Unformatted text preview: Homework #7 Key PGE331 20 points 1. Section 7-6.2 makes the distinction between small, constant oil compressibility and just oil constant compressibility. The "small" compressibility assumption linearizes the continuity equation and makes subsequent solution much easier than just "constant" compressibility. However, the constant compressibility case, Eq. 7-56, may be solved for steady state flow. Work example 7-10. . The steady-state version of Eq. (7-56) is
2 1 0= + It's helpful to return to a more primitive form of this equation. Replace the compressibility with its definition to give 0= 1 1 + which becomes 0= + or 0= A We are to derive the solution p( r) that satisfies the boundary conditions p( rw ) = and p( re ) = a. A first integration of A gives C1 = where C1 is a constant to be determined below Inserting the constant compressibility definition for oil phase density = gives C1 ( - ) = r ( - ) The subscript R denotes a reference pressure. This equation integrates to ( - ) C1 ln r = + 2 B For simplicity and without loss of generality we take the reference pressure to be the external pressure from here on, or pR = Now to evaluate the boundary conditions ( - ) C1 ln rw = + 2 = - + 2 C1 ln re = + 2 Subtracting the first from the second gives r C1 lne - - or = w r 1- - C1 = and, using the first boundary condition, 1 - - = - + 2 Solving this for the second constant gives 1- - C2 = - - 1- - - - C C2 = C2 = - - ( - ) - 1- - - C2 = D Using C and D to eliminate the constants from B gives 1 - - ( - ) = + - - Solving this for the pressure involves several steps
- 1- - ( - ) - = + co ( p- ) e = 1- - - - - e co ( p- ) - - - + - = - - - - + c ( p- ) eo = = = Taking logarithms and solving for p gives the desired result - 1 p( r) = + b. To show this replace all of the exponentials by a truncated Taylor's series as E e co = 1 + E becomes - (1 + ) 1 p( r) = + (1 + ) After cancellation + )(- ) - + (1 1 p( r) = + (1+ ) + 1 p( r) = + (1+ ) Several small compressibility approximations follow. + 1 1 p( r) = + = + 1 + and finally 1 1 p( r) = + 1 + + = p( r) = - which is the result for small compressibility flow. c. Form the error term as e r (r ) = 100 ( - ) The error is a function of one dimensionless variable and two parameters or e r (r ) = ; , 2. Inflow into a fractured well more closely approximates linear flow rather than radial. The objective of this exercise is to gain further understanding about stabilized flow into a hydraulically fractured well through development of the appropriate solutions in Cartesian coordinates. a. Solve the diffusivity equation for stabilized flow in Cartesian coordinates. The constant terminal rate condition for this coordinate system is P qmo = x 0 k oA x= where x = 0 is the location of the fracture face; the external boundary in this case is at x = L. The drainage area is A = LW. Derive additionally the following Start with the diffusivity equation in Cartesian coordinates and its corresponding boundary conditions for stabilized flow. 1 2 = 2 0 = =0 0 = = where f 0 0 is semi-steady state and f 1 is steady state flow. (1) 0 (2a) 0 (2b) Start by writing Eq. (1) as two ordinary differential equations 02 2 = -0 (3a) 0 0 = -0 Integrating Eq. (3b) twice gives (3b) 0 2 (, ) = - 0 + 1 () + 2 () 2 (4) where C1 and C0 are (at most) function of t only. Differentiating this equations with respect to time gives 0 0 = 1 () + () = -0 2 The primes in this equation indicate total time derivatives. Since the first term to the right of the first equal sign is a function of x we must have 1 () = 0 0 or C1 is a constant. This leaves 0 () = -0 2 which integrates to 0 2 () = -0 + 3 where C3 is also a constant. Introducing these equations into Eq. (4) gives the general form 0 2 (, ) = - 0 + 1 - 0 + 3 2 (5) To determine the constants use the boundary conditions (2). The first gives 0 0 = (-0 + 1 ) = 1 = =0 =0 and the second 0 0 = = (-0 + 1 ) = = -0 + 1 = 0 The first of these give C1 directly. 1 = from which we have 0 0 = 1 - = (1- ) 0 from the second boundary condition. The constants put into Eq. (5) give the particular solution to stabilized flow in Cartesian coordinates. (, ) = 2 + 3 - ) (1- + 2 (6) As for radial flow, we can't determine the constant C3 because of the absence of an initial condition. 0 b. Give the following end-members: P(x) for steady state flow, and P(x,t) for semi-steady state flow. Starting with steady-state flow (, ) = + 3 = 1, Eq. (6) is For semi-steady state flow f = 0 leading to 0 2 + (, ) = - + 2 3 To eliminate the constant, express this as the difference between pressures at two known points. Specifically (0, ) = 3 - ) (1- from which Eq. (6) gives 0 (, ) = (0, ) + - (1- ) 2 2 0 (7) 0 The elimination removes the time dependency. You should recognize the f = 1 case as probably the first form of Darcy's law you were ever exposed to. c. Expressions for the average pressure, P , for steady state flow and, P(t), for semi-steady state flow. The manipulations are the same for both types of stabilized flow. Start with the definition of average pressure =
= =0 = After insertion of Eq. (7) and integrating, this gives 0 = (0, ) + - (1- ) 2 6 2 (8) You can derive the ss and sss limits by setting f to the appropriate value. 0 d. Definitions of productivity index (in terms of P ) for both steady state and semi-steady state flow. The PI definition should include a skin factor. The idea of a skin factor in linear flow follows from Eq. (7). It is natural that = (0, ) - (9) which, combining with Eq. (8), gives 0 = + 2 1+ 2 - (1- ) ( ) 2 6 (10) The s in the equations is the skin factor. 0 Equation (10) used in the definition of the productivity index is = - 2 2 6 1+ 2 - (1- ) ) ( 0 = (1- ) 2 - (1+ ) 3 2 (11) You can easily show that the PI based on the total pressure drop is 0 = (1- ) 2 - (1+ ) 3 0 ...
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- Fall '07
- Thermodynamics, Darcy, 0 K