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Cauchy momentum equation

In convective or Lagrangian form:

\begin{aligned} \frac{Du}{Dt} = &\frac{1}{\rho} \nabla \cdot \sigma + \mathbf{f}\\[12pt] \because \space u&: \text{flow velocity} \quad (\text{unit: } m/s) \\ t &: \text{time} \quad (\text{unit: } s) \\ \frac{Du}{Dt} &: \text{material derivative of } \mathbf{u} = \partial_t \mathbf{u} + \mathbf{u} \cdot \nabla u \quad (\text{unit: } m /s^2) \\ \rho &: \text{density at given point of the continuum} \quad (\text{unit: } kg/m^3) \\ \sigma &: \text{stress tensor} \quad (\text{unit: Pa} = N/m^2 = \text{kg} \cdot m^{-1} \cdot s^{-2}) \\[8pt] \mathbf{f} &: \begin{bmatrix} f_x \\ f_y \\ f_z \end{bmatrix} \quad (\text{unit: } m/s^2) \\ \nabla \cdot \boldsymbol{\sigma} &= \begin{bmatrix} \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z} \\ \frac{\partial \sigma_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{zy}}{\partial z} \\ \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} \end{bmatrix} \quad (\text{unit: Pa}/m) \\ \end{aligned}

NOTE: $\mathbf{f}$ is the vector containing all accelerations caused by body force and $\nabla \cdot \boldsymbol{\sigma}$ is the divergence of stress tensor.

common annotation

We only use Cartesian coordinate system (column vector) for clarity, but equation often written using physical components (which are neither covariants (column) nor contra-variants (row))

differential derivation

generalized momentum conservation principles

The change in system momentum is proportional to the resulting force acting on this system

\vec{p}(t + \Delta t) - \vec{p}(t) = \Delta t \vec{\overline{F}}

where $\vec{p}(t)$ is momentum at time $t$ , and $\vec{\overline{F}}$ is force averaged over $\Delta t$

integral derivation

Applying Newton’s second law to a control volume in the continuum being gives

ma_i = F_i

Then based on Reynolds transport theorem using material derivative ¹ annotations:

\begin{align} \int_{\Omega} \rho \frac{D u_i}{D t} \, dV &= \int_{\Omega} \nabla_j \sigma_i^j \, dV + \int_{\Omega} \rho f_i \, dV \\ \int_{\Omega} \left( \rho \frac{D u_i}{D t} - \nabla_j \sigma_i^j - \rho f_i \right) \, dV &= 0 \\ \rho \frac{D u_i}{D t} - \nabla_j \sigma_i^j - \rho f_i &= 0 \\ \frac{D u_i}{D t} - \frac{\nabla_j \sigma_i^j}{\rho} - f_i &= 0 \end{align}

where $\Omega$ represents control volume.

conservation form

\frac{\partial j}{\partial t} + \nabla \cdot \mathbf{F} = \mathbf{s}

where $\mathbf{j}$ is the momentum density at given space-time point, $\mathbf{F}$ is the flux associated to momentum density, and $\mathbf{s}$ contains all body force per unit volume.

Assume conservation of mass, with known properties of divergence and gradient we can rewrite the conservation form of equations of motions

\frac{\partial}{\partial{t}}(\rho \mathbf{u}) + \nabla \cdot (\rho \mathbf{u} \otimes \mathbf{u}) = - \nabla p + \nabla \cdot \tau + \rho \mathbf{a}

where $\otimes$ is the outer product of the flow velocity $\mathbf{u}$ : $\mathbf{u} \otimes \mathbf{u} = \mathbf{u} \mathbf{u}^T$

convective form

\frac{D \mathbf{u}}{Dt} = \frac{1}{\rho} \nabla \cdot \sigma + \mathbf{f}

the definition of material derivative are as follow:

definition

For any tensor field $y$ that is macroscopic, or depends on ly on position and time coordinates $y=y(\mathbf{x}, t)$
$\frac{Dy}{Dt} = \frac{\partial y}{\partial t} + \mathbf{u} \cdot \nabla y$
where $\nabla y$ is the covariant dervative of the tensor, and $\mathbf{u}(\mathbf{x}, t)$ is the flow velocity

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Cauchy momentum equation

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differential derivation

integral derivation

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convective form

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Cauchy momentum equation

Étiquette

publié à

modifié à

durée

source

differential derivation

integral derivation

conservation form

convective form

Remarque

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