• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Unclear how heat interacts with Navier Stokes

+ 1 like - 0 dislike

I am playing around with an Navier stokes solver and I'm having trouble introducing heat.

Am I right in thinking this would be introduced in the ${\bf f}$ term of ${\partial{\bf u}\over\partial t} = -({\bf u}\cdot\nabla){\bf u}+v\nabla^2{\bf u}+{\bf f}$?

I find the lack of mass term disturbing. Please reassure me!

Also, how do I calculate the force vector, given that I have computed the scalar $dQ\over dt$ from Newton's law of cooling? Or is that the wrong approach?

This post imported from StackExchange Physics at 2014-03-17 04:25 (UCT), posted by SE-user spraff
asked Feb 22, 2013 in Theoretical Physics by spraff (5 points) [ no revision ]

3 Answers

+ 2 like - 0 dislike

The Navier-Stokes equation to which you refer is more generally the first moment of velocity of the Boltzmann equation. In order to get a proper connection to heating, you need a second-velocity-moment Navier-Stokes equation. The Boltzmann equation keeps track of distributions of particles. This changes the question from "What is the density and flow of a fluid at a point $\mathbf{x}$ at a time $t$?" to "What is the probability of finding a particle between $\mathbf{x}$ and $\mathbf{x}+\mathrm{d}\mathbf{x}$, with a velocity between $\mathbf{v}$ and $\mathbf{v}+\mathrm{d}\mathbf{v}$, at time $t$?" A nice transition between the two formalisms is discussed in these notes (although they include gravity as an external force the way plasma physicists include the Lorentz force... just imagine the equations without those terms for a plain fluid, keeping only the collisonal term).

It's worth noting that each moment depends on a term from the next higher moment ($d\rho/d t$ depends on $\mathbf{u}$, $d\mathbf{u}/d t$ depends on $\overleftrightarrow{P}$, $d E/d t$ (which is the same moment as $\overleftrightarrow{P}$) depends on the heat conduction, which is a 3rd order moment... Any equation that cuts off has to assume some kind of closure method. For example, to close at first order, you might assume that the pressure is isotropic. Or to close at second order, you might assume that the conduction is infinite (compared to the timescales of interest).

To answer your specific question, volumetric heating can result in a change in pressure, but you need an equation of state linking pressure, temperature and density. (Heating steam will have a very different response from heating water.) The modified pressure term can in turn couple to the first velocity moment (the Navier-Stokes equation you have written).

This post imported from StackExchange Physics at 2014-03-17 04:25 (UCT), posted by SE-user KDN
answered Feb 22, 2013 by KDN (20 points) [ no revision ]
+ 2 like - 0 dislike

To allow for heat effects in a fluid, you need to couple the Navier-Stokes equations, (momentum conservation) which BTW contain the continuity equation for mass conservation too, to the energy (or temperature) equation (energy conservation).

Momentum dissipation in the momentum equation

$$ \frac{\partial v}{\partial t} + (\vec{v}\cdot\nabla)\vec{v} = -\frac{\nabla p}{\rho} + \frac{1}{\rho}\nabla S $$

is more correctly described by the divergence of the symmetric stress tensor $S$

$$ S = \rho \,\nu (\nabla \circ\vec{v} + (\nabla \circ\vec{v})^{T}) + \rho \,\eta\, I (\nabla \cdot \vec{v}) $$

The coefficients $\mu$ and $\nu$ denote the dynamic and kinematic viscosity respectively, $\circ$ is the tensor (outer) product, and $I$ is the unity tensor.

In the energy equation, the momentum dissipation leads to a corresponding positive definite dissipation (frictional heating) $\epsilon$

$$ \epsilon = \frac{1}{\rho}(S \, \nabla) \cdot \vec{v} $$

The energy equation can, dependent on the system considered, have different sources of diabatic heating apart from the radiative heating (of which Newtonian cooling es a spacial case), such as latent heating due to phase transitions for example.

answered Feb 23, 2013 by Dilaton (6,240 points) [ revision history ]
+ 0 like - 0 dislike

In fluid dynamics, especially in modeling, there are different flavors for including heat.

First of all, heat can be a passive tracer which does not influence the flow, this basically means, that you solve the scalar transport-diffusion equation for the temperature (or energy if you which), which reads, in your notation

$${\partial{T}\over\partial t} +({\bf u}\cdot\nabla){T}=\alpha\nabla^2{T}+Q$$

where $\alpha$ is thermal diffusion coefficient and $Q$ are heat sources (which may be dependent on $T$ for radiation).

The density of fluids is temperature-dependent. You should incorporate this in the Navier-Stokes equations. If the density differences are only relevant for the buoyant terms (e.g. gravity), it is enough to include the effect in $\bf f$, by the Boussinesq approximation. This basically means that you linearize the $\rho(T)$ curve and include the force term as $\beta g (T-T_{ref})$ with $\beta$ a thermal expansion coefficient.

The next step in complexity, arises when $\rho$ changes such, that it takes over control for more terms. The Navier-Stokes equation you gave, assumes constant density, which therefor drops out of the equations. This is generally not true, and you have to take into account the variable density in all terms.

This post imported from StackExchange Physics at 2014-03-17 04:25 (UCT), posted by SE-user Bernhard
answered Feb 28, 2013 by Bernhard (0 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights