Diffusion, extensibility, flow, and orientation coupling in polymers filled with extensible and flexible fibers

Abstract

When a fiber is subjected to flow and mass transport, it deforms and swells and thus its orientation and length change accordingly. As a starting point, a semi-kinetic equation is proposed to describe the time evolution of the length and orientation of extensible fibers. Then a mesoscopic model, explicitly incorporating the coupling arising among flow, mass transport, and fibers extensibility and orientation in a mixture composed of a solvent and a fiber-reinforced polymer (FRP), is formulated. The derived governing and constitutive equations possess the GENERIC structure and are parameterized by mobility coefficients and the Helmholtz free energy density. The latter takes into account the orientational ordering of the fibers, the fiber-fiber topological interactions, and the flexible nature of the fibers. The unidirectional flow-free mass transport is thoroughly discussed via scaling analysis, numerical solutions, and comparison with sorption data selected from literature. The dynamics of the boundaries is also examined.

Appendices

Appendix A

Fiber length-orientation distribution function

Let ψ(p)dp designates the square of the length of a fiber whose unit vector p is in the solid angle dp. Under the transformation, the length increases by a factor (L ′ /L)², hence

$$ \psi^{\prime}\left(\boldsymbol{p}^{\prime}\right)d\boldsymbol{p}^{\prime }=\psi \left(\boldsymbol{p}\right)\boldsymbol{dp}{\left(\frac{L\prime }{L}\right)}^{\mathbf{2}} $$

(85)

The ratio (L ′ /L) is given by Eq. (13). The ratio (dp ’ /dp) can be obtained from the transformation rule of the vector p. To this end, we consider a cone with a height k and base area k²dp in the p space, and study how it is transformed when each of its points are transformed according to

$$ \boldsymbol{p}\to \boldsymbol{p}'=\boldsymbol{E}.\boldsymbol{p} $$

(86)

The volume of the cone is

$$ V=\frac{1}{3}{k}^3d\boldsymbol{p} $$

(87)

which it is transformed to

$$ V^{\prime }=\frac{1}{3}k{\prime}^3d\boldsymbol{p}^{\prime }=V\ \mathit{\det}\ \boldsymbol{E} $$

(88)

Now

$$ \frac{k'}{k}=\left|\boldsymbol{E}.\boldsymbol{p}\right|=\frac{1}{\left|{\boldsymbol{E}}^{-\mathbf{1}}.\boldsymbol{p}^{\prime}\right|} $$

(89)

From Eqs. (87) – (89), we have

$$ \frac{d\boldsymbol{p}^{\prime }}{d\boldsymbol{p}}=\frac{1}{\mathit{\det}\ \boldsymbol{E}}\frac{1}{{\left|{\boldsymbol{E}}^{-\mathbf{1}}.\boldsymbol{p}\prime \right|}^3} $$

(90)

Eq.(16) follows from Eqs.(85) and (90).

$$ \psi^{\prime}\left(\boldsymbol{p}^{\prime}\right)=\psi \left(\boldsymbol{p}\right)\frac{1}{\mathit{\det}\ \boldsymbol{E}}\frac{1}{{\left|{\boldsymbol{E}}^{-\mathbf{1}}.\boldsymbol{p}\prime \right|}^5} $$

(91)

In the case of inextensible fibers, det E = 1 and L ′  = L. Thus, one easily recovers the classical distribution function for the fiber’s orientation

$$ \psi^{\prime}\left(\boldsymbol{p}^{\prime}\right)=\psi \left(\boldsymbol{p}\right)\frac{1}{{\left|{\boldsymbol{E}}^{-\mathbf{1}}.\boldsymbol{p}\prime \right|}^3} $$

(92)

Appendix B

The time derivative of a regular real valued functional F = F(ρ, c, u, J a) is

$$ \frac{dF}{dt}=\int dr.\left(\left(\frac{\delta F}{\delta \rho}\right)\left(\frac{\partial \rho }{\partial t}\right)+\left(\frac{\delta F}{\delta c}\right)\left(\frac{\partial c}{\partial t}\right)+\left(\frac{\delta F}{\delta {u}_{\alpha }}\right)\left(\frac{\partial {u}_{\alpha }}{\partial t}\right)+\left(\frac{\delta F}{\delta {J}_{\alpha }}\right)\left(\frac{\partial {J}_{\alpha }}{\partial t}\right)+\left(\frac{\delta F}{\delta {a}_{\alpha \beta}}\right)\left(\frac{\partial {a}_{\alpha \beta}}{\partial t}\right)\right) $$

(93)

The time derivative of F contains both reversible and irreversible contributions and can be written as

$$ \frac{dF}{dt}={\left.\frac{dF}{dt}\right|}_{rev}+{\left.\frac{dF}{dt}\right|}_{irr} $$

(94)

Using Eq. (4), we can write explicitly the reversible part as

$$ {\displaystyle \begin{array}{l}{\left.\frac{\mathrm{dF}}{\mathrm{dt}}\right|}_{\mathrm{rev}}=\int d\boldsymbol{r}{\left(\frac{\delta F}{\delta \boldsymbol{\varkappa}}\right)}^t.\mathbf{\mathcal{L}}.\left(\frac{\delta \Phi}{\delta \boldsymbol{\varkappa}}\right)\\ {}=\int d\boldsymbol{r}\left(\frac{\delta F}{\delta \rho}\right)\left(\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta \rho}\right)\right)+\left(\frac{\delta F}{\delta c}\right)\left(\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta c}\right)\right)+\left(\frac{\delta F}{\delta {u}_{\alpha }}\right)\left(\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {u}_{\alpha }}\right)\right)+\left(\frac{\delta F}{\delta {J}_{\alpha }}\right)\left(\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {J}_{\alpha }}\right)\right)+\left(\frac{\delta F}{\delta {a}_{\alpha \beta}}\right)\left(\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {a}_{\alpha \beta}}\right)\right)\end{array}} $$

(95)

Using Eq. (6), we can express the irreversible part as

$$ {\displaystyle \begin{array}{l}{\left.\frac{dF}{dt}\right|}_{irr}=-\int d\boldsymbol{r}\left(\frac{\delta \mathrm{F}}{\delta \boldsymbol{\varkappa}}\right).\left(\frac{\delta \Xi}{\delta \left(\delta \Phi /\delta \boldsymbol{\varkappa}\right)}\right)\\ {}=-\int dr\left(\frac{\delta F}{\delta \rho}\right)\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta \rho \right)}\right)+\left(\frac{\delta F}{\delta c}\right)\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta c\right)}\right)+\left(\frac{\delta F}{\delta {u}_{\alpha }}\right)\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {u}_{\alpha}\right)}\right)+\left(\frac{\delta F}{\delta {J}_{\alpha }}\right)\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {J}_{\alpha}\right)}\right)+\left(\frac{\delta F}{\delta {a}_{\alpha \beta}}\right)\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {a}_{\alpha \beta}\right)}\right)\end{array}} $$

(96)

Now by using Eqs. (93–96), and by matching the different terms corresponding to the same multiplication factor (for example for the same multiplication factor, (δF/δρ), we obtain the term (∂ρ/∂t) appearing in (93), $\mathbf{\mathcal{L}}.\left(\delta \varPhi /\delta \rho \right)$ in (95) and (−δΞ/δ(δΦ/δρ)) in (96) and so on for the rest of the state variables), we can write the governing equations for the state variables as

$$ {\displaystyle \begin{array}{l}\frac{\partial \rho }{\partial t}=\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta \rho}\right)-\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta \rho \right)}\right)\\ {}\frac{\partial {u}_{\alpha }}{\partial t}=\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {u}_{\alpha }}\right)-\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {u}_{\alpha}\right)}\right)\\ {}\begin{array}{l}\frac{\partial c}{\partial t}=\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta c}\right)-\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta c\right)}\right)\\ {}\frac{\partial {J}_{\alpha }}{\partial t}=\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {J}_{\alpha }}\right)-\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {J}_{\alpha}\right)}\right)\\ {}\frac{\partial {a}_{\alpha \beta}}{\partial t}=\mathbf{\mathcal{L}}.\left(\frac{\delta \varPhi}{\delta {a}_{\alpha \beta}}\right)-\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {a}_{\alpha \beta}\right)}\right)\end{array}\end{array}} $$

(97)

Let us now determine each of the terms appearing in the governing equations. To this end, we start with the derived Poisson bracket (26), in which the function G is replaced by the free energy Φ , and then perform integration by parts. Thus the resulting Poisson bracket involves only derivatives of F and Φ with respect to the selected state variables. The next step is to factorize all the different terms that have the same Volterra derivative of F with respect to the same sate variable. For instance, for the state variable ρ the multiplication factor is (δF/δρ) and the associated term appearing in the Poisson bracket is −∂_α(δΦ/δu_α) which has in fact been designated previously by $\mathbf{\mathcal{L}}.\left(\delta \varPhi /\delta \rho \right)$. We follow the same reasoning for the remaining state variables, i.e. (δF/δu_α), (δF/δc), (δF/δJ_α) and (δF/δa_αβ). Therefore, we obtain the following reversible terms corresponding to the reversible kinematics

$$ {\displaystyle \begin{array}{l}\mathbf{\mathcal{L}}.\left(\frac{\updelta \Phi}{\updelta \uprho}\right)=-{\partial}_{\upalpha}\left(\frac{\delta \Phi}{\delta {u}_{\alpha }}\right)\\ {}\mathbf{\mathcal{L}}.\left(\frac{\updelta \Phi}{\updelta \mathrm{c}}\right)=-\uprho \left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upalpha}}\right){\partial}_{\upalpha}\mathrm{c}-{\partial}_{\upalpha}\left(\uprho \mathrm{c}\left(1-\mathrm{c}\right)\ \left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upalpha}}\right)\right)\\ {}\begin{array}{l}\mathbf{\mathcal{L}}.\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upalpha}}\right)=-{\partial}_{\beta}\left({u}_{\alpha}\left(\frac{\delta \Phi}{\delta {\mathrm{u}}_{\upbeta}}\right)\right)-\rho {\partial}_{\alpha}\left(\frac{\delta \Phi}{\delta \rho}\right)-{\mathrm{u}}_{\beta }{\partial}_{\alpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\beta }}\right)-{\mathrm{J}}_{\beta }{\partial}_{\alpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\beta }}\right)+\left(\frac{\delta \Phi}{\delta c}\right){\partial}_{\alpha }c\\ {}\kern6em -{a}_{\upbeta \upgamma}\ {\partial}_{\alpha}\left(\frac{\delta \Phi}{\delta {a}_{\upgamma \upbeta}}\right)-{\partial}_{\beta}\left({\mathrm{J}}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\beta }}\right)-2{\mathrm{a}}_{\upbeta \upgamma}\left(\frac{\updelta \Phi}{\updelta {\mathrm{a}}_{\upgamma \upalpha}}\right)\right)\\ {}\begin{array}{l}\mathbf{\mathcal{L}}.\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upalpha}}\right)=-{\partial}_{\upbeta}\left({\mathrm{J}}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upbeta}}\right)\right)-{\mathrm{J}}_{\upbeta}{\partial}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upbeta}}\right)+{\mathrm{c}\mathrm{\partial}}_{\upbeta}\left({\mathrm{J}}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\right)+{\mathrm{J}}_{\upbeta}{\partial}_{\upalpha}\left(\mathrm{c}\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\\ {}\ \begin{array}{l}\kern6.75em -{\partial}_{\upbeta}\Big(\left(1-\mathrm{c}\right)\left(\mathrm{c}{\mathrm{u}}_{\upalpha}+{\mathrm{J}}_{\upalpha}\right)\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)-\left(1-\mathrm{c}\right)\left(\mathrm{c}{\mathrm{u}}_{\upbeta}+{\mathrm{J}}_{\upbeta}\right){\partial}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\\ {}\begin{array}{l}\kern7em +\frac{{\mathrm{u}}_{\upalpha}}{\uprho}{\partial}_{\upbeta}\left(\uprho \mathrm{c}\left(1-\mathrm{c}\right)\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\right)+\uprho \mathrm{c}\left(1-\mathrm{c}\right){\partial}_{\upalpha}\left(\frac{{\mathrm{u}}_{\upbeta}}{\uprho}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\right)-\uprho \mathrm{c}\left(1-\mathrm{c}\right){\partial}_{\upalpha}\left(\frac{1}{\uprho}\frac{\updelta \Phi}{\updelta \mathrm{c}}\right)\\ {}\kern7em +\mathrm{c}{\mathrm{a}}_{\upgamma \upbeta}{\partial}_{\upalpha}\left(\frac{\updelta \Phi}{\updelta {\mathrm{a}}_{\upgamma \upbeta}}\right)-\mathrm{c}{\partial}_{\upbeta}\left(2{\mathrm{a}}_{\upalpha \upgamma}\ \left(\frac{\updelta \Phi}{\updelta {\mathrm{a}}_{\upgamma \upbeta}}\right)\right)\end{array}\end{array}\\ {}\begin{array}{l}\mathbf{\mathcal{L}}.\left(\frac{\updelta \Phi}{\updelta {\mathrm{a}}_{\upalpha \upbeta}}\right)=-{\partial}_{\upgamma}\left({\mathrm{a}}_{\upalpha \upbeta}\ \left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upgamma}}\right)\right)+{\mathrm{a}}_{\upalpha \upgamma}{\partial}_{\upgamma}\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upbeta}}\right)+{\mathrm{a}}_{\upbeta \upgamma}{\partial}_{\upgamma}\left(\frac{\updelta \Phi}{\updelta {\mathrm{u}}_{\upalpha}}\right)+{\partial}_{\upgamma}\left({\mathrm{a}}_{\upalpha \upbeta}\ \mathrm{c}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upgamma}}\right)\right)\\ {}\kern7.25em -{\mathrm{a}}_{\upalpha \upgamma}{\partial}_{\upgamma}\left(\mathrm{c}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upbeta}}\right)\right)-{\mathrm{a}}_{\upbeta \upgamma}{\partial}_{\upgamma}\left(\mathrm{c}\left(\frac{\updelta \Phi}{\updelta {\mathrm{J}}_{\upalpha}}\right)\right)\end{array}\end{array}\end{array}\end{array}} $$

(98)

We proceed in a similar way with the dissipation potential (27) to obtain the irreversible kinematics

$$ {\displaystyle \begin{array}{l}\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta \rho \right)}\right)=0\\ {}\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta c\right)}\right)=0\\ {}\begin{array}{l}\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {u}_{\alpha}\right)}\right)=-{\partial}_{\beta}\left(\eta (c)\left({\partial}_{\beta}\left(\frac{\delta \varPhi}{\delta {u}_{\alpha }}\right)+{\partial}_{\alpha}\left(\frac{\delta \varPhi}{\delta {u}_{\beta }}\right)\right.\right)+{\delta}_{\alpha \beta}\left(\left({\eta}_d-\frac{2}{3}\eta \right){\partial}_{\gamma}\left(\frac{\delta \varPhi}{\delta {u}_{\gamma }}\right)\right)\\ {}\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {J}_{\alpha}\right)}\right)=-\rho c\left(1-c\right){\varLambda}_J(c)\ \left(\frac{\delta \varPhi}{\delta {J}_{\alpha }}\right)\\ {}\left(\frac{\delta \varXi}{\delta \left(\delta \varPhi /\delta {a}_{\alpha \beta}\right)}\right)=-{\varLambda}_a(c){a}_{\alpha \gamma}\left(\frac{\delta \varPhi}{\delta {a}_{\gamma \beta}}\right)\end{array}\end{array}} $$

(99)

Replacing Eqs. (98) and Eqs. (99) into Eqs. (97) gives Eqs. (28–35).

Appendix C

Derivation of Eq. (45)

Our aim here is to derive Eq. (45). We start with Eqs. (43) and replace them into Eq. (34). The result is

$$ {\displaystyle \begin{array}{l}\frac{\partial {J}_{\alpha }}{\mathrm{\partial t}}=-{\partial}_{\upbeta}\left({J}_{\upalpha}{v}_{\beta}\right)-{J}_{\beta }{\partial}_{\upbeta}{v}_{\upalpha}+{\mathrm{c}\mathrm{\partial}}_{\upbeta}\left(\frac{{\mathrm{J}}_{\upalpha}{\mathrm{J}}_{\upbeta}}{\uprho \mathrm{c}\left(1-\mathrm{c}\right)}\right)-{\partial}_{\upbeta}\left(\frac{{\mathrm{J}}_{\upalpha}{\mathrm{J}}_{\upbeta}}{\uprho \mathrm{c}}\right)-\uprho \mathrm{c}\left(1-\mathrm{c}\right){\partial}_{\upalpha}\left(\frac{1}{\rho}\frac{\delta \varphi}{\delta c}\right)\\ {}+\mathrm{c}{\mathrm{a}}_{\upgamma \upbeta}{\partial}_{\alpha}\left(\frac{\delta \varphi}{\delta {a}_{\upgamma \upbeta}}\right)-\mathrm{c}{\partial}_{\upbeta}\left(2{\mathrm{a}}_{\upbeta \upgamma}\ \left(\frac{\delta \varphi}{\delta {a}_{\upgamma \upalpha}}\right)\right)-{\Lambda}_{\mathrm{J}}\left(\mathrm{c}\right)\ {J}_{\alpha}\end{array}} $$

(100)

The same reasoning gives for the pressure

$$ p=-\varphi +\rho \left(\frac{\delta \varphi}{\delta \rho}\right)+{a}_{\alpha \beta}\left(\frac{\delta \varphi}{\delta {a}_{\alpha \beta}}\right) $$

(101)

Using the assumption of absence of flow, i.e., v = 0, the symmetric second-order extra stress tensor is given by:

$$ {\sigma}_{\alpha \beta}=\frac{{\mathrm{J}}_{\upalpha}{\mathrm{J}}_{\upbeta}}{\uprho \mathrm{c}\left(1-\mathrm{c}\right)}-2{a}_{\alpha \gamma}\ \left(\frac{\delta \varPhi}{\delta {a}_{\gamma \beta}}\right) $$

(102)

The mechanical equilibrium assumption, i.e., ∂_αp + ∂_βσ_βα = 0, gives

$$ \rho {\partial}_{\alpha}\left(\frac{\delta \varphi}{\delta \rho}\right)-\left(\frac{\delta \varphi}{\delta c}\right){\partial}_{\alpha }c+{a}_{\beta \gamma}\ {\partial}_{\alpha}\left(\frac{\delta \varphi}{\delta {a}_{\gamma \beta}}\right)+{\partial}_{\beta}\left(\frac{J_{\alpha }{J}_{\beta }}{\rho c\left(1-c\right)}\right)-{\partial}_{\beta}\left(2{a}_{\alpha \gamma}\ \left(\frac{\delta \varphi}{\delta {a}_{\gamma \beta}}\right)\right)=0 $$

(103)

Now inserting Eq. (103) into Eq. (100) and using the fact that δφ/δρ_s = δφ/δρ + ((1 − c)/ρ)δφ/δc (since v = 0), we obtain a new simplified time evolution equation for the mass flux

$$ \frac{\partial {J}_{\alpha }}{\mathrm{\partial t}}=-{\partial}_{\upbeta}\left(\frac{{\mathrm{J}}_{\upalpha}{\mathrm{J}}_{\upbeta}}{\uprho \mathrm{c}}\right)-{\rho}_s{\partial}_{\alpha}\left(\frac{\delta \varphi}{\delta {\rho}_s}\right)-{\Lambda}_{\mathrm{J}}\left(\mathrm{c}\right)\ {J}_{\alpha } $$

(104)

Finally, using the incompressibility constraint ρ = Const, we have

$$ \frac{\partial {J}_{\alpha }}{\mathrm{\partial t}}=-{\partial}_{\upbeta}\left(\frac{{\mathrm{J}}_{\upalpha}{\mathrm{J}}_{\upbeta}}{\uprho \mathrm{c}}\right)-c{\partial}_{\alpha}\left(\frac{\delta \varphi}{\delta c}\right)-{\Lambda}_{\mathrm{J}}\left(\mathrm{c}\right)\ {J}_{\alpha } $$

(105)