Potential Flow Theory: Circulation and the Kutta-Jukowski Theorem

[Master1022]

Hi,

I just had a quick question about conventions in potential flow theory:
Question: What is the convention for $\Gamma$ for the streamline $\Psi = \frac{\Gamma}{2\pi} ln(\frac{r}{a} )$ and how can we interpret the Kutta-Jukowski Theorem $Lift = - \rho U \Gamma$?

Approach:
For the first part, am I correct in thinking that $\Gamma$ indicates a clockwise circulation? This is supported by the fact that
$$ v_{\theta} = - \frac{\partial \Psi}{\partial \theta} = - \frac{\Gamma}{2 \pi r} $$
where $\theta$ is defined to be defined as anti-clockwise from the +ve x axis. I was confused as I had seen diagrams on the internet where they have suggested that $+ \Gamma$ indicates an anticlockwise circulation. Another quick question I have is: is gamma the circulation of the fluid or the cylinder? If it's the former, does that mean that if the cylinder is rotating (e.g. CW) we take the circulation as being of equal 'strength' in the CCW direction?

For the second part, I have done the integration to get the result that $Lift = - \rho U \Gamma$ where Lift has been defined in the +ve y-direction. However, I don't think I really understand what this means. So this formula predicts that there will be a positive lift in the y-direction when there is an anticlockwise circulation?

Any help would be greatly appreciated

 

[Arjan82]

The stream function for circulatory flow is actually:
$$
\Psi = -\frac{\Gamma}{2 \pi} \ln \left( \frac{r}{a}\right)
$$
So with a minus sign. This means that $v_\theta = \frac{\Gamma}{2\pi r}$ and thus a positive $\Gamma$ means counter clockwise flow.

The stream function is valid for a flow around a cylinder of radius $a$. On this cylinder the flow direction and velocity is equal to the velocity of the edge of the cylinder. Therefore a positive $\Gamma$ denotes a flow around a cylinder spinning also in counter clockwise direction. Note that $\Gamma = 2\omega (\pi r^2)$ where $\omega$ is the rotational velocity of the cylinder. So $\Gamma$ and the rotation of the cylinder are directly related, not inversely as you suggest.

For your last question, a CW rotating cylinder with flow from left to right generates an upward force on the cylinder.

 

[Master1022]

The stream function for circulatory flow is actually:
$$
\Psi = -\frac{\Gamma}{2 \pi} \ln \left( \frac{r}{a}\right)
$$
So with a minus sign. This means that $v_\theta = \frac{\Gamma}{2\pi r}$ and thus a positive $\Gamma$ means counter clockwise flow.

The stream function is valid for a flow around a cylinder of radius $a$. On this cylinder the flow direction and velocity is equal to the velocity of the edge of the cylinder. Therefore a positive $\Gamma$ denotes a flow around a cylinder spinning also in counter clockwise direction. Note that $\Gamma = 2\omega (\pi r^2)$ where $\omega$ is the rotational velocity of the cylinder. So $\Gamma$ and the rotation of the cylinder are directly related, not inversely as you suggest.

For your last question, a CW rotating cylinder with flow from left to right generates an upward force on the cylinder.

Many thanks for the clarification @Arjan82 !

If we superimposed this circulation on a flow around a cylinder (with radius $a$) from left to right to get:
$$ \Psi = Ursin(\theta)\left( 1 - \frac{a^2}{r^2} \right) - \frac{\Gamma}{2\pi} ln \left( \frac{r}{a} \right) $$
Intuitively, I would expect $- \Gamma$ (pointing in the -ve CCW = CW direction) to push the stagnation points to the lower half of the cylinder. However, going through the mathematics, I end up with the condition for $v_{\theta} = 0$ occurs at:
$$ sin(\theta) = \frac{\Gamma}{4 \pi a U} $$
Is this technically a -ve value as $\Gamma$ is defined in the opposite direction to $\theta$? If that isn't the case, the a positive angle would suggest the stagnation points have been moved to the upper half of the cylinder (which doesn't agree with the diagram I have seen)

Thanks.

 

[Arjan82]

We need to be careful with definitions here. The stream function for a rotating cylinder in a horizontal flow is:
$$
\Psi = U\left(r-\frac{a^2}{r}\right)\sin(\theta) - \frac{\Gamma}{2\pi}\ln\left(\frac{r}{a}\right)
$$
as indeed you state, with U from left to right positive. This means that
$$
v_\theta = -\frac{\partial \Psi}{\partial r} = -U \sin(\theta)\left(1+\frac{a^2}{r^2}\right) + \frac{\Gamma}{2\pi r}
$$
And on a cylinder $r = a$, thus
$$
v_\theta = -2U \sin(\theta) + \frac{\Gamma}{2\pi a}
$$
At a stagnation point $v_\theta = 0$ such that
$$
sin(\theta) = \frac{\Gamma}{4 \pi a U}
$$
So, you have it correct. For the case when the flow is from left to right and the cylinder is rotating in clockwise direction, you have $\Gamma$ negative and $U$ positive. This means $\sin(\theta)$ must be negative as well, meaning the stagnation points are in the lower half as you would expect.

Note that in this case $U$ is defined positive for flow from left to right (in the positive x direction). Older text apparently have the habit to define $U$ (or sometimes called $V$) positive from right to left (so in the minus-x direction). Then the minus signs pop up in different places. This is where the confusion might come from.