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Reading J.N. Reddy - Introduction to Continuum Mechanics, the first law of thermodynamics is stated as

$$\frac{D}{Dt}(K+U)=W+H$$

Where $K$ denotes the kinetic energy, $U$ the internal energy, $W$ the power input and $H$ the heat input of the system.

Let's now focus on kinetic and internal energy terms

  • About $U$ is said that

    The kinetic energy associated with the (microscopic) motions of the molecules of the continuum is a part of the internal energy; the elastic strain energy and other forms of energy are also parts of internal energy, $U$.

    This suggests that we can write $U$ as

$$U=K_{micro}+U_ {other}$$

  • About $K$ is said that

    The kinetic energy of a system is the energy associated with the macroscopically observable velocity of the continuum

    Thus

$$K=K_{macro}$$

These reasonings lead to a reformulation of the first law, written above, as

$$\frac{D}{Dt}(K_{tot}+U_{other})=W+H$$

Having defined $$K_{tot} \equiv K_{macro}+K_{micro}$$

This latter formulation seems more natural to me since kinetic energy terms are just in one place, and, similarly to the work-energy theorem, $U$ now includes only potential energy terms.

So my question is:

  • Is there a formal way to write the general expression for the kinetic energy $K_{tot}$ in a continuum control volume and then splitting the two contributions into macro and micro kinetic energy, $K_{macro}$ and $K_{micro}$?

    My intuition would be that particles in the microscopic layer move in groups, and in the upper layer (i.e. the macroscopic layer) one of these groups is just considered as a point. Then, the mass of a group of particles in the micro-layer would be the infinitesimal mass associated with a point in the macro-layer, accordingly the velocity of the center of mass of a group in the micro-layer would be the velocity of a point in the macro-layer. But when i try to write this it into equations but i don't know where to start.

    Being, expression for $K$, making dependencies explicit, given by

$$K(t)=\frac{1}{2}\int_{\Omega(t)}\rho(\mathbf{x},t)\;\mathbf{v}(\mathbf{x},t)\cdot\mathbf{v}(\mathbf{x},t)\;d\Omega$$

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  • $\begingroup$ Any reason $D$ is being used in the derivative instead of '$\text d$'? $\endgroup$ – BioPhysicist Jan 23 '20 at 14:57
  • $\begingroup$ Just to highlight the fact that is the total or material derivative $\endgroup$ – Giorgio Pastasciutta Jan 23 '20 at 15:05
  • $\begingroup$ Here, first answer at researchgate.net/post/… a good explanation about microscopic and macroscopic kinetic energy $\endgroup$ – Giorgio Pastasciutta Jan 25 '20 at 14:59
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This is my solution attempt.

Main assumptions:

  • Continuum hypotesis is kept till molecular scale.
  • The following expression for total kinetic energy $K_{tot}$ will be used

$$K_{tot}=\int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}\cdot \mathbf{v}\,d\Omega$$

  • All quantities considered will be in Eulerian description (i.e. function of $(\mathbf{x},t)$).

Let's first consider a single particle, a region of material composed by millions of molecules. If we keep our continuum assumption till molecular scale, a particle could be simply considered as some region of material tracked by some control volume $\Omega_P(t)$.

We can next define particle mass $m_p$, particle center of mass displacement $\mathbf{r}_p$ and particle center of mass velocity $\mathbf{v}_p$ as

$$m_p\equiv\int_{\Omega_P(t)}\rho\,d\Omega$$

$$\mathbf{r}_p\equiv\frac{1}{m_p}\int_{\Omega_P(t)}\rho\,\mathbf{r}\,d\Omega$$

$$\mathbf{v}_p\equiv\frac{D\mathbf{r}_p}{Dt}=\frac{D}{Dt}\Big{(}\frac{1}{m_p}\int_{\Omega_P(t)}\rho\,\mathbf{r}\,d\Omega\Big{)}=\,...\,=\frac{1}{m_p}\int_{\Omega_P(t)}\rho\,\mathbf{v}\,d\Omega$$

Where last passage can be achieved through the use of Reynold's transport theorem and continuity equation, but for sake of brevity i omitted it.

Having defined the velocity of the center of mass of the particle, we can then refer any velocity inside the particle to it. The relative velocity for a molecule inside the particle will be

$$\mathbf{v}_{m}(\mathbf{x},t)\equiv\mathbf{v}(\mathbf{x},t)-\mathbf{v}_{p}(t)\;\;\;\;\;\;\;\;\forall \mathbf{x}\in\Omega_P(t)$$

Now comes the tricky part.

We want to apply this reasoning to all particles that compose the material. Imagining to fill all material volume $\Omega(t)$ with infinite $\Omega_P(t)$s, we will have a different $\mathbf{v}_{p}(t)$ for each $\Omega_P(t)$ considered. For this purpose, we can create $\mathbf{v}_{p}(\mathbf{x},t)$ as the function that, at time t, assigns to each $\mathbf{x}$ in $\Omega(t)$ a different $\mathbf{v}_{p}(t)$ depending on which specific region $\Omega_P(t)$ the coordinate $\mathbf{x}$ belongs to. Accordingly, the new definition for molecule relative velocity will be

$$\mathbf{v}_{m}(\mathbf{x},t)\equiv\mathbf{v}(\mathbf{x},t)-\mathbf{v}_{p}(\mathbf{x},t)\;\;\;\;\;\;\;\;\forall \mathbf{x}\in\Omega(t)$$

Having defined $\mathbf{v}_{m}$ and $\mathbf{v}_{p}$ for whole volume, we can then express the total kinetic energy $K_{tot}$ as

$$K_{tot}=\int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}\cdot \mathbf{v}\,d\Omega=\int_{\Omega(t)}\frac{1}{2}\rho\,(\mathbf{v}_m+\mathbf{v}_p)\cdot(\mathbf{v}_m+\mathbf{v}_p)\,d\Omega\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,$$

$$=\int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}_m\cdot\mathbf{v}_m\,d\Omega+\int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}_p\cdot\mathbf{v}_p\,d\Omega+\underbrace{\int_{\Omega(t)}\rho\,\mathbf{v}_m\cdot\mathbf{v}_p\,d\Omega}$$

Third integral can be split into the sum of the $\Omega_P(t)$ regions that compose it

$$\int_{\Omega(t)}\rho\,\mathbf{v}_m\cdot\mathbf{v}_p\,d\Omega=\sum_{P=1}^{+\infty}\int_{\Omega_P(t)}\rho\,\mathbf{v}_m\cdot\mathbf{v}_p\,d\Omega$$

Being $\mathbf{v}_p$ unvaried across the same region $\Omega_P(t)$, we can carry it outside of the integral, leading to

$$\sum_{P=1}^{+\infty}\mathbf{v}_p\cdot\int_{\Omega_P(t)}\rho\,\mathbf{v}_m d\Omega=\sum_{P=1}^{+\infty}\mathbf{v}_p\cdot\mathbf{0}=0$$

Since vectorial sum of molecules relative velocity $\mathbf{v}_m$ is $\mathbf{0}$ if carried across a whole volume $\Omega_P(t)$

Being third term null, we are left with the two terms that compose $K_{tot}$, which are

$$K_{macro} \equiv K_{p} = \int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}_p\cdot\mathbf{v}_p\,d\Omega$$

$$K_{micro} \equiv K_{m} = \int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}_m\cdot\mathbf{v}_m\,d\Omega$$

Where $\mathbf{v}_P$ is the particle velocity or the macroscopic velocity, which in texts will be simply referred to as $\mathbf{v}$, due to the fact that $K_{micro}$ is always included in internal energy $U$.

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  • $\begingroup$ Writing this to share my thoughts and also in the hope to receive some feedback about $\endgroup$ – Giorgio Pastasciutta Jan 26 '20 at 15:02
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Is there a formal way to write the general expression for the kinetic energy $𝐾_{π‘‘π‘œπ‘‘}$ in a continuum control volume and then splitting the two contributions into macro and micro kinetic energy, $𝐾_{π‘šπ‘Žπ‘π‘Ÿπ‘œ}$ and $𝐾_{π‘šπ‘–π‘π‘Ÿπ‘œ}$?

I'm not versant in continuum mechanics, so I really can't answer your question. I know what a control volume is in thermodynamics, as applicable to open systems, but I don't know what the difference (if any) is between the control volume of an open system, and a "continuum control volume".

But from what I know about thermodynamics $K_{macro}$ and $K_{micro}$ are the kinetic energies of the system with respect to two different reference frames. $K_{micro}$ is the system's kinetic energy with respect to the reference frame of the system, whereas $K_{macro}$ is the kinetic energy of the center of mass of the system in motion with respect to an external (to the system) frame of reference.

As for an open system control volume, $K_{micro}$ would be the kinetic energy of the working fluid entering and exiting the control volume and $K_{macro}$ would be the kinetic energy of the center of mass of the contents of the control as a whole with respect to some external frame of reference. I don't see how one could somehow mix them together into a general expression. It would seem any expression would necessarily be composed of two separate contributions. But then, as I said, I don't know the difference between a "continuum control volume" and the the control volume of an open system.

Though this doesn't answer your question, perhaps it will be of some help.

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  • $\begingroup$ Thanks for your answer, i just used the word "continuum control volume" to resume a bit informations indicating a control volume where continuum assumption holds (i.e. matter is distributed continuously without gaps or empty spaces - quoting J.N. Reddy again). The first problem regarding the writing of this general expression would be in fact the presence of a 'discrete' sub-layer formed by molecules. My idea is to not consider that as a discrete but as a continuum too, with some property that regions of it move together or 'in groups'. $\endgroup$ – Giorgio Pastasciutta Jan 23 '20 at 16:14
  • $\begingroup$ this would make now the layer above discrete, but we could again consider, from the upper layer poit of view, these 'groups' as infinitely small (and thus points) and filling all space again. In this way upper layer continuum hypotesis would be restored and lower layer continuum hypotesis kept. But also i do not know if there are matematical means to do what i tried to explain $\endgroup$ – Giorgio Pastasciutta Jan 23 '20 at 16:15
  • $\begingroup$ @GiorgioPastasciutta I am having trouble connecting "discrete sub layers" and the like with macroscopic and microscopic (internal) energy of a thermodynamic system. Like I said, I'm not conversant in continuum mechanics. Sorry could not be of more help. $\endgroup$ – Bob D Jan 24 '20 at 14:02

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