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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?

aghostinthefigures
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2 Answers2

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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.

Curiosity
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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.

Amey Joshi
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