Given. A Fetkovich-type isochronal-test IPR with the deliverability coefficient and exponent already fitted; the average reservoir pressure itself is not stated directly and must be recovered from the single stabilized test point.
Find. (a) $q_o$ at $P_{wf}=1800$ psig, current $\bar P_R$; (b) $q_o$ at $P_{wf}=1800$ psi once $\bar P_R$ has declined to 2400 psi with 40% water.
Approach. Back out $\bar P_R$ from the given $C$, $n$ and the single stabilized test point, then apply the same deliverability equation at each requested condition, converting liquid rate to oil rate through the stated water cut.
Recover the current average reservoir pressure. $1420=0.02\left[\bar P_R^2-2800^2\right]^{0.8}\Rightarrow\bar P_R^2=2800^2+\left(\dfrac{1420}{0.02}\right)^{1/0.8}=2800^2+71{,}000^{1.25}$. $\boxed{\bar P_R\approx3000\ \text{psig}}$ — a clean round number, a good check that $C$ and $n$ were read correctly.
(a) Oil rate at $P_{wf}=1800$ psig, current $\bar P_R=3000$ psig. $q_L=0.02\left[3000^2-1800^2\right]^{0.8}=0.02\,(5{,}760{,}000)^{0.8}$. With $f_w=0$, $q_o=q_L$. $\boxed{q_o=5120\ \text{STBO/day}}$.
(b) Later-life liquid rate at $\bar P_R=2400$ psi. The deliverability coefficient $C$ is a rock/fluid property from the isochronal test and does not change as $\bar P_R$ declines, so the same equation applies with the new $\bar P_R$: $q_L=0.02\left[2400^2-1800^2\right]^{0.8}=0.02\,(2{,}520{,}000)^{0.8}$. $\boxed{q_L=2643\ \text{STBL/day}}$.
(b) Convert to oil rate. With 40% water cut, $q_o=q_L(1-f_w)=2643\times0.60$. $\boxed{q_o=1586\ \text{STBO/day}}$ — markedly lower than part (a), reflecting both the depleted drive pressure and the water breakthrough.
Fig. 1 — Fetkovich IPR ($C=0.02$, $n=0.8$) with $\bar P_R\approx3000$ psig backed out of the isochronal test point, and the part (a) operating point at $P_{wf}=1800$ psig marked.
Fig. 2 — The same deliverability equation re-anchored to the depleted average reservoir pressure of 2400 psig for part (b); the whole curve collapses inward as the reservoir energy declines.
Quantity
Value
Backed-out current $\bar P_R$
≈3000 psig
(a) $q_o$ at $P_{wf}=1800$ psig, current $\bar P_R$
5120 STBO/day
(b) $q_L$ at $P_{wf}=1800$ psi, $\bar P_R=2400$ psi
