Date post: | 14-Apr-2018 |
Category: |
Documents |
Upload: | dheeraj-shukla |
View: | 273 times |
Download: | 10 times |
of 73
7/30/2019 Design data book (1).pdf
1/73
7/30/2019 Design data book (1).pdf
2/73
1INDEX
Sr. No. Content Page No.
1. Unit Conversion 2
2. Design of Process Vessel under Internal
Pressure
5
3. Design of Head & Closure 10
4. Design of Non-standard Flanges 17
5. Design of Process Vessel & Pipes under
External Pressure
27
6. Compensation for opening in Process Vessel 327. Design of Tall Vessel 39
8. Design of Supports for Process Vessel 47
9. Design of thick walled High Pressure Vessel 63
7/30/2019 Design data book (1).pdf
3/73
7/30/2019 Design data book (1).pdf
4/73
3
1 pdl1lbf1kgf
1tonnef
1tonf
1.3825*10-1N4.4482N9.8067N
9.8067*103N
9.9640*103
NVolume 1in
1 US gal1 UK gal
1ft31 barrel(petroleum
US)1 lube oil barrel
1 yd3
1.6387*10- m3.7853*10-3m3
4.546*10-3m32.8317*10-2m3
0.15898m30.20819m30.76455m3
Velocity 1 ft/h1 ft/min1ft/s
1 mile/h
8.4667*10-
m/s5.08*10-3m/s0.3048m/s
0.44704m/sViscosity(dynamic) 1 mN s/m2(cp)
1 lb/ft h1g/cm s (poise P)
1 lb/ft s
1*10-3N s/m24.1338*10-4N s/m2
0.1 N s/m21.4882N s/m2
Frequency 1 c/s 1Hz
Specific heat capacity 1 cal/gm C1 Btu/lb F1 Chu/l bC
4.1868*10 J/kg K4.1868*103J/kg K4.1868*103J/kg K
Temperature Difference 1 C1 C1R
1 K5/9 K5/9 K
Thermal Conductivity 1 Btu/h ft (F/in)1 kcal/h m C1 Btu/h ft F1 cal/s cm C
0.14423J/s m K1.163J/s m K1.7308J/s m K418.68J/s m K
Power 1 ftlbf/min1 ftlbf/s
1 m kgf/s1 hp (metric)1 hp (British)
2.2597*10- J/s1.3558J/s9.8065J/s
7.3548*102J/s7.457*102J/s
Pressure(Stress) 1 dyne/cm1 Pascal1 kgf/m
2
1 mm water
1lbf/ft2
0.1N/m1 N/m2
9.8067 N/m29.8067 N/m2
47.88 N/m2
7/30/2019 Design data book (1).pdf
5/73
4
1 cm water(gf/cm2)1 mbar1 matm
1torr(mmHg)
1 in water1 ft water1 in Hg
1 lbf/in2(psi)
1 m water1
at(kgf/cm2orkp/cm2)
1 bar1 atm
1N/mm
2
1tonf/in2
98.0671 N/m2
100 N/m2
101.33 N/m2
133.33 N/m2
249.09 N/m2
2.9891*103 N/m2
3.3866*103N/m2
6.8948*103N/m2
9.8067*103 N/m2
9.8067*104 N/m2
1*105 N/m2
1.0133*105N/m2
1*106N/m2
1.5444*10
7
N/m
2
Moment of inertia 1 lbft 0.04214 kg/m
Momentum 1 lbft/sAngular momentum 1 lb ft2/sViscosity(kinematic) 1 S(stokes)
1 ft2/h
7/30/2019 Design data book (1).pdf
6/73
5
Cylindrical Spherical
Thin wall thickness
(a)If 0.25 then thin wall thickness vessels is required.(b)If
1.5 then thin wall thickness vessels is required.
Thin wall thickness for cylindrical shell
(i) Internal pressure
P = or
(ii) Minimum wall thickness
t = or(iii) Circumferential stress
= (iv) Longitudinal stress
z =
Thin wall thickness for spherical shell
(i)Internal pressure
P = or
Design of process vessel under internal Pressure
7/30/2019 Design data book (1).pdf
7/73
6
(ii) Minimum thickness of wall
t = or
(iii)Circumferential and longitudinal stress
= z = or Where,
P = Internal design pressure
P = 1.05 * max working pressure
J = Joint efficiency factor
J= 0.85 double welded butt joint with fullpenetration
J= 0.8 single welded butt joint with backing
strip
F=design stress for the material specified
c = corrosion allowance
t=thickness of wall without corrosion allowance
t = t + c
t = thickness with corrosion allowance
D = mean diameter =
7/30/2019 Design data book (1).pdf
8/73
7
= outer diameter= inner diameter
THICK WALL VESSEL PRESSURE
There are three stresses applied
(i) Longitudinal stress(ii) Radial stress(iii) Hoop stress
(1)Longitudinal stress
=
(2)Radial stress=
Where,
D = diameter of shell where stress is to be
calculated
(3)Hoop stress=
7/30/2019 Design data book (1).pdf
9/73
8Stress at internal surface
(i) Longitudinal stress
= In this case Po = 0= , If Do = 0 then = Pi(ii) Radial stress
= Where, K =
(iii) Hoop stress
= Stress for external pressure
= 0 = zi
=
The maximum shear stress at any point in the cylinder
= =
=
7/30/2019 Design data book (1).pdf
10/73
9Theory of failure
(1)Maximum shear stress theory
= Where,
=yield stressPi = Internal pressure
K = (2)Maximum strain theory
=
[]
,
= Poisson ratio, is obtain by table 1(3)Maximum strain energy theory
=
7/30/2019 Design data book (1).pdf
11/73
7/30/2019 Design data book (1).pdf
12/73
11
TORI -SPHERICAL HEADS
Thickness of head
t =
C is the safe or stress concentration factor
Assume
(i)Safe factor C depend upon
or for the head without any opening or with fully compensated opening orreinforced opening
(ii) C depend upon
or
for uncompensated opening
If Sf is very less , he = ho
ho=
Unreinforced opening
ho=
Where,
he = Effective external height of head without straight flange
ho = outside height of flange
hi = inside height of flange
Sf= flange height
Sf= 40mm
7/30/2019 Design data book (1).pdf
13/73
12ri and ro inside and outside knuckle radius
Ri and Ro inside and outside crown radius
d is the diameter of the largest uncompensated opening in the head
For C
t/D0C =P/2fJ
now find value of C from he/D0 and t/D0by trial & error method using table 2
C for formed head without opening or fully compensated opening is given in
table 2
R0=D0
Blank diameter = D0+(D0/42)+(2/3)ri+2Sf For ,t25mm=D0+(D0/42)+(2/3)rI+2Sf+t ,For , t25mm
External height of the excluding straight flange
h0 =R0
V, Excluding straight flange = 0.0847 for ,ri= .06 Di=0.1313 , for ,2:1 ellipsoidal or deep dished head
For accuracy it is suggested to recalculate h0 by putting new value for he/D0
another method would we assumed some value of t & check the same from
t= PDOC/2fJ
7/30/2019 Design data book (1).pdf
14/73
13
ELLIPSOIDAL HEADS
Neglecting thinning effect
C = 2fJt / PD0 , J=1
D0 = outer diameter of shell
For, 2:1, ellipsoidal, he=h0=0.25D0
hi=0.25Di
he/D0=0.25 , by table 3 obtain d/Volume of elliptical dished head
Vn = (/4) (D0/6) = /24ELLIPTICAL HEAD
Thickness of elliptical head
tn= PDV/2fJwhere ,
P=internal pressure
D=major axis
V=stress intensification factor = (2+K2)
K=major axis =272.6HEMISPHERICAL HEAD
Neglecting thinning effect
C = 2fJt / PD0 , J=1
D0 = outer diameter of shell
he/D0 = 0.5 ,From table 3 obtain d/
7/30/2019 Design data book (1).pdf
15/73
14Volume of elliptical dished head
Vn = (/4) (D0/6) = /24Total volume contain in vessel where D is internal diameter
Vvessel=[{}2+{}2]Volume of torispherical dished head to straight flange
V=0.000049Where, di=inside diameter of vessel in inches
CONICAL HEAD
(1) Thickness of conical head at junctiont =
Where
De is the outer diameter
P is design pressure
J is the factor to be taken at joint =0.85
Where Z is the factor to be
20 30 45 60Z 1.00 1.35 2.05 3.20
Surface area , A = (1/2)
Volumetric capacity= (1/3)(h/4)(2) thickness away from the junction
t= ( ), P=design pressure
Dk=internal diameter of cone at a distance L
7/30/2019 Design data book (1).pdf
16/73
15
L=(1/2)t = thickness of shell+corrosion allowance
From the junction,
Dk=Di2LsinJ =0.85
Di=Do2t,Do and Di= external and internal diameter
Table 1
Material Specific weight( Poisson ratio= /E
AluminiumBrass
CopperIron
NickelSteel
2.65 8.35 8.74
7.74 8.74 7.70
0.340.350.350.280.360.30
Table 2
Stress concentration factor C for formed heads without opening or with ful ly
compensated opening t/D0
hE/D0 0.002 0.005 0.01 0.02 0.040.15 4.55 2.66 2.15 1.95 1.750.20 2.30 1.70 1.45 1.37 1.320.25 1.38 1.14 1.0 1.00 1.000.30 0.92 0.77 0.77 0.77 0.770.40 0.59 0.59 0.59 o.59 0.590.50 0.55 0.55 0.55 0.55 0.55
7/30/2019 Design data book (1).pdf
17/73
16Table 3
Stress concentration factor C for formed heads with uncompensated opening
d/h
E/D
00.5 1.0 2.0 3.0 4.0 5.0
0.15 1.67 1.86 2.15 2.65 3.10 3.600.20 1.28 1.45 1.85 2.30 2.75 3.250.25 1.00 1.15 1.60 2.05 2.50 2.950.30 0.83 1.00 1.45 1.88 2.28 2.700.50 0.60 0.80 1.10 1.50 1.85 2.15
Note: values can be interpolated
7/30/2019 Design data book (1).pdf
18/73
17
Gasket dimensionsdo/di = [ ( Ypm ) / { Yp (m+1) } ]
1/2
Where,
di = inside diameter of gasket
do= outside diameter of gasket
Y = minimum design gasket seating stress
p = internal design pressure
(Residual gasket force) = (gasket seating force)(hydrostatic pressure force)
TABLE1 Gasket thickness and width of gasket
Thickness (mm) Width (mm)
3 Up to 20
4 Over 20 and up to 30
5 Over 30
Thickness smaller than 3 mm can be used, if larger gasket seating stress is
desired.
The values of Y and m can be determined from table 1.1.Minimum Gasket width
N = (dodi)/2
2. Gasket seating widthbo = N/2
Design of Non-standard flanges
7/30/2019 Design data book (1).pdf
19/73
183.Effective gasket seating width
(a)b = bo when bo 6.3 mm(b)b = 2.5 ( bo )1/2 when bo> 6.3 mm
4.Diameter at location of gasket load G(a)G=di + N when b < 6.3 mm(b)G=do2b when b 6.3 mm
5.Maximum bolt space= [2d + {6t / (m + 0.5) } ]
Where,
d = bolt diameter,
m = gasket factor,
t = flange thickness.
The minimum bolt spacing should not be less than 2.5 d for smaller boltdiameter.
6.Minimum bolt circle diameterC = B + 2 (g + R)
or
C = n Bs/
Where,
C = bolt circle diameter,
B = inside diameter of flange,
g= thickness of hub at back of flange,
R = radial clearance from bolt circle to point of connection
of hub or nozzle and back of flange,
n = actual number of bolts,
Bs= bolt spacing.
7/30/2019 Design data book (1).pdf
20/73
19
Estimation of Bolt Loads:
1.Load due to design pressure:H=G
2p/4
Where;
H=Load due to design pressure, MN
G= Dia at location of gasket load reaction, m
P= design pressure, MN/m2
2.Load to keep joint tight under operation:Hp=G (2b) mp
Where;
p= design pressure
m= gasket factorG=Diameter at location of gasket load
b = effective gasket seating width
3. Total operating load:Wo = H + Hp
4.Load to seat gasket under bolting up condition :Wg= Gboy
y=minimum gasket seating stress (table 2)
7/30/2019 Design data book (1).pdf
21/73
205. Controlling load;
If WO > Wg, then controlling load = WO
If WO < Wg, then controlling load = Wg
6.Determination of minimum bolt area : Under operating condition
Am is :( Am= Ao)
Ao = Wo/So
Under bolting up conditionAg = Wg/Sg
Where:
Ao is the bolt area required under operating condition
Ag is the area required under bolting up condition
So is allowable stress for bolting material at design pressure (table 2)
Sg is allowable stress for bolting material at atmosphere temperature (table 2)
7. Calculation of flange outside diameter (A)
.
A= C+ bolt dia +0.02 meters
(Use Table 3)
7/30/2019 Design data book (1).pdf
22/73
21 Check for gasket width: -
To prevent damage to gasket during bolting up condition following condition to
be satisfied
Ab Sg/ GN < 2y (From table 3)
Determination of f lange moments
(a)Operating conditi on1. Total Load
Wo = W1 + W2 + W3
W1= ( B / 4) p
W2 = HW1
W3 = WOH =Hp
Where;
W1Hydrostatic end force on area inside of flange
H- Load due to design pressure
Hp- Load to keep joint tight under operation
2. Total flange momentMo = W1 a1+ W2a2 + W3a3
The values of a1, a2 and a3 for different flange type
Assume lap joint flange and use Table 5
7/30/2019 Design data book (1).pdf
23/73
22(b)Bolting up condition
1. Total flange momentMg = W a3
Where;
W = (Am + Ab) Sg / 2
a3 = (CG)/ 2
TABLE 2 Allowable Stresses for Bolting Materials in MN/m2
Material 50 100 200 250 300 350 400
Hot rolled carbon
steel
57.3 55.1 53.5 47.6
5% Cr Mo steel 138.0 138.0 138.0 138.0 138.0 138.0 138.0
13.8%Cr Ni steel 129.0 109.0 85.0 78.5 76.0 73.2 72.0
13% Cr Ni steel 176.0 162.0 140.5 134.0 126.5 119.0 104.5
18% Cr 2 Ni steel 212.0 195.0 170.0 161.0 152.0 144.0 127.0
Allowable stress MN/ m for design metal temperature not exceeding (0C)
7/30/2019 Design data book (1).pdf
24/73
23TABLE 3 FLANGES
Bolt size Root
area
Min.
no.of
bolts
Actu
al no
ofbolts
(n)
R(m) Bs
(m)
C=
nBs/
(m)
C=ID+2(1.4
15go+R) (m)
M 16 x 1.5 1.54 x 10-
4
50.8 52 0.025 0.07
5
1.24 1.0583
M 18 x 2 1.54 x 10-
4
43-7 44 0.027 0.07
5
1.05 1.0623
M 20 x 2 2 x 10- 33.7 36 0.030 0.07
5
0.86 1.0683
TABLE 4
I. loose type flange B g11.lap joint flange Outside dia. g1=g0
2. Raised face with hub Outside dia g1=0.5 g0
II. Integral type
1.ring only plane face Outside dia g1= g1
2. lap weld hub raised
face
Outside dia g1=21/2 g0
III. Optional type
1. plane face with weld
hub
Outside dia g1=21/2 g0
2, Ring only type raised
face
Outside dia g1=g0
7/30/2019 Design data book (1).pdf
25/73
24
Table 5 Moment arms for flange loads under operating conditions
Type of flange a1 a2 a3
Internal typeflanges
R + (g1/2) (R + g1 + a3) / 2 ( CG)/ 2
Loose type except
lap joint flanges
( CB)/ 2 (a+ a)/2 ( CG)/ 2Lap joint flanges ( CB)/ 2 ( CG)/ 2 ( CG)/ 2
Calculation of flange thickness:t2 = (MCFY/BST) =(MCFY/BSFO)
Where,
CF = bolt pitch correction factor.
CF = (BS/ (2d+t))1/2
M = MO
And
Y = Bt2SFO/M
Gasket material Gasket
factor
m
Min. design
seating
stress, Y,MN/m
Min. actual
gasket width
(mm)
Vulcanized rubber sheethardness above 70 IHRD
1.00 1.38 10
Asbestoswith a
suitablebinder for
operatingconditions
}3.2mmThick
1.6mm
2.00
2.75
11.00
25.50
10
10
7/30/2019 Design data book (1).pdf
26/73
250.8mm 3.50 44.85 10
Rubber with cotton fabricinsertion
1.25 2.76 10
Rubber withasbestos fabricinsertion, withor without wire
reinforcement
}3-
ply
2-ply
1-ply
2.25
2.50
2.75
15.25
20.00
25.50
10
10
10
Vegetable fibre 1.75 7.56 10Spiral-woundmetal,
asbestosfilled
} Carbonsteel
S.S.ormonelmetal
2.50
3.00
20.00
31.00
10
10
Corrugatedmetal, asbestos
inserted orasbestos filled
corrugatedmetal jacket
}Soft Al
SoftCu/brassIron/soft
steelMonel
metalS.S.
2.502.753.003.253.50
20.0025.0031.0038.0045.00
1010101010
Corrugatedmetal
} Soft AlSoft
Cu/brassIron/soft
steelMonel
metalS.S.
2.753.003.253.503.75
25.5031.0038.0045.0052.00
1010101010
7/30/2019 Design data book (1).pdf
27/73
26Asbestos filledflat metal jacket
} Soft AlSoft
Cu/brassIron/soft
steelMonelmetalS.S.
3.253.503.753.50
3.75
38.0045.0052.0055.00
62.50
10101010
10
Solid flat metal } Soft AlSoft
Cu/brassIron/soft
steel
MonelmetalS.S.
4.004.755.506.006.50
61.0090.00
125.00150.00180.00
66666
Ringjoint
} Iron/softsteel
MonelmetalS.S.
5.506.006.50
125.00150.00180.00
666
7/30/2019 Design data book (1).pdf
28/73
27
(1)CRITICAL LENGTH BETWEEN STIFFENERS:
= 0.3 for steel vessel
D= outer diameter of shell
t = thickness of shell
(2)OUT OF ROUNDNESS OF SHELLS (U):(a)For oval shape
(b)For dent or flat spots
Where,
a = depth of dent or flat spots (maximum value is to be taken).
U = Out of roundness factor, 1.5% for new vessel
(3)DETERMINATION OF SHELL THICKNESS WITH OUT STIFFENERRING:
Design is to be check for elastic instability or plastic deformation
Design of process vessel and pipes under external pressure
7/30/2019 Design data book (1).pdf
29/73
28If type of head and closures are not given then consider the vessel has tori-
spherical head (standard dished head) at the both end of shell having Ri =D0 and
ri=0.1D0 where Ri crown radius and ri knuckle radius, D0 outside shell diameter
Inside depth h for tori spherical head
Where:
Effective length of tower without stiffener
L= tangent to tangent length + 1/3(inside height of head) + 1/3(inside height of
closures)
Or
L= tangent to tangent length + 2/3h (inside height of head and closures)
Determination of safe pressure against elastic failure:
Where; p safe external pressure
E = modulus of elasticity at design temperature
t = shell thickness, D0=outside dia
7/30/2019 Design data book (1).pdf
30/73
29Value of K and m as a function of D0/L ratio of given in table below
D0/L K M0
0.10.20.30.40.60.81.01.52.03.04.05.0
0.733
0.1850.2240.2290.2460.5160.6600.8791.5722.3645.1449.03710.359
3.00
2.602.542.472.432.492.482.492.522.542.612.622.58
Checking for plastic deformation
If D0/L 5 or
, then
( )
= allowable compressive stressU= out of roundness factor
If D0/L>5, i.e. L external design pressure than calculated thickness from elastic instability
is correct otherwise thickness is not safe that calculated thickness against plastic
deformation.
7/30/2019 Design data book (1).pdf
31/73
30(4)DETERMINATION OF THICKNESS USING STIFFENER RING:
Given if stiffener is use effective length of tower will be the trace facing, else
the critical length between stiffeners is to be consider.
So again calculate D0/L and find the value of K and m and calculate the shell
thickness for elastic instability
Checking for Plastic deformation
If D/L5
If D/L>5 (5)DESIGN OF CIRCUMFERENTIAL STIFFENING RING:
Design of stiffening ring involve first to select a standard structure and then to
check for required moment of inertia of structure
Where, I= required moment of inertia
f=allowable stress
As=cross-section area of one circumferential stiffener
E= modulus of elasticity at temperature
L= distance between stiffener
t = corroded shell thickness
7/30/2019 Design data book (1).pdf
32/73
31Now, Select a 18 cm channel of following specification
Weight (Wt) =14.6 kg
As= 1.84
m
I=8.9 , No. of stiffener required = 5
Total weight of ring = DoWt no. of stiffener rings
Saving in shell material for using stiffening rings:
= D0 (t-ts) L = tangent to tangent length
7/30/2019 Design data book (1).pdf
33/73
32
1. Area to be compensated
Where, d = internal diameter of nozzle,m
= (outside dia of nozzle - 2thickness of nozzle)
c = corrosion allowance,m
reinforcement thickness.
Where,
D = outside diameter of the shell
2. Area available from Shell for reinforcement:
Where,is the actual shell thickness.
3. Area available from nozzle for reinforcement:
A = An ( no inside protution)
Where,
And
COMPENSATION FOR OPENINGS IN PROCESS VESSEL
7/30/2019 Design data book (1).pdf
34/73
33If nozzle length outside the vessel is larger than H1, the boundary limit the n
above value of H1 will be taken.
If, on the other hand, the nozzle length outside the vessel surface is less than or
equal to the height of the boundary limit, then,
H1 = actual length of nozzle
4. Area of the nozzle inside the vessel available for compensation:
Where, tn is the nozzle wall thickness.
And
If the inside protrusion of the nozzle goes beyond the boundary zone, then,
On the other hand if the inside protrusion is less or equal, then,
H2 = actual length of produced portion.
Excess area available in the nozzle for reinforcement:
5. Reinforcement area everywhere from shell and nozzle:
If it is found that , then no other external reinforcement necessary.
If
, the difference in area
is to be provided with
ring pad weldments
6. Area available from ring pad and weldmentsn within boundary limits:
2 ( d + 2c )(d + 2c + 2)} tp, tp= thickness of the ring pad,
Ring pad dimensions:
7/30/2019 Design data book (1).pdf
35/73
34Inner diameter = d
Outer diameter = 2( d + 2c )
7. Area of compensation within the boundary limit:
Area of compensation within the boundary limit should not be less than the
basic area removed from the shell during opening. i.e., A
A.
If material of construction for nozzle and ring pad having different allowablestress values for shell then area of compensation within the boundary limits:
Where,
is allowable stress for shell material.
is allowable stress for nozzle material. is allowable stress for ring pad material.
7/30/2019 Design data book (1).pdf
36/73
35
Uncompensated Opening
K factor:
Then, According to IS: 2825-1969, near the opening above equation becomes:
If or a little over 1, an opening diameter up to 0.05m need not be
compensated.
If , larger opening diameter up to 0.2m can remain unreinforceddepending upon the shell diameter.
Weakening factor:
Where, is pressure required to cause 0.2% permanent deformation near theopening.
Pressure required yielding the unpierced shell.
Theoretical shell thickness for uncompensated opening:
If opening is made away from welded joints, J=1.
UNCOMPENSATION FOR OPENINGS IN PROCESS VESSEL
7/30/2019 Design data book (1).pdf
37/73
36
Table 1
Where,
D0 is the shell outside diameter.d0 is the opening diameter.
ts is the actual shell thickness.
c is the corrosion allowance.
weakening factor.
Determination of Compensation Requirement for Openings in Heads
(a)For dished and hemisphere ends:If the opening and its compensation are located entirely within the spherical
portion of a dished end, tr is the thickness required for a sphere having a radiusequal to crown radius.
0.0 1.0000.25 0.9000.5 0.7850.75 0.7001.0 0.645
1.5 0.5452.0 0.4652.5 0.3903.0 0.3403.5 0.285
4.0 0.2454.5 0.2155.0 0.1805.5 0.155
6.0 0.1306.5 0.1157.0 0.0907.5 0.0808.0 0.075
7/30/2019 Design data book (1).pdf
38/73
37(b) For semi-ellipsoidal end:
When the opening and its compensation are in ellipsoidal end and are locatedentirely within a circle having a radius, measured from the centre of the end, of0.40 of the shell diameter, tr is the thickness required for a sphere having aradius R, derived from the following table:
Table 2
Compensation for Multiple Openings
Interaction between two openings, if their edge distance is roughly
:
Where L is the pitch andd is the inside diameter of the large opening.
Interaction between two openings is virtually negligible when,
As per IS: 2825-1969 the openings spaced apart a distance not less than
But in no case less than twice the diameter of the larger opening may be
regarded as isolated opening.
Effective cross sectional area for nozzle type reinforcement:
0.167 1.360.178 1.270.192 1.18
0.207 1.080.227 0.990.25 0.90
0.277 0.810.312 0.730.357 0.65
0.40 0.590.45 0.540.50 0.50
7/30/2019 Design data book (1).pdf
39/73
38
(a)If the openings are along longitudinal direction:
(b)If openings are along circumference or on sphere:
7/30/2019 Design data book (1).pdf
40/73
39
1. Determination of shell thickness:
ts = PDO / (2fJ+p)+C
Where, ts = thickness of shell
J= joint efficiency (0.85)
P= design pressure
f = allowable stress (from table A-1)
DO = outside diameter of shell
2. Determination of longitudinal stress:
a) The axial stress (tensile or compressive) due to pressure in:
Zp = P D2/4t(Di+t) , D = Di+t
And, Zp =PD/4t , [{DO= Di+t =Di}]
Where,D=Di for internal pressure
D=DO for vaccum(inclusive of insulation thickness)
t= corroded shell thickness(thickness w/o corrosion allowance)
b)The axial stress (compressive) due to dead loads:
a) The stresses induced by shell wt. at X meter height from top:
Zp=WS/ t(Di+t)
Where, WS= (DtXS)
Where, S=specific wt.(from table given below)
WS=wt. of shell of length X meter
t= shell thickness at the point under consideration
DESIGN OF TALL VESSELS
7/30/2019 Design data book (1).pdf
41/73
40Material Specific wt. N/m
3 Poissons ratio
Aluminium 2.65x104 0.34Brass 8.35x10 0.35
Copper 8.79x10 0.35
Iron 7.74x10 0.28Nickel 8.74x104 0.36
Steel 7.70x104 0.30
b) The stress induced in the shell due to a distance X meters from the top:
Zi= Wi/t(DI+t)
Where, Wi=(Dtinsins ){wt. of insulation up to a distance for a length of X
meters from the top}
tins= insulation thickness
ins= Specific weight of insulation (from table 1.1 given below)
c) The stress induced by the weight of the liquid supported by the inner
arrangement like tray for a distance X meter from the top is:
No. Of tray, N= [(X-top spacing)/tray spacing] + 1
Zi = Wl/t(Di+t)
Wl= (/4)D2(weir height)(sp. Gravity of water)(no. Of trays)
Wl=Wt. of liquid supported for distance X meters from the top
d) The axial stress due to the weight of attachments like trays,overhead
condensers, top head, platforms and ladders for a distance X meter from the tpo
is:
za= Wa/t(Di+t)
where, Wa= Wt of head + wt. of ladders + wt. of platform + wt. of liquid or trays
Wt. of ladder =37X
Wt. of platform = (/4)(dia. of platform)2(platform loading)
Wt. of trays = (/4)(tray loading)(no. Of trays)
7/30/2019 Design data book (1).pdf
42/73
41*for the design calculation weight of steel ladders plateforms,caged ladders
,plain ladders and trays (including liquid hold up)may be taken as given in the
following data:
steel ladder(caged)=37kgf per meter linear length
steel ladder (plane) = 15kgf per meter length
steel platform = 170 kgf per sq. Meter area
Distillation tray wt. (inclusive of liquid hold up) = 122kgf per sq. Meter area
Total dead load stress, Zw, acting along the axial direction of shell at the point
is given by:
Zw=Zs+ Zi+ Zl+ Za
for vessel which does not contain internal attachments like tray but consists only
of shell insulations, heads,minor attachments like nozzles,man holes,etc.the
additional load may be approx. Equal to 18% of the weight of a steel shell.
3. The longitudinal bending stresses due to dynamic loads:
a) The axial stress (tensile & compressive) due to wind loads in self-supporting
tall vessels:
The wind load on a vessel is given by:
PW = (1/2)CD V2 A
Where ,CD =drag coefficient = density of air
V= wind velocity
A = projected area normal to the direction of the wind
The wind load on tall cylinder vertical vessel can be calculated from the
following empirical formula (for a shape factor of ew =0.7)
pw= 0.05 V2
Where,pw= wind stress in N/m2 or min. Wind pressure
7/30/2019 Design data book (1).pdf
43/73
42V= wind velocity in Km/hour
The wind pressure for the bottom part & the rest of the upper part can be
directly obtained from the following table depending upon the zone of
insulation of the vessel.
TABLE.2
Wind pressure (kN/m2)
Region At, H =20m At, H=100m
Coastal area 0.7-1.0 1.5-2.0
Area withmoderate
wind
0.4 1.0
The total load due to wind acting on the bottom and upper parts of thevessel are determined from the following equations:
Pbw=k1k2p1h1D0
Puw= k1k2p2h2D0
Pbw=total force due to wind load acting on the bottom parts of the vessel with
height equal to or less than 20 meter
Puw=total force due to wind load acting on upper part above 20 meter
h1= height of the bottom part of the vessel equal to or less than 20 meter
h2= height of the upper part above 20 meter
p1= wind pressure for the bottom part of the vessel (from table 2,value given for
H=20 meter)
7/30/2019 Design data book (1).pdf
44/73
43p2= wind pressure for the bottom part of the vessel(to be determined from table
2 for mid point of upper part of vessel by interpolation of data given)
D0=outer dia. including insulation as the case may be
K1=coefficient depending upon shape factor
= 90 degree to the wind
= 0.7 for cylindrical surface
K2= coefficient depending upon the period of one cycle of vibration of the
vessel
=1 (if period of vibration T is 0.5 second or less)
= 2(if period exceeds 0.5 seconds)
The period of vibration T is given as
T=6.35 x 10-5(H/D)3/2(W/t)1/2
Where, H= tangent to tangent height +skirt height
W= total weight of shell
W= WS + Wi +Wl +Wa
Ws= weight of shell
Wi= weight of insulation
Wl= weight of liquid in tray
Wa= weight of attachments
If vessel height is less than 20 meter, then wind load Pw
Pw=k1k2 (pw)D0X
Where , pw= wind pressure or wind stress
D0= Di+(2XTins)
7/30/2019 Design data book (1).pdf
45/73
44* The bending moment at the base of the vessel due to wind load is determined
from the following equation:
a)If for the vessel H is less than or equal to 20 meter
Mw = Pbw(H/2)
b)for the vessel with H>20meter
Mw=Pbw (h1/2)+Puw(h1+h2/2)
The resulting bending stress in the axial direction is computed from thefollowing correlation:
zwm
= 4 Mw/t(D
i+t)D
i
Where, zwm=longitudinal stress due to wind moment(compressive on down
wind side & tensile on up upwind side),
Mw= bending moment due to wind load
Di=inner dia. of the shell
t= corroded shell thickness
4.Determination of resultant longitudinal stresses :
a)The resultant tensile stress (on upwind side)in the cross section of the vessel
at distance X meter from the top in absence of eccentric loads will be:
For internal pressure,
z = zp+ zwm - zw
z,tensile(maximum)= fJ
forv external pressure
z=( zwmor zsm )- zp - zw
b) The resultant compressive stress (on downwind side) is given by:
For internal pressure:-
z= zwm+ zw - zp
7/30/2019 Design data book (1).pdf
46/73
45z,compressive,(maximum) = 0.125E(t/Do)
Where,E =modulus of elasticity
For external pressure:-
z= (zwmor zsm) + zw + zp
Check : (safe design)
Equivalent stress , e= (e2 - ez+ z
2 )1/2
Here, e = hoop stress =P(Di+t)/2t or P(Do-t)/2t
z= tensile stress
Now calculate the value of z from:z= zp+ zwm - zw
And substitute the value of zp& e in the equation of equivalent stress and then
check it from, e = fJ
if e(calculated)< e (check design condition)then our calculated thickness is correct
here e is calculated by putting X = height of towerheight of skirt or tangent -
to-tangent height
Check for safe design:
At design conditions,
1) e < Fj2) (tensile)< fJ3) z(compressive)< 0.125E(t/Do)
At test conditions,
1) z < 1.3 fJ2) z(tensile)
7/30/2019 Design data book (1).pdf
47/73
46*Now calculate the value of z from:
z= zp+ zwm - zw
and substitute the value of zand e in equation of equivalent stress and then
check it from e= fJ
*if e(calculated) is < w (checked design condition),then our calculated
thickness is safe.
TABLE 1 Specific Weight of insulating Material
Material Apparent sp. Wt.KN/m3
Thermal conductivity
Asbestos 5.64 0.496Chalk 15.00 0.692
Plaster ,artificial 20.70 0.742Cotton wood 0.78 0.042Cork board 1.57 0.043
Cork ground 1.45 0.043Diatomaceous earth Powder
,fine2.70 0.069
Wool 1.08 0.036Felt , wool 3.26 0.052
Graphite, powdered 4.78 0.180Magnesia,molded& dry 12.20 0.432
Mineral,wood 7.84 0.605Rubber,hard 11.70 0.150
Sawdust 1.88 0.052Silk 0.99 0.045
D.Q Kern process heat transfer McGraw-Hill Book Co. Inc,New york
7/30/2019 Design data book (1).pdf
48/73
47
(1) Thickness of shell for internal pressure in given byt =
Where,
P= design pressure
f= allowable stress
j= joint efficiency factor= outer dia of shellC= corrosion allowance
SUPPORTS
1] Skirt Support
1) The tensile stress in the skirt will be maximum when the dead load(weight)is minimum i.e. the shell of the vessel is just erected and the shell is empty
without any internal attachment.
2) The compressive stress is to be determined when the vessel is filled up
with water for hydraulic test. Maximum load may be expected at any time and
this factor is always to be considered.
The maximum weight of the vessel with two heads and shell will be
= + Where,
Ws= Wt. of shell = () (H - 4)= Wt. of Head = 2(Wt. of each head) = outer dia. Of shell= shell thickness, with corrosion allowance
= specific weight of shell material from (table A-8) in
H= total height (tangent to tangent ht. + skirt height)
DESIGN OF SUPPORTS FOR PROCESS VESSELS
7/30/2019 Design data book (1).pdf
49/73
48
= Ws + Wi + Wl + WaThe value of Ws, Wi, Wl, Wa can be taken from Tall vessel
Now period of vibration at minimum dead weight is
Where,
H= Total height of vessel
D= Outer dia of vessel
ta = shell thickness with corrosion allowance
If Tmin 0.5 then K2 =1
Tmin > 0.5 then K2 = 2
Similarly,
Period of vibration at maximum dead weight is given by
Now if total height of vessel is 20m or less than 20 m, then the wind load is
determined as:
For maximum wind load:
Where ,
Pw =wind pressure = 0.05 Vw2
(Vw = wind velocity)
D0 = outer dia of shell + 2 (thickness of insulation)
K1 = 0.7 for cylindrical surface
= 1.4 for flat surface 90 to wind
7/30/2019 Design data book (1).pdf
50/73
49
For minimum wind load:
(PW) min = K1 K2 PW H D0
Minimum and maximum wind moments are given by:
(Mw)min = (Pw)min . H/2
(Mw)max = (Pw)max . H/2
If the total height of vessel is greater than 20m then
(Pbw) min = K1 K2 P1 h1 D
(Puw) min = K1 K2 P2 h2 D
Where,
P1 = wind pressure at height h1 = 20m
P2 = wind pressure at h2 > 20mD= outer dia of shell
Similarly,
(Pbw) max = K1 K2 P1 h1 D
(Puw) max = K1 K2 P2 h2 D
Where,K1, K2, h2, h1, P1, P2 are as before
D= outer dia of shell +2x insulation thickness
Maximum & minimum wind moment is given by:
(Mw) max = (Pbw) max (h1/2) + (Puw) max (h1 + h2 /2)
(Mw) min = (Pbw) min (h1/2) + (Puw) min (h1 + h2 /2)
7/30/2019 Design data book (1).pdf
51/73
50
As the thickness of skirt is excepted to be small assume Di = D0
Now, minimum longitudinal stress due to minimum wind moment is:
Where,
D = outer dia of skirt ( outer dia of shell when skirt support is cylindrical)
ts = thickness of skirt
Maximum longitudinal stress due to maximum wind moment is:
Minimum & maximum dead load stresses o the skirt is given by:
Where;
D = D0 when skirt is cylindrical
Now maximum tensile stress w/o any eccentric load
max max For safe tensile stress:
Where;
f = allowable stress
J = joint efficiency factor(0.7 for double welded butt joint forclass3 cons)
7/30/2019 Design data book (1).pdf
52/73
7/30/2019 Design data book (1).pdf
53/73
52l = outer radius of bearing plateouter radius of skirt
Z= l
= (
- l)/ 2
The allowable compressive strength of concrete foundation varies from 5.5
MN/ to 9.5MN/Substitute (max) = 5.5NM/and calculate l
By substituting the value of l again in same equation and calculate (max)
Thickness of bearing plate w/o gussets:= l = (max)M (max) = b l (
)
For b=1 M (max) =
If bearing plate thickness is equal to or less than 12 to 20mm , no gussets are
required otherwise gussets are required to reinforce the plate from table 10.1,1/b=1
M (max) = M y = -.119=
7/30/2019 Design data book (1).pdf
54/73
53ANCHOR BOLT DESIGN:
=
J = Wmin R / Mw (min)
If j < 1.5 then vessel is not steady by its own weight, Therefore anchor bolt are
used.
P bolt (n) = AWhere P bolt = load on bolt
N= no. of bolt
A= area of contact b/w bearing plate and foundation
(a rn) f =n P bolt
Where ar is root area of bolt
For f of bolt use table 7.5
SKIRT SUPPORT
1. Stress due to dead weight=
Where, = skirt thickness, = outer dia or dia of vessel2. Stress due to wind load =
Where = bending moment due to wind at base of vessel=
for height up to 20m
= + ( + ) for height > 20m
7/30/2019 Design data book (1).pdf
55/73
54Where, = k= K P2 h2 D0
=
D20 tsk
Where,h2= (height of Bessel + height of skirt)20mh1= ht. Up to 20m
K= 0.7P1&P2 = wind pressure
3. Stress due to seismic loadfsb=
Where,
Msb=
C W H & C =0.08
4. Maximum tensile stresssFmax= fwb - fd
Where,Fmax = permissible tensile stress
5. Max. Compressive stressfc(max)= fwb + fd
fc
yield point ( table A-1)
Calculate value of tskfrom above two formulae by equating the value of fc & ft
SADDLE SUPPORT
Horizontal cylindrical vessels are supported on saddles. A cylindrical vessel
with closure at the ends may be treated an equivalent cylinder having a
(1) Length Length
H = depth of closure
L = length of tangent lines
7/30/2019 Design data book (1).pdf
56/73
55(2) Load on support
w = uniformly distributed load
(3)
Bending moment at the support
Where,
A = distance between support nearest end of vessel.
H = height of headL = tangent to tangent length
R = radius of tank
(4) Bending moment at centre
(5) (a) If in case the stiffness is enough to maintain a circular cross section(i.e. A< 0.5R) the whole cross section is effective and therefore the stress due to
bending is given by:-
(i)At the topmost fibre of the cross-section
(ii)At the bottom most fibre of the cross-section
Where:
t = thickness of the shell
K1 = K2 = 1
7/30/2019 Design data book (1).pdf
57/73
56(b) For A > 0.5R the shell is not sufficiently stiffened by the end .
The value of the factor:-
K1 = 0.107 = 120K1 = 0.161
= 150
K2 = 0.192 = 120K2 = 0.279 = 150Stress in the shell at the mid span:-
The stress at the mid span
Arial stress in the vessel shell due to internal pressure
For design all these stresses are considerably than the permissible stress of
material.
And the combined stresses (fp+f1), (fpf2) and (fp+f3) should be within
permissible stress
BRACKET SUPPORT OR LUG SUPPORT
1) Maximum compressive load due to wind: Where K = K1K2 = const.
Pw can be calculated for height in same manner as in skirt support.
The main load on the bracket support is the dead weight of the vessel with its
contents & the wind load
The maximum total compressive load on the support is given by
} Where,
p = total forces due to wind load acting on vessel.
H=height of the vessel above foundation.
F= vessel clearance from foundation to vessel bottom.Db= diameter of the bolt circle.
7/30/2019 Design data book (1).pdf
58/73
57 W= max. wt. of vessel with attachments and its contents.
Wmax = Ws + Wi + W1 + Wa
n = no. of brackets.
2) Bracket (thickness of base plate) :From table 13.2
Vessel dia. (D) = (given) v/s A=?
No. of brackets = (given) v/s B=?
Where B= length of the base plate.
Average pressure on the base plate is given by
WhereP= total load
a = (140mm)
Maximum stress in a plate subjected to a pressure Pav & fixed at the edges is ..(1)Where f= bending stress (given)
In this case the load is only distributed on the surface of contact between the
base plate & the supporting beam; the actual stress may be taken as 40% more. .(2)For finding thickness of base plate T1 equation (2) is always used.
3) Thickness of web plates (gussets plates ):There are two web plates for each bracket.
The bending moment for each plate is = PC/2
C= (Adia of tank) / 2Stress at the edge of (3)
Where,
h=H(in c.m) from table 13.3 f=bending stress (given)
3PC=bending moment calculated above.
Calculate T2 from eq.(3)
T2 may be taken as 4 to 6mm.
7/30/2019 Design data book (1).pdf
59/73
584) Column support for bracket:It is proposed to use a channel section as column. The size chosen is ISMC 150
(from table c-3) bhatt)
Size =150
75
Area of cross section (A) =? (From table c-3)Modulus of section (Zyy) =90.4cm.
3
Radius of gyration (ryy) =? (From table c-3)
Weight (W) =164 N/m
Height from foundation (l) =given in question
Equivalent length for fixed ends (Ie) =
Slenderness ratio= Ie/ryyf = (P/A) + (P
width of flange/modules of section)
fc = (P/A)[1+(1/)(le/r)] + (Pwidth of flange/modules of section)Where, Density of material (steel)5) Base plate for column:
Size of column= _______150_______
_____20__________
Assume the base plate extend in mm. on either side of channelSide B= 0.8(width of flange) + 2(extend length)
Side C=0.95(depth of section) + 2(extend length)Extended length is always taken as 20mm.
Bearing pressure Pb = (P/number of brackets)(1/C)
(C=side C)
Pb should be less than the permissible bearing pressure for concrete.
Stress in the plate f = [(side C/2)
(extended lengtht2/10)]/(t2/6)
f= bending stress givenCalculate t (t is usually 4 to 6mm. thick.)
SADDLE SUPPORT FOR HORIZONTAL VESSEL
(1)Longitudinal bending moment at the support is
7/30/2019 Design data book (1).pdf
60/73
59Where
A = distance b/w support and its nerest end of vessel.
H = height of head.
L = tangent to tangent height.
R = radius of tank.The value of A, H is taken from table 13.3
W = uniformly distributed load.
Similarly the bending moment at the center of the span
()
(2)Longitudinal bending stress in shell at saddles
a) when supports are near the end of the vessel, so that A < 0.5 RThen,
(i)At the top most point of the cross sectionf2 =
(ii)At the bottom most fiber of cross sectionf2 =
Values of Factors K1 & K2
Condition Saddle Angle K1 K2
shell stiffened by end or rings 120 1 1
(i.e. A < R/2 or rings provided) 150 1 1
shell unstiffened by end or rings 120 0.107 0.192
(i.e. A > R/2 or no rings provided) 150 0.161 0.279
7/30/2019 Design data book (1).pdf
61/73
60
(3)Longitudinal bending stresses at midspan
(a)At the highest point of the cross section,f1 =
(b) At the lowest,
f1 =
Tangential shearing stresses
Case 1: shell not stiffened by vessel end (A > R/2)
Maximum tangential stress is given by:
It is not applicable if A > L/4
Value of K3 is depends on presence or absence of supporting rings and on
the saddle angle and is given by table below
Values of Factors K3 & K4
Condition Saddle Angle () K3 K4
A > R/2 and shell 120 1.171 unstiffened by rings 150 0.799
A > R/2 and shell 120 0.319
stiffened by rings in 150 0.319
plane of saddles
A > R/2 and shell 120 1.171
stiffened by rings 150 0.799 adjacent to saddles
7/30/2019 Design data book (1).pdf
62/73
61
Shell 120 0.880 0.401
stiffened 150 0.485 0.279
by end 120 0.880 0.880
of vessel 150 0.485 0.485
Circumferential stresses
(a) At the lowest point of the cross section,
(b)At the horn of the saddle,If L/R > 8, f4 =
If L/R < 8, f4 =
Stress can be reduced by welding a reinforcing backing plate.
If width of this plate > B + 10t and if its angle from the centre of cylinder> ( + 12) degree,
Then, substitute t with t + t1
Where, t1 = thickness of backing plate
Value of K5 & K6 are given below
K5 = 0.760 for = 1200
K5 = 0.673 for = 1500
Values of K6
A/R = 1200 = 1500
00.5 0.013 0.007
0.6 0.018 0.010
0.7 0.030 0.017
0.8 0.034 0.021
7/30/2019 Design data book (1).pdf
63/73
620.9 0.047 0.028
1.0 0.052 0.031
1.1- 3.0 0.055 0.033
Ring stiffeners
Ring stiffener is designed from the following correlation:
f = allowable compressive stress
Ar= cross section area of the stiffening ring,(thickness width of rectangular cross section),
Z = section modulus of ring cross section.
Values of K7 & K8as a function of saddle angle, , are given below
Values K7 and K8
Saddle Angle () K7 K8
120 0.0560 0.0528
150 0.0210 0.0316
Design of Saddle
Horizontal component of all radial loads may be determined by following
equations
F = K9 W1
Where, K9= 0.204 for = 1200
0.260 for = 1500
7/30/2019 Design data book (1).pdf
64/73
63
1.Stresses in a thick cylinder:-
= piDi2p0D0
2/ D02Di
2
= [piDi2p0D0
2/ D02Di
2][(pi-p0)Di2D0
2/D2(Do2-Di
2)] ,
=[piDi2p0D0
2/ D02Di
2]+[(pi-p0) Di2D0
2/D2(Do2-Di
2)]]
Where,
= y/F =stress,
D= any diameter where stress is evaluated
Di= internal diameter
Do= external diameter
P0=pressure acting inside the jacket
Pi=the inside shell pressure
And , D0=Di+2t
Or , D=Di, For maximum stress
, K=D0/Di
For external jacket thickness,
= p0(K2+1)/(K2-1)
Where,
D0=Jacket outside diameter
Di= jacket inner diameter
Again for,maximum stress ,D=Di,K= (D0
/Di)
DESIGN OF THICK WALLED HIGH PRESSURE VESSEL
7/30/2019 Design data book (1).pdf
65/73
642.Theories of elastic failure:-
a)Maximum principal stress theory:-
(max) = y/F=pi(k2+1)/(k2-1)
Where,
e=yield point of the material
P= internal pressure
And,K= (D0/Di)
Than calculate ,t(thickness).
b)Maximum strain theory:-
(max) = y/F=pi [(1-)+(1+ )k2/(k2-1)]
Where,
=poissions ratio
And, K= (D0/Di)
Than calculate ,t.
c)Maximum strain energy theory:-
= y/F=Pi(6+10k4)1/2/2(k2-1)
And, K= (D0/Di)
Than calculate ,t.
d) Maximum shear theory:-
(i) when maxi. Shear stress equals to the shear stress set up in the material at
elastic limit:-
=1/2(- r)=1/2 y
or, y/F = [2k2/(k2-1)]Pi=2 (max)
(ii) when elastic break down by maximum shear :-
= y/pi = (3)1/2k2/(k2-1)
7/30/2019 Design data book (1).pdf
66/73
65
SHELL DESIGN:
1.Head of liquid (or height of tank) is calculated as:
H = V/r2
Where,
H= head of liquid (m)
V = volume (capacity) of tank (m3)
R = Di/2 = inner radius (m)
[R can be calculated from table]
[if only V is given ,H and R can be calculated from the table]
2. Number of layers of plates in shell:
n = H/Width of plate
Where ,
H = height of tank (m)
Width of plate = 1.8 (from standard dimensions 6.3mx1.8mx1m)
3.Number of plates in a single layer:
=Di /length of plate
Where,
Di = inner dia. of shell(m)
Length of plate = 6.3m(standard)
DESIGN OF STORAGE TANK
7/30/2019 Design data book (1).pdf
67/73
664. Total number of plates used in shell:
=number of layers x plates used in a single layer
5. Internal pressure of shell
P= (H-0.3)g
Where,
P= internal pressure in N/m2
= density of liquid in Kgf/m3
H=height of tank in (m)
g= accelaration due to gravity(10 m/s2)
6.Thickness of plate
We can calculated the thickness of plates for each layer of the tank up to the
total height of tank by the following formula:-
t= [50 (He-0.3)x DiG/fJ] +C
Where, t=thickness
6. Average thickness of shell platestavg = (t1 + t2 + t3+ + tn) / n
Where,
t1, t2, t3, tn the thickness of the respective layers and n is the
number of layers
8.Stability check
If H1 > H then our calculated thickness is correct
Where,
H = height of tank in m
And
H1 = 1500 [tavg / P] [tavg /Di]
3/2
(m)
7/30/2019 Design data book (1).pdf
68/73
67Where,
P = superimposed load, wind load, sum of all external pressure acting on the
tank in kg/ m2.
Di = inner dia of tank in m.
Tavg =average thickness in mm.
BOTTOM DESIGN:
1. Thickness of bottom plate
From IS Code803-1976
Tank diameter thickness of bottom plate
>12 m 6 mm
370 Supprted Roof
7/30/2019 Design data book (1).pdf
69/73
68Where = roof curb angle
For determining act, first we should assume roof as self supporting conical
roof for which = 370
1. Thickness of Roof Plate
t = Di/ 5 sin
Where,
Di = inner dia of shell in m
= Roof curb angle (370)
t = Thickness of roof plate in m.
2. Dead load on Roof
Dead load = (Thickness of roof plate in m) . (density of plate material in kg/m3
3. Total load on Roof
Total load = super imposed load in kg / m2 + Dead lad in kg /m2
4. Actual Slope of Roof
sin act = [Di/ t] [P / 0.202 E]1/2
where ,
Di = inner dia of shell in m
t = thickness of roof plate in m
P = total load on roof in kg / m2
E = Modulus of elasticity in kg / m2 (table no.)
From here act can be calculated compare act that either > 37 or 37 and
decide what roof will be allowable as given in earlier conditions.
7/30/2019 Design data book (1).pdf
70/73
69SELF SUPPORTING ROOF DESIGN
1. Actual thickness of Roof plateIf actual is less than or equal to 37 then we calculate actual thickness of the
roof plate as given below
tact = Di/ 5 sin act
Where,
Di = inner dai of shell in m
act = actual Roof Curb Angle
tact = Actual Roof Plate thickness in mm
1. Roof LoadingFor self supporting roofs a uniform load of 125 kgf / m2is assumed2. Internal pressure
An internal pressure equivalent to
1. 75 kg / m2 for non pressure tanks2. 200 kg / m2 for class A tanks3. 550 kg / m2 for class B tanks3. ROOF SHAPE
The roof shape may have the following forms(a) Cone roof(b) Dome roof
(c) Umbrella roof
SUPPORTED ROOF DESIGNS
1) NO. of rafter on outer periphery = circumference / rafter spacing= d / 2 m ( assumed )
Where D = dia of shell in m
2) Actual spacing between two rafters = circumference / no. of rafters= D/ no. of rafters
3) Selection of central supportIS Code 8031976
7/30/2019 Design data book (1).pdf
71/73
70Diameter of tank type of central support6 -12.5 m one centre column
12.515 m circular15.20 m square
20.25 m pentagonal25.30 m hexagonal
4) No. of rafter plate girder= total no. of rafters / no. of girders (sides of polygon)
5) LENGTH OF SIDE OF POLYGON(a)a= D/cosec(180/n)
Where,
D= dia of the circle which contains the polygon
n= no. Of sides of polygon
6) Length of rafterAccording to IS Code 803-1976, we cannot take a rafter of length greater than
7.5m.
But in the case of tank of radius greater than 7.5 m it creates problem. Thereforewe have to spilt the rafter into two such parts that no one should be greater than
7.5 m. Hence choose the internal support under a circle which divides the radius
of tank into two parts that they are always less than 7.5m
1) Length of inner rafter =
2) Length of outer rafter =
Where
D = dia of tank in m
D1 = dia of inner circle in m
7) Perimeter of polygon= no. of sides length of a side
8)
Area of polygon A= n . a2 .Cot[ 180/n]
7/30/2019 Design data book (1).pdf
72/73
71Where ,
A= area of polygon in m2
n= number of sides (grider)
a= length of aside of a polygon in m
9) No. of inner rafter= periphery of polygon (n.a) / inner rafter spacing (2m)
10) Actual rafter spacing= periphery of polygon / no. of inner rafters
11) No. of inner rafter per girder
= no. of inner rafter / no. of sides of polygon
12) Total load on roof= surface area of cone density of roof material thickness of plate
= ( R L t) KGfSurface area of cone = R L
Where,
R= D/2 = radius of tank in mL= H = R/16 = height of cone roof
Density (p) of roof material in Kg/m3
Thickness (t) of roof plate = 6mm (from IS Code 803- 1976)
13) Load on polygon (Kgf)= area of polygon
density of roof material
thickness of roof plate
14) Load on outer rafter = total loadioad on polygon15) Load per outer rafter
= load on outer rafter / no. of inner rafter
16) Load on inner rafter =load on polygon17) Load per inner rafter = load on inner rafter / no. of inner rafter
7/30/2019 Design data book (1).pdf
73/73
7218) Load per grider = total load on roof (W) / 2n
Where, n = no. of sides of polygon
19) Bending moment (m) = WL2 / 8 (Kgf m2)Where,
W= total load on roof in Kgf
L= in m20) Section of modules (z) = B M / stress = M / F
With the help of Z and M by using steel plate we can find out the greater size.