24-Pet-A2 Petroleum Reservoir Fluids · December 2016
Question 5 of 7: Two-Phase Flash — Hexane/Methane in a PVT Cell
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 2016 · 3 hours, closed book, non-communicating calculator only · first five questions in the answer book are marked, all questions equal value, all parts of a multipart question equal weight.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (Ch. 1–2, PVT properties, reservoir/well-stream classification, material balance); 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. (reservoir fluid classification, material balance, well-stream gravity); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations).
Question 5: Two-Phase Flash — Hexane/Methane in a PVT Cell (20 marks)
Assumption (per exam Note 1): the question gives the cell split as 80% liquid and 20% gas by volume. Converting a volume split into the mole fractions $L$, $V$ used in the flash relations would need the molar volumes (densities) of both phases at cell conditions, and the paper gives neither. The split is therefore taken as the phase mole fractions $L=0.8$, $V=0.2$, the only reading the given data support.
Given. Basis: $n_{C6}=1$ mol hexane in the cell; phase split $L=0.8$, $V=0.2$ (taken as mole fractions, $L+V=1$); $K_{C6}=0.01$, $K_{C1}=10$. Formula sheet: $\sum_i\dfrac{z_i}{1+V(K_i-1)}=1$, $x_i=\dfrac{z_i}{1+V(K_i-1)}$, $y_i=K_ix_i$, $\sum x_i=\sum y_i=1$. Because $L=1-V$, these are equivalent to the Rachford–Rice form $\sum_i\dfrac{z_i(K_i-1)}{1+V(K_i-1)}=0$ with $x_i=\dfrac{z_i}{L+VK_i}$.
Find. (a) Moles of methane $n_{C1}$ charged to the cell; (b) $x_{C6},x_{C1},y_{C6},y_{C1}$.
Approach. For this binary system, Rachford–Rice with $z_{C1}=1-z_{C6}$ reduces to one linear equation in the feed composition for the stated $V$; solve it to get the feed mole fractions (hence, from the 1-mol hexane basis, the moles of methane charged), then recover the phase compositions from $x_i=z_i/(L+VK_i)$ and $y_i=K_ix_i$.
Part (a) — set up and solve for the feed composition. $\dfrac{z_{C6}(K_{C6}-1)}{1+V(K_{C6}-1)}+\dfrac{z_{C1}(K_{C1}-1)}{1+V(K_{C1}-1)}=0$. With $V=0.2$: hexane coefficient $=\dfrac{0.01-1}{1+0.2(-0.99)}=\dfrac{-0.99}{0.802}=-1.2344$; methane coefficient $=\dfrac{10-1}{1+0.2(9)}=\dfrac{9}{2.8}=3.2143$. So $-1.2344\,z_{C6}+3.2143\,z_{C1}=0 \Rightarrow z_{C1}/z_{C6}=0.3840$.
Solve with $z_{C6}+z_{C1}=1$. $z_{C6}(1+0.3840)=1\Rightarrow\boxed{z_{C6}=0.7225}$, $\boxed{z_{C1}=0.2775}$ (feed mole fractions).
Moles of methane. Since $z_{C6}=n_{C6}/n_{total}=1/n_{total}=0.7225$, total feed $n_{total}=1/0.7225=1.3840$ mol. Moles of methane charged $=n_{total}-n_{C6}=1.3840-1=\boxed{n_{C1}=0.384\ \text{mol methane}}$ per mole of hexane, to hit the specified $L=0.8,\,V=0.2$ split at these $K$-values.
Part (b) — liquid-phase composition. $x_{C6}=\dfrac{z_{C6}}{L+VK_{C6}}=\dfrac{0.7225}{0.8+0.2(0.01)}=\dfrac{0.7225}{0.802}=0.9009$. $x_{C1}=\dfrac{z_{C1}}{L+VK_{C1}}=\dfrac{0.2775}{0.8+0.2(10)}=\dfrac{0.2775}{2.8}=0.0991$. Check: $x_{C6}+x_{C1}=1.0000$. So $\boxed{x_{C6}=0.9009,\ x_{C1}=0.0991}$ — the liquid phase is, sensibly, overwhelmingly hexane.
Vapour-phase composition. $y_{C6}=K_{C6}x_{C6}=0.01(0.9009)=0.00901$. $y_{C1}=K_{C1}x_{C1}=10(0.0991)=0.9910$. Check: $y_{C6}+y_{C1}=1.0000$. So $\boxed{y_{C6}=0.00901,\ y_{C1}=0.9910}$ — the vapour phase is, equally sensibly, almost pure methane, consistent with $K_{C1}=10\gg1$ (very volatile) and $K_{C6}=0.01\ll1$ (very non-volatile) for this pairing.
