×661.5×0.9573 3×4×30×306×0.96= 2.13×10-4 in. 2) Period of vibrations T = 2π√yg = 2π√2.13 × 10−4 386 = 4.68×10 -4 sec.
3) Lateral force applied at top of structure
Ft=0.07TV or 0.25V =0 if T < 0.7 sec
Here, since T < 0.7 sec, Ft = 0 lb.
4) Horizontal force on vessel
Fh =Ah W =0.16×661.5 = 105.84 lb.
5) Vertical force on vessel:
Fv = (I+Cv) W = (0.957 + 0.5) 661.5 = 1124.55 lb.
6) Overturning moment at base:
M = L×F = 36.41 × 105.84 =3854 lb.
7) Overturning moments at tangent:
Mt = (L-l) × Fh + (H-1) Fr =5.71×105.84+0 = 604.35 lb.
8) Maximum eccentric load:
F1 = −𝐹𝑉 n − 4Mt nD = −1124.55 4 − 4×604.35 4×20.18 = −311.09 lb Negative sign indicates downward direction.
9) Horizontal load distribution:
Vn = VI ∑𝐼1 = 105.84 ×0.957 3.82 = 26.51 lb. ∑Vn = V =106.04 lb.
10) Bending moment in leg, M
M = F1× e ± Vn × l = -278.0× 76.93 25.4+26.51× 780 25.4 =-128.124 lb-in.
11) Vertical load distribution
FD = 𝐹𝑉 n = 1124.55 4 = 281.14 lb Fn = FD ± FLcos ∅ n a) Seismic force, FL = 4M nd = 4×128.124 4×26.22 = 4.88 lb. b) Wind force, FL = M 2𝑑1= 128.124 2×26.22= 2.44 lb. Fn1 = 281.14 + 4.88 × cos 0 = 283.50 lb Fn2 = 281.14 + 4.88× cos 90 = 281.14 lb Fn3 = 281.14 + 4.88 × cos 180 = 278.70 lb Fn4 = 281.14 + 4.88 × cos 270 = 281.14 lb
12) Axial stress in leg
fa = 𝐹𝑛
A = 283.50
1.124 = 252.22 Psi
13) Bending stress in leg
fb = M𝐶1 I = 128.124×2.087 0.957 = 279.40 Psi 14) Combined stress If 𝑓𝑎 𝐹𝑎 ≤ 0.15 then 𝑓𝑎 𝐹𝑎+ 𝑓𝑏 𝐹𝑏< 1 Here, 𝑓𝑎 𝐹𝑎= 0.012 and hence 252.22 20015.21+ 279.40 12009.13= 0.036 < 1 Thus the required condition is satisfied therefore it can be stated that design of supporting legs is safe.
15) Maximum compressive stress:
fc = −𝐹𝑉 3.1416 × 𝐷 × 𝑡− 4 𝑀𝑡 3.1416 × 𝐷2 × t = −1124.55 3.1416 × 20.18 × 0.2− 4 604.35 3.1416 × 20.182 × 0.2 = -98.13 Psi
Fc = Allowable compressive stress = factor B of ASME BPVC SECTION-VIII Division-1, external pressure considerations = 110 MPa = 15954 Psi
Since, fc is less than maximum allowable compressive stress; it can be stated that design is safe.
16) Local loads in shell:
Mx = Vn sin 𝜃 𝑙
Mx1 = 26.51× sin 0 × 30.71 = 0;
Mx2 = 26.51× sin 90 × 30.71 = 814.12 lb;
Mx3 = 26.51 sin 180 × 30.71 = 0;
Mx4 = 26.51× sin 270 × 30.71 = -814.12 lb
From above calculations it can be observed that all the stress are less than allowable stress i.e 20015.21 Psi; hence design is safe.
IX. DESIGN OF BASE PLATE
From Dennis Moss, Pressure vessel design manual, Third edition, 2004 following dimensions are selected for base plate: For 3in. × 3in. Legs
X = 1.75in.
M = 1.5 in.
Min plate thickness, t = 0.5 in.
Fig. 5: Base plate design X. RESULT
Sr.no. Parameters Dimensions
1. Thickness of shell 0.4 in
2. Vessel OD 20.18 in.
3. Thickness of head 0.4 in.
4. Knuckle radius 1.19 in.
5. Nozzle wall thickness 0.28 in.
6. Flange OD 12.6 in.
7. Gasket ID 6.61 in.
8. Gasket OD 9.88 in.
9. Flange thickness 1.38 in.
10. Flange rigidity 0.32
11. Leg support size 2.95in. X 2.95in. 12. Deflection in leg support 2.13 x 10-4 in.
13. Base plate size 6 in. × 6 in.× 0.5 in.
14. Bolt size ¾ in.
15. Stud length for flange 4.92 in. Table 4: Result
XI. CONCLUSION
Various dimensional parameters of pressure vessel are designed and checked according to the principles specified in American Society of Mechanical Engineers (A.S.M.E) Sec VIII Division-1, 2015. Supporting legs and base plate is designed successfully using Pressure vessel design manual by Dennis Moss, third edition. The designed pressure vessel is also then checked for various induced stresses and moments and it is concluded that various stresses and moments are below the permissible & allowable stress values. Hence the design of vessel is safe for the designed conditions.
REFERENCES
[1] ASME BPVC Section-VII Division-1, 2015. [2] ASME BPVC Section-II Part-2 properties, 2015 [3] ASTM standards, 2015.
[4] ASME B16.21, 2015. [5] ASME B16.5, 2015. [6] ASME B18-16M, 2004. [7] ASME B18-31-4M, 2009
[8] Dennis Moss, Pressure vessel design manual, Third edition, 2004.
[9] DUPONT Teflon PTFE, Properties handbook, 7/96 [10] Indian standard, IS1893-1,2002
