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24-Pet-A5 Petroleum Production Operations · December 2019

Question 5 of 5: Standing's Method — Future IPR for a Weak Water-Drive Reservoir

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams December 2019 — 17-Pet-A5 Petroleum Production Operations (3 hrs, open book). Reference texts: Golan & Whitson, Well Performance, 2nd ed.; Ahmed, Reservoir Engineering Handbook, 5th ed.; Brown, The Technology of Artificial Lift Methods, Vol. 2a–4.

The source NOTES state that the first four questions as answered constitute a complete paper; every question (1–5) is fully solved below.

Question 5: Standing's Method — Future IPR for a Weak Water-Drive Reservoir (25 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. The present Vogel AOF directly, plus PVT/relative-permeability data at both the present and the depleted future reservoir pressure.

Present $\bar P_R$ / future $\bar P_R$3000 psig / 1500 psig
Present $(q_o)_{max}$2000 STB/day
Present: $k_{ro},\mu_o,B_o$0.85, 3.0 cp, 1.18 bbl/STB
Future: $k_{ro},\mu_o,B_o$0.60, 3.70 cp, 1.12 bbl/STB

Find. The well's oil producing capacity (future AOF) at $\bar P_R=1500$ psi.

Approach. Standing's future-IPR method scales the productivity index by the change in the oil mobility ratio $k_{ro}/(\mu_o B_o)$ between the two pressures, then re-applies Vogel's relation $(q_o)_{max}=J^*\bar P_R/1.8$ at the future pressure.

  1. Present ideal productivity index. From Vogel's relation at $P_{wf}=0$: $J^*_{present}=1.8\,(q_o)_{max}/\bar P_R=1.8\times2000/3000$. $\boxed{J^*_{present}=1.2\ \text{STB/day/psi}}$.
  2. Oil mobility at present and future conditions. $\left(\dfrac{k_{ro}}{\mu_oB_o}\right)_{present}=\dfrac{0.85}{3.0\times1.18}=0.2401$; $\left(\dfrac{k_{ro}}{\mu_oB_o}\right)_{future}=\dfrac{0.60}{3.70\times1.12}=0.1448$.
  3. Scale the productivity index. $J^*_{future}=J^*_{present}\times\dfrac{0.1448}{0.2401}=1.2\times0.6030$. $\boxed{J^*_{future}=0.7236\ \text{STB/day/psi}}$.
  4. Future absolute open flow. $(q_o)_{max,future}=J^*_{future}\,\bar P_{R,future}/1.8=0.7236\times1500/1.8$. $\boxed{(q_o)_{max,future}=603\ \text{STB/day}}$ — the well's oil producing capacity once the reservoir has declined to 1500 psi.
Check: the future data set also reports GOR $=800$ scf/STB and $f_w=0.5$ at 1500 psi; these describe the future produced gas and water streams but are not inputs to Standing's oil-mobility scaling itself (the method needs only $k_{ro},\mu_o,B_o$), so they are not used further here — they would enter a full three-phase composite-IPR/nodal analysis if the well's total liquid or gas handling were being sized, which was not asked.
05501,1001,6502,20007501,5002,2503,000Oil rate, qo (STB/day)Flowing bottomhole pressure, Pwf (psig)Q5 -- Standing future IPR as reservoir depletesPresent, PR=3000 psi, qmax=2000Future, PR=1500 psi, qmax=603
Fig. 5 — Present ($\bar P_R=3000$ psi, $(q_o)_{max}=2000$ STB/day, blue) vs. future ($\bar P_R=1500$ psi, $(q_o)_{max}=603$ STB/day, red) Vogel IPR curves, showing how both the reservoir energy and the oil mobility loss (rising water saturation, falling $k_{ro}$) shrink the deliverability as the reservoir depletes.
QuantityValue
Present ideal productivity index, $J^*_{present}$1.20 STB/day/psi
Oil mobility ratio, future/present0.603
Future ideal productivity index, $J^*_{future}$0.724 STB/day/psi
Future oil producing capacity at $\bar P_R=1500$ psi603 STB/day
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