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24-Pet-A3 Fundamental Reservoir Engineering · May 2014

Question 4 of 7: Transient Pressure Response at an Observation Well

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

98-Pet-A3 — Fundamental Reservoir Engineering · National Exams, May 2014 · 3 hours, closed book, non-communicating calculator only · first five questions in the answer book are marked, all questions equal value, all parts of a multipart question equal weight. All seven questions are answered here as a complete study resource.

Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, transient well testing, radial flow); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, skin/productivity, relative permeability); McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, Z-factor); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (core analysis, capillary pressure).

Question 4: Transient Pressure Response at an Observation Well (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.

QuantityValue
Initial pressure $p_i$2000 psia
External radius $r_e$20,000 ft
Oil FVF $B_o$1.25 bbl/STB
Oil viscosity $\mu_o$1 cp
Total compressibility $c_t$$6\times10^{-6}$ psi$^{-1}$
Net pay $h$10 ft
Permeability $k$200 mD
Porosity $\phi$20%
Well radius $r_w$ / skin $S$0.3 ft / $-2$ (producing well only)
Rate history$q=500$ STBD, $0\le t\le10$ d; $q=0$, $10<t\le15$ d

Find. Pressure at an observation well 200 ft from the producer, 5 days after the producer is shut in (i.e. 15 days after production began).

Approach. Use the line-source (radial diffusivity) solution with superposition in time: model the production/shut-in history as a rate step of $+500$ STBD starting at $t=0$ plus a compensating rate step of $-500$ STBD starting at $t=10$ days. The observation well sees no skin (skin is a near-wellbore effect local to the producing well only) and is far enough from the producer, and well within $r_e$, that the infinite-acting line-source form applies throughout.

  1. Diffusivity and dimensionless time. $\eta=\dfrac{6.33k}{\phi\mu c_t}=\dfrac{6.33\times0.200}{0.20\times1\times6\times10^{-6}}=1.055\times10^{6}$ ft$^2$/day ($k=200$ mD $=0.200$ Darcy, per the formula sheet's units). Check infinite-acting: radius of investigation $\approx\sqrt{\eta t}=\sqrt{1.055\times10^6\times15}\approx3980$ ft, well inside $r_e=20{,}000$ ft, so the infinite-acting assumption is valid for the whole 15-day history.
  2. Dimensionless times at $r=200$ ft. For the full elapsed time $t=15$ d: $t_{D,1}=\eta t/r^2=1.055\times10^6\times15/200^2=395.6$. For the elapsed time since shut-in, $\Delta t=5$ d: $t_{D,2}=1.055\times10^6\times5/200^2=131.9$. Both exceed 100, so the log-approximation $p_D=\tfrac12(\ln t_D+0.809)$ applies to each.
  3. Dimensionless pressures. $p_{D,1}=\tfrac12(\ln 395.6+0.809)=\tfrac12(5.981+0.809)=3.395$. $p_{D,2}=\tfrac12(\ln131.9+0.809)=\tfrac12(4.882+0.809)=2.845$.
  4. Superpose the two rate steps. A producing rate of $+500$ STBD acting since $t=0$ contributes a drawdown $\propto p_{D,1}$; the compensating $-500$ STBD step acting since $t=10$ d (representing the well being shut in) contributes a partial build-up $\propto p_{D,2}$ with the opposite sign: $$\Delta p = \frac{0.141\,q\,\mu_o B_o}{kh}\left(p_{D,1}-p_{D,2}\right)=\frac{0.141\times500\times1\times1.25}{0.200\times10}\times(3.395-2.845)$$ $$\Delta p = 44.06\times0.550=24.2\text{ psi}$$
  5. Observation-well pressure. $p(r=200\text{ ft},t=15\text{ d})=p_i-\Delta p=2000-24.2$, so $\boxed{p=1975.8\text{ psia}}$.
QuantityValue
Diffusivity constant $\eta$$1.055\times10^6$ ft$^2$/day
$t_D$ at 15 d / 5 d395.6 / 131.9
Total pressure drop $\Delta p$ at observation well24.2 psi
Observation-well BHP after 5 d shut-in1975.8 psia