Correct Answer - Option 3 : Its streamlines are given by x = y .
Concept:
Continuity equation for a two – dimensional flow of an incompressible fluid is:
\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)
For the irrotational flow field:
\({\omega _z} = \frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0\)
Streamline equation:
\(\frac{{dx}}{u} = \frac{{dy}}{v}\)
Calculation:
\(\vec u = A\left( {x\hat i - y\hat j} \right)\)
u = Ax, v = -Ay
Continuity equation for a two – dimensional flow of an incompressible fluid is:
\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)
\(\frac{\partial }{{\partial x}}\left( {Ax} \right) + \frac{\partial }{{\partial y}}\left( { - Ay} \right) = A - A = 0;Satisfied\)
When x → 0 and y → ∞ ⇒
\(\vec u = - Ay\hat j;unidirectional\)
For the irrotational flow field:
\({\omega _z} = \frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0\)
\(\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial x}}\left( { - Ay} \right) - \frac{\partial }{{\partial y}}\left( {Ax} \right) = 0;Irrotational\)
Streamline equation:
\(\frac{{dx}}{u} = \frac{{dy}}{v}\)
\(\Rightarrow \frac{{dx}}{{Ax}} = \frac{{dy}}{{ - Ay}} \Rightarrow \frac{{dx}}{x} = \frac{{dy}}{{ - y}}\)
\(\Rightarrow \ln x = - \ln y + \ln C \Rightarrow \ln x + \ln y = \ln C\)
\(\Rightarrow \ln xy = \ln C \Rightarrow xy = c\)
So, x = y is not the equation of streamline.