24-Pet-A3 Fundamental Reservoir Engineering · Undated paper
Question 6 of 7: Undersaturated-Oil Material Balance — Initial Oil in Place and Free Gas Volume
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 6: Undersaturated-Oil Material Balance — Initial Oil in Place and Free Gas Volume (20 marks)
Given. Table above; $S_{wi}=0.25$; $p_i=3{,}000$ psia; $p_b=1{,}775$ psia; $c_f=c_w=3\times10^{-6}$ psi$^{-1}$; $B_g(1{,}000)=0.00265$ bbl/SCF; no water drive, no water production.
Find. (1) Initial oil in place $N$; (2) free (liberated) gas volume remaining in the reservoir at 1,000 psia.
Approach. While $p\ge p_b$ (the 2,500 psia data point, where $R_{so}=R_{soi}=845$ confirms the oil is still undersaturated), solve the exam's undersaturated-oil MBE directly for $N$ — a single equation, single unknown, including the rock/water expansion term. Then use that $N$ in an overall solution-gas balance at 1,000 psia (below $p_b$, where gas has come out of solution) to find the free gas remaining in the reservoir, and convert to reservoir barrels with the given $B_g$.
Initial oil in place from the 2,500 psia (still-undersaturated) point. Above $p_b$, $B_t=B_o$ and $R_p=R_{soi}$ (no free gas produced separately), so the $(R_p-R_{soi})B_g$ term vanishes: $N(B_o-B_{oi})+NB_{oi}\!\left[\dfrac{c_wS_{wi}+c_f}{1-S_{wi}}\right]\!\Delta p=N_pB_o$. With $B_{oi}=1.484$, $B_o=1.490$, $\Delta p=3{,}000-2{,}500=500$ psi: the rock/water term is $B_{oi}\!\left[\dfrac{(3\times10^{-6})(0.25)+3\times10^{-6}}{1-0.25}\right](500)=1.484(5\times10^{-6})(500)=3.71\times10^{-3}$. So $N\!\left[(1.490-1.484)+3.71\times10^{-3}\right]=33{,}000(1.490)$, i.e. $N(9.71\times10^{-3})=49{,}170$, giving $\boxed{N=5{,}063{,}852\ \text{STB}\approx5.06\ \text{million STB}}$.
Overall solution-gas balance at 1,000 psia. Total gas initially in solution, $NR_{soi}$, splits three ways at any later time: still dissolved in the remaining oil, $(N-N_p)R_{so}$; produced (as $N_pR_p$); or liberated but still in the reservoir as free gas, $G_f$: $G_f=NR_{soi}-(N-N_p)R_{so}-N_pR_p$. With $R_{soi}=845$, $N_p=868{,}505$, $R_{so}=570$, $R_p=1{,}447$: $G_f=5{,}063{,}852(845)-(5{,}063{,}852-868{,}505)(570)-868{,}505(1{,}447)$, so $\boxed{G_f=6.31\times10^{8}\ \text{SCF}}$.
Convert to reservoir volume. $\boxed{V_{free\,gas}=G_fB_g(1{,}000)=6.31\times10^{8}(0.00265)\approx1{,}671{,}800\ \text{bbl}}$.
Quantity
Value
(1) Initial oil in place, $N$
5,063,852 STB ($\approx$5.06 MMSTB)
Free gas liberated in reservoir, $G_f$
$6.31\times10^{8}$ SCF
(2) Free gas reservoir volume at 1,000 psia
$\approx$1,671,800 bbl
Check: applying the below-bubble-point (saturated) MBE directly to the 1,000 psia row instead gives $N\approx5.18$ million STB, about 2.3% higher — that simpler form (as given on the formula sheet) omits the small rock/water expansion term used above, which is the expected source of the discrepancy between the two check-points; the 2,500 psia (undersaturated) route is used here because it is the exam's own full equation, including that term explicitly.
