Question 4 of 4: Linear Waterflood — Buckley-Leverett / Welge Analysis

Question 4: Linear Waterflood — Buckley-Leverett / Welge Analysis (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. $B_o=B_w=1.0$; $h=20$ ft; $A=26{,}400$ ft²; $\phi=0.25$; $q_i=900$ bbl/day; $L=600$ ft; $\mu_o=2.0$ cp, $\mu_w=1.0$ cp; horizontal ($\alpha=0^\circ$); $S_{wc}=S_{wi}=0.20$; $S_{or}=0.25$; tabulated $f_w(S_w)$ and $df_w/dS_w$ above (no correlation to be used).

Find. (a) breakthrough time $t_{BT}$; (b) cumulative water injected $W_{i,BT}$; (c) cumulative oil produced $N_{p,BT}$; (d) the water-saturation profile $S_w(x)$ at $t=240$ days.

Approach. Locate the shock-front saturation $S_{wf}$ by Welge's tangent construction (the tabulated $S_w$ that maximizes the secant slope $f_w/(S_w-S_{wc})$, which coincides with the local $df_w/dS_w$ there); then apply $Q_i=1/f_w'(S_{wf})$, $t=Q_iV_p/q_i$, $N_p=V_p(\bar S_{w2}-S_{iw})$, and the frontal-advance equation $x(S_w,t)=\dfrac{5.615\,q_i\,t}{A\phi}\left(\dfrac{df_w}{dS_w}\right)_{S_w}$ for the saturation profile.

1. Pore volume. $$V_p=\frac{A\phi L}{5.615}=\frac{(26{,}400)(0.25)(600)}{5.615}=\boxed{705{,}254\text{ bbl}}$$
2. Welge tangent point. Scanning the secant slope $f_w/(S_w-0.20)$ across the table (0.25→1.24, 0.30→1.05, ..., 0.55→1.93, 0.60→1.97, 0.65→1.93, ...) shows it peaks at $S_w=0.60$, and the tabulated derivative there ($1.922$) matches that peak secant slope ($1.97$) far more closely than at any neighbouring row — confirming the front saturation $$S_{wf}=0.60,\qquad f_w(S_{wf})=0.788,\qquad f_w'(S_{wf})=1.922.$$
3. (a) Time to breakthrough. Pore volumes of water injected at breakthrough: $Q_{i,BT}=1/f_w'(S_{wf})=1/1.922=0.520$ PV. In bbl, $W_{i,BT}=Q_{i,BT}V_p=0.520\times705{,}254=366{,}937$ bbl. Since $q_i$ is constant, $$t_{BT}=\frac{W_{i,BT}}{q_i}=\frac{366{,}937}{900}=\boxed{407.7\text{ days}}$$
4. (b) Cumulative water injected at breakthrough. Already computed above: $$W_{i,BT}=\boxed{366{,}937\text{ bbl}}\approx 0.520\text{ PV}$$
5. (c) Oil produced at breakthrough. The average water saturation behind the front at the instant of breakthrough is $\bar S_{w2,BT}=S_{wf}+\dfrac{1-f_w(S_{wf})}{f_w'(S_{wf})}=0.60+\dfrac{1-0.788}{1.922}=0.60+0.110=0.710$. Then $$N_{p,BT}=V_p\left(\bar S_{w2,BT}-S_{iw}\right)=705{,}254\times(0.710-0.20)=705{,}254\times0.510=\boxed{359{,}900\text{ STB}}$$ (with $B_o=1.0$, reservoir bbl = STB). As a check, since no water has yet been produced at the instant of breakthrough, a simple volume balance ($W_i=N_pB_o+W_pB_w$, $W_p=0$) requires $N_{p,BT}\approx W_{i,BT}$; the two independent calculations agree to within about 2% (366,937 vs. 359,900 bbl), the residual gap coming from the table's coarse 0.05-$S_w$ spacing and its own rounding of $df_w/dS_w$ to three decimals, not from a modelling error.
6. (d) Saturation profile at $t=240$ days. The frontal-advance coefficient is $$C=\frac{5.615\,q_it}{A\phi}=\frac{5.615\times900\times240}{26{,}400\times0.25}=\frac{1{,}212{,}840}{6600}=183.8\text{ ft (per unit }df_w/dS_w\text{)}.$$ Multiplying by the tabulated derivative at each $S_w\ge S_{wf}$ gives the position of that saturation: $x(0.75)=183.8(0.501)=92.1$ ft, $x(0.70)=183.8(0.831)=152.7$ ft, $x(0.65)=183.8(1.313)=241.3$ ft, $x(0.60)=183.8(1.922)=\boxed{353.2\text{ ft}}$ (the shock front itself). Since $353.2\text{ ft} < L=600$ ft (and $t=240$ days $