$\newcommand{\Dev}{\operatorname{Dev}}$ $\newcommand{\dev}{\operatorname{dev}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\Grad}{\operatorname{Grad}}$ $\newcommand{\divg}{\operatorname{div}}$ $\newcommand{\Ei}{\operatorname{Ei}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\Divg}{\operatorname{Div}}$ $\newcommand{\cay}{\operatorname{cay}}$ $\newcommand{\rot}{\operatorname{rot}}$

The volume fraction of each constituent is given by $\varphi^{\alpha}=\rho^{\alpha}/\rho_{T}^{\alpha}$ . In a saturated mixture these volume fractions satisfy $\sum\limits _{\alpha}\varphi^{\alpha}=1$ . Substituting $\rho^{\alpha}=\varphi^{\alpha}\rho_{T}^{\alpha}$ into (2.10-1), dividing across by $\rho_{T}^{\alpha}$ (invariant for intrinsically incompressible constituents), and taking the sum of the resulting expression over all constituents produces $$\begin{equation} \divg\left(\sum\limits _{\alpha}\varphi^{\alpha}\mathbf{v}^{\alpha}\right)=\sum\limits _{\alpha}\frac{\hat{\rho}^{\alpha}}{\rho_{T}^{\alpha}}\,.\label{eq162} \end{equation}$$ This mass balance relation for the mixture expresses the fact that the mixture volume will change as a result of chemical reactions where the true density of products is different from that of reactants. Indeed, assuming that $\rho_{T}^{\alpha}$ is the same for all $\alpha$ would nullify the right-hand-side of (2.10.3-1) based on (2.10-2). We now adopt the assumption that solutes occupy a negligible volume fraction of the mixture ($\varphi^{\iota}\ll1)$ , from which it follows that $\varphi^{s}+\varphi^{w}\approx1$ and $\sum\limits _{\alpha}\varphi^{\alpha}\mathbf{v}^{\alpha}\approx\mathbf{v}^{s}+\mathbf{w}$ , where $\mathbf{w}=\varphi^{w}\left(\mathbf{v}^{w}-\mathbf{v}^{s}\right)$ is the volumetrix flux of solvent relative to the solid. Thus, the mixture mass balance may be reduced to $$\begin{equation} \divg\left(\mathbf{v}^{s}+\mathbf{w}\right)=\sum\limits _{\alpha}\frac{\hat{\rho}^{\alpha}}{\rho_{T}^{\alpha}}\,.\label{eq163} \end{equation}$$ In the special case of the solvent ($\alpha=w$ ), FEBio uses a solvent supply, $\hat{\varphi}^{w}=\hat{\rho}^{w}/\hat{\rho}_{T}^{w}$ , which may be incorporated in (2.10.3-2) as $$\begin{equation} \divg\left(\mathbf{v}^{s}+\mathbf{w}\right)=\hat{\varphi}^{w}+\sum\limits _{\alpha\ne w}\frac{\hat{\rho}^{\alpha}}{\rho_{T}^{\alpha}}\,.\label{eq163b} \end{equation}$$