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 0.0408212 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 1001 0.3875 61.27 stb/day b) The amount of oil produced during the first year: by 0.0408212 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 0.001342365 28,858 stb 0.001342 c) Yearly production for the next 5 years: N p,2 61.27 1 e 0.001342365 17,681 stb 0.001342 q2 qi e bt 100e 0.0408212( 2) 37.54 stb/day N p ,3 37.54 1 e 0.001342365 10,834 stb 0.001342 q3 qi e bt 100e 0.0408212(3) 23.00 stb/day N p,4 23.00 1 e 0.001342365 6,639 stb 0.001342 q4 qi e bt 100e 0.0408212( 4) 14.09 stb/day N p ,5 14.09 1 e 0.001342365 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 q1d 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 1a aq0 b 1 1 t . ba 1 a Combining Eqs Np a b q0 q1 t . ba 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 * * * * *