24-Pet-A3 Fundamental Reservoir Engineering · Undated paper
Question 5 of 7: Volumetric Check and Water Influx by p/Z Gas Material Balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Pet-A3 — Fundamental Reservoir Engineering · National Exams, May 2019 · 3 hours, closed book, approved Casio/Sharp calculator only · seven questions provided; five (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked), all questions equal value (20 marks each), all parts of a multipart question equal weight. All seven questions are answered below.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, Darcy flow, relative permeability); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, gas PVT, decline-curve analysis); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.; McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, capillary pressure).
Question 5: Volumetric Check and Water Influx by p/Z Gas Material Balance (20 marks)
Find. (1) Average water saturation at 3,000 psia; (2) the fraction of $G_p$ that is attributable to water influx rather than gas expansion.
Approach. Total reservoir pore volume ($V_b\phi$) is fixed; the gas-filled fraction of it shrinks both from pressure depletion (via $B_g$) and from water influx. Compute the gas-filled pore volume at 3,000 psia directly from $(G-G_p)B_g$, subtract from the fixed total pore volume to get the water volume and hence $S_w$; then use the gas MBE with water influx to isolate how much of $G_p$ is "extra" production caused by the shrinking gas space rather than pure gas expansion.
Gas formation volume factors. $B_g=0.02829\,zT/p$ ft³/SCF: $B_{gi}=0.02829(0.95)(660)/5{,}000=0.003548$ ft³/SCF; $B_g(3{,}000)=0.02829(0.90)(660)/3{,}000=0.005603$ ft³/SCF.
Total pore volume and water saturation at 3,000 psia. $V_b=5{,}000\ \text{ac-ft}\times43{,}560\ \text{ft}^3/\text{ac-ft}=2.178\times10^8$ ft³; total pore volume $=V_b\phi=0.18(2.178\times10^8)=3.9204\times10^7$ ft³ (constant — no rock/water compressibility given). Gas-filled volume at 3,000 psia: $(G-G_p)B_g=(7.5-3.52)\times10^9(0.005603)=2.230\times10^7$ ft³. Water volume there $=3.9204\times10^7-2.230\times10^7=1.690\times10^7$ ft³, so $\boxed{S_w(3{,}000\ \text{psia})=1.690\times10^7/3.9204\times10^7=0.431\approx43.1\%}$.
Water influx from the gas MBE. $G(B_g-B_{gi})+W_e=G_pB_g+W_pB_w\ \Rightarrow\ W_e=GB_{gi}-(G-G_p)B_g=7.5\times10^9(0.003548)-2.230\times10^7=2.661\times10^7-2.230\times10^7$, so $W_e=4.31\times10^6$ ft³.
Gas production attributable to water influx. Without any influx, the same pressure drop would require producing only $G_p'=G-\dfrac{GB_{gi}}{B_g(3{,}000)}=7.5\times10^9-\dfrac{2.661\times10^7}{0.005603}=2.751\times10^9$ SCF. The extra apparent production is $G_p-G_p'=W_e/B_g(3{,}000)=4.31\times10^6/0.005603$, so $\boxed{\Delta G_{p,\text{influx}}\approx0.77\ \text{MMM SCF},\ \text{about }21.9\%\text{ of the }3.52\ \text{MMM SCF total}}$.
Quantity
Value
(1) Average $S_w$ at 3,000 psia
0.431 (43.1%)
Water influx, $W_e$
$4.31\times10^6$ ft³
(2) Gas production from water influx
$\approx0.77$ MMM SCF ($\approx$21.9% of $G_p$)
Check: the given $z$-factor at 750 psia (0.95) is not needed for this two-point calculation (pressure only drops to 3,000 psia in the question asked) and is treated as reference data for a further depletion stage not required here.