Let's solve the decline curve analysis question step-by-step. --- # **Step 1: Identify a Suitable Decline ModelThere are three common decline models: **Exponential Decline:** \( q = q_i e^{-D t} \) - **Harmonic Decline:** \( q = \frac{q_i}{1 + D t} \) - **Hyperbolic Decline:** \( q = \frac{q_i}{(1 + b D t)^{1/b}} \) Let's plot (or analyze) the data to identify the best fit. ### **Quick Check for Decline Models** - **Exponential:** Plot \( \ln(q) \) vs. \( t \). Should be a straight line if exponential. - **Harmonic:** Plot \( 1/q \) vs. \( t \). Straight line if harmonic. - **Hyperbolic:** Plot \( \ln(q) \) vs. \( \ln(t) \). Straight line if hyperbolic. Let's use the exponential as a first check. --- # **Step 2: Linearize for Exponential Model** \[ \ln q = \ln q_i - D t \] Let's pick a few points from the table (first column for simplicity): | t (year) | q (100 STB/day) | ln(q) | |----------|------------------|-------| | .1 | 9.63 | 2.265 | | .2 | 9.28 | 2.229 | | .3 | 8.95 | 2.192 | | .4 | 8.64 | 2.154 | | .5 | 8.35 | 2.123 | | .6 | 8.07 | 2.089 | | .7 | 7.81 | 2.054 | | .8 | 7.55 | 2.022 | | .9 | 7.32 | 1.991 | Now plot \(\ln(q)\) vs \(t\) or fit a line: \[ \ln(q) = \ln(q_i) - D t \] The slope = \(-D\), intercept = \(\ln(q_i)\). --- # **Step 3: Calculate Decline Rate (D) and Initial Rate (\(q_i\))** Let’s use two points to estimate D: Using t=.1, q=9.63 (ln(q)=2.265) and t=.9, q=7.32 (ln(q)=1.991): \[ \text{Slope} = \frac{1.991 - 2.265}{.9 - .1} = \frac{-.274}{.8} = -.3425 \] Thus, \[ D = .3425 \ \text{year}^{-1} \] Intercept at \(t=\): \[ \ln(q_i) = 2.265 + (.3425 \times .1) = 2.265 + .03425 = 2.299 \] \[ q_i = e^{2.299} = 9.97 \text{ (approx 10,000 STB/day)} \] --- # **Step 4: Predict Time When \(q = 500\) STB/day** Given \(q = .5\) (since all rates are in 100 STB/day): \[ .5 = 10 \cdot e^{-.3425 t} \] \[ \frac{.5}{10} = e^{-.3425 t} \] \[ .05 = e^{-.3425 t} \] \[ \ln(.05) = -.3425 t \] \[ -2.996 = -.3425 t \] \[ t = \frac{2.996}{.3425} = 8.75 \ \text{years} \] --- # **Step 5: Calculate Reserves to be Recovered (Cumulative Production, Np)** For exponential decline: \[ N_p = \frac{q_i - q}{D} \] But for cumulative production up to time \(t\): \[ N_p = \int_^{t} q_i e^{-D t} dt = \frac{q_i}{D} \left[1 - e^{-D t}\right] \] Let’s calculate for \(t = 8.75\) years: \[ N_p = \frac{10}{.3425} \left[1 - e^{-.3425 \times 8.75} \right] \] First, calculate exponent: \[ -.3425 \times 8.75 = -2.996 \] \[ e^{-2.996} = .05 \] \[ N_p = \frac{10}{.3425} (1 - .05) = 29.2 \times .95 = 27.7 \text{ (100 STB)} \] --- # **Final Answers** ### **a) Decline Model:** **Exponential decline** model fits the data. ### **b) Model Parameters:** - \(q_i = 10,000\) STB/day - \(D = .3425\, \text{year}^{-1}\) ### **c) Time to Decline to 500 STB/day:** \[ \boxed{8.75 \text{ years}} \] ### **d) Reserves to be Recovered Before Marginal Rate:** \[ \boxed{27,700,000 \text{ STB}} \] --- ## **Summary Table** | Parameter | Value | |----------------------------------|--------------------------| | Decline Model | Exponential | | Initial Rate (\(q_i\)) | 10,000 STB/day | | Decline Rate (\(D\)) | .3425 year\(^{-1}\) | | Time to \(q=500\) STB/day | 8.75 years | | Reserves to \(q=500\) STB/day | 27.7 million STB | --- Let me know if you want to see for harmonic or hyperbolic models, or need calculation with the full set!