Page 35 - Fundamentals of Reservoir Engineering

CONTENTS XXXV

p
hg = ( + gz) (4.2) 102
ρ

dh
u = K (4.3) 102
dl

Kd p K d(hg)
u = + gz = (4.4) 102

gdl ρ g dl
p dp
Φ= + gz (4.5) 103
1atm ρ
−
p dp
−
Φ= + g(z z ) (4.6) 103
b
p b ρ
p
Φ= + gz (4.7) 103
ρ

kdΦ
ρ
u = (4.8) 103
µ dl
kρ dΦ
u =− (4.9) 104
µ dl

kρ dΦ
u = (4.10) 104
µ dr

k dp
u =− (4.11) 105
µ dl

ρ
kdΦ k dp dz
u = = − + ρ g (4.12) 107
µ dl µ dl dl

kdp ρ g dz
u =− + 6 (4.13) 107
µ dl 1.0133 10 dl
×
2
k(D) A(cm ) dp
q(cc / sec) =− (atm/ cm) (4.14) 107
µ (cp) dl
2
k(mD) A(ft ) dp
q(std/ d) =− (constant) (psi/ ft) (4.15) 107
µ (cp) dl

2
D 2 cm atm
kmD × A ft 2
stb r.cc / sec rb / d mD ft dp psi psi
q =− ×
d rb / d stb / d µ (cp) dl ft cm (4.16) 108

ft
D 1 cm atm 1
and since = ; = 30.48 and = ; equ.(4.16)
mD 1000 ft psi 14.7