PETE 411
Well Drilling
Lesson 21
Prediction of
Abnormal Pore Pressure
1
Prediction of
Abnormal Pore Pressure
 Resistivity of Shale
 Temperature in the Return Mud
 Drilling Rate Increase
 dc - Exponent
 Sonic Travel Time
 Conductivity of Shale
2
Read:
Applied Drilling Engineering, Ch. 6
HW #11
Slip Velocity
Due 10-28-02
3
Shale Resistivity
vs. Depth
1. Establish trend
line in normally
pressured shale
2. Look for
deviations from
this trend line
(semi-log)
4
EXAMPLE
Shale Resistivity
vs. Depth
1. Establish normal
trend line
2. Look for
deviations
(semi-log)
5
Shale Resistivity
vs. Depth
Pore Pressure
(lb/gal equivalent)
16 14 12 10
1. Establish normal
trend line
2. Look for
deviations
9 ppg
(normal)
3. Use OVERLAY
to quantify
pore pressure
(use with caution)
6
Depth, ft
Shale Density , g/cc
7
Depth, ft
Mud Temperature in flowline, deg F
8
Example
8.2 X
Why?
9
Example
8.8 X
Thermal conductivity, heat capacity, pore pressure... 10
11
DP = (P2 - P1)1,000
Effect of Differential Pressure
12
Typical Drilling Rate Profiles - Shale
Shale
The drilling rate in a normally
pressured, solid shale
section will generally
generate a very steady and
smooth drilling rate curve.
The penetration rate will be
steady and not erratic
(normally pressured, clean
shale).
13
Typical Drilling Rate Profiles - Sand
Sand
The drilling rate in a sand will
probably generate an erratic
drilling rate curve.
Sands in the Gulf Coast area
are generally very
unconsolidated. This may
cause sloughing, accompanied
by erratic torque, and
temporarily, erratic drilling
rates.
14
Typical Drilling Rate Profiles - Shaley Sands
Shaley Sands
This is generally the most
troublesome type drilling rate curve
to interpret.
Many times this curve will look
similar to a solid shale curve that
is moving into a transition zone.
Note: This is a prime example why you should not base
your decision on only one drilling parameter, even
though the drilling rate parameter is one of the better
parameters.
15
Typical Drilling Rate Profiles
Transition Zone
Shale
If you are drilling close to
balanced, there will probably
be a very smooth, (gradual)
increase in the drilling rate.
This is due to the difference
between the hydrostatic
head and the pore pressure
becoming smaller.
Dp
16
Typical Drilling Rate Profiles
Transition Zone
Shale
As the pressure Dp becomes
very small, the gas in the pores
has a tendency to expand which
causes the shale particles to pop
from the wall. This is called
sloughing shale.
The transition zone generally
has a higher porosity, making
drilling rates higher. In a clean
shale the ROP will increase in a
smooth manner.
17
Typical Drilling Rate Profiles
Note:
 If you are drilling overbalanced in a transition it
will be very difficult to pick up the
transition zone initially.
 This will allow you to move well into the
transition zone before detecting the problem.
18
Typical Drilling Rate Profiles
 This could cause you to move into a permeable
zone which would probably result in a kick.
 The conditions you create with overbalanced
hydrostatic head will so disguise the pending
danger that you may not notice the small
effect of the drilling rate curve change. This
will allow you to move well into that transition
zone without realizing it.
19
Determination of Abnormal Pore
Pressure Using the dc - exponent
From Ben Eaton:
P S  S  P   d c
      
D D  D  D  n  d cn
1.2



20
P S  S  P   d c
      
D D  D  D  n  d cn
Where
P
D
1.2



 formation pressure gradient, psi/ft
P
   normal water gradient in area
 D n
e.g., 0.433 or 0.465, psi/ft
S
D
 overburden stress gradient, psi/ft
dc
 actual d c  expon ent from plot
d cn
 d c  exp onent from the normal trend
21
Example
Calculate the pore
pressure at depth X using
the data in this graph.
Assume:
West Texas location with
normal overburden of
1.0 psi/ft.
X = 12,000 ft.
X
1.2 1.5
dc
22
Example
From Ben Eaton:
P
S  S  P   d c

     
D D  D  D  n  d cn
1.2



1.2
 1.2 
 1.0  [1.0  0.433]

 1.5 
P
 0.5662 psi/ft
D
23
Example
 P  0.5662 x 12,000  6794 psi
6794
EMW 
 10.9 lbm/gal
0.052 x 12,000
24
E.S. Pennebaker
 Used seismic field data for the
detection of abnormal pressures.
 Under normally pressured conditions the
sonic velocity increases with depth.
(i.e. Travel time decreases with depth)
(why?)
25
E.S. Pennebaker
 Any departure from this trend is an
indication of possible abnormal
pressures.
 Pennebaker used overlays to estimate
abnormal pore pressures from the
difference between normal and actual
travel times.
26
Interval Travel Time, msec per ft
27
Ben Eaton
also found a way to determine pore pressure
from interval travel times.
Example:
In a Gulf Coast well, the speed of sound is 10,000
ft/sec at a depth of 13,500 ft. The normal speed
of sound at this depth, based on extrapolated
trends, would be 12,000 ft/sec. What is the pore
pressure at this depth?
Assume: S/D = 1.0 psi/ft
28
Ben Eaton
From Ben Eaton,
P S  S  P   Dt n 
      

D D  D  D  n  Dt 
3.0
 10,000 
 1.0 - [1.0 - 0.465]

 12,000 
 0.6904 psi/ft
3
( Dt a 1/v )
29
Ben Eaton
From Ben Eaton
r = (0.6904 / 0.052) = 13.28 lb/gal
p = 0.6904 * 13,500 = 9,320 psig
Note: Exponent is 3.0 this time,
NOT 1.2!
30
Equations for Pore Pressure Determination
dC
 R 
log 
 60 N 

rNORMAL

 *


 r
 12 W 
 ACTUAL
log 
 10 6 D 

B 

S
P
S
P






D
D
 D n
D








 dc calculated

 d normal
c

S
P
S
P





D
D
 D n
D




S
P
S
P





D
D
D
D

n

S
P
S
P





D
D
D
D

n

 R obs

 R
n









 Cn

C
o

 Dt n

 Dt
o

1.2




1. 2




1.2








3.0
31
Pore Pressure Determination
32
EXAMPLE 3 - An Application...
Mud Weight = 10 lb/gal. (0.52 psi/ft)
Surface csg. Set at 2,500 ft.
Fracture gradient below surf. Csg = 0.73 psi/ft
Drilling at 10,000 ft in pressure transition zone
* Mud weight may be less than pore pressure!
DETERMINE Maximum safe underbalance
between mud weight and pore pressure if well
kicks from formation at 10,000 ft.
33
0.73 – 0.52 = 0.21 (psi/ft)
Casing Seat
Depth, ft
2,500
FractureGradient
= 0.73 psi/ft
Mud Wt. Grad
= 0.52 psi/ft
10,000
5,200
Pressure, psi
34
Example 3 - Solution
The danger here is fracturing the formation near
the casing seat at 2,500 ft.
The fracture gradient at this depth is 0.73 psi/ft,
and the mud weight gradient is 0.52 psi/ft.
So, the additional permissible pressure gradient
is 0.73 – 0.52 = 0.21 psi/ft, at the casing seat.
This corresponds to an additional pressure of
DP = 0.21 psi/ft * 2,500 ft = 525 psi
35
Example 3 – Solution – cont’d
This additional pressure, at 10,000 ft, is also
525 psi, and would amount to an additional
pressure gradient of:
525 psi / 10,000 ft = 0.0525 psi/ft
This represents an equivalent mud weight of
0.0525 / 0.052 = 1.01 lb/gal
This is the kick tolerance for a small kick!
36
Problem #3 - Alternate Solution
When a well kicks, the well is shut in
and the wellbore pressure increases
until the new BHP equals the new
formation pressure.
At that point influx of formation fluids
into the wellbore ceases.
 Since the mud gradient in the wellbore
has not changed, the pressure
increases uniformly everywhere.
37
Depth, ft
525
Casing Seat at 2,500 ft
After Kick and
Stabilization
Before Kick
Kick at 10,000 ft
DP
525
Wellbore Pressure, psi
38
At 2,500 ft
Initial mud pressure = 0.52 psi/ft * 2,500 ft = 1,300 psi
Fracture pressure = 0.73 psi/ft * 2,500 ft = 1,825 psi
Maximum allowable increase in pressure = 525 psi
At 10,000 ft
Maximum allowable increase in pressure = 525 psi
(since the pressure increases uniformly everywhere).
This corresponds to an increase in mud weight of
525 / (0.052 * 10,000) = 1.01 lb/gal
= maximum increase in EMW
= kick tolarance for a small kick size.
39
1,825 psi
Depth, ft
Casing Seat at 2,500 ft
1,300 psi
5,200 psi
5,725 psi
Kick at 10,000 ft
DP
Wellbore Pressure, psi
40
After Large Kick and Stabilization
Depth, ft
Casing Seat at 2,500 ft
After Small Kick
and Stabilization
Before Kick
Kick at 10,000 ft
DP
Wellbore Pressure, psi
41