This question was previously asked in

NPCIL ST ME (07/11/2019, Shift-2)

Option 3 : Irrotational

NPCIL 2020 Scientific Assistant Physics Mini Live Test

1344

30 Questions
30 Marks
40 Mins

__Explanation:__

__Velocity Potential function:__ This function is defined as a function of space and time in a flow such that the negative derivation of this function with respect to any direction gives the velocity of fluid in that direction.

If velocity potential (ϕ) exist, there will be a flow.

\(u = - \frac{{\partial ϕ }}{{\partial x}}\)

\(v = - \frac{{\partial ϕ }}{{\partial y}}\)

\(w = - \frac{{\partial ϕ }}{{\partial z}}\)

Now, Angular velocity is given by:

\({\omega _z} = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)\)

\({\omega _z} = \frac{1}{2}\left[ {\frac{\partial }{{\partial x}}\left( { - \frac{{\partial ϕ }}{{\partial y}}} \right) - \frac{\partial }{{\partial y}}\left( { - \frac{{\partial ϕ }}{{\partial x}}} \right)} \right]\)

\({\omega _z} = \frac{1}{2}\left[ { - \frac{{{\partial ^2}ϕ }}{{\partial x\partial y}}\; + \frac{{{\partial ^2}ϕ }}{{\partial y\partial x}}\;} \right]\)

Since ϕ is a continuous function

\(\frac{{{\partial ^2}\phi }}{{\partial x\partial y}} = \;\frac{{{\partial ^2}\phi }}{{\partial y\partial x}}\;\)

Therefore, ωz = 0

It implies that if velocity potential function exists for flow then the flow must be irrotational.