Question 2 of 7: Time to finite-acting and flowing pressure at 1 hr and 15 days
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Petroleum Engineering, 2015-Dec. 3 hours, closed book. This paper's own cover page reads “98-Pet-B5, Well Testing,” not Reservoir Mechanics, and every question below is pressure-transient/well-test analysis. NOTES item 4/5 state that five (5) questions constitute a complete exam and only the first five as they appear are marked; all seven questions on the paper are solved in full below. The one reference chart supplied with this paper (the page-7 “plot of dimensionless pressure versus dimensionless time”) is the generic exact line-source curve $p_D=0.5[-\mathrm{Ei}(-1/4t_D)]$, so every reading from it below is computed directly from that expression.
Reference texts: Lee, J., Well Testing, SPE Textbook Series Vol. 1 (diffusivity equation, radial flow, wellbore storage); Earlougher, R.C., Advances in Well Test Analysis, SPE Monograph Vol. 5 (Horner analysis, superposition in time, reservoir-limit test, multi-rate tests); Bourdet, D., Well Test Analysis: The Use of Advanced Interpretation Models, Elsevier (log-log diagnostic plots, wellbore storage); Warren, J.E. & Root, P.J., “The Behavior of Naturally Fractured Reservoirs,” SPE Journal, 1963.
Question 2: Time to finite-acting and flowing pressure at 1 hr and 15 days (20 marks)
Find. The time $t_{pss}$ at which the reservoir becomes finite-acting, and $p_{wf}$ at $t=1$ hr and $t=15$ days.
Approach. Use $t_{DA}=0.1$ (onset of pseudo-steady-state for a centered well in a circular drainage area) to get $t_{pss}$; then compute $p_{wf}$ at each requested time using whichever regime actually applies at that time – the infinite-acting line-source solution if $t
Time to reach pseudo-steady-state. With drainage area $A=\pi r_e^2$ and $t_{DA}=0.0002637\,kt/(\phi\mu c_tA)=0.1$:
$$t_{pss}=\frac{0.1\,\phi\mu c_t\,(\pi r_e^2)}{0.0002637\,k}=\frac{0.1(0.3)(2)(5\times10^{-5})(\pi\times1452^2)}{0.0002637(250)}$$
$$\boxed{t_{pss}\approx 301\text{ hr}\approx 12.6\text{ days}}$$
This is exactly the familiar rule-of-thumb form $t_{pss}\approx1191\,\phi\mu c_tr_e^2/k$, a useful self-check.
Regime check for each requested time. $t=1$ hr $\ll t_{pss}=301$ hr, so the well is still infinite-acting at 1 hr. $t=15$ days $=360$ hr $>t_{pss}=301$ hr, so by 15 days the reservoir has already gone pseudo-steady-state – the two parts require two different formulas, not the same one evaluated twice.
$p_{wf}$ at 1 hr (infinite-acting). $t_D=0.0002637kt/(\phi\mu c_tr_w^2)=35{,}160\gg100$, so the log approximation applies: $p_D=0.5(\ln t_D+0.809)=5.638$.
$$p_{wf}=p_i-\frac{141.2\,qB_o\mu}{kh}\,p_D=3000-\frac{141.2(500)(1.2)(2)}{250(20)}(5.638)$$
$$\boxed{p_{wf}(1\text{ hr})\approx 2809\text{ psia}}$$
$p_{wf}$ at 15 days (pseudo-steady-state). With pore volume $V_p=\pi r_e^2h\phi/5.615=7{,}077{,}577$ bbl, the PSS drawdown is the sum of a depletion term and a fixed steady-state (skin-free, $S=0$) geometry term:
$$p_{wf}=p_i-\frac{0.23395\,qB_ot}{c_tV_p}-\frac{141.2\,qB_o\mu}{kh}\Big(\ln\frac{r_e}{r_w}-0.75\Big)$$
$$p_{wf}=3000-142.8-268.3$$
$$\boxed{p_{wf}(15\text{ days})\approx 2589\text{ psia}}$$
Applying the infinite-acting formula from Step 3 at $t=360$ hr instead (ignoring that the boundary has already been felt) would give $\approx2709$ psia – about 120 psi too high, since it omits the depletion term entirely.
