Hydraulic Fracturing design

Tripoli University  PETROLEUM SEMINAR FOR

Hydraulic fracturing

CONTENT         

Introduction The objective of the hydraulic fracturing Fracture Mechanism Fracture Orientation Fracturing fluids Types of Hydraulic Fracturing Fluids Modeling of Hydraulic Fracturing Example problem References

Introduction 

Hydraulic fracture can be defined as process of creating a fracturing in a porous medium by injecting a fluid under pressure through a well bore in order to overcome native stresses and to cause material failure of the porous medium.



Hydraulic fracturing was introduced first in the Hugoton gas field in western Kansas in 1947. Fracturing techniques were developed in 1948 and the first commercial fracturing treatments were conduced in 1949. And the within a very few years thousands of wells per year were being stimulation by hydraulic fracturing treatment.

Continue…

Nature of hydr hydraulic aulic fracturing is a process applied to improve the ability of hydrocarbon fluids to flow to the hole and be recovered.   Hydraulic fracturing has been and will remain one of the primary engineering tools for improving well productivity in old and new wells.  Frac Fracturing turing has been used in some types of the formation such as sandstone and carbonates. 

The objective of the hydraulic fracturing: There are many different applications for hydraulic fracturing, such as :  Increase the flow rate of oil and/or gas from low permeability reservoir.  Increase the flow rate of oil and/or gas from wells that have been damaged.  Connect the natural fractures and/or cleats in a formation to the well bore.  Decrease the pressure drop around the well to minimize sand production.

Continue…

Decrease the pressure drop around the well to minimize problems with asphaltine and/or paraffin deposition. 6. Increase the area of drainage or the amount of formation in contact with the well bore. 7. Connect the full vertical extent of a reservoir to a slanted or horizontal well. 5.

Fracture Mechanism 

Fracture Mechanism can be divided into two steps:

1.

Fracture Initiation.

2.

Fracture Extension.

In-situ stress  Underground

formation are confined and under stress. Figure.1 illustrates the local stress state at depth for an element of formation. The stresses can be divided into 3 principal stresses. In Figure. 1, σz is the vertical stress, σx is the maximum horizontal stress, while σy is the minimum horizontal stress

σ 

 z 

 G  D ob

Where:  z = Over burden stress, psi. Gob = Overburden gradient, psi/ft. D = Depth, ft. 

Fracture Orientation  

 A hydraulic fracture will propagate perpendicular to the least principle stress Figure. (2).



 As a rule of thumb, if the fracture gradient is less than 0.8 psi/ft, the fracture will be vertical. If the fracture gradient is greater than 1.0 psi/ft, the fracture will be horizontal.

 Horizontal

fractures



 A vertical fracture

Fracturing fluids 

To select the proper fluid for a specific well it is necessary to understand the properties of fluids. The fluid design must be considered these characteristics:

Low leak off rate . 2. The ability of the fluid to carry the propping agent . 3. Friction loss . 4. Fluid viscosity. 1.

Types of Hydraulic Fracturing Fluids 1) 2)

3) 4) 5)

Water-Base fluids . Oil-base fluids: Napalm gels ,viscous refined oil and lease crude oils  Acid based fluids . Foams .  Emulsions .

Propping agents  

Propping agents are required to (prop-open) the fracture once the pumps are shut down and the fracture begins to close. The ideal propping agent will be strong, resistant to crushing, resistant to corrosion, have a low density, and readily available at low cost. The products that best meet these desired traits are silica sand, resin-coated sand, and ceramic proppant.

Modeling of the hydraulic fracturing 



The first fracture treatments were pumped just to see if a fracture could be created and if sand could be pumped into the fracture. In 1955, Howard and Fast published the first mathematical model that an engineer could use to design a fracture treatment. 2D fracture model: The Howard and Fast model was a twodimensional (2D) model. In the following years, other 2D models were published. When using a 2D model, the engineer fixes one of the dimensions (normally the fracture height), then calculates the width and length of the fracture.

This Figure is shows the PKN geometry and its normally used when the fracture length is much greater than the fracture height.

The KGD geometry is used if the length of fracture is less than the height.

3D fracture model  Pseudo three-dimensional (P-3D).  Planar three-dimensional (PL-3D).  Fully three-dimensional (3D). 

Example 



We will select KGD model to explain to the design of hydraulic fracture.  A fracturing treatment is intended to be conducted in an oil well completed in a tight limestone formation in order to increase the oil production rate from the well from 200 BPD to 600 BPD. Given the following information

DEPTH.

12,000 FT

Formation thickness

80 ft

Minimum horizontal stress

4185 psi

Over burden pressure gradient

1.1 psi/ft

Fracturing angle

30 degree

Reservoir oil compressibility

11×10-6 psi-1

Reservoir water compressibility

2 ×10-6 psi-1

Reservoir gas compressibility

5.30×10-4 psi-1

Gas saturation

0%

Oil saturation

75 %

Gas gravity

0.890

Oil formation volume factor 

1.17 resbbl/STB

Connate water saturation

25%

Formation porosity

25 %

Poisson's ratio

0.2

Fracturing fluid viscosity

6 cp

Fracturing fluid density (versa Gel)

12 ppg

Frictional pressure gradient inside tubing

0.1501 psi/ft

AVERAGE RESERVOIR PRESSURE (BHSP)

4000 PSI

Reservoir fluids viscosity

2 cp

Area of filter medium

45 cm2

Slop of fluid loss curve at lab

1.5 cm/min

Filtration pressure at lab

100 psi

Yong's modulus

3×107

Casing outer diameter 

9.6250 in

Casing inner diameter 

8.6810 in

Well bore radius

0.292 in

Drainage radius

745 ft

Proppant size and type (Z-proppant)

20/40 mesh

Porosity of packed proppant

35 %

Specific gravity of proppant

2.63

Bottom hole flowing pressure before fracturing

1200 psi

Biot constant

0.8

Initiation shear stress

650 psi

Fracturing fluid spurt loss

0.010 gal/ft2

Tubing outer diameter 

3.5 in

Tubing inner diameter 

2.9910 in

1/2

 Assume that: hf = h , qi =30 bbl/min ,Vi = 900 bbls Calculate: 1. The formation fracturing pressure. 2. The effective fracturing fluid coefficient. 3. The fracture volume. 4. The fracture efficiency. 5. The concentration of proppant in the fracturing fluid. 6. Well head injection pressure. 7. The well productivity ratio. 8. The bottom hole flowing pressure after fracturing. 9. The oil flow rate after fracturing 

Calculation of fracturing pressure (pf ): Pob (δz) = Gob X D = 1.1 X 1200 = 12,300Psi 

 f    P 

   ν    2  p   δ  P   z    1  ν    1  2ν    2  α   ν 1      



Where: ν = Poisson's ratio. τ ο = Initiation shear stress. α = Biot constant



  τ ο      

 p P 

Continue…    0.2       2   13200 - 4000   650   1 0.2     4000  7,750 Psi  P  f        1  2  0.2     2  0.8    1 0.2      

Calculation of fracturing fluid coefficient (CT):    C T     1    C ν

   1  1 1     Cc Cw 

 ∆P (Closure stress) = Pf -Pres = 7,750 – 4,000 = 3,750 Psi.  Viscosity control coefficient, Cν: C ν  0.0469  Where:

  k     ΔP        μ  ff      

 μ ff  = Fracturing fluid viscosity

C ν  0.0469 

 0.005 0.25  3750      6     

 0.04145  ft  /  min

Calculate total compressibility Ct: Ct = Sw x Cw + So x Co + Sg x Cg Ct = (0.25 x 2 x 10-6) + (0.75 8. x 11 x 10-6) + (530 x 10-6 x 0.0) Ct = 8.75 x 10-6 Psi -1 Compressibility control of reservoir fluids, Cc: Cc  0.0374  3750 

 0.005  0.25  8.750  10  6       2    

Wall building coefficient, Cw:

 0.0164    mact  Cw     f   A    

 0.01037  ft  /  min

mact 



mlab 

mact   1.5

 ΔP act   ΔP lab

3750  9.186cm / min 100

 0.0164    9.186  0.00335  ft  /  min Cw    45           1   0.00239  ft  /  min C T     1 1 1     0.04145 0.01037  0.00335    

Calculation of fracturing dimension ( L,Wf )  Vi    Pumping time  t    q    i  

,

t  V i  900  30 min qi 30

8  C T  π   t  α  π   ww  8  S   p         0.1856  8  C T  π  t   8  0.00239  π  30        α π  ww  8  S   8  0.01    0.262 ww  0.0107    p    π       ww         12  7.48        

The Fracture length (L): q  5.615   π   p     i     e  L  ww  8  S  2 12 7.48     64  π   h f   C T   

α

2

  .erfc(   )  2  α  1 π 

 

2   π  30  5.615 8 0.01    2  α   α   e .erfc(   )  w  L  w    1   2 7.48     π  64  π  80 0.00239  12  

  α 2  L  2571 0.262 ww  0.011  e .erfc(   )   

The Fracture area (A):  A = 4 x L x hf = 4 x 80 x L = 320 x L.    A          A    A      AQ        2 q 2 3   0  5.615   336 .9    i    

 Assume ww, Calculate: L, A, AQ.

2α

   1 π   

 Assume

ww, Calculate: L, A, AQ.

W W   ASSUM   E

CAL.

0.05

7.798

0.2

  α 2      2   α   1   e  .erfc (   α)         π       

 L CAL.

 A CAL.

 AQ CAL.

7.87087 

334

110,080

327 

2.941

2.50205

290

92,800

275

0.4

1.607 

1.11824

237 

75,840

225

0.6 

1.105

0.64740

199

63,680

189

0.8

0.842

0.42495

172

55,040

163

1.0

0.681

0.30190

151

48,320

143

α

Calculate Fracture width ( ww ). ww 

2     μ   ff     q   L i 0.350    '   h   f     E    

       

1

   E    '    E    2  1  ν  

  3  107      E '    2  1 0 . 2     

 3.125  107  Psi

4

1

2     4  40  15  L    ww  0.350   7    1.099  10  95      

 Assume L/re, Calculate: A, AQ, ww : L= L/re x re = L/re x 745  A = 4 x L x hf = 4 x 80 x L = 320 x L    A          A    A      AQ         2  qi 2 3 0 5.615 336.9               L/r e Assume

L Cal.

ww Cal.

A Cal.

AQ Cal.

0.2

149

0.069

47,680

142

0.4

298

0.099

95,360

283

0.6

447

0.121

143,040

425

0.8

596

0.140

190,720

566

1.0

745

0.156

238,400

708

(in)

300 (Min/ft)

Plotted (ww vs. AQ) on Log-Log scale. The intercept of two lines given solution: ww= 0.1 in , AQ = 300 min / ft From AQ equation:    A    → A=AQ  AQ   2  qi     

x 2 x qi x 5.615

 A = 300 x 2 x 30 x 5.615 = 101,070 ft2    A     L    4 h  f      

=

 101,070      4  80   316     

ft 

The fracturing dimension: (L= 316 ft, ww= 0.1 in).

Calculation of fracture volume (Vf ):  π   V   f      L  h f   ww   2   0.1   π   3    f      316  80   V  330.7   ft   12   2     

Calculation of fracture efficiency (Eff ):    V   f     330.7     100  6.54 %  E  ff        i  V  900 5.615        

Calculation of proppant weight needed (Wp):  p V   f   1  ρ V   f  (1  ) Sp.Gr 62.4 W 

W  p  330.7 (1  0.35) 2.63 62.4  35277 lbm

Calculation of proppant concentration (Cpp):   35,277    W   pp    0.933     i V  900 42         

C   pp   

lbm/gal (ppg)

Calculation of wellhead injection pressure: Pth = Pwh = Pf  – ∆Phydrostatic + ∆Pfric + ∆Pperf  Corrected fracturing fluid density ( ρmix ):          8.34   12   0.933   8.34    8.34  γ f   C     pp         ρmix.     12.405  1  0.0456 C    pp   1  0.0456  0.933         

lbm /  gal 

 A)Pf =7,750 Psi. B) ∆Phyd = 0.052 x ρmix x D = 0.052 x 12.405 x12,000 = 7,741 Psi C)∆Pfric= Gfric x D = 0.1501 x 12,000 =1,801 Psi D)∆Pperf = 0.0 Psi → open hole completion. Pth = Pwh = 7750  7741 + 1801 + 0 = 1810 Psi

3

10

Calculation of fracture conductivity (Fc): , Fc = kf x Wf . By using propped fracture permeability curves. From curves; by using closure stress = 3750 Psi & 20/40 mesh. The propped fracture permeability is 160 Darcy.

Kf =

160 x103 md  0.1

 F c  160  103  

 12

 

 π    1,047  md   ft  4 

Calculation of production increase: Relative conductivity,          0.1 π    160  103     4       5          

  J  f       J o 

 f   k   f    W      k e  

40  A

40  2225 51

L/re =316 / 745 =0.42 Entering the production increase curves with these values. Obtained the

          J  7.13  f         J  o      0.472 r  e      ln    

 3.2

Then,

  J  f       J o 

=

        3.2  ln 0.472 745      0.929        7.13        

 2.66 

Productivity ratio = 2.66 Construct of IPR curve before fracturing.  PI  

PI =

       3 7.08 10 k  h          r e    Bo ln  μ         r w      

       7.08  10  3  5  80       745  2  1.17  ln    0.292           

q     PI       P e  P wf  

q  PI (P e  P wf  )

=0.154 BPD/Psi

 0.154 ( 4,000  P wf  )

PWF, ASSUME

Q, (BPD)

4000

0

3,000

154

2,000

308

1,000

462

0

616

Construct tubing intake curve. By using Brown correlation Q

PTH

PWF

800

120

1,390

1,000

160

1,520

2,000

280

2,000

From the intersection of IPR curve with the tubing intake curve TPC. qo optimum before fracturing = 440 BPD Pwf optimum before fracturing = 1200 Psi Construct of IPR curve after fracturing. q f       J   f      J o  P e  P wf ' 

qo        P e  P wf  =2.66         q f          440     2.66     4000  P wf '   4000  1200      PWF' , ASSUME

QF , (BPD)

4000

0

3,000

544

2,000

1,088

1,000

1,632

0

2,176

From IPR curve and tubing intake curve after fracturing qf optimum after fracturing = 1270 BPD. Pwf ' optimum after fracturing =1640 psi.:

1640 1200

440

1270