1
$\begingroup$

enter image description here

I’m following Anderson’s Fundamentals Of Aerodynamics. However, the proof for the above theorem is not provided. Every time I try it, things get really lengthy and cumbersome. Is there some nice and short method to getting around to prove the time rate of change of circulation over a chose closed loop 0 in an inviscid, incompressible flow ?

$\endgroup$

1 Answer 1

1
$\begingroup$

The circulation is given by

$$\Gamma = \oint_{C(t)}\boldsymbol{u}d\boldsymbol{r}.$$

Now, take the material derivative of this expression (remember to use the product rule because our contour is time-dependent)

$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\dfrac{D \boldsymbol{u}}{Dt}d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\left[\dfrac{D\boldsymbol{r}}{Dt}\right].$$

Now, we assume that the flow in inviscid and all volume forces are conservative (have a potential $\Psi$). Such that we can use Euler's equation for the first integral

$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\left[-\dfrac{\nabla p}{\rho}-\nabla \Psi \right]d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\boldsymbol{u}.$$

The term on the right-hand side is integrable and has an anti-derivative of $\boldsymbol{u}^2$. As we are integrating on a circle (starting point = endpoint) it must be zero. If we assume that $\rho(p)$, which is called barotropic, then we can conclude that the closed contour integral on the left-side is also zero. Note, that the closed contour integral for conservative fields $-\nabla \Psi$ is zero, also called path independence. Hence, we conclude

$$\dfrac{D \Gamma}{Dt} =0.$$

This means that for inviscid, barotropic flow's and conservative volume forces the circulation $\Gamma$ is conserved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.