24-Pet-A3 Fundamental Reservoir Engineering · May 2018
Question 6 of 7: Decline-Curve Analysis from a Rate–Cumulative-Production Plot
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Pet-A3 — Fundamental Reservoir Engineering · National Exams, May 2018 · 3 hours, closed book, approved Casio/Sharp calculator only · five (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked), all questions equal value, all parts of a multipart question equal weight.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, well testing, relative permeability, Darcy flow); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, transient well testing, gas PVT, decline-curve analysis); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.; McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, Z-factor correlations).
Question 6: Decline-Curve Analysis from a Rate–Cumulative-Production Plot (20 marks)
Given. Scatter plot of cumulative gas production $G_p$ (MMSCF) vs. production rate $q$ (MMSCFD), boundary-dominated flow, five digitized points: $(q,\,G_p)=(120,\,5.1\times10^5),\,(165,\,4.1\times10^5),\,(220,\,3.2\times10^5),\,(250,\,2.2\times10^5),\,(290,\,1.2\times10^5)$. Economic rate limit $q_{lim}=10$ MMSCFD.
Find. (a) gas volume produced during the fifth year; (b) total production life to the economic limit.
Approach. A straight-line relationship between rate $q$ and cumulative production $G_p$ is the defining signature of exponential decline ($G_p=(q_i-q)/D$); least-squares fit the five points to that line to recover $q_i$ and $D$, then use the exponential-decline cumulative-production formula to get the fifth-year volume and the log-rate formula to get the time to the economic limit.
Fig. Q6: rate vs. cumulative production, with the least-squares straight-line fit whose slope and intercept give the exponential decline constant $D$ and initial rate $q_i$.
Fit the exponential-decline line. For exponential decline, $G_p=q_i/D-q/D$ — linear in $q$. Least-squares fit of the five points gives slope $=-2259.6$ MMSCF/(MMSCFD) and intercept $=788{,}257$ MMSCF. Since slope $=-1/D$: $\boxed{D=1/2259.6=4.426\times10^{-4}\ \text{day}^{-1}}$ (0.1615 yr$^{-1}$), and since intercept $=q_i/D$: $\boxed{q_i=788{,}257(4.426\times10^{-4})=348.8\ \text{MMSCFD}}$.
Cumulative production at 4 and 5 years. Since $q_i$ is a daily rate (MMSCFD) and $G_p$ must come out in MMSCF, the exponential-decline cumulative formula is evaluated with $t$ in DAYS and $D=D_{day}$ (the intercept of the fitted line is exactly $q_i/D_{day}$): $G_p(t)=\dfrac{q_i}{D_{day}}\left(1-e^{-D_{day}t}\right)=788{,}257\left(1-e^{-D_{day}t}\right)$. At $t=4\,\text{yr}=1460$ days: $G_p(4\,\text{yr})=788{,}257\left(1-e^{-0.0004426(1460)}\right)=375{,}156$ MMSCF. At $t=5\,\text{yr}=1825$ days: $G_p(5\,\text{yr})=788{,}257\left(1-e^{-0.0004426(1825)}\right)=436{,}774$ MMSCF.
Volume produced during the fifth year. $\Delta G_p=G_p(5\,\text{yr})-G_p(4\,\text{yr})=436{,}774-375{,}156$, so $\boxed{\Delta G_p\approx6.16\times10^{4}\ \text{MMSCF}\ (61.6\ \text{Bcf})}$.
Production life to the economic limit. $q(t)=q_ie^{-Dt}\ \Rightarrow\ t=\dfrac{\ln(q_i/q_{lim})}{D}=\dfrac{\ln(348.8/10)}{0.1615}$, so $\boxed{t\approx22.0\ \text{years}}$.
Check: the five (rate, cumulative-production) pairs are read directly off the exam's own scatter plot; a least-squares fit through all five (rather than a two-point chord) is used to reduce digitization error, and the resulting straight-line fit confirms the boundary-dominated exponential-decline assumption stated in the question.
