NivaarExam PrepOfficial exam papers ↗

24-Pet-A2 Petroleum Reservoir Fluids · December 2019

Question 3 of 7: Porosimetry and Undersaturated Oil PVT

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 3: Porosimetry and Undersaturated Oil PVT (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.

(a) Pore volume by Boyle's-law porosimetry

Given. Sample cell volume $V_{cell}=8\text{ cm}^3$; rock bulk volume $V_b=1\text{ cm}^3$; initial pressure $P_1=760$ mmHg; reference cell (empty, identical) volume $V_{ref}=8\text{ cm}^3$; final expanded pressure $P_2=360$ mmHg; isothermal (25°C constant).

Find. Pore volume $V_p$ of the rock sample.

Approach. The helium/gas space in the sample cell is the cell volume minus the rock's grain (matrix) volume, $V_{cell}-(V_b-V_p)$. Applying Boyle's law before and after opening the valve to the empty reference cell gives one equation in $V_p$.

  1. Set up Boyle's law. Gas-occupied volume before expansion $=V_{cell}-V_b+V_p=7+V_p$; after opening to the reference cell, gas volume $=7+V_p+V_{ref}=15+V_p$. $$P_1(7+V_p)=P_2(15+V_p)$$
  2. Solve for $V_p$. $$760(7+V_p)=360(15+V_p) \;\Rightarrow\; 5320+760V_p = 5400+360V_p \;\Rightarrow\; 400V_p=80$$ $$\boxed{V_p = 0.20\text{ cm}^3}$$
  3. Porosity (bonus check). $$\phi = V_p/V_b = 0.20/1.00 = 20.0\%$$
QuantityValue
Pore volume $V_p$0.20 cm³
Porosity $\phi$20.0%

(b) $B_o$, $R_s$, $B_g$ for undersaturated oil at 2800 psia, 250°F

Given. $P_i=2800$ psia $> P_b=1800$ psia (undersaturated); $T=250\,{}^{\circ}\text{F}$; reservoir liquid production $=100$ bbl/d; stock-tank oil $=65$ STB/d; combined (separator+stock-tank) gas $=50{,}000$ scf/d; gas gravity $\gamma_g=0.7$.

Find. $B_o$, $R_s$ and $B_g$ at reservoir conditions.

Approach. Because reservoir pressure exceeds the bubble point, all the produced gas was dissolved in the oil at reservoir conditions (no free gas cap in the reservoir), so $B_o$ and $R_s$ follow directly from the produced volumes; $B_g$ uses the real-gas law with $Z$ from the Dranchuk–Abou-Kassem (DAK) correlation applied to the formula sheet's pseudo-critical rules.

  1. Oil formation volume factor. $$B_o = \frac{V_{o,res}}{V_{o,STB}} = \frac{100\text{ bbl/d}}{65\text{ STB/d}} = \boxed{1.5385\ \text{bbl/STB}}$$
  2. Solution GOR. Since $P>P_b$, all produced gas came out of solution at the stock tank, so the total measured GOR equals $R_s$ at reservoir conditions: $$R_s = \frac{50{,}000\text{ scf/d}}{65\text{ STB/d}} = \boxed{769.2\ \text{scf/STB}}$$
  3. Pseudo-critical properties (formula sheet). $$T_{pc}=168+325(0.7)-12.5(0.7)^2=389.4\,{}^{\circ}\text{R}\qquad P_{pc}=677+15.0(0.7)-37.5(0.7)^2=669.1\text{ psia}$$ $$T_r=\frac{250+460}{389.4}=1.823\qquad P_r=\frac{2800}{669.1}=4.185$$
  4. Z-factor (DAK correlation) and $B_g$. Solving the DAK implicit equation at $(T_r,P_r)=(1.823,4.185)$: $Z=0.905$. $$B_g = 0.02827\frac{ZT}{p} = 0.02827\times\frac{0.905\times710}{2800} = \boxed{0.00649\ \text{ft}^3/\text{scf}}$$
QuantityValue
$B_o$1.5385 bbl/STB
$R_s$769.2 scf/STB
Z (DAK, reservoir cond.)0.905
$B_g$0.00649 ft³/scf
Check: because $P_i>P_b$, no free gas phase actually exists in this reservoir – $B_g$ above is the gas-phase property the produced associated gas would have if evaluated at reservoir $p,T$ (needed e.g. for a two-phase $B_t$ calculation once pressure later falls below $P_b$), not a statement that free gas is currently present downhole.