Given. A saturated-reservoir test point, with the bubble-point pressure above the average reservoir pressure.
Average reservoir pressure, $\bar P_R$
4350 psig
Bubble-point pressure, $P_b$
4400 psig
Stabilized test
$P_{wf}=3000$ psig, $q_o=680$ STB/day
Skin factor, $S$
0
Check: $P_b$ (4400 psig) is above $\bar P_R$ (4350 psig), so the reservoir is already saturated (two-phase) at every pressure encountered — Vogel's dimensionless equation applies over the full range with no above-$P_b$ straight-line segment needed.
Find. The complete Vogel IPR (rate as a function of $P_{wf}$) for this well.
Approach. Vogel's equation has one free constant, $(q_o)_{max}$; back it out from the single test point, then tabulate the curve.
Dimensionless drawdown at the test point. $R=P_{wf}/\bar P_R=3000/4350=0.6897$.
Vogel fraction at the test point. $\dfrac{q_o}{(q_o)_{max}}=1-0.2R-0.8R^2=1-0.2(0.6897)-0.8(0.6897)^2=0.4816$.
Solve for the absolute open flow. $(q_o)_{max}=\dfrac{680}{0.4816}$. $\boxed{(q_o)_{max}=1412\ \text{STB/day}}$.
Tabulate the IPR. With $(q_o)_{max}=1412$ STB/day and $\bar P_R=4350$ psig, $q_o(P_{wf})=1412\left[1-0.2\dfrac{P_{wf}}{4350}-0.8\left(\dfrac{P_{wf}}{4350}\right)^2\right]$ — e.g. at $P_{wf}=2175$ psig ($R=0.5$), $q_o=988$ STB/day.
Quantity
Value
Absolute open flow, $(q_o)_{max}$
1412 STB/day
$q_o$ at $P_{wf}=2175$ psig ($R=0.5$)
988 STB/day
(b) Fetkovich IPR from the four-point test, and comparison
Given. Four flow-after-flow test points on the same reservoir ($\bar P_R=4350$ psig).
$q_o=500$ STB/day
$P_{wf}=3400$ psig
$q_o=680$ STB/day
$P_{wf}=3000$ psig
$q_o=1000$ STB/day
$P_{wf}=2300$ psig
$q_o=1500$ STB/day
$P_{wf}=500$ psig
Find. The Fetkovich deliverability constants $C,n$, the resulting IPR/AOF, and a comparison with part (a).
Approach. Fetkovich's equation $q_o=C\left[\bar P_R^2-P_{wf}^2\right]^n$ linearizes in log-log space; fit $C,n$ to the four points by least squares, then evaluate the AOF at $P_{wf}=0$.
Linearize. $\log q_o=\log C+n\log\left[\bar P_R^2-P_{wf}^2\right]$. Computing $\bar P_R^2-P_{wf}^2$ for each point (with $\bar P_R^2=4350^2=18{,}922{,}500$) gives 7,362,500 / 9,922,500 / 13,632,500 / 18,672,500 psi² at the four rates.
Least-squares fit. Regressing $\log q_o$ on $\log\left[\bar P_R^2-P_{wf}^2\right]$ over the four points gives $\boxed{n=1.185}$, $\boxed{C=3.586\times10^{-6}}$.
Absolute open flow. $(q_o)_{max}=C\left(\bar P_R^2\right)^n=3.586\times10^{-6}\,(18{,}922{,}500)^{1.185}$. $\boxed{(q_o)_{max}=1497\ \text{STB/day}}$.
Compare with part (a). The two independent methods on the same reservoir give 1412 STB/day (Vogel, one point) and 1497 STB/day (Fetkovich, four points) — a 6% spread, which is good agreement for a single-point vs. multi-point fit and confirms both tests are sampling the same well deliverability.
Check: the fitted exponent $n=1.185$ slightly exceeds the textbook physical range $0.5\le n\le1.0$ for Fetkovich's equation; the four points are nonetheless very nearly log-log collinear (the regression $R^2$ is essentially 1), so this is read as scatter/rounding in the four-point data rather than a modelling error, and the least-squares result is reported as computed rather than artificially capped at $n=1$.
Fig. 1 — Vogel IPR from the single test point (part a, $(q_o)_{max}=1412$ STB/day) vs. Fetkovich IPR from the four-point test (part b, $(q_o)_{max}=1497$ STB/day), same reservoir. Test points marked.
