24-Pet-A6 Well Logging and Formation Evaluation · May 2017
Question 5 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. An undersaturated-to-saturated reservoir producing under solution-gas drive plus a small ("pot" / unsteady, single-tank) aquifer.
PVT and production history
$P$ (psia)
$B_o$
$R_s$
$B_g$
$N_p$ (MMSTB)
$G_p$ (MMscf)
3,000 ($p_i$)
1.316
650
–
0
0
2,500 ($p_b$)
1.324
650
0.00082
0.092
59.8
1,500
1.252
510
0.00135
0.850
490.0
1,300 (current)
1.231
450
0.00160
1.100
970.0
Find. OOIP ($N$) and cumulative water influx ($W_e$) at the current pressure (1,300 psi).
Approach. Cast the material balance in Havlena–Odeh straight-line form, $F=N\,E_o+W_e$, and combine it with the Pot (small, instantaneous-equilibrium) aquifer model $W_e=k\,\Delta p$ to get a straight line in $F/E_o$ vs. $\Delta p/E_o$, whose intercept is $N$ and slope is the aquifer constant $k$.
Underlying-drive functions.
$$F=N_pB_o+(G_p-N_pR_s)B_g,\qquad E_o=(B_o-B_{oi})+(R_{si}-R_s)B_g$$
with $B_{oi}=1.316$, $R_{si}=650$ (both at $p_i=3{,}000$ psi, above $p_b$).
Evaluate $F$ and $E_o$ at each pressure.
Havlena–Odeh terms
$P$
$E_o$ (rb/STB)
$F$ (res bbl)
$\Delta p=p_i-p$
$F/E_o$
$\Delta p/E_o$
2,500
0.0080
121,808
500
15,226,000
62,500
1,500
0.1250
1,140,475
1,500
9,123,800
12,000
1,300
0.2350
2,114,100
1,700
8,996,170
7,234
The 2,500 psi (bubble-point) row is excluded from the regression: $E_o$ there is only 0.008, so any small data error is amplified enormously in $F/E_o$ (its own value, 15.2 million, is wildly out of line with the other two rows) – this is the standard, well-documented Havlena–Odeh instability near $p_b$.
Fit the straight line through the 1,500 and 1,300 psi points. With the Pot aquifer model $W_e=k\,\Delta p$, the material balance becomes $F/E_o=N+k(\Delta p/E_o)$, a straight line in $(\Delta p/E_o,\,F/E_o)$:
$$k=\frac{8{,}996{,}170-9{,}123{,}800}{7{,}234-12{,}000}\approx 26.8\ \text{res bbl/psi}$$
$$N=\left(\frac{F}{E_o}\right)_{1{,}300}-k\left(\frac{\Delta p}{E_o}\right)_{1{,}300}=8{,}996{,}170-26.8(7{,}234)$$
$$\boxed{N\approx 8.80\text{ MMSTB (OOIP)}}$$
Cumulative water influx at the current pressure.
$$W_e(1{,}300\text{ psi})=k\,\Delta p=26.8\times1{,}700$$
$$\boxed{W_e\approx 45{,}525\text{ res bbl}}$$
