Question 2 of 7: Ternary Flash — Methane / CO 2 / n-Butane
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) $y_{nC_4}$, $x_{nC_4}$; (b) degrees of freedom $F$; (c) $K_i$ for each component; (d) vapour mole fraction $V$ given overall $z_{CO_2}=0.25$.
Approach. Close each phase's composition to 100% for part (a), apply Gibbs' phase rule $F=C-P+2$ for part (b), form $K_i=y_i/x_i$ for part (c), then use the formula-sheet flash relation together with an overall material balance on CO$_2$ to solve for the vapour fraction $V$.
Part (a) — n-butane composition by closure. Each phase's mole fractions sum to 1: $y_{nC_4}=1-y_{C_1}-y_{CO_2}=1-0.85-0.12$, so $\boxed{y_{nC_4}=0.03}$ (3% in the vapour). Likewise $x_{nC_4}=1-x_{C_1}-x_{CO_2}=1-0.15-0.30$, so $\boxed{x_{nC_4}=0.55}$ (55% in the liquid).
Part (b) — degrees of freedom. Gibbs' phase rule: $F=C-P+2$, with $C=3$ components and $P=2$ phases (liquid + vapour) in equilibrium: $F=3-2+2$, so $\boxed{F=3}$. Three independent intensive variables (e.g. $T$, $p$, and one composition) fully fix the state of this two-phase ternary system.
Part (c) — equilibrium K-values. $K_i=y_i/x_i$ for each component: $K_{C_1}=0.85/0.15=5.667$; $K_{CO_2}=0.12/0.30=0.400$; $K_{nC_4}=0.03/0.55=0.0545$. So $\boxed{K_{C_1}=5.667,\ K_{CO_2}=0.400,\ K_{nC_4}=0.0545}$. As expected for a mixture of a very light, a moderately volatile, and a much heavier component, $K_{C_1}\gg1$ (methane concentrates in the vapour), $K_{nC_4}\ll1$ (n-butane concentrates in the liquid), and $K_{CO_2}$ sits below 1 but closer to it.
Part (d) — vapour mole fraction from the CO2 balance. An overall material balance on any component splits between the two phases as $z_i=x_i(1-V)+y_iV$, where $V$ is the mole fraction of the total system that is vapour and $x_i,y_i$ are the (fixed) equilibrium compositions already measured. Applying this to CO$_2$: $z_{CO_2}=x_{CO_2}(1-V)+y_{CO_2}V=0.30-(0.30-0.12)V$. Substituting $z_{CO_2}=0.25$: $0.25=0.30-0.18V \Rightarrow V=\dfrac{0.30-0.25}{0.18}=\dfrac{0.05}{0.18}$, so $\boxed{V\approx0.278}$ (about 27.8% of the total moles in the cell are vapour). This is confirmed against the formula sheet's own flash form $x_{CO_2}=\dfrac{z_{CO_2}}{1+V(K_{CO_2}-1)}=\dfrac{0.25}{1+0.278(0.400-1)}=\dfrac{0.25}{0.833}=0.30$, which reproduces the measured liquid composition exactly.
