Question 3 of 7: Athabasca Bitumen — Solution GOR and Formation Volume Factor
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).
Question 3: Athabasca Bitumen — Solution GOR and Formation Volume Factor (20 marks)
Find. (a) mole fraction of dissolved methane in the live (reservoir) bitumen; (b) formation volume factor $B_o$ of the bitumen at reservoir conditions.
Approach. Take 1 Sm$^3$ of stock-tank bitumen as the basis; convert the GOR and the bitumen mass to moles using the given molar volume and molecular weights, then form the mole fraction directly. For part (b), the live oil at reservoir conditions is the same stock-tank bitumen mass plus the dissolved-gas mass; convert that combined mass to a reservoir volume via $V=m/\rho$ and divide by the (unit) stock-tank volume.
Moles of dissolved gas and bitumen (basis: 1 Sm3 bitumen). $4\ \text{Sm}^3$ gas $=4000$ L; at 22.4 L/mol, $n_{gas}=\dfrac{4000}{22.4}=178.6$ mol. The bitumen mass is $\rho_{std}\times1\ \text{m}^3=1005\ \text{kg}=1{,}005{,}000$ g, so $n_{bit}=\dfrac{1{,}005{,}000}{550}=1827.3$ mol.
Part (a) — mole fraction of methane. $x_{CH_4}=\dfrac{n_{gas}}{n_{gas}+n_{bit}}=\dfrac{178.6}{178.6+1827.3}=\dfrac{178.6}{2005.8}$, so $\boxed{x_{CH_4}\approx0.0890}$ (8.90 mol%) — consistent with a solution GOR this low (bitumen dissolves very little gas compared to a conventional black oil).
Part (b) — live-oil mass and reservoir volume. The reservoir (live) oil is the same 1005 kg of bitumen plus the mass of methane that stays in solution: $m_{gas}=n_{gas}\times MW_{gas}=178.6\times16=2857$ g $=2.857$ kg. Total live-oil mass $=1005+2.857=1007.9$ kg, unchanged from surface to reservoir (mass is conserved; only volume and density change). Using $V=m/\rho$ at reservoir density: $V_{res}=\dfrac{1007.9\ \text{kg}}{990\ \text{kg/m}^3}=1.0180\ \text{m}^3$.
Formation volume factor. $B_o=\dfrac{V_{res}}{V_{STO}}=\dfrac{1.0180\ \text{m}^3}{1\ \text{m}^3}$ (the stock-tank volume is the 1 Sm$^3$ basis), so $\boxed{B_o\approx1.018\ \text{res-m}^3/\text{std-m}^3}$. This is a very small formation volume factor for reservoir oil — direct evidence of how little gas an Athabasca bitumen holds in solution compared to a light or medium crude, whose $B_o$ is typically 1.1–1.5 or higher.
Quantity
Value
(a) Mole fraction dissolved methane, $x_{CH_4}$
0.0890 (8.90 mol%)
(b) Formation volume factor, $B_o$
1.018 res-m$^3$/std-m$^3$
Check: assumes the dissolved gas is ideal at standard conditions (22.4 L/mol, given) and that reservoir bitumen mass is exactly the sum of surface bitumen mass and dissolved-gas mass (no other components exchanged) — standard black-oil PVT assumptions, appropriate at this exam's level.
