24-Pet-A6 Well Logging and Formation Evaluation · May 2015
Question 5 of 8: Saturations from Material Balance and Layered Vertical Permeability
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Pet-A6 — Reservoir Mechanics · National Exams, May 2015 · 3 hours, closed book, Casio/Sharp approved calculator only · eight problems set (candidates answer Problems 1 and 2 plus any three of the remaining six per the exam's own instructions; all eight are solved in full below as a complete study resource), all questions equal value.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, decline curves, transient well testing, permeability averaging); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, decline-curve analysis, pressure buildup, PVT correlations); Golan, M. & Whitson, C.H., Well Performance, 2nd ed. (reserves methods, water/gas influx); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.
Check: this paper's own title page reads “98-PET-A6: Reservoir Mechanics”, not “Well Logging and Formation Evaluation” — the subject heading it is listed under does not match its content. Every problem below is answered as the paper actually printed it (material balance, decline-curve analysis, pressure-transient testing and permeability averaging — classic Reservoir Mechanics/Fundamental Reservoir Engineering topics), not well-logging.
Problem 5: Saturations from Material Balance and Layered Vertical Permeability (20 marks)
Given. HCPV $=3\times10^6$ rm³, $N_p=0.8\times10^6$ sm³, $S_{wi}=22\%$ (unchanged), $B_{oi}=1.333$, $B_{of}=1.49$ rm³/sm³; five-layer column with thickness/permeability pairs as listed.
Find. (i) $S_o$ and $S_g$ at 13 MPa; (ii) the vertical (harmonic, series-flow) average permeability of the layered column.
Approach. With $S_w$ constant and no water drive, the only pore volume not occupied by oil at any time is either water (fixed) or free gas — so $S_o$ follows directly from how much reservoir-barrel oil volume remains, without needing $R_s$ or $B_g$; $S_g$ is then the volumetric remainder. Vertical (series) flow through the layered column averages by the harmonic-mean formula weighted by thickness.
Fig. 3: Layered column (top to bottom) used for the series (vertical) permeability average.
(i) Oil saturation at 13 MPa. Initial oil in place from the HCPV: $N=\dfrac{\text{HCPV}\,(1-S_{wi})}{B_{oi}}=\dfrac{3\times10^6(0.78)}{1.333}$, giving $N=1.7554\times10^6$ sm³. Remaining stock-tank oil after $N_p=0.8\times10^6$ sm³: $N-N_p=0.9554\times10^6$ sm³. Its reservoir volume at 13 MPa: $(N-N_p)B_{of}=0.9554\times10^6(1.49)=1.4236\times10^6$ rm³. As a fraction of the (constant) HCPV: $S_o=\dfrac{1.4236\times10^6}{3\times10^6}$, so $\boxed{S_o=0.4745\ (47.5\%)}$.
Gas saturation. With $S_w$ unchanged and no water encroachment, the balance of the pore space is free gas: $S_g=1-S_w-S_o=1-0.22-0.4745$, so $\boxed{S_g=0.3055\ (30.6\%)}$.
(ii) Vertical (harmonic) average permeability. $\bar{k}_v=\dfrac{\sum h_i}{\sum(h_i/k_i)}$. $\sum h_i=10+1+4+2+8=25$ m. $\sum(h_i/k_i)=\dfrac{10}{350}+\dfrac{1}{0.5}+\dfrac{4}{1230}+\dfrac{2}{2.4}+\dfrac{8}{520}=0.0286+2.000+0.00325+0.8333+0.0154=2.8805$. So $\bar{k}_v=\dfrac{25}{2.8805}$, giving $\boxed{\bar{k}_v=8.68\ \text{md}}$.
Check: $B_g$ and the change in $R_s$ are extra PVT context (they would be needed for a produced-gas/free-gas-volume material balance if $G_p$ were also given); they are not required for the volumetric $S_o$/$S_g$ split above, which follows purely from the oil-volume shrinkage/expansion and the constant HCPV.
