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Chapter 8: Production Decline Analysis
8.1 Introduction
Production decline analysis is a traditional means of identifying well production
problems and predicting well performance and life based on real production data. It uses
empirical decline models that have little fundamental justifications. These models include
- Exponential decline (constant fractional decline)
- Harmonic decline, and
- Hyperbolic decline.
While the hyperbolic decline model is more general, the other two models are
degenerations of the hyperbolic decline model. These three models are related through
the following relative decline rate equation (Arps, 1945):
d bq dt
dq
q
where b and d are empirical constants to be determined based on production data. When d
= 0, Eq (8.1) degenerates to an exponential decline model, and when d = 1, Eq (8.1)
yields a harmonic decline model. When 0 < d < 1, Eq (8.1) derives a hyperbolic decline
model. The decline models are applicable to both oil and gas wells.
8.2 Exponential Decline
The relative decline rate and production rate decline equations for the exponential decline
model can be derived from volumetric reservoir model. Cumulative production
expression is obtained by integrating the production rate decline equation.
8.2.1 Relative Decline Rate
Consider an oil well drilled in a volumetric oil reservoir. Suppose the well’s production
rate starts to decline when a critical (lowest permissible) bottom hole pressure is reached.
Under the pseudo-steady state flow condition, the production rate at a given decline time
t can be expressed as:
s r
r B
kh p p q
w
e
c t (^) wf
- 2 ln
where p (^) t = average reservoir pressure at decline time t ,
c p (^) wf = the critical bottom hole pressure maintained during the production decline.
The cumulative oil production of the well after the production decline time t can be
expressed as:
dt
s r
r B
kh p p N
t
w
e o
c t p
wf ∫
0 0.^472
- 2 ln
The cumulative oil production after the production decline upon decline time t can also
be evaluated based on the total reservoir compressibility:
( (^) t o
t i p p p B
cN N = 0 − ) (8.4)
where ct = total reservoir compressibility,
N (^) i = initial oil in place in the well drainage area,
p (^) 0 = average reservoir pressure at decline time zero.
Substituting Eq (8.3) into Eq (8.4) yields:
( (^) t
o
t i
t
w
e o
c t p p B
cN dt
s r
r B
kh p p wf = −
∫ 0 0 0.^472
- 2 ln
) (8.5)
Taking derivative on both sides of this equation with respect to time t gives the
differential equation for reservoir pressure:
dt
dp cN
s r
r
kh p p t t i
w
e
c t (^) wf = −
- 2 ln
Since the left-hand-side of this equation is q and Eq (8.2) gives
dt
dp
s r
r B
kh
dt
dq (^) t
w
e ⎥ ⎦
141. 2 0 μ ln
Eq (8.6) becomes
dt
dq
kh
s r
r cN
q
w
e t i ⎥ ⎦
141. 2 μ ln
for exponential decline. As an exercise, this is left to the reader to prove.
8.2.3 Cumulative Production
Integration of Eq (8.16) over time gives an expression for the cumulative oil production
since decline of
∫ ∫
− = =
t bt i
t
Np qdt qe dt
0 0
i.e.,
( )
i bt p e b
q N
− = 1 −. (8.18)
Since , Eq (8.18) becomes
bt q qi e
−
( q q b
N (^) p = (^) i −
). (8.19)
8.2.4 Determination of Decline Rate
The constant b is called the continuous decline rate. Its value can be determined from
production history data. If production rate and time data are available, the b -value 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:
ln ( ) q = ln( qi ) − bt (8.20)
which implies that the data should form a straight line with a slope of -b on the log( q )
versus t plot, if exponential decline is the right model. Picking up any two points, ( t 1 , q 1 )
and ( t 2 , q 2 ), on the straight line will allow analytical determination of b -value because
( ) ( ) 1 1
ln q ln q bt
i
and
ln ( q (^) 2 ) = ln( q (^) i ) − bt 2 (8.22)
give
( )
2
1
2 1
ln
q
q
t t
b . (8.23)
If production rate and cumulative production data are available, the b -value can be
obtained based on the slope of the straight line on an Np versus q plot. In fact, rearranging
Eq (8.19) yields:
q = qi − bNp (8.24)
Picking up any two points, ( Np1 , q 1 ) and ( Np2 , q 2 ), on the straight line will allow analytical
determination of b -value because
q 1 = qi − bNp 1 (8.25)
and
q 2 = qi − bNp 2 (8.26)
give
2 1
1 2
p p
N N
q q
b
Depending on the unit of time t , the b can have different units such as month
The following relation can be derived:
a m d
b = 12 b = 365 b. (8.28)
where ba , bm , and bd are annual, monthly, and daily decline rates.
8.2.5 Effective Decline Rate
Because the exponential function is not easy to use in hand calculations, traditionally the
effective decline rate has been used. Since for small x -values based on
Taylor’s expansion, holds true for small values of b. The b is substituted by
, the effective decline rate, in field applications. Thus Eq (8.16) becomes
e x
x
−
e b
b
−
b '
(
t q = qi 1 − b ') (8.29)
Again, it can be shown that ...... 1 '
2 1
3
1
2 b q
q
q
q
q
q
n
n = = = = − −
Depending on the unit of time t , the can have different units such as month
1
. The following relation can be derived:
b '
( ) ( ) ( )
12 365
a m d
− b = − b = − b. (8.30)
where b ' (^) a , b ' (^) m , and b ' d are annual, monthly, and daily effective decline rates.
by = 0. 04082 ( 12 ) = 0. 48986 /year
365 28 , 858 stb
- 48986
0 1 100 61.^27
y
p b
q q N
or
( ) ( 1 ) 28 , 858 stb
- 001342
day
ln
- 001342365 , 1 = − =
− N e
b
p
d
c) Yearly production for the next 5 years:
( ) ( 1 ) 17 , 681 stb
- 001342
, 2 =^ − =
− N (^) p e
( ) 100 37. 54 stb/day
- 0408212 ( 2 ) 2 =^ = =
− − q qe e
bt i
( ) ( 1 ) 10 , 834 stb
- 001342
, 3 =^ − =
− N (^) p e
( ) 100 23. 00 stb/day
- 0408212 ( 3 ) 3 =^ = =
− − q qe e
bt i
( ) ( 1 ) 6 , 639 stb
- 001342
, 4 =^ − =
− N (^) p e
( ) 100 14. 09 stb/day
- 0408212 ( 4 ) 4 =^ = =
− − q qe e
bt i
( ) ( 1 ) (^4) , 061 stb
- 001342
, 5 =^ − =
− N (^) p e
In summary,
Year
Rate at End of Year
(stb/day)
Yearly Production
(stb)
0 1 2 3 4 5
8.3 Harmonic Decline
When d = 1, Eq (8.1) yields differential equation for a harmonic decline model:
bq dt
dq
q
which can be integrated as
bt
q q
0 (8.32)
where q 0 is the production rate at t = 0.
Expression for the cumulative production is obtained by integration:
∫
t
Np qdt
0
which gives:
( bt b
q N (^) p = ln 1 +
0 ). (8.33)
Combining Eqs (8.32) and (8.33) gives
[ ( q ) ( ) q b
q N (^) p ln 0 ln
0 = − ]. (8.34)
decline model may be verified by plotting the relative decline rate defined by Eq (8.1).
Figure 8-5 shows such a plot. This work can be easily performed with computer program
UcomS.exe.
q
t
Figure 8-1: A Semilog plot of q versus t indicating an exponential decline
q
p N
Figure 8-2: A plot of Np versus q indicating an exponential decline
q
t
Figure 8-3: A plot of log( q ) versus log( t ) indicating a harmonic decline
q
N p
Figure 8-4: A plot of Np versus log( q ) indicating a harmonic decline
- Select points ( t 1 , q 1 ) and ( t 2 , q 2 )
- Read t 3 at
- Calculate
- Find q 0 at t = 0
- Pick up any point ( t, q )
- Use
- Finally
q
t
1
2
q 3 (^) = q 1 q 2
12
2 3
1 2 23
t t t
t t t
a
b
−
⎞ ⎜ ⎝
⎛
a t a
b
q q
⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ ⎟ ⎠
⎞ ⎜ ⎝
⎛
=
0
(^1) ⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ ⎟ ⎠
⎞ ⎜ ⎝
⎛
⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛
=
0
log 1
log
t a
b
q
q
a
a a
b b (^) ⎟ ⎠
⎞ ⎜ ⎝
⎛
q 3
t 3
(t, q)
Figure 8-6: Procedure for determining a - and b -values
Computer program UcomS.exe can be used for both model identification and model
parameter determination, as well as production rate prediction.
8.7 Illustrative Examples
Example Problem 8-2:
For the data given in Table 8-1, identify a suitable decline model, determine model
parameters, and project production rate until a marginal rate of 25 stb/day is reached.
Table 8-1: Production Data for Example Problem 8-
t (Month) q (STB/D) t (Month) q (STB/D)
12.00 301.
11.00 332.
10.00 367.
9.00 406.
8.00 449.
7.00 496.
6.00 548.
5.00 606.
4.00 670.
3.00 740.
2.00 818.
1.00 904.
12.00 301.
11.00 332.
10.00 367.
9.00 406.
8.00 449.
7.00 496.
6.00 548.
5.00 606.
4.00 670.
3.00 740.
2.00 818.
1.00 904.
24.00 90.
23.00 100.
22.00 110.
21.00 122.
20.00 135.
19.00 149.
18.00 165.
17.00 182.
16.00 201.
15.00 223.
14.00 246.
13.00 272.
24.00 90.
23.00 100.
22.00 110.
21.00 122.
20.00 135.
19.00 149.
18.00 165.
17.00 182.
16.00 201.
15.00 223.
14.00 246.
13.00 272.
Solution:
A plot of log( q ) versus t is presented in Figure 8-7 which shows a straight line.
According to Eq (8.20), the exponential decline model is applicable. This is further
evidenced by the relative decline rate shown in Figure 8-8.
Select points on the trend line:
t1 = 5 months, q1 = 607 STB/D
t2 = 20 months, q2 = 135 STB/D
Decline rate is calculated with Eq (8.23):
( )
0. 1 1/month
ln
b =
Projected production rate profile is shown in Figure 8-9.
1
10
100
1000
10000
0 5 10 15 20 25 30
t (month)
q (STB/D
)
t (year) q (1000 STB/D) t (year) q (1000 STB/D)
Solution: A plot of relative decline rate is shown in Figure 8-10 which clearly indicates a harmonic decline model. q 1 = 5,680 stb/day at t = 2 years Therefore, Eq (8.40) gives:
- 38 1/year 2
b =.
- Table 8-2: Production Data for Example Problem 8-
- 2.00 5.
- 1.90 5.
- 1.80 5.
- 1.70 6.
- 1.60 6.
- 1.50 6.
- 1.40 6.
- 1.30 6.
- 1.20 6.
- 1.10 7.
- 1.00 7.
- 0.90 7.
- 0.80 7.
- 0.70 7.
- 0.60 8.
- 0.50 8.
- 0.40 8.
- 0.30 8.
- 0.20 9.
- 2.00 5.
- 1.90 5.
- 1.80 5.
- 1.70 6.
- 1.60 6.
- 1.50 6.
- 1.40 6.
- 1.30 6.
- 1.20 6.
- 1.10 7.
- 1.00 7.
- 0.90 7.
- 0.80 7.
- 0.70 7.
- 0.60 8.
- 0.50 8.
- 0.40 8.
- 0.30 8.
- 0.20 9. - 3.90 4. - 3.80 4. - 3.70 4. - 3.60 4. - 3.50 4. - 3.40 4. - 3.30 4. - 3.20 4. - 3.10 4. - 3.00 4. - 2.90 4. - 2.80 4. - 2.70 4. - 2.60 5. - 2.50 5. - 2.40 5. - 2.30 5. - 2.20 5. - 2.10 5. - 3.90 4. - 3.80 4. - 3.70 4. - 3.60 4. - 3.50 4. - 3.40 4. - 3.30 4. - 3.20 4. - 3.10 4. - 3.00 4. - 2.90 4. - 2.80 4. - 2.70 4. - 2.60 5. - 2.50 5. - 2.40 5. - 2.30 5. - 2.20 5. - 2.10 5.
- q 0 = 10,000 stb/day at t = On the trend line, select - 5 , - 10 ,
3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.
q (1000 STB/D)
- Δ
q/
Δ
t/q (year
-
)
Figure 8-10: Relative decline rate plot showing harmonic declineplot showing harmonic decline
0
2
4
6
8
10
12
0.0 1.0 2.0 3.0 4.0 5.0 6.
t (year)
q (1000 STB/D)
Figure 8-11:Figure 8-11: Projected production rate by a harmonic decline modelProjected production rate by a harmonic decline model
Example Problem 8-4:Example Problem 8-4:
For the data given in Table 8-3, identify a suitable decline model, determine model
rate till the end of the 5
th year.
Sol
ecline rate is shown in Figure 8-12 which clearly indicates a
line model.
Select poin
parameters, and project production
ution:
A plot of relative d
hyperbolic dec
ts:
3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.
- Δ
q/
Δ
t/q (year
-
)
q (1000 STB/D)
Figure 8-12: Relative decline rate rbolic decline
Figure 8-13: Relative decline rate plot showing hyperbolic decline
plot showing hype
0
2
4
6
8
10
12
0.0 1.0 2.0 3.0 4.0 5.
q (1000 STB/D)
t (year)
0
2
4
6
8
10
12
0.0 1.0 2.0 3.0 4.0 5.0 6.
t (year)
q (1000 STB/D)
Figure 8-14: Projected production rate by a hyperbolic decline model
Summary
This chapter presented empirical models and procedure of using the models to perform
production decline data analyses. Computer program UcomS.exe can be used for model
identification, model parameter determination, and production rate prediction.
References
Arps, J.J.: “ Analysis of Decline Curves,” Trans. AIME , 160 , 228-247, 1945.
Golan, M. and Whitson, C.M.: Well Performance , International Human Resource
Development Corp., 122-125, 1986.
Economides, M.J., Hill, A.D., and Ehlig-Economides, C.: Petroleum Production
Systems , Prentice Hall PTR, Upper Saddle River, 516-519, 1994.
Problems
8.1 For the data given in the following table, identify a suitable decline model, determine
model parameters, and project production rate till the end of the 10
th year. Predict
yearly oil productions:
