24-Pet-B2 Oil and Gas Evaluation and Economics · May 2016
Question 5 of 7: Exponential Decline
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2016, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (non-communicating calculator permitted), 7 questions of 20 marks each. NOTES item 5 states only the first five questions in the answer book are marked; all 7 are solved.
Reference texts: Katz et al., Handbook of Natural Gas Engineering; Lee & Wattenbarger, Gas Reservoir Engineering (SPE Textbook Series Vol. 5); Ahmed, Reservoir Engineering Handbook, 5th ed.; Mohitpour et al., Pipeline Design and Construction, 3rd ed. (ASME Press); McCain, The Properties of Petroleum Fluids, 3rd ed.
Check: the decline rate is stated per month while every question asks about years — converted using 1 month $=30.44$ days (365.25/12, the standard average-month convention) so the exponent $Dt$ carries no rounding drift when $t$ is expressed in exact calendar days.
Given.
Quantity
Value
Initial rate, $q_i$
10 MMSCFD
Decline rate, $D$
0.01/month $=3.285\times10^{-4}$/day
Decline type
Exponential
Find. (a) $q$ at 2 yr; (b) $G_p$ over those 2 yr; (c) time to reach 1 MMSCFD; (d) $G_p$ from end-Yr10 to end-Yr11.
Approach. Apply the exponential-decline rate-time and cumulative-time relations directly, $q(t)=q_ie^{-Dt}$ and $G_p(t)=(q_i-q(t))/D$, converting all times to days for consistency with $D$.
(a) Rate at 2 years. $t=2(365)=730$ days:
$$q(730)=10\,e^{-3.285\times10^{-4}(730)}$$
$$\boxed{q=7.87\ \text{MMSCFD}}$$
(b) Cumulative production over those 2 years.
$$G_p=\frac{q_i-q(t)}{D}=\frac{10-7.87}{3.285\times10^{-4}}$$
$$\boxed{G_p=6{,}491\ \text{MMSCF}=6.49\ \text{Bscf}}$$
(c) Time to decline to 1 MMSCFD. Solving $q_i e^{-Dt}=q$ for $t$:
$$t=\frac{1}{D}\ln\!\left(\frac{q_i}{q}\right)=\frac{1}{3.285\times10^{-4}}\ln(10)$$
$$\boxed{t=7{,}008\ \text{days}=19.2\ \text{years}}$$
(d) Cumulative production from end of Year 10 to end of Year 11. Evaluate $G_p(t)$ at $t=10$ and $11$ years and difference:
$$G_p(10\,\text{yr})=21{,}262\ \text{MMSCF}, \qquad G_p(11\,\text{yr})=22{,}299\ \text{MMSCF}$$
$$\boxed{\Delta G_p=1{,}037\ \text{MMSCF}}$$
Exponential rate-time decline $q(t)=q_ie^{-Dt}$ over the first 2,500 days (the answer to Question 5(c), 7,008 days, is far off this scale).
