Question

Starting with two coaxial cylinders with a small gap between the inner and outer cylinders and the outer clylinder is fixed while the inner cylinder rotates with an angular velocity w₁. The governing equation for the setup is 0 = d/dr ((1/r) (d/dr)(rv_thetha)). Show the velocity profile is v_thetha(r) = (w₁r₁²)/(r₂²-r₁²)*((r₂²/r )-r)

Answer 1

Given the governing differential equation:

$\frac{d}{dr}\left(\frac{1}{r} \frac{d}{dr}(rv_\theta) \right ) = 0\\ \\ \text{Integrating,}\\ \\ \frac{1}{r} \frac{d}{dr}(rv_\theta) = C_1 \\ \\ \text{Integrating again;} \\ \\ v_\theta = \frac{C_1 r}{2} + \frac{C_2}{r}$

The boundary conditions are;

$\text{At} \,\, r=r_1 \rightarrow v_\theta =\omega r_1\\ \text{At} \,\, r=r_2 \rightarrow v_\theta =0$

Now, at $r=r_1$

wri = + 2wr} = Ciri +2C

At,

= 1

$0 = \frac{C_1 r_2}{2} + \frac{C_2}{r_2} \\ \\ {\color{Red} 0= C_1 r_2^2 + 2 C_2}$

Solving simultaneously;

${\color{Red} C_1 = \frac{2\omega r_1^2}{(r_1^2 - r_2^2)} \, \text{and}\, \, \\ \\ C_2 = \frac{-\omega r_1^2 r_2^2}{(r_1^2 - r_2^2)}}$

On substitution;

$v_\theta = \frac{\omega r_1^2 r}{(r_1^2 - r_2^2)} - \frac{\omega r_1^2 r_2^2}{r(r_1^2 - r_2^2)}\\ \\ = \frac{\omega r_1^2 }{(r_1^2 - r_2^2)} \left(r-\frac{r_2^2}{r} \right )$

or

$v_\theta = \frac{\omega r_1^2 }{(r_2^2 - r_1^2)} \left(\frac{r_2^2}{r} - r\right )$