Chapter:

Properties-of-fluid

A solid cone of maximum radius R and vertex angle `2ta`is to rotate at angular velocity `omega`. An oil of viscosity `mu` and thickness t fills gap between cone and housing. Derive an expression for torque required and rate of dissipation of heat in bearing.

SOLUTION:

Given,

Maximum radius of the cone =R

Vertex angle = `2 theta`

Viscosity of the oil= `mu`

Thickness of the oil layer = t

A solid cone of maximum radius R and vertex angle `2theta`is to rotate at angular velocity `omega`. An oil of viscosity `mu` and thickness t fills the gap between the cone and the housing. Derive an expression for the torque required and the rate of dissipation of heat in the bearing.

Let us consider an elementary area `dA` at a radius r of the cone as shown in figure.Let ds be the inclined length of element and dy be the thickness of element. Now,the elementary area is 

`dA=2pir*ds`

or, `dA=2pi*r*(dr)/(sin theta)`

Now,

Shear stress,  `tau=mu*(du)/(dy)`

Let us assume the gap `t` to be small....Show More