In a rotating cylinder viscometer, radii of cylinders are 3.2 cm and 3 cm and outer cylinder is rotated steadily at 180 rpm. For a certain liquid filled in annular space to a depth of 7.5 cm, torque produced on inner cylinder is `10^(-4)`Nm. Assuming velocity distribution to be linear, calculate viscosity of liquid.

SOLUTION:

Given,

Radius of the inner cylinder, r=3 cm=0.03  m

Radius of outer rotating cylinder, R=3.2 cm =0.032 m

Speed of rotation of outer cylinder, N=180 rpm

Height of each cylinder, L=7.5 cm =0.075 m

Torque produced in inner cylinder, `T=10^(-4)` Nm.

Thickness of liquid layer, `dy=(0.032-0.030)=0.002 m`

Tangential speed of outer cylinder is,

`u=2pi*R*180/60`

or, `u=2pi**0.032**180/60 =0.603 m/s`

For linear velocity distribution, `du=u=0.603 m/s`

Now, from Newton's law of viscosity,

`tau=mu*(du)/(dy)`

or, `F/A=mu*(du)/(dy)`

or, `F=mu0.603/0.002**2pi**rL`

`= 301.5mu**2pi**0.03*0.075`

or,`F=4.26 mu`

Now,

Viscous Torque,`T=F*r`,

or,`10^(-4)=4.26mu**0.03=0.1278mu`

or,`mu=7.825**10^(-4)`Ns/`m^2`

Thus, `mu=78.2 **10^(-4)` Poise.

SOLUTION:

Given,

Radius of the inner cylinder, r=3 cm=0.03  m

Radius of outer rotating cylinder, R=3.2 cm =0.032 m

Speed of rotation of outer cylinder, N=180 rpm

Height of each cylinder, L=7.5 cm =0.075 m

Torque produced in inner cylinder, `T=10^(-4)` Nm.

Thickness of liquid layer, `dy=(0.032-0.030)=0.002 m`

Tangential speed of outer cylinder is,

`u=2pi*R*180/60`

or, `u=2pi**0.032**180/60 =0.603 m/s`

For linear velocity distribution, `du=u=0.603 m/s`

Now, from Newton's law of viscosity,

`tau=mu*(du)/(dy)`

or, `F/A=mu*(du)/(dy)`

or, `F=mu0.603/0.002**2pi**rL`

`= 301.5mu**2pi**0.03*0.075`

or,`F=4.26 mu`

Now,

Viscous Torque,`T=F*r`,

or,`10^(-4)=4.26mu**0.03=0.1278mu`

or,`mu=7.825**10^(-4)`Ns/`m^2`

Thus, `mu=78.2 **10^(-4)` Poise.