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Similarly to thermodynamic pressure, when we have:

$$P_{therm} = (\frac{\partial U}{\partial V})_{S}$$

Can we define the mechanical pressure for a fluid as:

$$P_{mech} = \frac{\partial \psi}{\partial(-\nabla \cdot \mathbf{u})}$$

Where $\psi$ is the strain energy density functional, which is defined as:

$$\psi = \mathbf{P} \cdot \mathbf{S}$$

Where $\mathbf{P}$ is the Cauchy stress tensor defined as for an incompressible Newtonian fluid: $$\mathbf{P} = -P_{mech}\mathbf{I}+2\mu\mathbf{S}$$

and $\mathbf{S}$ is the strain rate tensor defined as: $$\mathbf{S} = \frac{1}{2}(\nabla \mathbf{u}+(\nabla \mathbf{u})^{T})$$

I know it works fine for incompressible flow. But interestingly I didn't find this derivation in analogy to thermodynamic in any text book. Any idea or suggeston is appreciated.

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  • $\begingroup$ Incompressible homogeneous fluid has $\nabla\cdot\mathbf{u}=0$. How can you take a derivative w.r.t. that quantity? $\endgroup$ – Deep Sep 18 '18 at 5:44
  • $\begingroup$ @Deep You are right, I could think of it as this limit: $\lim_{\nabla \cdot \mathbf{u} \rightarrow 0} \frac{\partial \psi}{\partial (\nabla \cdot \mathbf{u})} = P_{mech}$, what's your opinion? $\endgroup$ – Mehrdad Yousefi Sep 18 '18 at 13:54
  • $\begingroup$ This way of defining pressure is new to me. May be you are on to something. $\endgroup$ – Deep Sep 19 '18 at 5:02
  • $\begingroup$ @Deep My intention is to understand what mechanical pressure really means. In fact, I know from solid mechanics that we could even derive the Cauchy stress tensor ($\mathbf{P}$) from strain energy density as: $\mathbf{P} = \frac{\partial \psi}{\partial \mathbf{S}}$ where $\mathbf{S}$ is the strain tensor. $\endgroup$ – Mehrdad Yousefi Sep 19 '18 at 13:09
  • $\begingroup$ I wanted to know two things here. First: Is it possible to use same concept for fluids instead of solids and replace $\psi$ with strain rate energy density and $\mathbf{S}$ with strain rate tensor to get Cauchy stress tensor. Second, what if I take derivative with respect to trace of strain rate tensor ($\nabla \cdot \mathbf{u}$) which implies its isotropic interaction to get isotropic mechanical pressure maybe?! Please let me know what you think. $\endgroup$ – Mehrdad Yousefi Sep 19 '18 at 13:14

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