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

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

Asked on 9th November 2019 in

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 $\frac{D\tau}{Dt}=0$