Test_1.key.2004 - PGE 323 Test #1 Key February 20, 2004...

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Unformatted text preview: PGE 323 Test #1 Key February 20, 2004 Pope 1 (a) The mass flux in each direction is given as follows: For x-direction: z uz [ Mass In] = t ( uxyz ) x, y, z [ Mass Out ] = t ( uxyz ) x + x , y, z For y-direction [ Mass In] = t ( uyzx ) x , y, z uy y u x x [ Mass Out ] = t ( uyzx ) x , y + y, z [ Mass In] = t ( uzxy ) x , y, z [ Mass Out ] = t ( uzxy ) x, y, z + z For z-driection Mass in - Mass out = Accumulated Mass xyz [( ) t + t - ( ) t ] = t{[( uxzy ) x , y, z - ( uxzy ) x + x , y, z ] + [( uyxz ) x , y, z - ( uyxz ) x , y + y, z ] + [( uzxy ) x , y, z - ( uzxy ) x , y, z + z ]} .(1) Dividing both the sides by xyzt and taking limits as x, y, z and t go to zero: () (ux ) (uy) (uz ) = - + + t x x x ----------------------(2) From Darcy's law: kx ux = - ----------------------------------------------(3) x ky uy = - ----------------------------------------------(4) y kz uz = - ----------------------------------------------(5) z Substituting equations (3), (4) and (5) into equation (2), gives the diffusivity equation 1.5-3 in the course notes: ky kx kz ( x ) ( y ) ( z ) () ------------------(6) = + + t x y z (b) Assuming steady state flow, equation (6) simplifies to: ky kx kz ( x ) ( y ) ( z ) 0= + + x y z Assuming constant density and viscosity, gives: ) (kz ) ) ( ky (kx y x + z 0= + y z x Assuming homogeneous and anisotropic medium (that is kx, ky, and kz are constant but not equal) and using the following transform: x X = kx y Y = ky Z = z kz We get, ) (kz ) (ky ) (kx (Y` ky ) (X` kx ) ( Z` kz ) 0= + + (X` kx ) ( Y` ky ) ( Z` kz ) Since kx, ky, kz are constant, they get cancelled out and we get the Laplace equation: 2 2 2 + + =0 X`2 Y`2 Z`2 (equation 1.5-46 of course notes). 2. Start with the potential function for line sources and sinks: =- 4h k x k y qi ln[(x - xi )2 + k x ( y - yi )2 ] + C y k k vx = - x x k y vy = - y =- x 2h k x k y q i (x - x i ) k (x - x i ) 2 + x (y - yi ) 2 ky k qi x ( y - yi ) ky =- k y 2h k x k y (x - x i ) 2 + x ( y - yi ) 2 ky vx = 1 kx qi (x - x i ) k 2h k y ( x - x i ) 2 + x ( y - yi ) 2 ky vy = kx 1 2h k y qi ( y - yi ) k (x - x i ) 2 + x ( y - yi ) 2 ky 3. The locations of wells: Injector (0, 0, 0); Producer 1: (-1000, -500, -100); Producer 2: (300, 200, 800) m3 D m3 Producer Rate: q1 = 2 m3 D = 2 = 2.3E - 5 D 86400 s s 3 m D m3 Injection rates: q2 = q3 = -1 m3 D = -1 = -1.16 E - 5 D 86400 s s 2 Reservoir permeability: k x = k y = k z = k = 1D 1E - 12 m a) ( x, y , z ) = 4 k qi x - xi ) + ( y - yi ) + ( z - zi ) + C ( i =1 2 2 2 1 2 -2 3 - 1 2 b) vx = - kx x 3 i =1 q1 ( x 2 + y 2 + z ) + 1 2 2 2 -2 = q ( + 2 x - x2 ) + ( y - y2 ) + ( z - z2 ) + C 4 k 1 2 2 2 -2 ( q3 x - x3 ) + ( y - x3 ) + ( z - z3 ) 1 2 2 2 -2 2.3E - ( x + y + z ) + 5 1 2 2 2 -2 = + ( ( -1.16 E - 5 ) x + 1000 ) + ( y + 500 ) + ( z + 100 ) + C 4 k 1 - x - 300 ) 2 + ( y - 200 ) 2 + ( z - 800 ) 2 2 ( ( -1.16 E - 5 ) 1 = 4 qi ( x - xi ) x - xi ) + ( y - yi ) + ( z - zi ) ( 2 2 2 - 3 2 3 - 2.3E - 5* x * ( x 2 + y 2 + z 2 ) 2 3 1 2 2 2 -2 = ( ( -1.16 E - 5 ) * ( x + 1000 ) * x + 1000 ) + ( y + 500 ) + ( z + 100 ) 4 3 2 2 2 -2 ( ( -1.16 E - 5 ) * ( x - 300 ) * x - 300 ) + ( y - 200 ) + ( z - 800 ) Similarly, we can get velocities in y and z directions as follows: 3 - 2.3E - 5* y * ( x 2 + y 2 + z 2 ) 2 3 1 2 2 2 -2 vy = ( ( -1.16 E - 5 ) * ( y + 500 ) * x + 1000 ) + ( y + 500 ) + ( z + 100 ) 4 3 2 2 2 -2 ( ( -1.16 E - 5 ) * ( y - 200 ) * x - 300 ) + ( y - 200 ) + ( z - 800 ) 3 - 2.3E - 5* z * ( x 2 + y 2 + z 2 ) 2 3 1 2 2 2 -2 vz = ( ( -1.16 E - 5 ) * ( z + 100 ) * x + 1000 ) + ( y + 500 ) + ( z + 100 ) 4 3 2 2 2 -2 ( ( -1.16 E - 5 ) * ( z - 800 ) * x - 300 ) + ( y - 200 ) + ( z - 800 ) c) Substitute location (100, 100, 100) into the velocity expression in x direction: 3 - 2.3E - 5*100* ( 1002 + 1002 + 1002 ) 2 3 1 2 2 2 -2 vx = ( ( 100 -1.16 E - 5 ) * ( 100 + 1000 ) * + 1000 ) + ( 100 + 500 ) + ( 100 + 100 ) 4 3 2 2 2 -2 ( 100 ( -1.16 E - 5 ) * ( 100 - 300 ) * - 300 ) + ( 100 - 200 ) + ( 100 - 800 ) 1 = * 4.426 E - 10 4*3.14*0.1 =3.524E-10 m s d) To make this calculation, evaluate 1 from part a at a distance R from the producer and 2 at a distance R from the injector to get: = 1 - 2 1 : x=0.1m, y=0 m, z=0. m 2 : x= -1000.1 m, y= -500 m, z= -100 m 1 - 2.3E - 5 ( 0.12 + 02 + 02 ) 2 + 1 - + 1000 ) 2 + ( 0 + 500 ) 2 + ( 0 + 100 ) 2 2 1 - 2 = ( ( 0.1 -1.16 E - 5 ) 4 k 1 2 2 2 - -1.16 E - 5 ) - 300 ) + ( 0 - 200 ) + ( 0 - 800 ) 2 0 ( ( + 1 2 2 2 - 2.3E - ( - 1000.1) + ( - 500 ) + ( - 100 ) 2 + 5 1 2 2 2 -2 - ( (- -1.16 E - 5 ) 1000.1 + 1000 ) + ( -500 + 500 ) + ( -100 + 100 ) + 4 k 1 2 2 2 -2 (- ( -1.16 E - 5 ) 1000.1 - 300 ) + ( -500 - 200 ) + ( -500 - 800 ) ( ) = 1.0E-3 4*3.14*1E-12 ( 2.3E - 4 + 1.16 E - 4 ) =2.75E+4 Pa e) The locations of real and image wells, unit in meters: Real injector: (0, 0, 0), Corresponding image injector: (0, 0, 2000) Sink 1: ( -1000, -500, -100), Corresponding image sink: ( -1000, -500, 2100) Sink 2: (300, 200, 800), corresponding image sink: (300, 200, 1200) 1 2 2 2 -2 2.3E - 5 x + y + z + 1 2 2 2 -2 + ( - 1.16E - 5) * ( x + 1000) + ( y + 500) + ( z + 100 ) 1 2 2 2 -2 ( - 1.16E - 5) * ( x - 300) + ( y - 200 ) + ( z - 800 ) + ( x , y, z ) = +C 1 4k - 2.3E - 5 x 2 + y 2 + ( z - 2000 ) 2 2 + 1 - ( - 1.16E - 5) * ( x + 1000 ) 2 + ( y + 500 ) 2 + ( z - 2100) 2 2 + 1 ( 2 2 2 -2 - 1.16E - 5) * ( x - 300) + ( y - 200 ) + ( z - 1200 ) [ [ ( ) [ [ [ ] 3 - 2.3E - 5 z x 2 + y 2 + z 2 2 + 3 2 2 2 -2 + ( - 1.16E - 5) * ( z + 100 ) ( x + 1000 ) + ( y + 500 ) + ( z + 100 ) 3 2 2 2 -2 1 ( - 1.16E - 5) * ( z - 800) ( x - 300 ) + ( y - 200 ) + ( z - 800 ) + z = 3 4 2 -2 2 2 2.3E - 5 ( z - 2000 ) x + y + ( z - 2000 ) + 3 - ( - 1.16E - 5) * ( z - 2100 ) ( x + 1000 ) 2 + ( y + 500 ) 2 + ( z - 2100 ) 2 2 + 3 ( 2 2 2 -2 - 1.16E - 5) * ( z - 1200) ( x - 300 ) + ( y - 200 ) + ( z - 1200 ) [ [ ( ) [ [ ] [ ] at z=1000 m, we have 3 2 2 2 -2 2.3E - 5 1000 x + y + 1000 + 3 2 2 2 -2 + ( - 1.16E - 5) * (1000 + 100) ( x + 1000 ) + ( y + 500) + (1000 + 100 ) 3 2 2 2 -2 1 ( - 1.16E - 5) * (1000 - 800 ) ( x - 300 ) + ( y - 200 ) + (1000 - 800 ) + z =1000 = 3 4 - 2.3E - 5 (1000 - 2000 ) x 2 + y 2 + (1000 - 2000) 2 2 + 3 - ( - 1.16E - 5) * (1000 - 2100) ( x + 1000 ) 2 + ( y + 500) 2 + (1000 - 2100 ) 2 2 + 3 ( 2 2 2 -2 - 1.16E - 5) * (1000 - 1200 ) ( x - 300) + ( y - 200 ) + (1000 - 1200 ) [ [ ( ) [ [ ] [ ] 3 2 2 2 -2 2.3E - 5 1000 x + y + 1000 + 3 2 2 2 -2 + ( - 1.16E - 5) * (1000 + 100 ) ( x + 1000) + ( y + 500 ) + (1000 + 100 ) 3 - 1 ( - 1.16E - 5) * (1000 - 800 ) ( x - 300) 2 + ( y - 200 ) 2 + (1000 - 800 ) 2 2 + = 3 4 2 -2 2 2 2.3E - 5 (1000 - 2000 ) x + y + (1000 - 2000 ) + 3 - ( - 1.16E - 5) * (1000 - 2100 ) ( x + 1000) 2 + ( y + 500 ) 2 + (1000 - 2100 ) 2 2 + 3 ( 2 2 2 -2 - 1.16E - 5) * (1000 - 1200) ( x - 300 ) + ( y - 200 ) + (1000 - 1200) [ [ ( ) [ [ ] [ ] =0 ...
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This note was uploaded on 05/15/2011 for the course PGE 323 L taught by Professor Johns during the Spring '10 term at University of Texas at Austin.

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