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Moving fluids are generally in a state of non-equilibrium. However, in fluid dynamics, people generally assume a state of local thermodynamic equilibrium and argue that in such a condition, equilibrium thermodynamic concepts such as pressure, temperature, entropy, internal energy etc. can be defined for a local fluid element. As such, the moving fluid can be considered as a continuum of local thermodynamic states and relations from equilibrium thermodynamics can be applied locally.

I am having trouble understanding this concept. For example, for a system in global equilibrium, pressure is defined as normal stress acting at a point, and it is equal in all directions.

Now consider a moving fluid. If we assume local equilibrium for this fluid, we can define a local property called pressure in the same way it was defined for a system in global equilibrium. This means that in a moving fluid, pressure at a point should be the normal stress at that point, and it should be equal in all directions (which is clearly not the case).

Perhaps my idea of local equilibrium or my conception of pressure is flawed. Can someone clarify my doubts? Thanks!

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Your question seems to be: normal stress inside a fluid in motion is not an isotropic quantity, but pressure as it is used in thermodynamic equations is an isotropic quantity, so how do we define pressure at a point inside a fluid in motion? In a flow, pressure at a point is defined as the average of the normal stresses (with an additional minus sign). In fact we just take the isotropic part of the stress tensor and define it to be pressure (see Fluid Mechanics by Kundu & Cohen). We use this pressure so defined in the thermodynamic equations. The justification for this definition, as far as I know, comes from its success in practical applications, but I am not sure if there is theoretical justification for it.

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  • $\begingroup$ I am not convinced that this is completely accurate: what if there is bulk viscosity? $\endgroup$
    – Quillo
    Commented Mar 9, 2022 at 8:44
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The local stress tensor for a Newtonian fluid is usually represented as the isotropic "thermodynamic pressure" tensor, plus the viscous stress tensor. The viscous stress tensor is not necessarily isotropic, and its components are proportional to the Newtonian viscosity, with an isotropic bulk viscosity stress tensor contribution. The local "thermodynamic pressure" is determined by the equation of state, applied locally in terms of the local specific volume and local temperature. For an incompressible fluid, the local "pressure" p at all locations is determined up to an arbitrary constant value that is obtained by applying the boundary conditions on the flow. The local thermodynamic equilibrium assumption is an approximation, but one that has been found to be exceedingly accurate, based on centuries of experimental and practical evidence.

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  • $\begingroup$ I think I may be struggling with the definition of "thermodynamic pressure". As far as I know, it is the normal stress acting at a point in a fluid that is in a state of global equilibrium. When we assume local equilibrium in a fluid and assign a local "thermodynamic pressure" at each point, shouldn't it have the same definition as for the case of global equilibrium? $\endgroup$
    – Shivam Sinha
    Commented Apr 30, 2017 at 16:15
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    $\begingroup$ No. For a deforming fluid, the stress tensor includes viscous stresses (which depend on the rate of strain, not just the strain). For the local thermodynamic pressure in a deforming fluid, the thermodynamic pressure is the value determined from the equation of state at the local temperature and specific volume. For an ideal viscous gas, this would be $RT_{loc}/V_{loc}$, where Vloc is the local specific volume. $\endgroup$
    – Chet Miller
    Commented Apr 30, 2017 at 17:42
  • $\begingroup$ Thank you, can you tell me about the conditions under which local equilibrium assumption can be said to be a good approximation? Is local equilibrium a good assumption when the bulk viscosity is large and there is a big difference between the mechanical and thermodynamic pressures? $\endgroup$
    – Shivam Sinha
    Commented Apr 30, 2017 at 20:14
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    $\begingroup$ I don't quite know how to answer this, so let me give you some data points: In freshman physics, when we studied heat conduction, they never once mentioned that any local equilibrium approximation was involved. In the engineering curriculum, when we studied transport processes, we were not told that a local thermodynamic equilibrium approximation was involved. In thermo, when we studied the open system (control volume) version of the first law, we were not informed of this. In my long engineering career, I never heard of anyone getting the wrong answer using this approximation. $\endgroup$
    – Chet Miller
    Commented Apr 30, 2017 at 22:23
  • $\begingroup$ @ChetMiller because that assumption is almost what defines hydrodynamics in its simplest form (navier stokes + heat). If it s not valid you use other theories beyond hydro. Good answer +1. $\endgroup$
    – Quillo
    Commented Mar 9, 2022 at 8:52

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