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The canonical process of determining the pressure, velocity, and density of a fluid under the influence (or not) of external forces is through simultaneously solving conservation of mass, conservation of momentum, and an equation of state for the pressure (or conservation of energy). For an incompressible Navier-Stokes fluid with constant density $\rho_0$, this would look like:

$$p = p(\rho, \vec{v}) = p(\rho_0,\vec{v})$$ $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = \nabla\cdot \vec{v} =0$$ $$\rho \left(\frac{\partial \vec{v}}{\partial t} + \vec{v}\cdot\nabla \vec{v}\right) = -\nabla p + \mu \nabla^2\vec{v} + \vec{f}$$

Given some external force, all three of the variables can be solved for with this system of equations.

However, when discussing flows in introductory fluid mechanics like pipe flow, we ignore the equation of state; we arbitrarily prescribe a pressure gradient and then use conservation of mass/momentum to obtain the resulting velocity for a fluid with a given density.

How do we not run the risk of prescribing a pressure gradient that violates the equation of state/conservation of energy? Surely the answer lies in the difference between thermodynamic and mechanical pressure, but is the mechanical pressure really just treated as a totally unconstrained variable?

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I think the role of pressure is to adjust itself immediately according to changes in the velocity field so that velocity is divergence-free at all times. In that sense, it does not have any thermodynamic meaning.

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    $\begingroup$ I think this is true, in the sense that this is how we use it, but we also make the implicit assumption that whatever pressure we get is "thermodynamically viable". That being said, one could get pressures in a solved mathematica system that make no thermodynamic sense, which has to be tracked by looking at the equation of state post-facto. $\endgroup$
    – aghostinthefigures
    Commented Nov 2, 2019 at 18:36
  • $\begingroup$ Good answer. For incompressible flows, an equation of state can not provide an additional equation. The pressure field is not the thermodynamic one (because of the compressibility assumption: the speed of sound is infinite). Pressure is "formal" and must follow from the velocity field (and vice versa), at every instant. Solving this issue is the main problem behind numerical incompressible flow solvers. See this answer for the meaning of "mechanical pressure" physics.stackexchange.com/a/427067/226902 $\endgroup$
    – Quillo
    Commented Mar 6, 2023 at 16:29
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You are correct in concluding that the pressure $p$ in the Navier-Stokes equations is not related to the equation of state. It is a purely mechanical quantity, defined as $p = -\sigma_{ii}/3$, where $\sigma_{ij}$ is the ($ij$-th component of) the stress tensor. The pressure that appears in the equation of state is a thermodynamic quantity and it is applicable only to equilibrium states. Elements of fluid in relative motion are not in exact thermodynamic equilibrium and therefore the thermodynamic and mechanical pressures are not identical in moving fluids.

You may refer to G K Batchelor's 'An Introduction to Fluid Dynamics' (pages 141, 142) for more information.

In section 3.4, Batchelor derives the equation (3.4.8) $p - p_e = -\kappa (\nabla\cdot\vec{u})$, where $p_e$ is the equilibrium, thermodynamic pressure and $\kappa$ s the bulk viscosity. For incompressible fluids, the distinction between the two is erased.

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    $\begingroup$ I’m familiar with Batchelor’s derivation—in fact, it is precisely why this is so confusing for me! If Batchelor is to be believed, the mechanical pressure is equal to the equilibrium, thermodynamic pressure in incompressible fluids. This means that the pressure in such flows can be uniquely determined from a thermodynamic equation of state /and/ from solving Navier-Stokes, suggesting that it’s over-constrained! $\endgroup$
    – aghostinthefigures
    Commented Oct 11, 2019 at 3:59

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