Volumetric Dry Gas Reservoir: 24-Pet-B2 Oil and Gas Evaluation and Economics, May 2016, Q4

Question 4: Volumetric Dry Gas Reservoir (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.

Given.

Quantity	Value
Initial pressure, $p_i$	3000 psia, $Z_i=0.80$
Abandonment pressure, $p_a$	500 psia, $Z_a=0.96$
Initial gas saturation, $S_{gi}$	0.30
Reservoir temperature, $T$	$200^\circ\text{F}$
Bulk volume, $V_b$	20,000 ac-ft
Porosity, $\phi$	0.30
Residual gas saturation, $S_{gr}$	0.25

Find. (a) OGIP $G$ and volumetric-depletion recovery factor to $p_a$; (b) recovery factor under a strong, pressure-maintaining water drive.

Approach. Get $B_{gi}$ and $B_g$ at abandonment from $B_g=0.02827\,ZT/p$, compute OGIP from the initial gas-filled hydrocarbon pore volume, get the volumetric-depletion RF from $B_g$ shrinkage, then get the water-drive RF from the trapped residual-gas saturation alone (no further $B_g$ shrinkage, since pressure is held at $p_i$).

Gas formation volume factors. $T=200+459.67=659.67\ ^\circ\text{R}$: $$B_{gi}=0.02827\,\frac{0.80(659.67)}{3000}=0.004973\ \text{ft}^3/\text{SCF}, \qquad B_{ga}=0.02827\,\frac{0.96(659.67)}{500}=0.03581\ \text{ft}^3/\text{SCF}$$
(a) Original gas in place. The initial gas-filled hydrocarbon pore volume is $V_b\phi S_{gi}$ (in ft$^3$, with $V_b=20{,}000\times43{,}560=8.712\times10^8$ ft$^3$); dividing by $B_{gi}$ converts reservoir-condition gas volume to standard-condition (surface) volume: $$G=\frac{V_b\phi S_{gi}}{B_{gi}}=\frac{8.712\times10^8(0.30)(0.30)}{0.004973}$$ $$\boxed{G=1.577\times10^{10}\ \text{SCF}=15.77\ \text{Bscf}}$$
(a) Volumetric-depletion recovery factor. With no aquifer support, the hydrocarbon pore volume is fixed, so $G_p/G=1-B_{gi}/B_g$ evaluated at abandonment: $$RF_a=1-\frac{B_{gi}}{B_{ga}}=1-\frac{0.004973}{0.03581}$$ $$\boxed{RF_a=0.861=86.1\%}$$
(b) Water-drive recovery factor. Holding $p=p_i$ throughout means $B_g$ never changes from $B_{gi}$, so there is no shrinkage-driven recovery mechanism at all — gas is instead swept out (and trapped) by the advancing water front. The volume fraction of the original gas-filled pore space left behind as immobile residual gas is $S_{gr}/S_{gi}$ (both saturations measured against the same original hydrocarbon pore volume, at the same $B_{gi}$ since pressure never drops), so: $$RF_b=1-\frac{S_{gr}}{S_{gi}}=1-\frac{0.25}{0.30}$$ $$\boxed{RF_b=0.167=16.7\%}$$

Quantity	Result
$B_{gi}$	0.004973 ft$^3$/SCF
$B_{ga}$	0.03581 ft$^3$/SCF
Original gas in place, $G$	15.77 Bscf
(a) RF, volumetric depletion	86.1%
(b) RF, strong water drive	16.7%

Topic: Volumetric material balance for a dry gas reservoir; water-drive gas trapping.

Key relations: $G=V_b\phi S_{gi}/B_{gi}$; volumetric $G_p/G=1-B_{gi}/B_g$; water-drive $RF=1-S_{gr}/S_{gi}$.

Why this works: Volumetric depletion recovers gas by letting it expand (shrinking $B_g$) into the same fixed pore space; a strong water drive instead physically displaces gas with an incompressible fluid that leaves a residual saturation behind — two entirely different recovery mechanisms sharing the same $G$.

Common pitfall: Assuming water drive always improves recovery because it maintains pressure — here it roughly halves ultimate recovery (16.7% vs. 86.1%) because trapped residual gas, not $B_g$ shrinkage, becomes the binding constraint; this counter-intuitive result is a classic teaching point in gas reservoir engineering.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (volumetric and water-drive gas reservoirs); Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 3.

Question 4 of 7: Volumetric Dry Gas Reservoir

Question 4: Volumetric Dry Gas Reservoir (20 marks)