Section 2 Design

of Process Vessels

Chapter
Chapter 22
Panorama
Panorama
 2.1
2.1 Structure
Structure and
and
Classification
Classification of
of Vessels
Vessels
1.Conception of Vessels:
Process Vessels are the various
equipments in the chemical
process.
2.Structure of vessels:
 3.Classification of vessels:

i. According to pressure and its type

(1)Internal Pressure Vessel
—— vessels where the media pressure
inside the vessel is larger than that outside
(gauge pressure).
Low pressure vessel (L):
0.1≤P < 1.6 MPa
Medium pressure vessel (M):
1.6 ≤P < 10 MPa
High pressure vessel (H):
10 ≤ P < 100 MPa
Ultra-high pressure vessel (U):
P ≥100 MPa
(2)External Pressure Vessel
——the media pressure inside the vessel
is lower than that outside
(gauge pressure). When the pressure
inside the vessel is less than 0.1 MPa (absolute
pressure), such vessels are called Vacuum
Vessel.
ii. According to temperature
Normal temperature vessel
-20< T ≤200℃
Medium temperature vessel
—— between normal T & high T vessels
Low temperature vessel
< -20 ℃
High temperature vessel
—— where the wall temperature is
above the creep temperature.
 High temperature vessel

Carbon steel & Low-alloy steel T> 420 ℃

Alloy steel (Cr-Mo steel) T> 450 ℃
Austenite stainless steel (Cr-Ni) T> 550 ℃
 iii. According to management
P, P*V Grade (I)
Factor Grade (II)
Media toxic and
combustible Grade (III)
importance
Degree of danger:
I < II < III
 2.2
Basic code and
Common Standard for
Design of Pressure Vessels
1.Conditions:
i. The maximum working pressure
Pw≥ 0.1 MPa
(neglecting the net pressure of liquid )
ii. Internal diameter Di (equal the maximum
dimension or size in the non-circular
sections) ≥ 0.15m, and V ≥ 0.025 m3
iii. With the medium as gas or liquefied gas,
or liquid whose maximum working
T ≥ standard b.p. (boiling point )
2.Basic Code and Common Standard:
i. 《 Security Technique Supervising Rules for
Pressure Vessel 》 1999
ii. GB 150 — 1998
《 Steel Pressure Vessel 》
iii. GB 151 — 1999
《 Tubular Heat Exchanger 》
iv. JB 4710 — 2000
《 Steel Tower Vessel 》
 2.3 Standardization of
Pressure Vessel Parts
1.Significance of Standardization:
i. It consolidates and harmonizes the various
activities in the manufacture and social life.
ii. It’s the important means to organize the
modernization production.
 iii. It’s the important component in scientific
management.
iv. Development of new products,
assures the interchangeability and common-
usability
Convenient to use and maintain.
v. It’s helpful in the interchange of international
science & technology, culture and economy.
2. Chemical Vessels
and Standards of
Equipments Components:

Cylinder Heads
Vessel Flange
Pipe Connecting (Nozzle) Flange
Support Stiffening Ring
Manhole Handhole
Sight (level) Glass Liquid Leveler (LG)

Expansion joint
Heat Exchanger Tube Tray
Float (floating) Valve (FV)
Bubble (bubbling) cap Packing etc.
3.Basic parameters of Standardization
i. Nominal Diameter —— DN (Dg)
is a typical dimension.
(1)Rolled cylinder and head
DN = Di (inside diameter)
DN Standard of pressure vessels:
300 350 400 450 500 …… 6000

48 grades in total.
(2)Seamless pipes
—— DN≠Di & DN ≠ Do,
but DN is a certain value that is smaller than
Do. When the DN is a certainty, Do is to be a
certainty, while Di depends on the thickness.
Denotation of seamless pipes:
such as 252.5 (outside × thickness)
 Check the standards according to DN.
Comparison of DN and Do
of seamless pipes/mm

DN 10 15 20 25 32 40 50 65

Do 14 18 25 32 38 45 57 76

δ 3 3 3 3.5 3.5 3.5 3.5 4

(3)Cylinder made by seamless pipes
—— DN = Do (outside diameter)
Six grades: 159 219 273 325 377 426
(4)DN of flanges
—— consistent to their suitable cylinders,
heads and tubes.
i.e. DN of flange = DN of cylinder
DN of head
DN of tube
ii. Nominal Pressure —— PN (Pg)
Prescriptive standard pressure grades
For example ——
PN (MPa) of pressure vessel flange:
0.25 、 0.6 、 1.0 、 1.6 、 2.5 、 4.0 、
6.4
iii. Application
In standard designing:
(1)Diameters of cylinders, heads and tubes
must be close to the standard grades.
e.p. Diameter of cylinder
should be 500 、 600 、 700 …
shouldn’t be 520 、 645 、 750 …
(2)When selecting the standard vessel parts,
the design pressure at the working
temperature should be regulated to a
certain grade of PN.
Then choose the parts according to PN
and DN.
4.Classification of standards:
i. Chinese Standard
Symbol: GB (Guo biao)
ii. Standard issued by Ministry
JB —— Ministry of Mechanical Industry
YB —— Ministry of Metallurgical Industry
HB —— Ministry of Chemical Industry
SY —— Ministry of Petroleum Industry
iii. Specialty Standard

iv. Trade Standard

v. Plant Standard
 2.4 Basic Requirements
& Contents of Vessel Design
1.Basic requirements:
i. Enough strength —— no breakage
ii. Enough rigidity ,Enough stability ——
limit deformation
iv. Durability —— assuring certain usage life
v. Tightness —— no leakage
vi. Saving materials and easy to manufacture
vii. Convenient to be installed, transported,
operated and maintained
viii. Rational technical economy
2.Basic Contents:
i. Selection of materials
Selecting the materials of
equipment according to the technical
indexes t, p, media and the principles of
material selection.
ii. Structure design
iii. Calculation of strength and thickness
(including the cylinders and heads)

iv. Strength verification in hydraulic

pressure test

v. Seal design; selecting or designing flanges

vi. Selection of support & the verification of
strength and stability
vii. Design and calculation of reinforcement
for opening
viii. Selection of other parts and accessories
ix. Other special design
x. Plotting the equipment drawings
xi. Compiling the equipment specifications
 Chapter 3 Stress Analysis of
Thin-walled Internal-P Vessel

3.1 Stress Analysis

of thin-walled Cylinders
Subjected to Internal Pressure
1.Thin-walled vessels
(1)Thin-walled vessels:
S / Di < 0.1 (Do / Di = K < 1.2)
(2)Thick-walled vessels:
S / Di ≥ 0.1
2.stress characteristics:
There are always two kinds of
stress in pressure vessels.
i. membrane stress
—— membrane (shell)
theory
P
ii. boundary stress
—— shell theory with
moments and conditions
of deformational
compatibility
 3.2 Membrane Theory ——
Rotary Shells’ Stress Analysis
1.Basic conceptions and hypothesis:
i. Basic conceptions
(1)rotary

curved-
surface
& shell

Cylinder
Cylinder ,, conical
conical shell,
shell, spherical
spherical shell
shell
(2)Axial Symmetry
Geometry figure, endured load
and restrictions of shell are all symmetry
to the revolving axis (OA).
Several basic conceptions
generatrix (generator), meridian,
normal, parallel circle (latitude), 母线，经
线，法线，纬线
longitudinal radius, tangential radius.
第一曲率半径，第二曲率半径
 O
(3)Generator (AB)
The plane curve B
which forms the curved
surface. B’
(4)Longitude (AB’)
Section passing OA
and intersecting with
shell, the cross-line is
AB’.
A
(5)Normal (n)
The line passing O
point M in meridian and
is vertical with midwall
surface.
The extension of ·
K2 ·
K1
normal must intersect with
K2’
·M
OA. ·
C D n
(6)Latitude (CND) N
The cross-line formed by the
conical surface passing point
A
K2’ intersects with the rotary
curved surface.
(7)Longitudinal radius (R1)
The radius of curvature of meridian
which passes point M in midwall surface
O
is called the longitudinal radius of point
M in meridian.The center K1
of curvature of the round
with diameter R1 must be ·
K1 K ·
in the extension of normal 2
·M
K2 ’
passing point M. ·
C D
For example: N
Longitudinal radius of point M:
R1 = M K1 A
(8)Tangential radius (R2)
*The plane which is vertical to the normal
passing the point M in meridian
intersects with the mid-wall surface, the
resulted cross line (EMF) is a curve, the
radius of curvature of this curve in point
M is called tangential radius.
*The center K2 of curvature of the
round with diameter R2 must be in the
extension of normal passing point M
and in the revolving shaft.
For example:
Tangential radius of point M:
R2 = M K 2
 exercise AA

Calculations BB
Calculations ofof
R1,R2
R1,R2 at
at different
different dd
points
points

CC

D
D
 Calculation of R1 and R2

1.Cylindrate shell
R1 = ∞ R2 = D / 2

δ
p
D
Spherical shell subjected to
uniform gas internal
pressure:
R 1 = R2 = D / 2 δ
D
3.Elliptical shell Known: Major semiaxis – a
Short semiaxis - b
(1)Find R1 and R2 of point A:
R1:
y
radius of curvature of A
 1  y   .
3 A(x,y)
' 2 2

b
  
y ''
.k a x

x 2
y 2 1
.k
 2 1
2

2
a b
1

R1  4 a 4  x 2 a 2  b 2
a b
  3
2
R2: R2 = K2 A = x / sin (b)
here: tg
sin  
1  tg 2 y
2
b x
tg  y   2
a y
'
.
A(x,y)

b

y b  2 x
2 b 2
2
2
.k a x
a
1
.k 2

putting them into (b), getting:

a4  x2  a2  b2 
1
R2 
b
 1
ab

4 2 2

R1  4 a  x a  b 2
 3
2

R2 
1 4
b
a  x a b
2 2
2

Special
Special Points:
Points:
x=0:
x=0: R11=R22=a /b
R =R =a 22
/b
x=a
x=a : R11=b /a
: R =b 22
/a ,, R
R22=a
=a
 Conical shell
Find R1 and R2 of point A:
R1 = ∞ D

R2 = A K2 = r / cos
k2
Small end: .A
r

R1 = ∞
R2 = 0
Dished shell rr11

For spherical
segment:
rr
R1=R2=R
For knuckle segment
of transition section: R1= r1, R2=
r  r1
 r1
sin 
For cylindrical shell: R1=∞, R2= r
ii. Basic hypothesis:

Small displacement hypothesis

Straight linear law hypothesis
Non-extrusion hypothesis
2.Free body balance equation( 区域平衡
方程式 )
—— calculation formula of
meridional stress （经向应力） :
i. Intercepting shell
—— uncovering the longitudinal
stress m
ii. Choosing separation body
iii. Analysis of stress
iv. Constitute balance equation

 Fz  0 Pz  N z  0 m
.
k2
R2
 2 C 
D p   m DS sin   0 C’
4
D/2 p
sin   putting it
R2
into the above equation, getting
z
pR2 D
m  (3 - 3)
2S
3.Infinitesimal balance equation
—— calculation formula of hoop
stress （微元体平衡方程式）
i. Intercepting shell
—— uncovering the meridional stress m
and circumferential stress σθ
ii. Choosing separation body
iii. Analysis of stress
iv. Constitute balance equation
k1 d1
k2
d2 Sdl1
 Fn  0 Pn  N m.n  N .n  0
p
d1
pdl1dl2  2 m Sdl2 sin 
2
d 2
 2  Sdl1 sin 0
2
 m  p
整理得：   (3 - 4)
R1 R2 S mSdl2
Basic calculation equation of membrane stress:

pR2  m  p
m   
2S R1 R2 S

Illustration of symbols:
m —— meridional stress of a random
point in rotary thin shell, MPa
 —— circumferential (hoop) stress of a
random point in rotary thin shell, MPa
P —— internal pressure, MPa
S —— thickness of wall, mm
R1 —— longitudinal radius of required stress
point in the mid-wall surface of the
rotary shell, mm
R2 —— tangential radius of required stress
point in the mid-wall surface of the
rotary shell, mm
v. Application range of membrane
theory
☆Applicable to axial symmetric thin-walled
shell without bending stress
☆No bending stress —— only normal stress
(tensile stress & compression stress)
☆Thin-walled shell
—— S / Di < 0.1 ( Do / Di = K < 1.2 )
☆Axial
symmetry and continuous
—— Geometry, loads, physical properties
☆Free supporting boundary
 3.3 Application
of Membrane Theory
Calculation equations:
pR2  m  p
m   
2S R1 R2 S
1.Cylindrical shell subjected to
uniform gas internal pressure:
∵ R1 = ∞ R2 = D / 2

s
Putting them into the
previous equations: p

D
pD pD
m   
4S 2S
2.Spherical shell subjected to
uniform gas internal pressure:
∵ R1 = R 2 = D / 2
Putting them into S
equations (3-3) and (3-4): D

pD pD
m   
4S 4S
3.Elliptical shell subjected to
uniform gas internal pressure:

Example:
Known:
Major semiaxis - a Short semiaxis - b
Thickness - S Internal Pressure - P
Find the m and  of a random point on the
elliptical shell.
 y
Solution:
(1)Find R1 and R2 of point A: .
A(x,y)

b

R1:
radius of curvature of A .k a x

 1   y' 
2

3
2
1
.k 2
 
y ''
(a)
x2 y2
By the elliptical equation: 2  2  1
a b
getting y’ and y’’, then put them into
(a).
R
result is: 1 
1
4
a b
a 4
 
x 2
a 2
 b 2
  3
2
R2: R2 = K2 A = x / sin (b)
here: tg
sin  
1  tg 2
2
b x
tg  y   2
'

a y
2
b 2
y b  2 x
2 2

a
putting them into (b), getting:
a  x a  b 
1
R2  4 2 2 2

b
(2)Find m and  of point A:
Putting R1 and R2 into (3-3) and (3-4), getting:

m 
p
2Sb

a  x a b
4 2 2 2

 
 
4
p a
  a  x a b
4 2 2 2
2  4 2 2 2 
2Sb 
 a  x a b  
Stress of special points on elliptical shell:
(1)x=0 (Top of elliptical shell)

pa  a 
m     
2S  b 
(2)x=a (Boundary or equator of elliptical shell)

pa pa  a2

m     2  2 
2S 2S  b 
Standard elliptical heads:
 The elliptical heads whose ratio of major and
short semiaxis a / b = 2 are called standard
elliptical heads.
 a / b = 2 ——
pa pD
x=0 (Top):  m    
S 2S
x=a (Boundary):
m   pa pD
  
 m
2S 4S

   pa   pD



S 2S
a/b=2
4.Conical shell subjected to
uniform gas internal pressure:
Example:
Known:
Diameter of tapered bottom - D
Half tapered angle - 
Thickness - S
Internal Pressure - P
Find the m and  of a random point on the
conical shell.
Solution: D
(1)Find R1 and R2 of point A:
k2
R1 = ∞ r
.A

R2 = A K2 = r / cos
(2)Find m and  of point A:
Putting R1 and R2 into (3-3) and (3-4)
respectively, getting:
pr 1 pr 1
m     
2 S cos S cos
Characteristics of stress distribution
of conical shell:
m 
5.Cylindrical shell subjected to
liquid static pressure:
i. Supporting along the boundary of bottom
Example: P o

Known:
.A

x
Gauge pressure – Po (Pa)
Liquid level – H (m)

H
D
Density of liquid -  (N/m3)
Find the m and  of a random
point on the wall of cylindrical shell.
Solution:
(1)Meridional stress:
Cutting along section B-B, taking the
lower part as the separation body.
Po
(Po+x)
m m
B
.A B
x

(H-x)
H

B-B
Establishing the balance equation of axial stress:
m (Po+x) m

(H-x)

 Fx  0
N
 2  2  2
( po    x) D   ( H  x) D   m D S    H D
4 4 4

po D
 m 
4S
(2) Circumferential (Hoop) stress:
Infinitesimal balance equation (3-4):
m  P
 
R1 R2 S
For point A:
R1 = ∞ R2 = D/2 P = Po +  x
Putting them into (3-4),
getting: when x=H:

 
 po   x  D
  max 
 po   H  D
2S 2S
ii. Supporting along the boundary of top
Example:
Known: P o

Gauge pressure – Po (Pa) .A

x
Liquid level – H (m)

H
D
Density of liquid -  (N/m )
3

Find the m and  of a random

point on the wall of cylindrical shell.
Solution:
(1)Meridional stress:
Cutting along section B-B, taking the
lower part as the separation body.
Po

B B
(Po+ x)
.A m m
x
H

(H-x)
D

H-x
B-B
 Establishing the balance equation of axial stress:

m (Po+ x) 
m

(H-x)

H-x
 Fx  0

 2  2
  H  x  D   po   x  D   m D S  0
4 4

 m 
 po   H  D
4S
(2)Hoop stress:
Infinitesimal balance equation (3-4):
m  P
 
R1 R2 S
For point A:
R1 = ∞ R2 = D/2 P = Po +  x
Putting them into (3-4),
getting: when x=H:
 
 po   x  D
  max 
 po   H  D
2S 2S
6.Examples:
A cylindrical vessel is with a spherical
upper head and a semi-elliptical lower
head a / b = 2. The average diameter D is
420mm. Thickness of all cylindrical shell
and heads are 8mm. The working pressure
P is 4MPa.
Calculating:
(1)Find m and  of the shell body.
(2)Find the maximum stresses on the
both the heads and their positions
respectively.
Solution:
(1)For cylinder m and  :
pD 4  420
m    52.5 (MPa )
4S 48
P
pD
   2 m 105 (MPa)
2S D S
(2)Upper head —— spherical
pD
 m     52.5 (MPa)
4S
(3)Lower head —— elliptical
When a / b = 2:
a = D/2 = 210 mm b = a/2 = 105 mm
x=0 (Top):
pa  a  pa pD
 m        105 (MPa)
2S  b  S 2S
pa pD
x=a (Bottom):  m    52.5 (MPa)
2S 4S

pa  a 
2
pa pD
   2  2       105 (MPa)
2S  b  S 2S
 3.4 Conception
of Boundary Stress
1.Forming of boundary stress:
Boundary
—— The joint and its vicinity of
two parts with different geometry
shape, load, material and physical
conditions, i.e. discontinuous point.
 Boundary stress forming not for
balancing the loads but for receiving
restrictions from self or exterior. It’s a
group of internal force with same value
but contrary direction occurring
between two parts which are forced to
realize transfiguration harmonization.
2.Characteristics of boundary stress:
i. Distributing along the wall non-evenly
ii. Different joint boundary forming different
boundary stress
iii. It’s local stress, i.e. only forming large stress
locally and decaying apparently
iv. Value of boundary stress can be 3~5 times
of that of membrane stress
v. Self-constrained
3.Treatments to boundary stress:
i. Treatments locally in structure
(1)Improving the structure of joint boundary
(2)Strengthening the boundary locally
(3)Assuring the quality of welding line at
boundary
(4)Decreasing the remnant stress at local and
processing the heat treatment to eliminate
the stress
(5)Avoiding the local stress added to the
boundary region overlap with connatural
stress
ii. Materials are of high plasticity
 Chapter 4 Strength Design
of Cylinders and Heads
subjected to Internal-Pressure
4.1 Basic Knowledge
of Strength Design
1.Criterions of elasticity failure:
 eq   t
s
Safety Allowance kept for the
requirements of safety:
o
    
eq n
eq —— equivalent stress
o —— limiting (ultimate) stress, can be
s 、  b 、  n 、  D, etc.
[] —— allowance stress
n —— safety coefficient
2.Strength Theory:
i. The first strength theory
—— the maximum tensile stress theory

 eqI   1  [ ] Applying to brittle materials

ii. The second strength theory
—— The maximum major strain theory
iii. The third strength theory
—— The maximum shear stress theory
   1  3
max
2
Shear limit: s
Shear limit :   0

Failure condition:
 max   0

1 s
i.e. ( 1 -  3 )  or  1   3   s
2 2
Strength condition:
Applying to the
   1   3  [ ]
III
eq plastic materials
iv. The fourth strength theory
—— the maximum deformation energy theory
1
 IV
eq  [( 1   2 )2  ( 2   3 )2  ( 3   1 )2 ]
2

Strength condition:
Applying to the
 IV
eq  [ ]
plastic materials
4.2 Strength Calculation of
Thin-walled Cylinder Subject
to Internal Pressure
1.Strength calculating equation:
i. Determining the major stress
pD pD
1    2 m 
2S 4S

 3  r 0
ii. Determining the equivalent stress

 eq  f ( 1 ,  2 ,  3 )
According to the third strength
theory:
pD pD
 eq   1   3 
III
0
2S 2S
iii. Strength condition
pD
 III
eq   [ ]t (4 - 3)
2S
iv. Strength calculation equation
pD
S  (4 - 4)
2 [ ]t

Solving equation (4-4) as following:

(1)Replacing medium diameter with internal
diameter: D + Di = S
Calculating pc from
Putting them into equation (4-4), getting:
p.
pc Di
S
2[ ]  pc
t
(2)Introducing the welded joint efficiency :
pc Di
S (4 - 5)
2[ ]   pc
t

This is the calculated thickness.

(3)Eliciting the corrosion allowable thickness
pc Di C2:
Sd   C2 (4 - 6)
2 [ ]   p c
t

This is the design thickness.

(4)Adding negative deviation C1:
pc Di
Sn   C2  C1  round  of value (4 - 7)
2[ ]   pc
t

Getting the nominal thickness

which indicated on the drawing.
(5)Calculating the effective
thickness:
S e  S n  (C1  C2 )
pc Di
  round  of value (4 - 8)
2[ ]   pc
t
v. Equation of strength verification

pc ( Di  Se )
 
t
 [ ] 
t
(4 - 9)
2 Se

vi. Calculating equation of pw

2[ ]t Se
[ pw ]  (4 - 10)
Di  Se
2.Strength calculating equation of
thin-walled spherical vessels:
pc Di
S
4[ ]   pc
t

pc Di
Sd   C2
4[ ]   pc
t

pc Di
Sn   C2  C1  round  of value
4[ ]   pc
t
 *Equation of strength verification:
pc ( Di  Se )
 t
 [ ] 
t

4 Se
*Equation of [pw] —— the maximum
allowable working pressure:
4[ ]t Se
[ pw ] 
Di  Se
*Scope of application of previous equation:
cylinder: P≤0.4 []t  (Do / Di ≤1.5)
spherical shell: P≤0.6 []t  (Do / Di ≤1.35)
Illumination of symbols:
Pc —— Calculated pressure MPa
Di 、 Do —— Internal & external diameters of
cylinder mm
S —— Calculated thickness mm
Sd —— Design thickness mm
Sn —— Nominal thickness mm
Se —— Efficient thickness mm
C1 —— Negative deviation mm
C2 —— Corrosion allowable thickness mm
C —— Additional value of wall thickness mm
 —— Welded joint efficiency
[]t —— Allowable stress at design temperature
MPa
t —— Calculated stress at design temperature
MPa
[Pw] —— The maximum allowable pressure at
design T MPa
3.Determination of design
parameters:
i. Pressure P
(1)Working pressure Pw
—— the maximum pressure at the top of
vessel and under normal operating
condition
(2)Design pressure P
—— the maximum pressure at the
specified top of vessel
The design pressure P and the
corresponding design temperature T are
conditions of designing load, and its value
is not less than working pressure.
(3)Calculated pressure Pc
—— the pressure which is used to determine
the thickness at corresponding
design temperature
Including the liquid (column) static
pressure, when the liquid (column) static
pressure ＜ 5% design pressure, it can be
neglected.
Choosing the value of design pressure

Illustrating in the following chart

 Conditions Evaluation of Design P
With safety valves P≤(1.05~1.1)Pw
Single vessel P≥Pw
(no safety devices)
With explosive media P≤(1.15~1.3)Pw
and rupture disk
With liquefied gas Determined by the charging
proportion and Tmax
External Pressure Vessel Under normal working
condition, P≥△P=P2-P1
Vacuum Pressure Vessel With safety valve: P=1.25△P
Without SV: P=0.1MPa
Jacketed Vessel As external P vessel
ii. Design Temperature T
—— the enacted temperature of metallic
components under normal operating
condition
Design P and design T both
are the design load condition.
iii. Allowable stress
limit Stress  0
[ ] 
t

Safe Coefficien t n
Normal T Vessel  b s 
[ ]   , 
 nb ns  min

Medium T Vessel   t
 t

[ ]  
t b
, s

 nb ns  min

High T Vessel  s
t
D
t
n 
t
[ ]  
t
, , 
 ns nD nn min
iv. Safe (Safety) coefficient n
Strength σ σ ts σtD σtn
b
Material

Performance
Safety nb ns nD nn
Coefficient

Carbon Steel ≥3.0 ≥1.6 ≥1.5 ≥1.0

Low Alloy Steel
High Alloy Steel ≥3.0 ≥1.5 ≥1.5 ≥1.0
v. Welded joint efficiency ( )
(1)Double welded butt or completely welded
butt which is the same as double one.
NDE 100%  = 1.0
NDE Local  = 0.85 Double welded butt
(2)Single welded butt
NDE 100%  = 0.9
NDE Local  = 0.8 Single welded butt
vi. Additional value of wall thickness C
(1)Negative deviation of plate and tube C 1
Referring to the teaching material page ,
figure 4-7 & 4-8, selecting according to
the nominal thickness Sn.
(2)Corrosion allowable thickness C2
C2 = K a B
Ka —— corrosion rate, mm/year
B —— design life of utility, year
Generally speaking:
Ka < 0.05 mm/year
Single corrosion C2 = 1 mm
Double corrosion C2 = 2 mm
Ka = 0.05~0.1 mm/year
Single corrosion C2 = 1 ~ 2 mm
Double corrosion C2 = 2 ~ 4 mm
For stainless steel,
when the media is little corrosive C2 = 0
4.Pressure Test and Strength
Verification of vessels:
i. Purpose
(1)Verifying the macro-strength and
deformation of vessels
(2)Verifying the tightness of vessels
ii. Time
(1)For new vessels, the Pressure Test and
Strength Verification should be proceeded
after completely welded and heat treatment.
(2)For used vessels, the Pressure Test and
Strength Verification should be proceeded
after examination and repair, and before
putting into production.
iii. Media in Test
(1)Water —— the most commonly used
Stipulation to T of water:
Carbon steels 、 16MnR 、
normalizing15MnVR —— T≮ 5 ℃
Other low alloy steels —— T≮ 15 ℃
Stainless steels ——
content of [Cl-] in water ≤ 25ppm
(2)For the vessels which cannot be filled
with liquid, something like dry and
clean air, nitrogen gas or other inert
gases can be used to fill these vessels.
iv. Determination of Pressure for Testing
(1)Internal Pressure Vessel
Hydrostatic Test [ ]
PT  1.25P
[ ]t
Pneumatic Test [ ]
PT  1.15P
[ ]t
(2)External Pressure Vessel
Hydrostatic Test PT  1.25P
Pneumatic Test PT  1.15P
 Interpretation of symbols:
P —— Design pressure, MPa
PT —— Test pressure, MPa
[] —— Allowable stress at test
temperature, MPa
[]t —— Allowable stress at design
temperature, MPa
v. Pressure Testing Methods
(1)Hydrostatic Test
*Filling the vessel in test with liquid.
*Slowly increasing P to the test pressure PT .
*Keeping this pressure more than 30
minutes.
*Decreasing P to 80% of PT .
*Checking the welded seam and connection,
reducing P to repair them if existing
leakage.
*Repeating the previous test until upping to
grade.
*After testing, discharging the liquid and
drying the vessel with compressed air.
(2)Pneumatic Test
*Slowly increasing P to 10% of PT as well as
≤0.05MPa.
*Keeping this P for 5 minutes and have an
primary inspection.
*If up to grade, continue to slowly increase P
to 50% PT, then by the △P=10% PT
degree difference increasing slowly P to P T.
*Keeping this P for 10 minutes.

*Decreasing P to 87% PT, then keeping it

and examining and repairing.

*Repeating the previous test until upping to

grade.
(3)Air (gas) Tight Test

*Slowly increasing P to PT.

*Keeping this P for 10 minutes.

*Decreasing P to the design pressure.

*Examining the sealing condition.

vi. Stress verification before pressure test
(1)Hydrostatic test
PT ( Di  S e )
T   0.9 s
2Se
(2)Pneumatic test
PT ( Di  S e )
T   0.8 s
2Se
T —— Calculating stress at testing pressure,
MPa
s —— Yielding point at testing temperature,
MPa
5.Examples:
i. There is a boiler barrel which Di=1300mm,
working pressure Pw=15.6MPa and it has
a safety valve. Also know that the design
T=350ºC, the material is 18MnMoNbR, it
is double welded butt with 100% NDE.
Try to design the thickness of this boiler
barrel.
Solution:
(1)Determining the parameters
Pc = 1.1PW = 1.1×15.6 = 17.16 MPa
(with the safety valve)
Di = 1300mm
[]t = 190MPa (Design T = 350ºC)
[] = 190 Mpa (At normal T, S > 60-100)
 = 1.0 (Double welded butt, 100% NDE)
C2 = 1 mm (Single corrosion, Low alloy steel)
(2)Calculating the thickness
pc Di
S
2[ ]t   pc
17.16 1300
  61.5 mm
2 190 1  17.16
Design thickness
Sd = S + C2 = 61.5 + 1 = 62.5 mm
Choosing C1 = 1.8 mm (P95 Figure 4-7)
Additional value of wall thickness
C = C1 + C2 = 1.8 + 1 = 2.8 mm
Nominal thickness
Sn = S + C + round-of value
= 61.5 + 2.8 + round-of value
= 65 mm
(3)Hydrostatic test for strength verification
*Parameters:
[ ]
PT  1.25 P
[ ]t
190
 1.25  17.6   21.45MPa
190
*Efficient thickness:
Se = Sn - C = 65 - 2.8 = 62.2 mm
s = 410 MPa
*Stress: PT ( Di  S e )
T 
2Se
21.45  1300  62.2 

2  62.2
 234.9 MPa
*Stress verification:
0.9 s  0.9  410 1  369 MPa
  T  0.9 S
That is to say the strength in
hydrostatic test is enough.
ii. There is a oxygen cylinder which has been
kept in storage for a long time, with
Do=219mm using 40Mn2A and made by
seamless steel pipe. The actual Sn= 6.5mm and
b = 784.8MPa, s = 510.12MPa, 5 = 18%,
the design T is normal T.
If the working pressure Pw=15MPa, is the
working stress less than the allowable stress? try to
find whether the thickness is enough or not. If
not, what is the maximum allowable working pressure
in this cylinder?
Solution:
it is the problem about strength verification
—— Whethert ≤ []t  or not
(1)Determining the parameters
Pc = 15MPa Do = 219 mm Sn = 6.5 mm
 b 784.8
[ ]    261.6 MPa
nb 3
 s 510.12
[ ]    318.8 MPa
ns 1.6
Choosing the little one: i.e. []t = 261.6MPa
 = 1.0 (for seamless steel)
C2 = 1 mm
C1 = 0 (for the minimum thickness,
negative deviation is neglected.)
Se = Sn - C = 6.5 - 1 = 5.5 mm
(2)Strength verification
pc ( Do  S e )
 
t

2Se
15   219  5.5
  291.1 MPa
2  5.5
Obviously, t > []t = 261.6 MPa
So, 15MPa is too large, should be reduced.
(3)Determining the maximum allowable
working P
2 [ ]  S e
t
[ p] 
Do  S e
2  261.6  1 5.5
  13.48 MPa
219  5.5
So, the maximum safety P for this
cylinder is 13.48 MPa
4.3 Designing Heads subject
to Internal Pressure
Classification according to the shape:
i. Convex heads
Semi-spherical head
Elliptical head
Dished head (spherical head with hem)
Spherical head without hem
ii. Conical heads
Conical head without hem
Conical head with hem

iii. Flat heads

1.Semi-spherical head

Di
Calculating equation for thickness
pc Di
S
4 [ ]   pc
t

pc Di
Sd   C2
4 [ ]   pc
t

pc Di
Sn   C2  C1  round  of value
4[ ]   pc
t
2.Thickness calculating equation
of elliptical head

ho hi (b)
i. Calculating equation
for thickness: Ri (a)
Di S
For the elliptical head
whose m = a / b ≤ 2
pa  a 
 max   m     
2S  b 
2. The maximum stress should be at the top point:
Putting m = a / b, a = D / 2 into the equation,
getting: mpD
 max 
4S
Under the condition about strength:
mpD
 max   [ ]t
4S
Then: pD m
S 
4 [ ]t
(1)Replacing P with Pc
(2)Multiplying []t with welded joint efficiency 
(3)Substituting D with Di, D = Di + S
(4) m = a / b = Di / 2 hi
Putting these conditions into the equation:
getting:
pc Di m pc Di  Di 
S    
 

2[ ]  
t mp c  2  2[ ]t
  0 .5 p c  4 hi 
2
m = a / b = Di / 2 hi
For the standard elliptical head whose m=2:
pc Di
S
2[ ]t   0.5 pc
For the elliptical head whose m>2:
 at boundary »  and m at the top point
Then introducing the stress strengthening
coefficient K to replace (Di / 4hi)
 K pc Di
S
2 [ ]   0.5 pc
t

In this equation:
1  
2
 Di 
K  2  
 2h 
 
6    
 i


For standard elliptical head: K=1

This is the common equation for calculating

the wall thickness of elliptical heads.

Beside these conditions:
for standard elliptical heads Se ≮ 0.15% Di
for common elliptical heads Se ≮ 0.30% Di

The straight side length of standard

elliptical heads should be determined
according to P103, Figure 4-11
iii. Working stress and the maximum
allowable working pressure
pc  KDi  0.5S e 
 
t

2Se

2[ ]  S e
t
[ p] 
KDi  0.5S e
3.Dished head
i. Structure

s
Containing three parts: r

i
R

ho
Sphere: Ri D
i

Transition arc (hem): r

Straightedge: ho (height)
ii. Calculating equation for thickness
M pc Ri and 1 Ri 
S M  
 3 

2[ ]   0.5 pc
t
4 r 

M —— Shape factor of dished head

iii. Working stress and the maximum
allowable working pressure
pc  MRi  0.5S e 
 
t

2Se

2[ ]  S e
t
[ p] 
MRi  0.5S e
iv. Dished head
When Ri = 0.9 Di & r = 0.17 Di
the dished head is standard dished head
and M = 1.325
So the equation is:
1.2 pc Di
S
2[ ]t   0.5 pc
4.Conical head
i. Structure
*without hem (suitable for  ≤ 30 o )
without local strength
with local strength
*with hem (suitable for  > 30 o )

—— Adding a transition arc and a

straightedge between the joint

of head and cylinder
ii. Calculating equation for thickness
The maximum stress is  in the main aspect
of conical head. m 

pD 1
 max    max  
2 S cos 

According to the strength condition:

pD 1
 max    max    [ ]t
2S cos 
Then pD 1
S 
2[ ] cos 
t

Replacing P with Pc, considering , and

changing D
p into
D Dc ，1D=Dc+S
S c c

2[ ]   pc cos 
t
 This equation only contains the membrane
stress but neglects the boundary stress at the
joint of cylinder and head. Therefore the
complementary design equation should be
established:
(1)Discriminating whether the joint of
cylinder and head should be reinforced
or not.
(2)Calculation for the local reinforcement.
Conical head without hem ( ≤ 30 o )
(1)Not require reinforcing
(consistent thickness for the whole head)
main aspect: pc Dc 1
S 
2[ ]   pc cos 
t

small aspect: pc Dis 1

S 
2[ ]   pc cos 
t
(2)Require reinforcing
(for the thickness of joint,
the reinforcement region)
Main aspect: Q pc Di
Sr 
2[ ]t   pc

Small aspect: Q pc Dis

Sr 
2[ ]   pc
t
Interpretation:
Dc —— inside diameter of large end
Di.s —— inside diameter of small end
Di —— inside diameter of cylinder
Q —— coefficient (Consulting the Figure
4-16 or 4-18 in book)
Conical head with hem ( > 30 o )
(1)Thickness of hem at the transition section
K pc Dis
S 
2[ ]   0.5 pc
t

(2)Thickness of conical shell at the joint

with
transition section f pc Dis
S 
[ ]   0.5 pc
t

K —— coefficient (Consulting Figure4-13)

f —— coefficient (Consulting Figure4-14)
4.Flat head
i. Structure
The geometric form of flat heads:
rotundity, ellipse, long roundness,
rectangle, square, etc.
ii. Characteristics of load
Round flat with shaft symmetry which is
subjected to uniform gas pressure
(1)There are two kinds of bending stress states,
distributing linearly along the wall.

(2)Radial bending stress r and hoop bending

stress t distributing along the radius.
▲For fixed egde
p
max = r.max

S
The maximum stress is
R
at the edge of disk.
 
2 t
D
 r .max   0.188P  0

r.max
S r
 R  PD 
 0.75  
 S  2 S 
▲ For simply supported ends
max = r.max = t.max P
The maximum stress is

S
in the center of disk. R
2
D
 r . max  0.31P   t
S

r.max
 R   PD  r
 1.24    
 S   2S  0
iii. Calculation equation for thickness
From the condition of strength max ≤ []t ,
getting:
fixed edge:
0.188 P
SD
[ ]t
simply supported ends:
0.31P
S D
[ ]t
 In fact, the supporting condition at boundary
of flat head is between the previous two.
After introducing the coefficient K which is
called structure characteristics coefficient and
considering the welded joint efficient , getting
the calculating equation for thickness of round disk:
K Pc
S p  Dc
[ ] 
t

Sn  S p  C2  C1  round  of value
5.Examples
Design the thicknesses of cylinder and heads
of a storage tank. Calculating respectively the
thickness of each heads if it’s semi-spherical,
elliptical, dished and flat head as well as
comparing and discussing the results.
Known: Di = 1200 mm Pc = 1.6Mpa
material: 20R []t = 133Mpa C2 = 1 mm
The heads can be punch formed by a complete
steel plate.
Solution:
(1)Determining the thickness of cylinder
S
pc Di 1.6  1200
  7.26 mm
2 [ ]   pc
t
2 133 1.0  1.6
 = 1.0 (Double welded butt, 100%
NDE)S d  S  C2  7.26  1.0  8.26 mm
C1 = 0.8 mm (Checking Figure 4-7)
Sd + C1 = 8.26 + 0.8 = 9.06 mm
Round it of, getting: Sn = 10 mm
Verify the strength under pressure test
(2)Semi-spherical head

S
pc Di 1 .6  1200
  3.62 mm
4 [ ]   pc
t
4 133 1.0  1.6
 = 1.0 (wholly punch forming)
S d  S  C2  3.62  1.0  4.62 mm
C1 = 0.5 mm (Checking Figure 4-7)
Sd + C1 = 4.62 + 0.5 = 5.12 mm
Round it of, getting: Sn = 6 mm
 (3)Standard elliptical head

S
pc Di 1.6  1200
  7.24 mm
2[ ]   0.5 pc 2 133 1.0  0.5 16
t

 = 1.0 (wholly punch forming)

S d  S  C2  7.24  1.0  8.24 mm

C1 = 0.8 mm (Checking Figure 4-7)

Sd + C1 = 8.24 + 0.8 = 9.04 mm
Round it of, getting: Sn = 10 mm
 (4)Standard dished head

S
1.2 pc Di 1 .2  1.6  1200
  8.69 mm
2[ ]   0.5 pc 2 133 1.0  0.5 1.6
t

 = 1.0 (wholly punch forming)

S d  S  C2  8.69  1.0  9.69 mm
C1 = 0.8 mm (Checking Figure 4-7)
Sd + C1 = 9.69 + 0.8 = 10.49 mm
Round it of, getting: Sn = 12 mm
(5)Flat head
K = 0.25; Dc = Di = 1200 mm; []t = 110 Mpa
K Pc 0.25 1.6
S p  Dc  1200   72.36 mm
[ ] 
t
110 1.0
 = 1.0 (wholly punch forming)
S d  S  C2  72.36  1.0  73.36 mm
C1 = 1.8 mm (Checking Figure 4-7)

Sd + C1 =73.36 + 1.8 = 75.16 mm

Round it of, getting: Sn = 80 mm
 Comparison:
Head-form Semi-sphe. Elliptical Dished Flat

Sn mm 6 10 12 80

kg 106 137 163 662

Selection:
It’s better to use the standard elliptical
head whose thickness is the same to that of
cylinder.
 Chapter 5 Design of Cylinders
and Formed Heads
subjected to External-Pressure

5.1 Summarization
1.Failure of External Pressure Vessel
Under the effect of external pressure, the
vessels may deform when the pressure is larger
than a certain value. This kind of damage is
called the failure of external pressure vessels.
2.Classification of Failure
Side bucking —— the main form of failure
Axial bucking
Local bucking
 5.2 Critical Pressure
1.Critical pressure and
critical compressive stress
The pressure that makes the external
pressure vessels fail is called the critical
pressure, indicating by Pcr.
At the moment that exists Pcr, the stress
inside the vessels is called the critical
compressive stress, indicating by cr .
2.Factors affect the critical pressure
i. Geometric dimension of cylinder

0.3 0.3
vacuum in failure

350
0.51 0.3

350
Degree of

 90  90
175
175

 90  90

(1) (2) (3) (4)

Pcr
mm HO2 500 300 120~150 300
 Comparison and analysis
for the experimental results
Figure (1) and Figure (2):
*When the value of L / D is equal, the larger the
value of S / D, the higher the Pcr.
Figure (2) and Figure (3):
*When the value of S / D is equal, the smaller
the value of L / D, the higher the Pcr.
Figure (3) and Figure (4):
*When the value of S / D and L / D are equal,
having the stiffening ring as well, high Pcr.
ii. Materials’ Performance of the cylinders
(1)The critical pressure (Pcr) hasn’t direct
relation with the strength ( s) of the
materials.
(2)The critical pressure (Pcr) depends of the
flexural rigidity of the cylinders in some
aspects.
The stronger the flexural rigidity, the more
difficult for the failure.
iii. The differential in the dimension at the
process of vessels’ manufacturing

Mainly reflecting on the “ellipticity” ( 椭圆度 ),

which is the processing differential in the
dimension of the cylindrical section.
 Ellipticity:
Dmax
Dmax  Dmin
e  100%
Dmin

DN
*Large ellipticity e can make the critical
pressure Pcr decrease and failure happen in
ahead.
*Regulated as in the engineering, ellipticity
e ≤ 0.5% when vessels subjected to the
external pressure are made.
3.Long cylinder, short cylinder
and rigid cylinder, the
calculating equations of their
critical pressure
 i. Long cylinder
—— cylinders with large L / Do
Calculating equation of the critical P:
3 3
2E  Se 
t
t  S e  For steel cylinders:
Pcr     2.2 E  
2 
1    Do  D
 o  = 0.3

Calculating equation of the critical

stress: 2
Pcr D Pcr Do t  Se 
 cr    1.1E  
2Se 2Se  Do 
ii. Short cylinder
—— cylinders with small L / Do
Calculating equation of the critical P:

P  2.59 E
' t  Se / Do  2.5

cr
 L / Do 
Calculating equation of the critical
stress:
 
'
'
Pcr Do
 1.3 E t  Se / Do 
1.5

cr
2Se  L / Do 
iii. Rigid cylinder
—— cylinders with small L / Do, large Se / Do
Designing criterion:
Only need to satisfy the strength condition:
compression ≤ [ ]tcompression
i.e.
Pc  Di  S e 
压 
t
 [ ]t压 
2Se
 4.Critical Length

Critical length —— which is used to classify

the long cylinder and short cylinder; and it is
the critical dimension of the short cylinder and
rigid cylinder.
L > Lcr Long cylinder
L’cr < L < Lcr Short cylinder
L < L’cr Rigid cylinder
i. Critical length Lcr of long and short cylinder:
Do
Lcr  1.17 Do
Se
’
ii. Critical length L of short and rigid cylinder:
cr

t
1.3E S e
L'
cr 
Do
  comp.
t

Se
5.3 Engineering Design of
External-P Vessels
1.Designing criterions
Pcr
Pc  [ P ] 
m
Pc —— Calculating Pressure, MPa
Pcr —— Critical Pressure, MPa
[p] —— Allowable External Pressure, MPa
m —— Stable safety coefficient
For cylinders, m = 3
at the same time, 椭圆度 e ≤ 0.5%
 Pcr DO 2
cr  Pcr D Pcr Do t  Se 
2 Se  cr    1.1E  
cr 2Se 2Se  Do 
 
Et
 '

P '
cr Do
 1. 3 E t  Se / Do  1.5

 L / Do 
t
Pcr 2E se cr
[ P]    2Se
m m DO
  f  Do Se , L Do 
2.Nomograph for the 2 t
thickness designing of Making : B  E 
the external-P cylinders m
i. Calculating Steps
Step 1: L 、 Do 、 Se →  So : [ P ]  B S e
Drawing the curve Do
Step 2: Find the relationship between  and [P]
2 t
Making : B  E 
m

2 t
For cylinder m=3 and B E
3
Then getting the relationship curve B = f ()
Se
So : [ P]  B
Do
ii. Steps of nomograph for the thickness
designing of the external-P cylinders (Tubes)
For the cylinders and tubes whose Do/Se ≥20:
(1)Supposing Sn, Se = Sn - C, calculating the
values of L / Do and Do / Se.
(2)Calculating the value of  (value of A),
checking the Figure (5-5).
If L / Do > 50, checking the figure using
L / Do = 50.
If L / Do < 0.05, checking the figure using
L / Do = 0.05.
(3)Calculating the value of B

According to the used material, choosing

the relevant graphs from Figure (5-7) and
Figure (5-14) and then finding the point A
from abscissa.
Two situations maybe encountered:
*Point A with that certain value lies at the right
of the curve and intersects with the curve,
then the value of B can be found directly in
the
figure.
*Point A with that certain value lies at the left
of the curve and has no joint with the curve,
then the value of B is calculated
2 tby the
B E A
following equation: 3
 (4)Calculating [P]

Putting the value of B into Equation (9) →[P]

Se B
[ P]  B  （9）
Do Do S e
(5)Comparing

~
~
If [ P ]  Pc i.e. the supposed Sn is usable, safe

If [ P]  Pc i.e. the supposed Sn is too large and

should be decreased
appropriately, repeating the previous calculating
steps until satisfying the first condition.
i.e. the supposed Sn is too small and
If [ P]  Pc
should be increased appropriately,
repeating the previous calculating steps until
satisfying the first condition.
3.Pressure test of external-P
vessels

i. Pressure test of external-P vessels and

vacuum vessels is processing as the
hydrostatic pressure test.
Testing pressure:
PT = 1.25 P
P —— design pressure
4.Example and discussion
Design the thickness of an external-P cylinder.
Known:
Calculating pressure: Pc = 0.2 MPa
Design temperature: t = 250℃
Inside diameter: Di = 1800 mm
Calculating length: L = 10350 mm
Additional value of wall thickness: C = 2 mm
Material: 16MnR; Et = 186.4 103 MPa
Solution:
(1)Assuming Sn = 14 mm
Then Do = Di + 2 Sn = 1828 mm
Se = Sn - C = 12 mm
Finding out:
L / Do = 10350  1828 = 5.7
Do / Se = 1828  12 = 152
(2)Calculating the value of  (A)
Checking the Figure 5-5, getting:
A = 0.000102
(3)Calculating the value of B
From Figure 5-9, we can see that
point A is at the left of the curve, then the
calculating equation is like following:

2 t 2
B  E A  186.4 103  0.000102  12.78 MPa
3 3
(4)Calculating [P]
B 12. 78
[ P]    0.0834 MPa
Do S e 152
(5)Comparing [P] and Pc
[P] < Pc = 0.2 MPa unsatisfied
Reassuming Sn, or setting the stiffening ring.
Calculation under the condition that supposes
there have two stiffening rings:
(1)Thickness is the same: Sn = 14 mm
After setting two stiffening rings,
the calculating length is like following:
Lorigin 10350
L   3450 mm
3 3

3450
Then L D o   1.9 (D o Se  152)
1828
(2)Calculating the value of  (A)
Checking the Figure 5-5, getting:
A = 0.00035

(3)Calculating the value of B

From Figure 5-9, we can see that
point A is at the right of the curve,
getting B = 42.5 MPa
(4)Calculating [P]

B 42.5
[ P]    0.28 MPa
Do S e 152
(5)Comparing [P] and Pc
[P] > Pc = 0.2 MPa satisfied
Calculation under the condition that supposes
to increase the thickness:
(1)Assuming: Sn = 20 mm
Then Do = Di + 2 Sn = 1840 mm
Se = Sn - C = 18 mm
Finding out:
L / Do = 10350  1840 = 5.6
Do / Se = 1828  18 = 102
(2)Calculating the value of  (A)
Checking the Figure 5-5, getting:
A = 0.00022

(3)Calculating the value of B

From Figure 5-9, we can see that
point A is at the right of the curve,
getting B = 27.5 MPa
(4)Calculating [P]

B 27 . 5
[ P]    0.27 MPa
Do S e 102
(5)Comparing [P] and Pc
[P] > Pc = 0.2 MPa and closing
So, we can use the steel plate with
Sn = 20 mm, whose material is 16MnR.
 5.4 Design of External-P
Spherical Shell and Convex Head

1.Design of external-P spherical

shell and semi-spherical head

i. Assuming Sn, and Se = Sn － C.

Calculating the value of Ro / Se.
ii. Calculating the value of  (A)
0.125
A
Ro S e

iii. Calculating the value of B and [P]

According the used material, choosing
the relevant graph from Figure 5-7 and
Figure 5-14 and finding out the point A at
the
abscissa.
Two situations maybe encountered:
(1)If point A is at the right of the curve, the
value of B can be found from the figure
directly.
B
Then [P] 
Ro S e

(2)If point A is at the left of the curve,

directly calculating:
t
0.0833E
[ P] 
( Ro S e ) 2
iv. Comparison
~~ i.e. the original assuming Sn is
If [ P ]  P c
usable, and safety.

i.e. the original assuming Sn is

If [ P]  Pc too small, S should be
n
increased appropriately,
repeating the previous
calculating steps until
satisfying
the first condition.
2.Design of external-P convex head

The method of designing the external-P

convex head is the same to that of designing
external-P spherical head. But the Ro in the
designing of spherical head should be adjusted
like following:
i. For elliptical head

Ro —— the equivalent spherical diameter of

elliptical head; Ro = K1Do
K1 —— coefficient; depending on a / b,
checking P141, Figure 5-3
ii. For dished head

Ro —— the equivalent spherical diameter

of the dished head; it’s the outside
diameter of the spherical part at
the dished head.
5.5 Design of the Stiffening Ring
in External-P Vessels
1.Function of stiffening ring
2.5
 Se 
3  
 Se   Do 
 Pcr  2.2 E 
t
 Pcr'  2.59 E t
 Do   L 
 
 Do 
Pcr
And [ P] 
m
From the previous equations, we can know the
methods to increase [P]:
i. Increasing S
ii. Decreasing the calculating length L

∴ Function of stiffening ring

—— decreasing calculating length to
increase [P]
2.Space length and number
of stiffening ring
Assuming the space length of stiffening ring is Ls
From the design criterions of external-P:
Pc ≤ [P] and [P] = Pcr / m
Making Pc = [P] then Pcr = m Pc (a)
From the equation for the critical pressure of
short cylinder:
Pcr  2.59 E
' t  S e Do 
2.5

 L Do 
Putting equation (a) in, getting:

P  2.59 E
' t  Se Do 
2.5
 m pc
cr
 Ls Do 

Then putting m=3 in, getting:

2.5
Do  S e 
( Ls ) max  0.86 Et
 
Pc  Do 
(Ls)max —— Under the condition that Do and
Se of the cylinder is determined,
the maximum space length
between the needed stiffening
rings working safely under the
calculating external pressure Pc,
mm.
The actual space length between stiffening
rings Ls ≤ (Ls)max is indicating safety.
The number of stiffening rings:
L
n  1
Ls

In the above equation:

L —— the calculating length of cylinder before
setting the stiffening rings, mm
Ls—— the space length between stiffening
rings, mm
3.Connection of stiffening rings
and cylinders
i. Connection Demands
Must assure all the cylinder and stiffening
ring are under the load together.
ii. Connection Methods
Welding —— Continuous Weld ( 连续焊接 )
Tack Weld ( 间断焊接 )
iii. The stiffening rings should not
be randomly crippled or cut
off. If those must be done, the
length of the arc that are crippled
or cut off should not be larger
than the values shown in Figure 5-19.
For example:
There is a horizontal external pressure vessel.
When the stiffening ring is set inside the
cylinder, in order not to affect the fluid flowing
or fluid discharging, we must leave a hole 豁口 at
the lowest position of the stiffening ring or set a
thoroughfare of fluid.
As illustrating like the following two figures
 Chapter 5 Components
and
Parts of Vessels

6.1 Flange Connection

1.The Sealing Theory ( 密封原理 )
and Connection Structure of
Flanges
i. Connection Structure
Three parts:
(1)Connected parts
—— a couple of flanges
(2)Connecting parts
—— several couples of
bolts and nuts
(3)Sealing parts
—— gasket
ii. Sealing Theory
Taking the bolts’ forced sealing as an example
to illustrate the Sealing Theory:

(1)Before butting (2)After butting (3)After charging

medium
2.The Structure and
Classification of Flanges
According to the connection ways of
flanges and equipment (pipelines)
(1)Integrated flange
—— S.O.flange (slip on flange)
W.N.flange (welding neck flange)
Pipeline Flange Vessel Flange
S.O.flange W.N.flange
(2)Simple [loose (type), lap joint, lapped] flange

On the welding ring Interlink on the

turn-down rims
(3)Screwed flange

Square flange Elliptical flange

3.Factors effect the sealing of
flanges
i. Bolt load under pretension condition
(bolt load for gasket sealing)
The bolt load is too small to seal specific
pressure ( 顶紧密封比压 ).
The bolt load is too large to avoid the gasket
being pressed or extruded.
Increasing the bolt load
appropriately can strengthen the
sealing ability of gasket.
So under the condition of certain
bolt load, decreasing the
diameter of bolts center circle or
increasing the number of them
are both beneficial for sealing.
ii. The types of sealing face
(1)plain (face) flange （ Raised Fac
e）
(2)M&FM (male and female)
(3)T&G (tongue and groove face)
(4)Conical face
(5)Trapezoidal groove face
iii. Properties of gasket
(1)The common-used materials of gasket
*Non-metal Material
—— Rubber, Asbestos, Synthetic resins.
Advantages: soft and corrosion resistant
Disadvantages: the properties of high-T
resistance and pressure
resistance is inferior to the
metallic materials.
Used in: Common and Medium T; Flange
sealing of Medium and Low P
devices and pipes.
*Metal (Metallic) Material
—— soft aluminum, copper, iron (soft
steel), 18-8 stainless steel.
Advantages: high-T resistant, with high strength
Demands: Excellent soft toughness
Used in: Medium and high T; Flange sealing
subjected to medium and high P
(2)Gasket Types (Classifying according to the
properties of materials)
*Non-metal Gasket
—— such as rubber gasket, asbestos-rubber
gasket.
*Compound Gasket (Metal and non-metal
compound gasket)
—— such as metal jacketed gasket ( 金属包
垫片 ) and Metal spirotallic
[spiral-
wound] gasket （金属缠绕垫）
Metal jacketed gasket ( 金属包垫片 ), i.e.
wrapping the metal slice around the asbestos
gasket or asbestos-rubber gasket

Metal spirotallic [spiral-wound] gasket ( 金属

缠绕垫片 ), i.e. making by alternately rolling
thin steel belt and asbestos
*Metal gasket
—— such as octagon ring gasket, elliptical
gasket, lens ring (washer) [grooved
metallic gasket]
(3)Selection of gasket
*Factors of working pressure and temperature
Medium and low P; common and medium T
—— Non-metal gasket
Medium P; Medium T
—— Metal and non-metal compound gasket
High P; high T —— Metal gasket
High vacuum; cryogenic —— Metal gasket
*Degree of demands for sealing
*Demands for the types of sealing face
*Properties of gasket

Concrete selection should be referred to

JB4704-92, JB4705-92, JB4706-92.
At the same time, the practical experience
should be taken into account.
iv. Rigidity ( 刚度 ) of flange
(1)If the rigidity of flange is not enough, there
will occur the serious buckling [ 翘曲 ]
deformation, as well the specific pressure will
decrease and the sealing face will be loose, as
a result, the sealing will fail.
(2)Measures to increase the rigidity of flange
 Bolt circle, Outer diameter, Thickness
(3)Strengthening the rigidity of flange to
increase the weight of flange as well the cost
of whole-flange’s sealing.
v. Effect of working conditions

Temperature
Temperature
Pressure
Pressure
Corrosive
Corrosive Characteristics
Characteristics
of medium
Penetrant
Penetrant Characteristics
Characteristics

Combined effect
Greatly affecting the sealing
 4.Standard and Selection
of Flanges

i. Standard number of pressure vessel flanges

JB / T 4701-2000 ∼ JB / T 4703-2000
 Standard types and marks
of pressure vessel flanges
Sealing Without lined ring With lined ring (C)
face
Code Plain M&FM T&G Plain M&FM T&G
Type M FM T G M FM T G
A-S.O.Flange
JB/T 4701-2000
P A T S C P A T S C

B-S.O.Flange
JB/T 4702-2000
P A T S C P A T S C

W.N.Flange P A T S C
P A T S C
JB/T 4703-2000
For example:
PN=1.6MPa, DN = 800mm, T&G C-S.O.Flange
with lined ring

T Flange: C-S 800 — 1.6 JB4702-92

G Flange: C-C 800 — 1.6 JB4702-92
 Nominal
Nominal Standard
Standard
Pressure
Pressure Code
Code
MPa
MPa
C-S 800 — 1.6 JB4702-92

Code
Code of
of Flange
Flange Type
Type
Nominal
Nominal
Diameter
Diameter
Code
Code of
of Sealing
Sealing mm
mm
face
face Type
Type
ii. Dimension of pressure vessel flanges

Dimension of flanges is only confirmed by

two standardized parameters PN and DN of
flanges.

Confirmation of Nominal Pressure PN of

flanges: JB4700-92 (Book, P160)
iii. Selection steps for pressure vessel flanges
(1)According to the design task, confirming
the types of flanges (S.O. or W.N.).
(Referring to P157 Table 6-2)
(2)According to the nominal diameter DN of
flanges , working temperature, design
pressure, material of flanges, confirming
the nominal diameter DN and nominal
pressure PN of flanges.
(Referring to P160 Table 6-4; P332 Appendix 12)
(3)Confirming the sealing face types of flanges
and the types of gaskets.
(Referring to P155 Table 6-1)

(4)According to the types of flanges, DN and

PN of flanges, checking and finding out the
dimension of flanges; number of bolts and
their specification.
(Referring to P336 Appendix 14)
(5)Confirming the material of bolts and nuts.
(Referring to P163 Table 6-6; P333
Appendix 13)

(6)Portraying the unit drawing of flanges.

Example:

There are flanges to connect the body of a

fractionating (rectifying) tower and the heads.
Knowns:
Inside diameter of tower: Di = 1000mm
Working temperature: t = 280℃
Design Pressure: P = 0.2MPa
Material of tower: Q235-AR
Solution:

(1)From P157 Table 6-2, A-S.O.Flange is selected.

(2)Confirming the nominal diameter DN and
nominal pressure PN
DN = 1000 mm
(Equal to the inside diameter of tower)
From P160 Table 6-4, choose the material of
tower as that of flanges, i.e. Q235-AR
t = 280℃
When PN = 0.25 Mpa,
Pallowable = 0.14 MPa < Pdesign = 0.2 Mpa
When PN = 0.6 Mpa,
Pallowable = 0.33 MPa > Pdesign = 0.2 Mpa

So, the nominal pressure of flanges is:

PN = 0.6 MPa
(3)Confirming the sealing face types of flanges
From P155 Table 6-1, choosing plain sealing
face, spirotallic [spiral-wound] gasket

(4)According to the DN and PN of flanges, from

Appendix 14, Table 32, finding out the
dimension of every part of flanges.
Specification of bolts: M20; Number: 36
(5)From P163 Table 6-6, finding out:
Material of bolts: 35 steel
Material of nuts: Q235-A

(6)Portraying the unit drawing of flanges

(Omitting)
Standard of tube flanges
(New Standard issued by Chemical Ministry)

European: HG 20592 — 97 ~ HG 20600 — 97

American: HG 20615 — 97 ~ HG 20621 — 97
6.2 Support for vessels

Support for horizontal vessels

Saddle support, ring support, leg, etc.
Support for vertical vessels
Skirt support, hanging support, etc.
1.Double-saddle support
i. The structure of double-saddle support
Gasket
120°

Web-plate

Sub-plate

Anchor bolt
ii. Position of support (A)
A≤Do/4 & < 0.2L. The maximum value < 0.25L
iii. Standard and selection of double-saddle
support
Type —— Stationary type: F Movable type: S
Model Type —— Light-duty: A Heavy-duty: B
Mark —— JB / T 4712-92 Support
Model
Model Type
Type Nominal
Nominal Diameter
Diameter Type
Type
2.Checking calculation of
stress in double-saddle
horizontal vessels

i. Load analysis for horizontal vessels

 q

A F F A
Shearing Force
Diagram
M1
M3 Bending Moment
Diagram
M2
ii. Reserved force to support
mg mg q  4 
F  or F   L  hi 
2 2 2 3 
In this equation:
q —— Mass load/unit length of vessels,
N / mm
L —— Distance between the T.L.
(tangent lines) of two heads, mm
hi —— Height of curved surface of heads,
mm
iii. The maximum radical bending moment
The section across the middle point of moment

 22 RRm22 hhi22
11
 
FL
FL  
m
L 22
i
4
4 A
A 
M11 
M  L   (Nmm)
(N mm)
44  1  44hhi i LL 
1
 33LL 
The section at support
 A Rm  hi 
2 2

 1  L  2 AL 
M 2   FA 1   (N  mm)
 4 h 
1 i
 3L 
iv. Calculation for stress of cylinder
—— to the vessels subjected to internal pressure
(1)Stress across the middle section
The most highest point in section (Point 1):

PPccRRmm MM1(Point
 point
The most lowest 
11 in 
section 1 2):
22SSee  RRmm SSee
22

PPccRRmm M
 22  M211
22SSee  RRmm2 SSee
(2)Stress in the section of support
The most highest point in section (Point 3):
PPccRRmm MM
 33  22
22SSee KK11 RRmm SSee
22

The most lowest point in section (Point 4):

PPccRRmm MM22
 44 
22SSee KK22 RRm2m2 SSee
v. Checking calculation for stress of
cylinder
tensile stress
comb . tensile 
comb . tensile max
max
   tt

compressive stress

  
   
tt
  the
thesmaller
smaller va
value
lue
 
 crcr  BB
comb
comb. comp
. comp. . max
max
In these two equations:
[]t —— The allowable stress of material at the
design T, MPa
[]c r —— The allowable compressive stress of
material, MPa
B —— Calculation method is the same with that
in design of external pressure, see P172
 6.3 Reinforcement for

opening of vessels

1.The phenomena and reason for

opening stress concentration
Stress concentration factor:

 max 
K
 

max —— The maximum stress

at the boundary of opening
* —— The maximum basic Small opening
in plate
stress of shell
 Reasons for stress concentration:
(1)Local Material of vessel wall is
decreased

(2)The continuity of structure is

damaged
2.Opening reinforcement’s
Designing

i. Designing Criterions
(1)Equi-area criterion of reinforcement
(2)Plastic failure criterion of reinforcement
ii. Reinforcement Structure
(1)Structure of Stiffening Ring
Nozzle (Connecting Tube)
Stiffening Ring

Shell
(2)Structure of 加强元件
Method —— Taking the
parts of nozzles or vicinity of
shells’ openings which need
to be reinforced as the 加强
元件 , then welding these
parts with nozzles or shells.
(3)Structure of Integral Reinforcement
Method —— Taking the connecting parts of
nozzles and shells as the integral forgings, at
the same time thickening them, then welding
them with nozzles and shells.
iii. Diameter Range of the openings that need
not to be reinforced
When the following requirements are all met,
the reinforcement is out of need.
(1)Design Pressure P ≤ 2.5 MPa
(2)The distance between two mid-points of two
nearby openings (taking length of are as the
length of curved surface) should be larger
than 2× (D1+D2), D1, D2 are the diameters of
the two openings respectively.
 (3)Nominal Outside Diameter of
connecting tubes ≤ 89 mm
(4)The minimum wall thickness δmin
of connecting tubes should meet
the following requirements:(mm)

 25 32 38 45 48 57 65 76 89

δmin 3.5 4.0 5.0 6.0

 3.Designing methods of
equi-area reinforcement

Metallic
Metallic areas
areas in
in local
local reinforcement
reinforcement
≥≥

the
the area
area of
of sections
sections which
which are
are
the
the position
position of
of openings
openings
i. Confirmation of the effective range of
opening and reinforcement areas

h1
h2
B

A A1 A2 A3 A4
Effective width:  B  2d 
 
 B  d  2 S n  2 S n. t  max
Effective [working] height:
Outside height

 h1  d S n.t 

 

 h1  actual overhang height of nozzle
 min
Inside height

 h2  d S n.t 

 

 h2  actual embedded height of nozzle 
 min
In these equations:
Sn —— Nominal thickness of cylinders

Sn.t —— Nominal thickness of connecting

tubes (nozzles)

d —— Diameter of openings d = di +2C

di —— Inside diameter of openings

C —— Additional value of wall thickness

Computation of metallic areas for
effective reinforcement
(1)The area of the sections on shell which
are the positions of openings A: A = S×d
(2)The unnecessary metallic area A1 on shell or
heads which is larger than calculating thickness S:
A1 = (B – d) (Se – S) – 2 (Sn.t – C) (Se – S) (1 – fr)
(3)The unnecessary metallic area A2 on nozzles
which is larger than the calculating thickness S:
A2 = 2 h1 ( Sn.t – St –C ) fr + 2 h2 ( Sn.t – C – C2 ) fr

(4)The metallic area of welding seam in the

reinforcement region A3:
A3 = according to the actual dimension
ii. Designing Steps in Reinforcement for openings
(1)Getting the following data from the strength
calculation:
Calculating wall thickness of cylinders or heads S
Nominal wall thickness of cylinders or heads Sn
Calculating wall thickness of nozzles St
Nominal wall thickness of nozzles Sn.t
Additional value of wall thickness C = C1+ C2
(2)Calculating the effective reinforcement range
B, h1, h2
(3)Calculating the necessary reinforcement area
A according to P183 Table 6-17
(4)Calculating the available reinforcement area
A1, A2, A3
(5)Judging whether it is necessary to add
some reinforcement area

If A1 + A2 + A3 ≥ A
reinforcement not required
If A1 + A2 + A3 < A
reinforcement required
(6)If reinforcement is required, calculating the
added reinforcement area A4
A4 = A －（ A1 + A2 + A3 ）

(7)Comparison
Finally getting A 1 + A 2 + A3 + A 4 ≥ A
6.4 Attachment of vessels

1.Man Hole and Hand Hole

i. Nominal Diameter of standard man-hole
DN ： 400 450 500 600
ii. Nominal Diameter of standard hand-hole
DN ： 150 250
2.Connecting Tubes (Nozzles) [ 接管 ]

3.Flg (Flange, Flanch) [ 凸缘 ]

4.Sight (Level) Glass [ 视镜 ]