29541844 6 motion compensation

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    1

    TAMU - Pemex

    Offshore Drilling

    Lesson 6

    Motion Compensation

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    2

    Motion Compensation

    Reentry

    Tensioners

    Heave Compensators

    Passive Motion Compensation

    Active and Semiactive Systems

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    3

    Re-entry

    It is possible to re-enter a borehole

    without using guidelines!

    1. Use land-based navigationequipment to get the vessel in

    the vicinity of the well

    or better still: Use GPS

    (Global Positioning System)

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    4

    Re-entry, contd

    2. Lower reentry string to a safe

    distance above the wellhead

    3.Use position location equipment

    to complete the job:

    (i) Television camera(ii) Acoustic device - pinger or transponder

    (iii) ROV - Remote Operated Vehicle

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    5

    Heave Compensation

    How do you maintain a constant tension

    on the marine riser - when the vesselheaves?

    How do you maintain a constant weight

    on the bit - when the vessel heaves?

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    6

    Deadweight Riser Tensioning System

    Dead Weight

    (~constanttension)

    Dead Weight

    Slip Joint

    Marine Riser

    Early design - OK up to 100,000 lbf. Attached to lower half of telescoping joint. Adds weight & is bulky...

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    Pneumatic Riser

    Tensioning System

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    Pneumatic/Hydraulic Heave

    Compensation System

    Pneumatic/Hydraulic Tensioners:

    Take up much less room thandead weights

    Facilitate changing the tension by

    changing the air pressure Can be used for the marine riser,

    the guidelines and the drill string

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    Pneumatic/Hydraulic Heave

    Compensation System

    Passive Systemsare

    the most popular Require essentially no energy input

    Use an air spring with a variable

    spring constant Can keep the tension within ~ 15%

    or even less.

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    Pneumatic/Hydraulic Heave

    Compensation System

    Active Systems

    Require external energy througheach load cycle

    Provide a highly consistent force

    But -- Have a high initial cost

    Have a high operating cost

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    Air reservoir

    reduces

    pressurechanges

    F = PA

    To support

    larger load,increase the

    pressure

    An AIR spring ...

    Passive

    System

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    Consider Change from P1

    and V1

    to P2

    and V2

    P1 and P2 are absolute pressures

    n = 1 for isothermal expansion or contraction

    n = 1.41 for adiabatic expansion or contraction

    1

    2

    1

    2

    P

    P

    F

    F

    n

    V

    V

    P

    P

    and

    =2

    1

    1

    2

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    14

    Change in Volume

    P1 and P2 are absolute pressures

    Isothermal Volume Change: PV = const.

    (slow; constant temperature)

    Adiabatic Volume Change: PV1.41

    = const.

    (fast; no heat flow)

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    15

    V

    V Relative

    Change involume

    37180

    411

    1

    1

    411

    2

    1

    1

    2 .V.

    V

    V

    V

    P

    P.. =

    =

    =

    Pres s

    ur

    eRatio,

    P2

    /P1

    ADIABATIC

    1.4

    1.0

    0.7

    -0.1 +0.2

    ISOTHERMAL

    25.

    8.

    1

    V0

    V

    V

    V

    P

    P

    1

    1

    2

    1

    1

    2 =

    =

    =

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    16

    Theoretical reservoirsize vs. pressure

    fluctuation.

    Allowable Pressure Fluctuation, %Vo l.o fR

    e s e rv

    o ir

    /Vo l .

    o f C

    y lind e r

    15 300

    15

    30

    Fig. 4-7

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    17

    Example

    Consider a 14-inch piston with a 10-foot

    stroke. We shall hook two units (cylinders)

    to the reservoir. Determine reservoir size

    for 15% force variation.

    Reservoir Volume = R m A L

    4)-7Figure(from5R

    cylinderofvolumereservoirofvolumelTheoretica

    =

    =R = 5

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    18

    Example contd

    Reservoir Volume = R m A L

    = 5 * 2 * 10 * 1.07 = 107 ft3

    m = 2 = number of cylinders

    L = 10 ft = piston stroke

    A = /4 (14/12)2

    = 1.07 ft2 = piston area

    VRES = R m A L

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    19

    Equation (on p. 170)

    Theoretical force variations as a function ofpiston position and heave can be

    determined by a modification of the

    previous pressure - volume equation:

    2

    LL

    2

    L-and

    1001LAmV

    V1001

    F

    FE

    n

    1

    2

    =

    =

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    20

    Where:

    E = theoretical percentage error in forceapplied to the load

    = force applied to the load relative to

    the force with the piston centeredA = piston area, ft.2

    L = full piston stroke, ft.

    V = reservoir volume + L = length of the stroke from the

    center of the piston, ft.

    3

    2 ft,

    mAL

    1

    2

    F

    F

    1001

    1001

    1

    2

    n

    LAmV

    V

    F

    FE

    2

    LL

    2

    L-

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    21

    Heave Down

    PercentC

    hangein

    Load

    0

    15

    -15-20 20

    15

    -15

    0

    010

    Heave Up, ft

    Reservoir Volume

    = 5 * vol. swept by piston

    Adiabatic

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    22Lower sheaves are attached to the cylinder (and vessel), upper sheaves are attached to the piston rod

    20 ft

    5 ft1,500 psig WP

    TypicalTensioner

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    23

    Motion

    Compensator

    Principles of

    Operation

    Purpose:

    Keep bit on

    bottomwith low bit

    weight change

    Air pressure can support entire weight of drillstring. e.g. 200,000 - 40,000 = 160,000lbf. Reduce air press.

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    24

    Rucker Heave

    Compensator

    Dual Pistons- on the

    travelling

    block

    Large air

    cylinders

    below deck

    Flexible hoses

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    25

    Figure 7-9

    Vetco dual

    piston HeaveCompensator

    L.P. hydraulic

    fluid throttledfor damping

    Piston

    balancing

    may be

    problem

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    26

    Figure

    7-10

    Split travelling block

    Hydraulicallyoperated pistons can

    be locked in any

    position with

    remotely operated

    valves.

    Air operated units must be mechanically locked in position because of compressible fluid in cylinders.

    Single

    Piston Rod

    Western Gear

    Heave

    Compensator

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    27

    Figure 7-13. Active Heave Compensator

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    28

    Semiactive Heave Compensator

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    Bumper Subs

    Fit into the drill collar string, but do

    not have the ruggedness of drill

    collars. To obtain a reasonableoperating life from bumper subs,

    action has to be taken on the

    following points of concern: 1. Always run the bumper subs at the

    neutral point in the string.

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    30

    Bumper Subs, contd

    2. Bumper subs are designed to

    stroke, and if operated at a single

    position, they will wear at that

    position.

    3. Running the subs in tension

    minimizes the area through which

    the torque will be transmitted, andwill cause excessive wear and fatigue.

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    31

    Bumper Subs

    4. Running the joint in compression

    increases torque reversals and fatigue

    in the tool. It also increases bitchatter and instantaneous peak

    torque, causing unnecessary wear to

    the tool and the string. This is the fault

    of the operation, not the sub.

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    32

    Bumper Subs

    5. From items 3 and 4 above, it is

    evident that once the string has

    been run, the weight on the bit isfixed. This weight should not be

    changed until the string has been

    pulled and the number of drill collarsbelow the bumper sub changed.

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    33

    Bumper Subs

    6.Do not separate the bumper subs in a

    drillstring for heavy drilling operations.

    7.Special care and maintenance are

    required to obtain a reasonable

    operating life.

    *Remember: bumper subs fit into the same

    part of the drillstring where drill collars and

    stabilizers have been destroyed. This is a

    severe test of workmanship & maintenance.

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    34

    Unbalanced

    BumperSubs

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    35

    Buoyant Riser Module

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    36

    Cantilever

    JackupRig

    Float out to

    location

    Then lower

    legs toseafloor

    Then jack up

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    37

    Jackup Rig

    Side View

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    38

    Jackup

    Rig

    Deck

    Plan

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    End of

    Lesson 6