Page 107 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 99 20.12.2006 10:36am

PRODUCTION DECLINE ANALYSIS 8/99
0 0 0
that is, where b , b , and b d are annual, monthly, and daily
m
a
effective decline rates, respectively.
q i bt
N p ¼ 1 e : (8:18)
b
Example Problem 8.1 Given that a well has declined from
Since q ¼ q i e bt , Eq. (8.18) becomes 100 stb/day to 96 stb/day during a 1-month period, use the
1 exponential decline model to perform the following tasks:
N p ¼ ð q i qÞ: (8:19)
b 1. Predict the production rate after 11 more months
2. Calculate the amount of oil produced during the first
year
8.2.4 Determination of decline rate
3. Project the yearly production for the well for the next 5
The constant b is called the continuous decline rate. Its
years
value can be determined from production history data. If
production rate and time data are available, the b value
Solution
can be obtained based on the slope of the straight line on a
semi-log plot. In fact, taking logarithm of Eq. (8.16) gives 1. Production rate after 11 more months:
ln (q) ¼ ln (q i ) bt, (8:20) 1
b m ¼ ln q 0m
which implies that the data should form a straight line with (t 1m t 0m ) q 1m
a slope of b on the log(q) versus t plot, if exponential 100
1
decline is the right model. Picking up any two points, ¼ ln ¼ 0:04082=month
(t 1 , q 1 ) and (t 2 , q 2 ), on the straight line will allow analyt- 1 96
ical determination of b value because Rate at end of 1 year:
ln (q 1 ) ¼ ln (q i ) bt 1 (8:21) q 1m ¼ q 0m e b m t ¼ 100e 0:04082(12) ¼ 61:27 stb=day
and If the effective decline rate b’ is used,
ln (q 2 ) ¼ ln (q i ) bt 2 (8:22) 0 q 0m q 1m 100 96
b ¼ ¼ ¼ 0:04=month:
give m q 0m 100

1 q 1 From
b ¼ ln : (8:23)
0
0
12
(t 2 t 1 ) q 2 1 b ¼ (1 b ) 12 ¼ (1 0:04) ,
y
m
If production rate and cumulative production data are one gets
available, the b value can be obtained based on the slope
0
of the straight line on an N p versus q plot. In fact, b ¼ 0:3875=yr
y
rearranging Eq. (8.19) yields
Rate at end of 1 year:
q ¼ q i bN p : (8:24)
0
q 1 ¼ q 0 (1 b ) ¼ 100(1 0:3875) ¼ 61:27 stb=day
Picking up any two points, (N p1 , q 1 ) and (N p2 , q 2 ), on the y
straight line will allow analytical determination of the b 2. The amount of oil produced during the first year:
value because
q 1 ¼ q i bN p1 (8:25) b y ¼ 0:04082(12) ¼ 0:48986=year

and 100 61:27
N p,1 ¼ q 0 q 1 ¼ 365 ¼ 28,858 stb
q 2 ¼ q i bN p2 (8:26) b y 0:48986
give or

q 1 q 2 100 1 1
b ¼ : (8:27) b d ¼ ln ¼ 0:001342
N p2 N p1
96 30:42 day
Depending on the unit of time t, the b can have different 100 0:001342(365)
1
units such as month 1 and year . The following relation N p,1 ¼ 0:001342 (1 e ) ¼ 28,858 stb
can be derived:
b a ¼ 12b m ¼ 365b d , (8:28) 3. Yearly production for the next 5 years:
where b a , b m , and b d are annual, monthly, and daily 61:27 0:001342(365)
decline rates, respectively. N p,2 ¼ 0:001342 (1 e ) ¼ 17; 681 stb
q 2 ¼ q i e bt ¼ 100e 0:04082(12)(2) ¼ 37:54 stb=day
8.2.5 Effective decline rate
Because the exponential function is not easy to use in hand 37:54
calculations, traditionally the effective decline rate has N p,3 ¼ (1 e 0:001342(365) ) ¼ 10,834 stb
been used. Since e x 1 x for small x-values based on 0:001342
Taylor’s expansion, e b 1 b holds true for small values bt 0:04082(12)(3)
q 3 ¼ q i e ¼ 100e ¼ 23:00 stb=day
of b. The b is substituted by b’, the effective decline rate, in
field applications. Thus, Eq. (8.16) becomes 23:00
N p,4 ¼ (1 e 0:001342(365) ) ¼ 6639 stb
0 t
q ¼ q i (1 b ) : (8:29) 0:001342
0
Again, it can be shown that q 2 ¼ q 3 ¼ ... ... ¼ q n ¼ 1 b . q 4 ¼ q i e bt ¼ 100e 0:04082(12)(4) ¼ 14:09 stb=day
q 1 q 2 q n 1
Depending on the unit of time t, the b’ can have different
1
units such as month 1 and year . The following relation 14:09 0:001342(365)
can be derived: N p,5 ¼ 0:001342 (1 e ) ¼ 4061 stb
0 0 12 0 365
(1 b ) ¼ (1 b ) ¼ (1 b ) , (8:30) In summary,
a m d