Find the magnitude of velocity at point \((2,3)\) for a two-dimensional flow field whose stream function is given by \(\psi = 2x^{2}-y^{2}\).

Find the magnitude of velocity at point \((2,3)\) for a two-dimensional flow field whose stream function is given by \(\psi = 2x^{2}-y^{2}\).

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    The stream function for the flow field is \[\psi =2x^{2}-y^{2}\]Velocity component \(u\) and \(v\) of the two-dimensional flow field is \[u=\frac{\partial \psi }{\partial y},v=-\frac{\partial \psi }{\partial x}\]Therefore,\[u=\frac{\partial \psi }{\partial y}=\frac{\partial }{\partial y}\left ( 2x^{2}-y^{2} \right )
    =\frac{\partial }{\partial y}\left ( 2x^{2} \right )-\frac{\partial }{\partial y}\left ( y^{2} \right )=0-2y=-2y\]\[v=-\frac{\partial \psi }{\partial x}
    =-\frac{\partial }{\partial x}\left ( 2x^{2}-y^{2} \right )
    =-\frac{\partial }{\partial x}\left ( 2x^{2} \right )+\frac{\partial }{\partial x}\left ( y^{2} \right )=-4x+0=-4x\] Therefore, at point \((2,3)\),\[u=-2y=-2\left ( 3 \right )=-6\\v=-4x=-4\left ( 2 \right )=-8\]Velocity, \(V=-6i-8j\)
    Magnitude of velocity = \[\sqrt{u^{2}+v^{2}}=\sqrt{\left ( -6 \right )^{2}+\left ( -8\right)^{2}}=\sqrt{36+64}=10\]

    Answered on 7th May 2021.
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