Velocity profile of a moving fluid over a flat (fixed) plate is given as: = from the fixed plate (yy = 0) in mm, u is the velocity in xx direction in mm/. Calculate the shear stress on the fluid at yy = 0.15 m if the viscosity of the fluid 8.6 poise (1 Poise= 1 yy.) 2

The question has some details missing which is the velocity profile .The velocity profile is given as u = (8y - 0.3y^2) mm/s

the shear stress on the fluid at y = 0.15 m = 6.80 x 10^-3N/m^2

Explanation:

The detailed steps is the use of the relationship between the shear stress, viscosity and the velocity profile. appropriate differentiation and substitution were made to calculate the shear stress as is shown in the attachment.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

c) power lifting, will be your answer.

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