Question

Verify the below velocity distribution describes a fluid in a state of pure rotation. What is the angular Velocity? (a)-Vx = -1/2bx (b)-Vy = -1/2by (c)-Vz = bz

Answer 1

Answer:

Angular velocity of fluid is zero.

Explanation:

Given that

$V_x=\dfrac{1}{2}bx$

$V_y=\dfrac{1}{2}by$

$V_z=-bz$

The angular velocity in Z direction

$\omega _z=0.5\times\left (\dfrac{\partial V_y}{\partial x}-\dfrac{\partial V_x}{\partial y} \right )$

$V_x=\dfrac{1}{2}bx$

$\dfrac{\partial V_y}{\partial x}=0$

$V_y=\dfrac{1}{2}by$

$\dfrac{\partial V_x}{\partial y}=0$

So

$\omega _z=0$

Similarly

$\omega _y=0.5\times\left (\dfrac{\partial V_z}{\partial x}-\dfrac{\partial V_x}{\partial z} \right )$

$\omega _x=0.5\times\left (\dfrac{\partial V_z}{\partial y}-\dfrac{\partial V_y}{\partial z} \right )$

$\omega _y=0$

$\omega _x=0$

$\omega =\omega_xi+\omega_yj+\omega_zk$

ω = 0

So the angular velocity of fluid is zero.