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Balance Equations: Mass Balance

Let \Omega_0 and \Omega represent the reference and the deformed configurations, respectively, of a body embedded in \mathbb{R}^3. Let x(X,t)\in\Omega represent the image of X\in\Omega_0 at a certain point in time t\in\mathbb{R} under the bijective smooth mapping f_t:\Omega_0\rightarrow\Omega. Notice that the set \Omega which is the range of f_t changes as time changes. The velocity vector is described by the vector valued function v=\dot{x}(X,t). F and L are the deformation and the velocity gradients, respectively, while J=\det(F). The density distribution of the continuum body in the reference configuration is given by the function \rho_0:\Omega_0\rightarrow\mathbb{R}. As the body deforms, the density at each material point changes with time and thus can be described by a function of both time and material points \rho_t:\Omega_0\times\mathbb{R}\rightarrow\mathbb{R}. The spatial density distribution of the continuum body in the deformed configuration is a function of the position x and of time t and is described by the function \rho:\Omega\times\mathbb{R}\rightarrow\mathbb{R}. The difference between \rho_t and \rho is merely a change of coordinates from X to x:

    \[\rho(x,t) = \rho_t(f_t(X),t) = \rho_t(X,t)\]

In this section, we will use the physical law of mass balance to establish the relationship between the mass densities \rho_0, \rho_t and \rho. If \mathrm{d}x and \mathrm{d}X represent an infinitesimal volume in the deformed and reference configurations, respectively, then, the relationship between the volume in the deformed versus the reference configurations is given by:

    \[JdX = dx\]

If m is the actual mass of both infinitesimal volumes and assuming that mass is preserved then the relationship between the densities is given by:

    \[\rho_0(X) = \frac{m}{dX} = \frac{m}{dx} \frac{dx}{dX} = \rho_t(X,t)J = \rho(x,t)J\]

This relationship can also be obtained using integrals over the whole continuum body as follows. The total mass M_0 of the reference configuration can be calculated by the integral:

    \[M_0 = \int_{\Omega_0} \rho_0 dX\]

The total mass M in the deformed configuration can be calculated by the integral:

    \[M = \int_{\Omega} \rho dx\]

The volume integral over \Omega can be replaced by the volume integral over\Omega_0 by using the relationship J\mathrm{d}X=\mathrm{d}x:

    \[M = \int_{\Omega_0} \rho J dX\]

Assuming that the total mass is conserved, i.e., M=M_0, and since \mathrm{d}X is arbitrary, the relationship between the densities is retrieved:

    \[\rho_0(X) = \rho (x,t)J\]

The Continuity Equation

The continuity equation describes the rates of change of the density functions. We first recall the relationship between the velocity gradient L and \dot{J} obtained in the velocity gradient section:

    \[\dot J = JTrace(L)\]

Assuming preservation of mass, i.e., the rate of change of \rho_0(X) is equal to zero leads to the following equivalent forms of the continuity equation:

Lagrangian Formulation

The continuity equation as a function of \rho_0(X) is given by:

    \[\frac{D \rho_0 (X)}{Dt} = 0\]

Alternatively, the continuity equation as a function of \rho_t (X,t) is given by:

    \[\frac{D \rho_0 (X)}{Dt} = \frac{D \rho_t (X,t)J}{Dt} = \frac{\partial \rho_t (X,t)}{\partial t}J + \rho_t(X,t)JTrace(L) = 0\]

Since J > 0 we reach the following form of the continuity equation:

    \[\frac{\partial \rho_t (X,t)}{\partial t} + \rho_t (X,t)Trace(L) = 0\]

Eulerian Formulation

The continuity equation as a function of \rho (x,t) is given by:

    \[\frac{D \rho_0(X)}{Dt} = \frac{d\rho(x,t)J}{dt} = (\frac{\partial \rho(x,t)}{\partial t} + \frac{\partial \rho(x,t)}{\partial x}\cdot v)J + \rho(X,T)JTrace(L) = 0\]

Since J > 0 we reach the following form of the continuity equation:

    \[\frac{\partial \rho(x,t)}{\partial t} + div(\rho(x,t)v) = 0\]

where, according to the results in the vector calculus section:

    \[div(\rho(x,t)v) = \frac{\partial \rho (x,t)}{\partial x} \cdot v + \rho(x,t)Trace(L)\]

Continuity Equation of Fluids

In fluids, the Eulerian formulation is adopted along with the approximation that the fluid is incompressible, i.e., J=1 and \dot{J}=J\mbox{Trace}(L)=0. Therefore, the continuity equation reduces to:

    \[Trace(L) = \frac{\partial v_1}{\partial x_1} + \frac{\partial v_2}{\partial x_2} + \frac{\partial v_3}{\partial x_3} = 0\]

Continuity Equation for Growth or Mass Transfer

In problems where there is growth or mass transfer, the density in the reference configuration is assumed to vary in time and its rate of change is equal to the source or the sink of the local mass change c(X,t) and the continuity equation is then derived from the equation:

    \[\frac{D \rho_0 (X)}{Dt} = c(X,t)\]


The position function of a 2units by 2units plate has the form:

    \[x_1 = X_1 + tX_2^2\cos 80^\circ\]

    \[x_2 = X_2 + tX_1^2\cos 80^\circ\]

    \[x_3 = X_3\]

where 0\leq t\leq 1 units. The density of the material in the reference configuration at t=0 is given by the function:

    \[\rho(X) = 1 + X_1X_2\]

The following tool draws the contour plots of \rho_0(X) and \rho_t(X,t) in the reference configuration and \rho(x,t) in the deformed configuration as a function of time. Vary the time to see the effect on the deformation and the effect on the density plots. Note that \rho_0(X) does not change when time evolves. The computations are based on first calculating the deformation gradient as follows:

    \[F =  \begin{pmatrix} 1 & 2tX_2 \cos 80^\circ & 0 \\ 2tX_1 \cos 80^\circ & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\]

thus, the determinant of F is given by:

    \[J = 1 - 4t^2X_1X_2 (\cos 80^\circ)^2\]

The density \rho_t (X,t) as time evolves is equal to:

    \[\rho_t (X,t) = \frac{\rho_0 (X)}{J}\]

Finding \rho(x,t) requires solving a nonlinear set of equations to express X in terms of x.

(Interactive Activity)


Quiz – Mass Balance

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