98740650-59764455-storage-tanks

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STORAGE TANKS 1. INTRODUCTION 2. GENERAL 3. DESIGN CODES 4. TYPE OF TANKS 5. SELECTION OF TANKS 6. MATERIAL SPECIFICATIONS 7. DESIGN OF COMPONENTS Shell design Bottom Plate design Wind girder design 8. SEISMIC ANALYSIS

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98740650-59764455-Storage-Tanks

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  • STORAGE TANKS

    1. INTRODUCTION

    2. GENERAL

    3. DESIGN CODES

    4. TYPE OF TANKS

    5. SELECTION OF TANKS

    6. MATERIAL SPECIFICATIONS

    7. DESIGN OF COMPONENTS

    Shell design

    Bottom Plate design

    Wind girder design

    8. SEISMIC ANALYSIS

  • 9. ANCHORAGE REQUIREMENT

    10. VENTING OF TANKS

    11. ROOF TO SHELL JOINT DETAIL

    12. REINFORCEMENT REQUIREMENT

    13. ROOF STRUCTURE DESIGN

    14. FOUNDATION DESIGN CONSIDERATION

    15. TYPE OF FLOATING ROOF AND ITS ACCESSORIES

    16. CALCULATION OF THICKNESS BY VARIABLE POINT METHOD

  • INTRODUCTION

    Throughout the chemical, petrochemical and refineries gases, liquids and

    solids are stored, accumulated or processed in vessels of various shapes

    and sizes.

    Such a large number of storage vessels or tanks are used by these

    industries that the design, fabrication and erection of these vessels have

    become a specialty of a number of companies.

    Only very few companies in process industries now design storage vessels

    having large volumetric capacity.

    However, the design of this equipment involves basic principles which are

    fundamental to the design of other types of equipment.

  • GENERAL

    Storage tanks are designed for internal pressure approximating atmospheric

    pressure, or internal pressure not exceeding the weight of the roof plates.

    Higher internal pressure upto 2.5 psi is permitted when additional

    requirement are met.

    Maximum operating temperature of 90 C is allowed however for higher

    temperature upto 260 C allowable stress modification to be done.

    Tank designed for one product can store other product of differing relative

    density(always of lesser density)

    Height-to-diameter ratio is often a function of the processing requirements,

    available land area and height limitations.

  • DESIGN CODES

    Following are the codes for the design of welded steel storage tanks,

    API 650

    BS 2654 For non-refrigerated, above ground storage

    IS 803

    API 12F For tanks for storage of Production liquids (upto 120m3)

    API 653 For tank inspection, repair, alteration and reconstruction

    API 620 For low-pressure storage tanks

    IS 10987 For under-ground/above-ground storage of petroleum

    products

    For design of reinforced tanks

    BS 4994 For tanks in reinforced plastics

  • TYPES OF TANKS

    The above ground storage tanks of large capacity are categorised based

    on the type of roof as follows,

    Simply SupportedInternal rafter typeInternal truss type

    Rafter with Central column

    Cone Roof Dome Roof

    Fixed roof

    Single DeckDouble Deck

    Open top

    Single DeckDouble Deck

    Closed top

    Floating roof

    Storage Tank

  • SELECTION OF TANK

    Selection of specific type of tank and type of roof depends upon the

    intended service conditions, such as the product being stored, its vapor

    pressure and corrosive nature and anticipated weather and loading

    conditions.

    Cone roof tanks are recommended for products with lower vapor

    pressure or with less emission control requirement.

    Even for products with higher vapor pressure, cone roof tanks can be

    used if the product capacity is less and necessary venting and

    blanketing arrangements are provided.

    Cone roof tanks are cheaper and easier to construct. Maintenance is

    very simple.

    Floating roof tanks are recommended for storing products of higher

    volatility. The steel deck provide good insulation over the entire

    surface of the liquid.

  • MATERIAL SPECIFICATIONS

    Following are the common plate material used for construction of tanks,

    A 36 upto 40 mm

    A 283 Gr C upto 25 mm

    A 285 Gr C upto 25 mm

    A 131 Gr A upto 12.5 mm

    A 131 Gr B upto 25 mm

    A 516 Gr 55,60,65,70 upto 40 mm

    A 537 Cl1, Cl2 upto 45 mm

    The minimum tensile strength of materials used in construction of tanks are

    between 55000 psi to 85000 psi.

    Carbon content between 0.15% to 0.25%

  • Low carbon steels are soft and ductile, easily sheared, rolled and formed into

    various shapes easily. Easy for welding and gives joints of uniform strength

    relatively free from localised stresses.

    Low alloy, high strength steels are also used but it is more difficult to

    fabricate, because they have low ductility.

    Plates shall be semi killed as minimum and fully killed and made to fine grain

    practice or normalised as required.

    For material with minimum tensile strength upto 80 ksi, the manual metal

    arc-welding electrodes shall conform to E60 and E70 classification series.

    For material with minimum tensile strength of 80 ksi to 85 ksi, electrodes

    shall conform to E80 series.

  • DESIGN OF TANK COMPONENTS

    SHELL DESIGN:As per Clause 3.6.3

    Shell thickness is calculated for two conditions, design conditions and

    hydrotest condition

    Design Condition:

    td = 2.6 D(H-1)G + CA

    Sd

    Hydrostatic Condition:

    tt = 2.6 D(H-1)G

    St

    where, td is the design thickness required in inches

    tt is the hydrostatic thickness required in inches

    D is the diameter of the tank in feet

    H is the height of the tank in feet

    G is the specific gravity of the product

    CA is the corrosion allowance in inches

    Sd is the allowable stress for design condition in psi

    St is the allowable stress for test condition in psi

  • Method explained above is one-foot method of calculating the shell

    thickness.

    This method is very conservative method and gives a higher thickness.

    For tanks above 60 m dia, this method is not preferred.

    Variable design point method is used for tanks above 60 m dia and if L/H

    ratio is less than or equal to 2, where L = (6Dt)0.5 where t is the thickness

    of bottom shell course.

    This method normally provides a reduction in shell-course thickness and

    hence total material weight. Variable design point method is explained

    separately.

    BOTTOM PLATE DESIGN:As per clause 3.4.1 of API 650

    All bottom plates shall have a minimum nominal thickness of 6 mm,

    exclusive of corrosion allowance.

    Annular bottom plate design:

    For calculating the thickness of annular bottom plate the hydrostatic test

    stress in first shell course shall be calculated as given below

    St = 2.6 D(H-1)G

    t

    where t is thickness of first shell course.

  • Thickness of annular plate shall be obtained from the table given below.

    Radial width of annular plate at any point around the circumference of the

    tank shall be either Aw1 or Aw2, whichever is greater

    Aw1= X + t + Y + L

    where, X = 24 or as per Appendix E.4.2 X=0.0274 WL/GH

    whichever is greater

    t = Provided thickness of the lowest course

    Y = Projection of annular plate outside the shell

    L = Annular-sketch plate lap

    WL = Weight of tank contents

    Aw2= 390 tb/ (HG)0.5

    where, tb = Thickness of the annular plate

  • WIND GIRDER DESIGN:As per Clause 3.9.7

    Tanks of larger diameter may not have the necessary inherent rigidity to

    withstand wind pressure without deforming and excessively straining the

    shell. To avoid this suitable stiffening or wind girders are provided.

    The maximum height of unstiffened shell H1 shall be calculated as follows:

    H1 = 600,000 t ((t/D)3)1/2 (100/V)2 (As per Clause 3.9.7)

    where, t = as ordered thickness of the top shell course(in.)

    D = nominal tank diameter(ft)

    V = wind velocity (mph)

    After the maximum height of the unstiffened shell, H1 has been determined,

    the height of the transformed shell shall be calculated as follows:

    Change the actual width each shell course into a transposed width of each

    shell course having the top shell thickness:

    Wtr = W((tuniform/tact)5)0.5

    where Wtr = Transposed width of each shell course, (in.)

    W = Actual width of each shell course (in.)

    tuniform = ordered thickness of top shell course (in.)

    tact = ordered thickness of shell course for which transposed width is

    calculated (in.)

  • Add the transposed widths of the courses. The sum of the transposed widths

    of the courses will give the height of the transformed shell

    If Wtr is greater than H1 an intermediate wind girder is required.

    For equal stability above and below the intermediate wind girder, the girder

    should be located at the mid height of the transformed shell.

    If half the height of the transformed shell exceeds the than H1 a second

    intermediate wind girder shall be used to reduce the height of unstiffened

    shell to a height less than the maximum.

    Overturning stability considering wind load shall be checked as follows:

    Overturning moment from wind pressure shall not exceed two-thirds of the

    dead load resisting moment.

    M less than or equal to 2/3(WD/2) for unanchored tanks

    where, M = overturning moment from wind pressure

    W = shell weight available to resist uplift, less any CA, plus dead weight

    supported by the shell minus simultaneous uplift from operating

    conditions such as internal pressure

  • The design procedure considers two response modes of the tanks and its

    contents:

    a. The relatively high-frequency amplified response to lateral ground motion

    of the tank shell and roof, together with the portion of the liquid contents that

    moves in unison with the shell.

    b. The relatively low-frequency amplified response of the portion of the liquid

    contents that moves in fundamental sloshing mode.

    The design requires the determination of the hydrodynamic mass associated

    with each mode and the lateral force and overturning moment applied to the

    shell as a result of the response of the masses to lateral ground motion.

    The overturning moment due to seismic forces applied to the bottom of the

    shell shall be determined as follows:

    M = ZI (C1WSXS + C1WrXt+ C1W1X1 + C2W2X2)

    where, Z = seismic zone factor

    I = Importance factor as per Appendix E

    SEISMIC ANALYSIS

  • C1 C2 = lateral earth quake force coefficients

    WS = Total weight of the tank shell(lb)

    XS = Height from the bottom of the tank shell to the shells CG(ft)

    Wr = Total weight of the tank roof(lb)

    Ht = Total height of tank shell(ft)

    W1 = Weight of the effective mass of the tank contents that move in

    unison with the tank shell(lb)

    X1 = Height from the bottom of the tank shell to the centroid of

    lateral seismic force applied to W1 (ft)

    W2 = Weight of the effective mass of the tank contents that move in

    unison in first sloshing mode(lb)

    X2 = Height from the bottom of the tank shell to the centroid of

    lateral seismic force applied to W2 (ft)

    Resistance to the over turning moment at the bottom of the shell may be

    provided by the weight of the tank shell and by anchorage of the tank shell or

    for unanchored tanks, the weight of a portion of the tank contents adjacent to

    the shell.

  • WL = 7.9tb(FbyGH)1/2

    where, WL = maximum weight of the tank contents that may be used to resist

    the shell overturning moment, in lb/ft of shell circumference.

    tb = thickness of the bottom plate under the shell(in.)

    Fby = minimum specified yield strength of the bottom plate under the

    shell (lb/in.2)

    G = design specific gravity of the liquid to be stored

    Now, calculate Wt, weight of tank shell & portion of fixed roof supported by

    the shell, in lb/ft of shell circumference.

    When M/[D2(Wt + WL )] is greater than 1.57 the tank is structurally unstable.

    When the tank is unstable any one of the following measures shall be carried

    out:

    a. Increase the thickness of the bottom plate tb under the shell.

    b. Increase the shell thickness, t.

    c. Change the proportions of the tank to increase the diameter and reduce the

    height.

    d. Anchor the tank.

  • Tank anchorage shall be provided if there exists a tendency for the shell and

    the bottom plate, close to the shell, to lift off its foundations due to the

    following reasons,

    Uplift on an empty tank due to internal design pressure counteracted

    by the effective weight of roof and shell.

    Uplift due to internal design pressure in combination with wind loading

    counteracted by effective weight of roof and shell, plus the effective

    weight of product considered .

    The anchorage shall not be attached to the bottom plate only but principally

    to the shell.

    The design shall accommodate movements of the tank due to thermal changes

    and hydrostatic pressure and reduce any induced stresses in the shell to a

    minimum.

    If an anchored tank is not properly designed, its shell can be susceptible to

    tearing.

    Care should be taken to ensure that the strength of the anchorage

    attachments is greater than the specified minimum yield strength of the

    anchors so that the anchors yield before the attachment fail.

    ANCHORAGE REQUIREMENT

  • The spacing between anchors shall not exceed 10ft. On tanks less than 50ft in

    diameter, the spacing between anchors shall not exceed 6ft.

    Minimum diameter of anchor bolts shall be 1in. excluding corrosion all.

  • Venting is required for all tanks.

    The venting system provided shall cater for the following:

    a) normal vacuum relief

    b) normal pressure relief

    c) emergency pressure relief

    Normal venting is accomplished by a pressure relief valve, a vacuum relief

    valve, a pressure vacuum(PV) valve or an open vent with or without a flame-

    arresting device.

    Emergency venting is by means of the following:

    Larger or additional valves or open vents.

    A gauge hatch that permits the cover to lift under abnormal internal

    pressure.

    A manhole cover that lifts when exposed to abnormal internal pressure.

    By means of frangible joint.

    VENTING OF TANKS

  • Vent sizing is as per API 2000

    Inbreathing (Vacuum Relief)

    Required venting capacity for liquid movement out of tank : iQ1

    5.6 ft3/hr for each 1 bbl/hr of maximum emptying rate

    Required venting capacity for thermal inbreathing : iQ2

    1 ft3/hr for each 1 bbl of tank capacity or 2ft3/hr for each 1 ft2 of total shell and

    roof area.

    Required total venting capacity for inbreathing : iQt = iQ1 + iQ2

    Outbreathing (Pressure Relief)

    Required venting capacity for liquid movement into tank : oQ1

    6 ft3/hr for each 1 bbl/hr of maximum filling rate

    Required venting capacity for thermal outbreathing : oQ2

    Should be 60% of the inbreathing requirement .

    Required total venting capacity for outbreathing : oQt = oQ1 + oQ2

  • Now select any standard venting device like open vent, open vent with flame

    arrester or Pressure/Vacuum valve

    Once the type and size of venting device is selected, the flow capacity of the

    device iQc, oQc will be known.

    Required number of device Ni shall be as follows:

    For inbreathing Ni = iQt/iQc

    For outbreathing No = oQt/oQc

    Sample design of Roof Vent:

    Condition:

    Tank capacity : 11,200 m3

    Tank diameter : 35,000 mm (114.8 ft)

    Tank height : 15,400 mm (50.5 ft)

    Max. filling rate : 2,600 m3/hr (16,354 bbl/hr)

    Max. emptying rate : 2,150 m3/hr (13,524 bbl/hr)

  • For Inbreathing(Vacuum relief)

    Required venting capacity for liquid movement out of tank : iQ1

    iQ1 should be 5.6 ft3/hr for each 1 bbl/hr of maximum emptying rate.

    iQ1 = 5.6 x 13,524 = 75,734 ft3/hr

    = 2,145 m3/hr

    Required venting capacity for thermal inbreathing : iQ2

    iQ2 should be 2 ft3/hr for each 1 ft2 of total shell and roof area.(A=55,450 ft2 )

    iQ2 = 2A = 2 x 55,450 = 110,900 ft3/hr

    = 3,140 m3/hr

    Required total venting capacity for inbreathing : iQt

    iQt = iQ1 + iQ2

    = 2,145 + 3,140

    = 5,285 m3/hr

  • For Outbreathing(Pressure relief)

    Required venting capacity for liquid movement out of tank : oQ1

    oQ1 should be 6.0 ft3/hr for each 1 bbl/hr of maximum filling rate.

    oQ1 = 6.0 x 16,354 = 98,124 ft3/hr

    = 2, 779 m3/hr

    Required venting capacity for thermal outbreathing : oQ2

    oQ2 should be 60% of the inbreathing requirement

    oQ2 = 0.6 x 110,900 = 66,540 ft3/hr

    = 1,884 m3/hr

    Required total venting capacity for outbreathing : oQt

    oQt = oQ1 + oQ2

    = 2,779 + 1,884

    = 4,663 m3/hr

  • Venting Device capacity

    Select a 12 Pressure/vacuum valve having the following capacity

    iQc = 3,366 m3/hr

    oQc = 5,600 m3/hr

    Required set of PV valve:

    For inbreathing : Ni

    Ni = iQt/iQc = 5,285/3,366 = 1.57

    For outbreathing : No

    No = oQt/oQc = 4,663/5,600 = 0.84

    Then, 2 sets of the above 12 venting device shall be provided for this tank.

  • Roof plates shall be attached to the top angle of the tank with a continuous

    fillet weld on the top side only.

    Frangible joint design:

    In the event of excessive internal pressure build up failure occurs first in the

    roof to shell joint protecting the bottom to shell joint.

    In most cases cone roofs are designed as frangible joints only.

    Following are the design conditions for a frangible joint:

    a. The continuous fillet weld between the roof plates and the top angle does

    not exceed 5 mm

    b. The roof slope at the top-angle attachment does not exceed 1:6

    c. The roof to compression-ring details are limited to those shown in figure.

    d. Cross-sectional area of the roof-to-shell junction,A should be less than

    value calculated by the following Aa=W/201,000 tan

    where W= Total weight of the shell & roof framing (but not the roof

    plate) supported by shell & roof

    ROOF TO SHELL JOINT

  • Openings in tank shells larger than a NPS 2 nozzle shall be reinforced.

    The minimum cross sectional area of the required reinforcement shall not be

    less than the product of the vertical diameter of the hole cut in the shell and

    the required plate thickness.

    Reinforcement may be provided by one or any combination of the following:

    a. The reinforcing plate

    b. The portion of the neck

    c. Excess shell-plate thickness.

    d. The material in the nozzle neck. The area in the neck available for

    reinforcement shall be reduced by the ratio of allowable stress in the neck to

    shell.

    The effective area of reinforcement provided by the neck is as follows,

    a. The portion extending outward from the outside surface of the tank shell

    plate to a distance equal to four times the neck-wall thickness

    REINFORCEMENT REQUIREMENT

  • b. The portion lying within the shell-plate thickness.

    c. The portion extending inward from the inside surface of the tank shell plate

    to a distance equal to four times the neck.

    Sample case:

    Manhole opening size : 511 mm

    Manhole neck thick : 10 mm

    Shell thickness required:5.958 mm, say 6 mm

    Shell thick provided : 10 mm

    Minimum c/s area of reinforcement required = 511 x 6

    = 3066 mm2

    Reinforcement provided:

    A. By excess plate thickness

    Reinforcement provided = 511 x (10-6) = 2044 mm2

    B. By manhole neck

    Neck thickness t = 10 mm, 4t = 40

    Shell thickness = 10 mm

  • Reinforcement provided by manhole neck = (40x10)+(10x10)

    = 500 mm2

    Reinforcement provided by A & B = 2044 + 500 = 2544 mm2

    Balance required reinforcement = 3066 - 2544 = 522 mm2

    Now the reinforcement plate OD can be selected such a way the total

    reinforcement area provided is higher than reinforcement required.

    OD of RF plate as per code = 1055 mm

    Selected OD of pad (to avoid fouling with weld seam) = 800 mm

    Reinforcement provided by RF pad = (800 - 511) x 6 = 1734 mm2

    which is greater than 522 mm2

    Therefore provided reinforcement is OK

  • All roofs and supporting structures shall be designed to support dead load

    plus a uniform live load of not less than 25 lb/ft2 of projected area.

    Rafters shall be spaced so that in the outer ring, their centers are not more

    than 2 feet apart measured along the circumference of the tank.

    Step 1:

    calculate the total load/unit area W acting on the roof.

    a. Live load & Vacuum load

    25 lb /ft2 as per API 650 + any vacuum load if any

    b. Dead load

    Weight of roof plate and roof structures

    Step 2:

    Now minimum number of rafters required shall be calculated.

    Say, a tank of 15 meter dia.

    Then minimum number of rafters required shall be ( x 15)/1.915 = 24.6 nos

    ROOF STRUCTURE DESIGN

  • Now say 30 rafters are provided.

    Provide 15 primary rafter and 15 secondary rafter.

    Step 3:

    Calculate the total load acting on the primary rafter

    Area of roof x W gives total load say P

    Now P/15 gives the load per rafter

    Since the roof is of cone type, loading is zero at the center and maximum at

    the periphery of the tank roof and is uniformly increasing nature from center

    to periphery.

    This condition can be considered as hinged end condition. Ra = P/15

    Step 4:

    C

    P/15 P/15

    ht

    Ha Hb

    Ra D/2=r Rb

  • Sum of the moments about end C equal to zero

    Ha x ht. + load x r x 2/3 = Ra x r

    Get value of Ha

    Maximum bending moment M = 1/3 x r x load - 1/3 Ha x ht

    Minimum section modulus required = M/All.Stress

    Now select a structural member with higher sectional modulus than required.

    Step 5:

    Now check for Induced compressive axial stress and bending stress

    Induced compressive axial stress =Ha/Ar

    where Ar = cross sectional area of the member selected

    Induced bending stress = M/Z

    where Z = section modulus of the member selected.

    If induced stress is less than allowable stress, then member size selected is OK

    For allowable stress values refer table 5.1 and 6.1 of IS-800

  • Typical Roof Structure Pattern:

  • Providing adequate foundations is an important part of ensuring an

    economical and safe liquid storage tank installation.

    Uneven foundation settlement on floating roof tank is a special problem as

    compared to fixed roof tank foundations.

    The seals of floating roof tanks will compensate for reasonable variation in

    the tank diameter such as out-of-roundness of the shell. Extreme conditions

    will impair roof seal efficiency or cause jamming of the roof, which can be

    corrected by releveling the tank. Proper foundation design will avoid this

    problem.

    At any tank site, the subsurface conditions must be known to estimate the soil

    bearing capacity and settlement that will be experienced.

    The subgrade must capable of supporting the load of the tank and its

    contents.

    The total settlement must not strain connecting piping or produce gauging

    inaccuracies, and the settlement should not continue to a point at which the

    tank bottom is below the surrounding ground surface.

    The tank grade or surface on which a tank bottom will rest should be

    constructed at least 0.3 m above the surrounding ground surface.

    TANK FOUNDATION DESIGN

    CONSIDERATIONS

  • Clean washed sand 75 to100 mm deep is recommended as a final layer

    because it can be readily shaped to the bottom contour of the tank to provide

    maximum contact area and will protect the tank bottom from coming into

    contact large particles and debris.

    The finished grade shall be crowned from its outer periphery to its center at a

    slope of one inch in ten feet.

    The crown will partly compensate for slight settlement, which is likely to be

    greater at the center.

    It will also facilitate cleaning and the removal of water and sludge.

    Typical foundation types are earth foundation without a concrete ring wall

    and earth foundation with a concrete ring wall.

    Foundation without a ring wall shall be adopted for small size tanks and on

    surface where adequate bearing capacity is available.

    Tanks with heavy or tall shells and/or self-supported roofs impose a

    substantial load on the foundation under the shell.

  • Advantages of Concrete ring wall

    Provides better distribution of the concentrated load of the shell to produce a

    more uniform soil loading under the tank.

    Provides a level, solid starting plane for construction of the shell.

    Provides a better means of leveling the tank grade, and it is capable of

    preserving its contour during construction.

    Disadvantages of Concrete ring wall

    It doesnt conform to differential settlements which may lead to high bending

    stress in the bottom plates adjacent to the ringwall.

    Ringwall shall not be less than 300 mm thick.

    The centerline diameter of the ringwall should equal the nominal diameter of

    the tank.

  • Pan type floating roof is the first type used in the industry. As the name

    indicates, this roof looked very much like a shallow pan. The single deck

    sloped to the centre for drainage.

    The pan roof could sink under heavy loads of water or snow or from leaks in

    the deck or drain. Since the single-deck was in direct contact with the stored

    liquid, the more volatile liquids would sometimes boil from the suns heat.

    Pontoon type floating roof has a single deck with an annular pontoon divided

    by bulkheads into liquid-tight pontoon compartments.

    The pontoon area was in excess of 50% of the total roof area. The top deck of

    the pontoon shaded the bottom deck which is in contact with the liquid.

    The single deck area was designed to balloon upward to contain vapors

    produced by boiling. This reduced considerably the heat input and further

    boiling.

    Double deck floating roof has two deck, one top and one bottom deck.

    These two decks are separated by rim plates and bulk heads to form liquid-

    tight pontoon compartments.

    TYPE OF FLOATING ROOFS AND ITS

    ACCESSORIES

  • The top deck provides an insulating air space over the entire area and boiling

    losses are held to a minimum.

    The deck slopes to one or more drainage points and open emergency overflow

    drains protect the roof from excessive water loads.

    Internal floating roofs is a fixed roof tank with a floating roof inside.

    The fixed roof provides a shade from the sun, protection from the wind and

    also keeps the rain and snow off the floating roof.

    Pontoon design:

    Floating roofs shall have sufficient buoyancy to remain afloat on liquid with a

    specific gravity of 0.7 and with primary drains inoperative for the following

    conditions:

  • a. 250 mm(10 in.) of rainfall in a 24-hour period with the roofs intact, except

    for double-deck roofs provided with emergency drains to keep water to a

    lesser volume that the roofs will safely support. Such emergency drains shall

    not allow the product to flow onto the roof.

    b. Single-deck and any two adjacent pontoon compartments punctured in

    single-deck pontoon roofs and may any two adjacent compartments

    punctured in double-deck roofs, both roof types with no water or live load.

    ACCESSORIES:

    Following are the accessories of floating roofs:

    Roof drain

    Emergency drain

    Bleeder vent

    Rim vent

    Foam seal

    Supporting legs

    Anti-rotation devices

    Automatic tank gauging

    Rolling ladder

  • Roof drain:

    Roof drains are for removing water from floating roofs in open top tanks.

    These drains are made out of pipes with swing joint assembly.

    These pipe drains are also called as flexible pipe drains as these pipes extends

    and shrinks with the varying level of the roof which depends on the product

    height.

    Emergency drain:

    Water automatically drains into the tank when it reaches a certain level on

    the roof. Rainwater cannot collect on the roof to endanger the safety of the

    floating roof .

    Bleeder vent:

    Vents the air from under a floating roof when the tank is being filled initially.

    After the liquid rises enough to float the roof off its supports the vent

    automatically closes. When the tank is being emptied the vent is automatically

    opened just before the roof lands on its support.

  • Rim vents:

    Rim vents are provided to release any excess pressure in the rim space after

    the roof is floating.

    Foam seals:

    One of the important component of a floating roof is the primary seal

    between the floating roof and the tank shell.

    A good seal closes the space effectively, yet permits normal roof movement

    while protecting against evaporation loses.

    Supporting legs:

    Floating roof shall be provided with supporting legs.

    Legs fabricated from pipe shall be notched or perforated at the bottom to

    provide drainage.

    The length of legs shall be adjustable from the top side of the roof.

    The operating and cleaning position levels of the supporting legs shall be

    specified of fixing the adjustable positions.

    The legs and attachments shall be designed to support the roof and a uniform

    live load of at least 1.2 kPa(25 lb/ft2)

    Steel pads shall be used to distribute the leg loads on the bottom of the tank.

  • Anti-rotation device:

    Required to prevent floating roofs from rotating and damaging rolling ladder,

    pipe drains and seal.

    A guided pole is used as anti-rotation device. The pole is fixed at the top and

    bottom and passes through a well. The guide pole can additionally used as

    gauging or sampling device.

    Rolling ladder:

    Rolling ladder provides safe and easy access from top of the tank to the

    floating roof.

    On floating roof a runway is provided, over this runway the ladder provided

    with spark proof wheels will travel.

    These ladders are provided with self-leveling treads.

  • CALCULATION OF THICKNESS BY VARIABLE POINT

    METHOD:

    Design by this method gives shell thickness at design points that

    results in the calculated stresses being relatively close to the actual

    circumferential shell stresses.

    To calculate the bottom-course thickness, preliminary values tpd and

    tpt for the design and hydrostatic test conditions shall first be

    calculated from the 1-foot method formula.

    The bottom shell course thickness t1d and t1t for the design and

    hydrostatic test condition shall be calculated using the following

    formulae:

    t1d = (1.06 - (0.463D/H)(HG/Sd)0.5 (2.6HDG/ Sd) + CA

    t1t = (1.06 - (0.463D/H)(H/St)0.5 (2.6HD/ St)

    To calculate the second-course thickness for both the design condition

    and the hydrostatic test condition, the value of the following ratio

    shall be calculated for the bottom course:

  • h1/(rt1)

    0.5

    where,

    h1 = height of the bottom shell course

    r = nominal tank radius

    t1 = actual thickness of the bottom shell course, less any

    thickness added for CA used to calculate t2

    If the value of the ratio is less than or equal to 1.375,

    t2 = t1

    If the value of the ratio is greater than or equal to 2.625,

    t2 = t2a

    If the value of the ratio is greater than 1.375 but less than 2.625,

    t2 = t2a + (t1 - t2 a)[2.1-(h1/1.25(rt1)0.5]

    where,

    t2 = minimum design thickness of the second shell course

    excluding any CA

    t2a = thickness of the second shell course as calculated for an

    upper shell course as calculated as described below

  • The preceding formula for t2 is based on the same allowable stress

    being used for the design of the bottom and second courses.

    For tanks where the value of the ratio is greater than or equal to

    2.625, the allowable stress for the second course may be lower than

    the allowable stress for the bottom course.

    To calculate the upper-course thickness for both the design condition

    and hydrostatic test condition, a preliminary value tu for the upper

    course thickness shall be calculated by the 1-foot method and then

    the distance x of the variable design point from the bottom of the

    course shall be calculated using the lowest value obtained from the

    following:

    x1 = 0.61(rtu)0.5 + 3.84 CH

    x2 = 12CH

    x3 = 1.22(rtu)0.5

    where,

    tu = thickness of the upper course at the girth joint

    C = [K0.5(K-1)]/(1+K1.5)

    K = tL/tu

  • The minimum thickness tx for the upper shell courses shall be

    calculated for both the design condition (tdx) using the minimum

    value of x obtained as explained above

    tdx = (2.6D(H-x/12)G)/ Sd) + CA

    ttx = (2.6D(H-x/12)G)/ St)

    The steps described above shall be repeated using the calculated

    value of tx as tu until there is little difference between the calculated

    values of tx in succession.

    Repeating the steps twice is normally sufficient.

  • Design Data

    Type of Tank : Double deck floating roof

    Diameter of Tank 'D' :92 m

    Height of Tank 'H' :20m

    Product Stored :Crude Oil

    Design specific gravity 'G': 0.9 Corrosion Allowance 'C.A.' :0.03937 inches 1 mm Course width : 2.5 m

    Capacity of Tank : 132952.2 cu.m. 836176.3 barrels

  • The bottom shell course thickness:

    t1d = [ 1.06 - (0.463 D/H) sqrt(HG/Sd) ] (2.6HDG/Sd

    = 1.59256 inches

    = 40.45102 mm

    Where,

    D = Tank Diameter = 301.8336

    H = Tank Height = 65.616

    G = Product Specific Gravity = 0.9

    Sd = Allowable stress for design condition

    of bottom course = 28000

    adding 0.03937 inches as corrosion allowance

    t1d = 1.63193 inches

    41.45102 mm

  • The second shell course thickness:

    h1/(rt1) 0.5 = 1.8327322

    Where,

    h1 = bottom course height = 98.424 inches

    r = nominal tank radius = 1811.002 inches

    t1 = bottom course thickness = 1.63193 inches

    From API 650 Cl.3.6.4.5

    h1/ (rt1)0.5 applicable formula

    1.375-2.625 t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1))

    > 2.625 t2 = t2a

    < 1.375 t2 = t1

    Since h1/(rt1) 0.5 = 1.8327322

    t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1)) where,

    t2 = minimum design thickness for second shell

    t2a = thickness for second shell course as calculated for an upper course

  • First trial:

    Course # 2

    H = 57.414 ft.

    tu = (2.6D (H-1) G)/Sd

    = 1.423024 inches

    tl = 1.59256 inches

    K = tl/tu = 1.119137

    C= k^0.5 (k-1) / (1 + K^1.5)

    = 0.05771

    a = (rtu)^0.5

    50.76514 inches

    x1 = 0.61a + 3.84CH

    = 43.69006 inches

    x2 = 12CH

    = 39.7604 inches

    x3 = 1.22a

    = 61.93347 inches

    x = min ( x1, x2, x3)

    = 39.7604 inches

    x/12 = 3.313367 inches

    tdx = (2.6D (H-x/12)G) / Sd

    = 1.36467 inches

  • Second trial: H = 57.414

    tu = 1.36467

    tl = 1.59256

    K = tl/tu

    1.166992

    C= k^0.5 (k-1) / (1 + K^1.5)

    0.079798

    a = (rtu)^0.5

    49.71338

    x1 = 0.61a + 3.84CH

    = 47.91824

    x2 = 12CH

    = 54.97837

    x3 = 1.22a

    = 60.65032

    x = 47.91824

    x/12 = 3.993187

    tdx = (2.6D (H-x/12)G) / Sd

    1.347522

  • Third trial:

    H = 57.414

    tu = 1.347522

    tl = 1.59256

    K = tl/tu

    1.181843

    C= k^0.5 (k-1) / (1 + K^1.5)

    0.086522

    a = (rtu)^0.5

    49.40005

    x1 = 0.61a + 3.84CH

    = 49.2095

    x2 = 12CH

    = 59.61084

    x3 = 1.22a

    = 60.26806

    x = 49.2095

    x/12 = 4.100791

    tdx = (2.6D (H-x/12)G) / Sd

    = 1.344808

  • t2a = 1.344808

    t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1))

    1.501837 inches

    38.14665 mm

    Adding 0.0625 inches corrosion allowance

    t2 = 1.541207 inches

    39.14665 mm

    The third shell course thickness:

    First trial:

    Course # 3

    H = 49.212

    tu = (2.6D(H-1)G)/Sd

    1.216132 tl = 1.501837

    As explained earlier repeat the steps and calculate the third shell course

    thickness. Similarly, shell thickness of other courses are calculated

  • ONE-FOOT METHOD DESCRIPTION:

    This method of calculating the thickness of the shell is based on the

    assumption that the tank is filled with water and the tension in each ring is

    calculated at a point 12 in. above the center line of the lower horizontal joint

    of the horizontal row of welded plates being considered.

    The hydrostatic pressure varies from a minimum at the top of the upper most

    course to a maximum at the bottom of the lowest course.

    In determining the plate thickness for a particular course, a design based

    upon the pressure at the bottom of the course results in over-design for the

    rest of the plate. A design based upon the pressure at the top of the course

    would result in under-design.

    However, some consideration should be given to the additional restraint

    offered by the plates adjoining a particular course.

    In the lowest course, the plates of the vessel bottom offer considerable

    restraint to the bottom shell course.This additional restraint of the bottom

    edge is effective for an appreciable distance or height from the bottom of the

    lowest course.

    In an intermediate course with a course of heavier plates below, the top of the

    heavier will be understressed.

    Therefore, a design based upon the pressure at a height of 1 ft from the

    bottom of the course may be considered conservative.