A process vessel is to be designed for maximum operating pressure@

of .$._01 kNl_m 2 • An understanding between purchaser and manufacturer

indicates hat the vessel should be over designed considering 6% extra to
maximum working pressure. Vessel has· outer diameter of 1.5m. The vessel
is made of IS 2002-1962 Grade 2A and its design temperature is 435°C.
Corrosion allowance of 2 mm is suggested for expected life of vessel. It is
fabricated according to either Class 1 or Class 2 of Indian Standard
specification where single welded butt joint with backing strip is used.
What will be standard plate thicknesses to fabricate the vessel for
cylindrical as well as spherical vessels?
 Maximum operating pressure (abs.) = 501 N/m 2
P abs = P alm +~Pe
P gau e = P abs - P at =501 - =
101 .325 399.675 kN/m2 - .
P = t5esign pre re = 1.06 x 399.675 = 423.6555 kN/m 2
= 4.32 kgf/cm 2
Do= 1.5 m = 1500 mm
IS 2002-1962 Grade 2A and its design temperature is 435°C
f = 4.3 kgf/mm2

9 111 100111! 'c"

111 0 1
Nl'" c
11111,cA110N gra de 2A and ·tI s des1gn
. t emperat ure .1s 435 .
Corrosion allowance of 2 mm is given. Now
 Cylindrical , ~ q
Class 1 \} Class 2 :l :.
=
tm,n POo/(200fJ +P) =8.3256 mm tm,n = 9.36 mm
trinal : 10.3256 mm tr,nal = 11 .36 mm
lstafldard = 11 mm ~taodard = 12 mm

Table B-1 teel Plates

Tbkkoess : 5, 5,6,7, ,9,I O,IJ,1 2,

(ffllO) 14, 16, I ·, :o, 22, 25, 2 ,
32, 36. 40, 4 5, 50, 56, 63,
71. 80. 0.8 in this case for class 2. So for class
1 we have minimum thickness as 8.3256, adding
 Spherical '-""'
Class 1 ✓ Class 2
J = 0.9 J = 0.8
~ = _e.Qo/(400fJ +P) = 4.17 mm tmm = 4.69 mm
tflnal : 6 .17 mm tfinal :6.69 mm V--
t standrud : 7 mm ~ tandard = 7 mm

TaQ B-1 S:eel Platts

Thickness : 5' 5 5. 6 .61

b. 9 1u. I I 12.
I I

(mm) 14. 16 • ._.

32. 36.
71. 80.
 Manufacturer has four different materials to construct the pressure vessel
having diameter and 4m len th. Materials are IS:2002-1962 Grade 28,
IS:2041-1962 20MoSS, IS:1570-1961 15Cr90MoSS and IS:3609-1966
1%Cr0.5%Mo. The maximum operating pressure (abs) and design
temperature are 450 kN/m 2 and 475° C, respectively. Corrosion allowance
of 3 mm is considered. Design a cylindrical vessel for each of these
materials for J=0.8. If manufacturer is havi ng 5 ton material of each
category, calculate the saving of material in shell for each design (if any).
Density of each material may be taken as 8000kg/m 3 •

Ci) Kl 1 001111 ~;~~?c"::~.. ,o we can consider this as a cylindrical shell.

Materials are IS: 2002-1962 grade 28. This
 Do= 2 m = 2000 mm , L = 4 m
Maximum Operating P (abs.) = 450 KN/m2
P9aug~ = 450-101 .325 348.675 KN/m2
P = Design Press = 1.05 x 348.675
= 366.10875 KN/m2
= 3.73 kgf/cm 2
(a) IS:2002 - 1962 Grade 28
J = 0.8, f = 3.6 kgf/mm 2
tm,n = PDo/(200fJ +P) = 12.868 mm
tfinal = 15.868 mm
tstandard = 16 mm

e 111 10011(!
H,tll 0 NUNI
cum,c..noH•
so here maximum
. operat·1ng pressure 450 .1s
given. We will first convert that into gauge
(b) IS: 2041 -1962 20Mo55
f = 7.7 kgf/mm 2
tmin = 6.037 mm
tr,nal = 9.037 mm
tstandard : 10 mm

(c) IS: 1570 - 1961 15Cr90Mo55

f = 11 .7 kgf/mm2
t,mn = 3.977 mm
tr,nal = 6.977 mm
tstandard = 7 mm
(d) IS: 3609 - 1966 1%Cr 0.5% Mo
f = 9.7 kgf/mm 2
t min = 4.795 mm
trina, = 7.795 mm
tstandard : 8 mm
A process vessel is to be operated at maximum operating pressure of 2
MN/m 2 {g). The vessel has diameter of 3 m and length of 5 m. This vessel is
made of steel having allowable design stress value of 120 MN/m 2 at design
temperature. The corrosion allowance is suggested to be 3 mm for life span
expected for the vessel which is to be fabricated with a weld joint efficiency
factor of~
{a) Determine the thickness of the shell to nearest integer. I_/"
{b) Determine the thickness of the flat head having plates welded to the
.. _ end of the shell {no inside welding).
-
Solution Pgauge(Maximum operating) = 2 MN/ m12
P= Design Pressure=2.1 MN/m 2 --

f=120 MN/~ D0 = 3 m --
J=1 _. C.A= 3mm
(a) tmin= 2;::p=26.022 mm --
trrna1=29.022 mm \./
Tstandard=30 mm v' Flat head having plates welded to the end of
the shell (no inside welding) C HOT LESS rttAN 1•2st,

(b) t=CDe v(;)~ D =D·

e I
C=0.7 .
---

De=3-0@*2 = 2.94 m
T min = 272.248 mm
No addition of
Tfinal = 272.248 If tmin > 30 mm corrosion
Tstandard = 273 mm allowance
Solution 120° apex angle (with sharp edges i.e. no knuckle part) both at and away
f rom junction .

(C) (i) At the junction 2a is apex angle

t= pDeC \./"" De = Do a=4J=6QO C = 3.2
2fj -
tmin = 84 mm
tstandard =t final =84 mm

~ 0-01 0·02 0·03 0·04 0·06 0·08 0·10 0·15 0·20 0·30 0·iO 0·50

10• 0·70 0·65 0·60 0·60 o·ss 0·55 0·SS 0·55 0·SS 0·55 0·55 0-55
20• 1·00 0·90 0·85 0·80 0·70 0-65 0·60 0·55 0·55 0·55 0·55 0·55
so· 1·35 1·2 l'l l ·0 0 ·90 o·as 0·80 0·70 0 ·55 o-,, 0·55 0·55
45• 2·05 1·85 1·65 1·5 1·3 1·2 J• l 0·95 0·90 0·70 0·55 O·S!
60• S·2 2·85 2·55 2·35 2·0 1•75 1·6 1·4 1·25 1·00 0·70 0-55
411\ ( NPIEL ONLINE 75• 6·8 S·8S !i·SS 4·75 S·85 S·S S· IS 2·7 2·4 l •65 l•OO 0·55
• IIT ROORKEE """ CERTIFICATIO
Solution 120° apex angle {with sharp edges i.e. no knuckle part) both at and away
from junction .
Actual thickness (at the junction)
Away from the junction
L=0.5 ~ -
✓~
L = 0.354965 m
Dk = D0 -2t S-2L sina
= 3 - 2 * 0.03 - 2 * 0.354965 sin 60 = 2.325183 m
O 1
t . =
min
P * - - = 41 .049
/i
2/J- P cosa
mm
= 41 .049 mm
t final

tstand = 45 mm
Solution 100° apex angle with ri=0.05 De and 4)=25°.

2 a =100° (apex angle)

a=50°
r 1=0.05 De 4'=25°

At the junction
ri/ De =0.05 4'=25° Now,
r;/ De =0.04 4'=200 C=0.8 r;/ De =0.06 4'=20° C=0.7
r;/ De=0 ..04 4):300 C=1 r;/ De=0.06 4'=30° C=0.9
ri/ De=0 . 04 4'=25° C=0.9 ri/ De=0.06 4'=25° C=0.8

D = 0 . 05
!J.. 4' = 25° C = 0.85
e
tmin=
( DC) = 22.3125mm
p2fj tfina1=25.3125 mm
 ;a06
-

0-01 0·02 0·03 - 0-08 0·10 O·l5 0·20 0•3() 0·40 0·SO

0·70 0·65 0·60 0·60 0·55 0·55 0·55 0·55 0·55 0·55 0·55 0-55
100° ape>
Solution ~
20•) 1·00 0·90 0·85 0·70 v 0-65 0·60 0-55 0•55 0-55 0·55 0·5.S
90• 1-, :; 1·2 I·I 1·0
~
0•90 ✓ 0·8$ 0·80 0·70 O·!,j o-,, O·,_, 0·3'

2 a =100° (apex angle) 45• 2·05 1·85 1·65 1·5 1·3 1·2 I· I 0-95 0·90 0·70 0·55 0·s.'
60• 5·2 2·85 2·55 2·35 2·0 1·75 1·6 1·4 1'25 1·00 0·70 0-55
a=5O°
75• 6·8 5·85 5•9.S 4·75 9·8S 3 ·S 9·15 2·7 2·♦ 1·5!i 1·00 O·.SS
r 1=O.O5 D0 '+'=25°

At the junction
r;/ De =0.05 '+'=25° Now,
v
✓r;/ De =0.04 4J=2Oo C=O.~ r;/De =0.06 4J=2O0 C=0 .7 v
v r;/ De=0.04 4):300 C=1 r;/ De=0.06 4):300 C=O.9 v
r;/De=0.04 '+'=25° C=O.9 V r;/ De=0.06 '+'=25° C=O.8 ~

!:.l.. = 0.05 4J = 25° C = 0.85 t' =26.83 mm

De
tmin=(p2i1c) = 22.3125mm tfina1=25.3125 mm
-
tstand-28 mm
. \._....,/
Solution

Away from the junction v

L= 0.5 ~ =0.15222 m V
✓~
Dk = D0 -2tS -2Lsina
= 3-2*0.03-2*0.15222 sin50
= 2.706785 m v
tmin = 37.1715 mm . \..,/

t stand = 40 mm
 Determine the thickness of the plate required to fabricate a dished head for
Solution
the vessel. Specification for the head are Ri= l.1D 0 ; ri=0.07D 0 ; St= 70 mm.

(d) Ri = 1.1D0
ri = 0.07D 0
St = 70 mm
;.= (:i)
=Ri =3.3 m
Take R0 rO = ri = 0.21 m

ho=Ro-j( R (~•)) •(Ro+C•)- Zro)

0 -
J¥ = 0.5612486m

= 0.4921538 m
D5/4Ro = 0.681818 m
hi =minimum of the above 3 =h 0
= 0.492 1538 mm
•I .ll2.
 Dete rmine the thickness of the plate requi red to fabricat e a dished head
Solution
for t he vessel. Speci fication for t he head are Ri=l.1D 0 ; ri=0.07D 0 ; Sr= 70mm .

Assuming opening is compensated:

h E =0.164
Do

..!....
Do
= (!!..£)
2[}
= 8.75*10·3 C
C = 1.85 (by interpolation)
t = 0.04856 mm
tmin= 48.56 mm
tfinal = 1.06*tmin=51 .4 76 mm
tstand = 56 mm
 .00875
Determ ine the th ickness of the plate requ ired or
Solution .005
to fabricate a dished head for the vesse l. .3912
3.92 0.002 Specification for the head are Ri=l.1D0 ; .002091
2.3912 0.005 ri=0.07D 0 ; Sr= 70mm . or
1.954 0.01
1.7876 0.02 .005118
; =0.164
1.6296 0.04 () .02
1.7876
.011188
Without Opening or With Fully Compcosalcd Opcoings 6 .015
t!Do 1.8708
0.005 0.01 0.02
.008018
It.J D. 0.002
4. 55
--
2.66 2.15 1.95
-01.75
04
.016
0. 15
0.20 2.30 1.70 1.45 1.37 1.32 1.85416
0.25 1.38 1.14 1.00 1.00 1.00 .008629
0.30 0.92 0.77 0.77 0.77 0.77 .0162
0.40 0.59 0.59 0.59 0.59 0.59
0.55 0.55 0.55 0.55
1.850832
0.50 0.55
.008753
 Det ermine the t hickness of the plat e requ ired t o fabricate a dished head
Solution
for the vessel. Specificat ion for the head are Ri=l.1 D0 ; ri=0.07D 0 ; Sf= 70mm .

Assuming opening is compensated:

~ E =0.164 v
0
✓

_t
Do
= (E.£)
2tJ
= 8.75*10· 3 C

C = 1.85 (by interpol ation ) ✓

t = 0.04856 mm
tmin= 48.56 mm
tfina,= 1.06*tmin ,51.476 mm
tstand = 56 mm
 Determine the th ickness of the plate requ ired to fabricate a dished head
Solution
for the vessel. Specification for the head are Ri= l.1D 0 ; ri=0.07D 0 ; Sr= 70mm .

Now since flt > 10 mm

R0 = Ri+t5 = 3.356 m 1
ro= ri +ts = 0.266 m )
✓ h 0 =0 . 523096 m
tmin=44.625 mm
DJ =Q.670441 m V /oJo =0.631664
✓z
tfina1=1 .06*tmin=47.3025 mm
/4Ro tstand=50 mm
hE=h 0 =0. 523096
;E=0.174 ; = 8.75*10·3 C ~t<10 mm hence no need of
0 o
extra calculation
C = 1.7
 Determine the blank diameter of the plate for form ing the dished head.
Solution

(e) Since t>25 mm

Blank diamet~r = D0 + (:;) + (~) r 1 + 2S1 + t
=3.4014286 m
Example-2
A vessel is being operated at maximum pressure of 2 MN/m 2 (g). It has outer
diameter of 2.Sm and length of Gm. The allowable stress of shell and corrosion
allowance is 150 MN/m 2 and 3mm, respectively. The vessel is of Class 2 where
double welded butt joint with full penetration is used for all joints. Compute the
followings:
(a) Thickness of plate used to fabricate a _lorjspherical head for: this vessel. In this
two fullv compensated openings are provided with diameters of 0.lm and 0.2m.
Consider Ri = 1.2 D0 , ri = 0.063D0 , St= 60mm. Carry out only one iteration.
(b} What is the maximum diameter of uncompensated opening can be made if 2:1
ellipsoidal head (hE=0.25D 0 ) of same thickness as computed in Part (a) is used.
(c) Thickness of flat formed head when flanged heads are butt welded to the shell.
 Design pressure = p = 2.1 MN/m 2
Solution Thickness of shell = t = 0.02042 m
t with corrosion allowance = 23.42 mm
t standard = 25 mm
Assumption Ri = Ro ri=ro
(a) Thickness of plate used
Calculation of hE Ri=l.2Do 3 Ro
to fabricate a torispherical
ri=0.063Do 0.1575 ro
head for this vessel. In this
0.52083 m
two fully compensated
6.88625
openings are provided with
0.37583 m
diameters of 0.lm and
0.19688
0.2m . Consider R-' = 1.2 DOf r-I
0.44371 m
= 0.06300 , St = 60mm . Carry
0.37583 m
out only one iteration.
0.15033 m
Solution
(a) Thickness of plate used
to fabricate a torispherical
head for thi s vessel . In this
two fully compensated
openings are provided with
diameters of 0.lm and 0.2m.
=
Consider Ri 1.2 0 0 , ri=
0.06300 , St= 60mm . Carry t = 0.041506 ✓
out only one iteration. 0.043996 ,1
t s = 45
 (b} What is the maximum diameter of uncompensated opening can be
Solution
made if 2:1 ellipsoidal head (hE=0.25O0 } of same thickness as computed in
Part (a) is used.

'f-- 200fJ
pDoC with Uncompensated Openlog5

0.008235 d/yt Do
t/DOC =
hE/Do = 0.25 hs/Do 0.5 1.0 2.0 3.0 4.0
-.- -5.0
C= 2.016 0.15 1.67 l.86 2.15 2.65 3.1(, ·3.60
12.23529 0.20 1.28 1.45 1.85 2.30 2.15 3.25
d/sqrt(t* D0 ) = 2.924444 ·0.25 J.00 I.] 5 l.60 2.05 2.50 2.95
d = 0.942037 m 0.30 0.83 1.00 1.45 J. 88 2.28 2.70
0.50 0.60 0.80 I.IO 1.50 l.85 2.15
 (c) Th ickness of flat formed head when flanged heads are butt welded
Solution
to the shell.

I
CD • /p
•7o 'VJ

C= 0.45
D= Di = ~
p/f = 0.014
t= 0.13045
0.13828
139 mm
An absorption tower is 5.2 m in outside diameter by 10 m in length from tangent
line to tangent line of closures. The tower is to be designed for full vacuum i.e.
p=0.1 MN/m 2 (g) at 400 °C. Assume a standard dished head at one end of the tube
and a flat head at the other end of the tower. Specification for the standard dished
'
head are ~ ;=~ ; ri=0.~ D0 ; S1=50 mm. The material of construction of the tower is
carbon steel having allowable stress value of 10 m 2 and E = 2 x 105 MN m 2•
(a) Determine the required thickness of the shel without stiffeners. L--""'
(b) Determine the required thickness of the shell with stiffeners located at 1 m
spacing. v
(c) Estimate the savings in shell material if a 18 cm channel is used as stiffeners
rjngs._Given: Weight of Channel=14.6 kg/m, Moment of Inertia of Channel =
~.9xl0·3 m4, U=l.5%, density of shell and stiffener material= 7850 kg/m 3 •
Solution
h;=R;-j(Ri - G')) • (Ri + (~;)- 2ri)
D0 =5.2m=D; R;=5.2m R;=D0 =5.2m r;=0.05*5.2=0.26m
St=0.05m ~~
- ~-
h;=0.8494m
L'=10+(1/3)*0.8494+0.05 = 10.333 m

D0 / L' =0.503
D0 / L' =0.4 k=0.246 m=2.43
D0 / L' =0.6 k=0.516 m=2.49
/ 0.4-0.503_ 0.246-/(
Do/ L =0.5o3 o.503-o.6 1<- 0.516
k=0.38506
m=2.4609
 I\.
-0.733 -m-
3 00
0.1 0. 165 2.60
0.224 2.54
0 229 2.47
Solution l .246
1
2.43
0. 516 2.49
0.660 2.48
0 879 2.49
1. 572 2.52
2.364 2.54
5. 144 2.6 1 ·
4.0 9,037 2.62 ·

D0 I L' =0.503 v 5.0 10.3 59 2.58

D0 I L' =0.4 k=0.246 m=2.43 LJ

D0 I L' =0.6 k=0.516 m=2.49
Do ILi =0 .503 0.4-0.503 0.246-K
0.50 3 -0.6 /(-0.516
k=0.38506
m=2.4609
Solution Determine the required thickness of the shell without stiffeners

Plastic failure
t 1
Elastic failure 2
P= f( Do) (1.s( 1-0.2(%)))
P=KE (...£_)m 1+ 100(t)
-Do
Do
P=0.1 MN/m 2
tfina1=21 .08 mm =2*1 ooc!:·zi) ( 1
+ 1.5•1.S(:-0.2•0.503))
100•(21.08)
tstand=22 mm 5200
=0.1353 MN/m 2 >0.1
Therefore, the calculated thickness is safe
against plastic deformation also
t=22 mm (without stiffness)
 Determine the required thickness of the shell with stiffeners located at
Solution
lm spacing
Do-5 2
L' =1 m L'- ~,._,,, \
.
k=10.623
m=2.588
Elastic failure
P=KE c;0
)m . tfina1=7.67 mm
tstand=8 mm
Plastic failure
Do=B >5
L'
P=2f(; ) = 0.295MN/m 2 >0.1 Therefore, the calculated thickness is safe
t=B mm (with stiffness) against plastic deformation also
 Estimate t he savings in shell material if a 18 cm channel is used as
Solution
st iffeners rings.

1=8.9*1o-3 m~ ✓
Weight=14.6 kg/m f=100 MN/m 2
As.=1.84*10-3 m2
1 94 10 3
D5L(ts+(~))t . . 2
5.2 •1 (o.008+( · •1 - ))•100
le= L (t with stiffness)=
12•£ 12•2•10 5

= 1.056*10-5 m2

le < I stiffness is enough to give the rigidity

Solution Estimate the savings in shell material if a 18 cm channel is used as
stiffeners rings
No of stiffness= (total length/spacing)-1
= ( 10/ 1 )-1 = 9
Total weight of rings
= no of stiffness *TT*D 0 *W
= 9* TT*5.2*14.6
= 2146.587 kg

Saving in shell material for using stiffness

= [TTDo[twithoutstiffness -twith stiffnessl*tangent to tangent length*p]-total weight of rings
= TT*5 .2*(zz- a)*10*7850-2146.587
1000
= 15806.99 kg
 ~~JLI,,o.-~f 3m outer diameter and Sm length is to be operated at
MN/m 2 (g). Flat heads are placed at both ends of the vessel.
Allowable stress and modulus of elasticity for material used are 90
MN m 2 and 2.lx105 MN/m 2, respectively.
a. Compute the standard thickness of shell if stiffeners are placed at
0.8m spacing.
b. Examine the suitability of ISLC 175 steel channel stiffener for this
vessel.
Solution
Design pressure - 0.06 MN/m 2
Outer diameter of shell= Do - 3 m
Length of shell= L - 8 m
Allowable stress = f -- 90 MN/m 2
E - 210000 MN/m 2
u - 1.5 %
Solution Compute the standard thickness of shell if stiffeners are placed at
0.8m spacing.
Effective length L' = 0.8 t.---
p =K E(t I D o)
111

Elastic failure Do/ L' = 3. 75 ~

D,,/l K m
- K = 8.06375 ~
0 0.733 3 00
m = 2.6175 ~
0.1 0. 1!IS 2.60
0.2 0 224 2.54 1/m = 0.382044
0.3 0 229 2.47
p/KE = 3.54E-08
0.4 l 246 2.43
0.5 16 2.49
= 0.001424
0.6
0.8 0.660 2.48 t = 0.004272 m
1.0 0 879 2 49
1.5 1.572 2 .52
2.0 2.364 2.54

3.0\ 5. 144 2.6 1 •

Solution Compute the standard thickness of shell if stiffeners are placed at
0 .8m spacing.
Plastic failure 3.949854
0.202026
p= ~ @ Notsatisfied

Plastic failure p =2f(-

'J l
D() I + 1.5 U(1- 0.2JJO I L )

final t = 0.00464 m
100(- J 1

Do
5 mm
 Examine the suitability of ISLC 175 steel channel stiffener for this
vessel
S,c1io,,o/ Dqtlt of Width of Tlt1ckMu Ctntrt of Momtnts of llltnlo
As 0.00224 m2 Dt1l1na1/on
Ortd ( A )
cm'
sttttan (It) /fllJtft (b) of ,.Yb (t, ) ,,a,tty (<..)
mm mm Ctn ,.. (cm' )
Rodi/ of l)'ratlo11
(cm)
!!!.!!.' ~ .!!!_ !i•_
I= 1.27E-06 m4 ISJ C 100 7.4 1 100 4~ 3.0 1.40 113 8 14 9 4.09 1.42
125 10.00 12S so 3.0 1.64 2700 2S.7 S. 18 1.60
ISO l26S ISO ss 3.6 1.66
Requi red I 2.0lE-06 m 175 14.24 17S 60 3.6 1.75
47 1.1
719.9
37 9
so.s
610
7 II
1.73
1.88

2" '"
200 17.77 200 10 4. 1 1,97 I 161.l B-42 I 08 2. 18
> 1.27E-06 100 1002
7S
100
40
so
37
40
I.JS
1.62
66. 1
164.7
11.S
14 I
3 02
4.06
1.16
125 1367 12S 4,4 I.S7
6S 2.0. 3S6. S7.2 S. 11 2.0S
Not suitable ISO ~
115
ISO
· 175
75
7S
◄.8
S. I
2.38
l.◄O I
697.2
141.4
103 l
126 S
6.16
7. 16
2.37
2.38
200
225
2 .12 200 7S s.s 2.3S I 72S S 146 9 I . II 2 .37
30.S3 22S 90 5.1 2.46 2 S47.9
lSO 3s.6S 209.S 9. 14 2.62
lSO 100 6. 1 2.10 3 617.9 249.4 10. 17 2.89
300 41. 11 300 JOO 6.1 2 ss 6 047.9 346.0 11.98 287
3SO 49.47 lSO 100 7.4 2.41
400 ,n! 9 312.6 39~.6 13.72 2.82
400 100 ao 1.36 13 919 S 460 4 IS. SO 281
ISMC 7S 8.67 15 40 4.4 I.J I 16.0 12.6 2.96 1.21
100 11.70 100 so 4.7 I.Sl 186.7 2S.9 4.00 1.49
l 2S
ISO
16. 19 l2S 6S s.o 1.94 416.4 S9.9 S.07 1.92
20.&8 ISO 7> S.4 2.22 779 4 102 3
l7S 24.38 l7S 6. 11 2.21
1S S.7 2. 20 I 223 3 121 0 7.08 2.23
200 28.21 200 7S 6.1 2.17
2lS I 819, 3 140,4 8.03 2.23
33.0J 22.$ 80 6,4 2.30 2 61M.6
2SO 38.67 117 2 9.03 2.38
2SO 10 7. 1 2.30 3 SIU
300 219. 1 9.94 2.38
◄S.64 )00 90 7.6 2.36 6 362.6
JSO SJ.66 )SO 310 8 11.81 2.61
100 8. 1 2.44 10 008 0 4306
400 13.66 2.83
IITROOUH ( NPlfl ONUNE 62.93 400 100 u 2 42 IS 082,1 504.8 IS.48 2.IJ
Example-1
Design a loose- e flange with a plain face for a reactor shell with 1.8m outside
diameter and O.O18 1ckness (g0). Other specifications are: Design temperature
-
: 200°c; Design pressure =2.2 MN/m 2; Allowable stress of flange material =120
MN/m 2; Allowable stress of bolting material = 120 MN/m 2; gasket materiaf
Corrugated soft Al metal, asbestos filled [Min. design seating stress (y) = 20
MN/m 2; Gasket factor (m) =2.5; and Min. actual gasket width =10mm]. Ratio of
gasket internal diameter to shell outside diameter is 1.01 ; corrosion allowance =
zero; weld joint efficiency factor = 1.
, Calculate effective gasket seating width
, Calculate minimum bolting area
, Which amongst the following bolts will be u ed for bolting the flange: M
36x3, M39x3, M42x3and M45x3? Give g1 1.41 g0~
, Estimate bolt-circle diameter. v
, Estimate flan e outside diameter after addition of 2 cm assumed gap
between end of bo c1rc ean e dof the fla~-9s
, Estimate various loads and moments under operating as well as bolting-
up conditions
, Estimate flange thickness (Poison'sratio =0.3)
Solution Calculate effective gasket seating width ~

Min gas ket width (N)

d o - di
N = - - =78mm
2
di = 1.01 x 1.8 = 1.818 m
N = 78 mm >10mm

di
00 =
( y - pm
y - p(m +
)½
1)
:. Nnnal = 78 mm; b 0 =
N
2 = 39 mm
Effective gasket width, b
20 - 2.2 X 2.5
do = ( 20 - 2.2(2.5 1) +
)½ x 1.818
b = b 0 ifb 0 < 63mm

= 2.5 ./brJ if b 0 > 6.3 mm

= 1.974 m
:. b = 15.61 mm
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 Calculate minimum bolting area ~ Allowa ble s tress fo r
Solution bolt n1 ate ria l at
Dia at bottom of gasket load reactio@
Wg = n(G x b) x y d es ign te n1p .
61
G = di + N if b 0 < 6.3 mm = TI (1.94278 X lS.
1000
) X 20

= do - 2b if b 0 > 6.3 mm = 1.0955 MN Allowable s tress

G = 1.974 - 2
x is.61 = 1 94278 Sg = s0 = 120 MN/m2 fo r bolt n1ate ria l
1000 · m at atn1. te mp.
Wo
Ao = ~ = 0.063083 m
2 2 2
H= 1tG P = 1t( 1.94278) x2.2 = 6 522 MN
4 4 .

Hp = n[G x 2b] x mP = 1.048 MN Wg 1.9055

Abe = S =
120
= 0.01588 m 2
w 0 = H + Hp g

= 6.522 + 1.048 = 7.57 MN Am = 0.063 m 2

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 Which amongst the following bolts will be used for bolting the fl ange:
Solution
M 36 x3, M 39 x3, M 42 x3 and M 45 x3?

g0 = 0.078 m
n= Actual No. of bolts = 92
= 0.02547 m
n8 5 92 x 80
For M 36x 3 C1 = - = 2.343 m
TI TI
Root Area = n (36 - 6) 2 C2 = ID+ 2(g1 + R)
4

= 706.858 mm 2 C2 = 1.8 + 2(0.02549 + 0.05) = 1.951 m

Minimum No. of bolts = Am

C1 - C2 = 0.392 m
Root Area
= 89.127
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 Which amongst the following bolts will be used for bolting the flange :
M 36 x3, M 39 x3, M 42 x3 and M 45 x3?
For M 39x3

Root area= n (bolt dia - 2t) 2 =855.2986 mm 2

4

Minimum No. of bolts R Am = 73.658

oot a rea
n =Actual no. of bolts=76 ·
n8 5 76 x 86
C1 = TT
- --
TT
- = 2.0805 m
C2 = 1.8 + 2(0.02547 + 0.052) = 1.9549 m
C1 - C2 = 0.1256 m
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Solution Which amongst the following bolts will be used for bolting the flange :
M 36 x3, M 39 x3, M 42 x3 and M 45 x3?

For~ 4~3
Root Area= 1T ( 42 - 6) 2 = 1017.876 mm~
4

Minimum No. of bolts=61.89

n = Actual no. of bolts= 64
64 X 91
C1 = - - - = 1.854 m
lT

C2 = 1.8 + 2(0.025471 + 0.055) = 1.961 m

C1 - C2 = -0.107 m

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 Which amongst the following bolts will be used for bolting the flange:
M 36 x3, M 39 x3, M 42 x3 and M 45x3?

For~
Root area =1194.59 mm 2
Minimum No. of bolts =52. 74
Actua l No. of bolts =56 ✓-
56 X 96
C1 = - - - = 1.711 m
TT

C2 = 1.8 + 2(0.02547 + 0.057) = 1.965 m

C1 - C2 = - 0.254 m

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Solution Which amongst the following bolts will be used for bolting the flange:
M 36 x3, M 39 x3, M 42 x3 and M 45 x3?

Bolt
M 36 x3 2.343 1.951 0.392
M 39 x3 2.0805 1.9549
M 42 x3 1.854 1.961 -0.107
M 45 x3 1.711 1.965 -0.254

Since for M 39x3 C1 - C2 is least and positive

:. M 39 x3 will be used for bolting the flange

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Solution Estimate bolt-circle diameter and Flange Outside Diameter

Bolt circle diameter C

C = C2 for the bolt for which difference is positive and minimum
C =1.9549 m
Flange Outside Diameter
A = C + 2 x bolt radius + 0.04
39
= 1.9549 + -1000 + 0.04 = 2.0339 m

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 Estimate various loads and moments under operating as
well as bolting-up condition_s_
- (C - 8)
nB 2 V al=
W1 = 4 P = 5.598 MN 2
1.9 549 - 1.8
TT
W2 =H- W1 = -4 ( G2 - B2 ) P 2
= 0.07745 m 9,--i I -:-rt
= 6.522 - 5.598 = 0.924 MN -f
~ (C-G) Wo
h
W 3 = Hp = 1.048 MN a3 =
2 _J_
WO = W2 + W3 + W3
= 7.57 MN -
(C - G)
2
= 0.00606 m e
~L
C
a, -
Solution Estimate various loads and moments under operating as
well as bolting-up conditions

(a1 + a3) ✓ 0.065 + 0.063

a2 =
2
= 0.041755 m w = _____ x 120
2
M0 = W1 a 1 + W2 a 2 + W3 a 3 = 7.68
= 0.4785 MNm
Mg = 7.68 X 0.00606
Bolting - up conditions
= 0.054654 MNm
Am +Ab
W= Sg M = Max of M 0 and Mg
2
Ab = No. of actual bolt x Root Area = 0.4785 MNm
= 0.065 m 2
Solution Estimate flange thickness (Poison's ratio =0.3)
MN t2 = 0.4785 x l _
Assume cf= 1, Sfo = 120 -2 , µ
m
= 0.3 1.8x120 X 15.9066 - 0.1877 m

0.080809
(0.078 + 0.1877) = 0 ·w~55133
k =A= 2.0339 = 1.13
8 1.8
0.955 k 2 log k
t = 0.139382463
Y= k-l [(1 - µ) + (1 + µ)4.605 k2 _
1
] CF = 0.609548193
t = 0.14655 7037
y = 0.955 [0.7 + (1.3 X 4.605) 1.132 log 1.1 3] CF = 0.599731647
0. 13 1. 1 32_ 1
t = 0 .145372124
= 15.9066 CF = 0.60132023
nC TT x 1.9549 t = 0.145564529
Bs = -;- = = 80.509 mm CF = 0.601061418
76
0.1455332
 A loose type flange is used toloin two parts of a shell with OD as 0.8 m. Design this
flange for following specifications: ----·
Flange face: Plain face; Design pressure (g)=2.S MN/m 2; Design temperature=400°C;
Allowable stress of shell material=120 MN/m 2 ; Allowable stress of flange material
at design temperature=130 MN/m 2; Bolts are made with IS:2002-1962 2A material;
Gasket material: soft aluminum solid flat metal; Ratio of gasket internal dia. to Shell
outside dia.=1.02; Corrosion allowance=0; Weld joint efficiency factor=l;
gn=0.0lSm; g 1 =1.415 gn, µ=0.3.
a. Calculate effective gasket seating width.
b. Select the suitable bolt for the flange: M 33x2, M 36x3, M 4Sx3, M 24x2?
C. Estimate flange outside diameter.
d. Estimate flange thickness.

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 Do -- 0.8 ·m
Design pressure (g) - 2.5 MN/ m 2
Allowable stress of shell _.. -- 120 MN/ m 2 67 MN
Allowable stress of fl ange = 130 MN/ m2
Allowable stress of bolt at atm temp Q
Allowable stress of bolt at des t emp --
96.10510N/ m 2
72.56921 MN/ m 2
9.8
7.4
,/j ~ "'-
')11N'I

y - 61 MN/ m 2
m -- 4
di/Do -- l .02
J - 1
go -- 0.015 m
~l - 0.3

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 ?lJl~U~ffl
Calculate effective gasket seating width

di
do/di
-
--
0.816
1.025449361
m d0 ( y- pm
d, = y - p(m + l)
r
do - 0.836766678 m
Min gasket w idth -- 0.010383339 >6mm
Min gasket width - 0.010383339 m
Actua l gasket width (N) - 0.010383339 m
Act ual do=di+2N -- 0.836766678 m
Basic gasket width (b0 ) -- 0.00519167
Effective gasket width (b) -- 0.00519167 m b=b0 if b0 $ 6.3 mm
G= d.+N
I
- 0.826383339 m

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.Solution Select the suitable bolt for the flange: M 33x2, M 36x3, M '- .x ,
M 24x2?
H - 1.34021 MN
Hp -- 0.26943 MN ✓
Wo=H+Hp -- 1.60964 MN Ao -- 72.569
W g=pGby - 0.82177 MN Ag -- 96.105
Am =M ax(Ao, Ag) -- 0.022 181 m2
.
.,,
I/Size Root area, m 2 Min. bolt Act. bolt R Bs C, c., C1 -C.,
I 33
36
"
2\ 0.000660185
3 0.0007065
33.6)
31.4 l,
36
32 '7
0.047
0.05
0.077
0.08
0.882356 0.93645 -0.05409
0.814874 0.94245 -0.12758
45 3 0.001193985 18.6 20 0.057 0.096 0.611155 0.95645 -0.34529
\ ~ 24 2/ 0.000314 70.6 72 0.035 0.075 1.718875 0.91245 0.806425
.....,_ /
✓ /

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• IITROOIK EE CERTIF IC ATION COURSE
 Estimate flange outside diameter

Size Root area, m 2 Min. bolt Act. bolt R Bs (1 c., C1-C.,

33 2 0.000660185 33.6 36 0.047 0.077 0.882356 0.93645 -0.05409
36 3 0.0007065 31 .4 32 0.05 0.08 0.814874 0.94245 -0.12758
45 3 0.001193985 18.6 20 0.057 0.096 0.611155 0.95645 -0.34529
( ~ 24 Z) 0.000314 70.6 72 0.035 0.075 1.718875 /{f.91245) 0.806425
/ ~

Fla nge diam et er = A = 0.95645 m =0.91245+(24/1000)+0.02

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Solution Estimate flange thickness

B=Do -- 0 .8 m
Wl -- 1.256
W 2=H-W 1 -- 0.084209743 K=A/ B - 1.1955625
W 3 =W 0 -H - 0 .269431423 y -- 10.96771161
a1=(C- B)/ 2 -- 0 .056225 --
CF 1
a3 =(C-G)/ 2 -- 0.04303333 t - 0.098829541 m
a 2=(a 1+a3 )/ 2 -- 0.049629165 -- 0.089534156
Bs
M o=W i a1+W 2a2+w3a3 -- 0.086392391 - - CF - 0.723957189
Ab - 0 .022608 t -- 0.084089781 m
w -- 2.152216317
M g=Wa 3 -- 0 .092617036 _,/
Controlling M -- @:092617030 MNm

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- -

Example-1
,/
A cylindrical vessel of 3 m outside diameter and 8 m length is supported by 4 lugs. The vessel is
filled with fluid having denslty870 kg/m 3 up to 5 m height and is being operated at 1.5 MN/m 2
pressure (g}. It is covered with flat heads from both ends having weight of 12 kN for each head.
Corrosion allowance is 2mm. It is Class 2 vessel having double welded butt joint with full
penetration for all joints. The vessel is made of IS: 2041-1962 20Mo55 and its design
-
temperature is 450°C.. The wind pressure on the vessel is 0.9 kN/m 2 • The height of vessel from
foundation i~Height of bracket from foundation is Sm and end diameter of anchor bolt
circle is 3.1 m. The steel channels are used to support lugs.
Given: Allowable stress of horizontal base and gusset plate: 100 MN/m 2; Allowable stress of
channel= 150 MN/m 2; Density of shell material and µ is 7600 ke/m 3 and 0.3, respectively.
a. Standard thickness of shell
b. Compute the minimum thickness of horizontal and gusset plates for I = b = h = 150 mm and
0 = 60 degree.
c. Determine the area of cross-section of channel column for section modulus of column as
'
200cm 3 • Also suggest the suitable size of column.

a
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-- - - - -

Solution
Do = 3 m
L = 8 m
Height of vessel from foundation = 1 m
Operating pressure(g) = 1.5 MN/m 2
Wt of each head = 12 kN
Corrosion = 0 .002 m
J = 0.85
Anchor bolt circle dia
Wind pressure, p
=
=
3 .1
0.9
m
kN/m 2
H Er a

No . of lugs = 4
z = 0.0002 m3
Height of lug from foundation = 5 m
I = 0.15 m
b = 0.15 m L

h = 0.15 ·m H,
th eta = 60 Degree
a
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 - -

Lug:
.
Support' y
~~

Solution Standard thickness of shell v---

✓
- 42673 .5 N/m 2
- 0.042674 ___. MN/m 2
Design pressure = p 1.575 MN/m 2
Thickness of shell = t - 0.026239 m
t with corrosion allowance = 0.028239 mm
t standard - 32 mm

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~ 11T ROORKEE "'" CERTIRCATION COURSE 17
 Support .:/'
-- - - - - - - - - _.,. '

Solution Compute the minimum thickness of horizontal

= h = 150 mm and 8 = 60 degree.
-
PW - 17.01 kN =0.7 x l x0.9 x9 x3 Pbw = K1 K2 Pl h1 D0
ts - 0.032 m w shell = TC (Do- ts) ts L p g
W shell
-- 177.8754901 kN
- 24 kN
p = 4 PW( H - H C ) + wm nx
Whead
11 C n
W liquid - 462.0191813 kN 0

c- D p = p H
wmax - 663.89467 14 kN (I = 0
"'' bl
2
p -- 209.870442 kN
a -- 0.05
-- 9327.575202 kN/m 2
P av
t hp - 0.000734 54 7 3P a L
f - - - -2x - - -
- 0.0271025 19 m tgh cos 0
tg -- 0.027982726 m
a
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 Determine th_g_ area of cross-section of channel column for section
Solution ....
modulus of column as 200cm~. Also suggest the suitable size of column.

Pa/Z - 52467.61051 J>/ X (Pal Z)+(~ ,L/2nZ) I

PwL/2nZ -- 53156.25 + 1
<
- 0.70415907 .f .f
- 0.29584093
X - 0.004729353 m2
- 47.29353307 cm 2
 'D,esign,of :Sad.die Support J"
Example-2
Design a saddle support for a horizontal drum (designed for internal
pressure of 2.1 MN/m 2) having inside radiu§. = 1.2m. tangent to tangent
length =r 15 m, depth of dish of head = 0.4m. The allowable stress
.
and yield
point of material is 150 MN/m2 and 240 MN/m2 , corrosion allowance = 0,
weld joint efficiency factor = 1, Modulus of elasticity = 2x105 MN/m 2 , load
on each saddle = 1.4x106 N, contact angle = 120°.

Check if the following assumed saddle design data [A/R=0.6 and B=0.5m]
satisfy all design stress conditions. If not, find the new values of A/R and B,
which satisfy all design stress conditions.
Solution
A
L=15 m -R =0.6
.
H=0.4 m A=0.72 m
Ri=1 .2 m B=0.5 m
 =20.656 MN/m 2
f{ = (::) + (n:~t) = 128.28 MN/m 2

< 150 MN/m2

f 2' =(PR)
2t
- ( k nR
2
Mt 2
2
) =76.07 MN/m 2 I •
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Solution
~q =("3w1) [t-ZA-H]
Rt l..+ H
=69.08 MN/m 2
q < O.Bf k3 = 1.171
< 0 .8*150
< 120 MN/m 2

k sw, ks=0.76
f3 t(B+lOt)
=94 _1OB MN/m2 < 0.5 yield stress
< 0.5*240
< 120 MN/m2
 Horizontal component of all radial load
F=kgW1 V - -~Q•
- I0!1 0 I
- -
k9 = 0.204 ✓
F=0.204*1.4*106
=0.2856 MN

0.2856 < (2)150

2
(\ )0.s - 3

1.428 MN/m 2 < 100 M~/m 2

a
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.,
1·
Example-1
A pressure vessel is being operated at maximum pressure of 2 MN/m 2 (g). It
has outer diameter of 2.5m and length of Gm. The allow~e stress at
working temperature and corrosion allowance is 150 MN/m2 and 3mm,
respectively. The vessel is of Class 2 where double welded butt joint with
full penetration is used for all joints.:. 21d +2

d
Determine thickness of ring pad (up to next
integer value) if it is required to compensate
opening of 0.2m diameter in shell. Use f
following details for calculations: ✓
Allowable stresses for nozzle and ring are 150
MN/m 2 • Nozzle wall thickness is 10 mm. Inside
protrusion of nozzle i~ and nozzl; length
above surface is 0.0Sm.

a
~ IIT ROORKEE
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CERTIFICATION COURSE Protruded nozzle
 Comge_
nsationifo_r Opening
Solution M.P = 2 MN/m 2 ✓
P = Design Pressure= 2 x 1.05 = 2.1 MN/m 2
f = 150 MN/m 2 \_./

J = 0.85 v
D = 2.Sm v

20.42 mm= tr

tfin al = 23.42 mm v
t sta nd ard = 25mm
 K'= PDo -0874 V
l.82f(t 5 - c) - ·
D0 = 0.2 1n ( outside dia)
On x-axis @x t 5 = 62500
Maximum opening =159 mm
Requirement of compensation as we
have 200 mm opening
d = insi.de dia of nozzle
= d 2tn = 180 mm
0 -

A
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.
Il
Solution A0 = 2H 1 (tn - tr' - c)
tn = 10 mm H1 = J(d + 2c)(t c) 0 -

H2 = 0.04mm = ✓ 186 x 7 = 36.08 mm

J A= (d + 2c)tr (H 1 )finai= 36.08 mm since it is less than 50mm
= (180 + 6) (20.42) A0 = 387.27 mm 2
= 3798.12 mm 2
A = (d + 2c)(t tr - c)
H2 = J(d + 2c)(tn - 2c)
5 5 -

= 293.88 mm 2 = ✓ 186 x 4 = 27.276 mm

(H 2)fin ai= 27.276 mm since it is less than 40mm
t , = Pdo,nozzle = 1.6336 mm
r 2fJ +P
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 ,Compe_
nsation1, for o·pening ,;;·
,

Solution A' = As + An = 605.48 + 293.88

✓
✓ Ai = 2H2 (tn - 2c) = 899.36 mm 2
= 2 X 27.276(10 - 6) A'< A
= 218.21 mm 2 (Ar= A -A'= 3798.12 - 899.36
✓ An= Ao+ Ai = 2898.76 mm 2
= 387.27 + 218.21 Ar= [2(d + 2c) - (d + 2tn)]tp
= 605.48 mm 2 tp = 16.85 mm
A5 = (d + 2c)(t 5 - tr - c) (tp) s = 18 mm
= 293.88 mm 2 ✓ (no need to add corros ion allowance)

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Solution Thickness of ring pad if allowable stress of nozzle and ring pad are
140 MN/m 2 and 130 MN/m 2, respectively.

An= Ao+ Ai
= 387.27 + 218.21
= 605.48 mm 2
fn = 140 MN/m 2
A5 = (d + 2c)(t5 - tr - c)
fr = 130 MN/m 2
= 293.88 mm 2 A'= 0.000858986
2
A=r 0.003391323
A= 3798.12 mm tp = 0.01972 m
20 mm
Example-2
A pressure vessel with outer diameter of 4m and length of 10m is to be
operated at maximum working pressure of 2 MN/m 2 (g) . The allowable
stress of shell material is 150 MN/m 2 at maximum design temperature .
Corrosion allowance is 2mm . It is Class 2 vessel having single welded butt
joint with backing strip for all joints. Vessel has opening of 0.22m.
2(d +2
Compute the thickness of ring pad if
d
compensation of opening is required .
Given : Allowable stresses for nozzle and
ring are 150 MN/m 2 • Inside protrusion
of nozzle is zero and nozzle length
t As
HI
j_

A
above surface is 0.02m.
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 Design pressure = p = 2.1 v MN/m 2
PJJ.I p 1)0
Thickness of shell = t = 0.034696 m V I= --
t with co rrosion allowance = 34.69641 mm v
2_(.J -P 2_f.J +P
t standard = 36 mm

Outer diameter of nozzle = 0.22 m Compensation must be required

Nozzle wa ll thickness = t /= 0 .001908 m

tr'+c = 0 .003908
tn = 5 mm

A
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 .Cq_mp_ .·fo.r Opem1
e.os.a~ion1 rr1g _:,

Solution
✓
Inside diamete r of nozzle d= 0.21
✓
Area to be compensated A= 0.0074 25 m2 A = (d + 2c)t,.
As= 0.000279 m2 A5 = (d + 2c)(t5 - tr - c)
(no corrosio n allowance)
H1-
- 0.000642 H, = ✓(d + 2c) (1 11
- c)
0.025338 m > 0.02 so H 1 is 0.02
A 0 = 4.37E-05 m2
A n = 4.37E-05 m2
A'= 0.000323 m2
A=
r
0.007102 m2
tp = 0.034146 m 0.036 m

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