Question

A viscometer (an instrument used to study characteristics of a non-ideal fluid) consists of a flat plate and a rotating cone. The cone has a large apex angle and the angle $\theta$ shown in figure is very small (typically less than 0.5° ). The apex of the cone just touches the plate and a liquid fills the narrow gap between the plate and the cone. The cone has a base radius $R$ and is rotated with constant angular speed $\omega$. Consider the liquid to be ideal and take its coefficient of viscosity to be $\eta$. Calculate the torque needed to drive the cone.

Answer 1

$\frac{2\pi \eta \omega R^3}{3\theta}$

Answer 2

To calculate the torque needed to drive the cone, we use the principles of viscous fluid flow.

1. Identify the geometry and fluid properties:

   • The liquid fills the narrow gap between a rotating cone and a stationary flat plate.
   • The cone has a base radius $R$ and rotates with constant angular speed $\omega$.
   • The angle between the cone and the plate is $\theta$, which is very small.
   • The liquid has a coefficient of viscosity $\eta$. (Note: The problem statement's "ideal fluid" contradicts "coefficient of viscosity $\eta$". We assume it means a viscous fluid with viscosity $\eta$.)

2. Determine the gap height: Consider a small annular element at a radial distance $r$ from the axis of rotation. The height of the gap, $h$, at this radius $r$ is given by $h = r \tan\theta$. Since $\theta$ is very small (typically less than 0.5°), we can approximate $\tan\theta \approx \theta$ (in radians). So, $h = r\theta$.

3. Calculate the velocity gradient: The flat plate is stationary, so the fluid layer at the bottom has a velocity of 0. The cone rotates, so the fluid layer adhering to the cone surface at radius $r$ has a tangential velocity $v = r\omega$. Assuming a linear velocity profile across the narrow gap (which is a valid approximation for small $\theta$), the velocity gradient $\frac{dv}{dy}$ is: $\frac{dv}{dy} = \frac{\text{change in velocity}}{\text{change in height}} = \frac{r\omega - 0}{h} = \frac{r\omega}{r\theta} = \frac{\omega}{\theta}$ Notice that the velocity gradient $\frac{\omega}{\theta}$ is constant throughout the fluid gap, independent of $r$.

4. Calculate the shear stress: According to Newton's law of viscosity, the shear stress $\tau_s$ is given by: $\tau_s = \eta \frac{dv}{dy} = \eta \frac{\omega}{\theta}$ Since $\frac{dv}{dy}$ is constant, the shear stress $\tau_s$ is also constant over the entire surface of the cone in contact with the fluid.

5. Calculate the torque on an elemental ring: Consider an annular element of radius $r$ and infinitesimal thickness $dr$. The area of this element is $dA = 2\pi r dr$. The viscous force $dF$ acting on this elemental area is: $dF = \tau_s \cdot dA = \left(\eta \frac{\omega}{\theta}\right) (2\pi r dr)$ The torque $d\tau$ exerted by this force about the axis of rotation is $r \cdot dF$: $d\tau = r \cdot dF = r \left(\eta \frac{\omega}{\theta}\right) (2\pi r dr) = \frac{2\pi \eta \omega}{\theta} r^2 dr$

6. Calculate the total torque: To find the total torque $\tau$ needed to drive the cone, integrate $d\tau$ from $r=0$ to $r=R$: $\tau = \int_0^R \frac{2\pi \eta \omega}{\theta} r^2 dr$ $\tau = \frac{2\pi \eta \omega}{\theta} \int_0^R r^2 dr$ $\tau = \frac{2\pi \eta \omega}{\theta} \left[ \frac{r^3}{3} \right]_0^R$ $\tau = \frac{2\pi \eta \omega}{\theta} \left( \frac{R^3}{3} - 0 \right)$ $\tau = \frac{2\pi \eta \omega R^3}{3\theta}$

The torque needed to drive the cone is $\frac{2\pi \eta \omega R^3}{3\theta}$.