thermodynamics - Explicit form of the entropy production in hydrodynamics

I'm trying to understand how hydrodynamics arise from a precise, mathematical formulation of thermodynamics, learning mostly from Landau's "Hydrodynamics".

So Landau starts from formulating the dynamical equations for the non viscous fluid, 5 equations, because there are 5 variables (pressure, density and a 3-component velocity field). The equations are: the mass continuity equation, the Euler equation (momentum continuity equation) and a statement of the fact, that there is no dissipation of energy, i.e. entropy is constant ($\frac{ds}{dt} = 0$).

Now, when describing a viscous flow we add dissipation through stress $-$ the Euler equation becomes the Navier-Stokes equation. My problem concerns the last equation, which should account for entropy production.

You see Landau doesn't say anything about entropy per se, but just finds an equation for energy transfer analogous to the other continuity equations, and finds a way of describing a lost in kinetic energy of a flow caused by viscosity.

My question is: how to state this equation as a direct generalization of the non viscous case, i.e. in terms of entropy? How to even approach this in terms of thermodynamics? What (thermodynamically speaking) is the process of dissipation? we need to know that, to calculate the $DQ = dE_{kin}$, don't we? Assuming we know the rate of change of the kinetic energy. Trusting Landau on that one, it has the form:

$$ \frac{dE_{kin} }{dt} = -\frac{\eta}{2} \int \left( \frac{\partial v_i}{\partial x_k}+ \frac{\partial v_k}{\partial x_i}\right)^2 dV $$

To get the $\frac{ds}{dt}$, do we just need to divide this by temperature? I'm trying to explain everything here in terms of thermodynamics, because I only know typical, very basic examples of thermodynamics processes and I have a difficulty in interpreting what is truly happening here (in terms of entropy).