Géomécanique avancée
Stabilité des puits de forage profonds
Jean Sulem
Laboratoire Navier/CERMES, Ecole des Ponts ParisTech [email protected] Borehole instabilities during drilling
Borehole stability problems: 5 to 10 % of the drilling costs (hundreds of million dollars per year) Challenges: Offshore drilling Deviated, multilateral, horizontal wells Deep and geologically complex formations Tectonically active areas Weak/swelling formations (mudstones, shales) High pore pressure … Borehole instabilities during drilling
Well design: Optimize the mud weight for stable drilling - In order to prevent influx of fluids (in particular gas) : Keep the mud weight above the pore pressure gradient. - In order to prevent loss of mud into fractures (“lost circulation”): Keep the mud weight below the fracture gradient. Optimize the mud chemistry For example, use of chemical additives (salt) when drilling in shales (presence of swelling clays) or to heal fractures to avoid mud loss in the formation. Borehole instabilities during drilling
Difficulties in borehole stability evaluation: - No visual observation of what is happening at great depth (thousand of metres away) - Large variations in the formation stresses and properties (poor information on in situ stresses, pore pressure, mechanical properties) - Complex coupled physical processes: redistribution of stresses, mud chemistry, temperature changes … Borehole Stability Analysis: Coupled THMC processes
Rock mechanical properties Earth stresses Well trajectory - Strength parameters - Vertical and horizontal stresses - Inclination - Elasticmoduli - Pore pressure - Azimut
Other rock properties Borehole stresses Other well parameters - Plasticity - Boundary conditions - Mudtype - Permeability - Poro-/Thermo-/Chemo- - Temperature - Ionic diffusivity Elastoplasticity - Membrane efficiency - Ionic exchange capacity Borehole failure criterion - Thermal diffusivity - Tensile failure - Anisotropy - Shear failure
Minimum permitted mud density Maximum permitted mud density - Hole collapse by shear / radial tensile failure - Fracture closure pressure - Pore pressure balance - Fracture initiation and propagation In-situ stress state
6 In-situ stress
Worldwide in-situ rock stress data Hoek and Brown (1980)
Vertical stress Average horizontal stress 7 Density
Sedimentary rocks (Schön 1996) 8 Borehole stability Stresses around boreholes Borehole failure Stresses in an infinite poroelastic hollow cylinder - Assumptions
Assumptions: Isotropic poroelastic saturated medium under isotropic loading σ 2 R R Internal radius 1 , External radius 2 R2 M( r,θ) σ Internal radial stress 1 External radial stress σ 2 R1 σ Hydraulic boundary conditions: 1 Imposed pore pressure OR imposed fluid flux Axisymmetry and no deformation in the direction of the cylinder axis (plane strain) = Initial condition: uniform pore pressure at t = 0, prf ( ) p f 0 Stresses in an infinite poroelastic hollow cylinder – Governing equations
Stresses, pore pressure and displacements depend only on r and are independent of z and θ ∂u ε 0 σ 0 r 0 ∂r σ = r ε = = , u is the radial displacement ε 0 σ 0 θ u θ 0 r ∂σ σ − σ - Equilibrium equation : r +r θ = 0 ∂r r ∂u u σ−bp =λ+()2 G +λ - Poroelastic constitutive equations: r f ∂r r (Biot effective stress) ∂u u σ−θ bp =λ +λ+()2 G f ∂r r ∂m d - Fluid mass balance: f +divq = 0 ∂t ∂m k d k f = ρ ∇ 2 () - Darcy’s law: q= −ρ grad() p fp f fµ f ∂t η f Stresses in an infinite poroelastic hollow cylinder – Pore pressure field
∂2 p1 ∂ p Steady state solution: ∇2 ()p =f + f = 0 w ∂r2 r ∂ r General solution: = + prf ( ) C1 ln rC 2
Boundary conditions:
- zero flux at the borehole wall (impervious mud cake) ∂p f =0 et pR ( ) =→ p prp ( ) = (uniform pore pressure ) ∂ f2 f 0 ff 0 r = r R 1 - Imposed pore pressure at the borehole wall (drained conditions)
p− p r pRp()= and pRp() =→ prp() =+ f0 f 1 ln f11 ff 20 f ff 1 R R ln 2 1 R1 Stresses in an infinite poroelastic hollow cylinder – Stress and displacement fields Equilibrium equations + poroelastic constitutive equations General solution: ∆ = − prf() pr f () p f 0
C b 1 r = +−4 ρ∆ρρ() uCr3 pdf rλ + 2 G r R1 r σ=() λ+−1 +G 1 ρ∆ρρ() r 2CGCG3 2 4 2 2 b 2 pdf rλ + 2 G r R1 r σ=() λ++1 −G 1 ρ∆ρρ+() G ∆ θ 222CGCGb3 4 2 2 pdbf 2 p f rλ+2 GrR1 λ+ 2 G
Boundary conditions : σ =σ σ =σ r=1 and r = 2 rR1 rR 2 Stresses in an infinite poroelastic hollow cylinder – Stress distribution with constant pore pressure Zero fluid flux at the borehole wall (undrained case) Notations: >> ↔ σ↔ σ↔σ R211 RRR, , 1 p w , 20 For isotropic loading and isotropic p= p f f 0 rock mass: R 2 Constant pore pressure σ =σ −() σ − p r0 0 w r Constant mean stress 2 Zero volumetric strain σ =σ + σ − R θ ()p 0 0 w r p= p At the borehole wall: f f 0 σ = σ z 0 σ = rp w 1 R 2 σ = σ − =() σ − θ 2 0 pw u0 p w 2G r σ = σ z 0 u 1 =() σ − p Maximum stress deviator at the borehole wall R2 G 0 w Stresses in an infinite poroelastic hollow cylinder – Stress distribution with varying pore pressure Imposed pore pressure at the borehole wall (drained case) p= p fr= R w
Steady state solution: It is assumed that beyond a drainage radius Re, the pore pressure is constant and equal to pf0
Re R2 R 2 ln σ=σ−σ−()p +η() p − p − r r0 0 w fw 0 R r r ln e R
Re R2 R 2 ln σ=σ+σ−()p −η() p − p + r θ 0 0w f 0 w R r r ln e R R 2ln e − ν − ν η =G = 1 2 σ=σ−η()p − p r with b b z0 f 0 w R λ+2G 2(1 −ν ) ln e R Stress distribution with heat flow
The mud temperature Tw is different from the temperature T0 of the surrounding formation. Stresses are affected. Correspondence between poroelasticity and thermoelasticity Poroelasticity Thermoelasticity
σ= ε+λεδ+ δ σ= ε+λεδ+α − δ ij2G ij kkij bp fij ij2G ij kkij 3() T KTT 0 ij
αΤ : coefficient of linear thermal expansion ↔ − p f T T 0 E b ↔ 3αK = α T 1− 2 ν T σ = σ = rp w rp w E σ=σ−2p −η 2 () p − p σ=σ−θ 2 p + α() T − T θ 0w f 0 w 0 w1 − ν T w 0 σ=σ−η2 ()p − p E z0 f 0 w σ=σ+ α()T − T z01 − ν T w 0 Stresses around boreholes – General elastic solution Vertical borehole with constant pore pressure
σ+σR2 σ−σ RR 42 R 2 σ=Hh −+ Hh +− θ+ r 1 13 4 cos2 pw 2r2 2 rr 42 r 2 σ+σR2 σ−σ R 4 R 2 σ=Hh +− Hh + θ− θ 1 13 cos2 pw 2r2 2 r 4 r 2 σ R2 v σ=σ−νσ−σ2() cos2 θ zV Hh r2 z σ − σ R4 R 2 τ=−H h −+ θ rθ 1 3 2 sin 2 2 r4 r 2 y σ τ =τ = h rzθ z 0 x σ σ ≠ σ H If H h the mean stress is not uniform
σ+σ+σ σ +σ+σ 2 R2 rθ zHhV= −+νσ−σ()()1 cos 2 θ 3 3 3 H h r2 Stresses at the borehole wall
σ σ = v rp w
σ=σ+σ−θ 2() σ−σ cos 2 θ− p Hh Hh w z σ=σ−νσ−σ() θ zV2 Hh cos 2 τ =τ =τ = y rθ rz θ z 0 σ h x
for θ= 0, σ=σθ θ =σ−σ 3 − p σ ,min h H w H θ=π σ=σ =σ−σ− for / 2,θ θ ,max 3 H hp w
Shear failure occurs in the direction of the minimum principal stress
Tensile failure occurs in the direction of the maximum principal stress Stresses around the borehole - Example
σ = V 88.2 MPa (depth 3213 m) σ = H 90 MPa (orientation east-west) σ = h 51.5 MPa = = pw p mud 31.5 MPa σ = c 45 MPa φ = π / 4
from Zoback, 2012 Stresses around the borehole - Example
σ = V 88.2 MPa (depth 3213 m) σ = H 90 MPa (orientation east-west) σ = h 51.5 MPa = = pw p mud 41.5 MPa σ = c 45 MPa φ = π / 4
The mud weight has been raised of 10 MPa. Raising the mud weight decreases the size of the breakouts considerably. The area in white shows the region where tensile stresses exist at the wellbore wall. Borehole failure criteria – Vertical hole, isotropic horizontal stresses and impermeable borehole wall Principal stresses at the borehole wall τ µσ=' + C σ = rp w σ = σ − θ 2 hp w σ = σ z V Mohr-Coulomb shear failure criterion C
σ = σ+σ Terzaghi effective stresses '1K p ' 3 c σ =σ− σ =σ− '11pf , ' 33 p f Uniaxial strength cos φ +φ πφ σ= = β =1 sin =2 + c 2C 2 C tan K tan 1− sin φ p 1− sin φ 4 2 π φ β = + angle of the normal to the failure plane with 4 2 respect to the direction of major principal stress Borehole failure criteria – Vertical hole, isotropic horizontal stresses and impermeable borehole wall Principal Terzaghi effective stresses at the borehole wall σ = − 'rp w p f σ =σ− − 'θ 2 hp w p f σ =σ − 'z Vp f Shear failure
σ− −σ 2(hp f ) c a σ≤σ≤σθ p ≤+ p r z w f 1+ tan 2 β σ − −σ Vp f c b σ≤σ≤σθ p ≤+ p rz wf tan 2 β σ − −σ Vp f c c σθ ≤ σ ≤σ p ≥+σ−− p 2( p ) rz wfhf tan 2 β
Tensile failure criterion Vertical fracturing occurs if: σ =−σ σ frac = σ− +σ 'θ T , T is the tensile strength pw,max 2 hfT p Borehole failure criteria – Vertical hole, isotropic horizontal stresses and impermeable borehole wall Example of graphical representation of conditions for borehole failure
σ= σ= c0 ; T 0 = σ p f0.4 V φ =30 °
from Fjaer et al, 2008 Borehole failure criteria – Vertical hole, isotropic horizontal stresses and permeable borehole wall Principal stresses at the borehole wall σ =p ; p = p rw fr= R w σ=σ− +η − θ 2hwp 2 () p wf p 0 σ=σ+η − zV2 ()p wf p 0 Shear failure σ− η −σ 2h 2 p f0 c For σ<σ<σθ , p = r z w ,min 2− 2 η σ−η −σ V2 p f0 c For σ<σ<σθ , p = r z w ,min 1− 2 η
Vertical fracturing occurs if: Horizontal fracturing occurs if: σ =−σ σ =−σ 'θ T 'z T 2σ− 2 ηp +σ σ−η2 p +σ pfrac = h f0 T pfrac = V f0 T w,max 2− 2 η w,max 1− 2 η Borehole failure criteria – Vertical hole, unequal far field stresses and constant pore pressure Principal stresses at the borehole wall σ = rp w σ=σ+σ−() σ−σ θ− θ Hh2 Hh cos2 p w σ=σ−νσ−σ() θ zV2 Hh cos2 Shear failure σ −σ− −σ 3H h 2 p f c For σ<σ<σθ , p = p + rz w,min f 1+ tan 2 β σ+νσ() −σ − −σ V2 Hhp fc For σ<σ<σθ , p = p + r zw,min f tan 2 β
Vertical fracturing occurs if: σ =−σ 'θ T frac =σ−σ − +σ pw,max 3 hHfT p Borehole breakouts Rupture modes for the failure of boreholes
Shear surfaces intersecting parallel to the Toroidal breakouts axial stress
Multishear surfaces intersecting parallel to Vertical hydraulic fracturing the radius from Fjaer et al, 2008 Borehole breakouts
Cylindrical cavity after failure
Borehole breakouts observed by a dowhole camera from Asquith and Krygowski (2004)
Wellbore breakouts appear in an ultrasonic borehole televiewer image as dark bands on either side of a well because of the low amplitude ultrasonic reflections off the wellbore wall
from Zoback, 2012 Pressure regimes from onshore to offshore Subsurface pressure and overpressure Mud weight and borehole failure
Shear Failure Fracture Gradient Gradient Wellbore breakout Wellbore breakouts small breakouts width
large breakouts width Other effects in borehole stability analysis
Non linear and anisotropic rock behaviour Time dependent effects: Establishment of pore pressure equilibrium Creep Temperature effects The drilling fluid is usually initially colder than the formation Chemical interaction between the drilling fluid and the formation For example use of oil-based mud rather than water-based mud when drilling in shales Environmental issues … APPENDIX 2D ELASTICITY FOR ISOTROPIC SOLIDS 2D ELASTICITY FOR ISOTROPIC SOLIDS
Strain tensor Stress tensor ∂u ∂u ε ε =x = y x, y , σ τ ∂x ∂ y x xy σ = ∂ στ 1 ∂u uy xy y ε =x + xy 2 ∂y ∂ x εε Equilibrium equations x xy ε = ∂σ ∂τ εε x +xy +X = 0 xy y ∂x ∂ y Compatibility condition for strains ∂ στ ∂ xy+ y +Y = 0 ∂2ε ∂ ε2ε ∂ 2 ∂x ∂ y x +y = 2 xy ∂y2 ∂ x 2 ∂∂ xy
Elastic constitutive relationships σ κσ κ ε =+( ) +−( ) 8G x 1 x 3 y σ κσ κ ε =−()() ++ νκ =3 − 4 for plane strain 8G y 3 x 1 y ετ = ν ν κ =()()3 − / 1+ for plane stress xy2G xy 35 Assumption 1 : Body forces derive from a potential V
∂2 ∂ 2 ∂V ∂ V 2 V V X=−, Y =− avec ∇=V + = 0 ∂x ∂ y ∂x2 ∂ y 2
Assumption 2 : There exists a function U called Airy function such that
∂2U ∂ 2 U ∂ 2 U τ σ σ = +V, = + V , =− x∂y2 y ∂ x 2 xy ∂∂ xy
The equilibrium equations are automatically fulfilled
The condition of compatibility for the strains are fulfilled if the Airy function U is the solution of the bi-harmonic equation:
∂4U ∂ 4 U ∂ 4 U ∇∇=2() 2 U +2 + = 0 ∂x4 ∂∂ xy 22 ∂ y 4
36 FORMULATION IN POLAR COORDINATES
Strain tensor Stress tensor ∂u1 ∂ u ε ε =r , =u + θ , σ τ r∂θ r ∂ θ σ = r r θ r r στ θrθ ∂u ∂ u ε =1 1 r − + θ θ rθ u 2 r∂θ ∂ r εε ε = r r θ εε θrθ
Equilibrium equations
σ∂ στ σ 1 ∂ − r+ θ rθ + r +=R 0 Condition of strains compatibility ∂r r ∂ θ r
τ σ θ∂ θ τ θ 1 ∂ 2 r+ + r +Θ= 0 ε ∂2 (rε ) ∂ εε ∂ 2 ∂ 2 ( r ) ∂r r ∂ θ r rθ − r r + r = 2 rθ ∂r2 ∂∂ r θ θ 2 ∂∂ r
37 For simplicity, we assume zero body forces
The equilibrium equations are automatically fulfilled for
∂∂2 ∂ 2 ∂∂ τ σ σ =+1UU 1 = U =− 1 U , θ θ , r rrr∂∂ θ 2θ 2 ∂ r 2 r ∂∂ rr
Strains compatibility condition is fulfilled if the Airy function U verifies the following bi-harmonic equation:
∂211 ∂ ∂∂ 22 U 11 ∂ U ∂ 2 U ∇∇=++2() 2 U ++ = 0 ∂rrrr2 ∂ θ 222 ∂∂θ rrrr ∂ 22 ∂
38 SOLUTION OF 2D ELASTICITY PROBLEMS USING COMPLEX VARIABLE POTENTIAL
The position of a material point in a solid is defined by a complex variable
z= x + iy
The general solution of the bi-harmonic equation ∇2( ∇ 2 U ) = 0 is of the following form U=Re { zz χφ ( ) + ( z )}
φσ σ + = ′( ) x y 4Re { z } ψ φ τ σ σ −+ =′′()() + ′ y x2i xy 2 zz z χψ ()()z= ′ z += ψκφ φ ( ) −′( ) − ( ) 2Guiu( x y ) zzz z
νκ =3 − 4 for plane strain ν ν κ =()()3 − / 1+ for plane stress 39 FORMULATION IN POLAR COORDINATES
φσ σ σ σ + = + = ′( ) rθ x y 4Re { z } τσ σ σ τ σ −+ = −+ 2iθ θ θ r2i r() y x 2 i xy e ′′ ′ 2iθ =2zz ψφ ()() + ze
−iθ 1 ′ − i θ uiu+=+θ () uiue =() ψκφ φ ()() zzz − − () ze r x y 2G
40 Circular opening in an infinite medium under uniaxial remote loading
1 A φ ()z= p z + 4 1 z θ = i y −1 B C z re ψ ()z= p z + + 2 1 z z 3 R p1 1 A A x φ φ ′()zp=1 − , ′′ () zp = 41 z2 2 z 3 1 −1B 3 C ψ ′()z= p 1 − − 2 1 z2 z 4 θ φσ σ σ σ +=+=′ =−Acos2 θ 4Re{}()z p 1 r x y 1 r 2 τσ σ σ τ σ −+ = −+ 2iθ θ θ r2i r() y x 2 i xy e ′′ ′ 2iθ =2zz ψφ ()() + ze
θ θ B A C − =p − e2i ++ + 3 e 2 i 1 r2 r 2 r 4 41 θ σ σ + = − Acos2 θ p 1 r 1 r 2 θ σ σ −=B −−− A C θ p 1 3 cos2 r 1 r2 r 2 r 4 A C θ 2τ θ =−p 1 + + 3 sin2 r 1 r2 r 4 Boundary conditions: =σ τ = = For r R , r r θ 0 2 2 4 σ =1 −+R 143 −+ R R θ 2 2 4 r p11 p 1 1 cos2 A= 2, RBRC = , =− R 2r2 2 r 2 r 4 2 4 θ σ =1 +−R 1 + 3 R θ p112 p 1 1 4 cos2 σ / r r θ p1 2 2 3 2 4 θ τ =−1 + 2R − 3 R rθ p1 1 sin 2 2 2 r2 r 4 1 for r= θ R , σ θ = p ( 1 − 2cos 2 ) 1 0 At the wall: − ≤σ ≤ p1θ 3 p 1 θ/π -1 0 0.5 42 Displacements
−iθ 1 ′ − i θ uiu+=+θ () uiue =( ψκφ φ ()() zzz − − () ze) r x y 2G
1A− 1 B C 2 2 4 ψ φ ()z= pz +, () z = pz ++ A=2 RBRC , = , =− R 41z 2 1 z z 3
2 θ8G κ =−+ θ ()κ rR + 2 ++− R ur 1 2cos2 1 12 cos2 Rp1 Rr r 2 8G=− 2 rR + 2 θ −−κ R uθ 12 sin 2 Rp1 Rr r
43 Circular opening in an infinite medium under biiaxial remote loading
2 2 4 p2 σ =1() + −+R 1() − −+ 43 R R θ r pp121 pp 12 1 cos2 2r2 2 r 2 r 4 y 2 4 σ =1() + +−R 1() − + 3 R θ θ pp121 pp 12 1 cos2 2r2 2 r 4 R 2 4 p1 θ τ =−1() − + 2R − 3 R rθ p1 p 2 1 sin 2 x 2 r2 r 4 = −≤≤−σ At the wall for rRpp , 321θ 3 pp 12
Displacements:
2 8G θ =+()() κ κ −++− rR 2() rRR + 2 +− uppr 121 pp 12 2 12 cos2 R Rr Rrr 2 8G=−() −+ 2 rRR 2 θ −−κ uθ p1 p 2 12 sin 2 R Rr r
44 Circular opening excavated in an elastic isotropic medium under initial anisotropic stress = σ =σ p1 K 00; p 2 0
Initial stress state in polar coordinates 1 σ=0 ()()()1 +K −− 1 K cos 2 θσ r 2 0 0 0
0 1 σ=θ ()()()1 +K +− 1 K cos 2 θσ 2 0 0 0
0 1 τ=θ ()()1 −K sin 2 θσ r 2 0 0 R2 R2 R 4 θ σ σ σ =1() + −+ 1() − −+ 43 r K00112 K 00 11 2 4 cos2 2r 2 r r σ 0 R2 R 4 θ σ σ σ =1() + +−− 1() + 3 θ ²00K112 K 00 11 4 cos2 y 2r 2 r R2 θ σ σν σ = −() − z K002 K 0 1 0 cos 2 2 R σ r K0 0 R2 R 4 x θ σ τ =−1() − + 2 − 3 rθ K01 0 1 sin2 2 r2 r 4 45 1 σ R2 R 4 R 2 =0 () + +−() −−ν θ ur 1 K0 1 K 0 3 4(1 ) cos 2 2 2 G r r r 1 σ R4 R 2 =−() 0 +−ν θ uθ 1 K 0 2(1 2 ) sin 2 2 2 G r3 r
At the wall r = R u 1 σ r =0 ()()() +−− −ν θ ()1K0 1 K 0 3 4 cos 2 R2 2 G u 1 σ θ =−() 0 () −νθ 1K0 34sin2 R2 2 G
for θ = 0 for θ = π/2 σ u 0 σ r =()() −ν −−ν ur 0 ()21K0 12 =()() −ν−−ν ()21 12 K0 R2 G R2 G 1− 2 ν if K > , convergence of the opening wall 0 2(1− ν ) Convergence of the opening wall 1− 2 ν ifK < , divergence of the opening wall 0 − ν 2(1 ) 46 σ 0
y E = 620MPa ν=0.2 R σ σ = K0 0 0 5 MPa x = K0 0.75 Radial displacement (m)
Tangential Hoop stress (MPa) displacement (m)
Stress concentration at the sidewalls 47 σ 0
y E = 620MPa
ν R σ =0.2 K0 0 σ = x 0 5 MPa
Hoop stress (MPa)
= K0 0.8 = K0 1.2 Stress concentration at the sidewalls Stress concentration at the crown and invert48