Drawdown test – permeability and skin from tabulated data: 24-Pet-B5 Reservoir Mechanics, May 2015, Q3

Question 3: Drawdown test – permeability and skin from tabulated data (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
Flow rate	$q$	90 STBD
Formation thickness	$h$	17 ft
Formation volume factor	$B_o$	1.1 bbl/STB
Initial reservoir pressure	$p_i$	1479 psia
Porosity	$\phi$	0.25
Total compressibility	$c_t$	$7.65\times10^{-6}\ \text{psi}^{-1}$
Oil viscosity	$\mu$	7.86 cP
Wellbore radius	$r_w$	0.21 ft

Drawdown pressure data
$t$ (hr)	$p_{wf}$ (psia)
1	284
3	215
5	183
7	162
9	147
11	134
13	124
15	115

Find. Formation permeability $k$ and skin factor $S$.

Approach. The blank semilog grid supplied with the exam is graph paper for the candidate to plot the table above; plotting $p_{wf}$ against $\log t$ and fitting the straight line gives the slope $m$ directly from the printed data (no chart digitization needed for this question, since the data points themselves are already tabulated numerically). $k$ follows from $m$, and skin from the extrapolated $p_{1hr}$ value together with $p_i$.

Fig. 2 – Semilog drawdown plot of the tabulated data, $p_{wf}$ vs. $\log t$, with the least-squares straight line.

Fit the semilog straight line. A least-squares regression of $p_{wf}$ against $\log_{10}t$ over all eight tabulated points gives $$p_{wf}=283.7-143.6\log_{10}t\qquad\Rightarrow\qquad m\approx 143.6\ \text{psi/cycle},\quad p_{1hr}\approx 283.7\ \text{psia}$$ ($p_{1hr}$ agrees almost exactly with the printed $t=1$ hr data point, 284 psia, confirming the eight points lie essentially on one straight line – there is no separate wellbore-storage-affected region to exclude.)
Permeability. Using the drawdown-test slope relation from the formula sheet, $$k=\frac{162.6\,q\mu B_o}{mh}=\frac{162.6(90)(7.86)(1.1)}{(143.6)(17)}$$ $$\boxed{k\approx 51.8\ \text{mD}}$$
Skin factor. With $p_i=1479$ psia and $p_{1hr}\approx283.7$ psia, the formula sheet's drawdown-test skin relation gives $$S=1.151\left[\frac{p_i-p_{1hr}}{m}-\log_{10}\!\left(\frac{k}{\phi\mu c_tr_w^2}\right)+3.23\right]$$ $$S=1.151\left[\frac{1479-283.7}{143.6}-\log_{10}\!\left(\frac{51.8}{(0.25)(7.86)(7.65\times10^{-6})(0.21)^2}\right)+3.23\right]$$ $$\boxed{S\approx 4.2}$$

Result	Value
Semilog slope, $m$	≈ 143.6 psi/cycle
$p_{1hr}$	≈ 283.7 psia
Permeability, $k$	≈ 51.8 mD
Skin factor, $S$	≈ 4.2

Topic. Semilog (MDH) drawdown-test analysis.

Key relations. $m=162.6q\mu B_o/(kh)$; $S=1.151[(p_i-p_{1hr})/m-\log(k/\phi\mu c_tr_w^2)+3.23]$.

Why this works. Once the near-wellbore transient has settled into infinite-acting radial flow, $p_{wf}$ falls linearly with $\log t$ (the Ei-function line-source solution, expanded for large argument); the slope of that line is set only by $k$ (everything else on the right-hand side is known), and the vertical offset between the ideal (zero-skin) line and the actual $p_{1hr}$ measures the extra, skin-driven pressure drop right at the sandface.

Common pitfall. A large positive skin (here $S\approx4.2$) means damage, not stimulation – it is easy to flip the sign convention and report the well as stimulated; a positive $S$ always corresponds to EXTRA pressure drop near the wellbore beyond the ideal radial-flow value, i.e. an under-producing, damaged well.

Source. Lee, Well Testing, ch.3–4; Earlougher, Advances in Well Test Analysis, ch.2.

Question 3 of 7: Drawdown test – permeability and skin from tabulated data

Question 3: Drawdown test – permeability and skin from tabulated data (20 marks)