Petroleum Engineering 405 Drilling Engineering
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Transcript Petroleum Engineering 405 Drilling Engineering
PETE 411
Well Drilling
Lesson 20
Abnormal Pressure
1
Abnormal Pressure
Normal Pore Pressures
Abnormal Pore Pressure Gradients
Fracture Gradients
Mud Weights
Casing Seat Depths
What Causes Abnormal Pressure?
Detection of Abnormal Pressure
Quantification of Abnormal Pressure
2
Read:
Applied Drilling Engineering, Ch. 6
HW #11
Slip Velocity
Due 10-28-02
3
Depth, ft
Normal and Abnormal Pore Pressure
10,000’
Normal Pressure Gradients
West Texas: 0.433 psi/ft
Gulf Coast: 0.465 psi/ft
Abnormal
Pressure
Gradients
??
Pore Pressure, psig
4
Pore Pressure vs. Depth
Depth, ft
0
0.433 psi/ft
8.33 lb/gal
0.465 psi/ft
9.00 lb/gal
Normal
Abormal
5,000
10,000
15,000
5
10
15
20
Pore Pressure, lb/gal equivalent
Density of mud required to control this pore pressure
5
Fracture Gradient
Pore Pressure
Gradient
6
* Pore
Pressure
Gradients
* Fracture
Gradients
•Casing
Setting
Depths
7
Some Causes of Abnormal Pressure
1. Incomplete compaction of sediments
Fluids in sediments have not
escaped and help support the
overburden.
2. Tectonic movements
Uplift
Faulting
8
Some Causes of Abnormal Pressure
3. Aquifers in Mountainous Regions
Aquifer recharge is at higher
elevation than drilling rig location.
4. Charged shallow reservoirs due to
nearby underground blowout.
5. Large structures...
9
HIGH PRESSURE
NORMAL PRESSURE
Thick, impermeable layers of shale (or salt) restrict the movement
of water. Below such layers abnormal pressure may be found.
10
HIGH PRESSURE
NORMAL PRESSURE
Hydrostatic pressure gradient is lower in gas or oil than in water.
11
When crossing faults it is possible to go from normal
pressure to abnormally high pressure in a short interval.
12
Well “A” found only Normal Pressure ...
13
sob
p
sz
sOB = p + sZ
14
?
15
Abnormal Pressure
cont’d
Detection of Abnormal Pore Pressures
Prediction of Abnormal Pore Pressures
D-Exponent
DC-Exponent
Example
Importance of Shale Density
16
Indications of Abnormal Pore Pressures
Methods:
1. Seismic data
2. Drilling rate
3. Sloughing shale
4. Gas units in mud
5. Shale density
6. Chloride content
17
Indications of Abnormal Pore Pressures
Methods, cont’d:
7. Change in Mud properties
8. Temperature of Mud Returns
9. Bentonite content in shale
10. Paleo information
11. Wire-line logs
12. MWD-LWD
18
Prediction and Detection of Abnormal
Pressure Zones
1. Before drilling
Shallow seismic surveys
Deep seismic surveys
Comparison with nearby wells
19
Prediction and Detection of Abnormal
Pressure Zones
2. While drilling
Drilling rate, gas in mud, etc. etc.
D - Exponent
DC - Exponent
MWD - LWD
Density of shale (cuttings)
20
Prediction and Detection of Abnormal
Pressure Zones
3. After drilling
Resistivity log
Conductivity log
Sonic log
Density log
21
–0.000085
DS
.
.
0.41 e
22
23
What is dexponent?
Decreasing ROP
24
D - Exponent
The
drilling rate
equation:
W
R K N
DB
E
D
Where
R = drilling rate, ft/hr
K = drillability constant
N = rotary speed, RPM
E = rotary speed expon.
W = bit weight, lbs
DB = bit diameter, in
D = bit wt. Exponent
or D - exponent
25
D - Exponent
W
R K N
DB
D
E
If we assume that K = 1
and E = 1
Then
R W
N DB
D
R
log
N
D
W
log
DB
26
D - Exponent
A modified version of this equation
follows:
R
log
60 N
d
12 W
log 6
10 DB
27
Example
Calculate the value of the d - exponent if
the drilling rate is 35 ft/hr, the rotary RPM is
100, and the weight on the 12 1/4” bit is
60,000 lbs.
R
log
60 N
d
12 W
log 6
10 DB
35
log
60 * 100
2.2341
12 * 60,000 1.2308
log
6
10 12.25
d = 1.82
28
Example
What happens to d if R doubles to 70 ft/hr?
70
log
60 * 100
1.9331
d
1.57
12 * 60,000 1.2308
log
6
10 12.25
Note that an increase in R resulted in a decrease in d.
Doubling R decreased d from 1.82 to 1.57
29
Example
d may be Corrected for density as
follows
mud weight for normal gradient (ppg)
dc d
actual mud weight in use(ppg)
9
9
e.g., dc d 1.82 * 1.37
12
12
30
Example 2
Calculate “d” if:
R = 20 ft/hr
N = 100 RPM
W = 25,000 lbf
DB = 9 7/8 in
R
log
60 N
d
12 W
log 6
10 DB
20
log
60 * 100
12 * 25,000
log 6
10 * 9.875
d = 1.63
31
Example 2
If the normal pore pressure gradient in the
area is 0.433 psi/ft, and the actual mud
weight is 11.2 #/gal, what is “dc”?
normal gradient (ppg)
8.33
1.63 *
dc d
11.2
actual mud weight (ppg)
dc = 1.21
32
33
Procedure for Determining Pore
Pressure From dc - Exponent
Calculate dc over 10-30 ft intervals
Plot dc vs depth (use only date from
Clean shale sections)
Determine the normal line for the
dc vs. depth plot.
Establish where dc deviates from the
normal line to determine abnormal
pressure zone
34
Procedure for Determining Pore
Pressure From dc - Exponent
Depth
Normal
Abnormal
dc - Exponent
35
Procedure for Determining Pore
Pressure From dc - Exponent
If possible, quantify the magnitude of the
abnormal pore pressure using
overlays, or Ben Eaton’s Method
P S S P dc calculated
D D D D n dc normal
Pore
Pressure
Grad.
Overburden
Stress Grad.
1 .2
Normal Pore
Pressure Grad.
36
In normally pressured
shales, shale
compaction increases
with depth
37
Shale Density - Mud Cup Method
1. Fill mud cup with shale until the weight is 8.33.
2. Fill to top with water, and record the reading Wtot.
8.33
Spec.Gravity
16.66 Wtot
Note: Dry sample carefully with towel.
Do not apply heat.
38
Alternate Method: Use variable density column.
See p. 270 in text
39
Pore Pressure from
Resistivity
Shale resistivity plots
may be developed
from (i) logs or
(ii) cuttings
What is the pore
pressure at the point
indicated on the plot?
[Assume Gulf Coast].
Depth=10,000 ft
10,000’
0.2
0.5
1
2 3
40
From plot,
EATON
Rn = 1.55 ohms
Robs = 0.80 ohms
P S S P R obs
D D D D n Rn
Depth
From Eaton:
1.2
1.2
P
0.80
0.95 0.95 0.465
D
1.55
10,000’
= 0.7307 psi/ft = 14.05 lb/gal
P = 0.7307 * 10,000 = 7,307 psi
0.2
0.5
1
2 3
41