Question 4 of 7: Black-Oil PVT Table — Bubble Point, F.V.F., Compressibility
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Pet-A2 — Petroleum Reservoir Fluids · National Exams, May 2018 · 3 hours, closed book, Casio/Sharp approved calculators only · a formula sheet is provided; FIVE (5) questions constitute a complete exam paper (the first five as submitted are marked); all questions equal value, all parts of a multipart question equal weight; oilfield-unit questions must be answered in field units.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (Ch. 1–2, PVT properties, reservoir/well-stream classification); 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. (phase behaviour, black-oil PVT laboratory data, gas hydrates/waxes/asphaltenes); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, well-stream gravity, pseudo-critical property correlations); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations, Gibbs' phase rule).
Find. (a) saturation state at $p_i=3000$ psia; (b) bubble point pressure $P_b$; (c) $B_{ob}$, $R_{sb}$; (d) $B_t$ at 1614.7 and 2780 psia; (e) $c_o$ above $P_b$; (f) $V/V_i$ from 3000 psia to $P_b$.
Approach. Read the bubble point directly off the table as the pressure where $R_s$ stops being constant (gas begins evolving) and $B_o$ reaches its maximum; apply $B_t=B_o+B_g(R_{sob}-R_{so})$ for part (d), a two-point secant on the undersaturated $B_o$ data for $c_o=-\frac{1}{B_{ob}}\left(\frac{dB_o}{dP}\right)_T$ for part (e), and integrate that (nearly-constant) compressibility for part (f).
Part (a) — saturated or undersaturated at 3000 psia. $R_s$ is constant at 740 SCF/STB for every pressure from 2780 psia down to 2144.7 psia — a clear sign that no free gas has evolved yet in that range (all the solution gas stays dissolved, so $R_s$ can't decline). Since the given initial pressure $3000$ psia is even higher than the top of that constant-$R_s$ range, $\boxed{\text{the oil is undersaturated at }p_i=3000\text{ psia}}$.
Part (b) — bubble point pressure. The bubble point is the pressure at which $R_s$ first starts to decline (gas first begins coming out of solution) — equivalently, where $B_o$ reaches its local maximum, since below $P_b$ the loss of dissolved gas outweighs the liquid's own pressure-expansion and $B_o$ turns over and falls. Both signatures point to the same row: $R_s$ drops from 740 to 702 SCF/STB, and $B_o$ falls from its peak of 1.473 to 1.456 rbbl/STB, immediately below 2144.7 psia. So $\boxed{P_b=2144.7\text{ psia}}$.
Part (c) — properties at the bubble point. Reading the $P_b=2144.7$ psia row directly: $\boxed{B_{ob}=1.473\text{ rbbl/STB}}$, $\boxed{R_{sb}=740\text{ SCF/STB}}$.
Part (d) — total (two-phase) F.V.F. $B_t=B_o+B_g(R_{sob}-R_{so})$, with $B_g$ converted from ft$^3$/SCF to rbbl/SCF using $1\ \text{bbl}=5.6146\ \text{ft}^3$. At 2780 psia (above $P_b$), no free gas exists yet, so $R_{so}=R_{sob}=740$ SCF/STB and the correction term vanishes: $B_t=B_o+B_g\times0=1.464$, so $\boxed{B_t(2780)=1.464\text{ rbbl/STB}}$ (identical to $B_o$, as expected above the bubble point). At 1614.7 psia (below $P_b$): $B_g=0.009714\ \text{ft}^3/\text{SCF}\div5.6146=0.0017301\ \text{rbbl/SCF}$, and $R_{so}=588$ SCF/STB, so $B_t=1.405+0.0017301\times(740-588)=1.405+0.0017301\times152=1.405+0.2630$, giving $\boxed{B_t(1614.7)\approx1.668\text{ rbbl/STB}}$.
Part (e) — isothermal oil compressibility above $P_b$. Using the two extreme undersaturated points, 2780 psia ($B_o=1.464$) and 2144.7 psia ($B_o=1.473$): $\left(\dfrac{dB_o}{dP}\right)_T\approx\dfrac{1.473-1.464}{2144.7-2780}=\dfrac{0.009}{-635.3}=-1.417\times10^{-5}\ \text{rbbl/STB/psi}$. Then $c_o=-\dfrac{1}{B_{ob}}\left(\dfrac{dB_o}{dP}\right)_T=-\dfrac{1}{1.473}\times(-1.417\times10^{-5})$, so $\boxed{c_o\approx9.62\times10^{-6}\ \text{psi}^{-1}}$ (a typical order of magnitude for undersaturated black-oil compressibility). Checking with the adjacent pair (2500→2144.7 psia) gives $c_o\approx9.55\times10^{-6}\ \text{psi}^{-1}$, confirming $B_o$ is essentially linear with $P$ over this undersaturated range.
Part (f) — relative volume expansion, 3000 psia → $P_b$. Integrating the (nearly constant) compressibility $c_o=-\frac{1}{V}\left(\frac{dV}{dP}\right)_T$ from the initial pressure down to the bubble point gives $\dfrac{V}{V_i}=\exp\!\big[c_o\,(p_i-P_b)\big]$. With $p_i-P_b=3000-2144.7=855.3$ psi: $\dfrac{V}{V_i}=\exp\!\big[9.62\times10^{-6}\times855.3\big]=\exp(0.00823)$, so $\boxed{V/V_i\approx1.0083}$ — the undersaturated oil expands by about 0.83% as reservoir pressure declines from 3000 psia to the bubble point, before any gas begins to evolve.
