**5. Navier-Stokes in rotational domain**

Navier-Stokes/fluid equation(s) are well known by most in the translational domain. However, as described earlier, a wind turbine exhibits rotational motion, and the fluid equations take a different form. At the outset, it is evident that extra forces will act on the fluid particle due to the rotational motion when viewed from the inertial frame of reference. These forces will have to be accounted for in the integral form of fluid equations as used by Fluent.

A vector is assumed to rotate with a radial velocity Ω [8]. From the perspective of the vector, the vector itself is static. However, from the inertial frame of reference, the vector is rotating. For a small time-period "t", the angle subtended by the new vector position with the old vector is:

$$
\theta = \Omega \Delta t \tag{15}
$$

The magnitude of change in the new vector from the old vector (position-wise) forms a sector of a circle [8]. This gives a net length of:

$$
\Delta i\_{new} - \Delta i\_{old} = r \Omega \Delta t \tag{16}
$$

Notice that the triangle formed by these three vectors is a right-angle triangle for a small change in vector in a short period "t." The (change in vector)/(new vector) = sin (*ϕ*)

Hence, we can write the net vector change with magnitude and direction as the following [8]. Note that the change vector is perpendicular to both the old vector as well as the rotational axis

$$
\Delta\hat{i} = |\hat{i}|\sin\left(\phi\right)\Omega\Delta t \frac{\overrightarrow{\Omega}\times\hat{i}}{|\overrightarrow{\Omega}\times\hat{i}|}\tag{17}
$$

By definition of cross product:

*Fluid Dynamics Simulation of an NREL-S Series Wind Turbine Blade DOI: http://dx.doi.org/10.5772/intechopen.107013*

$$|\overrightarrow{\Omega} \times \hat{i}| = |\hat{i}| \Omega \sin \left(\phi\right) \tag{18}$$

Therefore, substituting this in the previous equation

$$\frac{d}{dt}\hat{i} = \vec{\Omega} \times \hat{i} \tag{19}$$

This was for a stationary vector. One can extend the analogy to a vector rotating in a rotating frame of reference. As imaginable, the sum of the rate of change of individual vector rotation and the frame rotation will be the rate of change of net rotation [8, 9]. However, the rate of change in frame rotation has been defined earlier. Hence, we obtain Chasle's theorem [9].

$$\frac{d\overrightarrow{A}}{dt} = \frac{dA\_i}{dt} + \overrightarrow{\Omega} \times \overrightarrow{A} \tag{20}$$

where (*dAi dt* ) term is the vector rotation as seen by the observer in the rotating frame of reference [9].

For Navier-Stokes, the fluid velocity "u" is rotating and is viewed from a stationary frame of reference. Hence, by applying Chasle's theorem on the fluid velocity "u."

$$
\left(\frac{d\overrightarrow{u}\_{inertial}}{dt}\right)\_{inertial} = \left(\frac{d\overrightarrow{u}\_{inertial}}{dt}\right)\_{rotational} + \overrightarrow{\Omega} \times \overrightarrow{u}\_{inertial} \tag{21}
$$

Re-substituting Chasle's theorem twice in this equation, we obtain the final equation and assume constant velocity flow:

$$
\left(\frac{d\vec{u}\_{\text{rotational}}}{dt}\right)\_{\text{rotational}} = 2\vec{\Omega} \times \vec{u} + \vec{\Omega} \times \left[\vec{\Omega} \times \vec{x}\right] \tag{22}
$$

It is interesting to note that the following is known as Coriolis acceleration.

$$
\vec{\mathfrak{Q}} \vec{\mathfrak{Q}} \times \vec{\mathfrak{u}}\tag{23}
$$

The following is the centrifugal acceleration. These two equations put together define the Navier–Stokes in the rotational domain.

$$
\overrightarrow{\mathbf{\dot{Q}}} \times \left[ \overrightarrow{\mathbf{\dot{Q}}} \times \overrightarrow{\mathbf{x}} \right] \tag{24}
$$

The exact equation that will be solved in Fluent for our wind turbine is as follows. One can note the similarities between the earlier two equations and this:

$$\frac{\partial \rho}{\partial t} + \nabla .\rho \vec{v}\_r = \mathbf{0}$$

$$\nabla . \left(\rho \vec{v}\_r \vec{v}\_r\right) + \rho \left(2\vec{\Omega} \times \vec{v}\_r + \vec{\Omega} \times \vec{\Omega} \times \vec{r}\right) = -\nabla p + \nabla \tau\_r\tag{25}$$

The next section will be brief and devoted to CAD modeling of blade.
