Question

A submerged cylinder of diameter 1 m is rotating clockwise at 100 rpm, in a flow with a free stream velocity of 10 m/s. Assuming ideal flow, the number of stagnation points on the cylinder is

• A. 2
• B. 3
• C. 1
• D. 0

Hint:

For flow over a rotating cylinder, the key is the ratio of surface speed to free stream velocity. Just remember to compare \( v_t \) with \( 2U \): - \( v_t<2U \implies \) 2 stagnation points (on the cylinder). - \( v_t = 2U \implies \) 1 stagnation point (on the cylinder). - \( v_t>2U \implies \) 0 stagnation points (on the cylinder).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with the ideal (potential) flow of a fluid around a rotating cylinder, a classic case that demonstrates the Magnus effect. Stagnation points are points on the surface of the cylinder where the fluid velocity is zero. The number and location of these points depend on the ratio of the cylinder's surface speed to the free stream velocity.
Step 2: Key Formula or Approach:
The location of the stagnation points on the cylinder surface is given by the angle \( \theta \) (measured from the rear of the cylinder) where the tangential velocity is zero. The formula is: \[ \sin(\theta) = -\frac{\Gamma}{4\pi U R} \] where:
- \( \Gamma \) is the circulation, given by \( \Gamma = 2\pi R v_t = 2\pi R (R\omega) \). For a rotating cylinder, a simpler approach is to use the tangential velocity \(v_t\).
The circulation is \( \Gamma = 2\pi R^2 \omega \).
- \( U \) is the free stream velocity.
- \( R \) is the radius of the cylinder.
The condition for the number of stagnation points depends on the ratio of the cylinder's tangential surface speed, \( v_t = R\omega \), to the free stream velocity \(U\).
Let this ratio be \( \alpha = v_t/U \). The formula for the stagnation point location can be expressed as:
\[ \sin(\theta) = -\frac{2\pi R^2 \omega}{4\pi U R} = -\frac{R\omega}{2U} = -\frac{v_t}{2U} \] - If \( v_t / (2U)<1 \), there are two stagnation points on the cylinder surface.
- If \( v_t / (2U) = 1 \), there is one stagnation point at the bottom of the cylinder (\( \theta = -90^\circ \)).
- If \( v_t / (2U)>1 \), the stagnation point moves off the cylinder surface, and there are no stagnation points on the cylinder.
Step 3: Detailed Calculation:
Given values:
- Diameter \( D = 1 \) m, so radius \( R = 0.5 \) m.
- Rotational speed \( \omega = 100 \) rpm.
- Free stream velocity \( U = 10 \) m/s.
First, convert the rotational speed from rpm to rad/s:
\[ \omega = 100 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{200\pi}{60} = \frac{10\pi}{3} \approx 10.47 \text{ rad/s} \] Next, calculate the tangential surface speed of the cylinder:
\[ v_t = R\omega = 0.5 \text{ m} \times 10.47 \text{ rad/s} \approx 5.236 \text{ m/s} \] Now, calculate the critical ratio:
\[ \frac{v_t}{2U} = \frac{5.236}{2 \times 10} = \frac{5.236}{20} = 0.2618 \] Since the ratio \( \frac{v_t}{2U} = 0.2618<1 \), there are two stagnation points on the surface of the cylinder.
Step 4: Why This is Correct:
The calculation shows that the parameter determining the number of stagnation points, \( v_t / (2U) \), is less than 1. According to the theory of potential flow around a rotating cylinder, this condition corresponds to the existence of two distinct stagnation points located symmetrically on the lower half of the cylinder (for clockwise rotation).