Design data book (1).pdf

MADHAV INSTITUTE OF TECHNOLOGY & SCIENCE DEPARTMENT OF CHEMICAL ENGINEERING GWALIOR- 474005

DATA BOOK PROCESS EQUIPMENT & DESIGN

DHEERAJ SHUKLA CHEMICAL ENGINEERING, 3rd YEAR CM10008

1

INDEX

Sr. No. 1. 2. 3. 4. 5. 6. 7. 8. 9.

Content Unit Conversion Design of Process Vessel under Internal Pressure Design of Head & Closure Design of Non-standard Flanges Design of Process Vessel & Pipes under External Pressure Compensation for opening in Process Vessel Design of Tall Vessel Design of Supports for Process Vessel Design of thick walled High Pressure Vessel

Page No. 2 5 10 17 27 32 39 47 63

2

UNIT CONVERSION Conversion of units from existing units to S.I. Units: Quantity Length

Time

Mass

Area

Density

Acceleration Energy(Torque)

Force

Existing Unit 1inch 1ft 1yd 1mile 1Å 1min 1hr 1day 1year 1oz 1lbm 1cwt 1ton 1in.2 1mm2 1cm2 1ft2 1yd2 1kg/l 1lb/ft3 1lb/UK gal 1ib/US gal 1g/cm3 1cm/s2 1ft/s2 1erg 1ft pdl 1ft lbf 1cal 1kgf m 1Btu 1Chu 1hp-hr(metric) 1hp-hr(British) 1kW h 1 dyne

S.I. Unit 0.0254m 0.3048m 0.9144m 1.6093m 10-10m 60s 3600s 86.4*103s 31.5*106s 28.352*10-3kg o.454kg 50.8023kg 1016.06kg 645.16*10-6m2 1*10-6m2 1*10-4m2 9.2903*10-2m2 8.3613*10-1m2 1*103kg/m3 16.018kg/m3 99.779kg/m3 119.83kg/m3 103kg/m3 1*10-2m/s2 0.3048m/s2 1*10-7J 4.2139*10-2J 1.3558J 4.1868J 9.8067J 1055.1J 1.8991*103J 2.6477*106J 2.6845*106J 3.6*106J 1*10-5N

3

Volume

Velocity

Viscosity(dynamic)

Frequency Specific heat capacity

Temperature Difference

Thermal Conductivity

Power

Pressure(Stress)

1 pdl 1lbf 1kgf 1tonnef 1tonf 1in3 1 US gal 1 UK gal 1ft3 1 barrel(petroleum US) 1 lube oil barrel 1 yd3 1 ft/h 1 ft/min 1ft/s 1 mile/h 1 mN s/m2(cp) 1 lb/ft h 1g/cm s (poise P) 1 lb/ft s 1 c/s 1 cal/gm °C 1 Btu/lb °F 1 Chu/l b°C 1 °C 1 °C 1°R 1 Btu/h ft2(°F/in) 1 kcal/h m °C 1 Btu/h ft °F 1 cal/s cm °C 1 ftlbf/min 1 ftlbf/s 1 m kgf/s 1 hp (metric) 1 hp (British) 1 dyne/cm2 1 Pascal 1 kgf/m2 1 mm water 1lbf/ft2

1.3825*10-1N 4.4482N 9.8067N 9.8067*103N 9.9640*103N 1.6387*10-5m3 3.7853*10-3m3 4.546*10-3m3 2.8317*10-2m3 0.15898m3 0.20819m3 0.76455m3 8.4667*10-5m/s 5.08*10-3m/s 0.3048m/s 0.44704m/s 1*10-3N s/m2 4.1338*10-4N s/m2 0.1 N s/m2 1.4882N s/m2 1Hz 4.1868*103J/kg K 4.1868*103J/kg K 4.1868*103J/kg K 1K 5/9 K 5/9 K 0.14423J/s m K 1.163J/s m K 1.7308J/s m K 418.68J/s m K 2.2597*10-2J/s 1.3558J/s 9.8065J/s 7.3548*102J/s 7.457*102J/s 0.1N/m2 1 N/m2 9.8067 N/m2 9.8067 N/m2 47.88 N/m2

4

Moment of inertia Momentum Angular momentum Viscosity(kinematic)

1 cm water(gf/cm2) 1 mbar 1 matm 1torr(mmHg) 1 in water 1 ft water 1 in Hg 1 lbf/in2(psi) 1 m water 1 2 at(kgf/cm orkp/cm2) 1 bar 1 atm 1N/mm2 1tonf/in2 1 lbft2

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*103 N/m2 6.8948*103N/m2 9.8067*103 N/m2 9.8067*104 N/m2 1*105 N/m2 1.0133*105N/m2 1*106 N/m2 1.5444*107 N/m2 0.04214 kg/m2

1 lbft/s 1 lb ft2/s 1 S(stokes) 1 ft2/h

5

Design of process vessel under internal Pressure  Cylindrical  Spherical

Design of cylindrical and spherical vessels under internal pressure 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

σz =

Thin wall thickness for spherical shell (i)Internal pressure

P=

or

6

(ii) Minimum thickness of wall

t=

or

(iii)Circumferential and longitudinal stress

= σz =

or

Where, P = Internal design pressure P = 1.05 * max working pressure J = Joint efficiency factor J= 0.85 double welded butt joint with full penetration 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

= outer diameter = inner diameter

THICK WALL VESSEL PRESSURE

There are three stresses applied (i) (ii) (iii)

Longitudinal stress Radial stress Hoop stress

(1) Longitudinal stress

= (2) Radial stress –

= Where,

D = diameter of shell where stress is to be calculated (3) Hoop stress =

8

Stress at internal surface (i)

Longitudinal stress

=

In this case Po = 0

=

, If Do = 0 then =

(ii)

Pi

Radial stress

= Where, K = (iii)

Hoop stress

=

Stress for external pressure

= 0 =

zi

= The maximum shear stress at any point in the cylinder

= = =

σ

σ

9

Theory of failure (1) Maximum shear stress theory

= Where, =yield stress Pi = Internal pressure K= (2)Maximum strain theory

=

[

]

,

= Poisson ratio, is obtain by table 1 (3)Maximum strain energy theory

=

√

10

Design of Head and Closures Design of flat head Thickness of flat head t = C De√

, De = effective diameter =Di

F= allowable stress of material. Design of cylindrical and spherical vessels under C = design pressure constant internal

C depends on the method of attachments to the shell. Cases follow: I. II. III.

Flanged flat head butt welded to shell Plates welded to the inside of the shell Plates welded to the end of the shell (no inside welding)

IV.

Covers riveted or bolted with full face gaskets to shell flanges or side plates

V.

Covers with a narrow fane bolted flanged joint is placed within the bolts holes

C

(

)

Fb = bolt load VI.

Plates welded to the end of the shell with an additional fillet weld on the inside. This thickness is theoretically calculated to this 2mm thickness.

Corrosion allowance is to be added and another say 6% is to be added to take care of the reduction in thickness at the torus section. This gives a practically required min. thickness. The value of C is generally taken as 0.45.

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 or reinforced 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

12

ri 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/D0 by 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 ,t 25mm =D0+(D0/42)+(2/3)rI+2Sf+t ,For , t 25mm External height of the excluding straight flange h0 =R0 √(

)

V, Excluding straight flange = 0.0847 =0.1313

for ,ri= .06 Di

, 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

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) =

/24

ELLIPTICAL HEAD Thickness of elliptical head

tn= PDV/2fJ where , P=internal pressure D=major axis V=stress intensification factor = ¼(2+K2) K=major axis =272.6

HEMISPHERICAL HEAD Neglecting thinning effect C = 2fJt / PD0 , J=1 D0 = outer diameter of shell he/D0 = 0.5 ,From table 3 obtain d/√

14

Volume of elliptical dished head Vn = (

/4) (D0/6) =

/24

Total volume contain in vessel where D is internal diameter }2+{

Vvessel=[{

}2]

Volume of torispherical dished head to straight flange V=0.000049 Where, di=inside diameter of vessel in inches

CONICAL HEAD (1)

Thickness of conical head at junction t= 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 Z

20 1.00

30 1.35

45 2.05

Surface area , A = (1/2) Volumetric capacity= (1/3) (2)

(h/4)

thickness away from the junction t=

(

)

, P=design pressure Dk=internal diameter of cone at a distance L

60 3.20

15

L=(1/2)√ t = thickness of shell+corrosion allowance From the junction, Dk=Di 2Lsin J =0.85 Di=Do 2t ,Do and Di= external and internal diameter

Table 1 Material Aluminium Brass Copper Iron Nickel Steel

Specific weight ( ⁄ 2.65 8.35 8.74 7.74 8.74 7.70

Poisson ratio = /E 0.34 0.35 0.35 0.28 0.36 0.30

Table 2 Stress concentration factor C for formed heads without opening or with fully compensated opening t/D0 hE/D0 0.002 0.005 0.01 0.02 0.04 0.15 4.55 2.66 2.15 1.95 1.75 0.20 2.30 1.70 1.45 1.37 1.32 0.25 1.38 1.14 1.0 1.00 1.00 0.30 0.92 0.77 0.77 0.77 0.77 0.40 0.59 0.59 0.59 o.59 0.59 0.50 0.55 0.55 0.55 0.55 0.55

16

Table 3 Stress concentration factor C for formed heads with uncompensated opening hE/D0 0.15 0.20 0.25 0.30 0.50

0.5 1.67 1.28 1.00 0.83 0.60

1.0 1.86 1.45 1.15 1.00 0.80

d/√ 2.0 2.15 1.85 1.60 1.45 1.10

3.0 2.65 2.30 2.05 1.88 1.50

Note: values can be interpolated

4.0 3.10 2.75 2.50 2.28 1.85

5.0 3.60 3.25 2.95 2.70 2.15

17

Design of Non-standard flanges  Gasket dimensions 1/2

do/di = [ ( Y –pm ) / {spherical Y – p (m+1) } vessels ] Design of cylindrical and under Where, internal pressure 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 = (do – di)/2

2. Gasket seating width bo = N/2

18

3. Effective gasket seating width b = bo 1/2 (b) b = 2.5 ( bo )

(a)

when bo≤ 6.3 mm when bo> 6.3 mm

4. Diameter at location of gasket load G (a) G=di + N when b < 6.3 mm (b) G=do – 2b 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 bolt diameter.

6. Minimum bolt circle diameter C = B + 2 (g₁ + R) or C = n Bs/π Where, C = bolt circle diameter, B = inside diameter of flange, g₁= 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.

19

 Estimation of Bolt Loads:

1. Load due to design pressure: H=πG2p/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=πG (2b) mp Where; p= design pressure m= gasket factor G=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= πGboy y=minimum gasket seating stress (table 2)

20

5. 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 condition Ag = 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)

21

 Check for gasket width:To prevent damage to gasket during bolting up condition following condition to be satisfied

Ab Sg / πGN < 2y

(From table 3)

 Determination of flange moments

(a) Operating condition 1. Total Load Wo = W1 + W2 + W3 W1 = (π B² / 4) p W2 = H – W1 W3 = WO – H =Hp Where; W1 – Hydrostatic end force on area inside of flange H- Load due to design pressure Hp- Load to keep joint tight under operation 2. Total flange moment Mo = W1 a1+ W2a2 + W3a3 The values of a1, a2 and a3 for different flange type Assume lap joint flange and use Table 5

22

(b) Bolting up condition 1. Total flange moment Mg = W a3 Where; W = (Am + Ab) Sg / 2 a3 = (C – G)/ 2

TABLE 2 Allowable Stresses for Bolting Materials in MN/m2 Allowable stress MN/ m2 for design metal temperature not exceeding (0˚C)

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

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

85.0

78.5

76.0

73.2

72.0

23

TABLE 3 FLANGES Bolt size

Root area

M 16 x 1.5 1.54 x 10-

Min. no.of bolts

Actu al no of bolts (n)

R(m)

Bs (m)

C= nBs/ π (m)

C=ID+2(1.4 15go+R) (m)

50.8

52

0.025

0.07 5

1.24

1.0583

43-7

44

0.027

0.07 5

1.05

1.0623

33.7

36

0.030

0.07 5

0.86

1.0683

4

M 18 x 2

1.54 x 104

M 20 x 2

2 x 10-4

TABLE 4 I. loose type flange

B

g1

1.lap joint flange

Outside dia.

g1=g0

2. Raised face with hub

Outside dia

g1=0.5 g0

1.ring only plane face

Outside dia

g1= g1

2. lap weld hub raised face

Outside dia

g1=21/2 g0

1. plane face with weld hub

Outside dia

g1=21/2 g0

2, Ring only type raised face

Outside dia

g1=g0

II. Integral type

III. Optional type

24

Table 5 Moment arms for flange loads under operating conditions Type of flange Internal type flanges Loose type except lap joint flanges Lap joint flanges

a1 R + (g1/2)

a2 (R + g1 + a3) / 2

a3 ( C – G)/ 2

( C – B)/ 2

(a₁+ a₃)/2

( C – G)/ 2

( C – B)/ 2

( C – G)/ 2

( C – G)/ 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

Vulcanized rubber sheet hardness above 70 IHRD } Asbestos 3.2mm with a Thick suitable binder for 1.6mm operating conditions

Gasket factor m

Min. actual gasket width (mm)

1.00

Min. design seating stress, Y ,MN/m² 1.38

2.00

11.00

10

2.75

25.50

10

10

25

0.8mm

3.50

44.85

10

1.25

2.76

10

2.25

15.25

10

2.50

20.00

10

2.75

25.50

10

Vegetable fibre Spiral} Carbon wound steel metal, asbestos S.S.ormonel filled metal Corrugated } metal, asbestos Soft Al inserted or Soft asbestos filled Cu/brass corrugated Iron/soft metal jacket steel Monel metal S.S.

1.75 2.50

7.56 20.00

10 10

3.00

31.00

10

2.50 2.75 3.00 3.25 3.50

20.00 25.00 31.00 38.00 45.00

10 10 10 10 10

}

2.75 3.00 3.25 3.50 3.75

25.50 31.00 38.00 45.00 52.00

10 10 10 10 10

Rubber with cotton fabric insertion } Rubber with 3asbestos fabric ply insertion, with or without wire reinforcement 2ply

1ply

Corrugated metal

Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S.

26

Asbestos filled } flat metal jacket

3.25 3.50 3.75 3.50 3.75

38.00 45.00 52.00 55.00 62.50

10 10 10 10 10

Solid flat metal

4.00 4.75 5.50 6.00 6.50

61.00 90.00 125.00 150.00 180.00

6 6 6 6 6

5.50 6.00 6.50

125.00 150.00 180.00

6 6 6

Ring joint

}

Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. } Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. Iron/soft steel Monel metal S.S.

27

Design of process vessel and pipes under external pressure

(1) CRITICAL LENGTH BETWEEN Design of cylindrical andSTIFFENERS: spherical vessels under internal pressure √ = 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 STIFFENER RING: Design is to be check for elastic instability or plastic deformation

28

If type of head and closures are not given then consider the vessel has torispherical 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 hᵢ 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 + 2/3hᵢ (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

29

Value of K and m as a function of D0/L ratio of given in table below D0/L 0 0.1 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2.0 3.0 4.0 5.0

K 0.733 0.185 0.224 0.229 0.246 0.516 0.660 0.879 1.572 2.364 5.144 9.037 10.359

M 3.00 2.60 2.54 2.47 2.43 2.49 2.48 2.49 2.52 2.54 2.61 2.62 2.58

Checking for plastic deformation If D0/L ≤ 5 or (

)

, then

(

) (

)

= allowable compressive stress U= 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.

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˳/L≤5

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

31

Now, Select a 18 cm channel of following specification Weight (Wt) =14.6 kg As= 1.84 I=8.9

m² ,

No. of stiffener required = 5 Total weight of ring = πDoWt × no. of stiffener rings Saving in shell material for using stiffening rings: = D0× (t-ts) Lπ = tangent to tangent length

32

COMPENSATION FOR OPENINGS IN PROCESS VESSEL

1. Area to be compensated

Design of cylindrical and spherical vessels under internal pressure Where, d = internal diameter of nozzle,m

= (outside dia of nozzle - 2×thickness 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 √

33

If 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 If

, then no other external reinforcement necessary. , the difference in area ring pad weldments

is to be provided with

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:

34

Inner 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 allowable stress 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.

35

UNCOMPENSATION FOR OPENINGS IN PROCESS VESSEL Uncompensated Opening

Design of cylindrical and spherical vessels under  K factor: internal pressure 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 unreinforced depending upon the shell diameter.

 Weakening factor:

Where, is pressure required to cause 0.2% permanent deformation near the opening. Pressure required yielding the unpierced shell.  Theoretical shell thickness for uncompensated opening:

If opening is made away from welded joints, J=1.

36

Table 1 ⁄ √

⁄ 0.0 0.25 0.5 0.75 1.0 1.5 2.0 2.5 3.0 3.5

1.000 0.900 0.785 0.700 0.645 0.545 0.465 0.390 0.340 0.285

√ 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

0.245 0.215 0.180 0.155 0.130 0.115 0.090 0.080 0.075

Where, D0 is the shell outside diameter. d0 is the opening diameter. ts is the actual shell thickness. c is the corrosion allowance. Φ 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 radius equal to crown radius.

37

(b) For semi-ellipsoidal end: When the opening and its compensation are in ellipsoidal end and are located entirely within a circle having a radius, measured from the centre of the end, of 0.40 of the shell diameter, tr is the thickness required for a sphere having a radius R, derived from the following table: Table 2 ⁄

⁄

⁄

⁄

0.167 0.178 0.192 0.207 0.227 0.25

1.36 1.27 1.18 1.08 0.99 0.90

0.277 0.312 0.357 0.40 0.45 0.50

0.81 0.73 0.65 0.59 0.54 0.50

Compensation for Multiple Openings Interaction between two openings, if their edge distance is roughly

⁄ :

√ Where „L‟ is the pitch and„d‟ 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:

38

(

) (a)If the openings are along longitudinal direction:

(b)If openings are along circumference or on sphere:

39

DESIGN OF TALL VESSELS 1. Determination of shell thickness: ts = PDO / (2fJ+p)+C Design of cylindrical and spherical vessels Where, internal ts = thickness of shell under pressure 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: σZp = P D2/4t(Di+t) , D = Di+t And, σZp =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: σZp=WS/ πt(Di+t) Where, WS= (πDtXϒS) Where, ϒS=specific wt.(from table given below) WS=wt. of shell of length X meter t= shell thickness at the point under consideration

40

Material Aluminium Brass Copper Iron Nickel Steel

Specific wt. N/m3 2.65x104 8.35x104 8.79x104 7.74x104 8.74x104 7.70x104

Poisson’s ratio 0.34 0.35 0.35 0.28 0.36 0.30

b) The stress induced in the shell due to a distance X meters from the top: σZi= Wi/πt(DI+t) Where, Wi=(πDtinsϒins ){wt. of insulation up to a distance for a length of X meters from the top} tins= insulation thickness ϒins= 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 σZi = Wl/πt(Di+t) Wl= (π/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: σza= Wa/πt(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 = (π/4)(dia. of platform)2(platform loading) Wt. of trays = (π/4)(tray loading)(no. Of trays)

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, σZw, acting along the axial direction of shell at the point is given by: σZw=σZs + σZi + σZl + σZa 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 ρVω2 A Where ,CD =drag coefficient ρ = density of air Vω= 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 Vω2 Where,pw= wind stress in N/m2 or min. Wind pressure

42

Vω= 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 with moderate wind

0.4

1.0

 The total load due to wind acting on the bottom and upper parts of the vessel 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)

43

p2= 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)

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 the following correlation: σzwm= 4 Mw/πt(Di+t)Di Where, σzwm=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, σz = σzp + σzwm - σzw σz,tensile(maximum)= fJ forv external pressure σz=( σzwmor σzsm )- σzp - σzw b) The resultant compressive stress (on downwind side) is given by: For internal pressure:σz= σzwm + σzw - σzp

45

σz,compressive,(maximum) = 0.125E(t/Do) Where,E =modulus of elasticity For external pressure:σz= (σzwm or σzsm) + σzw + σzp Check : (safe design) Equivalent stress , σe = (σe2 - σe σz + σz2 )1/2 Here, σe = hoop stress =P(Di+t)/2t or P(Do-t)/2t σz= tensile stress  Now calculate the value of σz from: σz = σzp + σzwm - σzw And substitute the value of σzp & σe in the equation of equivalent stress and then check it from, σe = fJ  if σe (calculated)< σe (check design condition) then our calculated thickness is correct here σe is calculated by putting X = height of tower – height of skirt or tangent to-tangent height

Check for safe design: At design conditions, 1) σe < Fj 2) σ (tensile)< fJ 3) σz(compressive)< 0.125E(t/Do) At test conditions, 1) σz < 1.3 fJ 2) σz(tensile) 20m D= 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)

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)

51

(0.85 for double welded butt joint for class 2 cons) To find thickness, equate …………..(1) Maximum compressive stress due to maximum load is computed as: max

max

For safe compressive stress:

⁄

Where; E = young‟s modulus at design temp. ts = skirt thickness D0 = dia. Of skirt (outer dia of shell when skirt is cylindrical) α = half the top angle of conical skirt … (100maximum) (00 for cylindrical skirt) To find thickness Equate max

……………… (2)

The thickness which is maximum for (1) & (2) is considered as per IS 2852 1969, minimum corroded thickness of skirt is 7mm and taking 1mm as corrosion allowance. DESIGN OF SKIRT BEARING PLATE The maximum compressive stress b/w the bearing plate and the concrete foundation is given by (Max)=

+

A= π (

– l)l

Where, A = area of contact b/w bearing plate and concrete foundation is given by = outer dia of skirt

52

l = outer radius of bearing plate – outer radius of skirt Z= π 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√

=

M (max) = For b=1

(max)

bl( )

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 =√

53

ANCHOR 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) = A Where P bolt = load on bolt N= no. of bolt A= area of contact b/w bearing plate and foundation (a r n) 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,

2.

= skirt thickness,

= outer dia or dia of vessel

Stress due to wind load = Where

= bending moment due to wind at base of vessel = =

for height up to +

(

+

20m

) for height > 20m

54

Where, =k = K P2 h2 D0 = D20 tsk Where, h2= (height of Bessel + height of skirt) – 20m h1= ht. Up to 20m K= 0.7 P1&P2 = wind pressure 3.

Stress due to seismic load fsb=

(

)

Where, Msb= C W H & C =0.08 4.

Maximum tensile stresss Fmax= fwb - fd Where, Fmax = permissible tensile stress

5.

Max. Compressive stress fc(max)= fwb + fd

fc yield point ( table A-1) Calculate value of tsk from 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

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 head L = 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

56

(b) For A > 0.5R the shell is not sufficiently stiffened by the end . The value of the factor:K1 = 0.107 = 120 K1 = 0.161 = 150 K2 = 0.192 = 120 K2 = 0.279 = 150 Stress 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), (fp – f2) 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.

57

2)

∑ W= max. wt. of vessel with attachments and its contents. Wmax = Ws + Wi + W1 + Wa n = no. of brackets. 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

Where P= 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= (A – dia of tank) / 2 ⁄ ⁄ Stress 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.

58

4) 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/ryy f = (P/A) + (P width of flange/modules of section) fc = (P/A)[1+(1/ )(le/r)] + (P width 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 channel Side 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 given Calculate t (“t” is usually 4 to 6mm. thick.)

SADDLE SUPPORT FOR HORIZONTAL VESSEL (1) Longitudinal bending moment at the support is [

( )

]

59

Where 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 R Then, (i) At the top most point of the cross section f2 = (ii) At the bottom most fiber of cross section f2 ‟ = 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

(i.e. A > R/2 or no rings provided)

150

0.107 0.161

0.192 0.279

60

(3)Longitudinal bending stresses at mid – span (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 A > R/2 and shell unstiffened by rings A > R/2 and shell stiffened by rings in

A > R/2 and shell stiffened by rings

Saddle Angle ( )

K3

K4

1.171 0.799

… …

120 150 plane of saddles

0.319 0.319

… …

120 150 adjacent to saddles

1.171 0.799

… …

120 150

61

Shell stiffened by end of vessel

120 150 120 150

0.880 0.485 0.880 0.485

0.401 0.279 0.880 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 0 – 0.5 0.6 0.7 0.8

= 1200

= 1500

0.013 0.018 0.030 0.034

0.007 0.010 0.017 0.021

62

0.9 1.0 1.1- 3.0

0.047 0.052 0.055

0.028 0.031 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 & K8 as a function of saddle angle, , are given below Values K7 and K8

Saddle Angle ( ) 120 150

K7 0.0560 0.0210

K8 0.0528 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

63

DESIGN OF THICK WALLED HIGH PRESSURE VESSEL

1.Stresses in a thick cylinder:-

Design of cylindrical and spherical vessels σɵ = piDi2 – p0D02/ D02 –Di2 under internal pressure σ = [p D 2 – p D 2/ D 2 –D 2] –[(p -p )D 2D 2/D2(D 2-D 2)] , ɵ

i

i

0

0

0

i

i

0

i

0

o

i

σɵ=[piDi2 – p0D02/ D02 –Di2]+[(pi-p0) Di2D02/D2(Do2-Di2)]] Where, σɵ= σy/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, σɵ= p0(K2+1)/(K2-1) Where, D0‟=Jacket outside diameter Di‟= jacket inner diameter Again for,maximum stress ,D‟=Di‟,K= (D0‟/Di‟)

64

2.Theories of elastic failure:a)Maximum principal stress theory:σɵ(max) = σy/F=pi(k2+1)/(k2-1) Where, σe=yield point of the material P= internal pressure And,K= (D0/Di) Than calculate ,t(thickness). b)Maximum strain theory:σɵ(max) = σy/F=pi [(1- )+(1+ )k2/(k2-1)] Where, =poission‟s ratio And, K= (D0/Di) Than calculate ,t. c)Maximum strain energy theory:σɵ = σy/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:τ=1/2(σɵ- σr)=1/2 σy or, σy/F = [2k2/(k2-1)]Pi=2 τ(max) (ii) when elastic break down by maximum shear :σɵ= σy/pi = (3)1/2k2/(k2-1)

65

DESIGN OF STORAGE TANK

SHELL DESIGN:

Design of cylindrical and spherical vessels under internal pressure H = V/πr2

1.Head of liquid (or height of tank) is calculated as:

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: =πDi /length of plate Where, Di = inner dia. of shell(m) Length of plate = 6.3m(standard)

66

4. Total number of plates used in shell: =number of layers x plates used in a single layer

5. Internal pressure of shell P=ρ (H-0.3)g Where, P= internal pressure in N/m2 ρ= 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 plates tavg = (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)

67

Where, 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 Code – 803-1976 thickness of bottom plate

Tank diameter >12 m

6 mm

370

Self Supported Roof Supprted Roof

68

Where = 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.

69

SELF SUPPORTING ROOF DESIGN 1. Actual thickness of Roof plate If 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 Loading For self supporting roofs a uniform load of 125 kgf / m2is assumed 2. Internal pressure An internal pressure equivalent to 75 kg / m2 for non pressure tanks 200 kg / m2 for class A tanks 550 kg / m2 for class B tanks

1. 2. 3. 3. ROOF SHAPE

The roof shape may have the following forms (a) Cone roof (b) Dome roof (c) Umbrella roof

SUPPORTED ROOF DESIGNS

1)

2) 3)

NO. of rafter on outer periphery = circumference / rafter spacing = π d / 2 m ( assumed ) Where D = dia of shell in m Actual spacing between two rafters = circumference / no. of rafters = π D/ no. of rafters

Selection of central support IS Code 803 – 1976

70

Diameter of tank 6 -12.5 m 12.5 – 15 m 15.20 m 20.25 m 25.30 m

4)

type of central support one centre column circular square pentagonal hexagonal

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 rafter According 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. Therefore we 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) 2)

Length of inner rafter = Length of outer rafter =

Where D = dia of tank in m D1 = dia of inner circle in m

7)

8)

Perimeter of polygon = no. of sides

length of a side

Area of polygon A= ¼ n . a2 .Cot[ 180/n]

71

Where , 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 = ( R L t) KGf Surface area of cone = π R L

thickness of plate

Where, R= D/2 = radius of tank in m L= √ 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 load – ioad on polygon 15) Load per outer rafter = load on outer rafter / no. of inner rafter

16) Load on inner rafter =load on polygon 17) Load per inner rafter = load on inner rafter / no. of inner rafter

72

18) 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 m

20) 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.