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24-Pet-A2 Petroleum Reservoir Fluids · December 2015

Question 4 of 7: Bubble Point and Formation Volume Factors from Black-Oil PVT Data

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

Notes on this paper

98-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 2015 · 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.

Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (Ch. 1–2, PVT properties, reservoir/well-stream classification, material balance); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (Standing-Katz Z-factor correlation, gas properties); McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (black-oil PVT laboratory data, well-stream recombination); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (p/Z material balance, well-stream gravity); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations).

Question 4: Bubble Point and Formation Volume Factors from Black-Oil PVT 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. The PVT table above at $T=225\,{}^{\circ}\text{F}=685\,{}^{\circ}\text{R}$; formula sheet $B_t=B_o+B_g(R_{sob}-R_{so})$ and $c=-\dfrac{1}{B_{ob}}\left(\dfrac{dB_o}{dP}\right)_T$; $B_g=0.02827\,ZT/p$ (ft$^3$/SCF); 1 bbl $=5.6146\ \text{ft}^3$.

Find. (a) $p_b$; (b) $B_t$ at 2682 psia; (c) $c_o$ at 3500 psia; (d) $B_t$ at 800 psia.

Approach. Identify $p_b$ as the lowest pressure at which $R_s$ still sits on its constant plateau (and $B_o$ is at its maximum); above $p_b$, $B_t=B_o$; below $p_b$, add the liberated-gas volume via $B_g(R_{sob}-R_{so})$ after converting $B_g$ to bbl/SCF; estimate the undersaturated compressibility from the bracketing table points.

  1. Part (a) — Bubble point. $R_s$ holds constant at 632 SCF/STB for every pressure from 4500 down to 2682 psia (no free gas — single-phase liquid), then drops to 584 SCF/STB at 2500 psia once gas begins evolving. Correspondingly $B_o$ rises monotonically with declining pressure through this plateau (1.3474→1.3575→1.3686→1.3811→1.4040, pure liquid expansion) and only turns over and falls below it (1.4040→1.3782). The table's own listed break point is therefore $\boxed{p_b=2682\ \text{psia}}$.
  2. Part (b) — $B_t$ at 2682 psia. Since 2682 psia $=p_b$, the oil is exactly saturated and $R_{so}=R_{sob}=632$ SCF/STB, so the free-gas term vanishes and $B_t=B_o$. So $\boxed{B_t(2682)=1.4040\ \text{bbl/STB}}$.
  3. Part (c) — Compressibility at 3500 psia. 3500 psia is undersaturated and bracketed by the table points at 4000 psia ($B_o=1.3575$) and 3000 psia ($B_o=1.3811$); central-difference slope $\left(\dfrac{dB_o}{dP}\right)_T\approx\dfrac{1.3811-1.3575}{3000-4000}=\dfrac{0.0236}{-1000}=-2.360\times10^{-5}\ \text{bbl/STB per psi}$. Using $B_o(3500)=1.3686$ bbl/STB: $c_o=-\dfrac{1}{1.3686}\times(-2.360\times10^{-5})$, giving $\boxed{c_o=1.724\times10^{-5}\ \text{psi}^{-1}}$ at 3500 psia.
  4. Part (d) — $B_t$ at 800 psia. Below $p_b$: $B_o(800)=1.1791$ bbl/STB, $Z(800)=0.8808$, $R_s(800)=205$ SCF/STB. $B_g(800)=0.02827\dfrac{ZT}{p}=0.02827\times\dfrac{(0.8808)(685)}{800}=0.02827\times0.7541=0.02132\ \text{ft}^3/\text{SCF}$, i.e. $0.02132/5.6146=0.003797\ \text{bbl/SCF}$. Then $B_t=B_o+B_g(R_{sob}-R_{so})=1.1791+0.003797\times(632-205)=1.1791+0.003797\times427=1.1791+1.6215$. So $\boxed{B_t(800)=2.801\ \text{bbl/STB}}$.
QuantityValue
(a) Bubble-point pressure $p_b$2682 psia
(b) $B_t$ at 2682 psia1.4040 bbl/STB
(c) $c_o$ at 3500 psia$1.724\times10^{-5}\ \text{psi}^{-1}$
(d) $B_t$ at 800 psia2.801 bbl/STB