flare stack mechanical design calculations by abdel halim galala

DESIGN CALCULATIONS OF ELEVATED CYLINDRICAL FLARE STACKPage : 1 of 24

Designed by : Eng. Abdel Halim Galala, Design General Manager (Assistant) REV. : 0

Project : Design & procurement of Flare Stack Date : 4.4.2000

Job Name : Propylene Recovery Unit Location : Gulf of Seuz

Dwg. No. : Client : ABB/OPC

Stack Type : Self supported & Multiple Diameter Item : X-06-02

ELEVATED FLARE STACK

SELF SUPPORTED & MULTIPLE DIAMETER

Design General Mgr. BY

Eng. Abd El Halim Galala

Design General Manager Assistance

1st issue Dated : 6.6.1985

DESIGN CALCULATIONS OF ELEVATED CYLINDRICAL FLARE STACKAccording to ASME Code, Sec. VIII, Div. 1, Edition 95, Addenda 96. Page : 2 of 24






ContentsPage

A. Wind Loads as Computed in Accordance with ANSI A58.1. 3

B. Allowable Shell Buckling (Compression) Stress. 5

C. Shell Plate Thickness, Design Procedure. 7

1. Total Uncrodded Stack Weight. 7

2. Computation of the Projected Area. 8

3. Computation of the Wind Loads. 9

4. Computation of the Wind Moments. 11

5. Required Shell Plate Thickness. 12

6. Anchor Bolt Chair. 13

7. Width of Base Ring. 14

8. Base Plate Thickness. 15

9. Top Plate Thickness. 15

D. Vibration Analysis. 16

1. Cantilever Vibration. 16

- Analyzing Technique. 17

- Static Deflection. 18

- Dynamic Deflection. 19

2. Ovaling Vibration. 20

- Design of Ovaling Ring : 22

- Critical Wind Velocity. 22

- Required Ring Section Modulus. 22

E. References. 24







A. WIND LOADS AS COMPUTED IN ACCORDANCE WITH ANSI A58.1-1955.

The procrdure for calculation of the minimum design wind load normal to the surface is as follows :

1. The geographical area of the job site (Gulf of Suez & Mostorod) is located on the wind pressure map,

see Table-1. The basic wind pressure p is selected.

2. The wind design pressure pz, corresponding to tha basic wind pressure p, for various height

zones above the ground are given in Table 1.

3. To calculate design wind forces from wind pressures, shape factor B shall be used.The shape

factor for round objects is equal to 0.6 and is applied to the design pressure pz.

4. If the windward surface area projected on the vertical plane normal to the direction of the wind

is A ft2, then the resultant of the wind pressure load over the area pw is assumed to act at the

area centroid and is given by :

pw = A B pz, lb

The wind pressure forces are applied simultaneously, normal to all exposed windward surfaces of the

structure. The minimum net pressure B*pz in the above formula for cylindrical vertical vessels is not

less than : for L/D <= 10 13 PSF

and for L/D >= 10 18 PSF

Where L is the overall tangent-to-tangent length of the vessel, and D is the vessel nominal diameter.

Table 1.

Design Wind Pressure of the Job Site, p.

Height zone Basic wind

above grade pressure, p

Ft M PSF Kg/m2

30 9 15.3612 75

32 to 46 10 to 14 20.4816 100

49 to 62 15 to 19 25.602 125

over 65 over 20 30.7224 150

1 Kg/M2 = 0.2048159 lb/Ft

2

Computation of the Projected Area, A.

An approach to computing A which is often used and is recommended here is to increase the vessel

diameter D to the so called effective vessel diameter to approximate the combined design wind load :

De = (Vessel OD + Twice insulation Thickness) x Kd

The coefficient Kd is given in Table 2. The required projected area A will then be equal to :

A = De Ls

where Ls = length of the shell section in the zone of the uniform wind velocity.







* The method of determining wind loads on vessels of two or more Table 2.

diameters is the same as for a vessel of a uniform diameter. When Vessel OD including insulation Coefficient

the conical transition section is no more than 10% of the total INCH MM Kd

height, cylindrical sections can be assumed to extend to the mid- Less than 36 1.5

hight of the conical section. 36 to 60 1.4

Otherwise, the transition section should be considered as 60 to 84 1.3

separate section. 84 to 108 1.2

over 108 1.18

For Stack material ASTM A285 Grade C

Where,

E = Modulus of elasticity of plate material @ operating temp. 27600000 PSI

Y = Yield point stress of plate material @ operating temp. 30000 PSI

St = Allowable Tensile Strength of Plate Material @ operating temp. 15000 PSI

r = Steel Density 490 lb/ft3

7850 Kg/M3

V30 = 'Wind velocity at 30 ft height. 100 MPH

C = Corrosion allowance 0.125 INCH 3.175 MM

Flare stack segment lengths, Ls

Ls1-2 50 Ft 15240 MM

Ls2-3 16.8 Ft 5120.64 MM

*Ls3-4 17 Ft 5181.6 MM

*Ls4-5 11 Ft 3352.8 MM

*Ls5-6 10 Ft 3048 MM

Ls6-7 10 Ft 3048 MM

Total vessel height 114.8 Ft 34991 MM

Figure (1)







B. ALLOWABLE SHELL BUCKLING (COMPRESSION) STRESS.

Intially, some thicknesses at each section are assumed. Allowable compression stresses

at each level are determined from the following formula :

- For ta/d < 0.00425 Sc = (0.56 ta E)/d(1+0.004 E/Y)

- For higher ta/d ratios, the allowable compression stress used is that calculated for t/d = 0.00425

where ta = assumed corroded plate thickness at each level under consideration, in.

d = internal stack diameter at level under consideration, in.

Stack OD1-2 20 INCH 508 MM



Stack OD4-5 28 INCH 711.2 MM



By assuming corroded thicknesses as follows:

ta 1-2 0.166142 INCH 4.22 MM

ta2-3 0.30315 INCH 7.7 MM

ta 3-4 0.492126 INCH 12.5 MM

ta 4-5 0.307087 INCH 7.8 MM

ta 5-6 0.224409 INCH 5.7 MM

ta 6-7 0.275591 INCH 7 MM

Internal vessel dia. at each level, d 1-2 = OD1-2 - 2 ta1-2 19.66772 INCH 499.56 MM

d2-3 19.3937 INCH 492.6 MM

d 3-4 19.01575 INCH 483 MM

d 4-5 27.38583 INCH 695.6 MM

d 5-6 35.55118 INCH 903 MM

d 6-7 35.44882 INCH 900.4 MM

Since the calculated ratio, ta1-2 / d1-2 0.008447 > 0.00425

ta2-3 / d2-3 0.015631 > 0.00425

ta3-4 / d3-4 0.02588 > 0.00425

ta4-5 / d4-5 0.011213 > 0.00425

ta5-6 / d5-6 0.006312 > 0.00425

ta6-7 / d6-7 0.007774 > 0.00425







The ratio ta/d to be used in calculations shall be computed as follows :

Use ta1-2 / d1-2 = Min. (Calculated value, 0.00425) 0.00425

ta2-3 / d2-3 0.00425

ta3-4 / d3-4 0.00425

ta4-5 / d4-5 0.00425

ta5-6 / d5-6 0.00425

ta6-7 / d6-7 0.00425

Therefore, the allowable buckling (compression) stress Sc shall be :

Sc = (0.56 ta E) / d(1+ 0.004 E/Y) = (ta/d) (0.56 E) / (1+ 0.004 E/Y)

Final allowable buckling (compression) stress, Sc1-2 14035.9 PSI

Sc2-3 14035.9 PSI

Sc3-4 14035.9 PSI

Sc4-5 14035.9 PSI

Sc5-6 14035.9 PSI

Sc6-7 14035.9 PSI







C. SHELL PLATE THICKNESS, DESIGN PROCEDURE.

1. Total Uncrodded Stack Weights.

Weights at each level are calculated by adding corrosion allowance to the thicknesses

assumed above. [Uncorroded mean adding corrosion allowance ]

Wt1-2 = r (3.14) [OD1-22-(d1-2 - 2C)

2] Ls1-2 3067.04 lb

Wt2-3 = r (3.14) [OD2-32-(d2-3 - 2C)

2] Ls2-3 1504.943 lb

Wt3-4 = r (3.14) [OD3-42-(d3-4 - 2C)

2] Ls3-4 2173.824 lb

Wt4-5 = r (3.14) [OD4-52-(d4-5 - 2C)

2] Ls4-5 1400.718 lb

Wt5-6 = r (3.14) [OD5-62-(d5-6 - 2C)

2] Ls5-6 1331.634 lb

Wt6-7 = r (3.14) [OD6-72-(d6-7 - 2C)

2] Ls6-7 1524.498 lb

Total Weight 11002.66 lb Kg

By adding 15% of the calculated weight to compensate weight of piping, internals, platforms, ladders, etc., we get :

Wt1-2 3527.096 lb

Wt2-3 1730.685 lb

Wt3-4 2499.898 lb

Wt4-5 1610.826 lb

Wt5-6 1531.379 lb

Wt6-7 1753.173 lb

(Total Weight + 15%), Wt 12653.06 lb Kg




According to ASME Code, Sec. VIII, Div. 1, Edition 95, Addenda 96. Location : Gulf of Seuz



2. Computation of the Projected Area, A

An approach to computing A which is often used and is recommended here is to increase the vessel

diameter D to the so-called effective stack diameter De to approximate the combined design wind load :

De = Kd [(Vessel OD + 2 insulation thk.) + (pipe OD + 2 insulation thk. + (platform + ladder)]

The coefficient Kd is given in Table 2.

The required projected area A will then equal to : A = De * Ls

where Ls = Length of the shell section in the zone of the uniform wind velocity.

De = Kd [Stack OD + Fuel Gas Pipe OD + Steam Pipe OD]

N.B. In our case, the size of fuel gas and steam pipes can be neglected.

Stack OD

INCH MM

Coefficient Kd [fromTable 2] Kd1-2 [dia. < 36"] 1.5 20 508

Kd2-3 [dia. < 36"] 1.5 20 508

Kd3-4 [dia. < 36"] 1.5 20 508

Kd4-5 [dia. < 36"] 1.5 28 711.2

Kd5-6 [dia. = 36"] 1.4 36 914.4

Kd6-7 [dia. = 36"] 1.4 36 914.4

Effective Diameter, De = Kd * Stack OD, De1-2 = Kd1-2 * OD1-2 2.5 Ft MM

De2-3 2.5 Ft MM

De3-4 2.5 Ft MM

De4-5 3.5 Ft MM

De5-6 4.2 Ft MM

De6-7 4.2 Ft

Projected Area, A = De * Height Ls A1-2 = De1-2 - Ls1-2 125 Ft2

M2

A2-3 42 Ft2

M2

A3-4 42.5 Ft2

M2

A4-5 38.5 Ft2

M2

A5-6 42 Ft2

M2

A6-7 42 Ft2

M2







3. Computation of Wind Loads.

The resultant of the wind pressure load Pw over the area A is given by :

Pw = A * B * Pz

where

Pw = Wind pressure load over the projected area A, lb.

A = Windward surface area projected on the vertical plane normal to the direction of the wind, ft2.

B = Shape Factor, for round objects B = 0.6 0.6

Pz = Design wind pressure, psf, depends upon the geographical area of the job site (see Table 1).

B Pz = Min. net wind pressure, psf. For cylindrical vertical vessels :

- For L/D <= 10, BPz not less than 13 PSF

- For L/D >= 10, BPz not less than 18 PSF

where,

L is the overall tangent-to-tangent length of the vessel, 114.8 Ft 34991 MM

Davg. is the vessel average diameter.

=B451= (L1-2 D1-2 + L2-3 D2-3 + L3-4 D3-4 + L4-5 D4-5 + L5-6 D5-6 + L6-7 D6-7) / L 1.962834 Ft MM

L/Davg. 58.48686 > 10

Therefore, the net wind pressure BPz shall not be less than 18 PSF







Determining Design Wind Pressure B Pz for each Segment [based upon Table 1].

Table 3 From Table 1. Segment Design

Height zone Basic height, H wind

above wind above pessure,

grade pressure, P grade Pz

M PSF M PSF

Design Wind Pressure, Pz1-2 30.722385 34.991 30.722385

20

19

Pz2-3 25.601988 19.751 25.6019875

15

14

Pz3-4 20.48159 14.6304 20.48159

10

Pz4-5 9 9.4488 15.3611925

Pz5-6 15.361193 6.096 15.3611925

Pz6-7 0 3.048 15.3611925

Net wind pressure, NWP shall be as follows :B Pz1-2 18.43343 PSF

B Pz2-3 15.36119 PSF

B Pz3-4 12.28895 PSF

B Pz4-5 9.216716 PSF

B Pz5-6 9.216716 PSF

B Pz6-7 9.216716 PSF

Use the max. following wind pressure : B Pz1-2 = Max (NWP,18) 18.43343 PSF

B Pz2-3 18 PSF

B Pz3-4 18 PSF

B Pz4-5 18 PSF

B Pz5-6 18 PSF

B Pz6-7 18 PSF

Wind Load shall be as follows : Pw = A * B Pz

Pw1-2 = A1-2 B Pz1-2 2304.179 lb Kg

Pw2-3 756 lb Kg

Pw3-4 765 lb Kg

Pw4-5 693 lb Kg

Pw5-6 756 lb Kg

Pw6-7 756 lb Kg







4. Computation of Wind Moments.

In the following calculations we assumed that geographical location of

the job site does not required a moment calculation for the earthquake.

Therefore, the total wind moment shall be calculated as follows :

Figure (2)

Moment = Wind Load x Arm

M2 = Pw1-2 (0.5 L1-2) 57604.47 lb-ft

M3 =Pw1-2 (0.5 L1-2+L2-3)+Pw2-3(0.5L2-3) 102665.1 lb-ft

M4 =Pw1-2 (0.5 L1-2+L2-3+L3-4)+Pw2-3(0.5L2-3+L3-4)+ Pw3-4(0.5L3-4) 161190.6 lb-ft

M5 =Pw1-2 (0.5 L1-2+L2-3+L3-4+L4-5)+Pw2-3(0.5L2-3+L3-4+L4-5)+ Pw3-4(0.5L3-4+L4-5)+Pw4-5(0.5 L4-5) 207079.1 lb-ft

M6 =Pw1-2 (0.5 L1-2+L2-3+L3-4+L4-5+L5-6)+Pw2-3(0.5L2-3+L3-4+L4-5+L5-6)+ Pw3-4(0.5L3-4+L4-5+L5-6)+Pw256040.9 lb-ftM7 =Pw1-2 (0.5 L1-2+L2-3+L3-4+L4-5+L5-6+L6-7)+Pw2-3(0.5L2-3+L3-4+L4-5+L5-6+L6-7)+ Pw3-4(0.5L3-4+L 312562.7 lb-ft Table 4.

Wind Weight

Load Shear Moment @ section

W, lb Q, lb M, lb-ft WT, lb

2304.179

3527.09562

2304.1789 57604.5

756

5257.78058

3060.1789 102665

765

7757.67823

3825.1789 161191

693

9368.50446

4518.1789 207079

756

10899.8834

5274.1789 256041

756

12653.0565

Figure (3) 6030.1789 312563







5. Required Shell Plate Thickness.

The required shell plate thickness shall be computed as follows :

tr = (WT dr + 48 M) / 3.14 dr2 Sc

where WT = Total weight @ Section under consideration

By assuming corroded thicknesses as follows: ta 1-2 0.166142 INCH 4.22 MM

ta2-3 0.30315 INCH 7.7 MM

ta 3-4 0.492126 INCH 12.5 MM

ta 4-5 0.307087 INCH 7.8 MM

ta 5-6 0.224409 INCH 5.7 MM

ta 6-7 0.275591 INCH 7 MM

tr1-2 = (WT1-2 dr1-2 + 48 M2) / PI( ) dr1-22 Sc1-2 0.166173 INCH 4.2208 MM

N.B. The assumed thickness must be changed untill the calculated > the assumed thickness.

tr1-2 + C.A 0.291173 INCH 7.3958 MM

For construction, use (tr1-2 + C.A) 0.314961 INCH 8



tr2-3 + C.A 0.428282 INCH 10.8784 MM

For construction, use (tr2-3 + C.A) 0.433071 INCH 11 MM



tr3-4 + C.A 0.6195 INCH 15.7353 MM




tr4-5 + C.A 0.433321 INCH 11.0064 MM




tr5-6 + C.A 0.352475 INCH 8.95287 MM




tr6-7 + C.A 0.403855 INCH 10.2579 MM


N.B. Since there is no appreciable change in calculated thicknesses, the weights

based on the required thicknesses are almost the same as previouslyestimated.







6. Ancor Bolt Chair.

Anchor bolt material ASTM A36

Assume No. of anchor bolts, N 16 BOLTS

Bolt Circle Dia, DBC 43.93701 INCH 1116 MM

Max. Allowable stress of Anchor Bolt Material, SB 15000 PSI

The total tension in each anchor bolt is determined from the relationship:

WB = (48 M / N DBC) - Wt / N 20550.83 lb Kg

Thus, the anchor bolt area required at the root of the thread shall be:

AB = WB / SB 1.370055 INCH2

883.905 MM2

Diameter of the anchor bolt at the root of thread, D = (4 AB/3.14)0.5

1.320761 INCH 33.5473 MM

Use 16 holding down bolts with a dia. of , d 1.75 INCH 44.45 MM

Each bolt shall be provided with two hex heavy nuts and washer.

- Length of threaded portion, l INCH 150 MM

- Radius, R INCH 90 MM

- Length, L INCH 350 MM

- Bolt projection INCH MM

- Total length, M INCH 1085 MM

Figure (4) Figure (5)







7. Width of Base Ring, C.

Assume width of base ring, C 11.10236 INCH 282 MM

Bearing pressure on concrete foundation is calculated from

the following formula :

Pb = (48 M / PI Db2 C) + (Wt / 3.14 Db C) 341.9775 PSI < 500

SAFE

where Db = OD at bottom of stack 36 INCH 914.4 MM

Assume C, and note that the value of Pb is limited to :

- For 3000 lb concrete 750 PSI (Max.)

- For 2000 lb concrete 500 PSI (Max.)

N.B. If the calculated Pb exceeds the above limit, the value of C is modified

so that Pb falls within the allowable limit.







8. Base Plate Thickness, Tb

Base plate thickness Tb is calculated from the following equation:

Tb = e SQRT(3 Pb / 20000) 1.480195 INCH 37.5969 MM

where e = A + B 6.535433 INCH 166 MM

A 101 MM

B 65 MM

Use Tb 1.771654 INCH 45 MM

Figure (6)

9. Top Plate Thickness, TT

TT = SQRT[ 3 * WB * 6 / (4 * 20000 * e)] 0.841141 INCH 21.365 MM

where WB = Total tension in each bolt 20550.83 PSI

Use TT 1.181102 INCH 30 MM







D. VIBRATION ANALYSIS.

1. Cantilever vibration.

The following criteria is used to establish need for vibration analysis of stacks with Lc/L ratio not

exceeding 0.5 :

- W / L Dr2 <= 20 vibration analysis must be performed 9.443917 < 20 vibration analysisis required

- 20 < W / L Dr2 <= 25 vibration analysis should be performed

- 25 < W / L Dr2 vibration analysis need not be performed

Where Wt = Total corroded stack weight (see next page). 8071.981 lb

Le = Total length of stack. 114.8 ft

Dr = Average internal dia. of top half of stack. 2.72862 ft

Lc = Total length of conical section(s) of stack. 4.002625 ft

Conical section height 28"/20" 2.001312 ft 610 MM

Conical section height 36"/28" 2.001312 ft 610 MM

We note that Lc / L < 0.5 0.034866 OK

O.D. Stack OD1-2 20 INCH 508 MM






Corroded thickness tr1-2 0.189961 INCH 4.825 MM












Corroded internal diameter in inches : Corroded ID1-2 19.62008 INCH

Corroded ID2-3 19.38386 INCH





Flare stack segment heights, L1-2 50 Ft 15240 MM

L2-3 16.8 Ft 5120.64 MM

L3-4 17 Ft 5181.6 MM

L4-5 11 Ft 3352.8 MM

L5-6 10 Ft 3048 MM

L6-7 10 Ft 3048 MM

Total Corroded Stack Weights.

Weights at each level are calculated, without taking into account adding

corrosion allowance to the thicknesses assumed above.

Corroded mean neglecting corrosion allowance.

Wt1-2 = r (PI/4) [do2-di

2]1-2 L1-2 2011.418 lb 912.365 Kg

Wt2-3 = r (PI/4) [do2-di

2]2-3 L2-3 1089.511 lb 494.194 Kg

Wt3-4 = r (PI/4) [do2-di

2]3-4 L3-4 1788.879 lb 811.423 Kg

Wt4-5 = r (PI/4) [do2-di

2]4-5 L4-5 1129.776 lb 512.459 Kg

Wt5-6 = r (PI/4) [do2-di

2]5-6 L5-6 876.9458 lb 397.776 Kg

Wt6-7 = r (PI/4) [do2-di

2]6-7 L6-7 1175.45 lb 533.176 Kg

Total corroded stack Weight, Wt 8071.981 lb 3661.39 Kg

Analyzing technique.

Period of vibration is determined as follows :

T = 1.648 (Le2) / Dr E

0.51.515107 second

'where Le = effective length of stack 114.8 ft

Natural frequency of stack vibration, f = 1 / T 0.660019 cps

Critical wind velocity, Vc = 3 f Dr 5.402825 MPH







Max. wind velocity at the top of the stack is

Vw = V30 (L / 30)0.143

121.1556 MPH

V30 = Wind velocity at 30 ft height (basic wind velocity) 100 MPH

Max. gust velocity = 1.3 Vw 157.5022 MPH

Since the critical wind velocity, Vc falls within max. gust velocity, the stack must be

checked further for K. In that case, corroded stack weight must be > 15 times the wind force

at critical velocity or expressed as a formula, the ratio K should be less than 1/15.

K = Pc Dr Le / Ws = 0.0077 Dr5 E / Le

3 Ws 0.002632 < l / 15 0.06666667

Therefore, the stack is free from cantilever vibration.

N.B. For lined stacks, W can be used in place of Ws in order to reduce vibration.

Design modifications are required, if K in the above equation exceeds 1/15.

Static deflection.

The computed dynamic loading is applied as a stagnant pressure to the stack. Assuming it to be a

cantilever beam, amplitude at the top is approximated by :

Static Deflection, Ds = Pc Dr (Le)4 (12)

3 / 8 E I 0.303669 INCH 7.71321 MM

where, Pc = Total wind force at critical velocity

= C1 pair (1.467 Vc)2 / 2 0.074756 psf

C1 = Lift coefficient (usually taken as1) 1

pair = mass density of air 0.00238 lb.sec2/ft

4

I = Moment of inertia of top half of stack

= 3.14 * r3 * t 913.0678 INCH

438004.8 CM

4

r = Average internal radius of top half of stack 9.679528 INCH 245.86 MM

t = Average corroded plate thick. of top half of stack 0.320472 INCH 8.14 MM







Dynamic deflection.

At a critical wind velocity, the structure vibrates at resonant frequency, and thus the amplitude of

vibration is magnified greatly. The dynamic coefficient, which is a ratio of dynamic amplitude to static

amplitude, is called the magnification factor. This is a function of the lining, stiffness of the soil and

several other factors. The amount of static deflection must be multiplied by the magnification factor

to determine dynamic deflection. Approximate value of magnification factors for different types of stacks

as suggested by DeGhetto and Long are listed in Table 3.

Table 5. Magnification factors

Types of Spread Spread Piled

stacks footings footings foundations

on soft soil on medium soil and spread

(Bearing < (1500 psi < footings

1500 psi) Bearing > on stiff soil

3000 psi) and rocks

Lined

stacks 5 30 50

Unlined

stacks 30 60 90

Using a magnification factor of 30

Dynamic deflection = Static Deflection * Magnification factor Allowed deflection

= Ds * Magnification factor 9.110085 INCH > 8.036

NotT Good

N.B. For dynamic deflection, It is assumed to be within the allowable limits

when allowing about 7-inch deflection per 100 feet* of height of stack.

* "Shortcut method for calculating tower deflection", Hydrocarbon Processing 47. No. 11, Nov. 1968, p 230.

Allowed deflection of stack under consideration = 7 x L / 100 8.036 INCH







2. Ovaling vibration.

Natural frequency of the free ring is given by :

fr = (7.58 tr E0.5

) / 60 D2

Vortex shedding frequency is given by : fv = 0.2 V / D

Where V = wind velocity for vortex shedding is 45 MPH 45 MPH

or 66 fps as recommended by Dickey and 66 fps

Woodruff for most economical and safe stack

design as far as vibration is concerned.

Both these frequencies should be calculated at each level using the corresponding thicknesses and diameters.

Corroded internal diameters in feet, D D1-2 1.635007 Ft 498.35 MM

D2-3 1.615322 Ft 492.35 MM

D3-4 1.582513 Ft 482.35 MM

D4-5 2.275427 Ft 693.55 MM

D5-6 2.961778 Ft 902.75 MM

D6-7 2.948655 Ft 898.75 MM

At Section 1-2 : fr1-2 = (7.58 tr1-2 E0.5

) / 60 D2

1-2 47.16253 cps cps = cycle per second

fv1-2 = 0.2 V / D1-2 8.073362 cps

2 fv1-2 16.14672 cps

Since fr > 2 fv , no ovaling rings are required, and the stack is free from ovaling vibration.

N.B. t calculated upon corroded thk. & V = 66 fps.

At Section 2-3 : fr2-3 = (7.58 tr2-3 E0.5

) / 60 D2

2-3 78.36194 cps

fv2-3 = 0.2 V / D2-3 8.171748 cps

2 fv2-3 16.3435 cps


At Section 3-4 : fr3-4 = (7.58 tr3-4 E0.5

) / 60 D2

3-4 133.814 cps

fv3-4 = 0.2 V / D3-4 8.341163 cps

2 fv3-4 16.68233 cps


At Section 4-5 : fr4-5 = (7.58 tr4-5 E0.5

) / 60 D2

4-5 44.53771 cps

fv4-5 = 0.2 V / D4-5 5.80111 cps

2 fv4-5 11.60222 cps








At Section 5-6 : fr5-6 = (7.58 tr5-6 E0.5

) / 60 D2

5-6 17.3512 cps

fv5-6 = 0.2 V / D5-6 4.456782 cps

2 fv5-6 8.913564 cps


At Section 6-7 : fr6-7 = (7.58 tr6-7 E0.5

) / 60 D2

6-7 23.51664 cps

fv6-7 = 0.2 V / D6-7 4.476618 cps

2 fv6-7 8.953235 cps








Design of Ovaling Rings.

If, at any section, fr < 2fv, ovaling rings are required to stiffen that section.

Section modulus of the rings, whenever required, can be determined as follows :

- Critical wind velocity at the section under consideration is :

Vo = ( 60 fr D ) / 2 S FPM

where fr = Natural frequency cps

S = Strouhal number, is 0.2 over a wide range of 0.2

Reynolds numbers.

D = Intenal vessel dia. (corroded) at level under consideration. Ft

Critical wind velocity at Section, Vo1-2 = ( 60 fr1-2 D1-2 ) / 2 S 11566.66 FPM

Vo2-3 = ( 60 fr2-3 D2-3 ) / 2 S 18986.96 FPM

Vo3-4 = ( 60 fr3-4 D3-4 ) / 2 S 31764.36 FPM

Vo4-5 = ( 60 fr4-5 D4-5 ) / 2 S 15201.34 FPM

Vo5-6 = ( 60 fr5-6 D5-6 ) / 2 S 7708.563 FPM

Vo6-7 = ( 60 fr6-7 D6-7 ) / 2 S 10401.37 FPM

Required Section Modulus of the Ring.

Now, the section modulus of stiffeners at section under consideration can be found from the formula

as used by Moody :

Sm = [ (7)10-7

(Vo)2 D

2 (Hr) ] / St INCH

3

Where, Vo = Critical wind velocity at section of consideration FPM

Hr = Spacing throughout the length of the section under consideration0.5 Ls Ft

St = Allowable tensile strength of stiffener material 15000 PSI

Sm1-2 = [ (7)10-7

(Vo1-2)2 D1-2

2 Hr1-2) ] / St 0.417255 INCH

3

Sm2-3 = [ (7)10-7

(Vo2-3)2 D2-3

2 Hr2-3) ] / St 0.368735 INCH

3

Sm3-4 = [ (7)10-7

(Vo3-4)2 D3-4

2 Hr3-4) ] / St 1.002306 INCH

3

Sm4-5 = [ (7)10-7

(Vo4-5)2 D4-5

2 Hr4-5) ] / St 0.307085 INCH

3

Sm5-6 = [ (7)10-7

(Vo5-6)2 D5-6

2 Hr5-6) ] / St 0.121627 INCH

3

Sm6-7 = [ (7)10-7

(Vo6-7)2 D6-7

2 Hr6-7) ] / St 0.219485 INCH

3

Stiffeners having section modulus equal to or greater than Sm should be provided at spacing

Hr throughout the length of the section under consideration. If stiffeners are required for more

than one section, different size and spacing should be used for economy, if possible.







Notes







E. REFERENCES.

1. ANSI A58.1-1955.

2. ASCI 7-1988.

3. Pressure Vessel Design Handbook-2nd Edition, by Henry H. Bednar.

4. Flare stack paper research.

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flare stack mechanical design calculations by abdel halim galala

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