Traditional Decline Curves
Decline Curves that plot flow rate vs. time are the most common tools
for forecasting production and monitoring well performance in the field.
These curves quickly show by graphic means which wells or fields are
producing as expected or not. Also used to predict future production
rates and total recoveries that can be used in economic evaluations of the
well or field. Mainly used because they are easy to set up and to use in
the field. They are not based on any of the physics of the flow of oil and
gas through the rock formations, empirical in nature. The most common
forms are daily or monthly flow rates vs. the month. Water and gas rates
are commonly plotted along with the oil rate, or GOR and WOR.
Cumulative production vs. the months is also very common, both oil and
water can be plotted.
These plots are plotted both on linear plots and semi-log plots with the q
on the log scale.
There are three models of decline curves
 Exponential decline (constant fractional decline)
 Harmonic decline, and
 Hyperbolic decline.
These are empirical in nature. And are related by the equation
1 dq
 bq d
q dt
b and d are empirical constants to be determined from the production
data. When d = 0 the model is a exponential decline, d = 1 it is
harmonic. When 0 < d < 1 it is a hyperbolic decline model. These are
applicable to both oil and gas wells.
Exponential decline
The exponential decline well has a constant fractional decline. It
declines at a constant percentage during the life of the well. It will plot
as a straight line on a semi log plot.
The equation that describes this decline
where qi is the initial production at t0.
Cumulative Production
To calculate the cumulative production
Np 
qi

1  e bt 
b
Determination of the Decline rate
Since this model is a straight line on the semi log plot
or
Effective Decline Rate
Because the exponential function is not easy to use the following
equation may be used
q  qi 1 b'
t
Example:
Given that a well has declined from 100 stb/day to 96 stb/day during a one-month
period, use the exponential decline model to perform the following tasts:
a) Predict the production rate after 11 more months
b) Calculate the amount of oil produced during the first year
c) Project the yearly production for the well for the next 5 years.
Solution:
a) Production rate after 11 more months:
bm 
t1m
q
1
ln  0 m
 t 0 m   q1m



 1   100 
   ln 
  0.04082/month
 1   96 
Rate at end of one year
q1m  q0m e bmt  100e 0.0408212  61.27 stb/day
If the effective decline rate b’ is used,
b' m 
q0 m  q1m 100  96

 0.04/month .
q0 m
100
From
1  b' y  1  b' m   1  0.04
12
one gets
b' y  0.3875/year
12
Rate at end of one year
q1  q0 1  b' y   1001  0.3875  61.27 stb/day
b) The amount of oil produced during the first year:
by  0.0408212  0.48986/year
N p ,1 
q0  q1  100  61.27 

365  28,858 stb
by
 0.48986 
or
  100  1 
1
bd  ln 

  0.001342
day
  96  30.42 
N p ,1 


100
1  e 0.001342365  28,858 stb
0.001342
c) Yearly production for the next 5 years:
N p,2 
61.27
1  e 0.001342365   17,681 stb
0.001342
q2  qi e bt  100e 0.0408212( 2)  37.54 stb/day
N p ,3 
37.54

1  e 0.001342365   10,834 stb
0.001342
q3  qi e bt  100e 0.0408212(3)  23.00 stb/day
N p,4 
23.00

1  e 0.001342365   6,639 stb
0.001342
q4  qi e bt  100e 0.0408212( 4)  14.09 stb/day
N p ,5 
14.09

1  e 0.001342365   4,061 stb
0.001342
In summary,
Year
Rate at End of
Year (stb/day)
0
1
2
3
4
5
100.00
61.27
37.54
23.00
14.09
8.64
Yearly
Production
(stb)
28,858
17,681
10,834
6,639
4,061
68,073
Harmonic Decline
When d = 1, yields differential equation for a harmonic decline model:
1 dq
 bq
q dt
which can be integrated as
q
q0
1  bt
where q0 is the production rate at t = 0.
Expression for the cumulative production is obtained by integration:
t
N p   qdt
0
which gives:
Np 
q0
ln 1  bt  .
b
(
Combining
Np 
q0
ln q0   ln q  .
b
Hyperbolic Decline
When 0 < d < 1
q
t
dq
q q1d  0 bdt
0
which results in
q
or
q0
1 dbt 1/ d
q
q0
 b 
1  t 
 a 
a
where a = 1/d.
Expression for the cumulative production is obtained by integration:
t
N p   qdt
0
which gives:
Np 
1a
aq0   b  
1

1

t
 .
 
ba  1   a  
Combining Eqs
Np 
a 
 b 
q0  q1  t  .

ba  1 
 a 
8.5 Model Identification
Production data can be plotted in different ways to identify a representative
decline model. If the plot of log(q) versus t shows a straight line (Figure 1), the
decline data follow an exponential decline model. If the plot of q versus Np shows
a straight line (Figure 2), an exponential decline model should be adopted. If the
plot of log(q) versus log(t) shows a straight line (Figure 3), the decline data
follow a harmonic decline model. If the plot of Np versus log(q) shows a straight
line (Figure 4), the harmonic decline model should be used. If no straight line is
seen in these plots, the hyperbolic decline model may be verified by plotting the
relative decline rate. Figure 5 shows such a plot.
q
t
Figure 1: A Semilog plot of q versus t indicating an exponential decline
Np
q
Figure 2: A plot of Np versus q indicating an exponential decline
q
t
Figure 3: A plot of log(q) versus log(t) indicating a harmonic decline
Np
q
Figure 4: A plot of Np versus log(q) indicating a harmonic decline

q
q t
H ar
ic
mon
oli
Hyperb
Dec
line
e
c Declin
Exponential Decline
q
Figure 5: A plot of relative decline rate versus production rate
Determination of Model Parameters
Once a decline model is identified, the model parameters a and b can be
determined by fitting the data to the selected model. For the exponential
decline model, the b-value can be estimated on the basis of the slope of the
straight line in the plot of log(q) versus t . The b-value can also be determined
based on the slope of the straight line in the plot of q versus Np.
For the harmonic decline model, the b-value can be estimated on the basis of
the slope of the straight line in the plot of log(q) versus log(t) shows a straight
line, or Eq (8.32):
q0
1
q1
b
t1
The b-value can also be estimated based on the slope of the straight line in the
plot of Np versus log(q).
For the hyperbolic decline model, determination of a- and b-values is of a little
tedious. The procedure is shown in Figure 6.
1. Select points (t1, q1)
and (t2, q2)
2. Read t3 at
q3  q1 q 2
q
 b  t  t  2t 3
3. Calculate    1 2 2
t 3  t1 t 2
a
4. Find q0 at t = 0
5. Pick up any point (t*, q*)
6. Use
q* 
q0
 b 
1   t* 
 a 
7. Finally
a
b
b   a
a
q 
log  0 
 q* 
a
 b 
log 1   t* 
 a 
1
q3
(t*, q*)
2
t3
t
Figure 6: Procedure for determining a- and b-values
.
Illustrative Examples
Example Problem 2:
For the data given in Table 1, identify a suitable decline model, determine
model parameters, and project production rate until a marginal rate of 25
stb/day is reached.
Table 1: Production Data for Example Problem 8-2
t (Month) q (STB/D)
t (Month) q (STB/D)
1.00
904.84
13.00
272.53
2.00
818.73
14.00
246.60
3.00
740.82
15.00
223.13
4.00
670.32
16.00
201.90
5.00
606.53
17.00
182.68
6.00
548.81
18.00
165.30
7.00
496.59
19.00
149.57
8.00
449.33
20.00
135.34
9.00
406.57
21.00
122.46
10.00
367.88
22.00
110.80
11.00
332.87
23.00
100.26
12.00
301.19
24.00
90.72
Solution:
A plot of log(q) versus t is presented in Figure 7 which shows a straight line, 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:
b
1
 135 
ln 
  0.1 1/month
5  20  607 
Projected production rate profile is shown in Figure 9.
10000
q (STB/D)
1000
100
10
1
0
5
10
15
t (month)
20
25
30
Figure 7: A plot of log(q) versus t showing an exponential decline
0.15
-q/t/q (Month-1)
0.13
0.11
0.09
0.07
0.05
3
203
403
603
803
1003
q (STB/D)
Figure 8: Relative decline rate plot showing exponential decline
1000
900
800
q (STB/D)
700
600
500
400
300
200
100
0
0
10
20
30
40
t (month)
Figure 9: Projected production rate by an exponential decline model
Example Problem 3:
For the data given in Table 2, identify a suitable decline model, determine
model parameters, and project production rate till the end of the 5th year.
Table 2: Production Data for Example Problem 3
t (year) q (1000 STB/D)
t (year) q (1000 STB/D)
9.29
2.10
5.56
8.98
2.20
5.45
8.68
2.30
5.34
0.50
8.40
2.40
5.23
0.60
8.14
2.50
5.13
0.70
7.90
2.60
5.03
7.67
2.70
4.94
0.90
7.45
2.80
4.84
1.00
7.25
2.90
4.76
1.10
7.05
3.00
4.67
1.20
6.87
3.10
4.59
1.30
6.69
3.20
4.51
1.40
6.53
3.30
4.44
1.50
6.37
3.40
4.36
1.60
6.22
3.50
4.29
1.70
6.08
3.60
4.22
1.80
5.94
3.70
4.16
1.90
5.81
3.80
4.09
2.00
5.68
3.90
4.03
0.20
0.30
0.40
0.80
Solution:
A plot of relative decline rate is shown in Figure 10 which clearly indicates a
harmonic decline model.
On the trend line, select
q0 = 10,000 stb/day at t = 0
q1 = 5,680 stb/day at t = 2 years
gives:
10,000
1
5,680
b
 0.38 1/year .
2
Projected production rate profile is shown in Figure 11.
0.4
-q/t/q (year-1)
0.35
0.3
0.25
0.2
0.15
0.1
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
q (1000 STB/D)
Figure 10: Relative decline rate plot showing harmonic decline
12
q (1000 STB/D)
10
8
6
4
2
0
0.0
1.0
2.0
3.0
t (year)
4.0
5.0
6.0
Figure 11: Projected production rate by a harmonic decline model
Example Problem 4:
For the data given in Table 3, identify a suitable decline model, determine
model parameters, and project production rate till the end of the 5th year.
Solution:
A plot of relative decline rate is shown in Figure 12 which clearly indicates a
hyperbolic decline model.
Select points:
t1 = 0.2 year , q1 = 9,280 stb/day
t2 = 3.8 years, q2 = 3,490 stb/day
q3  (9,280)(3,490)  5,670 stb/day
Read from decline curve t3 = 1.75 yaers at q3 = 5,670 stb/day.
 b  0.2  3.8  2(1.75)
 0.217
 
2
 a  (1.75)  (0.2)(3.8)
Read from decline curve (Figure 13) q0 = 10,000 stb/day at t0 = 0.
Pick up point (t* = 1.4 yesrs, q* = 6,280 stb/day).
 10,000 
log

6,280 

a
 1.75
log 1  0.217 (1.4) 
Projected production rate profile is shown in Figure 14.
b  0.217(1.758)  0.38
Table 3: Production Data for Example Problem 4
t (year) q (1000 STB/D)
t (year) q (1000 STB/D)
0.10
9.63
2.10
5.18
0.20
9.28
2.20
5.05
0.30
8.95
2.30
4.92
0.40
8.64
2.40
4.80
0.50
8.35
2.50
4.68
0.60
8.07
2.60
4.57
0.70
7.81
2.70
4.46
0.80
7.55
2.80
4.35
0.90
7.32
2.90
4.25
1.00
7.09
3.00
4.15
1.10
6.87
3.10
4.06
1.20
6.67
3.20
3.97
1.30
6.47
3.30
3.88
1.40
6.28
3.40
3.80
1.50
6.10
3.50
3.71
1.60
5.93
3.60
3.64
1.70
5.77
3.70
3.56
1.80
5.61
3.80
3.49
1.90
5.46
3.90
3.41
2.00
5.32
4.00
3.34
0.38
-q/t/q (year-1)
0.36
0.34
0.32
0.3
0.28
0.26
0.24
0.22
0.2
3.00
5.00
7.00STB/D)
q (1000
9.00
11.00
Figure 12: Relative decline rate plot showing hyperbolic decline
12
q (1000 STB/D)
10
8
6
4
2
0
0.0
1.0
2.0
3.0
4.0
5.0
t (year)
Figure 13: Relative decline rate plot showing hyperbolic decline
12
q (1000 STB/D)
10
8
6
4
2
0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t (year)
Figure 14: Projected production rate by a hyperbolic decline model
*
*
*
*
*