24-Pet-A2 Petroleum Reservoir Fluids · December 2019
Question 5 of 7: Flash and Differential Liberation PVT Data
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 5: Flash and Differential Liberation PVT Data (20 marks)
(a) Relative volume at 4000 psig from the flash-vaporization chart
Given. Digitized chart points (pressure, total volume): the bubble point (knee of the curve) at ≈3000 psig, $V_{sat}=67$ cc; and at 4000 psig, $V=65$ cc.
[Figure not reproduced: Flash-vaporization total-volume curve, drawn from the paper's own annotated data points. The "knee" where the slope changes marks the bubble point. See the official exam paper.]
Find/Approach. $V/V_{sat}$ is defined as barrels (or cc) at the indicated pressure per unit volume at the saturation (bubble-point) pressure.
Check: this chart's own visually-read bubble point (≈3000 psig) is a rough reading and need not exactly coincide with the more precise 2620 psig bubble point tabulated later for the same 220°F sample in the differential-liberation table (part b / Question 7b) – the calculation above stays self-contained within this chart's own data.
(b) Differential liberation table – properties at 1100 psig
Given. Differential vaporization on a black oil at 220°F; extract at/around 1100 psig:
Differential vaporization data (relevant rows)
P, psig
Gas removed @220°F, cell P, cc
Gas removed @60°F/14.65 psia, scf
Oil volume, cc
1350
5.705
0.01618
55.876
1100
6.891
0.01568
54.689
850
8.925
0.01543
53.462
600
12.814
0.01543
52.236
350
24.646
0.01717
50.771
159
50.492
0.01643
49.228
0
—
0.03908
42.540
Residual (60°F)
—
—
39.572
Find. $R_{sd}$, relative total volume $B_{td}$, gas Z-factor, and $B_g$, all at 1100 psig.
Approach. $R_{sd}(p)$ is the cumulative scf of gas still to be evolved between $p$ and the stock tank, normalized to the residual-oil basis (1 bbl $=158{,}987.3$ cc); $B_{td}$ adds the free gas still occupying the cell at $(p,220\,{}^{\circ}\text{F})$ to the oil volume; the liberated-gas Z-factor at each step follows from the real-gas law applied to the SAME mass of gas measured at cell conditions and at standard conditions.
Solution GOR at 1100 psig. Sum the "gas removed @ 60°F" column for every pressure step below 1100 psig (850, 600, 350, 159, 0): $$\Delta V_g = 0.01543+0.01543+0.01717+0.01643+0.03908 = 0.10354\text{ scf}$$ $$R_{sd}(1100) = \Delta V_g\times\frac{158{,}987.3\text{ cc/bbl}}{39.572\text{ cc}} = \boxed{416\ \text{scf/STB}}$$
Relative total volume. The free gas still in the cell at 1100 psig occupies exactly the "gas removed @220°F, cell P" volume for that step (its volume just before expulsion): $$B_{td}(1100) = \frac{V_{oil}+V_{gas,cell}}{V_{residual}} = \frac{54.689+6.891}{39.572} = \boxed{1.556}$$
Z-factor of the liberated gas at this step. Applying the real-gas law to the same gas mass at cell conditions (1, subscript "1": $220\,{}^{\circ}\text{F}=680\,{}^{\circ}\text{R}$, $p=1100+14.65=1114.65$ psia) and standard conditions (2: $60\,{}^{\circ}\text{F}=520\,{}^{\circ}\text{R}$, $14.65$ psia, $Z_{sc}\approx1$): $$Z_1 = \frac{P_1V_1T_2}{P_2V_2T_1},\qquad V_2 = 0.01568\text{ scf}\times28{,}316.85\text{ cc/scf} = 444.1\text{ cc}$$ $$Z_1 = \frac{1114.65\times6.891\times520}{14.65\times444.1\times680} = \boxed{0.903}$$
Gas formation volume factor at 1100 psig. $$B_g(1100) = 0.02827\frac{Z_1T_1}{P_1} = 0.02827\times\frac{0.903\times680}{1114.65} = \boxed{0.01557\ \text{ft}^3/\text{scf}}$$
