Harmonic Decline-Curve Analysis: 24-Pet-A3 Fundamental Reservoir Engineering, Undated paper, Q7

Question 7: Harmonic Decline-Curve Analysis (20 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. Linear regression of the harmonic decline plot: $1/q_t=0.0000343\,t+0.0028571$ (t in months, $q_t$ in STB/D), $R^2=1.0000000$.

[Figure not reproduced: Fig. Q7: harmonic decline plot $1/q_t$ vs. $t$ with its linear regression (source figure), reproduced from the given fit equation. See the official exam paper.]

Find. (1) Initial rate $q_i$; (2) initial decline rate $d_i$; (3) decline rate at $t=36$ months.

Approach. The harmonic-decline linearization $1/q_t=1/q_i+(d_i/q_i)t$ matches the given regression term-by-term: the intercept is $1/q_i$ and the slope is $d_i/q_i$; the instantaneous decline rate at any later time then follows from the harmonic decline-rate relation $d_t=d_i/(1+d_it)$.

Initial rate from the intercept. $1/q_i=0.0028571\ \Rightarrow\ \boxed{q_i=1/0.0028571=350.0\ \text{STB/D}}$.
Initial decline rate from the slope. Slope $=d_i/q_i=0.0000343\ \Rightarrow\ d_i=0.0000343(350.0)=0.012005\ \text{month}^{-1}$. Converted to an annual (nominal) rate, $\boxed{d_i=0.012005\times12=0.1441\ \text{yr}^{-1}=14.41\%/\text{yr}}$ (in month$^{-1}$: $d_i=0.01201\ \text{month}^{-1}$).
Decline rate at $t=36$ months (3 years). Harmonic decline: $d_t=\dfrac{d_i}{1+d_it}=\dfrac{0.012005}{1+0.012005(36)}=\dfrac{0.012005}{1.4322}=0.008383\ \text{month}^{-1}$. As a check, the rate itself is $q_{36}=q_i/(1+d_it)=350.0/1.4322=244.4$ STB/D, consistent with reading $1/q_{36}=0.0040919$ directly off the fit line. Annualized, $\boxed{d_{36}=0.008383\times12=0.1006\ \text{yr}^{-1}=10.06\%/\text{yr}}$.

Quantity	Value
(1) Initial oil rate, $q_i$	350.0 STB/D
(2) Initial decline rate, $d_i$	0.01201 month$^{-1}$ (14.41%/yr)
(3) Decline rate at $t=36$ mo	0.008383 month$^{-1}$ (10.06%/yr)

Check: "decline rate in %/year" is taken as the nominal decline rate scaled linearly (month$^{-1}\times12$), matching how the exam's own formula sheet defines $d$ additively in time with no distinction between nominal and effective decline — the standard treatment for this introductory decline-curve formula set.

Topic: Harmonic decline-curve analysis ($b=1$ special case of Arps' hyperbolic decline).

Key relations: $1/q_t=1/q_i+(d_i/q_i)t$ (linearizes harmonic decline in $1/q_t$ vs. $t$); $q_t=q_i/(1+d_it)$; $d_t=d_i/(1+d_it)$.

Why this works: harmonic decline is the special case of Arps' hyperbolic model with exponent $b=1$, which is exactly the case where $1/q_t$ is linear in time — so a straight-line regression on $1/q_t$ vs. $t$ directly hands back $q_i$ (intercept) and $d_i$ (slope$\times q_i$) without needing any nonlinear curve fit, and the instantaneous decline rate always falls over time as the well depletes, never staying constant as it would under exponential decline.

Common pitfall: forgetting to divide the fitted slope by the fitted intercept to recover $d_i$ (the slope alone is $d_i/q_i$, not $d_i$), and reporting the initial decline rate $d_i$ instead of the smaller, correctly time-decayed $d_t$ when a rate at a later time is asked for.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 8 (decline-curve analysis); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 9.

Question 7 of 7: Harmonic Decline-Curve Analysis

Question 7: Harmonic Decline-Curve Analysis (20 marks)