24-Pet-A2 Petroleum Reservoir Fluids · December 2018
Question 7 of 7: Laboratory PVT Data — Formation Volume Factor and Solution GOR
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 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); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, pseudo-critical property correlations, gas/oil PVT relations); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations, Gibbs' phase rule).
Question 7: Laboratory PVT Data — Formation Volume Factor and Solution GOR (20 marks)
Find. (a) $B_o$ (resbbl/STB), $R_{so}$ (SCF/STB); (b) $B_t$ and $R_{so}$ at 2253 psia.
Approach. Part (a) is a single differential liberation: $B_o$ is the reservoir-to-stock-tank volume ratio, and $R_{so}$ is the released gas volume per unit stock-tank oil volume, converted from cc/cc to SCF/STB with the 5.6146 ft$^3$/bbl factor. Part (b) first reads $R_{sob}$ and $B_{ob}$ directly at the bubble point (all the charged gas is, by definition, dissolved there), then uses the formula-sheet total-FVF relation $B_t=B_o+B_g(R_{sob}-R_{so})$ to back out $R_{so}$ at the lower, two-phase pressure.
Part (a) — oil formation volume factor. $B_o=\dfrac{V_{res}}{V_{STB}}=\dfrac{400\ \text{cc}}{274\ \text{cc}}$, so $\boxed{B_o\approx1.4599\ \text{resbbl/STB}}$ (a volume RATIO, so it is the same number in cc/cc as in resbbl/STB).
Part (a) — solution gas-oil ratio. Converting the 274 cc stock-tank oil volume to STB ($1\ \text{bbl}=158{,}987\ \text{cc}$): $V_{STB}=274/158{,}987=1.7234\times10^{-3}\ \text{STB}$. Then $R_{so}=\dfrac{1.21\ \text{SCF}}{1.7234\times10^{-3}\ \text{STB}}$, so $\boxed{R_{so}\approx702\ \text{SCF/STB}}$.
Part (b) — bubble-point $R_{sob}$ and $B_{ob}$. At the bubble point, ALL 37,500 cc of the charged standard-condition gas has just dissolved into the 274 cc of standard-condition oil, so the cc/cc solution ratio there is $R_{sob,raw}=37{,}500/274=136.86\ \text{scc/scc}$; converting with the same 5.6146 ft$^3$/bbl factor, $\boxed{R_{sob}\approx768\ \text{SCF/STB}}$. The bubble-point oil FVF is $B_{ob}=400/274$, so $\boxed{B_{ob}\approx1.4599\ \text{resbbl/STB}}$ — identical to part (a)'s $B_o$, confirming this is the SAME fluid sample being characterized by two independent PVT experiments.
Part (b) — total formation volume factor at 2253 psia. Below the bubble point the cell holds both oil and evolved gas, so $B_t$ is the TOTAL cell volume per unit stock-tank oil: $B_t=\dfrac{V_{o}+V_g}{V_{STB}}=\dfrac{388+42}{274}=\dfrac{430}{274}$, so $\boxed{B_t\approx1.5693}$. (The liquid-only FVF at this pressure is $B_o=388/274\approx1.4161$, needed for the next step.)
Part (b) — solution GOR at 2253 psia. Rearranging the formula sheet's $B_t=B_o+B_g(R_{sob}-R_{so})$ for $R_{so}$, working in the same raw cc/cc units as the given $B_g=0.01$: $R_{sob,raw}-R_{so,raw}=\dfrac{B_t-B_o}{B_g}=\dfrac{1.5693-1.4161}{0.01}=15.33$, so $R_{so,raw}=136.86-15.33=121.53\ \text{scc/scc}$. Converting to field units with the same 5.6146 factor: $\boxed{R_{so}\approx682\ \text{SCF/STB}}$ at 2253 psia — correctly LESS than $R_{sob}\approx768\ \text{SCF/STB}$, since gas has come out of solution as pressure dropped below the bubble point.
