24-Pet-A2 Petroleum Reservoir Fluids · December 2016
Question 4 of 7: Composition — Mole to Mass Fraction and Stream Recombination
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 4: Composition — Mole to Mass Fraction and Stream Recombination (20 marks)
Check: the source table lists the C6+ (hexanes-plus) molecular weight as 72.15 lb$_m$/lb-mol — identical to the pentane rows above it. A lumped C6+ fraction normally carries a higher MW (typically 90–115+, since it bundles hexane through the heaviest tail components); it is used exactly as printed, with this note flagging the anomaly.
Given. Reservoir-fluid composition (mole %) with molecular weights, part (a) table above (sums to 100.00%); second gas stream composition (mole %) for part (b), part (b) table above (sums to 100.00%); mixing basis 0.7 mol reservoir fluid + 0.3 mol second gas.
Find. (a) Mass fraction of each component; (b) mole-fraction composition of the 0.7:0.3 mixture.
Approach. For part (a), convert each component's mole fraction to a mass contribution via $m_i\propto y_iM_i$, then normalize by the total mixture mass. For part (b), since both streams are metered in moles, the resultant composition is simply the mole-fraction-weighted average of the two streams, $z_i=0.7\,y_{1,i}+0.3\,y_{2,i}$, with any component absent from a stream (H$_2$S, CO$_2$ in stream 2) taken as zero there.
Part (a) — mass contributions and normalization. Compute $y_iM_i$ for each component (per 100 mol basis) and sum: $M_{av}=\sum y_iM_i = 0.83(28.01)+0.12(34.08)+0.93(44.01)+24.62(16.04)+6.26(30.07)+6.77(44.10)+1.86(58.12)+4.71(58.12)+2.72(72.15)+1.55(72.15)+49.63(72.15)$, all divided by 100 for the mole-fraction basis, giving $\boxed{M_{av}=52.21\ \text{lb}_m/\text{lb-mol}}$. Each mass fraction is then $w_i=y_iM_i/\sum(y_jM_j)$.
Part (a) — result. Because C6+ carries both the largest mole fraction (49.63%) and a molecular weight roughly 4.5× methane's, it dominates the mass basis even more than the mole basis: $\boxed{w_{C6+}=68.59\%}$ of the total mass, versus $\boxed{w_{C1}=7.56\%}$ for methane (24.62% by mole, but light) and under 1% each for the three non-hydrocarbons. Full mass-fraction table below.
Part (b) — weighted recombination. For each component, $z_i=0.7\,y_{1,i}+0.3\,y_{2,i}$. E.g. methane: $z_{C1}=0.7(24.62)+0.3(70.27)=17.234+21.081=38.315\%$; C6+: $z_{C6+}=0.7(49.63)+0.3(1.02)=34.741+0.306=35.047\%$; CO$_2$ (absent from stream 2): $z_{CO_2}=0.7(0.93)+0.3(0)=0.651\%$. Every other component follows the same weighted sum; the full result sums to $100.00\%$, confirming the mole-fraction basis is preserved.
