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Let $Y(t)$ be the porous thermal front and $X(t)$ the polymer front. With the literal stated convention for Darcy velocity,Therefore $Y=\beta X$ with $\beta =\Gamma \varphi$. This is thermal-front retardation in Darcy flow. The region $0<x<Y$ is cold injected fluid of dynamic viscosity $\mu$; $Y<x<X$ is heated injected fluid of dynamic viscosity $b\mu$; $X<x<L$ is oil of dynamic viscosity ${\mu }_h$. The pressure drop is consequentlywhere ${\mu }_{\mathrm{eff}}=\mu [\beta +b(1-\beta )]$. Apply the preceding fixed-pressure planar Darcy displacement calculation with $\mu$ replaced by this effective dynamic viscosity. The polymer-front speed isIts position solves ${\mu }_{h}LX+({\mu }_{\mathrm{eff}}-{\mu }_h)X^2/2=k\Delta Pt/\varphi$, and $Y=\Gamma \varphi X$. These expressions hold while both interfaces remain in the field, the porous thermal front is sharp and stable, and $0\le \Gamma \varphi \le 1$. For $b=1$ the original constant-viscosity answer is recovered.

Sometimes a thermal retardation factor is defined relative to the pore velocity instead. Under that alternative definition one substitutes $\beta =\Gamma$ throughout. The factor of $\varphi$ in the boxed answer follows specifically from the PDF's phrase “fraction of the Darcy speed.”

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.