PETE 411
Well Drilling
Lesson 4
Drilling Cost & Drilling Rate
1
Contents







The AFE
Drilling Cost and Bit Change
Factors Affecting Drilling Rate
Bit Weight, Rotary Speed
Bottom-hole Cleaning
Mud Properties, Solids Content
Hydrostatics
2
Assignments
Read:
ADE, Ch. 1 (All)
Learn: Rig Components - Definitions
HW #2.
ADE 1.12, 1.13, 1.14
Due Friday, Sept. 13, 2002
3
Before getting approval to drill a well the
Drilling Engineer must prepare an AFE
- a detailed cost estimate for the well
DRY
HOLE
COMPLETED
INTANGIBLE COSTS
$
$
TANGIBLE COSTS
$
$
TOTAL COST
$
$
4
AUTHORIZATION FOR EXPENDITURE (AFE)
EXPENDITURE
DRY HOLE
COMPLETED
(24.5 DAYS)
(32.5 DAYS)
INTANGIBLE COSTS
LOCATION PREPARATION
DRILLING RIG AND TOOLS
DRILLING FLUIDS
RENTAL EQUIPMENT
CEMENTING
SUPPORT SERVICES
TRANSPORTATION
SUPERVISION AND ADMIN.
SUB-TOTAL
TANGIBLE COSTS
TUBULAR EQUIPMENT
WELL HEAD EQUIPMENT
COMPLETION EQUIPMENT
SUB-TOTAL
30,000
298,185
113,543
77,896
49,535
152,285
70,200
23,282
814,928
406,101
SUB-TOTAL
+ CONTINGENCY (15% ??)
65,000
366,613
116,976
133,785
54,369
275,648
83,400
30,791
1,126,581
846,529
16,864
0
422,965
156,201
15,717
1,018,447
1,237,893
1,423,577
2,145,028
2,466,782
5
Drilling Cost vs. Time
DEPTH
ft
TD
DAYS or DOLLARS
6
Drilling Cost Analysis
The Drilling Engineer:
 Recommends drilling procedures that
will safely drill and complete the well
at the lowest cost possible
 Makes recommendations concerning
routine rig operations:
7
The Drilling Engineer
Examples of routine rig operations
 drilling fluid treatment
 pump operation
 bit selection
 handling problems during the
drilling process
8
The Drilling Cost Equation:
Cb  Cr( tb  tc  tt )
Cf 
D
Cf = drilling cost, $/ft
Cb= cost of bit, $/bit
$
ft
Eq. 1.16
D = footage drilled
with bit, ft/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
9
Example 1.5
A recommended bit program is being prepared for
a new well using bit performance records from
nearby wells.
Drilling performance records for three bits are
shown for a thick limestone formation at 9,000 ft.
Determine which bit gives the lowest drilling cost if
the operating cost of the rig is $400/hr, the trip time
is 7 hours, and connection time is 1 minute per
connection.
10
Example 1.5 cont’d
Assume that each of the bits was operated at
near the minimum cost per foot attainable for
that bit.
Bit
Bit
Cost
($)
Rotating
Time
(hours)
A
B
C
800
4,900
4,500
14.8
57.7
95.8
Connection
Time
(hours)
0.1
0.4
0.5
Mean
Penetration
Rate
(ft/hr)
13.8
12.6
10.2
Which bit would you select?
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Solution:
The cost per foot drilled for each bit type
can be computed using Eq. 1.16. For Bit
A, the cost per foot is
Cb  Cr( tb  tc  tt )
Cf 
D
$
ft
800  400(14.8  0.1  7)
Cf 
 $46.81/ft.
13.8(14.8)
12
Solution:
Similarly, for Bit B,
Cb  Cr( tb  tc  tt )
Cf 
D
$
ft
4,900  400(57.7  0.4  7)
Cf 
 $42.56/ft.
12.6(57.7)
13
Solution, cont’d
Finally, for Bit C,
Cb  Cr( tb  tc  tt )
Cf 
D
$
ft
4,500  400(95.8  0.5  7)
Cf 
 $46.89/ft.
10.2(95.8)
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Solution, cont’d
Bit A:
Bit B:
Bit C:
$46.81 /ft
$42.56 /ft
$46.89 /ft
The lowest drilling cost was obtained
using Bit B. - Highest bit cost …but intermediate bit life and ROP...
15
Drilling Costs
Tend to increase exponentially with
depth. Thus, when curve-fitting drilling
cost data, it is often convenient to
assume a relationship between total
well cost, C, and depth, D, given by
C = aebD
…………………..(1.17)
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Drilling Costs, cont’d
C = aebD
Constants a and b depend primarily on the
well location.
Shown on the next page is a least-squares
curve fit of the south Louisiana completed
well 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.
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Fig. 1-65. Least-square curve fit of 1978 completed well
costs for wells below 7,500 ft in the south Louisiana area.18
Penetration Rate
When major variations are not present in
the subsurface lithology, the penetration
rate usually decreases exponentially with
depth. Under these conditions, the
penetration rate can be related to depth,
D, by
dD
 2.303a 2 D
 Ke
,........(1.18)
dt
where K and a2 are constants.
WHY?
19
Drilling Time
The drilling time, td , required to drill to a
given depth can be obtained by separating
variables and integrating. Separating
variables gives
K
0
td
D
dt   e
2.303 a 2D
dD
0
Integrating and solving for td yields
1
td 
(e 2.303 a2D  1)......... .....( 1.19)
2.303a 2K
20
Drilling Time cont’d
Plotting depth vs. drilling time from past
drilling operations:
A. Allows more accurate prediction of time and
cost for drilling a new well
B. Is used in evaluating new drilling
procedures (designed to reduce drilling
time to a given depth).
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EXAMPLE - Cost per ft
t
hr
R
fph
D
ft
Total Cost
$
Cf
$/ft
5
10
20
25
30
35
40
90
80
60
50
40
30
20
475
900
1,600
1,875
2,100
2,275
2,400
36,950
47,800
69,200
79,750
90,200
100,550
110,800
77.80
53.10
43.30
42.50
43.00
44.20
46.20
These cost data are plotted below.
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Cost per ft for one entire bit run
80
70
60
Minimum Cost
50
40
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Economic Procedure 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 foot
was $43.00 after 30 hr.
 Primarily applicable to tooth-type bits
where wear rate is predictable.
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Economic Procedure in above Table
 Also used with tungsten carbide insert
bits when inserts are broken or
pulled out of the matrix.
 Unfortunately, wear rate with insert
bits is unpredictable.
 Economically, the insert bit should be
pulled when the cost in $/ft begins to
increase.
25
Economic Procedure in Table
 Bits pulled for economic reasons make
it hard to obtain wear information.
 Operator might pull bit after 120 hr of
use but part of bit might get left in
hole. Recovery is very difficult. Avoid!
 75% of rock bits are pulled green or
before the bit is worn out.
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An increase in
TORQUE may
indicate that a bit
should be pulled.
Experience often
dictates when to
pull bit (footage or
hours).
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Factors that affect Penetration Rate
Variables:
 Type of Drill bit
 Bit weight
 Rotary speed
 Bottom-hole cleaning
 Mud properties
Fixed Factors:
 Rock hardness
 Formation pore pressure
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Bit Selection is based on
 Past bit records
 Geologic predictions of lithology
 Drilling costs in $/bit...
 Drilling cost in $/ft
29
Bit Weight and Rotary Speed
 Increasing bit weight and rotary speed
boosts drilling rate
 These increases accelerate bit wear
 Field tests show that drilling rate
increases more or less in direct
proportion to bit weight
30
40,000 lbf
Consider 10” hole
(don’t overdo!!)
Bit Weight x 1,000 lb/in
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Don’t overdo!
Casing wear,
bit life ...
Rotary Speed, RPM
32
EFFECT OF BACK PRESSURE
Keep P_bit = const.= 550 psi
33
EFFECT OF BACK PRESSURE
0 - 5,000 psi
Hydrostatic Pressure, 1,000’s of psi
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EFFECT OF DRILLING FLUID
mud vs. gas
Drilled with mud
Drilled with gas
Drilling Time, days
35
EFFECT OF DRILLING FLUID
water vs. air
Rotating Time, hours
36
Old style water course bits
37
EFFECT OF SOLIDS IN THE MUD
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Hydrostatic Pressure Gradient
Fresh Water Pressure Gradient = 0.433 psi/ft
Density 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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Hydrostatic Pressure
with 14 lb/gal mud:
p = GM * Depth
 Mud 

 * 0.433 psi/ft * Depth
 8.33 
Pressure  0.052 * Mud Weight * Depth
= 0.052 *14.0 *12,000
= 8,736 psig
(5,196 psi with water)
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Hydrostatic Pressure Required
What mud weight is required to
balance a pore pressure of 10,000 psig
at a vertical depth of 12,000 ft?
Pressure
Required Mud Weight 
0.052 * Depth
10,000
Required Mud Weight 
0.052 *12,000
MW = 16.03 lb/gal
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