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Author: Admin | 2025-04-28
Tempted to guess that this is an irrotational flow. However the calculation shows that \[\begin{split}\underline{\omega}&= \underline{\nabla}\times\underline{u}=(\partial_y\underline{u}_z-\partial_z\underline{u}_y, \partial_z\underline{u}_x-\partial_x\underline{u}_z, \partial_x\underline{u}_y-\partial_y\underline{u}_x)\\&=(0,0,-\partial_y\underline{u}_x)=(0,0,-\alpha)\neq \underline{0}.\end{split} \tag{12.14}\] So there is a non-zero vorticity – the flow is not irrotational. To see this intuitively, imagine placing an extended object into the flow. If you want to do this in your bathtub, I recommend gluing two matches together to form a cross and place this on the water surface. If you now create a flow, this object may start to rotate. It will do so in the flow in this example because the upper end of the object finds itself in an area of faster flow while the lower end is in an area of slower flow, so the object will rotate clockwise. The evolution of vorticityTo derive an equation which governs the evolution of vorticity, we take the \(curl\) of Euler’s equation in the form given in Eq. 12.7. \[\underline{\nabla}\times\partial_t\underline{u}+\underline{\nabla}\times(\underline{\omega}\times\underline{u})=-\underline{\nabla}\times\underline{\nabla} H.\tag{12.15}\] Using the fact that \(curl\) of \(grad\) is zero, we obtain \[\partial_t\underline{\omega} + \underline{\nabla}\times(\underline{\omega}\times \underline{u}) = \underline{0}.\tag{12.16}\] We will now use the general vector calculus identity \[\underline{\nabla}\times(\underline{a}\times\underline{b})=(\underline{\nabla}\cdot\underline{b})\underline{a}+(\underline{b}\cdot\underline{\nabla})\underline{a}-(\underline{\nabla}\cdot\underline{a})\underline{b}-(\underline{a}\cdot\underline{\nabla})\underline{b}\tag{12.17}\] which holds for any two differentiable vector fields \(\underline{a}\) and \(\underline{b}\). Applying this with \(\underline{a}=\underline{\omega}\) and \(\underline{b}=\underline{u}\) gives \[\partial_t\underline{\omega} + (\underline{\nabla}\cdot\underline{u})\underline{\omega}+(\underline{u}\cdot\underline{\nabla})\underline{\omega}-(\underline{\nabla}\cdot\underline{\omega})\underline{u}-(\underline{\omega}\cdot\underline{\nabla})\underline{u}= \underline{0}.\tag{12.18}\] Using the incompressibility condition \(\underline{\nabla}\cdot\underline{u}=0\) and the fact that the \(div\) of a \(curl\) is zero and thus \(\underline{\nabla}\cdot\underline{\omega}=0\), the above simplifies to \[\partial_t\underline{\omega} + (\underline{u}\cdot\underline{\nabla})\underline{\omega} = (\underline{\omega}\cdot\underline{\nabla})\underline{u}.\tag{12.19}\] Using the definition of the material derivative we obtain the\[\frac{D\underline{\omega}}{Dt} = (\underline{\omega}\cdot\underline{\nabla})\underline{u}.\tag{12.20}\]Note that the pressure \(p\) does not appear in
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