Notes on this paper
National Exams December 2014 — 98-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; Beggs, Production Optimization Using Nodal Analysis , 2nd ed.; Craft & Hawkins, Applied Petroleum Reservoir Engineering , 3rd ed.
Question 5: Standing's Method — Predicting Future IPR (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.
Present $\bar P_r$ / $(q_o)_{max}$ 3000 psig / 2000 STB/day Future $\bar P_r$ 1500 psig Present: $\mu_o$, $B_o$, $k_{ro}$ 3.0 cp, 1.18 bbl/STB, 0.85 Future: $\mu_o$, $B_o$, $k_{ro}$ 3.70 cp, 1.12 bbl/STB, 0.60 Future water cut, $f_w$ 0.5
Find. The well's oil producing capacity when $\bar P_r$ declines to 1500 psig.
Approach. Standing's method scales the present tangent productivity index by the ratio of oil relative-mobility $(k_{ro}/\mu_oB_o)$ at future vs. present conditions, then rebuilds the future AOF from the scaled $J^*$.
Present tangent PI. At low drawdown Vogel's slope gives $J_p^*=1.8\,(q_o)_{max,p}/\bar P_{r,p}=1.8\times2000/3000=\boxed{J_p^*=1.2\ \text{STB/day-psi}}$.Relative-mobility ratio. $(k_{ro}/\mu_oB_o)_p=0.85/(3.0\times1.18)=0.2401$; $(k_{ro}/\mu_oB_o)_f=0.60/(3.70\times1.12)=0.1448$. Ratio $=0.1448/0.2401=0.6030$.Future tangent PI. $J_f^*=J_p^*\times0.6030=1.2\times0.6030=\boxed{J_f^*=0.7236\ \text{STB/day-psi}}$.Future AOF (oil producing capacity). $(q_o)_{max,f}=J_f^*\,\bar P_{r,f}/1.8=0.7236\times1500/1.8=\boxed{(q_o)_{max,f}=603\ \text{STB/day}}$.
Check: the future $k_{ro}=0.60$ already reflects the oil relative-permeability reduction caused by the rising water saturation ($f_w=0.5$) as the weak water drive advances — Standing's mobility-ratio method carries that effect through the $k_{ro}$ term itself, so no separate water-cut correction is applied on top of the 603 STB/day oil-rate answer; that figure is the well's oil (not total liquid) producing capacity.
3000.0 2250.0 1500.0 750.0 0.0 0.0 500.0 1000.0 1500.0 2000.0 q_o, STB/day P_wf, psig Present, P̄r=3000 psig Future, P̄r=1500 psig Fig. 5 — Present ($\bar P_r=3000$ psig) and future ($\bar P_r=1500$ psig, weak water drive) IPR curves generated by Standing's method.
Quantity Present ($\bar P_r=3000$) Future ($\bar P_r=1500$) $(k_{ro}/\mu_oB_o)$ 0.2401 0.1448 Tangent PI, $J^*$ 1.20 STB/day-psi 0.7236 STB/day-psi AOF, $(q_o)_{max}$ 2000 STB/day 603 STB/day
Topic: Standing's method for predicting future IPR curves as reservoir pressure declines.
Key relations: $J^*=1.8\,(q_o)_{max}/\bar P_r$ (Vogel tangent PI); $J_f^*=J_p^*\left[(k_{ro}/\mu_oB_o)_f/(k_{ro}/\mu_oB_o)_p\right]$; $(q_o)_{max,f}=J_f^*\bar P_{r,f}/1.8$.
Why this works: As reservoir pressure depletes, both fluid properties ($\mu_o$ rises, $B_o$ falls as solution gas comes out) and rock properties (oil relative permeability falls as gas/water saturation rises) change the well's flow capacity; scaling by the ratio of the combined mobility term isolates exactly that combined effect while keeping Vogel's dimensionless curve shape intact.
Common pitfall: Applying $q_o=J_f^*(\bar P_{r,f}-P_{wf})$ as a straight line at the future condition — the future IPR is still Vogel-shaped (curved), not linear; only the tangent PI at $P_{wf}=\bar P_r$ is a straight-line quantity.
Source: Ahmed, Reservoir Engineering Handbook , 5th ed., Ch. 6 (Standing's future-IPR method); Golan & Whitson, Well Performance , 2nd ed., Ch. 2.
