Original Gas in Place (Volumetric Method): 24-Pet-B2 Oil and Gas Evaluation and Economics, Undated paper, Q7

Question 7: Original Gas in Place (Volumetric Method) (10 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.

Find. Original gas in place, $G$ (SCF).

Approach. Compute the initial gas formation volume factor $B_{gi}$ from the exam's own formula, then divide the hydrocarbon pore volume by $B_{gi}$ to convert reservoir gas volume into surface (standard) gas volume. This is the classic appraisal-stage calculation: it needs only the static rock-and-fluid snapshot at discovery, not any production history, which is why it is the first OGIP figure computed for any newly delineated dry-gas pool before a single well has been put on stream.

Initial gas formation volume factor. Convert temperature to Rankine, $T_i=160+459.67=619.67\ ^\circ\text{R}$, then apply the formula sheet's $B_g$ relation at initial conditions: $$B_{gi}=0.02827\frac{Z_iT_i}{p_i}=0.02827\frac{0.88(619.67)}{2000}=\boxed{0.007708\ \text{ft}^3/\text{SCF}}$$ This says each standard cubic foot of gas occupies only 0.0077 ft$^3$ at the 2000-psia reservoir condition — the small number is expected at this pressure and is what makes a modest-looking pore volume translate into tens of billions of surface cubic feet below.
Hydrocarbon pore volume. Multiply areal extent by net pay, porosity, and hydrocarbon (non-water) saturation, using the standard 43,560 ft$^2$/acre conversion: $$\text{HCPV}=43{,}560\,A\,h\,\phi\,(1-S_{wi})=43{,}560(8000)(8)(0.20)(1-0.22)$$ $$\boxed{\text{HCPV}=4.349\times10^{8}\ \text{ft}^3}\ \ (=9984\ \text{ac-ft of hydrocarbon-filled rock})$$
Original gas in place. Convert this reservoir-condition hydrocarbon volume to a surface (standard-condition) gas volume by dividing through by $B_{gi}$: $$G=\frac{\text{HCPV}}{B_{gi}}=\frac{4.349\times10^{8}}{0.007708}$$ $$\boxed{G=5.642\times10^{10}\ \text{SCF}=56.42\ \text{Bscf}}$$

Quantity	Result
Initial gas FVF, $B_{gi}$	0.007708 ft$^3$/SCF
Hydrocarbon pore volume	$4.349\times10^{8}$ ft$^3$ (9984 ac-ft)
Original gas in place, $G$	56.42 Bscf

Topic: Volumetric estimation of original gas in place for a dry-gas reservoir.

Key relations: $B_{gi}=0.02827\,Z_iT_i/p_i$; $G=43{,}560\,Ah\phi(1-S_{wi})/B_{gi}$.

Why this works: The volumetric method converts a static rock-volume estimate (from maps, cores, logs) directly into a surface gas volume using only the initial PVT state — no production history is needed, unlike a material-balance estimate, which makes it the appraisal-stage tool of choice before any wells have produced.

Common pitfall: Forgetting the $(1-S_{wi})$ term, or using porosity/saturation as fractions inconsistently with the percentage values given — both silently move the answer by tens of percent.

Source: Ahmed, Reservoir Engineering Handbook, Ch. 5 (volumetric gas-in-place); Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 1–2.

Question 7 of 7: Original Gas in Place (Volumetric Method)

Question 7: Original Gas in Place (Volumetric Method) (10 marks)