# A Review of Mixture Theory for Deformable Porous Media and Applications
## 3 Applications of Mixture Theory
### Mixture Theory Based Modelling and Simulation of Composite Manufacturing Processes  
The application of mixture theory to model the manufacturing process of composite materials by compression moulding is discussed in this section. This is the process in which deformable porous material is pre-heated and then compressed. In particular, number of fibers are pre impregnated with certain quantity of liquid matrix possibly distributed in piles in a unidirectional or multidirectional way and finally placed among the porous mold. This generated mixture of solid and liquid is then heated and compressed by a piston which generates an increase in solid volume fraction and produces a flow in a deformable porous material. In this complex and fast process of compression moulding, the focus is to reduce the fabrication cost and should be accurate enough to reduce the chance of creating any inhomogeneities or damage to the solid. In general this process can be defined as a process of liquid flow through a deformable porous material and mixture theory equations for flow in deformable porous media can be helpful. The mathematical model and the associated examples are taken from the research article of [PERSON] et al. [22]. The geometry of the flow model is given in Figure 4.  
Here the set of mathematical equations are first written in Eulerian frame from Equations (7), (8) and (15) and then performed necessary mathematical manipulation steps to write down a partial differential equation in Eulerian frame on a time-varying domain along with moving boundary. These equations are difficult to hand numerically, therefore, Lagrangian coordinates are used to write down the following partial differential equation on a fixed solid matrix  
\[\frac{\partial e}{\partial t}-\frac{(1+e^{*})^{2}}{\mu}\frac{\partial}{ \partial X}\left[\frac{K(e)}{(1+e)}\Sigma(e)\frac{\partial e}{\partial X} \right]=0 \tag{22}\]  
Figure 4: The schematic of a compression moulding process. Figure 3: Solid displacement as a function of radial distance for various values of time \(t\) and permeability parameter \(m\). where \(e=\frac{1-u}{u}\) and \(u\) is solid volume fraction. Here \(K(e)=\alpha e^{(\frac{-\beta}{1+e})}\), \(\sigma(e)=\gamma[e^{(\frac{\beta}{1+e})}-e^{(\frac{\beta}{1+e^{*}})}]\) and \(\Sigma(e)=\left|\frac{d\sigma(e)}{de}\right|\), where \(\gamma=0.3\) Pa, \(\alpha=1.5\times 10^{-2}\) m\({}^{2}\), \(\delta=25\) and \(\beta=15\) are used in the simulation below. Initial and boundary conditions for velocity and pressure driven dynamics are written as given by [PERSON] et al. [22]  
\[e(X,0)=e^{*},\frac{\partial e(0,t)}{\partial X}=0 \tag{23}\] \[e(L^{*},t)=\sigma^{-1}(P_{0}(t))\quad\text{(for pressure driven dynamics)}\] (24) \[\frac{\partial e(L^{*},t)}{\partial X}=-\frac{\mu}{K(e(L^{*},t)) \Sigma(e(L^{*},t))}\frac{1+e(L^{*},t)}{(1+e^{*})}v_{p}(t)\quad\text{(for velocity driven dynamics)} \tag{25}\]  
where \(\mu\) is the liquid viscosity, \(L^{*}\) the initial height of wet prepreg, \(P_{0}(t)\) the pressure applied on piston and \(v_{p}(t)\) the piston velocity. The above equations are then solved numerically after nondimensionalization and the results for solid volume fraction \(u(Y,T)\) are shown in Figures 5 and 6. In particular, Figure 5 shows solid volume fraction \(u\) as a function of Lagrangian coordinate \(Y\) and normalized time \(T\) for velocity driven dynamics when \(v_{p}(t)=3[1-\cos(20\pi T)]\). Note that \(Y=0\) indicates the piston surface whereas \(Y=1\) corresponds to the draining surface. The Figure 5 shows the compression of the porous material starts from the draining surface which we represent by \(Y=1\). When this part feels the load, the material under the piston shows compression after sometime. It is important to note that the solid volume fraction of the porous material attains homogeneous state after certain time. Figure 6 shows solid volume fraction \(u\) as a function of Lagrangian coordinate \(Y\) and normalized time \(T\) for pressure profile \(P_{0}(t)=1.25\sin(\frac{50\pi}{6}T)\).