Answer to Question #184632 in Mechanical Engineering for Ram Kumar

Question #184632

The Velocity Distribution for flow over a flat plate is given by u=(2/3)y-y2, Where u is the point velocity in meter per second at a distance y meter above the plate.

Determine the shear stress at y=0 and y=15 cm. Assume dynamic viscosity as 8.63 poises


1
Expert's answer
2021-04-29T07:57:46-0400

Given "u=\\dfrac{2}{3}y-y^2." Then


"\\dfrac{du}{dy}=\\dfrac{2}{3}-2y"


Shear stress is given


"\\tau=\\mu\\cdot\\dfrac{du}{dy}"

Assume dynamic viscosity as "\\mu=8.63\\ \\text{poise}=0.863\\ N\\cdot s\/m^2"


At "y=0"

"\\dfrac{du}{dy}=\\dfrac{2}{3}-2(0)=\\dfrac{2}{3}(s^{-1})"


Then


"\\tau=0.863\\ N\\cdot s\/m^2\\cdot\\dfrac{2}{3}s^{-1}=0.575 N\/m^2"

The shear stress at y = 0 is "0.575N\/m^2."


At "y=15 cm=0.15m"

"\\dfrac{du}{dy}=\\dfrac{2}{3}-2(0.15)=\\dfrac{1.1}{3}(s^{-1})"


Then


"\\tau=0.863\\ N\\cdot s\/m^2\\cdot\\dfrac{1.1}{3}s^{-1}=0.316 N\/m^2"

The shear stress at y = 15 cm is "0.316N\/m^2."



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