Question

A circular plate with diameter of 20 cm is placed over a fixed bottom plate with a 1 mm gap between two plates filled with Kerosene at 40 degree C, shown in the following figure. Find the torque needed to rotate the top plate at 5 rad/s. The velocity distribution in the gap is linear and the shear stress on the edge of the rotating plate can be neglected.

Answer 1

T = 1.17 x
`10^(-3)` N-m

Step-by-step explanation:

Given :

Gap between the two plates , dy = 1 mm

dy = 1 x
`10^(-3)` m

Angular velocity of the top plate , ω = 5 rad/s

Diameter of the plate, D = 20 cm

Radius of the plate, R = 10 cm

= 0.1 m

Temperature of the kerosene = 40°C

Viscosity of kerosene at 40°C = 0.0015 Pa-s

Now let us take a small elemental ring of thickness dr at a radius r.

Therefore, area of this elemental ring of dr = 2πrdr

Now linear velocity at radius r = ω x r

5r m/s

Now applying Newtons law of viscosity we get,

Shear stress, τ = μ.
`(du)/(dy)`

`\Rightarrow (F_(s))/(A)=\mu .(du)/(dy)`

`\Rightarrow (F_(s))/(A)=\mu .(5r-0)/(1* 10^(-3))`

`\Rightarrow (F_(s))/(A)=\mu * 10^(3)* 5r`

`F_(s)=\mu * 10^(3)* 5r* 2\pi rdr`

`F_(s)=5 * 10^(3)* \mu * r* 2\pi rdr`

`F_(s)=(18849)/(400)* r^(2)dr`

Now we know torque due to small strip,

dT =
`F_(s)` x r

dT =
`(18849)/(400)* r^(3)dr`

Therefore total torque for r=0 to r=R can be calculated. So by integrating,

`\int dT=\int_(0)^(R)(18849)/(400)* r^(3)dr`

`T = (18849)/(400)* (R^(4))/(4)`

`T = 47.1225* (0.1^(4))/(4)`

T = 1.17 x
`10^(-3)` N-m