Question 9.2: A rotating shaft, as illustrated in Figure 9.6, causes the f......

As the pressure along the free surface will be constant, we may observe that the free surface is perpendicular to the pressure gradient. Determination of the pressure gradient, therefore, will enable us to evaluate the slope of the free surface.

Rearranging equation (9-20), we have

\nabla P=\rho\mathbf{g}-\rho{\frac{D\mathbf{v}}{D t}}              (9-20)

The velocity \mathbf{v}=A\mathbf{e}_{\theta}/r, where A is a constant, when using the coordinate system shown in Figure 9.7. Assuming that there is no slip between the fluid and the shaft at the surface of the shaft, we have

v(R)=\omega R={\frac{A}{R}}

and thus A=\omega R^{2} and

\mathbf{v}={\frac{\omega R^{2}}{r}}\mathbf{e}_{\theta}

The substantial derivative Dv/Dt may be evaluated by taking the total derivative

{\frac{d\mathbf{v}}{d t}}=-{\frac{\omega R^{2}}{r^{2}}}\mathbf{e}_{\theta}{\dot{r}}+{\frac{\omega R^{2}}{r}}{\frac{d\mathbf{e}_{\theta}}{d t}}

where d\mathbf{e}_{\theta}/d t=-{\dot{\theta}}\mathbf{e}_{r}. The total derivative becomes

{\frac{d{\mathbf{v}}}{d t}}=-{\frac{\omega R^{2}}{r^{2}}}\dot r\mathbf{e}_{\theta}-{\frac{\omega R^{2}}{r}}{\dot{\theta}}\mathbf{e}_{r}

Now the fluid velocity in the r-direction is zero, and \dot θ for the fluid is v/r , so

\left({\frac{d\mathbf{v}}{d t}}\right)_{\mathrm{fluid}}={\frac{D\mathbf{v}}{D t}}=-{\frac{\omega R^{2}}{r^{2}}}v\mathbf{e}_{r}=-{\frac{\omega^{2}R^{4}}{r^{3}}}\mathbf{e}_{r}

This result could have been obtained in a more direct manner by observing that Dv/Dt is the local fluid acceleration, which for this case is -v^{2}\mathbf{e}_{r}/r, The pressure gradient becomes

\nabla P=-\rho g\mathbf{e}_{z}+\rho{\frac{\omega^{2}R^{4}\mathbf{e}_{r}}{r^{3}}}

From Figure 9.8, it can be seen that the free surface makes an angle β with the r axis so that

\tan\beta={\frac{\rho\omega^{2}R^{4}}{r^{3}\rho g_{}}}

={\frac{\omega^{2}R^{4}}{dr^3}}