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Question 15.18: A cylinder 20 cm in diameter and 10 m long is made to turn 1......

A cylinder 20 cm in diameter and 10 m long is made to turn 1200 revolution per minute with its axis perpendicular in a stream of air having uniform velocity of 20 m/s. Find (a) the circulation, (b) the lift force experienced by the cylinder and (c) the position of stagnation points. Take density of air as 1.22 kg/m³.

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Given data:

Diameter of cylinder                                     D = 20 cm = 0.2 m

Radius of cylinder                                         R = 0.1 m

Speed of cylinder                                            N = 1200 rpm

Velocity of air                                                  U_{\infty}=20 \mathrm{~m} / \mathrm{s}

Density of air                                                    \rho=1.22 \mathrm{~kg} / \mathrm{m}^3

Tangential velocity of cylinder is given by

ν_\theta=\frac{\pi D N}{60}=\frac{\pi \times 0.2 \times 1200}{60}=12.57 \mathrm{~m} / \mathrm{s}

(a) The circulation is found from Eq. (15.22) as

ν_{\theta}={\frac{\Gamma}{2\pi R}}  (15.22)

\Gamma=2 \pi R ν_\theta=2 \pi \times 0.1 \times 12.57=7.898 \mathrm{~m}^2 / \mathrm{s}

(b) The lift force is given by Eq. (15.27) as

F_L=\rho L U_{\infty} \Gamma=1.22 \times 10 \times 20 \times 7.898=1927.1 \mathrm{~N}

(c) Position of stagnation points are obtained using Eq. (15.29) as

\sin \theta=-\frac{\Gamma}{4 \pi U_{\infty} R}

=-\frac{7.898}{4 \pi \times 20 \times 0.1}=-0.3142=-\sin 18.31^{\circ}

= sin (180° + 18.31°) and sin (360 – 18.31°)
= sin 198.31° and sin 341.69°

or                        θ = 198.31° and 341.69°

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