Page 43 - Fundamentals of Reservoir Engineering

CONTENTS XLIII

0.0189 (3500 p ) = α − ½ p D(MBH) (t ) (7.47) 185
−
DA
wf
kh t +∆ t 4t
t
7.08 10 -3 (p − p ) 1.151 log + p () − ½ ln D (7.48) 189
=
×
qB i ws(LIN) t ∆ D D γ
µ
o
kt
t DA = 0.000264 (t-hours) (7.49) 189
φµ cA
qB
µ
m = 162.6 o psi/log.cycle (7.50) 190
kh
kh
7.08 10 -3 (p − p ) = p (t ) + S (7.51) 190
×
qB o i wf D D
µ
(p − p ) k
S = 1.151 ws(LIN) 1 -hr wf − log + 3.23 (7.52) 191
m φµ cr 2
w

( )
k = k (abs ) × k ro S w (7.53) 191

qB t +∆ t
µ
p = p − 162.2 o log (7.54) 192
i
ws
kh t ∆
kh n ∆ q j
7.08 10 -3 (p − p ws ) = p D ( D t ) − p D ( t∆ D ) (7.55) 193
×
t + ∆
i
qB o j1 q n n j D 1
−
µ
n
=
kh t +∆ t
7.08 10 -3 ( i p ) = 1.151 log n
×
p −
qB o ws(LIN) t ∆
µ
n
n ∆ q 4t (7.56) 193
t
+ j p D ( D + t ∆ ) − ½ ln D n
j1 q n n j D 1 γ
−
=
* * t n
p − p = m log (7.57) 194
n t t
kh *
p D(MBH) (t )0.01416 qB o (p - p) = 2.303 log (C t ) (7.58) 195
DA
A
DA
µ
n
µ
* qB o
n
p − p = 162.6 log (C t ) = m log (C t ) (7.59) 195
A DA
A DA
kh
*
p − p = m log(C t ) (7.60) 195
n t n t A DA n
( p − p * ) ( p − p = m log t t n (7.61) 195
*
)
−
n t
n t
*
( p − p )
t +∆ t
log s = (7.62) 198
t ∆ s m
t +∆ t
log s = log (C t DA ) (7.63) 198
A
t ∆ s