Question 1 of 4: Line-Drive Waterflood — Fractional Flow, Breakthrough Time, Saturation Profile

Problem 1: Line-Drive Waterflood — Fractional Flow, Breakthrough Time, Saturation Profile (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $f_w(S_w)$ and the Welge tangent construction; (b) water breakthrough time $t_{BT}$; (c) the 1-D water-saturation profile $S_w(x)$ at 0.20 pore volumes injected.

Approach. Build $f_w(S_w)=1/[1+(k_{ro}\mu_w)/(k_{rw}\mu_o)]$ from the given Corey-type curves, locate the shock (front) saturation $S_{wf}$ by Welge's tangent-line construction from the initial condition $(S_{wc},0)$, then apply the Buckley-Leverett frontal-advance equation $x(S_w,t)=\dfrac{q_i t}{A\phi}\dfrac{df_w}{dS_w}$ for breakthrough and for the saturation profile.

1. Fractional-flow curve. With $S_e=(S_w-S_{wc})/(1-S_{wc}-S_{or})=(S_w-0.20)/0.50$, the given curves reduce to $k_{rw}=0.20\,S_e^2$ and $k_{ro}=0.80(1-S_e)^2$ (since $S_o-S_{or}=(1-S_{wc}-S_{or})(1-S_e)$). Substituting into $f_w=\left[1+\dfrac{k_{ro}\mu_w}{k_{rw}\mu_o}\right]^{-1}$ gives $$f_w(S_w)=\left[1+\frac{4}{3}\left(\frac{1-S_e}{S_e}\right)^{2}\right]^{-1},\qquad 0.20\le S_w\le 0.70.$$ Evaluating point-by-point (e.g. $S_w=0.40$: $S_e=0.40$, $k_{rw}=0.032$, $k_{ro}=0.288$, $f_w=0.357$; $S_w=0.60$: $S_e=0.80$, $k_{rw}=0.128$, $k_{ro}=0.032$, $f_w=0.923$) traces the S-shaped curve plotted in Fig. 1.
2. Welge tangent-line construction. Since the reservoir is at connate water ($S_{wi}=S_{wc}=0.20$, no mobile water ahead of the flood), the tangent is drawn from $(S_{wc},f_w)=(0.20,0)$ to the point on the $f_w$ curve that maximizes the secant slope $f_w(S_w)/(S_w-S_{wc})$ — equivalently, the point where the tangent line just touches the curve. A numerical sweep of the curve locates this at $$\boxed{S_{wf}=0.578,\quad f_w(S_{wf})=0.878,\quad \left.\frac{df_w}{dS_w}\right|_{S_{wf}}=2.323}.$$ This is the water-saturation value that propagates as a stable shock front; saturations between $S_{wc}$ and $S_{wf}$ are not physically realized (they would require the curve to be multi-valued in $x$).
3. Breakthrough time. The dimensionless pore volumes injected at breakthrough is the reciprocal of the tangent slope, $W_{iD,BT}=1/f_w'(S_{wf})=1/2.323=0.4305$ PV. The swept pore volume is $$PV=A\,\phi\,L=3000\times0.25\times2700=2{,}025{,}000\ \text{ft}^3=\frac{2{,}025{,}000}{5.615}=360{,}641\ \text{bbl}.$$ With $q_i=250$ STB/day (taken as 250 res bbl/day, $B_w\approx1$), $$t_{BT}=\frac{PV\cdot W_{iD,BT}}{q_i}=\frac{360{,}641\times0.4305}{250}=\boxed{621\ \text{days}}.$$
4. Saturation profile at 0.20 PV injected. Since $W_{iD}=0.20\lt W_{iD,BT}=0.4305$, the front has not yet broken through. Each saturation $S_w\in[S_{wf},\,1-S_{or}]$ behind the front travels to $x(S_w)=W_{iD}\,L\,f_w'(S_w)$ (a rearrangement of the frontal-advance equation using $PV/(A\phi)=L$). The front itself sits at $$x_f=W_{iD}\,L\,f_w'(S_{wf})=0.20\times2700\times2.323=\boxed{1254\ \text{ft}}\ (\lt L=2700\text{ ft, confirming pre-breakthrough}).$$ Ahead of the front ($x_f\lt x\le L$) the formation is undisturbed at $S_w=S_{wc}=0.20$; behind it, $S_w$ decreases smoothly from $1-S_{or}=0.70$ at the injector ($x=0$, where $f_w'\to0$) down to $S_{wf}=0.578$ at $x=x_f$, where it drops discontinuously (the shock) to $0.20$. The full profile is plotted in Fig. 2.