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To start with I know thermodynamics deals with processes at equilibrium. Hence the thermodynamic pressure should most likely be the pressure of a fluid at equilibrium.

I'm not sure if a fluid flow (in general unsteady) is in thermodynamic equilibrium (say flow in a channel which has a pressure gradient) and so would the static pressure at a point in the channel be different from the thermodynamic pressure?

What does this entail about the ideal gas law $p = \rho RT$? can it be used for moving flow? What is the pressure in the equation referring to; mechanical or thermodynamic?

EDIT: To clear up any confusion- In a given flow we can measure the pressure at any point, say using a pitot tube to get the stagnation and static pressure. My question is then, is the static pressure we measure (which is by definition an "$F/A$" (force / area) quantity any different from the thermodynamic pressure? The pressure in $P = \rho RT$ must be referring to the thermodynamic pressure, since the equation is derived purely from the laws of thermodynamics. However, in all literature I have encountered, compressible flows use the ideal gas equation to as a link between the incompressible variables ($p, \mathbf{V}$) and the full set of compressible variables ($p, \mathbf{V}, \rho, T $). So it seems the two pressures are equivalent?

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  • $\begingroup$ I'm not sure if a fluid flow (in general unsteady) is in thermodynamic equilibrium (say flow in a channel which has a pressure gradient)... thermodynamic equilibrium requires mechanical equilibrium, which is most usually not present when fluids move; the motion of the fluid is accompanied by dissipation and the flow has to be either sustained by external supply of energy or decays. The equation $p=\rho RT$ is usable even for gas out of equilibrium, if it is not too far from it. See the comment in D.W.'s answer. $\endgroup$ – Ján Lalinský Apr 17 '14 at 19:09
  • $\begingroup$ To answer the part of your question pertaining to whether or not thermodynamic force is "the same" and mechanical force: they have the same units $\dfrac{J}{m^3}=\dfrac{kg m^2s^-2}{m^3}=\dfrac{kg}{m s^2}=\dfrac{F}{m^2}$, and that's sufficient for me to say they're the "same"... $\endgroup$ – D. W. Apr 18 '14 at 2:22
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I don't think I quite understand your question, but I'll do my best.

In Thermodynamics, pressure is defined in a bevy of ways. If we look at the Thermodynamic Identity: $$ dU = TdS - PdV + \mu dN$$ (where $U$ is the Energy, $T$ is the Temperature, $S$ is the Entropy, $P$ is the Pressure, $V$ is the Volume, $\mu$ is the Chemical Potential, and $N$ is the Number of Particles) we can see that pressure is: $$ P = -\left( \dfrac{dU}{dV} \right)_{\text{constant } S,N} = T \left( \dfrac{dS}{dV} \right)_{\text{constant } U,N} = \mu \left( \dfrac{dN}{dV} \right)_{\text{constant } S,U}.$$

However, there are even more identities for pressure (derived in the same way) if we use the Helmholtz Free Energy: $$ F = U - TS \to dF = -S dT - PdV + \mu dN. $$

Mechanical Pressure - at least in the way I think you're thinking of it - is pretty simple, at least relatedly. Pressure is just: $$ P = \dfrac{F}{A}, $$ Force per unit Area.

I'm not the person to ask about pressure in fluid flow. I don't know much about fluid dynamics.

I do know this though, the equation you mentioned: $$ P = \rho RT $$ is called the (monatomic) Ideal Gas Law, and is derived under the assumption that the gas is at equilibrium and is non-interacting, - along with a few other assumptions that I don't remember - so you generally can't apply it to dynamic fluids (though, as others have pointed out, there are various situations in which you can apply it). The pressure in the equation is the Pressure the gas exerts on it's surroundings (i.e. the pressure the gas inside a balloon exerts outwards on the balloon).

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    $\begingroup$ The equation $P=\rho RT$ is not restricted to monatomic gases, but to gases raried enough. For dry air in atmospheric conditions it applies quite well. It can be applied to moving air, which is a fluid, if the difference in velocities in the situation considered is much smaller than the speed of sound in the air. $\endgroup$ – Ján Lalinský Apr 17 '14 at 19:06
  • $\begingroup$ @JánLalinský Thanks for the comment. I interpret this then that the ideal gas equation only holds for incompressible flow? Is there a way of showing this? $\endgroup$ – Dipole Apr 17 '14 at 23:38
  • $\begingroup$ @D.W. Thanks for your answer. Please see the edited version of my question. $\endgroup$ – Dipole Apr 17 '14 at 23:41
  • $\begingroup$ @Jack, I would not say that. Compressible flow of air may be described by local ideal gas equation if the pressure and temperature is meaningful locally. The ideal gas equation fails if the quantities not even locally meaningful description of the medium, or if the air gets too dense/wet so that mutual forces between molecules become considerable. $\endgroup$ – Ján Lalinský Apr 18 '14 at 19:23
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The difference has to do with the fact that when you sum the normal stress on each face of a differential fluid element using the Newtonian constitutive law, you get something different from the thermodynamic pressure, which is what you normally think of as "pressure". There's a good explanation of it in Viscous Fluid Flow by Frank White.

So the constitutive law for a fluid (or any continuum) is what connects the stress to the strain. For a Newtonian fluid, the constitutive law is:

$$\tau_{ij} = -p\delta_{ij}+\mu(u_{i,j}+u_{j,i}) + \delta_{ij}\lambda u_{k,k}$$

Where $\mu$ is the dynamic viscosity and $\lambda$ is the bulk viscosity, both properties of the fluid. When you sum this over all the faces of the fluid element, you get:

$$\tau_{ii}=-3(p-u_{i,i}(\frac{2}{3}\mu+\lambda))$$

Divide by -1/3 to get:

$$p_{mech} = p_{therm}-u_{i,i}(\frac{2}{3}\mu+\lambda)$$

The original pressure term was the thermodynamic pressure and I added a subscript to make it a little clearer in the last equation. These two pressures are different by the product of the divergence of the velocity and a term related to the material properties. If you're talking about incompressible flow, then there's no difference at all because the divergence of the velocity is zero. If you're talking about compressible flow, then the difference is still small, but depends on how compressible the fluid is and how big this fluid property term is. Stokes basically assumed away the problem by saying that $$\frac{2}{3}\mu+\lambda=0$$ Which is "Stokes Hypothesis"

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Basically we know pressure as force over area: $$p = \dfrac {F}{A} = \dfrac {F x}{A x} = \dfrac {Work}{Volume} = \dfrac{Energy}{Volume}.$$

In continuum mechanics, pressure will be evaluated through the derivation of the strain energy functional with respect to Jacobian of the deformation gradient as: $$ p = \dfrac{\partial(\psi)}{ \partial J}, $$ where $\psi$ is the strain energy functional and $J=det(F)$. This pressure could be a function of $J$ and also temperature, depending on the energy functional of the employed constitutive model.

$\mathbf{Fluid~wise~ talking}$, I think the best way to distinguish different definitions for pressure is to look at Bernoulli equation regarding the fact this is valid only for incompressible fluids as reads:

$$\dfrac{p(static)}{\gamma} + \dfrac{1}{2} \rho v^2 (dynamic~pressure) + Z (related ~to ~hydrostatic ~pressure) = Cte,$$

where $\gamma = \rho g$.

  • Static pressure: Pressure at any given point of a fluid (either compressible or incompressible)

  • Hydrostatic pressure: The pressure at any given point of a non-moving (static) >incompressible< fluid. For instance, in a Barotropic fluid, the static pressure and hydrostatic pressure are the same.

  • Piezometric (head) or Hydraulic pressure: $$h= Z + \dfrac{p(static)}{\gamma}$$ for incompressible fluids.

  • Stagnation pressure: The pressure that fluid exerts when it is forced to stop moving:
    $$p_0 = p (static pressure) + \dfrac{1}{2} \rho v^2 (dynamic~ pressure)$$

  • Mechanical(total) pressure: $$p(mech) = p (static) + \dfrac{1}{2} \rho v^2$$

  • Thermodynamic pressure: The definition of this pressure depends on whether the flow is incompressible (divergence free) or compressible.

---> incompressible: $$p(mech) = p(thermo)$$

---> compressible: $$p(mech) = p(thermo) + \nabla \cdot v * A,$$

where $A$ is a term related to material properties of the flow like bulk and shear modulus.

Also the pressure can evaluated through equation of state (EOS) and in general, it is the rate of internal energy with respect to volume as is: $$p = \dfrac{\partial U}{\partial V}$$

It is clear from the above formulation that the equivalency of thermodynamic and mechanical pressure emerges when the flow is divergence free ($\nabla \cdot v=0$)

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  • $\begingroup$ You should use the Latex formatting, without that your posts don't seem very well. $\endgroup$ – peterh Dec 19 '16 at 12:50
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The mechanical and thermodynamic pressure have different definition and roots. In general they are not the same. P(mech)= P(static) + P(dynamic). In this equation P(static) can be assumed to be P(th). So for a fluid at rest P(mech)=P(static)=P(th) but for a fluid in motion P(mech)=P(th)+P(dynamic). you can see this fact in Bernolli equation: P/gamma + (u^2)/2g + Z = C. the sum of first two therms is mechanical or total pressure. the first is thermodynamic or static pressure and the second dynamic pressure.

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