Correct Answer - Option 3 : P - III,Q - I, R - IV, S - II
Explanation:
\(\nabla \times \vec V = 0 - Irrotational\;flow\;\)
\(\vec V = \overrightarrow {{V_x}} \widehat {\dot i} + \overrightarrow {{V_y}} \widehat {\dot J} + \overrightarrow {{V_z}} \hat k\)
\(\nabla = \frac{\partial }{{\partial x}}\widehat {\dot i} + \frac{\partial }{{\partial y}}\widehat {\dot J} + \frac{\partial }{{\partial z}}\hat k\)
\(\nabla \times \vec V = Curl\left( {\vec V} \right) = \left| {\begin{array}{*{20}{c}} i&{\dot J}&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{V_x}}&{{V_y}}&{{V_z}} \end{array}} \right|\;\)
Curl of any vector is zero, which means the flow is not rotating inside the domain or the flow is imitational flow.
\(Curl.\vec V = 0,\) i.e Flow is irrotational
\(\nabla .\vec V = 0\) - Incompressible continuity equation
The general continuity equation for any flow field is
\(\frac{{\partial \rho }}{{\partial t}} + \nabla \left( {\rho .\vec V} \right) = 0\)
A fluid is said to be incompressible if the density of the fluid is not varying with time or space i.e. constant.
So, \(\frac{{\partial \rho }}{{\partial t}} = 0\;\& \;\nabla \left( {\rho .\vec V} \right) = 0\)
\(Or,\;\rho \left( {\nabla .\vec V} \right) = 0\;\)
Or, \(\nabla \vec{.V} = 0\), incompressible continuity equation.
\(\frac{{D\vec V}}{{D\rho }} = 0\;\) - Zero acceleration of the fluid particle
D – Total derivative
\(\vec V = {\vec V_x}\widehat {\dot i} + {\vec V_y}\hat J + {\vec V_z}\hat k\)
\(\frac{{D\vec V}}{{Dt}} = \frac{{D{V_x}\;}}{{Dt}}\widehat {\dot i} + \frac{{D{V_y}}}{{Dt}}\widehat {\dot J} + \frac{{D{V_z}}}{{Dt}}\hat k\)
\(\vec a = {a_x}\widehat {\dot i} + {a_y}\widehat {\dot J} + {a_z}J\hat k\) – total acceleration
\({a_x} = \frac{{D{V_x}}}{{Dt}} = \frac{{\partial u}}{{\partial t}} + \frac{{\partial u}}{{\partial x}}.\frac{{\partial x}}{{\partial t}} + \frac{{\partial u}}{{\partial y}}\;\frac{{\partial y}}{{\partial t}} + \frac{{\partial u}}{{\partial z}}\frac{{\partial z}}{{\partial t}}\;\;\)
\( = \underbrace {\frac{{\partial u}}{{\partial t}}}_{Local\,\,acceleration} + \,\,\underbrace {u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} + w\frac{{\partial u}}{{\partial z}}}_{connective\,\,acceleration}\)
\(\frac{{D\vec V}}{{Dt}}\) – Rate of change of velocity w.r.t. time is acceleration.
As per the given conditions in the question
\(\frac{{D\vec V}}{{Dt}} = 0,\) So total acceleration i.e. local acceleration + Connective acceleration of fluid-particle is zero.
\(\frac{{\partial \vec V}}{{\partial t}} = 0\) – Steady flow.
Steady flow is defined as that type of flow in which the fluid characteristics like velocity, pressure, density, etc. at a point do not change w.r.t. time.
Here velocity is not varying with time i.e. \(\frac{{\partial \vec V}}{{\partial t}} = 0\)
So, the flow is steady flow.
