1

PETE 411

Well Drilling

Lesson 20

Abnormal Pressure

2

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

3

HW #11 Slip Velocity

Due 10-28-02

Read:

Applied Drilling Engineering, Ch. 6

4

Abnormal Pressure Gradients

Normal Pressure Gradients West Texas: 0.433 psi/ft Gulf Coast: 0.465 psi/ft

Normal and Abnormal Pore Pressure

Pore Pressure, psig

Dep

th,

ft

10,000’ ? ?

5

Pore Pressure vs. Depth0

5,000

10,000

15,000

5 10 15 20Pore Pressure, lb/gal equivalent

De

pth

, f

t

Normal Abormal

Density of mud required to control this pore pressure

0.433 psi/ft 8.33 lb/gal

0.465 psi/ft 9.00 lb/gal

6

Pore Pressure Gradient

Fracture Gradient

7

* Pore

Pressure Gradients

* Fracture Gradients

•Casing Setting Depths

8

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

9

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...

10

HIGH PRESSURE

NORMAL PRESSURE

Thick, impermeable layers of shale (or salt) restrict the movement of water. Below such layers abnormal pressure may be found.

11

HIGH PRESSURE

NORMAL PRESSURE

Hydrostatic pressure gradient is lower in gas or oil than in water.

12

When crossing faults it is possible to go from normal pressure to abnormally high pressure in a short interval.

13

Well “A” found only Normal Pressure ...

14

OB = p + Z

ob

pz

15

?

16

Abnormal Pressure cont’d

Detection of Abnormal Pore Pressures

Prediction of Abnormal Pore Pressures

D-Exponent

DC-Exponent

Example

Importance of Shale Density

17

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

18

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

19

Prediction and Detection of Abnormal Pressure Zones

1. Before drilling Shallow seismic surveys Deep seismic surveys Comparison with nearby wells

20

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)

21

Prediction and Detection of Abnormal Pressure Zones

3. After drilling Resistivity log Conductivity log Sonic log Density log

22

. – . SD000085.0e41.0

23

24

Decreasing ROP

What is d-exponent?

25

D - Exponent

The

drilling rate

equation:

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

D

B

E

D

WNKR

26

D - Exponent

If we assume that K = 1

and E = 1

Then

D

B

E

D

WNKR

BDW

log

NR

logD

D

BD

W

N

R

27

D - Exponent

A modified version of this equation follows:

B6 D10W12

log

N60R

log

d

28

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.

B6 D10W12

log

N60R

log

d2308.1

2341.2

25.1210000,60*12

log

100*6035

log

6

d = 1.82

29

Example

What happens to d if R doubles to 70 ft/hr?

Note that an increase in R resulted in a decrease in d.

Doubling R decreased d from 1.82 to 1.57

57.12308.1

9331.1

25.1210000,60*12

log

100*6070

log

d

6

30

Example

d may be Corrected for density as follows

)ppg(useinweightmudactual

)ppg(gradientnormalforweightmudddc

37.112

9*82.1

12

9dd.,g.e c

31

Example 2

Calculate “d” if: R = 20 ft/hr

N = 100 RPM

W = 25,000 lbf

DB = 9 7/8 in

875.9*10000,25*12

log

100*6020

log

6

B6 D10W12

log

N60R

log

d

d = 1.63

32

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”?

2.11

33.8*63.1

)ppg(weightmudactual

)ppg(gradientnormalddc

dc = 1.21

33

34

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

35

Procedure for Determining Pore Pressure From dc - Exponent

dc - Exponent

Dep

th Normal

AbnormalN

ormal Trend

36

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

Pore Pressure Grad.

Overburden Stress Grad.

Normal Pore Pressure Grad.

2.1

c

c

n normald

calculatedd

D

P

D

S

D

S

D

P

37

In normally pressured shales, shale

compaction increases with depth

38

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.

Note: Dry sample carefully with towel. Do not apply heat.

totW66.16

33.8Gravity.Spec

39

Alternate Method: Use variable density column.

See p. 270 in text

40

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

0.2 0.5 1 2 3

10,000’

Dep

th

41

From plot, Rn = 1.55 ohms

Robs = 0.80 ohms

From Eaton:

EATON

2.1

n

obs

n R

R

D

P

D

S

D

S

D

P

2.1

55.1

80.0465.095.095.0

D

P

= 0.7307 psi/ft = 14.05 lb/gal

P = 0.7307 * 10,000 = 7,307 psi 0.2 0.5 1 2 3

10,000’

Dep

th