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  Why hydrodynamic turbulence without heat terms matters?

+ 3 like - 0 dislike

A lot of research is made on turbulence in "pure" Navier-Stokes equations (NSE). There is a notion of energy cascade when energy comes from larger scales to lower scales and than dissipate.
However *physically* energy does not just vanish, it turns into heat.

And (if I understand right) presence of heat itself produces a lot of movement. So taking heat into account should change drastically the behavior of the system.

So my question is -- why so much effort is spent on studying pure NSE if they are unphysical and this effort might be useless for a study of actual behavior (when one includes heat terms in NSE)? I know that people study NSE with heat as well but my impression is that this field is much less active (at least from fundamental viewpoint) the one for the case of pure NSE.

I am thinking about a system with a constant force producing turbulence and a heat bath.

UPD: by "pure" NSE I mean the incompressible case represented by system of only two equations -- for momentum and the continuity one.

asked Mar 25, 2015 in Theoretical Physics by demitau (15 points) [ revision history ]
edited Mar 31, 2015 by demitau

The full Navier-Stokes equations are quite realistic for the dynamics of real fluids. Heat is properly taken into account; see dilaton's answer. The dissipation is in terms of entropy production, not loss of energy - the ''dissipated'' energy simply goes into higher and higher frequencies, mking it less and less accessible.

What do you mean by "less accessible"? In the full system these high frequencies produce heat which itself produces movements of the liquid. In the heat-free system energy in higher modes is just destroyed by viscosity.

The incompressible case is of course an approximation to the full NSE, valid only if the divergence of the flow velocity field can be assumed to be negligible. This is equivalent to the assumptions that the material density is essentially constant within each infinitesimal volume that moves with the flow velocity; see http://en.wikipedia.org/wiki/Incompressible_flow . As a consequence of this approximation, the energy balance of the full NSE is satisfied only approximately - which means that small amounts of energy remain unaccounted for. But it simplifies the system of equations enough that it is a frequently used approximation. - Note that the equations must be solved numerically anyway, and it is enough to use approximations that are accurate only to the accuracy of the discretization used.

You mean that if I consider compressible flow without heat terms, the energy balance would be saisfied precisely? I was always thinking that to have it one has to have both compressibility and heat terms.

2 Answers

+ 2 like - 0 dislike

To describe the evolution of a (in this example non-relativistic) fluid system, the evolution equations for all relevant variables, conservation laws, the second law of thermodynamics, and appropriate (a)n appropriate equation(s) of state have to be considered.

The evolution equation for momentum is the Navier-Stokes equation which in a geophysical context can be written as

\[\frac{d u}{d t} + (u\cdot\nabla)u = -\frac{\nabla p}{\rho} -\nabla\Phi +\frac{1}{\rho}\nabla S\]

$\Phi$ is the geopotential and $S$ is the stress tensor. Conservation of angular momentum is taken into account by imposing the contraint that the stress tensor

\[S = \rho \nu \{ \nabla \circ u + (\nabla \circ u)^T\} + \rho \eta I (\nabla \cdot u)\]

where $\nu$ and $\eta$ denote the dynamic and kinematic viscosity respectively, is symmetric.

The evolution equation for internal energy (or equivalently temperature) can be written as

\[\frac{d e}{d t} = \frac{p}{\rho^2}\frac{d\rho}{d t} + Q_{rad} + Q_{lat} -\frac{1}{\rho}\nabla J + \frac{1}{\rho}(S\nabla)\cdot u\]

The second law of thermodynamics is considered by demanding that the last term in the above equation

\[\epsilon = \frac{1}{\rho} (\nabla S)\cdot u\]

which describes the frictional heating or dissipation is positive definite.

Conservation of mass is considered by including the continuity equation

\[\frac{d\rho}{d t} = \frac{\partial\rho}{\partial t} +\nabla(\rho u)\]

into the relevant system of equations.

As you can see, this is a coupled system of equations. The kinetic energy dissipated due to the friction term in the Navier-Stokes equation reappears as dissipative heating $\epsilon$ in the internal energy (or temperature) equation, so the (kinetic) energy can not disappear.

A nice example of a study which includes the temperature equation in addition to the Navier-Stokes equation, is the stuy of Sukorianski et al., who investigate stochastically forced turbulent flows with stable stratification by making use of renormalization-group like methods. By deriving a coupled system of RG equations that describes the scale-dependence of the anisotropic diffusivities of velocitie as well as of (potential) temperature fluctuations, they are by assuming the presence of a Kolmogorov scale invariant subrange able to repoduce the correct kinetic energy cascade and by slightly extending their work, it should be possible to derive the corresponding scale-dependence of the spectrum of temperature fluctuations (or available potential energy) too.

If this answer is not exactly what you wanted, I hope that it helps a bit at least.


answered Mar 26, 2015 by Dilaton (6,240 points) [ revision history ]

This is an interesting article, I should invest more time in reading it, but it does not seem to answer my question (or maybe I should just read it really carefully) --  "why people so often consider turbulence in pure NSE if they are unphysical?". In other words, are there situations when one can consider turbulence in pure NSE (incompressible, without heat terms -- when you have only two equations -- for momentum with $\Phi=0, \frac{1}{p}\nabla{S} = \nu \nabla^2 u$, and divergence = 0) and still get results relevant for physics? I mean usully in physics you neglect something that has small influence and consider a simpler system. I do not see when heat terms might have small influence here.

@demitau as I understand it, neglecting the temperature (or energy) equation is in principle only allowed in an (approximately) isothermal situation or if all processes that change the internel energy (or temperature) are negligible or cancel each other.

@Dilaton yes, it clarifies the picture. Probably this situation can be guaranteed to occur by imposing some assumptions on the interaction with the heat bath.

+ 2 like - 0 dislike

We neglect thermal effects when the fluid is incompressible. This is often a good approximation. For example, you need to go to extreme conditions for the compressibility of water to make a difference.

It looks like turbulence in incompressible fluids exhibits Kolmogorov scaling and intermittency. This indicates that the interesting (and not yet understood) properties of turbulence are not the result of thermal effects, but are a consequence of the non-linear dynamics inherent to Navier-Stokes equations.

The equations of an incompressible fluid are relatively easily to write and contains few variables. As compared to the answer of Dilaton, you can simply set $\Phi$ and $S$ to zero and constrain the pressure through the incompressibility condition $\vec{\nabla} \cdot \vec{v} = 0$. The only degrees of freedom are the pressure and velocity fields. This provides a simple (yet complex enough) set-up to understand turbulence before we include more complicated thermal effects.

answered Apr 4, 2015 by Steven Mathey (350 points) [ no revision ]

Sorry, I feel a bit confused here. I do not doubt that incompressible equations are a good limit for the compressible ones. The thing I do not understand is why these intermittency and Kolmogorov scaling observed in incompressible fluids are believed to be a consequence of only two equations (incompressible without heat terms) rather than a consequence of the full system (incompressible including heat terms).

Through experiments and analytical as well as numerical calculations, we have observed that incompressible turbulence exhibits the energy cascade and Kolmogorov scaling. If you want a nice introduction to this, you can read the book of Uriel Frisch on turbulence.

I believe that thermal effects do not enter the study of turbulence because the energy cascade and Kolmogorov turbulence happens in a range of scales where energy is conserved. There the equation of state does not matter. If there is an energy cascade, thermal effects will be important at either small or large scales. This is the definition of the cascade dynamics.

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