Question #142984
In a stream of glycerine in motion, at a certain point the velocity gradient is 0.25 metre per sec per metre . The mass density of fluid is 1268.4 kg per cubic metre and kinematic viscosity is 6.30 x 10^-4 square metre per second. Calculate the shear stress at the point.
1
Expert's answer
2020-11-09T20:29:19-0500

By the Newton's Law of Viscosity

τ=μdudy\tau=\mu\dfrac{du}{dy}

Kinematic viscosity is defined as the ratio between the dynamic viscosity and and density of fluid.


ν=μρ\nu=\dfrac{\mu}{\rho}

Substitute


τ=νρdudy\tau=\nu\cdot\rho\cdot\dfrac{du}{dy}

Given

ρ=1268.4kg/m3,\rho=1268.4kg/m^3,

ν=6.3×104m2/s,\nu=6.3\times10^{-4}m^2/s ,

dudy=0.25/s\dfrac{du}{dy}=0.25/s

τ=1268.4kg/m36.3×104m2/s0.25/s=\tau=1268.4kg/m^3\cdot6.3\times10^{-4}m^2/s\cdot0.25/s==0.199773N/m20.2N/m2=0.199773N/m^2\approx0.2N/m^2

The shear stress at the point is 0.2N/m2.0.2N/m^2.



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