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24-Pet-A2 Petroleum Reservoir Fluids · December 2015

Question 5 of 7: Two-Phase Flash of an Oil–Solvent Mixture

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

98-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 2015 · 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. (black-oil PVT laboratory data, well-stream recombination); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (p/Z 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 of an Oil–Solvent Mixture (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.

Check: the question assigns $K_o=10$ (oil) and $K_s=0.01$ (solvent) — the reverse of the usual light-solvent/heavy-oil intuition ($K>1$ normally marks the more volatile component). The values are used exactly as given; the algebra below is self-consistent with them regardless of which physical species they are meant to represent.

Given. Target phase split $L=0.8$, $V=0.2$ (mole basis, $L+V=1$); $K_o=10$, $K_s=0.01$. Formula sheet: Rachford–Rice $\sum_i\dfrac{z_i(K_i-1)}{1+V(K_i-1)}=0$; $x_i=\dfrac{z_i}{L+VK_i}$; $y_i=K_ix_i$.

Find. (a) the feed mole fraction of solvent $z_s$ and the resulting moles of solvent per mole of oil $n_s/n_o$; (b) $x_o,x_s,y_o,y_s$.

Approach. For a binary system, the Rachford–Rice equation with $z_o=1-z_s$ reduces to one linear equation in $z_s$ for the stated $V$; solve it to get the feed composition (hence the mixing ratio), then recover phase compositions from $x_i=z_i/(L+VK_i)$ and $y_i=K_ix_i$.

  1. Part (a) — Set up and solve for the feed composition. $\dfrac{z_o(K_o-1)}{1+V(K_o-1)}+\dfrac{z_s(K_s-1)}{1+V(K_s-1)}=0$. With $V=0.2$: coefficient for oil $=\dfrac{10-1}{1+0.2(9)}=\dfrac{9}{2.8}=3.2143$; coefficient for solvent $=\dfrac{0.01-1}{1+0.2(-0.99)}=\dfrac{-0.99}{0.802}=-1.2344$. So $3.2143\,z_o=1.2344\,z_s$, i.e. $z_o/z_s=0.3840$.
  2. Solve with $z_o+z_s=1$. $z_s(1+0.3840)=1 \Rightarrow \boxed{z_s=0.7225}$, $\boxed{z_o=0.2775}$ (feed mole fractions of solvent and oil).
  3. Moles of solvent per mole of oil. $\dfrac{n_s}{n_o}=\dfrac{z_s}{z_o}=\dfrac{0.7225}{0.2775}$, giving $\boxed{n_s/n_o = 2.604\ \text{mol solvent per mol oil}}$ — i.e. about 2.6 moles of solvent must be charged per mole of oil to hit the target $L=0.8$, $V=0.2$ split at these $K$-values.
  4. Part (b) — Liquid-phase composition. $x_o=\dfrac{z_o}{L+VK_o}=\dfrac{0.2775}{0.8+0.2(10)}=\dfrac{0.2775}{2.8}=0.0991$. $x_s=\dfrac{z_s}{L+VK_s}=\dfrac{0.7225}{0.8+0.2(0.01)}=\dfrac{0.7225}{0.802}=0.9009$. Check: $x_o+x_s=1.000$. So $\boxed{x_o=0.0991,\ x_s=0.9009}$.
  5. Vapour-phase composition. $y_o=K_ox_o=10(0.0991)=0.9910$. $y_s=K_sx_s=0.01(0.9009)=0.00901$. Check: $y_o+y_s=1.000$. So $\boxed{y_o=0.9910,\ y_s=0.00901}$.
QuantityValue
Feed $z_o$, $z_s$0.2775, 0.7225
Moles solvent per mole oil2.604
Liquid $x_o$, $x_s$0.0991, 0.9009
Vapour $y_o$, $y_s$0.9910, 0.00901