Section 2 Design of Process Vessels Chapter 2 Panorama

Section 2 Design Section 2 Design of Process Vesselsof Process Vessels

Chapter 2 Chapter 2 PanoramaPanoramaChapter 2 Chapter 2 PanoramaPanorama

1.Conception of Vessels:1.Conception of Vessels:

Process Vessels are the various Process Vessels are the various equipments in the chemical equipments in the chemical process.process.

2.1 Structure and 2.1 Structure and

ClassificationClassification of Vesselsof Vessels

2.1 Structure and 2.1 Structure and

ClassificationClassification of Vesselsof Vessels

2.2.Structure of vessels:Structure of vessels:

3.3.Classification of vessels:Classification of vessels:

i. According to pressure and its typei. According to pressure and its type

(1)Internal Pressure Vessel(1)Internal Pressure Vessel

—— —— vessels where the media pressure vessels where the media pressure

inside the vessel is larger than that outsideinside the vessel is larger than that outside

(gauge pressure).(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 the media pressure inside the vessel is lower than that outsideis lower than that outside

(gauge pressure). When the pressure(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

Grade (I)

Grade (II)

Grade (III)

Factor

P, P*V

Media toxic and combustible

importance Degree of danger:

I < II < III

2.2

Basic code and

Common Standard for

Design of Pressure Vessels1.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 252.5 (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 Chapter 3 Stress Analysis of Thin-walled Internal-P VesselThin-walled Internal-P Vessel

3.1 Stress Analysis 3.1 Stress Analysis

of thin-walled Cylinders of thin-walled Cylinders Subjected to Internal PressureSubjected to Internal Pressure

1.1.Thin-walled vesselsThin-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:2.stress characteristics: There are always two kinds of

stress in pressure vessels.

i. membrane stress

—— membrane (shell)

theory

ii. boundary stress

—— shell theory with

moments and conditions

of deformational

compatibility

P

3.2 3.2 Membrane Theory ——Membrane Theory ——

Rotary Shells’ Stress AnalysisRotary Shells’ Stress Analysis

(1)rotary curved-surface & shell

1.Basic conceptions and 1.Basic conceptions and hypothesis:hypothesis: i. Basic conceptions

Cylinder , conical shell, spherical shellCylinder , conical shell, spherical 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

A

O

B

B’

(3)Generator (AB)

The plane curve which forms the curved surface.

(4)Longitude (AB’)

Section passing OA and intersecting with shell, the cross-line is AB’.

(5)Normal (n) The line passing point M in meridian and is vertical with midwall surface. The extension of normal must intersect with OA.(6)Latitude (CND) The cross-line formed by the

conical surface passing point K2’ intersects with the rotary curved surface.

A

O

M

n

K2K1

··

·K2’

·C

ND

(7)Longitudinal radius (R1) The radius of curvature of meridian which passes point M in midwall surface is called the longitudinal radius of point

M in meridian.The center K1

of curvature of the round

with diameter R1 must be in the extension of normal passing point M.For example:Longitudinal radius of point M:

R1 = M K1

A

O

MK2

K1

··

·K2’

·C

ND

(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 K2

exerciseexercise AA

BB

CC

DD

dd

Calculations of R1,R2 at different

points

Calculations of R1,R2 at different

points

Calculation of RCalculation of R11 and R and R22

1.Cylindrate shell R1 = ∞ R2 = D / 2

δ

D

p

Spherical shell subjected to uniform gas internal

pressure:R1 = R2 = D / 2

D

δ

3.Elliptical shell Known: Major semiaxis – a Short semiaxis - b

1

''

2

32'

y

y

(1)Find R1 and R2 of point A:

R1:

radius of curvature of Ay

xa

b

A(x,y).

..

k2

k1

12

2

2

2

b

y

a

x

232224

41 1

baxaba

R

R2: R2 = K2 A = x / sin (b)

here:

2 1

sintg

tg

ya

xbytg

2

2'

22

222 x

a

bby

22242

1baxa

bR

putting them into (b), getting:

y

xa

b

A(x,y).

..

k2

k1

232224

41 1

baxaba

R

22242

1baxa

bR

Special Points:

x=0: R1=R2=a2/b

x=a : R1=b2/a , R2=a

Special Points:

x=0: R1=R2=a2/b

x=a : R1=b2/a , R2=a

Conical shellConical shell

Find R1 and R2 of point A:

R1 = ∞

R2 = A K2 = r / cos

Small end:

R1 = ∞

R2 = 0

D

r

.A

k2

For spherical

segment:

R1=R2=RFor knuckle segment

of transition section: R1= r1, R2=

For cylindrical shell: R1=∞, R2= r

Dished shellDished shell

rr

r1r1

11

sinr

rr

ii. Basic hypothesis:ii. Basic hypothesis:

Small displacement hypothesisSmall displacement hypothesis

Straight linear law hypothesisStraight linear law hypothesis

Non-extrusion hypothesisNon-extrusion hypothesis

2.2.Free body balance equation(Free body balance equation( 区域平区域平衡方程式衡方程式 ))

—— —— calculation formula of calculation formula of meridional stressmeridional stress （经向应力）（经向应力） ::

i. i. Intercepting shell Intercepting shell —— —— uncovering the longitudinal uncovering the longitudinal

stress stress m

ii. Choosing separation bodyii. Choosing separation body iii. Analysis of stressiii. Analysis of stress

iv. Constitute balance equationiv. Constitute balance equation

0 0 zzz NPF

0sin 4

2 DSpD m

getting equation, above theinto

it putting 2/

sin2R

D

3)-(3 2

2

S

pRm

z

D

m

p

C C’

.k2R2

3.3.Infinitesimal balance equationInfinitesimal balance equation —— —— calculation formula of hoop calculation formula of hoop stressstress （微元体平衡方程式）（微元体平衡方程式） i. Intercepting shell i. Intercepting shell

—— —— uncovering the meridional stress uncovering the meridional stress m and circumferential stress σθ

ii. Choosing separation bodyii. Choosing separation body iii. Analysis of stressiii. Analysis of stress

iv. Constitute balance equation iv. Constitute balance equation

0P 0 ..n nnmn NNF

2

sin2 1221

dSdldlpdl m

02

sin2 21

d

Sdl

4)-(3 21 S

p

RRm

：整理得

k2k1

d2

d1

p

mSdl2

Sdl1

Basic calculation equation of membrane stress:

2

2

S

pRm

21 S

p

RRm

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, MPaS —— 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

S

pRm 2

2

S

p

RRm

21

Calculation equations:

1.Cylindrical shell subjected to uniform gas internal pressure: ∵ R1 = ∞ R2 = D / 2

Putting them into the

previous equations:

S

pD

S

pDm 2

4

s

D

p

2.Spherical shell subjected to uniform gas internal pressure: ∵ R1 = R2 = D / 2

Putting them into

equations (3-3) and (3-4): D

S

S

pD

S

pDm 4

4

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.

Solution:


R1:

y

xa

b

A(x,y).

..

k2

k1

1

''

2

32'

y

y

12

2

2

2

b

y

a

x

(a)

232224

41 1

baxaba

R

radius of curvature of A

By the elliptical equation: getting y’ and y’’, then put them into (a).result is:

R2: R2 = K2 A = x / sin (b)

here:

2 1

sintg

tg

ya

xbytg

2

2'

22

222 x

a

bby

22242

1baxa

bR

putting them into (b), getting:

(2)Find m and of point A:

Putting R1 and R2 into (3-3) and (3-4), getting:

2224

42224

2224

22

2

baxa

abaxa

Sb

p

baxaSb

pm

Stress of special points on elliptical shell:(1)x=0 (Top of elliptical shell)

b

a

S

pam 2

2

2

22

2 b

a

S

pa

S

pam

(2)x=a (Boundary or equator of elliptical shell)

Standard elliptical heads: The elliptical heads whose ratio of major and

short semiaxis a / b = 2 are called standard elliptical heads.

a / b = 2 ——

x=0 (Top):

x=a (Boundary):S

pD

S

pam 2

2

42

S

pD

S

paS

pD

S

pam

m

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:


R1 = ∞

R2 = A K2 = r / cos (2)Find m and of point A:

Putting R1 and R2 into (3-3) and (3-4)

respectively, getting:

D

r

.Ak2

cos

1

cos

1

2

S

pr

S

prm

Characteristics of stress distribution

of conical shell:

m

5.Cylindrical shell subjected to liquid static pressure: i. Supporting along the boundary of bottom Example: Known: Gauge pressure – Po (Pa) Liquid level – H (m) Density of liquid - (N/m3)

Find the m and of a random

point on the wall of cylindrical shell.

D

H

.A

x

Po

Solution:

(1)Meridional stress:

Cutting along section B-B, taking the

lower part as the separation body.

(H-x)

(Po+x)m m

N

D

H

.A

x

Po

B B

B-B

Establishing the balance equation of axial stress:

0 xF

222

4

4)(

4)( DHSDDxHDxp mo

S

Dpo

4 m

(H-x)

(Po+x)m m

N

(2) Circumferential (Hoop) stress:

Infinitesimal balance equation (3-4):

S

P

RRm

21

S

Dxpo

2

S

DHpo

2

max

For point A:

R1 = ∞ R2 = D/2 P = Po + x

Putting them into (3-4),

getting: when x=H:

ii. Supporting along the boundary of top Example: Known: Gauge pressure – Po (Pa) Liquid level – H (m) Density of liquid - (N/m3)

Find the m and of a random

point on the wall of cylindrical shell.

D

H

.A

x

Po

Solution:

(1)Meridional stress:

Cutting along section B-B, taking the

lower part as the separation body.

H-x

(H-x)

mm(Po+ x)

Po

D

H

.A

x

B B

B-B

Establishing the balance equation of axial stress:

0 xF

0 4

4

22 SDDxpDxH mo

S

DHpo

4

m

H-x

(H-x)

mm(Po+ x)

(2)Hoop stress:

Infinitesimal balance equation (3-4):

S

P

RRm

21

S

Dxpo

2

S

DHpo

2

max

For point A:

R1 = ∞ R2 = D/2 P = Po + x

Putting them into (3-4),

getting: when x=H:

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 :

D S

P

)MPa( 5.5284

4204

4

S

pDm

(MPa) 10522

mS

pD

(MPa) 5.524

S

pDm

(2)Upper head —— spherical

(3)Lower head —— elliptical

When a / b = 2:

a = D/2 = 210 mm b = a/2 = 105 mm

x=0 (Top):

(MPa) 10522

S

pD

S

pa

b

a

S

pam

(MPa) 5.5242

S

pD

S

pam

(MPa) 105 2

22 2

2

S

pD

S

pa

b

a

S

pa

x=a (Bottom):

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 Chapter 4 Strength Design Design

of Cylinders and Heads of Cylinders and Heads

subjected to Internal-subjected to Internal-PressurePressure

4.1 Basic Knowledge 4.1 Basic Knowledge

of Strength Designof Strength Design

1.1.Criterions of elasticity failure:Criterions of elasticity failure:ts eq

n

o

eq

eq —— equivalent stresso —— limiting (ultimate) stress, can be

s 、 b 、 n 、 D, etc.[] —— allowance stressn —— safety coefficient

Safety Allowance kept for the

requirements of safety:

2.2.Strength Theory:Strength Theory: i. The first strength theory —— the maximum tensile stress theory

231

max

][1 Ieq

ii. The second strength theory —— The maximum major strain theory iii. The third strength theory —— The maximum shear stress theory

Applying to brittle materials

Shear limit:

Failure condition:

Strength condition:

Applying to the plastic materials

2 :limitShear 0 s

0max

ss 3131 or

2 )-(

2

1 i.e.

][31 IIIeq

iv. The fourth strength theory

—— the maximum deformation energy theory

Applying to the plastic materials

])()()[(2

1 213

232

221 IV

eq

][ IVeq

Strength condition:

4.2 4.2 Strength Calculation ofStrength Calculation ofThin-walled Cylinder Subject Thin-walled Cylinder Subject to Internal Pressureto Internal Pressure

S

pD

21 S

pDm 42

03 r

1.Strength calculating equation: i. Determining the major stress

ii. Determining the equivalent stress

),,( 321 feq

S

pD

S

pDIIIeq 2

0231

3)-(4 ][2

tIIIeq S

pD

iii. Strength condition

According to the third strength theory:

iv. Strength calculation equation

4)-(4 ][ 2

t

pDS

Calculating pc from p.Putting them into equation (4-4), getting:

ct

ic

p

Dp

][2S

Solving equation (4-4) as following:

(1)Replacing medium diameter with internal

diameter: D + Di = S

(2)Introducing the welded joint efficiency :

5)-(4 ][2 c

tic

p

DpS

6)-(4 ][ 2 2C

p

DpS

c

tic

d

This is the design thickness.

(3)Eliciting the corrosion allowable thickness C2:

This is the calculated thickness.

(4)Adding negative deviation C1:

Getting the nominal thickness which indicated on the drawing.

(5)Calculating the effective thickness:

7)-(4 valueofround][2

S 12n

CCp

Dp

ct

ic

8)-(4 valueofround][2

)( 21

ct

ic

ne

p

Dp

CCSS

v. Equation of strength verification

9)-(4 ][2

)( t

e

eict

S

SDp

10)-(4 ][2

][ei

et

w SD

Sp

vi. Calculating equation of pw

2.Strength calculating equation of thin-walled spherical vessels:

ct

ic

p

Dp

][4S

2d ][4S C

p

Dp

ct

ic

valueofround][4

S 12

CCp

Dp

ct

icn

*Equation of strength verification:

*Equation of [pw] —— the maximum allowable working pressure:

*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)

t

e

eict

S

SDp][

4

)(

ei

et

w SD

Sp

][4][

Illumination of symbols:Pc —— Calculated pressure MPa

Di 、 Do —— Internal & external diameters of cylinder mmS —— Calculated thickness mm

Sd —— Design thickness mm

Sn —— Nominal thickness mm

Se —— Efficient thickness mm

C1 —— Negative deviation mmC2 —— Corrosion allowable thickness mmC —— Additional value of wall thickness mm —— Welded joint efficiency[]t —— Allowable stress at design temperature MPat —— 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

(no safety devices)

P≥Pw

With explosive media

and rupture disk

P≤(1.15~1.3)Pw

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

min

s , ][

sb

b

nn

min

s , ][

s

t

b

tbt

nn

min

D , , ][

n

tn

D

t

s

tst

nnn

nt Coefficien Safe

Stresslimit ][

0 t

High T Vessel

Medium T Vessel

Normal T Vessel

iv. Safe (Safety) coefficient n

Material

Strength

Performanceσb σt

s σtD σt

n

Safety

Coefficient

nb ns nD nn

Carbon Steel

Low Alloy Steel

≥3.0 ≥1.6 ≥1.5 ≥1.0

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

Single welded butt

(2)Single welded butt

NDE 100% = 0.9

NDE Local = 0.8

vi. Additional value of wall thickness C

(1)Negative deviation of plate and tube C1

Referring to the teaching material page , figure 4-7 & 4-8, selecting according to

the nominal thickness Sn.

(2)Corrosion allowable thickness C2

C2 = Ka 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 tT PP

][

][25.1

tT PP][

][15.1

PPT 25.1

PPT 15.1

(2)External Pressure Vessel

Pneumatic Test

Pneumatic Test

Hydrostatic Test

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% P△ T

degree difference increasing slowly P to PT.

*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

s0.9 2

)(

e

eiTT S

SDP

s0.8 2

)(

e

eiTT S

SDP

T —— Calculating stress at testing pressure,

MPa

s —— Yielding point at testing temperature,

MPa

(2)Pneumatic test

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

][2 c

tic

p

DpS

mm 5.6116.1711902

130016.17

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:

tT PP][

][25.1

aMP45.21190

1906.1725.1

*Efficient thickness:

Se = Sn - C = 65 - 2.8 = 62.2 mm

s = 410 MPa

2

)(

e

eiTT S

SDP

MPa 9.2342.622

2.62130045.21

MPa 36914109.0 9.0 s

ST 9.0

*Stress verification:

That is to say the strength in hydrostatic test is enough.

*Stress:

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

—— Whethert ≤ []t or not

(1)Determining the parameters

Pc = 15MPa Do = 219 mm Sn = 6.5 mm

MPanb

b 6.2613

8.784][

MPans

s 8.3186.1

12.510][

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

2

)(

e

eoct

S

SDp

MPa 1.291

5.52

5.521915

Obviously, t > []t = 261.6 MPa

So, 15MPa is too large, should be reduced.

(3)Determining the maximum allowable working P

][ 2

][eo

et

SD

Sp

MPa 48.135.5219

5.516.2612

So, the maximum safety P for this

cylinder is 13.48 MPa

4.3 4.3 Designing Heads subject Designing Heads subject

to Internal Pressureto 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

S

Calculating equation for thickness

][ 4 c

tic

p

DpS

][ 4 2C

p

DpS

ct

icd

valueofround][4

S 12

CCp

Dp

ct

icn

2.Thickness calculating equation of elliptical head

b

a

S

pam 2max

hi (

b)

ho

Di S

Ri (a)

i. Calculating equation

for thickness:

For the elliptical head

whose m = a / b ≤ 2

2. The maximum stress should be at the top point:

Putting m = a / b, a = D / 2 into the equation,

getting:

S

mpD

4max

t

S

mpD][

4max

t

mpDS

][ 4

Then:

Under the condition about strength:

(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:

i

i

ct

ic

ct

ic

h

D

p

Dpmmp

DpS

45.0 ][2

22

][2

m = a / b = Di / 2 hi

For the standard elliptical head whose m=2:

ct

ic

p

DpS

5.0 ][2

For the elliptical head whose m>2:

at boundary » and m at the top pointThen introducing the stress strengthening

coefficient K to replace (Di / 4hi)

In this equation:

2

22

6

1

i

i

h

DK

ct

ic

p

DpKS

5.0 ][ 2

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

e

eict

S

SKDp

2

5.0

ei

et

SKD

Sp

5.0

][2][

3.Dished head i. Structure Containing three parts:

Sphere: Ri

Transition arc (hem): r

Straightedge: ho (height)

Di

r

s

ho

R i

ct

ic

p

RpMS

5.0 ][2

r

RM i3

4

1

M —— Shape factor of dished head

and

ii. Calculating equation for thickness

iii. Working stress and the maximum allowable working pressure

e

eict

S

SMRp

2

5.0

ei

et

SMR

Sp

5.0

][2][

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:

ct

ic

p

DpS

5.0 ][2

2.1

4.Conical head i. Structure

*without hem (suitable for ≤ 30 o )

without local strengthwith 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.

cos

1

2 maxmax S

pD

m

tmaxmax ][

cos

1

2

S

pD

According to the strength condition:

Then cos

1

][2

t

pDS

cos

1

][2

ct

cc

p

DpS

Replacing P with Pc, considering , and

changing D into Dc ， D=Dc+S

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:

cos

1

][2

ct

cc

p

DpS

cos

1

][2

ct

sic

p

DpS

small aspect:

(2)Require reinforcing

(for the thickness of joint,

the reinforcement region)

Main aspect:

ct

icr p

DpQS

][2

ct

sicr p

DpQS

][2

Small aspect:

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

ct

sic

p

DpKS

5.0 ][2

ct

sic

p

DpfS

5.0 ][

K —— coefficient (Consulting Figure4-13)

f —— coefficient (Consulting Figure4-14)

(2)Thickness of conical shell at the joint with

transition section

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

max = r.max

The maximum stress is

at the edge of disk.

S

R

p

0

t

r.m

ax

r

2

max. 188.0

S

DPr

S

PD

S

R

275.0

▲ For simply supported ends

max = r.max = t.max

The maximum stress is

in the center of disk.

S

R

P

t

r.m

ax

0r

2

max. 31.0

S

DPr

S

PD

S

R

2 24.1

iii. Calculation equation for thickness

From the condition of strength max ≤ []t ,

getting:

fixed edge:

t

PDS

][

188.0

t

PDS

][

31.0

simply supported ends:

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:

tc

cp

PKDS

][

valueofroundS 12 CCS pn

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

][ 2 c

tic

p

DpS

mm 26.76.10.11332

12006.1

mmCSSd 26.80.126.72

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

= 1.0 (Double welded butt, 100% NDE)

(2)Semi-spherical head

][ 4 c

tic

p

DpS

mm 62.36.10.11334

12006.1

mmCSSd 62.40.162.32


Sd + C1 = 4.62 + 0.5 = 5.12 mm


= 1.0 (wholly punch forming)

(3)Standard elliptical head

ct

ic

p

DpS

5.0 ][2

mm 24.7165.00.11332

12006.1

mmCSSd 24.80.124.72


Sd + C1 = 8.24 + 0.8 = 9.04 mm



(4)Standard dished head

ct

ic

p

DpS

5.0 ][2

2.1

mm 69.86.15.00.11332

12006.12.1

mmCSSd 69.90.169.82


Sd + C1 = 9.69 + 0.8 = 10.49 mm



(5)Flat headK = 0.25; Dc = Di = 1200 mm; []t = 110 Mpa

tc

cp

PKDS

][

mm 36.72

0.1110

6.125.01200

mmCSSd 36.730.136.722

C1 = 1.8 mm (Checking Figure 4-7) Sd + C1 =73.36 + 1.8 = 75.16 mm



Comparison:

Head-form

Sn mm

kg

Semi-sphe. Elliptical Dished Flat

6

106

10 12 80

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 Chapter 5 Design of Cylinders Cylinders

and Formed Heads and Formed Heads subjected to External-subjected to External-PressurePressure

5.1 Summarization5.1 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

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 .

5.2 Critical Pressure

2.Factors affect the critical pressure i. Geometric dimension of cylinder

Degree of

vacuu

m in

failu

re

90175

0.51

(1)

90175

0.3

(2)

90

350

0.3

(3)

90350

0.3

(4) Pcr

mm HO2500 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.


*When the value of S / D is equal, the smaller the value of L / D, the higher the Pcr.


*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.

*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.

DmaxD

min %100minmax

DN

DDe

Ellipticity:

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:

2

1.122

o

et

e

ocr

e

crcr D

SE

S

DP

S

DP

3

21

2

o

et

cr D

SEP

3

2.2

o

et

D

SE

For steel cylinders: = 0.3

Calculating equation of the critical stress:

ii. Short cylinder

—— cylinders with small L / Do

Calculating equation of the critical P:

o

oetcr DL

DSEP

/

/ 59.2

5.2'

o

oet

e

ocrcr DL

DSE

S

DP

/

/ 3.1

2

5.1''

Calculating equation of the critical stress:

iii. Rigid cylinder

—— cylinders with small L / Do, large Se / Do Designing criterion: Only need to satisfy the strength condition:

compression ≤ [ ]tcompression

i.e. ][ 2

t

e

eict

S

SDP压压

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:

e

oocr S

DDL 17.1

ii. Critical length L’cr of short and rigid cylinder:

e

otcomp

et

cr

S

D

SEL

3.1

.

'

5.3 Engineering Design of

External-P Vessels

][ m

PPP cr

c

1.Designing criterions

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%

2.Nomograph for the thickness designing of the external-P cylinders

i. Calculating Steps

Step 1: L 、 Do 、 Se →

Drawing the curve

oeo DLSDf ,

2

1.122

o

et

e

ocr

e

crcr D

SE

S

DP

S

DP

o

oet

e

ocrcr DL

DSE

S

DP

/

/ 3.1

2

5.1''

2

B :Making tEm

o

e

D

SBPSo ][:

O

et

cr

t

cr

e

Ocrcr

Ds

mE

mP

P

E

SDP

2][

2

Step 2: Find the relationship between and [P]

2

B :Making tEm

3

2 tEB For cylinder m=3 and

o

e

D

SBPSo ][:

Then getting the relationship curve B = f ()

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 by the

following equation:A

3

2 tEB

(4)Calculating [P]

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

）（9 ][eoo

e

SD

B

D

SBP

(5)Comparing

cPPIf ~~ ][

cPPIf ][

cPPIf ][ i.e. the supposed Sn is too small and

should be increased appropriately,

repeating the previous calculating steps until

satisfying the first condition.

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 usable, safe

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


From Figure 5-9, we can see that point A is at the left of the curve, then the calculating equation is like following:

A 3

2 tEB MPa 78.12000102.0104.186

3

2 3

(4)Calculating [P]

][ eo SD

BP MPa 0834.0

152

78.12

(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:

mmL

L origin 34503

10350

3

)152SD( 9.11828

3450DL eoo Then



A = 0.00035


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]

][ eo SD

BP MPa 28.0

152

5.42


[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



A = 0.00022


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]

][ eo SD

BP MPa 27.0

102

5.27


[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)

eo SRA

125.0

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.

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

eo SR

BThen [P]

2)(

0833.0][

eo

t

SR

EP

iv. Comparison

cPPIf ~~ ][

cPPIf ][

i.e. the original assuming Sn is usable, and safety.

i.e. the original assuming Sn istoo small, Sn should be increased appropriately, repeating the previouscalculating steps until satisfyingthe 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

3

2.2

o

etcr D

SEP

o

o

e

tcr

DL

DS

EP

5.2

'

59.2

m

PP crAnd ][

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 ringAssuming 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:

o

oetcr DL

DSEP

5.2'

59.2

Putting equation (a) in, getting:

Then putting m=3 in, getting:

c

os

oetcr pm

DL

DSEP

59.2

5.2'

5.2

max 86.0)(

o

e

c

ots D

S

P

DEL

(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:

In the above equation:

L —— the calculating length of cylinder before

setting the stiffening rings, mm

Ls—— the space length between stiffening

rings, mm

1sL

Ln

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 Chapter 5 Components and and

Parts of Parts of VesselsVessels

6.1 Flange Connection6.1 Flange Connection1.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 chargingmedium

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)

S.O.flange

Pipeline Flange Vessel Flange

W.N.flange

(2)Simple [loose (type), lap joint, lapped] flange

Interlink on the turn-down rims

On the welding ring

(3)Screwed flange

Square flange Elliptical flange

3.Factors effect the sealing of flangesi. 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 Face ）

(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 flangeBolt circle, Outer diameter, ThicknessBolt 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

of medium

TemperatureTemperatureTemperatureTemperature

PressurePressurePressurePressure

Corrosive CharacteristicsCorrosive CharacteristicsCorrosive CharacteristicsCorrosive Characteristics

Penetrant CharacteristicsPenetrant CharacteristicsPenetrant CharacteristicsPenetrant Characteristics

Combined effectGreatly affecting the sealing

4.Standard and Selection

of Flanges

i. Standard number of pressure vessel flanges

JB / T 4701-2000 JB / T 4703-2000 ∼

A-S.O.FlangeJB/T 4701-2000

B-S.O.Flange

JB/T 4702-2000

W.N.Flange

JB/T 4703-2000

Plain PlainM&FM M&FMT&G

M FM T G

P

P

P

P

P

P

A

A

A

A

A

A

T

T

T

T

T

T

S

S

S

S

S

S

C

C

C

C

C

C

Type

Sealing

Code

Without lined ring With lined ring (C)face

Standard types and marks of pressure vessel flanges

T&GM FM T G

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

C-S 800 — 1.6 JB4702-92

NominalPressure

MPa

NominalPressure

MPa

Code of Flange TypeCode of Flange Type

Code of Sealing face Type

Code of Sealing face Type

NominalDiameter

mm

NominalDiameter

mm

StandardCode

StandardCode

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 vessels6.2 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

120°

Gasket

Web-plate

Anchor bolt

Sub-plate

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 TypeModel Type Nominal DiameterNominal Diameter TypeType

2.Checking calculation of stress in double-saddle horizontal vessels

i. Load analysis for horizontal vessels

Shearing ForceDiagram

Bending Moment Diagram

M1

M2

M3

F FA A

q

ii. Reserved force to support

In this equation:q —— Mass load/unit length of vessels, N / mmL —— Distance between the T.L. (tangent lines) of two heads, mm

hi —— Height of curved surface of heads, mm

2

mgF

ihL

qmgFor

3

4

22

iii. The maximum radical bending moment

The section across the middle point of moment

The section at support

mm)(N

4

3

41

21

4

2

22

1

L

A

L

hL

hRFL

Mi

im

mm)(N 4

3

41

21

4

2

22

1

L

A

L

hL

hRFL

Mi

im

mm)(N

34

1

21

1

22

2

Lh

ALhR

LA

FAMi

im

mm)(N

34

1

21

1

22

2

Lh

ALhR

LA

FAMi

im

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):

The most lowest point in section (Point 2):eme

mc

SR

M

S

RP

2

21

1

eme

mc

SR

M

S

RP

2

21

1

eme

mc

SR

M

S

RP

2

21

2

eme

mc

SR

M

S

RP

2

21

2

(2)Stress in the section of support

The most highest point in section (Point 3):

The most lowest point in section (Point 4):eme

mc

SRK

M

S

RP

2

21

23

eme

mc

SRK

M

S

RP

2

21

23

eme

mc

SRK

M

S

RP

2

22

24

eme

mc

SRK

M

S

RP

2

22

24

v. Checking calculation for stress of cylindertensile stress

compressive stress

ttensilecomb max. t

tensilecomb max.

luesmaller va the

max..

Bcr

t

compcomb

luesmaller va the

max..

Bcr

t

compcomb

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 —— The maximum stress

at the boundary of opening

* —— The maximum basic

stress of shell

max

K

Small opening in plate

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)

Shell

Stiffening Ring

(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)

δmin

25 32 38 45 48 57 65 76 89

3.5 4.0 5.0 6.0

3.Designing methods of equi-area reinforcement

Metallic areas in local reinforcementMetallic areas in local reinforcement

≥≥

the area of sections which are

the position of openings

the area of sections which are

the position of openings

i. Confirmation of the effective range of opening and reinforcement areas

A1 A3 A4A2A

B

h1

h2

d

Effective width:

Effective [working] height:

Outside height

Inside height

max .22

2

tnn SSdB

dB

min1

.1

nozzle ofheight overhang actual

h

Sdh tn

min2

.2

nozzle ofheight embedded actual

h

Sdh tn

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 A1 + A2 + A3 + A4 ≥ A

6.4 Attachment of vessels

1.Man Hole and Hand Holei. 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 ]

Documents

Section 2 Design of Process Vessels Chapter 2 Panorama