I learned about the integral form of the navier stokes equation and I am trying to find an explanation for this viscous term at the end but I've searched everywhere and I can't find anything. The equation I am talking about is this: $$\frac{\partial }{\partial t} \bigg( \iiint\limits_{V} \rho \vec{u}_i dV \bigg)= -\oint _{\partial S} \vec{u}_i \big( \rho u(\vec{r},t) \cdot dS \big) - \oint _{\partial S} P dS +\text{viscous term}$$ What is this +viscous term is it a number, an equation yielding viscosity or something else? Also from my knowledge $\vec{u}_i$ is an eigenvector but I don't know if it has to do with the velocity.

Edit: I came back to this question and saw that there is a $\partial S$ limit on a closed line integral $\oint$ but there is also a $dS$ on that same integral. The $dS$ is supposed to represent a 2D surface but the $\partial S$ is supposed to represent the boundary of that surface. So should we use the $\oint$ or the $\oiint$ symbol in the formula? (I don't know how to make the oiint version appear).


2 Answers 2


Momentum equation - integral form. Let's start from integral balance, for a fixed control volume $V$ \begin{equation} \dfrac{d}{dt} \int_V \rho \mathbf{u} + \oint_{\partial V} \rho \mathbf{u} \mathbf{u} \cdot \mathbf{\hat{n}} = \int_V \rho \mathbf{g} + \oint_{\partial V} \mathbf{t_n} \ , \end{equation} saying that the time derivative of the momentum equals the external forces, written as the sum of the volume force and the surface forces.

The stress vector $\mathbf{t_n}$ can be written as a function of the unit normal $\mathbf{\hat{n}}$ and the stress tensor $\mathbb{T}$ (second order tensor, usually symmetric), by the Cauchy equilibrium of an infinite element as,

\begin{equation} \mathbf{t_n} = \mathbf{\hat{n}} \cdot \mathbb{T} \ . \end{equation}

Momentum equation - differential form. Introducing this relation in the surface integral on the RHS of the integral balance of momentum, applying the divergence theory to transform surface integrals into volume integrals, since the equation must hold for every volume $V$, we get the differential form of the momentum equation,

\begin{equation} \partial_t (\rho \mathbf{u}) + \nabla \cdot (\rho \mathbf{u} \otimes \mathbf{u}) = \rho \mathbf{g} + \nabla \cdot \mathbb{T} \ . \end{equation}

Newtonian fluids - constitutive equation. A Newtonian fluid is defined as a linear isotropic fluid. For this kind of fluids, the stress tensor can be written as the sum of a pressure and a viscosity contributions,

\begin{equation} \mathbb{T} = -p \mathbb{I} + \mathbb{S} \ , \end{equation}

and the tensor of the viscous stress $\mathbb{S}$ has a linear and isotropic relation with the gradient of the velocity field, that can be written as

\begin{equation} \mathbb{S} = 2 \mu \mathbb{D} + \lambda (\nabla \cdot \mathbf{u}) \mathbb{I} \ , \end{equation}

being $\mu$ and $\lambda$ the viscosity coefficients, $\mathbb{I}$ the identity tensor, and $\mathbb{D}$ the deformation velocity tensor,

\begin{equation} \mathbb{D} = \dfrac{1}{2} \left[ \nabla \mathbf{u} + \nabla^T \mathbf{u} \right] \ . \end{equation}

Inserting the constitutive equation of a Newton fluid into the momentum equation, assuming constant and uniform values of the viscosity coefficients, you get

\begin{equation} \partial_t (\rho \mathbf{u}) + \nabla \cdot (\rho \mathbf{u} \otimes \mathbf{u}) = \rho \mathbf{g} - \nabla p + \nabla \cdot \mathbb{S} \ . \end{equation}

Viscous stress contribution - explicit expression. Now, if you want the explicit version of the viscous vectors stress acting on a surface, you can use the Cauchy relation to discern pressure from viscous contributions

\begin{equation} \mathbf{t_n} = \mathbf{\hat{n}} \cdot \mathbb{T} = \mathbf{\hat{n}} \cdot (-p \mathbf{I} + \mathbf{S} ) = - p \mathbf{\hat{n}} + \mathbf{s_n} \ , \end{equation}

having defined the viscous stress vector as \begin{equation} \mathbf{s_n} = \mathbf{\hat{n}} \cdot \mathbb{S} \ , \end{equation} that you can write for a Newtonian fluid as \begin{equation} \begin{aligned} \mathbf{s_n} = \mathbf{\hat{n}} \cdot \mathbb{S} & = \mathbf{\hat{n}} \cdot \left( \mu \left[ \nabla \mathbf{u} + \nabla^T \mathbf{u} \right] + \lambda (\nabla \cdot \mathbf{u}) \mathbb{I} \right) \\ & = \mu \mathbf{\hat{n}} \cdot \left[ \nabla \mathbf{u} + \nabla^T \mathbf{u} \right] + \lambda (\nabla \cdot \mathbf{u}) \mathbf{\hat{n}} \ , \end{aligned} \end{equation} so that the viscous stress contribution in the integral equation becomes \begin{equation} \mathbf{F}^{visc} = \oint_{\partial V} \mathbf{s_n} = \oint_{\partial V} \mathbf{\hat{n}} \cdot \mathbb{S} = \oint_{\partial V} \left\{ \mu \mathbf{\hat{n}} \cdot \left[ \nabla \mathbf{u} + \nabla^T \mathbf{u} \right] + \lambda (\nabla \cdot \mathbf{u}) \mathbf{\hat{n}} \right\} \ . \end{equation}

If you need its contribution in the differential equation, you need to compute the divergence $\nabla \cdot \mathbb{S}$, that reads

\begin{equation} \nabla \cdot \mathbb{S} = \mu \nabla^2 \mathbf{u} + (\mu + \lambda)\nabla (\nabla \cdot \mathbf{u}) \ . \end{equation}

  • $\begingroup$ anonymous downvoting on this answer = LOL, sucker $\endgroup$
    – basics
    Sep 23, 2023 at 15:59
  • $\begingroup$ Well to be fair, you don't really answer why there's this mysterious "+viscous term" in the equation, you just outline mathematical forms of viscosity. $\endgroup$
    – Kyle Kanos
    Sep 23, 2023 at 16:02
  • $\begingroup$ Also, I should have been more clear with my statements. I had not downvoted your post. I actually added my own answer because you, for at least a second time that I've noticed, didn't even bother trying to answer the question asked but when on a mildly-related tangent. $\endgroup$
    – Kyle Kanos
    Sep 23, 2023 at 20:36

What is this +viscous term is it a number, an equation yielding viscosity or something else? Also from my knowledge $\vec{u}_i$ is an eigenvector but I don't know if it has to do with the velocity.

I'll answer the second half of this first: $\vec{u}_i$ is not an eigenvector, it is the $i$th component of the velocity vector. More concretely, given $\vec{u}=(u_x,\,u_y,\,u_z)$, then $\vec{u}_i$ indicates one of the three terms (i.e., $i\in(x,\,y,\,z)$). The equation you've written is a normally written as the vector expression, $$\frac{\partial}{\partial t}\iiint_V\rho\vec{u}\,\mathrm{d}V\equiv\frac{\partial}{\partial t}\iiint_V\rho\left(\begin{array}{c}u_x \\ u_y \\ u_z\end{array}\right)\,\mathrm{d}V$$ By adding the subscript $i$, the equation is looking at a particular dimension (e.g., $x$).

Now the viscous term must yield some sort of vector, in order to match the dimensions of $\vec{u}$ on the left hand side. The particular form varies mildly between different authors, but it generally takes the form of a second gradient of the velocity, $$\text{viscosity}\propto\nabla\cdot\nabla\vec{u}\tag{1}$$ where $\nabla\vec{u}$ is a dyadic product and yields a tensor; taking the divergence of that yields a vector, which is what we require.

I suspect that part of the reason that authors write the integral form as "plus viscous terms" is that the argument of the integral, $\nabla\vec{u}$, is a tensor and the tensorial version of the divergence theorem is a little more cluttered than the vectorial version, $$\iiint_V\frac{\partial T_{i_1i_2\cdots i_n}}{\partial x_{i_q}}\,\mathrm{d}V={\subset\!\supset} \llap{\iint}_ST_{i_1i_2\cdots i_n}n_{i_q}\,\mathrm{d}S$$ which may raise more questions by students than writing it down as words.

  • $\begingroup$ Oh alright that makes sense. Yeah this can cause a lot of confusion. Also I noticed that you guys use the $\oint$ symbol with subscripts about volume and surfaces. Shouldn't those require $\oiiint$ and $\oiint$ respectively? $\endgroup$
    – Mitsos YT
    Sep 23, 2023 at 20:26
  • $\begingroup$ @MitsosYT I think the volume should be $\iiint$, but the surface should use the o-int versiont (which unfortunately, is not supported by mathjax on this site, but there are ways around that) $\endgroup$
    – Kyle Kanos
    Sep 23, 2023 at 20:34

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