Page 38 - Fundamentals of Reservoir Engineering

CONTENTS XXXVIII

p = p e = constant, at r = r e (5.14) 132
∂ p
and = 0 for all r and t (5.15) 132
t ∂

1 ∂ k ∂ p k ∂ ρ ∂ p kρ ∂ p kρ ∂ 2 p ∂ p
r ρ + r + + r 2 = φ cρ (5.16) 133
r ∂ r µ
r ∂ µ r ∂ r ∂ µ r ∂ µ r ∂ t ∂

∂ p ∂ ρ
cρ = (5.17) 133
r ∂ r ∂

1 ∂ k ∂ p k ∂ p 2 kρ ∂ p kρ ∂ 2 p ∂ p
r ρ + c r ρ + + r = φ cρ (5.18) 133
r ∂ r µ
r ∂ µ r ∂
µ r ∂ µ r ∂ 2 t ∂

∂ 2 p + 1 p = φµ c ∂ p (5 19) 134
∂
r ∂ 2 r ∂ r k t ∂
1 ∂ ∂ p = φµ c ∂ p (5.20) 134
r

r ∂ r r ∂ k t ∂

cp << 1 (5.21) 134
(5.22) 134
c t = c o S o + c w S wc + c f

φ ab s ol ut e × (c o S o + c w S wc + c f ) (5.23) 134

(c S + c S + c )
φ absol ute (1 − S wc) × o o w wc f (5.24) 135
(1 S )
−
wc
cV (p i − p) = qt (6.1) 136

1 ∂ ∂ p qµ
r =− (6.2) 137
2
rr ∂ r ∂ π rkh
e
µ
∂ p q r 2
=− + C 1 (6.3) 137
2
π
t ∂ 2r kh
e
∂ p qµ 1 r
= − 2 (6.4) 137
π
r ∂ 2 kh r r e
2
p r qµ r r
2
p p wf = 2kh lnr − 2 r e r w (6.5) 137
π
2
qµ r r
p − p wf = ln − 2 (6.6) 137
r
2kh r w 2r e
π
qµ r 1
p − p wf = ln e − + S (6.7) 137
e
2kh r w 2
π