Page 40 - Fundamentals of Reservoir Engineering
P. 40
CONTENTS XL
qµ 1 4A
pp wf = ln + S (6.22) 144
−
2kh 2 γ C r 2
π
Aw
a) p = p at t = 0, for all r
i
b) p = p at r = ∞ , for all t (7.1) 150
i
∂ p qµ
c) lim r = , for t > 0
π
r → 0 r ∂ 2 kh
∂ s φµ cr
= (7.2) 150
t ∂ 2k t
∂ s φµ cr 2
= 2 (7.3) 150
r ∂ 4k t
dp′
p′ + s = − sp′
ds (7.4) 151
dp′ (s1 )
+
=− ds
p′ s
e − s
p′ = C 2 s (7.5) 151
∞ − s
qµ e
p = p − ds (7.6) 152
r,t
i
4kh s
π
φµ cr 2
x =
4k t
∞ e − s ∞ e − s
s ds = s ds (7.7) 152
x φµ cr 2
x =
4kt
ei(x) ≈− ln x − 0.5772 (7.8) 152
ei(x) ≈− ln (γ ) x for x < 0.01 (7.9) 153
qµ 4kt
p = p − ln + 2S (7.10) 153
wf
i
4kh γ φ µ cr w 2
π
2
qµ φ µ cr
p = p − ei (7.11) 154
i
r,t
4kh 4k t
π
cAhφ ( i ) p = qt (7.12) 156
p −
qµ 4A kt
p = p − ½ ln + 2π + S (7.13) 156
wf
i
2kh γ Cr 2 φ µ cA
π
Aw
