A Newtonian fluid having a specific gravity of 0.93 and a kinematic viscosity of

Question

A Newtonian fluid having a specific gravity of 0.93 and a kinematic viscosity of3 10-4m2/sflows past a fixed surface. Due to the no-slip condition, the velocity at the fixed surface is zero, and the velocity profile near the surface is shown in the figure above. Determine the magnitude and direction of the shearing stress developed on the plate. Express your answer in terms ofUand?, withUand?expressed in units of meters per second and meters, respectively.

I have tried solving the equation for u and differentiating to get du/dy then using the sg, density of water, and kinematic viscosity (viscosity= sg*density of H20*kinematic viscosity) to get the absolute viscosity and plugging that in to the shear stress (shear stress= viscosity * du/dy) but it is incorrect.

A Newtonian fluid having a specific gravity of 0.93 and a kinematic viscosity of 3 ½ 10-4m2/s flows past a fixed surface. Due to the no-slip condition, the velocity at the fixed surface is zero, and the velocity profile near the surface is shown in the figure above. Determine the magnitude and direction of the shearing stress developed on the plate. Express your answer in terms ofUand?, withUand?expressed in units of meters per second and meters, respectively. I have tried solving the equation for u and differentiating to get du/dy then using the sg, density of water, and kinematic viscosity (viscosity= sg*density of H20*kinematic viscosity) to get the absolute viscosity and plugging that in to the shear stress (shear stress= viscosity * du/dy) but it is incorrect.

Explanation / Answer

ur formula and approach is right viscosity=0.279 since figure is not given its difficult to find du/dy if possible give the equation of U and rate this answer :)

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