Given. A linear "shoestring" waterflood with explicit relative-permeability functions (no chart reading required). (the explicit text value $S_{wi}=0.30$ is used throughout, per the given rel-perm formulas.)
Given data
$W\times h\times L$
300 × 20 × 1,000 ft
$\phi$
0.15
$S_{wi}$
0.30
$S_{orw}$
0.20
$\mu_o,\ \mu_w$
2.0, 1.0 cp
$q_{inj}$
350 bbl/day
Fig. Q4 – fractional-flow curve $f_w(S_w)$ with the Welge tangent from $S_{wi}$, marking the shock front $S_{wf}$ and the producing-well saturation $S_{w2}=0.66$.
Find. (a) $N_p$ at $S_{w2}=0.66$; (b) $W_p$ at the same condition; (c) areal sweep after 25,000 bbl injected.
Approach. Build the fractional-flow curve from the given relative-permeability functions, apply Welge's method (average saturation behind the front from the tangent slope at $S_{w2}$) for parts (a)–(b), and use the Buckley–Leverett frontal-advance relation for the shock front to answer the pre-breakthrough sweep in part (c).
Fractional-flow function (no gravity, horizontal bed).
$$f_w(S_w)=\frac{1}{1+\dfrac{\mu_w}{\mu_o}\dfrac{k_{ro}}{k_{rw}}}=\frac{1}{1+\dfrac12\dfrac{(1-S_{wD})^3}{S_{wD}^4}}$$
At $S_{w2}=0.66$: $S_{wD}=\dfrac{0.66-0.30}{1-0.20-0.30}=0.72$, giving $f_{w2}=0.9608$ and (by numerical differentiation of $f_w$) $\left.\dfrac{df_w}{dS_w}\right|_{0.66}=1.227$.
Average saturation behind the front (Welge).
$$\bar S_w=S_{w2}+\frac{1-f_{w2}}{\left(df_w/dS_w\right)_{S_{w2}}}=0.66+\frac{1-0.9608}{1.227}$$
$$\boxed{\bar S_w\approx0.692}$$
Cumulative oil production at $S_{w2}=0.66$.
$$N_p=V_p(\bar S_w-S_{wi})=160{,}285(0.692-0.30)$$
$$\boxed{N_p\approx 62{,}830\text{ STB}}$$
Cumulative water production at the same condition. Material balance (water injected = water retained + water produced), using $Q_{iD}=1/f_w'(S_{w2})$ for the dimensionless cumulative injection:
$$W_p=V_p\!\left[\frac{f_{w2}}{\left(df_w/dS_w\right)_{S_{w2}}}-(S_{w2}-S_{wi})\right]=160{,}285\left[\frac{0.9608}{1.227}-0.36\right]$$
$$\boxed{W_p\approx 67{,}828\text{ STB}}$$
Shock-front saturation and areal sweep after 25,000 bbl injected. The shock (leading front) saturation $S_{wf}$ is found from the Welge tangent condition $f_w(S_{wf})/(S_{wf}-S_{wi})=(df_w/dS_w)_{S_{wf}}$, solved numerically: $S_{wf}\approx0.626$, giving $(df_w/dS_w)_{S_{wf}}\approx2.747$. In this linear, uniform-cross-section "shoestring" system the whole width is flooded uniformly, so the areal sweep efficiency equals the fractional distance the front has advanced, $x_{D}=(df_w/dS_w)_{S_{wf}}\,Q_{iD}$:
$$Q_{iD}=\frac{25{,}000}{160{,}285}=0.1560\text{ PV}$$
$$x_D=2.747\times0.1560$$
$$\boxed{\text{Areal sweep}\approx 42.8\%\ (\text{pre-breakthrough; breakthrough occurs at }Q_{iD}=1/2.747=0.364\text{ PV})}$$
