# Continuum mechanics, physical interpretation of terms in balance eq.

1. Aug 24, 2014

### Telemachus

Hi there. I'm reading Gurtin's 'the mechanics and thermodynamics of continua', and working some exercises of his book. In the section 21: 'The first law: balance of energy', after the derivation of the balance equation, he uses an identity to rewrite the balance of energy.

The balance of energy can be written:

$\rho \dot \varepsilon=T:D -div(\vec q)+q$

The dot over the epsilon represents the material time derivative, epsilon is the internal energy, rho the mass density, T is the Cauchy stress tensor, D the stretching tensor, and $\vec q$ is the heat flux vector, while the scalar $q$ represents the heat transferred to the region by agencies external to the deforming body (e.g. radiation).

Then, due to a relation previously derived in the book, the given material time derivative is related to the spatial time derivative through:

$\rho \dot \varepsilon=(\rho \varepsilon)'+div (\rho \varepsilon \vec v)$

Then, the balance of energy in the local form can be written:

$(\rho \varepsilon)'=T:D-div(\vec q+\rho \varepsilon \vec v)+q$

The book then asks to provide a physical interpretation for the quantity: $\vec q+\rho \varepsilon \vec v$.

What I did was to rewrite the localized form in the global form for the divergence:

$\displaystyle -\int_{\cal P(t)} div(\vec q+\rho \varepsilon \vec v) dv=-\int_{\partial \cal P(t)}\vec q \cdot \hat n da-\int_{\partial \cal P(t)} \rho \varepsilon \vec v \cdot \hat n da$

(The differential is over volume, but the field $\vec v$ represents the velocity field)

So, the q term clearly represents the heat influx through the boundary $\partial \cal P(t)$, and the term containing $\rho \varepsilon \vec v$ I think that represents the inflow of internal energy in $\cal P(t)$ due to convection. Is this right? I'm in doubt because it is not a control volume which we are working with, but the convecting region. So it is not so clear to me that this flux of internal energy due to convection is actually possible, and I was in hope to clarify this through this discussion.

Bye there and thanks in advance.

Last edited: Aug 24, 2014
2. Aug 24, 2014

### Staff: Mentor

Yes. Your interpretation is correct, and you did it correctly (assuming you have a fixed macroscopic control volume and v represents to flow velocity relative to the frame of reference of that fixed macroscopic control volume). On the other hand, if the system boundary is moving, then the flow velocity relative to the boundary will not be v, and you must use the relative velocity at the boundary to get the net convection of internal energy across the boundary.

Chet

3. Aug 24, 2014

### Telemachus

Hi Chet. Thanks for your answer. I worked it as if the boundary is moving ($\cal P(t)$ represents the deformed region convecting with the body, I should've been more clear about it). The identities I've used are for deformed bodies, I don't know if it hold for a control volume, I should check it. My doubt is if even for the deformed region convecting with the body there is an inflow of internal energy due to convection itself, because in the deformed body the boundary is always the boundary, so it wasn't clear to me, but the relation seems to suggest than that's the way it is (so maybe I could find a better interpretation than what I'm giving to the physical situation).

I have that, for the deformation $\chi$:

$x=\chi(\rm X,t)$, where x is the spatial point, and X the material point.

Then, the deformed region at time t is defined as: $\cal{P_t}=\chi_t( \rm P )$, where P is the material region (the subscript indicates a fixed time t). Same notation is used for the boundary $\partial \cal P_t$.

This is the book I'm using (page 185): http://books.google.com.ar/books?id... of continua&pg=PA185#v=onepage&q=185&f=false

Great book btw.

Last edited: Aug 24, 2014
4. Aug 24, 2014

### Staff: Mentor

No. If the boundary is a material surface (moving with the fluid velocity), then there is no convection of internal energy across the material surface into the volume. To show this, you need to integrate the entire equation over the volume and properly apply the 3D version of the Leibnitz rule, which is called Reynolds Theorem. The convection will cancel out from the final result (the term involving the partial derivative of internal energy with respect to time is involved).

However, if you do your integration over a small control volume fixed in space, then the term you were referring to will be equal to the rate of convection of internal energy across its surface. In short, the control volume must be fixed in space.

Chet

5. Aug 24, 2014

### Telemachus

What I've got from the Reynolds transport theorem, and using the relation for the conservation of mass is:

$\displaystyle \frac{D}{Dt} \int_{\cal P_t} \rho \varepsilon dv= \int_{\cal P_t} \rho \dot \varepsilon dv$

$\frac{D}{Dt}$ denotes the material time derivative.

If then I use $\rho \dot \varepsilon=(\rho \varepsilon)'+div(\rho \varepsilon \vec v)$

I have:

$\displaystyle \int_{\cal P_t} \rho \dot \varepsilon dv= \int_{\cal P_t} T:Ddv- \int_{\cal P_t} div \vec q dv+ \int_{\cal P_t} qdv=\int_{\cal P_t} [(\rho \varepsilon)'+div(\rho \varepsilon \vec v)]dv$

Last edited: Aug 24, 2014
6. Aug 24, 2014

### Staff: Mentor

This was not a correct application of Reynolds Transport theorem. See the link: http://en.wikipedia.org/wiki/Reynolds_transport_theorem. When it is done correctly, the last term will be cancelled by another identical term of opposite sign (as your intuition tells you it must).

Chet

7. Aug 24, 2014

### Telemachus

Thank you very much Chet. I'll let you see the steps I did in more detail. I think you are right, because intuitively I know that what you said must hold, but its late, and perhaps I'm making some silly mistake.

First, in my notation, Reynolds Transport Theorem states:

$\displaystyle \frac{D}{Dt} \int_{\cal P_t} \phi dv= \int_{\cal P_t} (\dot \phi + \phi div \vec v) dv$

From the conservation of mass $\dot \rho \varepsilon=-\rho \varepsilon div \vec v$

Perhaps I'm not starting with the right foot. Then what I did was:

$\displaystyle \frac{D}{Dt} \int_{\cal P_t} \rho \varepsilon dv= \int_{\cal P_t} (\dot {(\rho \varepsilon)}+\rho\varepsilon div \vec v) dv=\int_{\cal P_t} \rho \dot \varepsilon dv$

And then I've used that $\rho \dot \varepsilon=(\rho \varepsilon)'+div(\rho \varepsilon \vec v)$ and equated to the other part of the energy balance:

$\displaystyle \int_{\cal P_t} [(\rho \varepsilon)'+div(\rho \varepsilon \vec v)]dv=\int_{\cal P_t} T:Ddv- \int_{\cal P_t} div \vec q dv+ \int_{\cal P_t} qdv$

Perhaps I'm missing something, but I think the procedure is quite right. I know the term you say has to appear, but I'm not seeing from where I should get it.

Now I was thinking, perhaps I can get a term like that by developing the spatial time derivative $(\rho \varepsilon)'$, I'll try that later.

8. Aug 24, 2014

### Telemachus

Yes, I think that was it :D thank you very much Chet.

I have this identity: $(\rho \phi)'=\rho\phi'+\rho\vec v \cdot grad \phi-div(\rho \phi \vec v)$

So, in my case I would have: $(\rho \varepsilon)'=\rho\varepsilon'+\rho\vec v \cdot grad \varepsilon-div (\rho \varepsilon \vec v)$

Now, do you have any physical interpretation for the term $\rho \vec v \cdot grad \varepsilon$?

9. Aug 24, 2014

### Staff: Mentor

This equation:

$\displaystyle \int_{\cal P_t} [(\rho \varepsilon)'+div(\rho \varepsilon \vec v)]dv=\int_{\cal P_t} T:Ddv- \int_{\cal P_t} div \vec q dv+ \int_{\cal P_t} qdv$

when combined with this equation:

$\displaystyle \frac{D}{Dt} \int_{\cal P_t} ρε dv= \int_{\cal P_t} ((ρε)'+ div(ρε\vec v)) dv$

just gives:

$\displaystyle \frac{D}{Dt} \int_{\cal P_t} ρε dv=\int_{\cal P_t} T:Ddv- \int_{\cal P_t} div \vec q dv+ \int_{\cal P_t} qdv$

This is exactly what you expect, and doesn't involve any convective internal energy flux across the material boundary surface.

Chet

10. Aug 24, 2014

### Telemachus

Yes, you are right. Thank you.