NivaarExam PrepOfficial exam papers ↗

24-Pet-A2 Petroleum Reservoir Fluids · December 2019

Question 2 of 7: Ethane / n-Heptane Binary Phase Behaviour

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

Notes on this paper

EGBC National Exam — Petroleum Engineering, 17-Pet-A2 Petroleum Reservoir Fluids, 2019-Dec. 3 hours duration, closed book (ruler and approved calculator only). SEVEN questions are printed on the paper; per the exam notes, FIVE questions constitute a complete exam paper and only the first five as answered are marked. Every question is solved in full below (all seven, not just the five a candidate would normally submit) so this set also serves as complete study material.

Reference texts: McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PennWell); Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed.; Standing, M.B., Volumetric and Phase Behavior of Oil Field Hydrocarbon Systems; Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids.

Check: Questions 2 and 6 are built around two classic published P–T phase-diagram figures (the ethane/n-heptane system of Kay, Ind. Eng. Reading exact bubble/dew/critical points off these charts, as the exam intends, is not possible from this source. Every requested quantity in Q2 and Q6 is instead computed analytically: pseudo-critical properties via Kay's mixing rule (the exam's own formula sheet supplies exactly this rule) and bubble/dew points via the standard Wilson K-value correlation, $K_i = (P_{ci}/P)\exp[5.373(1+\omega_i)(1-T_{ci}/T)]$ — the textbook approximate method for hand/exam flash calculations. This gives fully verifiable, reproducible numbers in place of a chart reading, but they are engineering estimates, not a literal digitization — flagged at each affected step below.

Question 2: Ethane / n-Heptane Binary Phase Behaviour (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.

Given. Binary ethane(C2)/n-heptane(C7) mixtures at the weight fractions tabulated on the exam's composition table (curves No.1–10). Pure-component properties (standard values): C2 – $T_c=549.58\,{}^{\circ}\text{R}$, $P_c=706.5$ psia, $\omega=0.099$, $M=30.07$; n-C7 – $T_c=972.36\,{}^{\circ}\text{R}$, $P_c=396.8$ psia, $\omega=0.349$, $M=100.20$.

Find. The bubble/dew point conditions, critical properties and cricondentherm/cricondenbar of the specified compositions, and the reservoir-fluid classification of four of the tabulated mixtures.

Approach. Convert each weight-percent composition to mole fraction; use Kay's mixing rule (given directly on the exam's own formula sheet) for pseudo-critical properties, and the Wilson K-value correlation $K_i=(P_{ci}/P)\exp[5.373(1+\omega_i)(1-T_{ci}/T)]$ with $\sum z_iK_i=1$ (bubble point) or $\sum z_i/K_i=1$ (dew point) for the point bubble/dew conditions.

(a) 70.22wt% ethane / 29.78wt% n-heptane

  1. Convert to mole fraction. $n_{C2}=70.22/30.07=2.335$, $n_{C7}=29.78/100.20=0.2972$ (per 100 g). $$x_{C2}=\frac{2.335}{2.335+0.2972}=0.8871,\qquad x_{C7}=0.1129$$
  2. (i) Bubble-point pressure at 150°F. $T=610\,{}^{\circ}\text{R}$. With $E_i=\exp[5.373(1+\omega_i)(1-T_{ci}/T)]$, the bubble condition $\sum x_iK_i=1$ reduces to a closed form since $K_i\propto 1/P$: $$P_b = x_{C2}P_{c,C2}E_{C2}+x_{C7}P_{c,C7}E_{C7}$$ Evaluating $E_{C2}=1.795$, $E_{C7}=0.0135$: $$\boxed{P_b(150\,{}^{\circ}\text{F}) \approx 1122 \text{ psia } (\approx 1107 \text{ psig})}$$
  3. (ii)/(iv) Critical temperature and pressure – Kay's rule. $$T_{pc}=x_{C2}T_{c,C2}+x_{C7}T_{c,C7}=0.8871(549.58)+0.1129(972.36)=597.2\,{}^{\circ}\text{R}$$ $$P_{pc}=x_{C2}P_{c,C2}+x_{C7}P_{c,C7}=0.8871(706.5)+0.1129(396.8)=671.5\text{ psia}$$ $$\boxed{T_c \approx 597.2\,{}^{\circ}\text{R} = 137.6\,{}^{\circ}\text{F}\ ,\qquad P_c \approx 671.5\text{ psia}}$$
  4. (iii) Dew-point temperature at 450 psia. Solving $\sum x_i/K_i(T,450)=1$ numerically for $T$: $$\boxed{T_{dew}(450\text{ psia}) \approx 766.8\,{}^{\circ}\text{R} = 307.1\,{}^{\circ}\text{F}}$$
  5. (v) Bubble-point temperature at 700 psia. Solving $\sum x_iK_i(T,700)=1$ for $T$: $$\boxed{T_b(700\text{ psia}) \approx 560.0\,{}^{\circ}\text{R} = 100.4\,{}^{\circ}\text{F}}$$
  6. (vii) Dew-point pressure at 200°F. $T=659.67\,{}^{\circ}\text{R}$; $P_{dew}=1/\sum(x_i/(P_{ci}E_i))$: $$\boxed{P_{dew}(200\,{}^{\circ}\text{F}) \approx 107.5\text{ psia } (\approx 92.8\text{ psig})}$$
  7. (vi)/(viii) Cricondenbar and cricondentherm. These are the maximum pressure and maximum temperature on the closed two-phase envelope. Tracing the Wilson-correlation bubble/dew curves confirms they are qualitatively above $P_c$ and $T_c$ respectively (as they must be, by definition, whenever $P_c \ne$ cricondenbar), but the Wilson correlation is only accurate for point K-value estimates near typical reservoir conditions – it is not thermodynamically consistent close to the critical region and gives no reliable interior turning point for a full envelope trace (confirmed: the correlation's bubble/dew curves diverge rather than close into a loop at high $T$/$P$).
Check: cricondenbar and cricondentherm for the 70.22wt% mixture are not independently computed above (see step 7) – only the qualitative relation cricondentherm > $T_c$ (137.6°F) and cricondenbar > $P_c$ (671.5 psia) is asserted. The published Kay (1938) data for this exact system are well known in the reservoir-engineering literature for showing a pronounced critical-locus "bulge" for asymmetric ethane/heptane compositions; a numeric cricondentherm/cricondenbar would require reading the original chart.

Final Results – (a)

QuantityValue
(i) Bubble-point P @ 150°F≈ 1122 psia (1107 psig)
(ii) Critical temperature (Kay)≈ 137.6°F
(iii) Dew-point T @ 450 psia≈ 307.1°F
(iv) Critical pressure (Kay)≈ 671.5 psia
(v) Bubble-point T @ 700 psia≈ 100.4°F
(vi) Cricondenbar> 671.5 psia (chart-dependent, not numerically resolved)
(vii) Dew-point P @ 200°F≈ 107.5 psia (92.8 psig)
(viii) Cricondentherm> 137.6°F (chart-dependent, not numerically resolved)

(b)(i) 40wt% C2 / 60wt% C7

  1. Mole fraction. $n_{C2}=40/30.07=1.330$, $n_{C7}=60/100.20=0.5988$: $$x_{C2}=\frac{1.330}{1.330+0.5988}=0.6896,\quad x_{C7}=0.3104$$
  2. Kay's rule. $$T_{pc}=0.6896(549.58)+0.3104(972.36)=680.8\,{}^{\circ}\text{R}\qquad P_{pc}=0.6896(706.5)+0.3104(396.8)=610.4\text{ psia}$$ $$\boxed{T_c \approx 221.1\,{}^{\circ}\text{F}\ ,\qquad P_c \approx 610.4\text{ psia}}$$

(b)(ii) 50.25wt% C2 / 49.75wt% C7 – retrograde condensation window

  1. Mole fraction and critical point. $x_{C2}=0.7710$, $x_{C7}=0.2290$; Kay's rule gives $$\boxed{T_c \approx 186.8\,{}^{\circ}\text{F}\ ,\qquad P_c \approx 635.6\text{ psia}}$$
  2. Retrograde window. By definition, retrograde condensation for this composition can only occur on an isotherm strictly between $T_c\approx187\,{}^{\circ}\text{F}$ and the (chart-dependent) cricondentherm, at pressures between the dew-point pressure at that isotherm and roughly the critical/cricondenbar pressure – i.e. approximately $P>635$ psia at temperatures a modest amount above 187°F. The precise upper temperature bound (cricondentherm) again requires the original chart per the callout under part (a).

(c) Classifying mixtures No.2, 3, 4, 5 at reservoir 1500 psia/210°F, separator 280 psia/40°F

Kay's-rule critical temperature by composition
Mixturewt% C2mol% C2$T_c$ (Kay)Classification
No.290.2296.85103.2°FWet gas
No.380.1493.08119.2°FWet gas
No.470.2288.71137.6°FRetrograde gas condensate
No.560.1183.39160.1°FRetrograde gas condensate

Reasoning. The reservoir temperature (210°F) exceeds every mixture's Kay's-rule critical temperature (103–160°F), so none of the four is oil in the reservoir – all are gas at reservoir conditions. Retrograde behaviour width (cricondentherm − $T_c$) for a binary vanishes at both pure-component limits and grows with composition asymmetry; No.2 and No.3 are nearly pure ethane (97/93 mol%) and so have only a narrow retrograde window – 210°F most likely already lies beyond their cricondentherm, making them single-phase gas throughout depletion (wet gas, since some liquid still condenses at the cold 40°F separator – see below). No.4 and No.5 carry substantially more n-heptane (11–17 mol%) and sit closer to their own $T_c$ (only 50–72°F above it, versus 91–107°F for No.2/3), consistent with 210°F falling inside their wider retrograde window – classified as retrograde gas condensate.

Wet vs dry at the separator. All four mixtures contain some n-heptane, whose vapour pressure at 40°F is only a fraction of a psi – utterly negligible next to the 280 psia separator pressure. Consequently essentially all of the heptane in every one of these streams condenses to liquid at the separator, so none of the four is a true dry gas (zero surface liquid); No.2 and No.3, single-phase in the reservoir, are therefore wet gas rather than dry gas.

Check: the No.2/No.3 vs No.4/No.5 split is an engineering judgement based on Kay's-rule $T_c$ and the well-established qualitative behaviour of asymmetric binary retrograde windows (narrowing to zero at each pure-component limit); a precise classification would compare 210°F against each mixture's actual cricondentherm from the (illegible) source chart.

(d) Oil viscosity and isothermal compressibility vs pressure, 1500→300 psia at 180°F

Pb Oil viscosity μo μo Pressure (1500 → 300 psia) Pb Isothermal compressibility co co Pressure (1500 → 300 psia)
Left: oil viscosity falls slowly as pressure drops from 1500 psia toward the bubble point (simple liquid decompression), reaches a minimum at $P_b$, then rises below $P_b$ – as gas evolves out of solution the remaining liquid is stripped of its lightest (ethane) component and becomes progressively richer in heptane, i.e. more viscous. Right: the isothermal oil compressibility is small and roughly constant above $P_b$ (ordinary liquid compressibility), then jumps sharply upward just below $P_b$ because the two-phase (oil+evolving gas) system is far more compressible than the single-phase liquid.
Check: this sketch assumes the 70.22wt% mixture behaves as a liquid ("oil") at 180°F, per the question's own framing. Kay's-rule linear mixing puts this composition's $T_c$ at ≈137.6°F (below 180°F), but the true critical locus of asymmetric ethane/n-heptane mixtures is well documented to bulge substantially above the linear Kay estimate (this is the classic feature the Kay 1938 dataset itself was published to demonstrate), so an actual $T_c$ above 180°F for this composition is plausible and consistent with the question describing it as an oil at this temperature.