= Solution
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> gives
$$
\boxed{T_2-t_s=
\int_0^{T_2}\left(1-\frac{u_2(t)}{u_2^0}\right)dt>0}
$$
whenever 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 \b[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.
Back to article page
