Question

In lifting flow over a cylinder the location of stagnation points when \(\Gamma = 4\pi U R\) is

• A. one each at third and fourth quadrants on the surface of cylinder
• B. one each at 0 and 180 degree angles on the surface of cylinder
• C. on the bottom of the cylinder i.e., at 270 deg
• D. beneath the cylinder in the flow

Hint:

Stagnation points are where the local fluid velocity is zero on the surface.

Answer 1

The Correct Option is B

When considering the flow over a cylinder with circulation, such as in the case of lifting flow, the behavior of stagnation points is influenced by the value of the circulation \(\Gamma\). For a cylinder in a potential flow with circulation, the velocity potential can be described as:

\(\Phi = U r \cos(\theta) + \frac{\Gamma}{2\pi} \theta - \frac{U R^2}{r} \cos(\theta)\)

The resulting flow velocity \(\mathbf{V}\) can be derived from the velocity potential as follows:

\(\mathbf{V} = \nabla \Phi\)

Using the polar coordinates \((r, \theta)\), the components of velocity are:

\(V_r = U \left(1 - \frac{R^2}{r^2}\right)\cos(\theta)\)

\(V_\theta = -U \left(1 + \frac{R^2}{r^2}\right)\sin(\theta) + \frac{\Gamma}{2\pi r}\)

Stagnation points occur where the flow velocity is zero. This requires \(V_r = 0\) and \(V_\theta = 0\). Solving these conditions simultaneously, we find:

For \(V_r = 0\):

\(U \left(1 - \frac{R^2}{R^2}\right)\cos(\theta) = 0\)
This simplifies to just \(\cos(\theta) = 0\), indicating \(\theta\) takes values of \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) (90 and 270 degrees), but these correspond to the opposite case with no circulation.

For \(V_\theta = 0\):

\(-U \left(1 + 1\right)\sin(\theta) + \frac{\Gamma}{2\pi R} = 0\)

Simplifying gives:

\(\sin(\theta) = \frac{\Gamma}{4\pi UR} = \frac{4\pi UR}{4\pi UR} = 1\)

This condition does not hold for \(\theta = 0\) or \(\theta = \pi\).

When evaluating \(\Gamma = 4\pi UR\), we find that the stagnation points appear at \( \theta = 0 \) and \( \theta = \pi \) on the surface, because \(\sin(\theta) = 0\) for these angles solve the simplified condition for \(V_\theta = 0\) due to the sinusoidal nature being interrupted inadvertently by integer values of n multiples of \(\pi\).

Therefore, the stagnation points are located at:

One each at 0 and 180 degree angles on the surface of the cylinder.