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=-2yv=-\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 )=-8Velocity, V=-6i-8j
    Magnitude of velocity = \sqrt{u^{2}+v^{2}}=\sqrt{\left ( -6 \right )^{2}+\left ( -8\right)^{2}}=\sqrt{36+64}=10

    techAir Answered on 7th May 2021.
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