Given. Fourteen rate–time observations spanning 78 months of production, with the semilog plot showing a curving (non-exponential) decline.
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. Q3 – 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 integration of $q(t)$ over the same interval, which agrees to within 1 STB.)
