Question

If the equation for the velocity profile is given by: v = 4y^2/3. 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

At y = 0.25 ft velocity gradient will be 1.5866

At y = 0.5 ft velocity gradient will be 1.257

Step-by-step explanation:

We have given velocity
`v=4y^{(2)/(3)}`

We have to find velocity gradient

Velocity gradient is nothing but rate of change of velocity

So velocity gradient
`=(dv)/(dy)=(2)/(3)y^{(-1)/(3)}`

(a) Now velocity gradient at y = 0.25 ft

`=(dv)/(dy)_(y=0.25)=(2)/(3)0.25^{(-1)/(3)}=1.5866`

(b) Velocity gradient at y = 0.5 ft

`=(dv)/(dy)_(y=0.5)=(2)/(3)0.5^{(-1)/(3)}=1.257`