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

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

      1 Answers

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

      Answered by techAir on 9th November 2019..

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