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 Δ-400 ft, 4y-650 ft, and h = 60 ft. Reservoir fluid properties are B = 1 RBSTB and μ = 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.

Answer 1

Δx = 400ft; Δy = 650 ft; h = 60 ft; k = 270 md; φ = 0.27; B = 1 RB/STB; µ = 1 cp; P_bw = 3000 psia;

dp/dx|_bE = -0.2 psi/ft

Solution:

The gridlocks have same dimensions and properties. So T_1,2 = T_2,3 = T_x

T_x = β_c A_xk_x / (µBΔx) where β_c is transmissibility conversion factor = 0.001127

T_x = 0.001127 * (650*60)*270/(1*1*400) = 29.6683 STB/D-psi

T_x = 29.6683 STB/D-psi

q_sc1 = q_sc2 = q_sc3 = 0

For reservoir 1,

q_{sc bw-1} = [β_c A_xk_x / (µBΔx/2)]₁ [(p_bw - p₁ ) - γ( Z_bw - Z₁ )]

= [0.001127*650*60*270/(1*1*400/2)][(3000-p₁) -  γ*0] = 59.3366(3000-p₁) STB/D

q_{sc bw-1} = 59.3366(3000-p₁) STB/D

For reservoir 3,

  q_{sc bE-3} = [β_c A_xk_x / (µB)]₃ [dp/dx|_bE - γ(dZ/dx|_bE )]

= [0.001127*650*60*270/(1*1)] [-0.2-(γ*0)] = -2373.462 STB/D

q_{sc bE-3} = -2373.462 STB/D

The general flow equation for gridblock 1,

T_1,2(p₂-p₁)+q_{sc bw-1}+q_sc1 = 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,

  T_1,2(p₁-p₂)+T_2,3(p₃-p₂)+q_sc2 = 0

  29.6683(p₁-p₂) + 29.6683(p₃-p₂) +0 = 0

  29.6683p₁ - 59.3366p₂ + 29.6683p₃ = 0  ------------------------------------------------- 2

For gridblock 3,

T_2,3(p₂-p₃)+q_{sc bE-3}+q_sc3 = 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;

q_{sc bw-1} = 59.3366(3000-p₁) STB/D

= 59.3366(3000-2960) = 2373.464

q_{sc bw-1} = 2373.464 STB/D