number of baffles, N
b
+1=LL
S
=4.880.350=14 N
b
=13
Thickness of baffles, t
b
=6mm
4.Tie Rods and spacers:
Tie rods are provided to retain all cross baffles and take support plates accurately. For shell diameter, 300-500mm
Diameter of Rod = 9mm Number of rods=4
5.Flanges
Design pressure=0.11 Nmm
2
Flange material IS:2004-1962,class 2 Bolting steel :5 Cr-Mo steel
Gasket material: asbestos composition Shell inside diameter = 438mm.
Shell thickness: 8mm=g
o
Outside diameter of shell: 446 mm Allowable stress of flange material : 100MNm
2
Allowable stress of bolting material = 138 MNm
2
Shell thickness of flange = 10 mm. Outside diameter of flange = 325 mm.
6. Determination of gasket width
d
O
d
i
= [y-Pmy-Pm+1]
0.5
Assume a gasket thickness of 10 mm y = minimum design yield seating stress = 25.5 MNm
2
m = gasket factor = 2.75 d
O
d
i
= [44.85-0.11 x2.7544.85-0.112.75+1]
0. 5
d
O
d
i
= 1.001=1.001 d
O =
1.001 x 0.438= 0.4385 m Minimum gasket width = 0.4385 -0.4382 = .00075.
Taking gasket width of N= 0.010m d
o
= 0.458 m. Basic gasket seating width, b
o
= 5mm Diameter of location of gasket load reaction is
G= d
i
+ N =0.438+0.01
= 0.448 m
7.Estimation of Bolt loads .
Load due to design pressure H= πG
2
P4 = 3.14 x0.448
2
x0.114 = 0.01756 MN
Load to keep joint tight under operation
b = 2.5 b
0.5
= 6.12mm. H
p
= π G2bmp = 3.14 x0.448 x 2 x0.00612 x2.75 x0.11
= 0.00656 MN Total operating load
W
o
=H+ H
p
=0.01755 + 0.00656
=0.0241 MN.
Load to seat gasket under bolting condition W
g
= πGby = 3.14 x 0.448 x 6.12 x 10
-3
x 25.5 = 0.862MN.
W
g
W
o
,controlling load=0.8620 MN
8.Calculation of optimum bolting area
A
m
=A
g
=W
g
S
g
= 0.862
138 =
6.246×10
-3
m
2
Calculation of optimum bolt size Bolt size,M18 X 2
Actual number of bolts =20 Radial clearance from bolt circle to point of connection of hub or nozzle and back of
flange = R = 0.027 m C =ID + 21.415g + R
=325 +2[1.41 x0.008+0.027] = 0.726m
Bolt circle diameter = 0.40163 m. Using bolt spacing Bs = 45mm
C = n Bs 3.14 =44 x 0.045 3.14 = 0.63 Hence C = 0 .726
Calculation of flange outside diameter Let bolt diameter = 18 mm.
A= C+ bolt diameter +0.02 =0.716 +0.018+0.02
= 0.764m.
Check for gasket width A
b
S
G
πGN =1.54×10
-4
x44 x1383.14 x0.4486
2
= 66.43 2 xy. where S
G
is the Allowable stress for the gasket material
9.Flange moment computation:
a For operating condition
W
o
= W
1
+W
2
+W
3
W
1
= ∏B
2
P4 = ∏ x0.446
2
x0.114 = 0.0173 MN
W
2
=H-W1 =0.01756-0.0173
= 1.6×10
-4
MN. W
3
=W
o
-H=H
p
= 0.00672MN. M
o
=Total flange moment M
o
=W
1
a
1
+ W
2
a
2
+ W
3
a
3
a
1
=C-B2=0.726-0.4462 a
1
=0.14 m a
3
=C-G2=0.726-0.4482 a
3
=0.1395m a
2
=a
1
+ a
3
2= 0.14 +0.139 0.2 =0.139m
M
o
=0.01739 x0.140 +1.60×10
-4
x0.1395 x0.00672 +0.139 M
o
= 3.391×10
-3
MN-m
b For bolting condition M
g
=Wa
3
W=A
m
+A
b
xS
g
2
W= 6.246×10
-03
+6.76×10
-3
x1382 W=
0.897 MN
M
g
= 0.0.897 x 0.139 = 0.125 MN-m
M
g
M
o
,Hence moment under operating condition M
g
is controlling, M
g
=M
10.Calculation of flange thickness
t
2
= M C
F
Y B S
F
, S
F
is the allowable stress for the flange material K =AB = 0.7640.446 = 1.71
For K = 1.71 Y = 4.4 Assuming C
F
=1 t
2
= 0.125 x 1 x4.4 0.446 x 100 t= 0.11m
Actual bolt spacing B
S
= πCn = 3.140.77644 = 0.052m
11.Bolt Pitch Correction Factor
C
F
= [B
s
2d+t]
0.5
= 0.0522 x0.018+0.11
12
= 0.596 √C
F
=0.772
Actual flange thickness = √C
F
xt = 0.11 x0.772
= 0.085 m = 85mm.
12.Channel and channel Cover
t
h
=G
c
√KPf = 0.446 x√0.25 x0.1195
= 0.00767m =7.67mm t
h
= 8mm including corrosion allowance
13.Tube sheet thickness
t
ts
=FG√0.25Pf = 1 x0.448√0.3 x0.1195
= 0.0084=8.84 mm t
ts
= 9mm including corrosion allowance.
14. Saddle support
