 Solution

Let $X(t)$ denote the polymer–oil material interface, measured from the injection well, and let $U(t)$ be the Darcy velocity. Incompressibility and uniform cross-sectional area make $U$ independent of $x$ in this planar model. Integrating Darcy's law through the injected layer and the oil layer givesThe material interface moves at the pore velocity, so $X^{\prime }=U/\varphi$. Put $a=\mu -{\mu }_h$ and $A={\mu }_{h}L$. Then $(A+aX)X^{\prime }=k\Delta P/\varphi$, which integrates from $X(0)=0$ toThus the fixed-pressure planar Darcy displacement hasFor $a\ne 0$, $X=[\sqrt{A^2+2ak\Delta Pt/\varphi }-A]/a$; its continuous $a=0$ limit is $X=k\Delta Pt/(\varphi {\mu }_{h}L)$. Positive layer dynamic viscosities ensure the resistance stays positive until breakthrough, which occurs at $t_b=\varphi L^2(\mu +{\mu }_h)/(2k\Delta P)$. The formulas stop there, since the two-layer geometry then changes.