PGE 323 Test _1.key

PGE 323 Test _1.key - Key PGE 323 Test #1 February 14, 2006...

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Unformatted text preview: Key PGE 323 Test #1 February 14, 2006 1. 30 pts (a) 20 pts Starting with: rrz t +t - rrz t = z ( u r ) + rr ( u z ) + rz ( u ) t r r z t - z ( u r ) r + rr ( u z ) + rz ( u ) r +r z +z + Divide through by r rz and t rrt u z ) - ( u z ) rrz ) - rzt u r ) - ( u r ) ( ( r ( z t +t ( ) t r +r z +z = + rrzt rrzt rrzt rzt u ) - ( u ) ( + + rrzt Simplify and pull out a (-1) from each term on the right hand side ) u - - ru r ) u z ) ( (r ( z +z - ( u z ) z t +t ( ) t - 1 r ) r +r ( r = - t r r z ( 1 u ) + - ( u ) - r Take the limit of each term as the denominator . 0 ( ) X% - t 1 = - r ( r { u r } ) + ( u z ) + 1 ( u ) r z r (b) 10 pts Darcy's Law for Cylindrical Coordinates (neglecting gravity): k ur = - r k uz = - z k u = - .P r .P z P r ( ) Inserting into the final equation from (a) ( ) = - -(( rk r P k z P 1 k 1 R - + - + - t r r z z r F r Ideal gas law: = PM RT P ( r ) Inserting ideal gas law into above: PM R 2 1 1 RT - rPMk r P PMk z P PMk = - + - + - t r r z RT z RTF r r RT The term M cancels out, leaving RT ( P ) = - ((2 rPk r P Pk z P Pk 1 R 1 - + - + - t r r z z F r r . P 1 P2 = r 2 r P ( r ) P ( r ) Now put in terms of pressure squared as done in the course notes: P P 1 P2 = t 2P t Final Solution: 2 rk P 2 k P 2 k P 2 1 P 1R2 1 = - r - + - + - z 2 P t r r z z F r r ( ) 2. 40 pts (a) 15 pts Checked to make sure that the sketches had 2 wells located at different depths, with the first and last perforations labeled, well type indicated, and well rate indicated. (x1 ,y1 ,z1 ) z injector (q) (x11 ,y11 ,z11 ) producer (-q) y (x10 ,y10 ,z10) x (x20 ,y20 ,z20 ) The particular form of the potential equation is: = -x ) (x i (y - yi ) (z - zi ) - qi + + ky kz 4 k x k y k z i kx 2 2 1 - 2 2 where qi = xi = yi = zi = ;i . 10 . q 10 - q 10 ; 20 i > 10 xI xP yI yP ;i 10 ; 20 > 10 i ;i 10 ; 20 > 10 i (i - 1) L 10 + z1 ;i 10 (i - 11) L 10 + z11 ; 20 > 10 i (b) 5 pts Checked sketch to see if there were pathlines that reached the producer well and others that didn't, and if the front accurately followed the pathlines. (c) 10 pts The magnitude of the velocity equation is: 2 v = v2 + v2 + vz x y The velocities vx, vy, and vz are calculated at some point R away from the top perforation. The coordinates of R are said to be (xR,yR,zR) and the location of the top perforation is (x11,y11,z11). This means that the equations for the 3 velocities are: 3 -q - 2 2 2 2 x - x ) ( ( y -y ) (z -z ) 10 vx = ( x R - x11 ) R 11 + R 11 + R 11 ky kz 4 k x k y k z kx 3 -q - 2 2 2 2 x - x ) ( ( y -y ) (z -z ) 10 vy = ( yR - y11 ) R 11 + R 11 + R 11 ky kz 4 k x k y k z kx 3 -q - x - x ) 2 ( y - y ) 2 ( z - z ) 2 2 ( 10 vz = ( z R - z11 ) R 11 + R 11 + R 11 ky kz 4 k x k y k z kx (d) 10 pts The magnitude of the darcy flow along a streamline is calculated by k . u =- s Rearrangement gives u =- s k Since u = v The potential gradient along the streamline can be calculated using the result from (c) in the equation v =- s k 3. 10 pts Starting equation: = - 2 k 2 qi ln x - x i ) + x ( y - yi ) C + ( ky 4h k x k y i Differentiate with respect to x: 2 ( x - xi ) - = qi k x 4h k x k y i ( x - x i ) 2 + x ( y - yi ) 2 ky Definition of vx is: vx = -k x x Combining the two equations gives: 2 ( x - xi ) -k x - vx = qi 2 kx 2 h k x k y i 4 ( x - x i ) + ( y - yi ) ky Simplification gives the final solution: vx = kx ( x - xi ) 1 qi k 2h k y i ( x - x i ) 2 + x ( y - yi ) 2 ky 4. 20 pts a) 6 pts k . ux = - x x ky uy = - y k uz = - z z b) 4 pts Darcy's Law is restricted to use in situations when the Reynolds Number is below 1. c) 4 pts In the differential form, Darcy's Law can be applied to a wide range of problems with varying boundary conditions. Once the boundary conditions are specified, however, the cases for which the law can be applied are reduced. Starting with the equation in differential form means that we are able to correctly apply the equation to our specific set of boundary conditions. d) 6 pts Variable k q A (P) L u Darcy Units Darcy cP cm3/sec cm2 atm cm cm/sec SI Units m2 Pa-sec m3/sec m2 Pa m m/sec Field Units milliDarcy cP Barrels/day ft2 psia ft ft/day ...
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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.

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