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

INDEX

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 32

7. Design of Tall Vessel 39

8. Design of Supports for Process Vessel 47 9. Design of thick walled High Pressure Vessel 63

Conversion of units from existing units to S.I. Units:

Quantity Existing Unit S.I. Unit

Length 1inch 1ft 1yd 1mile 1Å 0.0254m 0.3048m 0.9144m 1.6093m 10-10m Time 1min 1hr 1day 1year 60s 3600s 86.4*103s 31.5*106s Mass 1oz 1lbm 1cwt 1ton 28.352*10-3kg o.454kg 50.8023kg 1016.06kg Area 1in.2 1mm2 1cm2 1ft2 1yd2 645.16*10-6m2 1*10-6m2 1*10-4m2 9.2903*10-2m2 8.3613*10-1m2 Density 1kg/l 1lb/ft3 1lb/UK gal 1ib/US gal 1g/cm3 1*103kg/m3 16.018kg/m3 99.779kg/m3 119.83kg/m3 103kg/m3 Acceleration 1cm/s2 1ft/s2 1*10-2m/s2 0.3048m/s2 Energy(Torque) 1erg 1ft pdl 1ft lbf 1cal 1kgf m 1Btu 1Chu 1hp-hr(metric) 1hp-hr(British) 1kW h 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 Force 1 dyne 1*10-5N

UNIT CONVERSION

1 pdl 1lbf 1kgf 1tonnef 1tonf 1.3825*10-1N 4.4482N 9.8067N 9.8067*103N 9.9640*103N Volume 1in3 1 US gal 1 UK gal 1ft3 1 barrel(petroleum US)

1 lube oil barrel

1 yd3 1.6387*10-5m3 3.7853*10-3m3 4.546*10-3m3 2.8317*10-2m3 0.15898m3 0.20819m3 0.76455m3 Velocity 1 ft/h 1 ft/min 1ft/s 1 mile/h 8.4667*10-5m/s 5.08*10-3m/s 0.3048m/s 0.44704m/s Viscosity(dynamic) 1 mN s/m2(cp) 1 lb/ft h 1g/cm s (poise P) 1 lb/ft s 1*10-3N s/m2 4.1338*10-4N s/m2 0.1 N s/m2 1.4882N s/m2 Frequency 1 c/s 1Hz

Specific heat capacity 1 cal/gm °C 1 Btu/lb °F 1 Chu/l b°C 4.1868*103J/kg K 4.1868*103J/kg K 4.1868*103J/kg K Temperature Difference 1 °C 1 °C 1°R 1 K 5/9 K 5/9 K

Thermal Conductivity 1 Btu/h ft2(°F/in) 1 kcal/h m °C 1 Btu/h ft °F 1 cal/s cm °C 0.14423J/s m K 1.163J/s m K 1.7308J/s m K 418.68J/s m K Power 1 ftlbf/min 1 ftlbf/s 1 m kgf/s 1 hp (metric) 1 hp (British) 2.2597*10-2J/s 1.3558J/s 9.8065J/s 7.3548*102J/s 7.457*102J/s Pressure(Stress) 1 dyne/cm2 1 Pascal 1 kgf/m 2 1 mm water 1lbf/ft2 0.1 N/m2 1 N/m2 9.8067 N/m2 9.8067 N/m2 47.88 N/m2

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 at(kgf/cm 2 orkp/cm2) 1 bar 1 atm 1N/mm2 1tonf/in 2 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 Moment of inertia 1 lbft2 0.04214 kg/m2 Momentum 1 lbft/s Angular momentum 1 lb ft2/s Viscosity(kinematic) 1 S(stokes) 1 ft2/h

 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

σ

z

=

Thin wall thickness for spherical shell (i)Internal pressure

P =

or

Design of process vessel under internal Pressure

Design of cylindrical and spherical vessels under

internal pressure

(ii) Minimum thickness of wall

t =

or

(iii)Circumferential and longitudinal stress

= σ

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 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

=

= 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 =

Stress at internal surface

(i) Longitudinal stress

=

In this case Po = 0

=

, If Do = 0 then

= P

i

(ii) 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

=

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

=

√

Design of flat head

Thickness of flat head

t = C De√ , De = effective diameter =Di

F= allowable stress of material. C = design constant

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

I. Flanged flat head butt welded to shell

II. Plates welded to the inside of the shell

III. 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.

Design of Head and Closures

Design of cylindrical and spherical vessels under

internal pressure

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

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 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

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

t

n= 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

Volume of elliptical dished head

Vn = ( /4) (D0/6) = /24

Total volume contain in vessel where D is internal diameter Vvessel=[{ }

2

+{ }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

20 30 45 60

Z 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

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 Specific weight

( ⁄ Poisson ratio = /E Aluminium Brass Copper Iron Nickel Steel 2.65 8.35 8.74 7.74 8.74 7.70 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

Table 3

Stress concentration factor C for formed heads with uncompensated opening d/√ hE/D0 0.5 1.0 2.0 3.0 4.0 5.0 0.15 1.67 1.86 2.15 2.65 3.10 3.60 0.20 1.28 1.45 1.85 2.30 2.75 3.25 0.25 1.00 1.15 1.60 2.05 2.50 2.95 0.30 0.83 1.00 1.45 1.88 2.28 2.70 0.50 0.60 0.80 1.10 1.50 1.85 2.15

 Gasket dimensions

do/di = [ ( Y –pm ) / { Y – p (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 = (do – di)/2

2. Gasket seating width

bo = N/2

Design of Non-standard flanges

Design of cylindrical and spherical vessels under

internal pressure

3. 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=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,

 Estimation of Bolt Loads:

1. Load due to design pressure:

H=πG2

p/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

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)

 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

(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

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

TABLE 3 FLANGES Bolt size Root

area 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) 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 10-4 33.7 36 0.030 0.07 5 0.86 1.0683 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

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

Table 5 Moment arms for flange loads under operating conditions

Type of flange a1 a2 a3

Internal type flanges

R + (g1/2) (R + g1 + a3) / 2 ( C – G)/ 2

Loose type except lap joint flanges

( C – B)/ 2 (a₁+ a₃)/2 ( C – G)/ 2

Lap joint flanges ( 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 Gasket

factor m Min. design seating stress, Y ,MN/m² Min. actual gasket width (mm) Vulcanized rubber sheet

hardness above 70 IHRD

1.00 1.38 10 Asbestos with a suitable binder for operating conditions } 3.2mm Thick 1.6mm 2.00 2.75 11.00 25.50 10 10

0.8mm 3.50 44.85 10

Rubber with cotton fabric insertion 1.25 2.76 10 Rubber with asbestos fabric insertion, with or 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 10 Spiral-wound metal, asbestos filled } Carbon steel S.S.ormonel metal 2.50 3.00 20.00 31.00 10 10 Corrugated metal, asbestos inserted or asbestos filled corrugated metal jacket } Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. 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 Corrugated metal } Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. 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

Asbestos filled flat metal jacket

} Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. 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 } Soft Al Soft Cu/brass Iron/soft steel Monel metal S.S. 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 Ring joint } Iron/soft steel Monel metal S.S. 5.50 6.00 6.50 125.00 150.00 180.00 6 6 6

(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 STIFFENER RING:

Design is to be check for elastic instability or plastic deformation

Design of process vessel and pipes under external pressure

Design of cylindrical and spherical vessels under

internal pressure

If 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 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

Value of K and m as a function of D0/L ratio of given in table below D0/L K M 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 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 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<0.2D0, then

If p > external design pressure than calculated thickness from elastic instability is correct otherwise thickness is not safe that calculated thickness against plastic

(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

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

As= 1.84 m²

I=8.9 ,

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

1. Area to be compensated

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 √

COMPENSATION FOR OPENINGS IN PROCESS VESSEL

Design of cylindrical and spherical vessels under

internal pressure

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 , 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,

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.

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 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.

UNCOMPENSATION FOR OPENINGS IN PROCESS VESSEL

Design of cylindrical and spherical vessels under

internal pressure

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. 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.

√ ⁄ 0.0 1.000 0.25 0.900 0.5 0.785 0.75 0.700 1.0 0.645 1.5 0.545 2.0 0.465 2.5 0.390 3.0 0.340 3.5 0.285 √ ⁄ 4.0 0.245 4.5 0.215 5.0 0.180 5.5 0.155 6.0 0.130 6.5 0.115 7.0 0.090 7.5 0.080 8.0 0.075

(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

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:

⁄ ⁄ 0.167 1.36 0.178 1.27 0.192 1.18 0.207 1.08 0.227 0.99 0.25 0.90 ⁄ ⁄ 0.277 0.81 0.312 0.73 0.357 0.65 0.40 0.59 0.45 0.54 0.50 0.50

( )

(a)If the openings are along longitudinal direction:

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

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:

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

DESIGN OF TALL VESSELS

Design of cylindrical and spherical vessels

under internal pressure

Material Specific wt. N/m3 Poisson’s ratio Aluminium 2.65x104 0.34 Brass 8.35x104 0.35 Copper 8.79x104 0.35 Iron 7.74x104 0.28 Nickel 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: 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)

*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 ρVω

2

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

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

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

* 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,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 = (σe

2

- σe σz + σz 2

)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) <faJ 3) σZ(compressive) <0.125Ea(ta/Do)

*Now calculate the value of σz from:

σz = σzp + σzwm - σzw

and substitute the value of σz and σe in equation of equivalent stress and then

check it from σe= fJ

*if σe (calculated) is < σw (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 0.496 Chalk 15.00 0.692 Plaster ,artificial 20.70 0.742 Cotton wood 0.78 0.042 Cork board 1.57 0.043 Cork ground 1.45 0.043

Diatomaceous earth Powder ,fine 2.70 0.069 Wool 1.08 0.036 Felt , wool 3.26 0.052 Graphite, powdered 4.78 0.180 Magnesia,molded& dry 12.20 0.432 Mineral,wood 7.84 0.605 Rubber,hard 11.70 0.150 Sawdust 1.88 0.052 Silk 0.99 0.045

(1) Thickness of shell for internal pressure in given by t = Where, P= design pressure f= allowable stress j= joint efficiency factor

= outer dia of shell

C= 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 = π ( ) (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

Design of cylindrical and spherical vessels

under internal pressure

= Ws + Wi + Wl + Wa

The 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 Vw

2

(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

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 > 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)

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

(0.85 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 … (100

maximum)

(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

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√ = (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

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, = skirt thickness, = outer dia or dia of vessel

2. Stress due to wind load

=

Where = bending moment due to wind at base of vessel

= for height up to 20m

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

(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

(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.

∑ 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

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)

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

[ ( )

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 0.107 0.192

(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 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

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 = 120 0 K5 = 0.673 for = 1500 Values of K6 A/R = 1200 = 1500 0 – 0.5 0.013 0.007 0.6 0.018 0.010 0.7 0.030 0.017 0.8 0.034 0.021

0.9 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 & K8 as 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

1.Stresses in a thick cylinder:- σɵ = piDi2 – p0D02/ D02 –Di2 σɵ = [piDi 2 – p0D0 2 / D0 2 –Di 2 ] –[(pi-p0)Di 2 D0 2 /D2(Do 2 -Di 2 )] , σɵ=[piDi 2 – p0D0 2 / D0 2 –Di 2 ]+[(pi-p0) Di 2 D0 2 /D2(Do 2 -Di 2 )]] 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(K 2

+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

Design of cylindrical and spherical vessels

under internal pressure

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+ )k 2 /(k2-1)] Where, =poission‟s ratio And, K= (D0/Di) Than calculate ,t.

c)Maximum strain energy theory:-

σɵ = σy/F=Pi(6+10k

4

)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 = [2k

2

/(k2-1)]Pi=2 τ(max)

(ii) when elastic break down by maximum shear :- σɵ= σy/pi = (3)1/2k2/(k2-1)

SHELL DESIGN:

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

H = V/πr2

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)

DESIGN OF STORAGE TANK

Design of cylindrical and spherical

vessels under internal pressure

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)

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

Tank diameter thickness of bottom plate

>12 m 6 mm

<12 m 8 mm

2. bottom diameter (Db)

Since diameter of the bottom of tank extends deyond the shell by65 mm

⁄

3. Circumference of Bottom = πDb

4. Number of stiffening rings

For increasing the thickness of shell stiffeners provided, around the shell

No. of rings = π Db / 6.3

ROOF DESIGN

There are two cases in the roof designing

Case1 → When ≤370 Self Supported Roof

Case2 → When > 370

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.

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

1. 75 kg / m2 for non pressure tanks

2. 200 kg / m2 for class A tanks

3. 550 kg / m2 for class B tanks

3. 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

= π d / 2 m ( assumed )

Where D = dia of shell in m

2)

Actual spacing between two rafters = circumference / no. of rafters

= π D/ no. of rafters

3)

Selection of central support

Diameter of tank type of central support 6 -12.5 m one centre column

12.5 – 15 m circular 15.20 m square 20.25 m pentagonal 25.30 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 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) 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

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 thickness of plate = ( R L t) KGf

Surface area of cone = π R L 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

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