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Engineering Fluids and Thermpdynamics Questions

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ENPE 450 WELL TESTING Ezeddin Shirif Petroleum Systems Engineering University of Regina WELL TESTING Scope of the Course: The course is designed to cover the fundamental concepts of well test interpretation and design for oil wells. WHAT ARE THE EXPECTED OUTCOMES? • • • • • Understand the importance of well testing in formation evaluation Understand well test types and objectives Develop basic understanding of well test analysis methods Use well testing to identify productivity problems Design of well tests • • • MAIN OBJECTIVES OF WELL TESTING Reservoir Evaluation Obtain formation properties such as permeability and initial pressure Reservoir Management Use well testing to monitor well condition and average reservoir pressure Reservoir Description Infer geologic features which influence pressure behavior FORMATION EVALUATION Main task is to define the basic properties of a reservoir. • Reservoir limits: Drainage area How large is the reservoir? • Storage capacity: ϕ h Pore volume • Hydrocarbon contents: Sw • Productivity: kh, pi What will be the flow rate, Q? • Economic value APPRAISAL (EXPLORATION) WELL TESTING List of information desired from a well test in order of priority. • Rate determination (productivity) • Well condition (skin factor) • Collection of fluid samples (PVT) • Formation characteristics (k, pi, layering, fractures) • Reservoir boundaries (no flow, constant pressure, faults) DEVELOPMENT WELL TESTING List of information desired from a well test in order of priority. • Average reservoir pressure • Well condition (skin factor) • Formation characteristics (k, layering, fractures) STEPS OF WELL TESTING • • • Design of the well test Execution of the well test and data collection Analysis or interpretation of the well test WHY WELL TESTING? • • • Easy to perform Cheap to run for the amount of information that can be obtained In-situ measurement of properties DERIVATION OF THE DIFFUSIVITY EQUATION WELL TESTING CONCEPT During a well test, the pressure response of a reservoir to changing production (or injection) is monitored. Time Output Reservoir Pressure Rate Input Time KEY CONCEPTS TO BE USED • • • Conservation of mass: mass balance Darcy’s law: equation of motion (relationship between flow rate, formation and fluid properties and pressure gradient) Equation of state: relationship between compressibility, density and pressure DERIVATION OF THE DIFFUSIVITY EQUATION Basic assumptions: • • • • • • • • Single phase radial flow One dimensional horizontal radial flow Constant formation properties (k, h, ϕ) Constant fluid properties (µ, B, co) Slightly compressible fluid with constant compressibility Pore volume compressibility is constant (cf) Darcy’s law is applicable Neglect gravity effects RADIAL SYSTEM r+∆r r RADIAL SYSTEM r mass in mass out r+∆r volume element CONSERVATION EQUATION (Mas flow in) r – (Mass flow out) r+∆r = Mass accumulation M ass flow rate = Area × velocity × density × time Mass accumulation = change of mass over tim e ∆ t Bulk volum e b etween r and r+ ∆ r = π [(r+ ∆ r) 2 -r 2 ] h = π [r 2 +2r∆ r+(∆ r) 2 -r 2 ] h 2 π r∆ r h (Mass flow rate in) r = [2π rh v ρ ]r ∆ t (Mass flow rate out) r+∆r = (2π rhv ρ ) r + ∆r ∆ t Mass accumulation = ( 2π r ∆ rhφρ )t + ∆t − ( 2π r ∆ rhφρ )t CONSERVATION EQUATION (Mas flow in) r – (Mass flow out) r+ ∆ r = Mass accumulation ( 2π rhv ρ ) r ∆ t − (2π rhv ρ ) r + ∆ r ∆ t = ( 2π r ∆ rhφρ )t + ∆t − ( 2π r ∆ r hφρ )t Divide both sides by 2 π r∆ rh ∆ t (φρ )t + ∆t − (φρ )t 1 ( rv ρ ) r – (r v ρ ) r + ∆ r [ ]= ∆t ∆r r 1 ∂ ( rv ρ ) ∂ (ϕ ρ ) − = ∂t r ∂r The above equation is call the continuity equation. DARCY’S LAW k ∂p µ ∂r Substitute Darcy’s law into the continuity equation ∂p ∂ (ϕρ ) 1 k ∂ [r ρ ]= ∂r ∂t r µ ∂r Expand the derivat ives of the products of terms ∂ ∂p ∂ρ ∂p ∂ρ ∂ϕ 1 k +ρ [ρ (r ) + r ( )( )] = ϕ ∂r ∂r ∂r ∂r ∂t ∂t r µ Apply the chain rule to th e derivatives of ρ and ϕ 1 k ∂ ∂p ∂ρ ∂p 2 ∂ρ ∂ϕ ∂p [ρ (r ) + r ( )( ) ] = [ϕ ] +ρ r µ ∂r ∂r ∂p ∂r ∂p ∂p ∂t v=− EQUATION OF STATE The relationship between compressibility, density and pressure is given by: 1 ∂ρ co = ρ ∂p The definition of pore volume compressibility is: 1 ∂ϕ cf = ϕ ∂p Use the above equations to replace the derivatives of ρ a nd ϕ 1 k ∂ ∂p ∂p ∂p [ρ (r ) + rco ρ ( ) 2 ] = [ϕ co ρ + ϕ c f ρ ] r µ ∂r ∂r ∂r ∂t DIFFUSIVITY EQUATION Divide both sides of the above equation by k ρ and multiply by µ to obtain: 1 ∂ ∂p ∂p 2 ϕ µ ct ∂p (r ) + co ( ) = ∂r r ∂r ∂r k ∂t ct = co + c f The above equation is called the diffusivity equation. It is non-linear partial differential equation due to the ∂p 2 term co ( ) . To be able to solve it using analytical ∂r techniques it needs to be linearized. LINEARIZATION OF THE DIFFUSIVITY EQUATION To linearize the diffusivity equation we need to make the assumption that the pressure gradients are small and use the assumption that fluid compressibility is small. ∂p 2 Therefore the term co ( ) is small compared to other ∂r terms and can be neglected. The linearized diffusivity equation becomes: 1 ∂ ∂p ϕ µ ct ∂p (r ) = r ∂r ∂r k ∂t OIL FIELD UNITS k = permeability (md) h = formation thickness (ft) ϕ = porosity (fraction) µ = fluid viscosity (cp) r = radial distance (ft) rw = wellbore radius (ft) t = time (hours) p = pressure (psig or psia) pi = initial reservoir pressure (psig or psia) q = flow rate (STB/D) B = formation volume factor (res. bbl/STB) ct = total system compressibility (psi-1) DIFFUSIVITY EQUATION IN OIL FIELD UNITS ϕ µ ct 1 ∂ ∂p ∂ 2 p 1 ∂p ∂p (r ) = 2 + = −4 ∂r r ∂r ∂r r ∂r 2.637 ×10 k ∂t The above equation is second order (highest derivative with respect to space is two) linear partial (it has derivatives with respect to two independent variable r and t) differential equation. To solve it we need two boundary conditions and an initial condition. BOUNDARY CONDITIONS Rate 1. The flow rate at the wellbore is given by Darcy’s law: kh ∂p q= (r ) r=rw at r = rw for t ; 0 . 141.2 B µ ∂ r q ∂p 141.2qB µ (r ) r=rw = ∂r kh Time p = p i as r → ∞ for any value of any time. Pressure 2. If we consider a system large enough so that it will behave like an infi nite acting system, then far away from the wellbore the pressure will remain at p i . pi Radius INITIAL CONDITION Pressure Since the promlem under investigation is time dependant, the initial pressure should be known at time t = 0. Using uniform pressure distribution throughout the system at t = 0. p(r,0) = pi at t = 0 for all values of r . pi Radius DIMENSIONLESS VARIABLES T o m ak e th e eq u atio n an d its so lu tio n m o re g en eral fo r an y flu id an d reservo ir system it, is m o re co n ve n ien t to ex p ress it in d im en sio n less varialb les. D im e n sio n less p r essu r e: kh ( p i − p ) 1 4 1 .2 q B µ D im en sio n less tim e: pD = tD 2 .6 3 7 × 1 0 − 4 kt = ϕ µ c t rw2 D im en sio n less rad iu s: r rD = rw DIMENSIONLESS VARIABLES Differential equation: ∂ 2 pD 1 ∂pD ∂pD + = 2 rD ∂rD ∂rD ∂t D Boundary conditions: ∂p D (rD ) rD =1 = -1 ∂rD (1) (2) p D = 0 as rD → ∞ for any value of t D (3) Inital condition: p D (rD ,0) = 0 at t D = 0 for all values of rD ( 4) SOLUTION OF THE DIFFUSIVITY EQUATION DIMENSIONLESS EQUATIONS Differential equation: ∂ 2 pD 1 ∂pD ∂pD + = 2 rD ∂rD ∂rD ∂t D Boundary conditions: ∂p D (rD ) rD =1 = -1 ∂rD (1) (2) p D = 0 as rD → ∞ for any value of t D (3) Inital condition: p D (rD ,0) = 0 at t D = 0 for all values of rD ( 4) SOLUTION OF THE DIFFUSIVOTY EQAUTION To be able to solve the partial differential equation (PDE) we need to convert it to an ordinary differ ent ial equation ( ODE). Boltzmann Transformation: rD2 Let η = 4t D 2rD rD ∂η = = ∂ rD 4t D 2t D rD2 η ∂η = − 2 =∂t D 4t D tD BOLTZMANN TRANSFORMATION 1 ∂pD 1 ∂pD ∂η 2η ∂pD 1 ∂pD = = 2 = rD ∂rD rD ∂η ∂rD rD ∂η 2t D ∂η ∂ 2 pD ∂ 2 pD ∂η 2 ∂pD ∂ 2η ∂ ∂pD = ( )= ( ) + 2 2 ∂rD ∂rD ∂rD ∂η ∂rD ∂η ∂rD2 ∂ 2 pD rD2 ∂pD 1 η ∂ 2 pD 1 ∂pD = + = + 2 2 2 ∂η 4t D ∂η 2t D t D ∂η 2t D ∂η ∂pD ∂pD ∂η η ∂pD = =− t D ∂η ∂t D ∂η ∂tD BOLTZMANN TRANSFORMATION η ∂ pD 1 ∂pD 1 ∂pD η ∂pD + + =− 2 2t D ∂η 2t D ∂η t D ∂η t D ∂η 2 η ∂ 2 pD 1 ∂pD η ∂pD + + =0 2 t D ∂η t D ∂η t D ∂η Multiply the equation by t D and divide by η : d 2 pD 1 dpD + (1 + ) =0 2 η dη dη Second order ordinary differential equation. (5) BOLTZMANN TRANSFORMATION Boundary condition, Eqn (2) yields: 2 rD2 ∂ p D ∂p D ∂pD ∂η ∂pD rD = rD = = 2η = −1 4t D ∂η ∂ rD ∂ η ∂ rD ∂η If we assume that the well can be represented by a li ne source (a w ell with very small radius) we get: li m (rD rD → 0 ∂p D ) = -1 ∂ rD ∂p D 1 lim ( η )=− η →0 2 ∂η Boundary condition, Eqn (3) and the initial ( 6) condition, Eqn (4) g ive: p D = 0 as η → ∞ (7) SOLUTION OF THE ODE EQUATION To solve the second order differential equation we need to reduce it to a first order differential equation. Reduction of order: dpD Let y = = pD’ dη dpD2 d dp D d dy ‘ = ( )= ( pD )= 2 dη dη dη dη dη Second order ODE ⇒ d 2 pD 1 dpD + (1 + ) =0 2 dη η dη ⇒ First order ODE dy 1 + (1 + ) y = 0 dη η SOLUTION OF THE ODE EQUATION S ep arate th e variab les: dy 1 = − (1 + ) d η y η In teg rate b o th sid es o f th e eq u ati o n : l n ( y ) = −η − ln (η ) + c ln ( p D’ ) = −η − l n (η ) + c ln (η p D’ ) = −η + c η p D’ = e − η + c = e c e − η dpD = c1 e − η η dη SOLUTION OF THE ODE EQUATION U sing condition (6): 1 c1 = − 2 dp D 1 −η =− e η 2 dη dp D 1 −η = − ηe 2 dη S eparate the variab le s :
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