5. Frac Gradient - Lecture notes 5 PDF

Preview

Summary

5. Frac Gradient - Lecture notes 5...

Description

Fracture Gradient

1

Fracture Gradient Determination • Hydraulic fracturing occurs when the wellbore pressure is high enough to split the grains of surrounding formation apart • Reasons to find fracture gradient:    

Avoid lost circulation Hydraulics, casing and mud programs Design hydraulic fracturing treatments Well control situations (avoid underground blowout)

Introduction • Need to determine the pressure required to create and propagate a fracture.  Fracture propagates perpendicular to minimum stress

• Minimum horizontal stress (fracture gradient) can be determined through:  empirical correlations  field tests • LOT – Leak-Off Test • XLOT – Extended Leak-Off Test

3

Wellbore pressure operational window (minimum and maximum mud weight) Compressive failure

Tensile failure

Stable well

Wellbore collapse (hole instability) kick

Lost circulation

12

Wellbore pressure operational window (Wellbore Stability and Pore Pressure Can Switch) Compressive failure

Tensile failure

Stable well

Wellbore collapse (hole instability) kick

Lost circulation

13

Wellbore pressure operational window (minimum and maximum mud weight)

14

Fracture Gradient Prediction Methods So what does a drilling engineer really care about? •

Answer: the wellbore pressure containment or the maximum wellbore pressure to not induce lost circulation

Commonly called “fracture gradient”, we can find this pressure several ways. The calculated approaches correlate the vertical stress to the horizontal stress to finally obtain a frequently so-called minimum horizontal stress and the “fracture gradient”.

15

Formation stress determines 2 major categories of behavior Instability: Formation compressive stress determines stability of the borehole. If the stress around the hole exceeds the strength of the rock, the rock fails and the hole enlarges. Lost Circulation: Formation compressive stress determines the wellbore opening pressure. If the stress holding the hole closed is less than the pressure trying to open it, it opens (lost returns).

Where Does Stress Come From? Or how an engineer can ruin anything, including s’mores, for his family

Overburden Gravity is the primary force acting on the rock. The vertical force from gravity causes the rock to try and expand. Because there is other rock in the way, it can’t, and the vertical load creates stress instead

Poison’s Ratio Determines How Rock Distributes Stress Rock attempts to expand horizontally due to its vertical load (overburden). Because it’s constrained by the adjacent rock the attempted expansion becomes stress. The amount of lateral stress that results from a given overburden load varies with the Poisson’s Ratio

Overburden (Gravity) Vertical Stress & Strain Poisson’s Ratio ( )=

L/L lateral L/L Vertical

Lateral Expansion K0 is the ratio of horizontal to vertical effective stress. If Poisson’s is 0.5, K0 is 1.0 and horizontal equals vertical. K0 = /(1- )

Conventions for Principle  y Stresses 

v > Hmax > 1 > 2 > 3

Hmin

v or 1

Hmax or 2

x

min or 3

z

If described as v , Hmax, Hmin , the v is vertical. These values are the total stress and include pore pressure If described as 1 , 2, 3 , the 1 may not be vertical (i.e. tectonics). It is simply whatever is largest, then 3 is the smallest and 2 is the intermediate value. It is also usually effective stress

Tectonics May Increase or Reduce the Original Stress There are very few formations that have no horizontal stress asymmetry at all (Hmin ≠ Hmax ). It’s only a matter of severity. Tectonic plate movement is the primary cause. Movement can increase (compressional movement) or decrease (extensional movement) stress in any given direction. So tectonics can either help us or hurt us, depending on whether we are concerned with lost returns or instability (more on that later) Other example causes include; salt domes, mud diapirs, river delta deposits, coastline deposits that are unsupported on the ocean side, differential depletion, and differential overburden

How Do We Determine the Principle Stresses? Vertical Stress (Sv) Integrate a density log to obtain the total vertical weight above the zone of interest. Formation density does not change much within an area, so offset logs provide a close approximation. On land wells it is usually around 1 psi/ft TVD Nominal Horizontal Stress Calculate the nominal theoretical stress ( H) from the overburden ( v), a reasonable Poisson’s ratio, and a known pore pressure (PP) using the Soil Mechanics Equation:

H 

  V  PP   PP 1 

LOT

Minimum Horizontal Stress (Hmin) Conduct leakoff tests (LOT). The borehole opening pressure will usually be within 100 psi of the minimum far field stress in shales. We have a full lesson on LOT interpretation later Maximum Horizontal Stress (Hmax) Cannot be measured directly. It is usually calculated from the difference in the breakout on two sides (degrees of breakout) because this difference is caused by the difference in Hmin and Hmax. Fairly inaccurate

Degrees of Breakout

Hmin

Dego

Hmax

Hmin

How Increased/Decreased Pressure Modifies Stresses Fluid transmits 100% of any change in vertical pressure to horizontal pressure, while grains transmit only 20-70% of any change in vertical effective stress (Poisson’s). If we reduce pressure, we’re moving the vertical load from a phase that is very efficient at creating horizontal stress to one that’s very inefficient. For every psi of decline in pore pressure, the horizontal stress declines 1/3 to 2/3, depending on the Poisson’s of the grains Grain Contact

20-70% of Grain Contact Stress 



Pore Pressure

= Overburden

H 

PSI

 1

 V  PP   PP

100% of Pore Pressure = Total Horizonal Stress (  H  Integrity)

Pore Pressure Vertical Grain Stress Vertical Grain Stress Transferred To Horizontal, K0 = /(1- )

Soil Mechanics Equation σH = Ko (σV -PP) + PP Horizontal Grain Contribution σH

= Total Horizontal Stress (Integrity)

σV

= Total Overburden Stress (psi)

PP = Pore Pressure (psi) Ko

= Effective Stress Ratio  /(1- )



= Poisson’s Ratio 0.15 - 0.5

 Soft Sand Medium Sand Hard Sand Medium Limestone Hard Limestone Soft Shales Salt

.20 .17 .15 .25 .23 .27- .35 0.5

K .25 .20 .18 .33 .30 .37-.54 1.0

Horizontal Fluid Press Contribution

Don’t trust these numbers. Calculate your Poisson’s from local experience or testing Normal Pressure (8.4ppg)* Integrity (psi/ft) Integrity (ppg) .71 13.7 .69 13.2 .67 12.9 .76 14.6 .74 14.2 .78-.87 15.0-16.8 1.0 19.2

* σ H = K(1.0 - 0.437) + 0.437 in normal pressure regime. But reality can vary greatly (i.e., tectonics, pore pressure)

Poisson’s Determines Horizontal Stress for Given OB OB

Shale

Shales transmit more of the overburden into horizontal stress (higher Poisson's)

 OB

Lost Returns Fracture

Sand

Sand transmits less of the overburden into horizontal stress (lower Poisson’s)



Methods of Predicting Fracture Pressure Gradient

31

The Hubbert & Willis Equation Hubbert and Willis equation (1957) •

equates the average horizontal stress to the fracture pressure

•

assumes overburden stress to be 1 psi/ft and  equal to 0.25. p frac 

 1 



ob

 p p  p p

p    ob   p   D D   D p  p 1 1 g frac   1 . 0  p   p  3 D  D 3 p frac



0 . 25 1  0 . 25

pp D 2 pp    1 . 0   D  

Hubbert and Willis equation predicts values that are usually too low when compared with field data 32

The Matthews & Kelly Equation Matthews and Kelly equation (1967) introduced the concept of a variable horizontalto-vertical stress ratio. Their equation is the following: p frac D

 K MK

34

  Ve  p p   D  D 

The Matthews and Kelly Equation Matthews and Kelly assumed a 1.0-psi/ft overburden gradient in developing their curves. To calculate a fracture gradient by this method follow this procedure: •

Obtain the pore pressure

•

Determine the effective (matrix) vertical stress, Ve  1 . 0 D  p p

•

Determine the depth Deq for which the effective vertical stress would be the normal value:  Ve D eq 

1 . 0  0 . 465

•

Use the value of Deq to find KMK from the curve

•

Calculate the fracture gradient using Matthews and Kelly equation. They call this breakdown gradient and say it is higher than the fracture extension gradient.

35

The Matthews and Kelly Equation Example: The table lists shale resistivities and computed pore pressures for a well located in the East Cameron Block, offshore Louisiana. Estimate the fracture pressure at 8,110 and 15,050 ft using the Matthews and Kelly correlation

36

The Matthews and Kelly Equation Solution: The first depth is under normal compaction and KMK is found as 0.69 from the Louisiana curve. Hence, 5,204

p frac D

8,110

 0 . 69 1 . 0  0 . 465   0 . 465  0 . 834 psi / ft

For the undercompacted interval at 15,050 ft, the equivalent depth is determined by: D eq 

1 . 0  0 . 815  15 , 050 1 . 0  0 . 465

 5 , 204 ft

The matrix stress coefficient at 5,204 ft is found to be 0.57. Hence, p frac D

 0 . 57 1 . 0  0 . 815  0 . 815  0 . 92 psi / ft

37

The Eaton’s Gulf Coast Equation • Eaton’s correlation is based on offshore LA data in moderate water depths (S = overburden): p frac D



E 1  E

pp  pp  S     D D D  

• Eaton back-calculated Poisson’s ratios using field fracturing pressures data and his variable overburden curve. (His Poisson’s ratio thus calculated differs from the definition of Poisson’s ratio giving the subscript E)

38

The Eaton’s Gulf Coast Equation The Poisson’s ratio trend for the Gulf Coast area is plotted on the right side of the figure.

39

The Eaton’s Gulf Coast Equation Picking points graphically doesn’t make sense if we can have an equation to describe the curve…

Which Mitchell realized: 2

 D   D   D  g ob  0.84753  0.01494     1.199 *10 5    0.0006   1,000   1,000   1,000  

2

3

...............2.9 a  3

D   D   D   D     0 .00668    0 .00035    6 . 71  10  6   1, 000   1, 000   1, 000   1, 000 

 E  0 .23743  0 .05945  ( eq 3 . 70 )

40

4

The Eaton’s Gulf Coast Equation Example: The table lists shale resistivities and computed pore pressures for a well located in the East Cameron Block, offshore Louisiana. Estimate the fracture pressure at 8,110 and 15,050 ft using the Eaton correlation

41

The Eaton’s Gulf Coast Equation Solution: First thing is to determine the overburden pressure at 8,110 ft using the Mitchell approximation: 2

3

2

3

 D   D   D  g ob  0.84753  0.01494    0.0006    1.199  105    1,000   1,000   1,000 

 8,110   8,110   8,110  g ob  0.84753  0.01494    0.0006    1.199  10 5   1 , 000 1 , 000 1 , 000       g ob  0.936 psi / ft

Now determine the Eaton’s Poisson’s ratio using the Mitchell approximation: 2

E E

3

 8 ,110   8 ,110   8 ,110   8 ,110   0 .23743  0 .05945     6 . 71  10  6    0 .00035    0 .00668   1, 000   1, 000   1, 000   1, 000   0 . 438

42

4

The Eaton’s Gulf Coast Equation The fracture gradient at 8,110 ft is given by: g frac 

0 . 438 0 . 936  0 . 465   0 . 465  1 0 . 438

g frac  0 . 832 psi / ft

Repeating the same procedure for the depth of 15,050 ft we found: g ob  0 . 977 psi / ft ;

 E  0 . 468 ;

g

frac

 0 . 958 psi / ft

43

The Eaton’s Gulf Coast Equation Results from Eaton’s correlation usually agree more with field data.

44

What if I’m not drilling in the GoM? The data necessary to develop the Eaton’s technique for other tectonically relaxed areas of the earth are as follows: 1. Overburden stress gradient vs. depth. Such data can be derived from bulk densities taken from logs, seismic data or shale density measurements 2. Actual fracture pressure gradient for several depths. These can be lost-circulation or squeeze data or actual fracturing data 3. Formation pressures that apply to the data in item 2 (In items 2 and 3 the depths must correspond) 4. With these data and the equation below, the Poisson’s ratio can be back-calculated and plotted vs depth p frac

E  D  S 1 E



pp D

p  p  D D

   45

Field Determination of Fracture Pressure Gradient

46

Field Determination of Fracture Gradient • Leak-off test, LOT, - pressure test in which pumping continues only until the maximum borehole pressure that can be applied without mud loss, which is called the leak-of pressure (LOP), is reached. • Formation Integrity Test, FIT, - pressure test in which we only want to determine if a formation can withstand a defined amount of pressure without mud loss. 47

Field Determination of Fracture Gradient • Extended Leak-off test, XLOT, - is an extended version of LOT test.  Pumping continues until the surface pressure peaks at formation breakdown pressure (FBP).  Pumping is then continued for a few more minutes to ensure stable fracture propagation into the undisturbed rock formation with the pumping pressure stabilizing to the value called fracture propagation pressure (FPP)

• U.S. onshore personnel will recognize this as similar to a “mini-frac”

48

Field Determination of Fracture Gradient Schematic borehole configuration during a leak-off test (LOT) or extended-leak-off test (XLOT)

49

Field Determination of Fracture Gradient Idealized relationship between pumping pressure and time or volume of injected fluid during an XLOT

50

Why Do LOT? • Test the cement job  channeled sheath will require a squeeze job

• Confirm FG estimate  If not, it may be necessary to modify well plan and casing program

• Input for prudent decision making during a well-control problem 51

Field Determination: LOT • Run and cement casing • Drill out ~ 10 to 30 ft below the casing seat • Circulate to clean the hole • Pull the bit out to shoe • Close the BOPs • Pump slowly and monitor the standpipe pressure

52

Flowrate

Leak-off Test: typical results

Pump stopped

53

0.25 to 0.50 bpm

Leak-off Test: Fluid Compression • After the conclusion of the LOT, release slowly the pressure and monitor the volume of drilling fluid bled from the well • Compare this volume with the pumped volume during the LOT • Should be approximately equal if only filtration losses occurred

54

Leak-off Test: Example After drill the casing shoe at 4,000 ft, the operator performs a leak-off test and the following data are registered:

Pumped Volume (bbl)

Surface Pressure (psi)

0

0

0.5

45

1.0

125

1.5

230

2.0

350

2.5

470

3.0

590

3.5

710

4.0

830

4.5

950

5.0

990

5.5

1010 55

Leak-off Test: Example If the mud weight is 9 ppg, determine the fracture gradient, in ppg, at 4,000 ft.

Solution: 1. Plot the data in a Cartesian graph

56

Surface Pressure, psi

Leak-off Test: Example 1100 1000 900 800 700 600 500 400 300 200 100 0

0

2 4 Pumped volume, bbl

Last point on the line is 950 psi.

57

6

Leak-off Test: Example Ps = 950 psi

Leak-off pressure = PS + PHYD = 950 + 0.052 * 9 * 4,000 = 2,822 psi

PLEAKOFF 2,822  D 4, 000

psi ft

Fracture gradient = 0.706 psi/ft EMW = 0.706/0.052 = 13.6 ppg 58

Leak-off Test Interpretation Pressure decline is large and does not stabilize: it might be attributed to the existence of natural fractures or to ineffective cementing.

59

LOT Interpretation

60

Experimental Determination: FIT Same as procedure as LOT but: • Test pressure brought up to predetermined value, then stopped & held. • Quick, very simple to do • Does not tell us much about the formation or cement job. 61

Formation Integrity Test: typical results predicted pressure-volume relation

Pump pressure

predicted leak-off pressure x

pumping stopped x x

x

x

Volume pumped

62

Formation Integrity Test: Example ?????

MW = 10.0 ppg

How much surface pressure will be required to test the casing seat to 14.0 ppg equivalent?

TVDshoe = 4,000 ft

63

Casingwassetat8,000ft inawell.Theoperatorperformsaleak‐offtestat9000ft.If themudweightinthewellis10lb/gal,whatistheequivalentfracturegradientatthe casingshoe,roundedtothenearesttenth?

Volumepumped (bbl) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

Pressureat surface (psi) 0 50 110 180 300 420 540 660 780 900 1020 1100 1080 1000

(bbl) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

Pressureat surface (psi) 0 50 110 180 300 420 540 660 780 900 1020 1100 1080 1000

DeltaP 0 50 60 70 120 120 120 120 120 120 120 80 ‐20 ‐80

LOT 1200 1000 800

Pressure

Volumepumped

600 400 200 0 0

1

2

3

4

Volumepumped

5

6

7

Field Determination of Fracture Gradient Wellbore pressure containment is often described with a fracture gradient measured by LOT. Since LOT is only about the near-wellbore effect, it is clear that the drilling engineer is most likely referring to the near-wellbore effect and the geo-mechanist the far-field. The drilling engineer’s fracture gradient could be much different from the geo-mechanist’s. This is also why directly using a LOT value to calibrate a fracture gradient makes the geo-mechanist uncomfortable. The XLOT value, which is truly for the fa...