Jonathan T. answered • 10/29/23
10+ Years of Experience from Hundreds of Colleges and Universities!
(a) To locate the stagnation points around a rotating circular cylinder in a uniform stream, we need to consider the relative motion between the fluid and the cylinder. Stagnation points are the points where the fluid velocity relative to the cylinder becomes zero.
The velocity of the fluid far away from the cylinder is \(U_\infty = 10 \, \text{m/s}\), and the cylinder is rotating at \(90 \, \text{rpm}\). To find the stagnation points, we need to consider the superimposed velocities due to the rotation of the cylinder and the uniform stream.
Let's consider a point on the front face of the cylinder. The velocity at this point (\(V\)) is the vector sum of the uniform stream velocity (\(U_\infty\)) and the velocity due to the rotation of the cylinder (\(U_r\)).
1. The velocity \(U_\infty\) of the uniform stream is in the direction of the flow, which is horizontal.
2. The velocity \(U_r\) due to the rotation of the cylinder can be calculated as:
\[U_r = \omega \cdot r\]
where
- \(\omega\) is the angular velocity of the cylinder in radians per second. We need to convert \(90 \, \text{rpm}\) to radians per second:
\[\omega = \frac{90 \, \text{rpm}}{60} \cdot 2\pi \, \text{rad/s} = 3\pi \, \text{rad/s}\]
- \(r\) is the distance from the axis of rotation to the point on the front face of the cylinder. Since the diameter is \(0.4 \, \text{m}\), \(r = 0.2 \, \text{m}\).
Now, we can calculate \(U_r\):
\[U_r = 3\pi \cdot 0.2 \, \text{m/s} \approx 1.884 \, \text{m/s}\]
So, at the front face of the cylinder, the velocity \(V\) is the vector sum of \(U_\infty\) and \(U_r\):
\[V = U_\infty + U_r = 10 \, \text{m/s} + 1.884 \, \text{m/s} \approx 11.884 \, \text{m/s}\]
At the front face of the cylinder, the velocity is \(11.884 \, \text{m/s}\).
The stagnation point occurs when \(V = 0\). Therefore, to locate the stagnation point, we set \(V\) to zero:
\[0 = U_\infty + U_r\]
\[0 = 10 \, \text{m/s} + 1.884 \, \text{m/s} - V_{\text{stagnation}}\]
Solving for \(V_{\text{stagnation}}\):
\[V_{\text{stagnation}} = 10 \, \text{m/s} + 1.884 \, \text{m/s} = 11.884 \, \text{m/s}\]
So, the stagnation point is located at a velocity of \(11.884 \, \text{m/s}\) on the front face of the cylinder.
(b) To calculate the rotational speed when the stagnation point moves off the cylinder, we can use the concept of circulation. Circulation (\(\Gamma\)) is the line integral of the velocity around a closed path, and it is conserved in an inviscid, incompressible flow. When the stagnation point moves off the cylinder, the circulation remains constant.
The formula for circulation is:
\[\Gamma = \oint V \cdot dl\]
Where \(V\) is the velocity vector and \(dl\) is an infinitesimal element along a closed path. In our case, the closed path is a circle around the cylinder.
When the stagnation point is on the front face of the cylinder, the circulation is:
\[\Gamma_1 = \oint V_1 \cdot dl = 11.884 \, \text{m/s} \cdot 2\pi \cdot 0.2 \, \text{m} = 15.023 \, \text{m}^2/\text{s}\]
Now, when the stagnation point moves off the cylinder and reaches the top or bottom of the cylinder (where the velocity is the same as the free stream velocity \(U_\infty\)), the circulation remains constant:
\[\Gamma_2 = \oint V_2 \cdot dl = U_\infty \cdot 2\pi \cdot 0.2 \, \text{m} = 10 \, \text{m/s} \cdot 2\pi \cdot 0.2 \, \text{m} = 12.566 \, \text{m}^2/\text{s}\]
Since circulation is conserved, \(\Gamma_1 = \Gamma_2\). We can now find the rotational speed (\(U_r\)) when the stagnation point moves off the cylinder by equating the two circulations:
\[\Gamma_1 = \Gamma_2\]
\[15.023 \, \text{m}^2/\text{s} = 12.566 \, \text{m}^2/\text{s} + U_r \cdot 2\pi \cdot 0.2 \, \text{m}\]
Now, solve for \(U_r\):
$$ U_r \cdot 2\pi \cdot 0.2 \, \text{m} = 15.023 \, \text{m}^2/\text{s} - 12.566 \, \text{m}^2/\text{s} $$
\[U_r \cdot 2\pi \cdot 0.2 \, \text{m} = 2.457 \, \text{m}^2/\text{s}\]
\[U_r = \frac{2.457 \, \text{m}^2/\text{s}}{2\pi \cdot 0.2 \, \text{m}}\]
\[U_r \approx 1.949 \, \text{m/s}\]
So, the rotational speed when the stagnation point moves off the cylinder is approximately \(1.949 \, \text{m/s}\).
