Radius of investigation at the end of wellbore storage: 24-Pet-B5 Reservoir Mechanics, December 2014, Q3

Question 3: Radius of investigation at the end of wellbore storage (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Oil production rate	$q$	400 STB/D
Initial reservoir pressure	$p_i$	2000 psia
Formation thickness	$h$	20 ft
Porosity	$\phi$	0.25
Wellbore radius	$r_w$	0.25 ft
Oil formation volume factor	$B_o$	1.2 bbl/STB
Total compressibility	$c_t$	$1\times10^{-6}\ \text{psi}^{-1}$
Oil viscosity	$\mu$	1 cp

Check: the paper prints $c_t=1\times10^{6}\ \text{psi}^{-1}$ (positive exponent); a compressibility of $10^{6}\ \text{psi}^{-1}$ is physically impossible (typical oilfield $c_t\sim10^{-6}$ to $10^{-5}\ \text{psi}^{-1}$), so this is treated as a misprint for $1\times10^{-6}\ \text{psi}^{-1}$.

Find. The radius of investigation $r_{inv}$ at the moment wellbore-storage effects end, read from the log-log and semi-log plots of the drawdown test.

Approach. Read the permeability off the slope of the semilog straight line, identify the time at which wellbore storage ends as the point where that straight line begins (confirmed against the log-log plot), then substitute both into the formula-sheet radius-of-investigation relation.

Fig. 1 – Semilog drawdown plot, $p_{wf}$ vs. $\log t$ (read from the printed figure). The straight-line (infinite-acting radial-flow) region runs from about $t=0.1$ hr to $t\approx30$–40 hr; the late-time droop beyond $\sim$100 hr reflects the closed outer boundary, not measurement error.

Fig. 2 – Log-log plot of $|p_i-p_{wf}|$ vs. $t$. The early, steep rise flattens into the same infinite-acting trend by $t\approx0.1$ hr, corroborating the semilog plot's straight-line onset.

Read the semilog slope $m$. Four points lying on the straight-line portion of Fig. 1 – $(0.1\text{ hr},1570)$, $(1\text{ hr},1535)$, $(10\text{ hr},1495)$, $(100\text{ hr},1450)$ psia – fit a least-squares line of slope $$m\approx 40\ \text{psi/cycle}\quad\text{(Check: read from the printed chart)}$$
Permeability from the semilog slope. Using the drawdown-test slope relation from the formula sheet, $$k=\frac{162.6\,q\mu B_o}{mh}=\frac{162.6(400)(1)(1.2)}{(40)(20)}$$ $$\boxed{k\approx 97.6\ \text{mD}}$$
Identify the end of wellbore storage. Wellbore storage distorts the data until the true infinite-acting radial-flow (semilog straight-line) response begins. Both Figs. 1 and 2 show that onset at essentially the same time – the semilog curve straightens out and the log-log curve's early steep rise flattens into the same trend – consistently by $$t_{wbs}\approx0.1\ \text{hr}$$ (the two independent plots agreeing on this time is itself the self-check that the reading is right).
Radius of investigation at that time. Substituting $k$, $t_{wbs}$ and the given fluid/rock properties into the formula sheet's radius-of-investigation relation, $$r_{inv}\approx\sqrt{\frac{kt}{948\,\phi\mu c_t}}=\sqrt{\frac{(97.6)(0.1)}{948(0.25)(1)(1\times10^{-6})}}$$ $$\boxed{r_{inv}\approx 203\ \text{ft}}$$

Result	Value
Semilog slope, $m$	≈ 40 psi/cycle
Permeability, $k$	≈ 97.6 mD
Time wellbore storage ends, $t_{wbs}$	≈ 0.1 hr
Radius of investigation, $r_{inv}$	≈ 203 ft

Topic. Wellbore storage and the radius of investigation.

Key relations. $m=162.6q\mu B_o/(kh)$; $r_{inv}\approx\sqrt{kt/(948\phi\mu c_t)}$.

Why this works. A drawdown response only carries true reservoir information once wellbore storage has died out – before that, ALL of the surface rate is simply unloading/loading the wellbore's own fluid column, and the measured $p_{wf}$ says nothing about $k$. The onset of the semilog straight line is exactly the diagnostic that marks this transition, and substituting that time back into the radius-of-investigation formula answers "how far had the transient actually reached the reservoir by the time useful data started."

Common pitfall. Using the FIRST data point on the plot (here $t=0.01$ hr) as "the" time for this calculation – that point sits deep in the wellbore-storage-dominated region and would understate $r_{inv}$ by a factor of $\sqrt{10}\approx3$; the question specifically asks for the radius at the end of wellbore storage, not at the start of the test.

Source. Lee, Well Testing, ch.3 (wellbore storage) & ch.2 (radius of investigation); Earlougher, Advances in Well Test Analysis, ch.2.

Question 3 of 7: Radius of investigation at the end of wellbore storage

Question 3: Radius of investigation at the end of wellbore storage (20 marks)