Question 5 of 22

Question 5

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. The volumetric data below describe a conventional reservoir; the oil resource in place is required from N = V_R × φ × (1/B_o) × (1 - S_w).

Find. The oil resource in place N, in cubic metres and in barrels.

Approach. Convert the areal extent to a consistent unit, form the gross rock pay volume, then scale by porosity, hydrocarbon saturation and the formation volume factor to get stock-tank oil in place.

1. Convert area and form the pay volume. $A = 40\text{ km}^2 = 40\times10^{6}\text{ m}^2$, so $$V_R = A \cdot h = 40\times10^{6}\times 20 = \boxed{8.0\times10^{8}\text{ m}^3}$$This is the gross bulk rock volume of the pay zone, before any porosity or saturation correction.
2. Apply the volumetric OOIP formula. $$N = V_R\cdot\phi\cdot\frac{1}{B_o}\cdot(1-S_w) = 8.0\times10^{8}\times0.10\times\frac{1}{0.9}\times(1-0.15)$$$$N = \boxed{75{,}555{,}556\text{ stm}^3}$$Dividing by B_o (rather than multiplying) is essential here — B_o is defined as reservoir volume per stock-tank volume (>1 for a live oil), so reservoir pore volume of oil must be divided by B_o to shrink it down to stock-tank (surface) volume.
3. Convert to barrels. Using 1 m³ = 6.2898 bbl, $$N = 75{,}555{,}556 \times 6.2898 = \boxed{475.4\text{ million bbl}}$$This is the volumetric oil in place (a resource estimate), not a reserve — a recovery factor (not asked for here) would still need to be applied to get recoverable reserves.