Balance laws

Learning objectives

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Integral continuity and Transport

Consider a specific quantity \(\Psi\) in an arbitrary volume \(\mathcal C\). \(\Psi\) : mass per unit volume, or energy per unit volume

Assuming continuity or local conservation means

1. \(\Psi\) can increase (decrease) according to influx (outflux) \(\mathbf F\)

2. \(\Psi\) can increase (decrease) according to production (decay) \(\mathbf S\)

3. there is no other mechanism

in equations

\[ \begin{aligned} \underbrace{\frac{d}{dt} \int_{\mathcal C} \Psi d \mathbf x}_{ \text{ rate of change of $\Psi$ in $\mathcal C$}} &= \underbrace{- \int_{\partial \mathcal C} \mathbf F \cdot \mathbf n d \mathbf \sigma}_{ \text{ in-/outflow across the surface (1.)}} + \underbrace{ \int_{\mathcal C} \mathbf S d \mathbf x}_{ \text{ production/decay in (2.)}} \end{aligned} \]

Divergence Theorem (DT)

\[ \begin{align} \int_{\mathcal C} \nabla \cdot \mathbf F d \mathbf x = \int_{\partial \mathcal C } \mathbf F \cdot \mathbf n d \mathbf \sigma \end{align} \]

Reynold’s transport theorem (RTT)

\[ \begin{align} \frac{d}{dt} \int_{\mathcal C(t)} \mathbf \Psi d \mathbf x = \int_{\mathcal C(t)} \frac{\partial \mathbf \Psi}{\partial t} d \mathbf x + \int_{\partial \mathcal C(t) } \Psi \left( \mathbf v_c \cdot \mathbf n \right) \mathbf d \mathbf \sigma \end{align} \]

Are we now moving or not?

We use the following:

From the point of view of a constant control volume: \(\mathbf v_c = \mathbf 0\). An advective transport term appears as part of \(\mathbf F\).

Alternative viewpoint:

From the point of view of a material point moving through the fluid with constant mass, \(\mathbf v_c = \frac{dX_i}{dt} = \mathbf v\). No additional advective transport term appears in \(\mathbf F\). The transport naturally appears from the movement of the control volume.

Applying DT and RTT

\[ \begin{aligned} \underbrace{\frac{d}{dt} \int_{\mathcal C} \Psi d \mathbf x}_{ \text{RTT}} &= \underbrace{- \int_{\partial \mathcal C} \mathbf F \cdot \mathbf n d \mathbf \sigma}_{ \text{DT}} + \int_{\mathcal C} \mathbf S d \mathbf x \end{aligned} \]

Integral form

\[ \begin{aligned} \int_{\mathcal C} \frac{\partial}{\partial t} \Psi d \mathbf x & = - \int_{\mathcal C} \nabla \cdot \mathbf F d \mathbf x + \int_{\mathcal C} \mathbf S d \mathbf x \end{aligned} \]

Differential/local form (strong form)

\[ \begin{aligned} \partial_t \Psi & = - \nabla \cdot \mathbf F + \mathbf S \end{aligned} \]

• We assumed some regularity of \(\Psi, \mathbf F\), and \(\mathbf S\). Basically we have to demand that they are continuous and differentiable. Otherwise the strong formulation doesn’t make any sense.

Types of transport

The two basic types of transport are: \(\mathbf F = \Psi \mathbf v + \mathbf J\)

1. Advective transport at velocity \(\mathbf v\), hence \(\Psi \mathbf v\), and

2. Diffusive transport with a diffusive flux \(\mathbf J\),

Example

gradient-driven transport \(\mathbf J = - D \nabla \Psi\), in which \(D\) stands for the diffusion coefficient.

Application of gradient-driven transport

• Fourier’s law (heat conduction)

• Fick’s law (mass diffusion)

Generic balance law

\[ \begin{aligned} \partial_t \Psi + \nabla \cdot \left( \Psi \mathbf v \right) & = - \nabla \cdot \mathbf J + S \end{aligned} \]

Remark

• The specific quantity \(\Psi\) being scalar implies \(\nabla \Psi\) to be a vector.

• The diffusion coefficient can be a scalar \(D\) or a second order tensor \(\mathbf D\) (anisotropic diffusion)

• Both \(D \nabla \Psi\) and \(\mathbf D \nabla \Psi\) will be a vector.

• \(\nabla \cdot D \nabla \Psi\) and \(\nabla \cdot \mathbf D \nabla \Psi\) will be a scalar

• It has the same dimension as \(\Psi\).

Component-notation of diffusive transport

Scalar \(D\)

\[ \begin{aligned} \partial_t \Psi &+ \partial_x \left( \Psi v_x \right) + \partial_y \left( \Psi v_y \right) + \partial_z \left( \Psi v_z \right) \\[1em] & = \partial_x \left( D \partial_x \Psi \right) + \partial_y \left( D \partial_y \Psi \right) + \partial_z \left( D \partial_z \Psi \right) + S \end{aligned} \]

Tensor \(D_{ij}\)

\[ \begin{aligned} \partial_t \Psi + \partial_x \left( \Psi v_x \right) + \partial_y \left( \Psi v_y \right) + \partial_z \left( \Psi v_z \right) = \partial_x & \left( d_{xx} \partial_x \Psi + d_{xy} \partial_y \Psi + d_{xz} \partial_z \Psi \right) \\ + \partial_y & \left( d_{yx} \partial_x \Psi + d_{yy} \partial_y \Psi + d_{yz} \partial_z \Psi \right) \\ + \partial_z & \left( d_{zx} \partial_x \Psi + d_{zy} \partial_y \Psi + d_{zz} \partial_z \Psi \right) + S \end{aligned} \]

Index notation

\[ \begin{aligned} \partial_t \Psi + \partial_i \Psi v_i & = \partial_i d_{ij} \partial_j \Psi + S \end{aligned} \]

Examples anisotropic transport

Whenever diffusion differs with the orientation

Example

flow in a layered porous medium

Mass balance

Set \(\Psi = \rho\) being mass per unit volume. \[ \mathbf J = 0 \quad \text{and} \quad \mathbf S = 0. \]

This yields the well known mass balance:

\[ \partial_t \rho + \nabla \cdot \left( \rho \mathbf v \right) = 0 \]

Momentum balance

Set \(\Psi = \rho \mathbf v\), hence being momentum per unit volume. \[ \mathbf J = \mathbf \sigma \quad \text{and} \quad \mathbf S = \rho \mathbf b, \]

in which \(\mathbf b\) is a body force and \(\mathbf \sigma\) is the Cauchy stress tensor.

All in all, this yields the well known momentum balance:

\[ \partial_t ( \rho \mathbf v ) + \nabla \cdot \left(\rho \mathbf v \otimes \mathbf v \right) = \nabla \cdot \mathbf \sigma + \rho \mathbf b \]

The terms corresponds from left to right:

1. \(\partial_t ( \rho \mathbf v )\) : local change of momentum per unit volume

2. \(\nabla \cdot \left( \rho \mathbf v \otimes \mathbf v \right)\) : influx/outflux of momentum into control volume due to advective transport

3. \(\nabla \cdot \mathbf \sigma\) : force action of the ambient continuous medium through its boundary (examples: stretched rod, fluid at rest)

4. \(\rho \mathbf b\) : total mass force acting on the medium, e.g. gravitational force

Momentum balance in velocity form

Use the mass balance to write

\[ \partial_t \mathbf v + \left( \mathbf v \cdot \nabla \right) \mathbf v = \tfrac{1}{\rho} \nabla \cdot \mathbf \sigma + \mathbf b \]

Identifying the total derivative yields

\[ \frac{D}{Dt} \mathbf v = \tfrac{1}{\rho} \nabla \cdot \mathbf \sigma + \mathbf b, \]

which mimicks Newton’s second law.