5. Pore Pressure 4v2456

PETE 625 Well Control Lesson 5 Pore Pressure

Contents  Normal Pore Pressure  Subnormal Pore Pressure  Abnormal Pore Pressure  Origins of Pore Pressure  Origins of Pore Pressure  Origins of Abnormal Pore Pressure  Bulk Density and Porosity vs. Depth 2

Assignments Homework # 3: Ch 2, Problems 1 - 10 due Wednesday, Sept 22, 2004 Read:

Chapter 2 to p. 60

Depth, ft

Normal and Abnormal Pore Pressures

10,000

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

Subnormal

Abnormal Pressure Gradients

??

Pore Pressure, psig

4

Pore Pressure vs. Depth

Depth, ft

0

5,000

10,000

0.433 psi/ft

8.33 lb/gal

0.465 psi/ft

9.00 lb/gal

Normal

Abormal

15,000 5

10

15

20

Pore Pressure, lb/gal equivalent

Density of mud required to control this pore pressure 5

Lost Returns

Kicks 6

7

Pore Pressure  = formation pressure  = formation fluid pressure  = pressure in fluid contained in the pore spaces of the rock

8

Pore Pressure Normal pressure gradients correspond to the hydrostatic gradient of a fresh or saline water column Example 2.1. Determine the pore pressure of a normally pressured formation in the Gulf of Mexico at 9,000’ depth. 9

Pore Pressure TABLE 2.1 -

pn = gnD = 0.465 psi/ft * 9,000 ft pn = 4,185 psig

10

Normal Pressure

11

Subnormal Pressures Formation pressure gradients less than normal gradients for a given area. Lost circulation problems and differential sticking are common problems in these areas 12

Subnormal pressures due to faulting

8,000’ 9,000’

13

Aquifer outcrops below rig

14

Production of oil or gas

15

Abnormal Pressures Abnormal Pressures are formation pressures greater than normal pressures Can cause severe drilling problems There are many possible causes of abnormal pressure 16

Abnormal Pressure All abnormal pressures require some means of sealing or trapping the pressure within the rock body. Otherwise hydrostatic equilibrium back to a normal gradient would eventually be restored.

17

Abnormal Pressure Massive shales provide good pressure seals, but shales do have some permeability, so, given sufficient time, normal pressures will eventually be established. It may take tens of millions of years for a normal pressure gradient to re-occur. 18

Pressure Seals

19

Abnormal pressures Dense rocks should always be a warning to a driller that the pore pressure may be changing Many abnormal pore pressure processes are simply the reverse of those which effect subnormal pressures

20

Abnormal pressures For example, the converse to a low piezometric water level is abnormal pressure resulting from an Artesian source. A thick gas sand that is normally pressured at the bottom of the sand will be abnormally pressured at the top of the sand. 21

Pore pressures do not always increase with depth

22

Causes of abnormal pressure TABLE 2.2 -

23

Aquifer

24

Thick gas sand 2 P = 605 - 0.05 * 300 = 605 - 15 = 590 psig

1 p = 0.465 * 1,300 = 605 psig

3 g = 590/1,000 = 0.590 psi/ft EMW = 0.590/0.052 11.3 ppg

25

Normal Faulting

9,000 ft

4,650 0.052 * 9,000 = 9.94 ppg

10,000 ft

0.465

psi * 10,000 ft ft = 4,650 psi 26

Downfaulting Top of Transition Zone

Pressure may increase

27

Salt Diapirs Salt diapirs plastically “flow” or extrude into the previously deposited sediment layers. The resulting compression can result in overpressure.

28

Salt formations Normally pressured

Salt Pressure at the bottom of the salt is often extremely overpressured 29

Erosion

EMW =

p 0.052 * Depth

30

Caprock Mineral Deposition

Possible precipitation of carbonate and silica minerals 31

Man-Made Abnormal Pressures Underground blowout

Casing leaks

Faulty cement job

32

Compaction Theory of Abnormal Pressure Best fits most naturally occurring abnormal pressures In new areas, geologic and geophysical interpretations along with analogy to known areas are always important 33

Compaction Theory During deposition, sediments are compacted by the overburden load and are subjected to greater temperatures with increasing burial depth. Porosity is reduced as water is forced out. 34

Compaction Theory Hydrostatic equilibrium within the compacted layers is retained as long as the expelled water is free to escape If water cannot escape, abnormal pressures occur

35

Compaction Theory Undercompacted Shales

Water is expelled from the shales

Pore water expelled because of increasing overburden

If the expelled water is not free to escape, abnormal pressures may result. Sufficient compaction cannot occur so the pore fluids carry more of the overburden 36

The overburden load is ed by the vertical stress in the grain framework and by the fluid pore pressure

Compaction Theory

σ

ob

=σ

σ

ob

= overburden stress

σ

eV

= matrix stress

eV

+ pp

pp = pore pressure

37

Compaction Theory The average porosity in sediments, generally decreases with increasing depth - due to the increasing overburden This results in an increasing bulk density with increasing depth, and increasing rock strength

38

Compaction Theory From a porosity log, we can construct a plot of bulk density vs. depth From this (or directly from a density log, we can calculate overburden stress vs. depth.

39

Compaction Theory TABLE 2.4 -

40

Bulk Densities - Santa Barbara Channel φ = 0.37e

−0.0001609 D

− Kφ D

φ = φ0 e ρ = f (φ )

41

GOM Bulk Densities

42

Pore Pressure Prediction Overburden Pressure vs. Depth Porosity vs. Depth Pore Pressure Prediction By Analogy By Seismic Methods From Drilling Rate Changes

Factors that Affect Drilling Rates 43

Overburden Stress σ ob = ∫ ρb gdD D

σ ob = 0.052 ∫ [ ρma (1 −φ ) + ρ f φ ]dD 0

setting setting

φ = φ0 e

−kφ D

andintegrating int egrating and

σ ob

 ( ρma − ρ f )φ0 −kφ D = 0.052 ρma D − 1−e kφ 

(

)

  

44

Example 2.5 Calculate the overburden stress at a depth of 7,200 ft in the Santa Barbara Channel. Compare to Eaton’s prediction. Assume φ o = 0.37 ρ

ma

kφ ρ

= 2.6 gm/cc

= 0.0001609 ft-1 f

= 1.044 gm/cc 45

Solution

σob

 ( ρ − ρf ) φ0 1 − e −k φD = 0.052 ρmaD − ma kφ 

(

)

  

( 2.6 − 1.044 ) 8.33 * 0.37 ( − 0.0001609*7,200 )   σ ob = 0.052 2.6 * 8.33 * 7,200 − * 1− e  0.0001609   σ ob = 7,032 psig

Eaton’s Fig. 2.21 shows a value of : gob = 0.995 psi/ft So, (σ

)

ob eaton

= 0.995 * 7,200 = 7,164 psig { Difference = 132 psi or 1.9% }

46

Overburden stress depends upon porosity, and porosity depends on overburden stress Shales are more compactible than sandstones. Young shales are more compactible than older shales. Limestones and dolomites are only slightly compactible. 47

Rule of Thumb A common assumption for sedimentary deposits is gob = 1.0 psi/ft This is not a good assumption in young sediments Eaton predicts that an overburden stress gradient of 1 psi/ft be achieved at a depth of 20,000 ft in the GOM Eaton predicts that an overburden stress gradient of 1 psi/ft be achieved at a depth of 7,400 ft in the Santa Barbara Channel 48

0.84 psi/ft

Eaton’s ob stress gradient for GOM

1 psi/ ft at 20,000’

0.89 psi/ft

Eaton’s ob stress gradient for Santa Barbara Channel 1 psi/ ft at 7,400’

49

Shale porosity depends not only on depth e.g. At 6,000’ depth φ varies from 3% to 18%

Note the ~ straight line relationship on semilog paper

50

Eaton’s porosities from the Santa Barbara Channel. The straight line is a plot of the equation: φ = 0.37e-0.0001609D At D = 0, φ = 0.37 At D = 10,000 ft φ = 0.074 51

52