RE: 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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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

techAir Answered on 9th November 2019.
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