Answer:
At y = 0.25 ft velocity gradient will be 1.5866
At y = 0.5 ft velocity gradient will be 1.257
Explanation:
We have given velocity [tex]v=4y^{\frac{2}{3}}[/tex]
We have to find velocity gradient
Velocity gradient is nothing but rate of change of velocity
So velocity gradient [tex]=\frac{dv}{dy}=\frac{2}{3}y^{\frac{-1}{3}}[/tex]
(a) Now velocity gradient at y = 0.25 ft
[tex]=\frac{dv}{dy}_{y=0.25}=\frac{2}{3}0.25^{\frac{-1}{3}}=1.5866[/tex]
(b) Velocity gradient at y = 0.5 ft
[tex]=\frac{dv}{dy}_{y=0.5}=\frac{2}{3}0.5^{\frac{-1}{3}}=1.257[/tex]
