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

Question 4: Bubble Point and Formation Volume Factors from 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; formula sheet relations $B_t=B_o+B_g(R_{sob}-R_{so})$ and $c=-\dfrac{1}{B_{ob}}\left(\dfrac{dB_o}{dP}\right)_T$; unit conversion 1 bbl $=5.615\ \text{ft}^3$.

Find. (a) bubble-point pressure $p_b$; (b) $B_t$ at $p_b$; (c) isothermal oil compressibility at 4000 psia; (d) $B_t$ at 2000 psia.

Approach. Locate $p_b$ as the pressure at which $R_s$ stops declining (all gas still in solution) and $B_o$ is at its maximum; apply the two-phase FVF formula at and below $p_b$, converting the table's $B_g$ from ft$^3$/SCF to bbl/SCF so it is dimensionally consistent with $B_o$; estimate the compressibility slope from the bracketing table points.

1. Part (a) — Bubble point. Reading down the table, $R_s$ is constant at 950 SCF/STB for $p=4500,4000,3500$ psia and then drops to 810 SCF/STB at 3000 psia — free gas can only appear once $p$ drops below $p_b$, so $p_b$ is the lowest pressure at which $R_s$ is still at its plateau value. This is confirmed by $B_o$: it rises with declining pressure above $p_b$ (pure liquid expansion: 1.31→1.32→1.33) and only turns over and falls once gas begins evolving (1.33→1.30). The peak $B_o=1.33$ bbl/STB at $p=3500$ psia locates $\boxed{p_b = 3500\ \text{psia}}$.
2. Part (b) — $B_t$ at bubble point. At $p_b$ the oil is exactly saturated, so $R_{so}=R_{sob}=950$ SCF/STB and the free-gas term vanishes: $B_t=B_o+B_g(R_{sob}-R_{so})=1.33+B_g(0)$. So $\boxed{B_t(p_b) = 1.33\ \text{bbl/STB}}$ — at the bubble point the total and oil FVF are identical, by definition.
3. Part (c) — Compressibility at 4000 psia. 4000 psia sits between the two undersaturated table points 4500 psia ($B_o=1.31$) and 3500 psia ($B_o=1.33$); estimate the slope by central difference: $\left(\dfrac{dB_o}{dP}\right)_T \approx \dfrac{1.33-1.31}{3500-4500}=\dfrac{0.02}{-1000}=-2.0\times10^{-5}\ \text{bbl/STB per psi}$. Using $B_{ob}=1.33$ bbl/STB from Part (b): $c=-\dfrac{1}{1.33}\times(-2.0\times10^{-5})$, giving $\boxed{c = 1.50\times10^{-5}\ \text{psi}^{-1}}$ at 4000 psia.
4. Part (d) — $B_t$ at 2000 psia. Below $p_b$: $B_o(2000)=1.23$ bbl/STB, $R_{so}(2000)=550$ SCF/STB, $B_g(2000)=0.0070185\ \text{ft}^3/\text{SCF}$. Convert $B_g$ to bbl/SCF: $B_g=0.0070185/5.615=0.0012500\ \text{bbl/SCF}$. Then $B_t=B_o+B_g(R_{sob}-R_{so})=1.23+0.0012500\times(950-550)=1.23+0.0012500\times400=1.23+0.500$. So $\boxed{B_t(2000\ \text{psia}) = 1.730\ \text{bbl/STB}}$.