A(0.02in ) = 100 ×
π 4
× (0.02) 2 = 3.142 × 10 − 2 in 2
ID = 0.04in(number= 50) K (0.04in ) = 20 × 10 6 × (0.04) 2 = 32000darcy
A(0.04in) = 50 ×
k avg =
π 4
× (0.04) 2 = 6.283 × 10 −2 in 2
(8000 × 3.142 × 10 −2 ) + (32000 × 6.283 × 10−2 )
π
4
× (2) 2
= darcy
PROBLEM 7.14 Suppose, after cementing, an opening 0.01 in. wide is left between the cement and an 8 in. diameter hole. If this circular fracture extends from the producing formation through an impermeable shale 20 ft thick to an underlying water sand, at what rate will water enter the producing formation (well) under a 100 psi pressure drawdown? The water contains 60,000 ppm salt and the bottom-hole temperature is 150°F.
ANSWER:
q = 8.7 × 109
w2 Ac ( p1 − p2 ) μBL (
q = 8.7 × 10
9
8 º 0.01 2 ª 8.01 2 ) − ( ) 2 » × 100 ) × π × «( 12 ¼ 12 ¬ 12 = 215BPD (0.49)(1)(20)
PROBLEM 7.15 A high water-oil ratio is being produced from a well. It is thought that the water is coming from an underlying aquifer 20 ft from the oil producing zone. In between the aquifer and the producing zone is an impermeable shale zone. Assume that the water is
120