24-Pet-B2 Oil and Gas Evaluation and Economics · December 2014
Question 4 of 7: Decline Curve Analysis from a Gp–q Crossplot
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2014, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (Casio/Sharp approved calculators only), 7 questions of 20 marks each. NOTES item 5 states only the first five questions in the answer book are marked; all 7 are solved.
Reference texts: Katz et al., Handbook of Natural Gas Engineering; Lee & Wattenbarger, Gas Reservoir Engineering (SPE Textbook Series Vol. 5); Ahmed, Reservoir Engineering Handbook, 5th ed.; Mohitpour et al., Pipeline Design and Construction, 3rd ed. (ASME Press); McCain, The Properties of Petroleum Fluids, 3rd ed.
Question 4: Decline Curve Analysis from a Gp–q Crossplot (20 marks)
Given. Seven ($q$, $G_p$) points read off the exam chart (q in MMSCFD, $G_p$ in MMSCF). Each printed dot sits on a grid intersection: (150, 400,000), (180, 340,000), (210, 280,000), (240, 220,000), (270, 160,000), (300, 100,000), (330, 40,000).
Find. (a) $G_p$ after 10 years of production; (b) reserves at an economic-limit rate of 25 MMSCFD.
Approach. A straight-line relationship between $G_p$ and $q$ is the diagnostic signature of exponential decline ($G_p=(q_i-q)/D$, from the formula sheet), so a least-squares fit of the seven points gives $q_i$ and $D$ directly; part (a) then uses the exponential decline equation forward to $t=10$ yr, and part (b) simply reads the same fitted line at $q=25$ MMSCFD.
Linear fit. The seven points lie exactly on one straight line (every 30 MMSCFD step in $q$ adds 60,000 MMSCF to $G_p$), so the least-squares fit is exact: $G_p=700{,}000-2000\,q$ (MMSCF), $R^2=1.000$. That is the exponential-decline signature.
Recover $q_i$ and $D$. At $G_p=0$ (start of production), $q=q_i$: $q_i=\dfrac{700{,}000}{2000}=\boxed{q_i=350.0\ \text{MMSCFD}}$. Since the fitted slope is $-1/D$: $D=1/2000=\boxed{D=5.000\times10^{-4}\ \text{day}^{-1}}$.
(a) Rate and cumulative at 10 years. $t=10\times365=3650$ days. $q(t)=q_ie^{-Dt}=350.0\,e^{-5.000\times10^{-4}(3650)}=350.0\,e^{-1.825}=56.43$ MMSCFD. $G_p(t)=\dfrac{q_i-q(t)}{D}=\dfrac{350.0-56.43}{5.000\times10^{-4}}$: $\boxed{G_p(10\ \text{yr})=587{,}148\ \text{MMSCF}}$.
(b) Reserves at the economic limit. Reading the same fitted line directly at $q=25$ MMSCFD: $G_p=700{,}000-2000(25)$: $\boxed{\text{Reserves}=650{,}000\ \text{MMSCF}}$. This is an extrapolation below the lowest charted rate (150 MMSCFD), which is legitimate only because exponential decline keeps the $G_p$–$q$ relation linear all the way down.
Fig. 1 — Cumulative gas production vs. rate; the linear fit is the exponential-decline diagnostic $G_p=q_i/D-q/D$. The 10-year point (a) and the economic-limit reserve point (b) at $q=25$ MMSCFD both lie on the same extrapolated line.
