ABET CRITERION (E, K) 7-14 A single-phase fluid reservoir is described by three
ID: 2074554 • Letter: A
Question
ABET CRITERION (E, K) 7-14 A single-phase fluid reservoir is described by three equal gridblocks as shown in Fig. 7-16. The reservoir is horizontal and has homogeneous and isotropic rock properties, k-270 md and 0.27. Gridblock dimensions are -400 ft, 4y-650 ft, and h = 60 ft. Reservoir fluid properties are B = 1 RBSTB and = 1 cp. The reservoir left w boundary is kept constant at 3000 psia, and the reservoir right boundary is kept at a pressure gradient of -0.2 psi/ft. Two 7-in vertical wells were drilled at the center of gridblocks 1 and 3. The well in gridblock I injects 300 STB/D of fluid and the well in gridblock 3 produces 600 STB/D of fluid. Both wells have zero skin. Assume that the reservoir rock and fluid are incompressible. Find the pressure distribution in the reservoir es 2 'sc3 '-600 STB/D qsc1-300 STB/D 60 ft A,=3000 psia 400 ft 400 ft Fig. 7-16. Discretized ID reservoir in Exercise 7-14.Explanation / Answer
x = 400ft; y = 650 ft; h = 60 ft; k = 270 md; = 0.27; B = 1 RB/STB; µ = 1 cp; Pbw = 3000 psia;
dp/dx|bE = -0.2 psi/ft
Solution:
The gridlocks have same dimensions and properties. So T1,2 = T2,3 = Tx
Tx = c Axkx / (µBx) where c is transmissibility conversion factor = 0.001127
Tx = 0.001127 * (650*60)*270/(1*1*400) = 29.6683 STB/D-psi
Tx = 29.6683 STB/D-psi
qsc1 = qsc2 = qsc3 = 0
For reservoir 1,
qsc bw-1 = [c Axkx / (µBx/2)]1 [(pbw - p1 ) - ( Zbw - Z1 )]
= [0.001127*650*60*270/(1*1*400/2)][(3000-p1) - *0] = 59.3366(3000-p1) STB/D
qsc bw-1 = 59.3366(3000-p1) STB/D
For reservoir 3,
qsc bE-3 = [c Axkx / (µB)]3 [dp/dx|bE - (dZ/dx|bE )]
= [0.001127*650*60*270/(1*1)] [-0.2-(*0)] = -2373.462 STB/D
qsc bE-3 = -2373.462 STB/D
The general flow equation for gridblock 1,
T1,2(p2-p1)+qsc bw-1+qsc1 = 0
29.6683(p2-p1) + 59.3366(3000-p1) + 0 = 0
29.6683p2 - 29.6683p1 + 178009.8 - 59.3366p1 = 0
-89.0049p1 + 29.6683p2 + 178009.8 = 0 ----------------------------------------------- 1
For gridblock 2,
T1,2(p1-p2)+T2,3(p3-p2)+qsc2 = 0
29.6683(p1-p2) + 29.6683(p3-p2) +0 = 0
29.6683p1 - 59.3366p2 + 29.6683p3 = 0 ------------------------------------------------- 2
For gridblock 3,
T2,3(p2-p3)+qsc bE-3+qsc3 = 0
29.6683(p2-p3) -2373.462 + 0 = 0
29.6683p2 - 29.6683p3 - 2373.462 = 0 -------------------------------------- 3
Solving equation 1, 2 and 3 gives
p1 = 2960 psia;
p2 = 2880.0002 psia;
p3 = 2800.0003 psia;
qsc bw-1 = 59.3366(3000-p1) STB/D
= 59.3366(3000-2960) = 2373.464
qsc bw-1 = 2373.464 STB/D
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