24-Pet-A6 Well Logging and Formation Evaluation · May 2017
Question 1 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2017 – 98-Pet-A6, Reservoir Mechanics (3 hours, open book, non-communicating calculator permitted). The exam consists of EIGHT questions; candidates respond to 100 marks of their choice out of 120. All eight questions are answered in full below.
Reference texts: Craft, B.C. and Hawkins, M. (rev. Terry & Rogers), Applied Petroleum Reservoir Engineering, 3rd ed.; Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.
If doubt exists..." "2. This is an OPEN BOOK EXAM..." "3. The exam consists of EIGHT (8) questions..." "4. Clarity and organization...") immediately above Question 1. every real "Question N" heading on this paper parses cleanly (checked by direct read of the source), so it does not affect the answers below.
Given. A well produces at a constant rate from the centre of a circular drainage area; reservoir and fluid properties are tabulated below.
Given data
$r_e$
1,000 ft
$k$
600 md
$p_i$
2,500 psi
$h$
32 ft
$r_w$
4 in. = 0.3333 ft
$\phi$
0.15
$q$
1,000 STB/day
$\mu_o$
2 cP
$c_t$
$12\times10^{-6}$ psi$^{-1}$
$B_o$
1.333 res bbl/STB
Find. (a) the time at which the boundary is first felt (end of infinite-acting/transient flow); (b) $p_{wf}$ at $t=1$ hr and $t=48$ hr.
Approach. Compare the radius of investigation against $r_e$ to find the finite-acting time, then apply the correct flow-regime solution – the Ei-function (transient) solution while $t$ is less than that time, and the pseudosteady-state (bounded-circle) solution once it is exceeded.
Radius of investigation vs. time. The transient radius of investigation grows as $r_{inv}=0.0325\sqrt{\dfrac{kt}{\phi\mu c_t}}$ (field units, $t$ in hours). Setting $r_{inv}=r_e$ and solving for $t$:
$$t_{fa}=\left(\frac{r_e}{0.0325}\right)^2\frac{\phi\mu c_t}{k}=\left(\frac{1{,}000}{0.0325}\right)^2\frac{0.15\times2\times12\times10^{-6}}{600}$$
$$\boxed{t_{fa}\approx 5.68\text{ hr}\ (0.237\text{ days})}$$
Beyond this time the pressure disturbance has reached $r_e$ and the well is no longer infinite acting.
Classify $t=1$ hr and $t=48$ hr against $t_{fa}$. Since $1\text{ hr}\lt t_{fa}=5.68\text{ hr}$, the 1-hour pressure is still in the infinite-acting (transient) regime and is found from the exact line-source (Ei-function) solution. Since $48\text{ hr}\gg t_{fa}$, the well has long since become finite acting and the pseudosteady-state solution for a bounded circular drainage area applies instead – using the transient Ei-solution at 48 hr would overstate the pressure because it ignores the depletion of the closed boundary.
Transient pressure at $t=1$ hr.
$$p_{wf}=p_i-\frac{70.6\,q\mu_oB_o}{kh}\,E_1\!\left(\frac{948\,\phi\mu_o c_t r_w^2}{kt}\right)$$
With $t=1$ hr the Ei-argument is $x=\dfrac{948\times0.15\times2\times12\times10^{-6}\times0.3333^2}{600\times1}=6.32\times10^{-7}$, giving $E_1(x)\approx13.16$. Then
$$p_{wf}(1\text{ hr})=2{,}500-\frac{70.6(1{,}000)(2)(1.333)}{(600)(32)}(13.16)$$
$$\boxed{p_{wf}(1\text{ hr})\approx 2{,}365.7\text{ psi}}$$
Pseudosteady-state pressure at $t=48$ hr. For a closed circular reservoir,
$$p_D=2\pi t_{DA}+\ln\!\left(\frac{r_e}{r_w}\right)-\frac34,\qquad t_{DA}=\frac{0.0002637\,kt}{\phi\mu_o c_t\,(\pi r_e^2)}$$
At $t=48$ hr, $t_{DA}=0.6715$, $\ln(r_e/r_w)=\ln(3{,}000)=8.006$, so $p_D=2\pi(0.6715)+8.006-0.75=11.48$. Then
$$p_{wf}(48\text{ hr})=p_i-\frac{141.2\,q\mu_oB_o}{kh}\,p_D=2{,}500-\frac{141.2(1{,}000)(2)(1.333)}{(600)(32)}(11.48)$$
$$\boxed{p_{wf}(48\text{ hr})\approx 2{,}275.0\text{ psi}}$$
As a consistency check, the Ei- and pseudosteady-state formulas return almost the same pressure (within 1 psi) when both are evaluated exactly at $t=t_{fa}$, confirming the two solutions join smoothly at the finite-acting time.
