Question

If the equation for the velocity profile is given by: 3/2 = 4yv . Assuming v is in ft/s, what is the velocity gradient at the boundary and at y = 0.25 ft and 0.5 ft from boundary?

Answer 1

When y = 0.25 ft, velocity gradient = -6 ft/s

When y = 0.5 f, velocity gradient = -1.5 ft/s

Step-by-step explanation:

Given;

equation for the velocity profile, 3/2 = 4yv

Rearrange this equation, you will get;

`4yv = (3)/(2)\\\\v = (3)/(2) *(1)/(4y) \\\\v = (3)/(2)((1)/(4) )((1)/(y) )\\\\v = (3)/(8)y^(-1)\\\\ gradient \ of \ velocity \ = (dv)/(dy) \\\\(dv)/(dy) = -1((3)/(8))y^(-2)`

When y = 0.25 ft

`(dv)/(dy) = -1((3)/(8))(0.25)^(-2)\\\\(dv)/(dy) = -6 \ ft/s`

When y = 0.5 ft

`(dv)/(dy) = -1((3)/(8))(0.5)^(-2)\\\\(dv)/(dy) = -1.5 \ ft/s`