Question

A thin horizontal disc of radius $R=10cm$ is located within a cylindrical cavity filled with oil whose viscosity $\eta =0.08P$ (figure shown above). The clearance between the disc and the horizontal planes of the cavity is equal to $h=1.0mm$. Find the power developed by the viscous forces acting on the disc when it rotates with the angular velocity $\omega =60rad/s$. The end effects are to be neglected.

Answer 1

When the disc rotates the fluid in contact with, corotates but the fluid in contact with the walls of the cavity does not rotate. A velocity gradient is then set up leading to viscous forces. At a distance $r$ from the axis the linear velocity is $\omega r$ so there is a velocity gradient $\frac{\omega r}{h}$ both in the upper and lower clearance. The corresponding force on the element whose radial width is $dr$ is
$\eta 2\pi rdr\frac{\omega r}{h}$ (from the formular $F=\eta A\frac{dv}{dx}$)
The torque due to this force is
$\eta 2\pi rdr\frac{\omega r}{h}r$
and the net torque considering both the upper and lower clearance is
$2{\int }_0^R\eta 2\pi r^{3}dr\frac{\omega }{h}$
$=\pi R^4\omega \eta /h$
So power developed is
$P=\pi R^4{\omega }^2\eta /h=9.05W$ (on putting the values).
(As instructed end effects i.e. rotation of fluid in the clearance $r>R$ has been neglected.)