Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 / 2

Let $k_1>k_2$ and let both original and injected fluids initially have dynamic viscosity $\mu$. Assume sharp displacement fronts, constant porosity and negligible exchange between layers. The imposed pressure difference gives Darcy velocities $q_i^0=k_i\Delta P/(\mu L)$ and interstitial velocities $u_i^0=q_i^0/\varphi$. The two arrival times are thereforeIf $A_i$ is the flow cross-section of layer $i$, its volume discharge is $Q_i^0=A_i{q}_i^0$. Equal viscosities make the hydraulic resistance independent of front position, so the total fluid discharge $Q_1^0+Q_2^0$ stays constant. The injected-fluid component at the outflow well isIt jumps from zero to the high-permeability discharge at the first breakthrough, then to the total discharge at the second. Before a layer breaks through, its outflow is original reservoir fluid. Cross-sections are needed to state an absolute volume discharge or its partition; the arrival times do not require them.

A model must track polymer age as well as its position. Polymer enters during $s\in [t_a,t_a+\Delta t]$ and activates at age $\tau$. The first activation in laboratory time is $t_a+\tau$, not $\tau$ measured from initial water injection. Assume activation permanently changes the local dynamic viscosity from $\mu$ to $\lambda \mu$, with $\lambda >1$, and neglect polymer diffusion, retention and shear-dependent rheology.

Let $X_i(t)={\int }_0^t{u}_i(r)dr$ be cumulative pore-fluid displacement. A fluid particle entering at time $s$ has position $X_i(t)-X_i(s)$ until it exits. At time $t\ge t_a+\tau$, the youngest activated polymer was injected at $s_y=\min (t_a+\Delta t,t-\tau )$. The length of activated polymer still inside layer $i$ is exactlySet ${\ell }_i=0$ when $t<t_a+\tau$. This accounts both for gradual activation of the slug and its later exit. Hydraulic resistances add in series, so delayed polymer diversion in parallel porous layers is described byThese history-dependent equations close the model and can be integrated forwards, retaining previously computed values of $X_i$. The overall volume discharge and low-permeability inflow fraction areThe original displacement front reaches the distal well when $X_i(t)=L$. Consequently the injected-fluid outflow is ${\sum }_i{Q}_i(t)1_{X_i(t)\ge L}$, and the complementary sum gives original-fluid production.

The mechanism is especially transparent if $\Delta t<\tau$, so the entire slug is injected before activation begins, and it activates before leaving the high-permeability layer. Before activation, its lengths are $u_i^0\Delta t$. While each complete activated slug remains in its layer, those lengths stay fixed andThe fractional resistance increase is larger in the fast layer. An effective selective regime is $\Delta t<t_a$ and $\Delta t<\tau <t_a$, with $(\lambda -1)\Delta t/t_a$ appreciable while $(\lambda -1)\Delta t/t_s$ is small; the full delay model handles the activation ramps and the later exits. The relative share entering the slow layer then increases, although its absolute discharge cannot increase at fixed pressure. The total discharge decreases during polymer occupancy and returns to its original value once the activated slugs leave. A fixed-total-injection-rate experiment would have a different pressure response and cannot be substituted for this fixed-pressure model.

Before the unmodified low-permeability breakthrough, the only original-fluid discharge is from layer 2. Polymer raises its resistance, so its rate is never greater than $Q_2^0$ at the same fixed pressure. Selective diversion therefore does not increase the instantaneous original-fluid discharge during this initial interval.

It can increase original-fluid production at a later specified time by delaying completion of the sweep. Let $T_2$ be the treated low-permeability breakthrough time. Integrating its interstitial velocity giveswhenever activated polymer occupies that layer for a positive time before breakthrough. For $t_s<t<T_2$, the unmodified reservoir produces no original fluid, whereas the treated reservoir still produces $Q_2(t)>0$. Thus the strategy can give greater original-fluid flux at a given time after unmodified breakthrough, through delayed sweep rather than increased absolute slow-layer flow. A sufficient activation condition is $t_a+\tau <t_s$ with a nonzero slug that has not already exited the slow layer. If polymer only affects layer 1 after its breakthrough and never affects layer 2 before its own breakthrough, this particular benefit does not occur. Integrated recovery over a chosen finite operating window must be distinguished from a comparison at one time.