24-Pet-A6 Well Logging and Formation Evaluation · May 2015
Question 6 of 8: Exponential and Harmonic Decline — Producing Life and Cumulative Production
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 6: Exponential and Harmonic Decline — Producing Life and Cumulative Production (20 marks)
Find. For both exponential and harmonic decline: total producing life $t_a$, decline rate $D$, and yearly cumulative production.
Approach. The ultimate-recovery formula for each decline type gives $D$ directly (since $q_i$, $q_a$ and $N_p$ at abandonment are all known); $t_a$ then follows from the same type's time formula, and yearly cumulative production is tabulated from each type's own $N_p(t)$ equation.
Exponential decline rate and life. $N_p=\dfrac{q_i-q_a}{D}\ \Rightarrow\ D=\dfrac{q_i-q_a}{N_p}=\dfrac{425-30}{795{,}000}$, so $\boxed{D_{exp}=4.969\times10^{-4}\text{/day}=0.1815\text{/yr}}$. Then $t_a=\dfrac{\ln(q_i/q_a)}{D}=\dfrac{\ln(14.167)}{4.969\times10^{-4}}$, giving $\boxed{t_{a,exp}=5335\text{ days}\approx14.6\text{ yr}}$.
Harmonic decline rate and life. $N_p=\dfrac{q_i}{D}\ln\!\dfrac{q_i}{q_a}\ \Rightarrow\ D=\dfrac{q_i\ln(q_i/q_a)}{N_p}=\dfrac{425(2.651)}{795{,}000}$, so $\boxed{D_{harm}=1.417\times10^{-3}\text{/day}=0.5176\text{/yr}}$. Then $t_a=\dfrac{q_i/q_a-1}{D}=\dfrac{13.167}{1.417\times10^{-3}}$, giving $\boxed{t_{a,harm}=9291\text{ days}\approx25.4\text{ yr}}$ — harmonic decline takes almost twice as long to reach the same abandonment rate because its long "tail" of low-rate production is much flatter than the exponential's.
Yearly cumulative production. $N_p^{exp}(t)=\dfrac{q_i-q_i e^{-Dt}}{D}$ and $N_p^{harm}(t)=\dfrac{q_i}{D}\ln\!\dfrac{q_i}{q_i/(1+Dt)}$, evaluated at $t=1,2,\dots$ yr (365.25-day years), tabulated below and capped at the ultimate recovery once each type reaches its own $t_a$.
Fig. 3: log(rate) vs. time. Exponential decline plots as the straight line on this axis pair.
Fig. 4: log(rate) vs. cumulative production. Harmonic decline plots as the straight line on this axis pair.
