7a. Pore Pressure Prediction.ppt 2b424n

TAMU - Pemex

Well Control Lesson 7 Pore Pressure Prediction

Contents  Porosity  Shale Compaction  Equivalent Depth Method  Ratio Method  Drilling Rate  dC-Exponent  Moore’s Technique  Comb’s Method

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Pore pressure prediction methods Most pore pressure prediction techniques rely on measured or inferred porosity. The shale compaction theory is the basis for these predictions. 3

Pore pressure prediction methods Measure the porosity indicator (e.g. density) in normally pressured, clean shales to establish a normal trend line. When the indicator suggests porosity values that are higher than the trend, then abnormal pressures are suspected to be present. The magnitude of the deviation from the normal trend line is used to quantify the abnormal pressure. 4

Porosity should decrease with depth in normally pressured shales

1. Establish “Normal” Trend Line in good “clean” shale

Transitio n 2. Extrapolate normal trend line

3. Determine the magnitude of the deviation 5

Older shales have had more time to compact, so porosities would tend to be lower (at a particular depth). Use the trend line closest to the transition. Lines may or may not be parallel.

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Equivalent Depth Method

De

The normally compacted shale at depth De has the same compaction as the abnormally pressured shale at D. Thus, 

V

= Ve

i.e., ob - pp = obe - pne pp = pne + (ob - obe) D 

ob

=  V + pp 7

Example 2.6 Estimate the pore pressure at 10,200’ if the equivalent depth is 9,100’. The normal pore pressure gradient is 0.433 psi/ft. The overburden gradient is 1.0 psi/ft. At 9,100’, pne = 0.433 * 9,100 = 3,940 psig At 9,100’, obe = 1.00 * 9,100 = 9,100 psig At 10,200’, ob = 1.00*10,200 = 10,200 psig

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Solution  pp = pne + (ob - obe) ……………. (2.13) = 3,940 + (10,200 – 9,100) pp = 5,040 psig  The pressure gradient, gp = 5,040/10,200 = 0.494 psi/ft  EMW = 0.494/0.052 = 9.5 ppg 9

The Ratio Method uses (Xo/Xn) to predict the magnitude of the abnormal pressure We can use: Depth

• drilling rate • resistivities • conductivities Xn

Xo

• sonic speeds

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Shale Porosity Indicator

Pore pressures can be predicted:  Before drilling (planning)  During drilling.  After drilling

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Before drilling the well (planning)  Information from nearby wells  Analogy to known characteristics of the geologic basin  Seismic data

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Table 2.6 – Cont’d

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Seismic Surveys, as used in conventional geophysical prospecting, can yield much information about underground structures, and depths to those structures. Faults, diapirs, etc. may indicate possible locations of abnormal pressures 15

Typical Seismic Section

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Under normal compaction, density increases with depth. For this reason the interval velocity also increases with depth, so travel time decreases t = tma(1-) + tf 

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Sound moves faster in more dense medium In air at sea level, Vsound = 1,100 ft/sec In distilled water, Vsound = 4,600 ft/sec In low density, high porosity rocks, Vsound = 6,000

ft/sec

In dense dolomites, Vsound = 20,000 ft/sec

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Example 2.7 Use the data in Table 2.7 to determine the top of the transition zone, and estimate the pore pressure at 19,000’ using the equivalent depth method using Pennebaker’s empirical correlation

Ignore the data between 9,000’ and 11,000’. Assume Eaton’s Gulf Coast overburden gradient. 19

Solution Plot interval travel time vs. depth on semilog paper (Fig. 2.31) Plot normal trend line using the 9,000 data.

6,000-

From Fig. 2.20, at 19,000’, gob = 0.995 psi/ft (ob)19,000 = 0.995 * 19,000 = 18,905 psig 20

Equivalent Depth Method: Use Ignore

From the vertical line, De = 2,000’ obe = 0.875 * 2,000 =1,750 (Fig. 2.20) But, pne = 0.465 * 2,000 = 930 psig

tn

to

pp = 930 + (18,905-1,750) pp = 18,085 psig

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Fig. 2.30

Pennebaker’s correlation for Gulf Coast sediments Higher travel time means more porosity and higher pore pressure gradient Example 2.7 (Table 2.7) to = 95 sec/ft @ 19,000’ tn = 65 sec/ft @ 19,000’ to/ tn = 95/65 = 1.46

0.95

pp = 0.95 * 19,000 = 18,050 psig 22

Comparison Pore Pressure at a depth of 19,000 ft:  Pennebaker: 18,050 psi or 0.950 psi/ft or 18.3 ppg  Equivalent Depth Method: 18,085 psi or 0.952 psi/ft or 18.3 ppg 23

While Drilling  dc-exponent  MWD & LWD  Kicks  Other drilling rate factors (Table 2.5)

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TABLE 2.5 -

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Penetration rate and abnormal pressure Bits drill through overpressured rock faster than through normally pressured rock (if everything else remains the same). When drilling in clean shales this fact can be utilized to detect the presence of abnormal pressure, and even to estimate the magnitude of the overpressure. 26

TABLE 2.8 -

Note, that many factors can influence the drilling rate, and some of these factors are outside the control of the operator. 27

Effect of bit weight and hydraulics on penetration rate

Inadequate hydraulics or excessive imbedding of the bit teeth in the rock

Drilling rate increases more or less linearly with increasing bit weight. A significant deviation from this trend may be caused by poor bottom hole cleaning

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Effect of Differential Pressure on Drilling Rate

Decrease can be due to: • The chip hold down effect

Differential pressure is the difference between wellbore pressure and pore fluid pressure

• The effect of wellbore pressure on rock strength

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Drilling underbalanced can further increase the drilling rate.

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The chip hold-down effect The mud pressure acting on the bottom of the hole tends to hold the rock chips in place

Important hold-down parameters: Overbalance

Drilling fluid filtration rate

Permeability

Method of breaking rock (shear or crushing) 31

TABLE 2.9 -

• Drilling rates are influenced by rock strengths. • Only drilling rates in relatively clean shales are useful for predicting abnormal pore pressures.

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ob is generally the maximum in situ principal stress in undisturbed rock

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Stresses on Subsurface Rocks ob, H1, H2 and p all tend to increase with depth ob is in general the maximum in situ principal stress. Since the confining stresses H1 and H2 increase with depth, rock strength increases. 34

Stresses on Subsurface Rocks The pore pressure, p, cannot produce shear in the rock, and cannot deform the rock. Mohr-Coulomb behavior is controlled by the the effective stresses (matrix). When drilling occurs the stresses change. ob is replaced by dynamic drilling fluid pressure. 35

The degree of overbalance now controls the strength of the rock ahead of the bit.

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Rock failure caused by roller cone bit. The differential pressure from above provides the normal stress, o

Formation fracture is resisted by the shear stress, o, which is a function of the rock cohesion and the friction between the plates. This friction depends on o.

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Fig. 2.41 - Differential Pressure 0.1 in below the bit. Vertical Stress Horizontal Stress Pore Pressure Wellbore Pressure

= 10,000 psi = 7,000 psi = 4,700 psi = 4,700 psi

(Induced Differential Pressure in Impermeable rock. FEM Study)

When ob is replaced by phyd (lower) the rock immediately below the bit will undergo an increase in pore volume, associated with a reduction in pore pressure. In sandstone this pressure is increased by fluid loss from the mud. 38

Drilling Rate as a Pore Pressure Predictor Penetration rate depends on a number of different parameters. R = K(P1)a1 (P2)a2 (P3)a3… (Pn)an A modified version of this equation is:

 W  R  K 3 N   db 

d

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Drilling Rate as a Pore Pressure Predictor Or, in its most used form:

 

d    

R log 60 N 12W log 6 10 d b

 W  R  K 3 N   db 

     

R  ft/hr N  rpm d  d  exponent W  Bit Weight , lbf d b  Bit Diameter, in 40

d

d-exponent The d-exponent normalizes R for any variations in W, db and N Under normal compaction, R should decrease with depth. This would cause d to increase with depth. Any deviation from the trend could be caused by abnormal pressure. 41

d-exponent Mud weight also affects R….. An adjustment to d may be made:

dc = d (n /c) where dc = exponent corrected for mud density n = normal pore pressure gradient c = effective mud density in use 42

Example  While drilling in a Gulf Coast shale, R = 50 ft/hr W = 20,000 lbf N = 100 RPM ECD = 10.1 ppg

(Equivalent Circulating Density)

db = 8.5 in  Calculate d and dc 43

Solution 

50  log    2.079 60 * 100   d   1.554  12 * 20,000  log   6  10 * 8.5  d  1.34 

0.465  dc  1.34    0.052 * 10.1  dc  1.19



 log R 60N d  12 W log  10 6 d b 

     

 n   d c  d   c 

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Example 2.9  Predict pore pressure at 6,050 ft (ppg): from data in Table 2.10 using:  Rhem and McClendon’s correlation  Zamora’s correlation  The equivalent depth method

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TABLE 2.10 d-EXPONENT AND MUD DENSITY DATA FOR A WELL LOCATED OFFSHORE LOUISIANA

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Step 1 is to plot the data on Cartesian paper (Fig. 2.43). Transition at 4,700 ft? …or is it a fault? Seismic data and geological indicators suggest a possible transition at 5,700 ft.

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Fig. 2.43 Slope of 0.000038 ft-1

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Rehm and McClendon  gp = 0.398 log (dcn-dco) + 0.86 = 0.398 log (1.18 - 0.95) + 0.86 gp = 0.606 psi/ft

p = 0.606 / 0.052 = 11.7 ppg 49

Zamora From Fig. 2.44  gp = gn (dcn/dco)

= 0.465 * (1.18/.95) gp = 0.578 psi/ft

0.95

1.18

 p = 0.578/0.052

 p = 11.1 ppg 50

Equivalent Depth Method  From Fig. 2.20, at 6,050 ft, gob = 0.915 psi/ft  ob = 0.915 * 6,050 = 5,536 psi

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Equivalent Depth Method  From Fig. 2.43, Equivalent Depth = 750 ft  At 750 ft,



= 0.86 * 750 = 645 psi

obe

pne = 0.465 * 750 = 349 psig 52

Equivalent Depth Method  From Eq. 2.13, at 6,050 ft  pp = pne + (ob - obe) pp = 349 + (5,536 - 645) = 5,240 psig p = 19.25 * (5,240 / 6,050) = 16.7 ppg Perhaps the equivalent depth method is not always suitable for pp prediction using dc !! 53

Overlays such as this can be handy, but be careful that the scale is correct for the graph paper being used; the slope is correct for normal trends; the correct overlay for the formation is utilized.

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To improve pore pressure predictions using variations in drilling rate: Try to keep bit weight and rpm relatively constant when making measurements Use downhole (MWD) bit weights when these are available. (Frictional drag in directional wells can cause large errors) Add geological interpretation when possible. MWD can help here also. 55

Improved pore pressure predictions Keep in mind that tooth wear can greatly influence penetration rates. Use common sense and engineering judgment. Use several techniques and compare results. 56

Moore’s Technique Fig. 2.45 Moore proposed a practical method for maintaining a pore-pressure overbalance while drilling into a transition.

Drilling parameters must be kept constant for this technique to work. 57

Comb’s Method  Combs attempted to improve on the use of drilling rate for pore pressure by correcting for:  hydraulics  differential pressure  bit wear

 in addition to W, db, and N 58

Comb’s Method 

W   R  R d   3,500 db 

aW

 N     200 

aN



q     96 db dn 

aq

f  pd  f  tN 

q = circulating rate dn = diameter of one bit nozzle f(pd) = function related to the differential pressure f(tN) = function related to bit wear aW = bit weight exponent = 1.0 for offshore Louisiana aN = rotating speed exponent = 0.6 for offshore Louisiana aq = flow rate exponent = 0.3 for offshore Louisiana

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Tooth wear factor Correction would depend upon bit type, rock hardness, and abrasiveness

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Differential pressure factor

Method is too complicated and too site specific.

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