6.1.5.6 Viscous Torque

In fluid systems, viscosity introduces additional torque requirements when rotating one surface over another, separated by a fluid film. Below are examples of viscous torque calculations for various geometries.

1. Rotating Cylinder

A cylinder of radius \( r \) and height \( l \) rotates at angular speed \( \omega \), separated by a fluid layer of thickness \( h \). The tangential velocity is \( r\omega \). Using Newton’s law of viscosity:

$$ T = \mu \frac{r\omega}{h} \cdot 2\pi r l \cdot r = \frac{2\pi \mu \omega l r^3}{h} $$

2. Rotating Disc

A disc of radius \( r \), rotating over a fluid film of thickness \( h \) with angular speed \( \omega \), has tangential velocity at radius \( x \) as \( x\omega \). The torque is calculated by integration:

$$ T = \int_0^r x \cdot \mu \frac{x\omega}{h} \cdot 2\pi x \, dx = \frac{1}{2} \cdot \frac{\pi \mu \omega r^4}{h} $$

3. Rotating Cone

A cone of base radius \( r \), half-angle \( \theta \), and fluid film thickness \( h \), rotates at angular speed \( \omega \). The shear acts on an elemental disc of area \( da = \frac{2\pi x \, dx}{\sin\theta} \). Torque is given by:

$$ T = \int_0^r x \cdot \mu \frac{x\omega}{h} \cdot \frac{2\pi x \, dx}{\sin\theta} = \frac{1}{2} \cdot \frac{\pi \mu \omega r^4}{h \sin\theta} $$

4. Rotating Hemisphere

For a hemisphere of radius \( r \), rotating at angular speed \( \omega \) with a fluid film of thickness \( h \), torque is calculated using polar coordinates where the radius element is \( r\sin\theta \) and thickness is \( rd\theta \).

Figure 6.5 (placeholder): Rotating Hemisphere Schematic

$$ T = \int_0^{\pi/2} (r\cos\theta) \cdot \mu \frac{r\cos\theta\omega}{h} \cdot 2\pi (r\cos\theta)r \, d\theta $$ $$ = \frac{2\pi \mu \omega r^3}{h} \int_0^{\pi/2} \cos^3\theta \, d\theta $$

Using Gamma functions or definite integral:

$$ \int_0^{\pi/2} \cos^3\theta \, d\theta = \frac{2}{3} $$

T = \frac{4}{3} \cdot \frac{\pi \mu \omega r^4}{h}