You have the right starting point with energy basically, but I'm finding your homework hint more useful than where you go from energy of a differential unit. It says "The energy density should be easy to identify." The energy density is:
$$ \frac{\text{energy}}{\text{volume}} = \frac{\text{mass}}{\text{volume}} \frac{\text{energy}}{\text{mass}} = \rho \frac{v^2}{2} + \rho e $$
For your problem, this is practical because we're already seeing this form in the answer.
Let's look at the components mentioned in the hint and see if those help us.
- internal energy
- kinetic energy
- energy lost to heat
- energy lost to work against pressure
- energy lost to work against viscous stress
Looking at these, and looking at the equation, we have a vivid picture painted of what all the terms mean. A fast deconstruction makes that look like the following if I denote the fluid energy (internal plus kinetic) as $w$ (I would rather call it $e$, but they took that).
$$ \frac{d}{dt} \rho w = \frac{d}{dx} \left( -v \rho w + \text{conduction - expansion + kinetic} \right) $$
These are all in the exact same order as what was given in the hint. It sounds like you're job is to walk backward from this view to some fundamental principle, specifically, conservation of energy. So start with a completely theoretical statement of that (say, equation 26) and start matching things up. Starting from the equation in that linked pdf, the first term already matches. That's easy, because you know what direction the fluid is flowing in. Next term, you have a Del dot-product with the same term. You should be able to work that out.
Moving on, Del dot q is also exactly what you're looking for (to get the conduction term). You'll need some finesse arguing why the signs should be the way they are. Then, volumetric heat contribution can be equated to frictional heating.
The equation,
$$
\nabla\cdot (\rho \textbf v \otimes \textbf v),
$$
can be written in index notation as,
$$
\partial_i (\rho v_i v_j),
$$
where the dot product becomes an inner product, summing over two indices,
$$
\textbf a \cdot \textbf b = a_i b_i,
$$
and the tensor product yields an object with two indices, making it a matrix,
$$
\textbf c \otimes \textbf d = c_i d_j =: M_{ij}.
$$
Now we differentiate using the product rule,
$$
\partial_i (\rho v_i v_j)=(\partial_i \rho) v_i v_j + \rho (\partial_i v_i) v_j + \rho v_i (\partial_i v_j).
$$
Let’s look at the terms separately:
$\bullet (\partial_i \rho) v_i v_j $: assuming $\rho=\rho(\textbf x)$, the expression within the brackets is the vector $(\partial_x\rho, \partial_y\rho, \partial_z\rho)$, which then gets dot multiplied with the vector $\textbf v$. This yields a number, say $c_1$, which gets multiplied to every component of the vector $v_j$. So the result here is a vector. If $\rho$ is constant, this term vanishes.
$\bullet\rho (\partial_i v_i) v_j$: Here we calculate the divergence of $\textbf v$,
$$
\partial_i a_i = \nabla \cdot \textbf a = \text{div }\textbf a,
$$
and multiply this number with $\rho$, yielding another number, say $c_2$. This gets multiplied onto every component of $v_j$. The resulting thing here is again a vector.
$\bullet\rho v_i (\partial_i v_j)$: Here we construct a matrix with the composition rule,
$$ M_{ij} := \partial_i v_j,
$$
that is for example $M_{13}=\partial_x v_z$. We then multiply a (row)vector $v_i$ to this matrix, yielding a different vector. Finally, every component of this new vector gets multiplied by $\rho$, so we have a vector again.
Best Answer
The Navier Stokes equations are non-linear in velocities, and thus have multiple solutions, some of which are time-dependent turbulent solutions. In practice, the most stable solution prevails. In turbulent flow, the fluid velocity and pressure vary rapidly with time and spatial position, even for so-called "steady flows." In "steady" turbulent flows the time average velocities and pressure are constant, however. The time averages of velocity component products give rise to the so-called turbulent stresses, which are typically much larger than the viscous stresses based on the average velocities. These play a major role in determining the time-average flow and pressure.