# conservation of energy for a viscous fluid

I have a pair of questions about conservation of energy and momentum in a viscous fluid. First, energy.

Symon's Mechanics states an equation for conservation of energy in a droplet of a non-viscous fluid as ${d \over dt}({1 \over 2} \rho v^2 \delta V)=\vec{v} \cdot (\vec{f} - \nabla p) \delta V$ (2nd edition, page 327, eq. 8-149). This is perfectly sensible; it says that d/dt (kinetic energy)=force*velocity, where $\rho$ is the fluid density in $kg \over {m^3}$, $\vec v$ is the fluid's velocity, $\vec f$ is the body force in $nt \over {m^3}$, $p$ is the fluid pressure and $\delta V$ is the volume of the droplet.

But then he updates it to a viscous fluid merely by replacing $\nabla p$ with $\nabla \cdot P$; i.e., replacing simple hydrostatic pressure with the stress tensor (page 441, eq. 10-174). I don’t understand this – it seems to neglect the energy that must be continuously converted to heat by viscous friction inside the droplet. Am I missing something? (Note that I’m using the 2nd edition; perhaps the 3rd edition has changed this).

Similar question: Symon gives the conservation-of-momentum equation for a droplet of non-viscous fluid as ${d \over dt} (\rho \vec v \delta V)=(\vec f - \nabla p) \delta V$. My guess is that for this case, we can update this to a viscous fluid simply by replacing $\nabla p$ with $\nabla \cdot P$. The stress tensor will correctly model all external viscous forces on the droplet. It will not model internal forces, but internal forces do not affect momentum anyway. Correct?

First of all: This is an old, and by now moderately obscure text book. If you would like to learn about fluid mechanics at the undergraduate level there are better texts to study.

Second: Your first equation is not the equation of energy conservation in the fluid. It is not written in conservative form $dw/dt=-\vec{\nabla}\vec{\jmath}$, where $w$ is a suitable density, and $\jmath$ is a current. It also does not include the internal energy density, which explains your question about viscous heating. Yes, there is viscous heating, but your equation does not include it, because it only includes the kinetic energy density. This equation is simply an equation for the kinetic energy density, so in the presence of dissipation all you have to do is replace ideal forces $\nabla_i p$ by the full stress tensor, $\nabla_j P_{ij}$.

Regarding your second question: The dissipative stress tensor $$P=p\delta_{ij}-\eta(\nabla_i v_j +\nabla_j v_i +\delta_{ij}(\zeta-2\eta/3)(\nabla\cdot v)).$$ describes the internal forces in the fluid, in particular viscous friction between fluid layers moving at different velocity.

The first equation you posted is incorrect in general, but for reasons other than those given above. Let me show you why this is so.

To simplify things, suppose the fluid is viscous and subject to a conservative force field. Let's suppose also that no heat is added to it, and that the fluid is in a vessel of rigid walls such that its velocity field there is zero, i. e. $u_i|_{S}=0$, where $S$ is the surface of the vessel. The vessel's volume is $V$. Integrating your first equation over $V$, and using the Reynolds transport theorem, it follows that:

$\iiint_{V}\rho \dfrac{D}{Dt}(\dfrac{1}{2} u_i u_i )d^3x = \dfrac{\partial}{\partial t} \iiint_{V} \dfrac{1}{2} \rho u_i u_i d^3x \ +\ \iiint_{V}\dfrac{\partial}{\partial x_i}(\dfrac{1}{2} \rho u_i u_i )d^3x= \iiint_{V} u_i f_i d^3x \ -\ \iiint_{V}u_i \dfrac{\partial p}{\partial x_i}d^3x$

Where the material derivative operator $\dfrac{D}{Dt}$, is defined as: $\dfrac{D}{Dt}:=\dfrac{\partial}{\partial t}+u_i \nabla_i$

From this equation, integrating by parts the second term on the middle, and using the boundary condition $u_i|_{S}=0$, it can be shown that this term vanishes and that the equation is equivalent to:

$\dfrac{\partial}{\partial t} \iiint_{V} \dfrac{1}{2} (\rho u_i u_i +\phi)d^3x= -\ \iiint_{V}u_i \dfrac{\partial p}{\partial x_i}d^3x$

Where $\phi$ is the potential energy density of the mass forces $f_i$, given by $u_i f_i=-\dfrac{\partial \phi}{\partial t}$

This result clearly violates energy conservation, for it says that the total mechanical energy of a non viscous fluid, given by $\iiint_{V} \dfrac{1}{2} (\rho u_i u_i +\phi)d^3x$, diminishes at a rate proportional to the quantity $\iiint_{V}u_i \dfrac{\partial p}{\partial x_i}d^3x$ which only vanishes if the fluid is at rest or if it is incompressible.

Thus, the correct expression of the second term on the right hand side of your first equation must be $\dfrac{\partial (u_i p)}{\partial x_i}$ and not $u_i \dfrac{\partial p}{\partial x_i}$. In general, it must be $\dfrac{\partial (u_i\tau_{ij})}{\partial x_j}$, where $\tau_{ij}$ is the stress tensor, to preserve energy conservation.

P.D. Sorry for my bad english grammar