24-Pet-A6 Well Logging and Formation Evaluation · May 2017
Question 6 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2017 – 98-Pet-A6, Reservoir Mechanics (3 hours, open book, non-communicating calculator permitted). The exam consists of EIGHT questions; candidates respond to 100 marks of their choice out of 120. All eight questions are answered in full below.
Reference texts: Craft, B.C. and Hawkins, M. (rev. Terry & Rogers), Applied Petroleum Reservoir Engineering, 3rd ed.; Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.
If doubt exists..." "2. This is an OPEN BOOK EXAM..." "3. The exam consists of EIGHT (8) questions..." "4. Clarity and organization...") immediately above Question 1. every real "Question N" heading on this paper parses cleanly (checked by direct read of the source), so it does not affect the answers below.
Given. Fourteen rate–time observations spanning 78 months of production.
Production history (STB/day)
$t$=0
1,568
$t$=42
191
$t$=6
970
$t$=48
161
$t$=12
664
$t$=54
137
$t$=18
478
$t$=60
118
$t$=24
363
$t$=66
103
$t$=30
285
$t$=72
90
$t$=36
230
$t$=78
80
Fig. Q6 – production history (points) with the fitted hyperbolic decline curve, on a log rate axis.
Find. (a) $q$ at $t=114$ months; (b) incremental $N_p$ from $t=66$ to $t=114$ months.
Approach. Fit Arps' hyperbolic decline $q(t)=q_i(1+bD_it)^{-1/b}$ to the full history by choosing $b$ so that $q^{-b}$ is most linear in $t$ (least-squares $R^2$), then use the fitted model to extrapolate rate and integrate for cumulative production.
Linearizing transform. For hyperbolic decline, $q^{-b}=q_i^{-b}+q_i^{-b}bD_i\,t$ is linear in $t$ for the correct $b$. Scanning $b$ from 0.01 to 2.00 and taking a linear regression of $q^{-b}$ on $t$ (in days) for each trial, the best linear fit ($R^2=0.99998$) occurs at
$$b\approx0.519,\qquad q_i\approx1{,}565.4\text{ STB/day},\qquad D_i\approx0.002985\text{ /day}$$
recovered from the fitted intercept $q_i^{-b}$ and slope $q_i^{-b}bD_i$. The fit reproduces the table closely (e.g. 1,565 vs. 1,568 STB/day at $t=0$; 160.3 vs. 161 at $t=48$ mo).
Projected rate at $t=114$ months (end of 1995).
$$q(t)=\frac{q_i}{(1+bD_it)^{1/b}}$$
With $t=114\text{ mo}=3{,}469.8$ days,
$$q(114\text{ mo})=\frac{1{,}565.4}{\left(1+0.519(0.002985)(3{,}469.8)\right)^{1/0.519}}$$
$$\boxed{q(114\text{ mo})\approx 44.1\text{ STB/day}}$$
Incremental cumulative production, Month 66 to Month 114. Integrating the hyperbolic rate equation,
$$N_p(t)=\frac{q_i^{\,b}}{(1-b)D_i}\Big[q_i^{\,1-b}-q(t)^{1-b}\Big]$$
Evaluating at $t=66$ mo ($N_p=796{,}233$ STB) and $t=114$ mo ($N_p=894{,}443$ STB) and subtracting:
$$\Delta N_p=894{,}443-796{,}233$$
$$\boxed{\Delta N_p\approx 98{,}210\text{ STB}}$$
(Confirmed by direct numerical (Simpson) integration of $q(t)$ over the same interval, which agrees to within 1 STB.)
