profile pic
⌘ '
raccourcis clavier
Search

Navier-Stokes equations

partial differential equations describing the motion of fluid substances. One of seven $1M problems in mathematics.

  • Étiquette

    • publié à

    • modifié à

    • durée

      3 min de lecture

    • source

      llms.txt

    Express momentum balance for Newtonian fluids making use of conversation of mass.

    derivations

    derived as a particular form of the Cauchy momentum equation

    • url: thoughts/Cauchy-momentum-equation
    • description: convective form

    convective form

    DuDt=1ρ∇⋅σ+f\frac{D \mathbf{u}}{Dt} = \frac{1}{\rho} \nabla \cdot \sigma + \mathbf{f} Lien vers l'original

    By setting Cauchy stress tensor σ\sigma to viscosity term τ\tau (deviatoric stress) and pressure term −pI-p \mathbf{I} (volumetric stress), we have

    ρDuDt=−∇p+∇⋅τ+ρa\rho \frac{D\mathbf{u}}{Dt} = - \nabla p + \nabla \cdot \tau + \rho \mathbf{a}

    where:

    assumption upon Cauchy stress tensor

    1. stress is Galilean invariant 1, or it doesn’t depend directly on the flow velocity, but the spatial derivatives of the flow velocity

      tensor gradient ∇u\nabla \mathbf{u}

      rate-of-strain tensor: ε(∇u)≡12∇u+12(∇u)T\boldsymbol{\varepsilon} (\nabla \mathbf{u}) \equiv \frac{1}{2} \nabla \mathbf{u} + \frac{1}{2} (\nabla \mathbf{u})^T

    1. Deviatoric stress is linear in this variable σ(ε)=−pI+C:ε\sigma (\varepsilon) = -p \mathbf{I} + \mathbf{C} : \varepsilon,

      • where pp is independent on the strain rate tensor
      • C\mathbf{C} is the fourth-order tensor for constant of proportionality (viscosity tensor)
      • :: is the double-dot product

    2. fluid is assumed to be isotropic, and consequently C\mathbf{C} is an isotropic tensor.

      Furthermore, the deviatoric stress tensor is symmetric by Helmholtz decomposition, expressed in terms of two Lamé parameters, second viscosity λ\lambda and dynamic viscosity μ\mu:

      σ(ε)=−pI+λtr(ε)I+2με\sigma (\varepsilon) = -p \mathbf{I} + \lambda \text{tr}(\varepsilon)\mathbf{I} + 2 \mu \varepsilon

      Where I\mathbf{I} is the identity tensor and tr(ε)\text{tr}(\varepsilon) is the trace of the rate-of-strain tensor. Thus we can rewrite as:

      σ=−pI+λ(∇⋅u)I+μ(∇u+(∇u)T)\sigma = -p \mathbf{I} + \lambda (\nabla \cdot \mathbf{u}) \mathbf{I} + \mu (\nabla \mathbf{u} + (\nabla \mathbf{u})^T)

      Given trace of the rate of strain tensor in three dimension is the divergence of the flow (rate of expansion):

      tr(ε)=∇⋅u\text{tr}(\varepsilon) = \nabla \cdot \mathbf{u}

      • trace of the stress tensor then becomes tr(σ)=−3p+(3λ+2μ)∇⋅u\text{tr}(\sigma) = -3p + (3 \lambda + 2 \mu) \nabla \cdot \mathbf{u} (trace of identity tensor is 3)

      • alternatively decomposing stress tensor into isotropic and deviatoric part in fluid dynamic:

      σ=−[p−(λ+23μ)(∇⋅u)]I+μ(∇u+(∇u)T−23(∇⋅u)I)\boldsymbol{\sigma} = -\left[ p - \left( \lambda + \frac{2}{3} \mu \right) (\nabla \cdot \mathbf{u}) \right] \mathbf{I} + \mu \left( \nabla \mathbf{u} + (\nabla \mathbf{u})^T - \frac{2}{3} (\nabla \cdot \mathbf{u}) \mathbf{I} \right)

      Introduce bulk viscosity ζ\zeta:

      ζ≡λ+23μ\zeta \equiv \lambda + \frac{2}{3} \mu

      We now have the following linear stress equation:

      linear stress constitutive equation

      σ=−[p−ζ(∇⋅u)]I+μ[∇u+(∇u)T−23(∇⋅u)I]\boldsymbol{\sigma} = -\left[ p - \zeta (\nabla \cdot \mathbf{u}) \right] \mathbf{I} + \mu \left[ \nabla \mathbf{u} + (\nabla \mathbf{u})^T - \frac{2}{3} (\nabla \cdot \mathbf{u}) \mathbf{I} \right]

    Compressible flow

    Convective form

    ρDuDt=ρ(∂u∂t+(u⋅∇)u)=−∇p+∇⋅{μ[∇u+(∇u)T−23(∇⋅u)I]}+∇[ζ(∇⋅u)]+ρa.\begin{aligned} &\rho \frac{D \mathbf{u}}{D t} = \rho \left( \frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} \right) \\ &= -\nabla p + \nabla \cdot \left\{ \mu \left[ \nabla \mathbf{u} + (\nabla \mathbf{u})^T - \frac{2}{3} (\nabla \cdot \mathbf{u}) \mathbf{I} \right] \right\} + \nabla \left[ \zeta (\nabla \cdot \mathbf{u}) \right] + \rho \mathbf{a}. \end{aligned}

    With index notation:

    ρ(∂ui∂t+uk∂ui∂xk)=−∂p∂xi+∂∂xk[μ(∂ui∂xk+∂uk∂xi−23δik∂ul∂xl)]+∂∂xi(ζ∂ul∂xl)+ρai.\begin{aligned} \rho \left( \frac{\partial u_i}{\partial t} + u_k \frac{\partial u_i}{\partial x_k} \right) &= -\frac{\partial p}{\partial x_i} \\ &+ \frac{\partial}{\partial x_k} \left[ \mu \left( \frac{\partial u_i}{\partial x_k} + \frac{\partial u_k}{\partial x_i} - \frac{2}{3} \delta_{ik} \frac{\partial u_l}{\partial x_l} \right) \right] \\ &+ \frac{\partial}{\partial x_i} \left( \zeta \frac{\partial u_l}{\partial x_l} \right) \\ &+ \rho a_i. \end{aligned}

    Conservation form

    ∂∂t(ρu)+∇⋅(ρu⊗u+[p−ζ(∇⋅u)]I−μ[∇u+(∇u)T−23(∇⋅u)I])=ρa.\begin{equation} \begin{aligned} \frac{\partial}{\partial t} (\rho \mathbf{u}) &+ \nabla \cdot \Bigg( \rho \mathbf{u} \otimes \mathbf{u} + \Big[ p - \zeta (\nabla \cdot \mathbf{u}) \Big] \mathbf{I} \\ &\quad - \mu \Big[ \nabla \mathbf{u} + (\nabla \mathbf{u})^T - \frac{2}{3} (\nabla \cdot \mathbf{u}) \mathbf{I} \Big] \Bigg) \\ &= \rho \mathbf{a}. \end{aligned} \end{equation}

    Incompressible flow

    Remarque

    1. Implies the laws of motion are the same in all inertial frames of references

      Often refers to this principle as applied to Newtonian mechanics, that is Newton’s laws of motion hold in all frames related to one another by a Galilean transformation. ↩

    Vous pourriez aimer ce qui suit

  • data
  • being.
  • Merkle DAG
  • Questions about Apology
  • DB design, concurrency and transaction
  • Vector calculus
  • Gaussian elimination, LU decompositions, and errors LS solving