Micropolar fluids are fluids with microstructures. They
belong to a class of fluids with a nonsymmetric stress tensor.
Micropolar fluids consist of rigid, randomly oriented
or spherical particles with their own spins and microrotations,
suspended in a viscous medium.
Physical examples of micropolar fluids can be seen in
ferrofluids, blood flows, bubbly liquids, liquid
crystals, and so on, all of them containing intrinsic polarities.
The following (including notations) is based on textbook
G. Lukaszewicz, Micropolar Fluids: Theory and Applications, Birkhauser, Boston, 1999
http://books.google.ru/books?id=T3l9cGfR9o8C
We start with Cauchy momentum equation
$$
\rho \frac{D\vec{v}}{Dt} = \rho \vec{f} + \nabla \cdot \hat{T},
$$
where $\vec{f}$ is body force and $\hat{T}$ is stress tensor.
If we assume that fluid is polar, that is it has its own internal angular momentum (independent of the motion of fluid as a whole) then we need an additional
equation expressing conservation of angular momentum (for nonpolar fluid conservation of angular momentum is a consequence of Cauchy equation):
$$
\rho \frac{D}{Dt}(\vec{l} + \vec{x}\times\vec{v}) = \rho \vec{x} \times \vec{f} + \rho\vec{g} + \nabla \cdot (\vec{x} \times \hat{T} + \hat{C}).
$$
Here $\vec{l}$ is an intrinsic (internal) angular momentum per unit mass, $\vec{g}$ is body torque and $\hat{C}$ is a new object called couple stress tensor
(this equation could be seen as its definition).
Now in order to close this system of equation we need to express the stress tensor and couple stress tensor through the characteristics of fluid dynamics.
For this we need to make certain assumptions: the absence of preferred direction and position, reduction to hydrostatic pressure in the absence of deformations and
linear dependence on the velocity spatial derivatives $v_{i,j}$ (deformation). For polar fluid we also define the vector field $\vec{\omega}$ -- microrotation
which represents the angular velocity of rotation of particles of the fluid. We further assume that the fluid is isotropic and $\vec{l} = I \,\vec{\omega}$ with $I$ a scalar called the microinertia coefficient. Couple stress tensor should be a linear function of the spatial derivatives of the microrotation field: $\omega_{i,j}$. All this assumptions allow us to specify the general form for stress tensor:
$$
T_{ij}=(- p +\lambda v_{k,k})\,\delta_{ij}+\mu\,(v_{i,j}+v_{j,i})+\mu_r\,(v_{j,i}-v_{i,j}) - 2 \mu_r\, \epsilon_{mij}\omega_m,
$$
and couple stress tensor:
$$
C_{ij} = c_0\, \omega_{k,k} \delta_{ij} + c_d\, (\omega_{i,j}+\omega_{j,i}) + c_a \,(\omega_{j,i}-\omega_{i,j}).
$$
Note: we have three parameters $c_0$, $c_d$ and $c_a$ (called coefficients of angular
viscosities) because there are three irreducible representations for the action of $SO(3)$ group on rank 2 tensor: scalar (times $\delta_{ij}$), traceless symmetric tensor, antisymmetric tensor. The same for the triplet $(\lambda,\mu, \mu_r)$ (which are called second viscosity coefficient, dynamic
Newtonian viscosity and dynamic microrotation viscosity).
Substituting this expressions for $T_{ij}$ and $C_{ij}$ into Cauchy equation and angular momentum equation we obtain the equations from the question.
The notations correspondence is: $$ \vec{\omega} \to \vec{N}^{*}, \qquad \mu_r \to k_1^{*}, \qquad I\to j^{*} \qquad c_a+c_d \to \gamma^{*}.$$
Additionally we see that the equations in the question assume $\mathrm{div} \vec{v} =0 $ (incompressible flow) and $\mathrm{div} \vec{\omega} =0 $ -- this
one is generally not true, but symmetries of the system could make it so. Also we see that there are no body torque and the body force term corresponds to Lorentz force.
The error seems to lie near the end of your derivation, when you claim
$$-\partial_{\mu}(u^{\mu}\rho)=\partial_{t}(v^{0}\rho)-\boldsymbol{\nabla}\cdot\left(\boldsymbol{v}\,\rho\right)$$
(Where I am using bold text to refer to vectors now.) This is, in fact, not the case. Recall that $\partial_{\mu}=(\partial_{t},\boldsymbol{\nabla})$, and requires no minus signs, since derivatives are naturally covariant. On the other hand, $u^{\mu}=(u^{0},\textbf{u})$ naturally transforms like coordinate vectors, and thus naturally has an upstairs index. Thus, the real expression would be
$$-\partial_{\mu}\left(u^{\mu}\rho\right)=-\left(\partial_t(v^{0}\rho)+\boldsymbol{\nabla}\cdot(\boldsymbol{v}\,\rho)\right)$$
With this, the continuity equation is immediately in the form you want it.
The lesson is that, when doing these kinds of calculations, it is important to recall how vectors like $\partial$ and $v$ are defined. Do they naturally have upstairs or downstairs indices? You get negatives when you contract two of the same type of vector, but none when you contract a vector and a covector.
I hope this helps!
Best Answer
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.