Starting with the definition of circulation, derive Kelvin’s circulation theorem.

Starting with the definition of circulation, derive Kelvin’s circulation theorem.

Worldtech Asked on 9th November 2019 in Aerodynamics.
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    Kelvin’s circulation theorem tells  that the time rate of change of circulation around a closed curve consisting of the same fluid elements is zero,that is \(\frac{D\tau}{Dt}=0\).

    Circulation is defined as \[\tau=\oint_{c}\overrightarrow{V}\cdot \overrightarrow{ds}\]

    \[  \frac{D\tau}{Dt}=\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}+\oint_{c}\overrightarrow{V}\cdot \overrightarrow{ds} \]

    \[\frac{D\overrightarrow{ds}}{Dt}=\overrightarrow{dV}\]

    \[\oint_{c}\overrightarrow{v}\cdot \overrightarrow{dV}=\oint_{c}d\left ( \frac{V^{2}}{2} \right )=0\]

    \[\frac{D\overrightarrow{V}}{dt}=-\frac{1}{\rho}\nabla p\]

    \[\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}=-\oint_{c}\frac{1}{\rho}\nabla p\cdot \overrightarrow{ds}=-\oint_{c}\frac{dp}{\rho}\]

    when \(\rho\)=constant or\( \rho\)=\(\rho(p)\),then

    \[-\oint_{c}\frac{dp}{\rho }=0\]

    Therefore \[\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}=0\]

    or \[\frac{D\tau}{Dt}=0\]

    techAir Answered on 9th November 2019.
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