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Continuum expression of first law of thermodynamics:

$$\frac{D E_t}{D t}=\nabla\cdot({\bf \sigma\cdot v}) - \nabla\cdot{\bf q}$$

(I've seen it in my physics book)

How this equation is derived?

Where the $\bf q$ is Heat flux vector, $\sigma$ is stress tensor and $\bf v$ is velocity vector field.

(First law of thermodynamics: $$\frac{DE_t}{D t}=\frac{D W}{D t}+\frac{D Q}{D t}$$)

Achmed
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    Can you please give the definition of your variables? – Sanya Jun 27 '16 at 19:33
  • Please note that to give us any chance of helping you, you need to give us more information. Context of the equation, where did you get it from, etc. – Blazej Jun 27 '16 at 21:24
  • Closely related: https://physics.stackexchange.com/q/67163/226902 – Quillo Jun 25 '23 at 12:52
  • For a continuum version of the SECOND law, see Clausius–Duhem inequality, a way of expressing the second law in continuum mechanics (also useful in determining whether the constitutive relation of a material is thermodynamically allowable). – Quillo Jul 26 '23 at 15:19

2 Answers2

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The starting point for the development of this equation is an energy balance on a fixed control volume of fluid: $$\int{\frac{\partial E}{\partial t}dV}=-\int{E(\vec{v}\cdot \vec{n})dA}+\int{(\vec{v}\cdot \vec{\sigma} \cdot \vec{n})dA}-\int{\vec{q}\cdot\vec{n}}dA$$ where dV is differential volume, dA is differential surface of the control volume, and $\vec{n}$ is a unit outwardly directed normal from the control volume at its surface. The term on the left hand side represents the rate of change of energy within the control volume. The first integral on the right hand side represents the net rate at which energy is leaving the control volume by by flow into and out of the control volume. The second integral on the right hand side represents the rate at which the surroundings are doing work on the material on the control volume. The third integral on the right hand side represents the rate at which heat is leaving the control volume. If we apply the divergence theorem to the area integrals on the right hand side, we obtain: $$\int{\left(\frac{\partial E}{\partial t}+\nabla \cdot(E\vec{v})\right)dV}=\int{(\nabla \cdot{(\vec{\sigma}\cdot \vec{v})}-\nabla \cdot \vec{q})}dV$$ This then yields:$$\frac{\partial E}{\partial t}+\nabla \cdot(E\vec{v})=\nabla \cdot{(\vec{\sigma}\cdot \vec{v})}-\nabla \cdot \vec{q}\tag{1}$$ Eqn. 1 is identical to the relationship given in Transport Phenomena by Bird, Stewart, and Lightfoot, with$$E=\rho (u+\frac{v^2}{2}+\phi)$$where $\rho$ is the fluid density, u is the internal energy per unit mass, v is the magnitude of the velocity, and $\phi$ is the gravitational potential.

For an incompressible fluid, the left hand side reduces to:$$\frac{\partial E}{\partial t}+\nabla \cdot(E\vec{v})=\frac{\partial E}{\partial t}+\vec{v}\cdot \nabla E=\frac{DE}{Dt}$$

ALTERNATE METHOD: An alternate method of arriving at these same results is to use a moving control volume (rather than a fixed control volum employed in the development above), which moves and deforms with the fluid, and which contains a fixed mass of fluid. So this system is essentially the closed system encountered is standard thermodynamics. The starting equation for this development is: $$\frac{d}{dt}\left[\int{EdV}\right]=\dot{Q}-\dot{W}\tag{2}$$ where the left hand side is the rate of change of energy within the moving volume of fluid (system), $\dot{Q}$ is the rate of addition of heat to the system and $\dot{W}$ is the rate at which the system is doing work on the surroundings: $$\dot{Q}=-\int{\vec{q}\cdot\vec{n}}dA$$ $$\dot{W}=-\int{(\vec{v}\cdot \vec{\sigma} \cdot \vec{n})dA}$$ If we apply the Reynolds Transport Theorem (i.e., the 3D generalization of the Leibnitz rule for differentiation under the integral sign) to the left hand side of Eqn. 2, we obtain: $$\frac{d}{dt}\left[\int{EdV}\right]=\int{\frac{\partial E}{\partial t}dV}+\int{E(\vec{v}\cdot \vec{n})dA}$$ The remainder of the development is the same as above.

Chet Miller
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  • @Achmed What is it about my answer that is causing you difficulty? – Chet Miller Jun 28 '16 at 12:11
  • your Answer is ok but... I still can't understand (for example) how exactly $\frac{D Q}{D t}$ becomes $\nabla\cdot\bf q$ – Achmed Jun 28 '16 at 14:00
  • or how exactly $\frac{D W}{D t}$ becomes $\nabla\cdot({\bf \sigma\cdot v})$ – Achmed Jun 28 '16 at 14:02
  • D/Dt represents the material time derivative of a quantity (for an observer traveling with a material point in the fluid). DQ/Dt is not included in my derivation, and has no meaning. As I said, the integral term involving q in the control volume heat balance represents the net rate of heat leaving the control volume. Application of the divergence theorem produces the $\nabla \cdot \vec{q}$. Similarly, there is no such thing as DW/Dt. D/Dt is defined as $\frac{\partial}{\partial t}+\vec{v} \cdot \nabla$. – Chet Miller Jun 28 '16 at 14:06
  • Note that you do not need incompressibility for the material time derivative to appear if you look at $\frac{D}{Dt}E_s$ with $E_s = E/ \rho$ (see Bird, Armstrong, Hassager: Dynamics of Polymeric Liquids, Chapter 1) – Sanya Jun 28 '16 at 14:20
  • Actually, you do. Taking into account the continuity equation, the term in question becomes $\rho \frac{DE_s}{Dt}$, with the $\rho$ out front. Incidentally, did you notice the reference to my paper in Bird et al from 1967 with Joe Goddard regarding the Goddard-Miller model? – Chet Miller Jun 28 '16 at 14:40
  • @Achmed I have appended my Answer with an alternative development that you may find more intuitively satisfying. Hope this works better for you. – Chet Miller Jun 29 '16 at 14:18
  • @ChesterMiller that was my point - the expression $\rho \frac{D}{Dt}E_s$ has the material time derivative and does not need any incompressibility :D for the material time derivative of $E$ it is of course needed, no arguing there. And yes, the Goddard-Miller modell was one of the first things I read in Bird et al because I was interested in viscoelasticity :) – Sanya Jun 29 '16 at 14:29
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Ok, let's take the 1st law (in any reference frame) as $$\frac{D}{Dt}E_t(V) = P + Q $$ where $V$ is a part of the body, $P$ is the work rate by mechanical processes and $Q$ is the change of heat content. They will hopefully become more clear when I say how they can be calculated: $$P= \int_V dV \rho \vec{v} \vec{b} + \int_{\partial V} da \vec{v} \mathbf{T}\vec{n}$$ Here $\vec{v}$ is the velocity field, $\vec{b}$ is any force having an effect on the bulk of the body, e.g. gravitation, $\rho$ is the density, \vec{n} is the surface normal and $\mathbf{T}$ is the total stress tensor (including pressure!). So there are two kind of force types working on the continuum assumed: forces that work on the surface of the body element in question and forces that work everywhere. $$Q= \int_V dV \rho r - \int_{\partial V} da \vec{q} \vec{n}$$ Here $r$ is the body bulk heat production and $\vec{q}$ is the heat flux, the same principle as above applies.
So the assumption in continuum mechanics is that there is no work rate source apart from $r$, $\vec{q}$, $\vec{b}$ and $\mathbf{T}$ - or, that any work rate source can be expressed as one of them.
If we now apply Gauss' theorem to have only volume integrals and neglect $r$ and $\vec{b}$ and set $\mathbf{T} \approx \sigma$, we arrive at a local formula - exactly your formula above.
Did your physics book say anything about assumptions or approximations?

Sanya
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  • I guess No, It doesn't say anything about that. – Achmed Jun 28 '16 at 10:17
  • ok :| I certainly miss those two mechanisms of changing total energy in the formula above ... My derivation is taken from Liu, Continuum Mechanics, Berlin, Heidelberg: Springer, 2002 if you want to read more on the topic ... – Sanya Jun 28 '16 at 10:20