24-Pet-A2 Petroleum Reservoir Fluids · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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.
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. A single composite chart plots three reservoir-fluid properties against pressure (2000–6000 psia) on three different vertical scales, with the bubble-point pressure marked near 4200 psia and the initial reservoir pressure near 5500 psia; the three curves are unlabelled as to which property uses which scale.
Find. (a) Which curve is Rs, which is Bg, which is Bo; (b) the freehand μo-vs-P trend during depletion; (c) the most-mobile pressure and why; (d) the correct word in each of the five underlined pairs.
| Curve shape | Property | Reasoning |
|---|---|---|
| Rises from a low value, then goes exactly flat above ~4200 psia | Rs – Solution GOR | Below the bubble point, rising pressure dissolves more gas into the oil, so Rs increases; once all the available gas is in solution (at Pb), further pressure increase cannot dissolve more gas, so Rs is exactly constant above Pb. |
| Rises to a peak exactly at the bubble point, then declines gently above it | Bo – Oil formation volume factor | Below Pb, more dissolved gas swells the liquid, so Bo rises as pressure rises toward Pb. Above Pb no more gas dissolves, so the only remaining effect is liquid compressibility, which shrinks the oil slightly as pressure increases – giving the characteristic peak exactly at the bubble point. |
| Falls hyperbolically (concave-up) as pressure rises across the whole range | Bg – Gas formation volume factor | From the real-gas law, $B_g \propto ZT/p$; at roughly constant reservoir temperature $B_g$ falls approximately as $1/p$, giving the smooth hyperbolic decline with no break at the bubble point (a property of the free gas, not the oil). |
Oil viscosity follows a V-shaped (checkmark) trend with its minimum exactly at the bubble point. From the initial pressure (5500 psia, undersaturated) down to Pb (≈4200 psia), μo decreases slightly as pressure falls, because the oil is simply decompressing (less tightly packed liquid, lower internal friction) with no compositional change. Below Pb, as pressure continues to fall to 2000 psia, gas evolves out of solution and the remaining liquid is progressively stripped of its lightest, most mobile components – so μo increases again, more steeply than it fell above Pb. Plotted on the same pressure axis as part (a), this trend is two intersecting straight-ish segments crossing at the bubble-point pressure, mirroring the template shown on the source graph.
Most mobile at the bubble-point pressure, ≈4200 psia. Mobility is defined as $\lambda_o = k/\mu_o$ (permeability divided by viscosity); for a fixed rock permeability, mobility is highest exactly where viscosity is lowest. From part (b), the V-shaped μo-vs-P trend has its minimum precisely at the bubble point – it is the pressure at which the oil is as decompressed as possible without yet losing any of its light ends to a free gas phase, so it is at its least viscous, most mobile state.
| # | Correct choice | Reasoning |
|---|---|---|
| 1 | divide | Reservoir barrels are always the larger volume for a liquid (dissolved gas swells it), so converting down to stock-tank volume means dividing by a number greater than 1. |
| 2 | Bo | Bo is defined precisely as reservoir-barrels of oil per stock-tank barrel; dividing the reservoir oil volume by Bo gives stock-tank volume by definition. |
| 3 & 4 | reservoir / stock-tank (both blanks) | By definition, $B_o = V_{\text{reservoir}}/V_{\text{stock-tank}}$ – the ratio is always reservoir-condition volume over stock-tank (surface) volume, never the other way round. |
| 5 | greater | Dissolved gas and thermal expansion always make a liquid occupy more space at reservoir conditions than at the surface, so $B_o,B_w>1$ always. |
