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Properties-of-fluid

A circular disc of diameter D is slowly rotated in a liquid of large viscosity `mu` at a small distance h from a fixed surface. Derive an expression of torque (T) necessary to maintain an angular velocity `omega`.

SOLUTION:

Let us consider an elementary ring of disc at a radius r and having width dr as shown in the figure. Let the linear velocity at this radius is `omega*r`

A circular disc of diameter D is slowly rotated in a liquid of large viscosity `mu` at a small distance h from a fixed surface. Derive an expression of torque (T) necessary to maintain an angular velocity `omega`.

Now,

Shear stress is, `tau=mu*(du)/(dy)`

And, torque is, T= shear stress * area * radius

or, `T=tau*2pi*r*dr*r`

`=mu*(du)/(dyl*2pi*r*dr*r`

Let us assume the gap h to be small such that the velocity distribution can be assumed to be linear.

Now, for linear velocity profile,

`(du)/(dy)=(omega*r)/h`

Thus, the torque on the element is given by,

`dT=mu*(omega*r)/....Show More

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Properties of fluid

Fluid-Mechanics