Air at \(25^{\circ} \mathrm{C}\) flows normal to the axis of an infinitely long cylinder of 1.0 -m radius. The approach velocity is \(130 \mathrm{km} / \mathrm{hr}\). Find the rotational velocity of the cylinder so that a single stagnation point occurs on the lower-most part of the cylinder. Assume potential flow.

Short Answer

Expert verified
The rotational velocity of the cylinder in rad/s can be obtained by calculating the conversion of velocity from km/hr to m/s and solving the equation of velocity potential for a rotating and translating cylinder in a fluid stream.

Step by step solution

01

Convert Parameter Units

The given approach velocity is in km/hr. However, for the calculations we need to convert it into m/s since the radius of the cylinder is given in meters. This conversion is done using the formula \[ v = \frac{130 \times 1000}{60 \times 60} \]
02

Understand the Conditions

In order for a stagnation point to appear at the lower-most part of the cylinder, both the velocities due to rotation and translational motion must cancel out each other. The direction of velocities due to translational motion and rotation are opposite at the lower-most point of the cylinder.
03

Apply the Velocity Potential Formula

The formula of velocity potential for a rotating and translating cylinder in a fluid stream is given by: \[ u_r = v - r\omega \cos(\theta) \] and \[ u_{\theta} = r\omega \sin(\theta) \], where \(\omega\) is the rotational speed, \(r\) is the radial distance (which is the radius of the cylinder), \(v\) is the translational velocity and \(\theta\) is the angle. For a stagnation point at the bottom of the cylinder (\(\theta = 180\degree = \pi\)), these become \[ u_r = v - r\omega \] and \[ u_{\theta} = 0 \]. By setting \(u_r=0\), we get the equation \(v = r\omega \). Solving this equation, we find the rotational speed \(\omega\) of the cylinder.
04

Calculate the Rotational Speed

Upon substituting the known values of r = 1.0 m and v = 36.11 m/s (from step 1) into the equation from step 3 \(v = r\omega \), we find the value of \(\omega \).

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