24-Pet-B2 Oil and Gas Evaluation and Economics · May 2015
Question 4 of 7: Harmonic Decline from a ln(q)–Gp Crossplot
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2015, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (non-communicating calculator permitted), 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: Harmonic Decline from a ln(q)–Gp Crossplot (20 marks)
Given. Nine ($G_p$, $\ln q$) pairs read from the field chart (see figure): $G_p\approx$ 343,000–1,112,000 (chart x-axis labelled MSCFD, a printed unit slip — treated as cumulative production in MSCF, consistent with the physical scale of the plot; see the check note), $\ln q$ from 7.17 down to 6.21.
Find. (a) Time to produce $G_p=2$ MMMSCF $=2{,}000{,}000$ MSCF; (b) reserves at an economic-limit rate of 100 MSCFD.
Approach. A straight line on a $\ln q$ vs. $G_p$ crossplot is the diagnostic signature of harmonic decline: combining $q=q_i/(1+Dt)$ with $dG_p/dt=q$ gives $\ln q=\ln q_i-(D/q_i)G_p$, linear in $G_p$ with slope $-b=-D/q_i$. Fit the nine points by least squares to get $q_i$ and $b$, then use $q=q_ie^{-bG_p}$ together with the time integral $t=(1/b)(1/q-1/q_i)$ for part (a), and the fitted line directly (solved for $G_p$ at $q=q_{min}$) for part (b).
Linear fit. Least-squares regression of $\ln q$ on $G_p$ over the nine digitized points gives $\ln q=7.5683-1.2074\times10^{-6}\,G_p$, with $R^2=0.996$ confirming the harmonic-decline diagnosis (a straight line on this specific crossplot, not on $q$ vs. $G_p$ or $q$ vs. $t$).
Recover $q_i$ and $D$. At $G_p=0$: $q_i=e^{7.5683}=\boxed{q_i=1936\ \text{MSCFD}}$. The fitted slope is $-b=-D/q_i$, so $D=b\,q_i=(1.2074\times10^{-6})(1936)=\boxed{D=2.337\times10^{-3}\ \text{day}^{-1}}$.
(a) Time to produce 2 MMMSCF. $G_p=2\times10^6$ MSCF. From the fitted line, $q=q_ie^{-bG_p}=1936\,e^{-1.2074\times10^{-6}(2\times10^6)}=173.0$ MSCFD. Since $dG_p/dt=q_ie^{-bG_p}$ integrates to $t=(1/b)(1/q-1/q_i)=\dfrac{1}{1.2074\times10^{-6}}\left(\dfrac{1}{173.0}-\dfrac{1}{1936}\right)$: $\boxed{t=4358\ \text{days}=11.9\ \text{years}}$.
(b) Reserves at the 100 MSCFD economic limit. Reading the fitted line directly at $q=q_{min}=100$ MSCFD: $G_p=\ln(q_i/q_{min})/b=\ln(1936/100)/(1.2074\times10^{-6})$: $\boxed{\text{Reserves}=2.454\times10^6\ \text{MSCF}=2.454\ \text{MMMSCF}}$.
Fig. 1 — $\ln q$ vs. cumulative production $G_p$, read from the printed chart (solid markers) with the least-squares harmonic-decline fit (line). The straight-line shape on these axes is the diagnostic for harmonic (not exponential) decline.
Quantity
Value
Initial rate, $q_i$
1936 MSCFD
Decline rate, $D$
$2.337\times10^{-3}$ day$^{-1}$
(a) Time to produce 2 MMMSCF
4358 days (11.9 yr)
(b) Reserves at 100 MSCFD limit
2.454 MMMSCF
Check: the nine ($G_p$, $\ln q$) points were read from the printed chart (page 3) — individual point positions carry roughly ±1–2% uncertainty; the fitted line’s $R^2=0.996$ indicates this does not materially affect the regression. The chart’s own x-axis is labelled “Cuumulative production (MSCFD)” — MSCFD is a *rate* unit, not a cumulative-volume unit; treated as a source labelling slip and read as MSCF (cumulative), consistent with the plotted numerical scale (up to ~1.2 million) against gas rates in the hundreds of MSCFD. Part (a)’s target ($G_p=2$ MMMSCF) lies beyond the chart’s own plotted range (max ~1.2 MMSCF) and is reached by extrapolating the fitted straight line.
