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24-Pet-A2 Petroleum Reservoir Fluids · Undated paper

Question 3 of 6: P-T Diagram Interpretation, Real-Gas Limitations & API Blending

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, National Examinations May 2019. 3 hours duration, closed book, formula sheet supplied, personal scientific calculator permitted. SIX questions are printed on the paper; per the exam's own instructions, any FIVE constitute a complete answer paper. Every question is solved in full below (all six, 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.; Standing, M.B., Volumetric and Phase Behavior of Oil Field Hydrocarbon Systems; Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed.

Check: two graphical-correlation questions (4B's Carr et al. gas-viscosity chart, and 6a's Standing-Katz z-factor chart) are answered using the standard published Lee-Gonzalez-Eakin (1966) gas-viscosity correlation and the Dranchuk & Abou-Kassem (1975) z-factor equation of state respectively – both are the accepted numerical proxies for reading these two charts precisely, and are flagged again at each point of use.

Question 3: P-T Diagram Interpretation, Real-Gas Limitations & API Blending (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.

i) P-T diagram interpretation (8 marks)

Pressure, psia Temperature, °C Critical Point A B D
Composite P-T phase envelope: A sits outside the dome on the low-temperature (liquid) side, B outside the dome on the high-temperature (gas) side, D inside the two-phase envelope. Left branch of the envelope (lower-T side of the critical point) = bubble-point curve; right branch (higher-T side) = dew-point curve.
Question 3(i) – corrected statements
RowCorrect answerReasoning
(a) single-phase oilA, classified undersaturatedA lies outside the envelope on the liquid side – single-phase liquid above its bubble-point pressure is by definition undersaturated.
(a) two-phase reservoirD, classified saturatedD lies inside the dome – any point in the two-phase region is, by definition, at a saturated state (liquid and vapor coexisting in equilibrium).
(b) curve toward lower TBubble-point curveThe branch bounding the liquid region (left of the critical point) is the locus of bubble points.
(b) curve toward higher TDew-point curveThe branch bounding the vapor region (right of the critical point, including the retrograde zone) is the locus of dew points.
(c) A reaches D due toIsothermal pressure depletion during productionReservoir temperature is essentially fixed; producing the well drops reservoir pressure along a near-vertical (constant-T) path, carrying the fluid state from A straight down across the bubble-point curve into the two-phase region at D.
(d) highest-API oilReservoir AA is undersaturated and still holds its full original solution gas / light-end content. D has already lost gas (and with it, light ends) to a free gas phase, so its remaining liquid flashes to a heavier, lower-°API stock-tank oil. (B is a single-phase gas reservoir, not an oil reservoir at all, though its surface condensate – if any – could itself be very light.)

ii) Ideal-gas-law limitations and the van der Waals fix (7 marks)

Two major limitations of the ideal gas law:

  1. Zero molecular volume assumed. The ideal gas law $pV=nRT$ treats gas molecules as dimensionless point masses with no physical volume of their own. Real gas molecules occupy finite volume, which becomes significant at the high pressures typical of petroleum reservoirs, where molecules are packed much closer together.
  2. No intermolecular forces assumed. The ideal gas law assumes molecules neither attract nor repel one another. Real molecules exert van der Waals attractive forces, which become significant at high pressure and low temperature (exactly reservoir conditions) and reduce the pressure the gas actually exerts on its container compared with the ideal prediction.

How van der Waals corrected the equation (1873): $\left(p+\dfrac{an^2}{V^2}\right)(V-nb)=nRT$. The term $an^2/V^2$ is added to the measured pressure to compensate for intermolecular attraction, which otherwise pulls molecules away from the container walls and makes the measured pressure lower than the ideal prediction. The term $nb$ is subtracted from the total volume to exclude the finite volume $b$ physically occupied by the molecules themselves, leaving only the volume actually available for molecular motion.

iii) API gravity of a three-component blend (5 marks)

Given. 52% of a 32°API oil, 45% of a 48°API oil, 3% (balance) of a 10°API oil, blended by volume percent.

Find. The blend's API gravity.

Approach. Convert each component's °API to specific gravity, take the volume-weighted average SG of the blend, then convert the blend SG back to °API.

  1. Convert each °API to SG using $SG=\dfrac{141.5}{{}^\circ API+131.5}$. $$SG_1=\frac{141.5}{32+131.5}=0.8654,\qquad SG_2=\frac{141.5}{48+131.5}=0.7883,\qquad SG_3=\frac{141.5}{10+131.5}=1.0000$$
  2. Volume-weighted average SG. $$SG_{blend}=0.52(0.8654)+0.45(0.7883)+0.03(1.0000)=\boxed{0.8348}$$
  3. Convert back to °API. $${}^\circ API_{blend}=\frac{141.5}{0.8348}-131.5=\boxed{38.0^\circ API}$$
Question 3(iii) – results
QuantityValue
Blend specific gravity0.835
Blend API gravity38.0 °API