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DrillCostRate-1

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    1

    PETE 411Well Dri l l ing

    Lesson 4

    Dri ll ing Cos t & Dri ll ing Rate

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    2

    Contents

    The AFE

    Drilling Cost and Bit Change

    Factors Affecting Drilling Rate Bit Weight, Rotary Speed

    Bottom-hole Cleaning

    Mud Properties, Solids Content Hydrostatics

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    3

    HW #2. ADE 1.12, 1.13, 1.14Due Friday, Sept. 13, 2002

    Read: ADE, Ch. 1 (All)

    Learn:Rig Components - Definitions

    Ass ignments

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    Before gett ing app roval to d r i l l a wel l the

    Dri ll ing Engineer mus t prepare an AFE

    - a detai led cost est imate for the wel l

    DRY COMPLETED

    HOLE

    INTANGIBLE COSTS $ $

    TANGIBLE COSTS $ $

    TOTAL COST $ $

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    5

    AUTHORIZATION FOR EXPENDITURE (AFE)

    EXPENDITURE DRY HOLE COMPLETED(24.5 DAYS) (32.5 DAYS)

    INTANGIBLE COSTSLOCATION PREPARATION 30,000 65,000DRILLING RIG AND TOOLS 298,185 366,613DRILLING FLUIDS 113,543 116,976RENTAL EQUIPMENT 77,896 133,785CEMENTING 49,535 54,369

    SUPPORT SERVICES 152,285 275,648TRANSPORTATION 70,200 83,400SUPERVISION AND ADMIN. 23,282 30,791

    SUB-TOTAL 814,928 1,126,581

    TANGIBLE COSTS

    TUBULAR EQUIPMENT 406,101 846,529WELL HEAD EQUIPMENT 16,864 156,201COMPLETION EQUIPMENT 0 15,717

    SUB-TOTAL 422,965 1,018,447

    SUB-TOTAL 1,237,893 2,145,028

    + CONTINGENCY (15% ??) 1,423,577 2,466,782

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    Dril ling Cost vs . Time

    DEPTHft

    DAYS or DOLLARS

    TD

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    Dril ling Cost Analys is

    The Drilling Engineer:

    Recommends drilling procedures that

    will safely drill and complete the wellat the lowest cost possible

    Makes recommendations concerningroutine rig operations:

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    The Dri l l ing Eng ineer

    Examplesof routine rig operations

    drilling fluid treatment

    pump operation

    bit selection

    handling problems during thedrilling process

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    The Dri l l ing Cos t Equat ion :

    Cf= drilling cost, $/ftCb= cost of bit, $/bit

    Cr= fixed operating cost of rig, $/hr

    tb = total rotating time, hrs

    tc = total non-rotating time, hrs

    tt = total trip time(round trip), hrs

    D

    ft$

    D)ttt(CCCtcbrb

    f

    Eq. 1.16

    = footage drilledwith bit, ft/bit

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    Example 1.5

    A recommended bit program is being prepared fora new well using bit performance records fromnearby wells.

    Drilling performance records for three bits areshown for a thick limestone formation at 9,000 ft.

    Determine which bit gives the lowest drilling cost ifthe operating cost of the rig is $400/hr, the trip timeis 7 hours, and connection time is 1 minute perconnection.

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    Assume that each of the bits was operated atnear the minimum cost per foot attainable forthat bit.

    Mean

    Bit Rotating Connection PenetrationCost Time Time RateBit ($) (hours) (hours) (ft/hr)

    A 800 14.8 0.1 13.8

    B 4,900 57.7 0.4 12.6C 4,500 95.8 0.5 10.2

    Example 1.5 contd

    Which bit would you select?

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    Solut ion:

    The cost per foot drilled for each bit typecan be computed using Eq. 1.16. For Bit

    A, the cost per foot is

    /ft.81.46$)8.14(8.13

    )71.08.14(400800Cf

    ft

    $

    D

    )ttt(CCC

    tcbrbf

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    Solut ion:

    Similarly, for Bit B,

    /ft.56.42$)7.57(6.12

    )74.07.57(400900,4Cf

    ft$

    D)ttt(CCC tcbrbf

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    Solution, contd

    Finally, for Bit C,

    /ft.89.46$)8.95(2.10

    )75.08.95(400500,4Cf

    ft$

    D)ttt(CCC tcbrbf

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    Solution, contd

    Bit A: $46.81 /ft

    Bit B: $42.56 /ft

    Bit C: $46.89 /ft

    The lowest drilling cost was obtained

    using Bit B. - Highest bit cost but -intermediate bit life and ROP...

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    Dril ling Costs

    Tend to increase exponentially withdepth. Thus, when curve-fitting drilling

    cost data, it is often convenient toassume a relationship between totalwell cost, C, and depth, D, given by

    C= aebD ..(1.17)

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    Drilling Costs, contd

    Constants aand b depend primarily on thewell location.

    Shown on the next page is a least-squares

    curve fit of the south Louisiana completedwell data given in Table 1.7.

    Depth range of 7,500 ft to 21,000 ft.

    For these data,a =1 X 105 dollars

    b =2 X 10-4 ft -1.

    C= aebD

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    Fig. 1-65. Least-square curve fit of 1978 completed wellcosts for wells below 7,500 ft in the south Louisiana area.

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    Penetration Rate

    When major variations are not present inthe subsurface lithology, the penetrationrate usually decreases exponentially with

    depth. Under these conditions, thepenetration rate can be related to depth,D, by

    where K anda2

    are constants.

    )18.1,........(Kedt

    dD Da303.2 2

    WHY?

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    Dri ll ing Time

    The drilling time,td, required to drill to agiven depth can be obtained by separating

    variables and integrating. Separatingvariables gives

    dDedtKDa303.2D

    0

    t

    0

    2d

    Integrating and solving fortdyields

    )19.1.....().........1e(Ka303.2

    1t Da303.2

    2

    d2

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    Drilling Time contd

    Plotting depth vs. drilling time from pastdrilling operations:

    A. Allows more accurate prediction of time andcost for drilling a new well

    B. Is used in evaluating new drilling

    procedures (designed to reduce drillingtime to a given depth).

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    EXAMPLE - Cos t per ft

    t R D Total Cost Cfhr fph ft $ $/ft

    5 90 475 36,950 77.80

    10 80 900 47,800 53.1020 60 1,600 69,200 43.3025 50 1,875 79,750 42.5030 40 2,100 90,200 43.0035 30 2,275 100,550 44.2040 20 2,400 110,800 46.20

    These cost data are plotted below.

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    Cost per f t for one ent i re bi t run

    Minimum Cost

    80

    70

    60

    50

    40

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    Econom ic Procedu re in above Table

    Can pull bit after about 25 hr. ($42.50/ft)

    - the precise pulling time is not critical

    Note that the cost in dollars per footwas $43.00 after30 hr.

    Primarily applicable to tooth-type bitswhere wear rate is predictable.

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    Econom ic Procedu re in above Table

    Also used with tungsten carbide insertbits when inserts are broken orpulled out of the matrix.

    Unfortunately, wear rate with insertbits is unpredictable.

    Economically, the insert bit should bepulled when the cost in $/ft begins toincrease.

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    Econom ic Procedu re in Table

    Bits pulled for economic reasons makeit hard to obtain wear information.

    Operator might pull bit after120 hrofuse but part of bit might get left inhole. Recovery is very difficult. Avoid!

    75% of rock bits are pulled green orbefore the bit is worn out.

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    An increase in

    TORQUE mayindicate that a bitshould be pulled.

    Experience oftendictates when to

    pull bit (footage orhours).

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    Fac to rs that affec t Penetrat ion Rate

    Type of Drill bit

    Bit weight

    Rotary speed

    Bottom-hole cleaning

    Mud properties

    Rock hardness

    Formation pore pressure

    Variables:

    Fixed Factors:

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    B it Select ion is based on

    Past bit records

    Geologic predictions of lithology

    Drilling costs in $/bit...

    Drilling cost in $/ft

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    B it Weigh t and Ro tary Speed

    Increasing bit weight and rotary speedboosts drilling rate

    These increases accelerate bit wear

    Field tests show that drilling rateincreases more or less in directproportion to bit weight

    40 000 lbf

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    Consider 10 hole

    (dont overdo!!)

    Bit Weight x 1,000 lb/in

    40,000 lbf

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    32Rotary Speed, RPM

    Dont overdo!Casing wear,

    bit life ...

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    EFFECT OF BACK PRESSUREKeep P_bit = const.= 550 psi

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    34Hydrostatic Pressure, 1,000s of psi

    EFFECT OF BACK PRESSURE0 - 5,000 psi

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    35Drilling Time, days

    Drilled with gas

    Drilled with mud

    EFFECT OF DRILLING FLUIDmud vs. gas

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    36Rotating Time, hours

    EFFECT OF DRILLING FLUIDwater vs. air

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    37Old style water course bits

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    EFFECT OF SOLIDS IN THE MUD

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    Hydrostat ic Pressu re Grad ient

    Fresh Water Pressure Gradient = 0.433 psi/ftDensity of Fresh Water = 8.33 lb/gal

    Hydrostatic Pressure (at 12,000 ft depth):

    with water: p = Gw

    * Depth (vertical depth)

    = 0.433 psi/ft * 12,000 ft

    = 5,196 psi

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    Hydrostat ic Pressure

    with 14 lb/gal mud: p = GM * Depth

    Depth*psi/ft433.0*8.33

    Mud

    Depth*WeightMud*0.052Pressure

    = 0.052 *14.0 *12,000

    = 8,736 psig

    (5,196 psi with water)

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    Hydrostat ic Pressure Requ ired

    Depth*0.052

    PressureWeightMudRequired

    What mud weight is required tobalance a pore pressure of 10,000 psigat a vertical depth of 12,000 ft?

    12,000*0.05210,000WeightMudRequired

    MW = 16.03 lb/gal