Find. The complete Vogel IPR (rate vs. $P_{wf}$) for this well.
Check: $P_b$ (4400 psig) is above $\bar P_r$ (4350 psig), i.e. the entire reservoir already sits at or below its bubble point (fully saturated, two-phase flow at every radius). There is therefore no above-$P_b$ straight-line segment to build — Vogel's equation is applied directly over the whole drawdown range using $\bar P_r$ itself as the reference pressure, exactly as printed on the exam's own formula.
Approach. Since the reservoir is saturated everywhere, use Vogel's dimensionless IPR directly with the single test point to back out $(q_o)_{max}$, then tabulate the curve.
Vogel's equation. $\dfrac{q_o}{(q_o)_{max}}=1-0.2\left(\dfrac{P_{wf}}{\bar P_r}\right)-0.8\left(\dfrac{P_{wf}}{\bar P_r}\right)^2$. With $P_{wf}/\bar P_r=3000/4350=0.6897$: $1-0.2(0.6897)-0.8(0.6897)^2=1-0.1379-0.3805=0.4816$.
Solve for AOF. $(q_o)_{max}=q_o/0.4816=680/0.4816=\boxed{(q_o)_{max}=1412\ \text{STB/day}}$.
Tabulate the IPR. Substituting $(q_o)_{max}=1412$ STB/day back into Vogel's equation for a range of $P_{wf}$ generates the complete curve (Fig. 1, solid blue). The table below reproduces the test point exactly ($q_o=680$ STB/day at $P_{wf}=3000$ psig), confirming the fit.
$P_{wf}$, psig
$q_o$, STB/day
4350
0
4000
197
3500
454
3000 (test)
680
2500
877
2000
1043
1500
1180
1000
1287
500
1365
0 (AOF)
1412
(b) Fetkovich IPR (flow-after-flow test)
Given.
$\bar P_r$ (same reservoir as part a)
4350 psig
Test 1: $q_o$, $P_{wf}$
500 STB/day, 3400 psig
Test 2: $q_o$, $P_{wf}$
680 STB/day, 3000 psig
Test 3: $q_o$, $P_{wf}$
1000 STB/day, 2300 psig
Test 4: $q_o$, $P_{wf}$
1500 STB/day, 500 psig
Find. The Fetkovich IPR curve and a comparison with part (a).
Approach. Fetkovich's empirical deliverability equation $q_o=C(\bar P_r^2-P_{wf}^2)^n$ linearizes on a log-log plot of $q_o$ vs. $(\bar P_r^2-P_{wf}^2)$; fit $n$ (slope) and $C$ (intercept) to the four stabilized points by least squares, then generate the curve.
Linearize. $\log q_o=\log C+n\log(\bar P_r^2-P_{wf}^2)$. Computing $\bar P_r^2-P_{wf}^2$ for each test (with $\bar P_r^2=18{,}922{,}500$ psi$^2$):
$q_o$, STB/day
$P_{wf}$, psig
$\bar P_r^2-P_{wf}^2$, psi$^2$
500
3400
7,362,500
680
3000
9,922,500
1000
2300
13,632,500
1500
500
18,672,500
Least-squares fit. Regressing $\log_{10}q_o$ on $\log_{10}(\bar P_r^2-P_{wf}^2)$ across the four points gives $\boxed{n=1.185}$ and $\boxed{C=3.586\times10^{-6}\ \text{STB/day-psi}^{2n}}$. (The fit is not exact through every point — real flow-after-flow data rarely falls on a perfect line — but it reproduces the trend within about 3%.)
AOF and curve. $(q_o)_{max}=C\,\bar P_r^{2n}=3.586\times10^{-6}\times(18{,}922{,}500)^{1.185}$: $\boxed{(q_o)_{max}=1497\ \text{STB/day}}$. Substituting back for a range of $P_{wf}$ generates the Fetkovich curve (Fig. 1, dashed red).
Fig. 1 — IPR comparison: Vogel (part a, using the single test at 680 STB/day) vs. Fetkovich (part b, fitted to all four flow-after-flow points), same reservoir and $\bar P_r=4350$ psig.
$P_{wf}$, psig
$q_o$ Vogel, STB/day
$q_o$ Fetkovich, STB/day
4000
197
164
3000
680
697
2000
1043
1130
1000
1287
1404
0 (AOF)
1412
1497
Comparison. The two methods agree to within about 6% at the AOF (1412 vs. 1497 STB/day) and track closely across the whole drawdown range, which is expected since both are fitted through the shared 680 STB/day @ 3000 psig test point. Fetkovich sits slightly above Vogel at every $P_{wf}$ here because the fitted exponent ($n=1.185$) is a little steeper than Vogel's built-in shape near mid-range drawdowns, given the scatter in the four-point test data; Vogel's single-point method is the simpler of the two but leans on the theoretical 0.2/0.8 curve shape, while Fetkovich lets the data itself set the curvature via $n$.
