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24-Pet-B3 Petroleum Geology · May 2016

Question 5 of 22

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2016 — 98-Pet-B3, Oil and Gas Evaluation and Economics (3 hours, closed book, approved non-programmable calculator only). The exam's own cover page is titled "Oil and Gas Evaluation and Economics" and every question is property valuation / reserves & production economics / DCF-NPV screening content — no geology anywhere.

Reference texts: Thompson & Wright, Oil Property Evaluation; Canadian Oil and Gas Evaluation Handbook (COGEH), Vol. 1 (Society of Petroleum Evaluation Engineers, Calgary Chapter); National Instrument 51-101, Standards of Disclosure for Oil and Gas Activities (Canadian Securities Administrators); SPE/WPC/AAPG/SPEE Petroleum Resources Management System (PRMS); Economides & Nolte, Reservoir Stimulation; Ahmed, Reservoir Engineering Handbook.

The exam's own instructions ask for only 7 of the 10 short-answer questions and note the Cash-Flow/Future-Value tables are graded by column; for "choose N of M" exams, every item below is answered in full as a study resource. Questions 1–10 correspond to the exam's printed Short-Answer items 1–10; Questions 11–20 correspond to the printed Multiple-Choice items 1–10; Question 21 is the Future Value table; Question 22 is the Cash Flow table.

Question 5 (5 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.

Given. The volumetric data below describe a conventional reservoir; the oil resource in place is required from N = VR × φ × (1/Bo) × (1 - Sw).

Given data
QuantitySymbolValue
Area of the reservoirA30 km²
Average pay thicknessh25 m
Porosityφ8%
Formation volume factorBo0.9 m³/stm³
Water saturationSw15%

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

Approach. Compute the gross reservoir bulk volume VR = A×h, then apply the volumetric oil-in-place formula directly with the given φ, Bo and Sw.

  1. Reservoir bulk volume. $$V_R = A \times h = (30 \times 10^6\ \text{m}^2) \times 25\ \text{m} = 7.5 \times 10^8\ \text{m}^3$$
  2. Oil in place, cubic metres. Substituting into N = VR×φ×(1/Bo)×(1-Sw): $$N = (7.5\times10^8)\times0.08\times\frac{1}{0.9}\times(1-0.15) = \boxed{5.667\times10^7\ \text{m}^3}$$
  3. Oil in place, barrels. Converting at 6.2898 bbl/m³: $$N_{bbl} = (5.667\times10^7\ \text{m}^3)\times 6.2898\ \tfrac{\text{bbl}}{\text{m}^3} = \boxed{3.564\times10^8\ \text{bbl}\ (356.4\ \text{MMbbl})}$$
QuantityValue
Reservoir bulk volume VR7.5×108 m³
Oil in place N5.667×107 m³ (56.7 million m³)
Oil in place N356.4 MMbbl