Respuesta :
Answer:
T = 1.17 x [tex]10^{-3}[/tex] N-m
Explanation:
Given :
Gap between the two plates , dy = 1 mm
dy = 1 x [tex]10^{-3}[/tex] 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.[tex]\frac{du}{dy}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{du}{dy}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{5r-0}{1\times 10^{-3}}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu \times 10^{3}\times 5r[/tex]
[tex]F_{s}=\mu \times 10^{3}\times 5r\times 2\pi rdr[/tex]
[tex]F_{s}=5 \times 10^{3}\times \mu \times r\times 2\pi rdr[/tex]
[tex]F_{s}=\frac{18849}{400}\times r^{2}dr[/tex]
Now we know torque due to small strip,
dT = [tex]F_{s}[/tex] x r
dT = [tex]\frac{18849}{400}\times r^{3}dr[/tex]
Therefore total torque for r=0 to r=R can be calculated. So by integrating,
[tex]\int dT=\int_{0}^{R}\frac{18849}{400}\times r^{3}dr[/tex]
[tex]T = \frac{18849}{400}\times \frac{R^{4}}{4}[/tex]
[tex]T = 47.1225\times \frac{0.1^{4}}{4}[/tex]
T = 1.17 x [tex]10^{-3}[/tex] N-m
