Page 46 - Fundamentals of Reservoir Engineering

CONTENTS XLVI

()
1 ∂ kρ µΖ ∂ mp µΖ ∂ m () p
c
r = φρ (8.10) 248

rr ∂ µ 2p r ∂ 2p t ∂
()
1 ∂ r ∂ mp = φµ c ∂ m () p (8.11) 248
r ∂ r r ∂ k t ∂

()
∂ mp 2p ∂ p 2p q
= ⋅ = − ⋅ (8.12) 249
2
t ∂ µΖ t ∂ µΖ π rh c
φ
e
()
1 ∂ r ∂ mp =− 2 pq (8.13) 249
2
r ∂ r r ∂ π r kh Ζ res
e
()
1 ∂ r ∂ mp =− 2p q sc ⋅ T (8.14) 249
sc
2
r ∂ r r ∂ π rkh T sc
e
1422 QT r 3
(
mp mp wf ) = kh ln r w e − 4 + S (8.15) 250
() −
711 QT 4 .000264kt
(
mp i mp wf ) = ln + 2S (8.16) 250
() −
kh γ φ µ i w 2
( ) c r
( pp+ wf ) ( p p− wf ) p pdp
or equivalent to (8.17) 251
2 µΖ p wf µΖ

dp µ u (8.18) 252
dr = k

dp µ u 2 (8.19) 253
dr = k u βρ+

e r 2p q 2
() nD
∆ mp = βρ dr (8.20) 254
µΖ 2 π rh
w r
e r pq 2 β Tγ
∆ m () p = constant × g dr (8.21) 254
nD 22
w r ΖΤ µ rh
2
q
βγ gsc e r dr
Τ
()
∆ mp = constant × (8.22) 254
nD h 2 r µ 2
w r
2
q
βγ gsc 1 1
Τ
()
∆ mp = constant × − (8.23) 254
nD h 2 r w r e
Τ
βγ g Q 2
∆ m () p = 3.161 × 10 − 12 = FQ 2 (8.24) 254
2
nD hr
µ wp w