Correct Answer - Option 1 : Existence of velocity potential is an indication of irrotational nature of the flow
Concept:
Velocity Potential function: This function is defined as a function of space (3-Dimensional) 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.
\(u = - \frac{{\partial ϕ }}{{\partial x}}\)
\(v = - \frac{{\partial ϕ }}{{\partial y}}\)
\(w = - \frac{{\partial ϕ }}{{\partial z}}\)
From continuity equation
\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)
⇒ \(\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} + \frac{{{\partial ^2}\phi }}{{d{z^2}}} = 0\;\)
If velocity potential (ϕ) satisfies the Laplace equation, it represents the possible steady incompressible irrotational flow
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.
