I thought I would post an answer to this, to piece together and summarize some of the helpful information in the comments:
I think the key misconception is that, in the Eulerian case, momentum is considered as a continuous field that is a function of both time and space, i.e. $m\mathbf{v}=m\mathbf{v}(\mathbf{x},t)$, whereas in the Lagrangian case, we are considering an individual parcel of fluid, that is moving with the flow. Therefore, momentum is a function only of time ($m\mathbf{v}=m\mathbf{v}(t)$), but the parcel also has a position, which is also a function of time ($\mathbf{x}=\mathbf{x}(t)$).
So, looking at the Eulerian case, as SMeznaric's comment states:
$$\mathbf{F}=\frac{d(m\mathbf{v})}{d t}$$
not
$$\mathbf{F}=\frac{\partial(m\mathbf{v})}{\partial t}$$
Force is the total derivative of momentum, from Newton's Second Law. Because momentum is a function of both time and space ($m\mathbf{v}=m\mathbf{v}(\mathbf{x},t)$), the partial derivatives with respect to position and time have to be expanded, as shown in the textbook. This then results in:
$$\mathbf{F}=\Delta x\Delta y\Delta z\Bigl[\frac{\partial(\rho\mathbf{v})}{\partial t}+\mathbf{v}\cdot\nabla(\rho\mathbf{v})\Bigr]$$
which, on the right-hand side, gives the transient and advective terms of the Navier-Stokes momentum equation. The $\mathbf{F}$ on the left-hand-side is made up of contributions to the force on the volume from the pressure gradient, gravity and viscous forces.
In the Lagrangian case, Newton's Second Law still holds, i.e.:
$$\mathbf{F}=\frac{d(m\mathbf{v})}{d t}$$
But the key difference is that now we are considering a parcel of fluid moving with the flow; therefore, as explained above, momentum is now a function only of time. Mass of the parcel is constant, so that can be brought out of the derivative:
$$\mathbf{F}=m\frac{d(\mathbf{v})}{d t}$$
So, now those same forces from the pressure gradient, gravity and viscosity on the left-hand-side are imposed by the wider flow field on the parcel and are equal to the mass times the total derivative of the parcel velocity. So here, if we know the forces being imposed, then we can simply integrate to find the velocity of the parcel and track its path through the flow field.
I hope this helps.
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.
Best Answer
The missing equation is energy conservation $$ \frac{\partial}{\partial t} {\cal E} + \vec\nabla \cdot\vec\jmath^{\,\cal E}=0 $$ where ${\cal E}$ is the energy density and $\vec\jmath^{\,\cal E}$ is the energy current $$ \vec\jmath^{\,\cal E} = \vec{u}\left[ {\cal E}+P \right] -\eta u\cdot\sigma-\kappa\vec\nabla T\, . $$ Now the equations close if you have an equation of state, $P=P({\cal E}^0,\rho)$, where ${\cal E}^0={\cal E}-\frac{1}{2}\rho u^2$ is the internal energy density. Note that the equation of state also fixes $T({\cal E}^0,\rho)$ using thermodynamic identities (although this is tedious in practice; for a non-interacting gas things are simple, $T=mP/\rho$). The energy equation can be rewritten in various ways, for example as an equation for entropy production.