Question 15.21: A circular cylinder of diameter 2 m and length 10 m rotates ......

Given data:
Diameter of cylinder                              D= 2 m

Radius of cylinder                                  R= 1 m

Length of cylinder                                L = 10 m

Velocity of air                                        U_{\infty}=50 \mathrm{~m} / \mathrm{s}

Lift force per m length \frac{F_L}{L}=5 \mathrm{~kN}=5000 \mathrm{~N}

Density of air ρ= 1.22 kg/m³

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

F_L=\rho L U_{\infty} \Gamma

or                              \frac{F_L}{L}=\rho U_{\infty} \Gamma

or                                5000 =1.22 x 50 x F

or                                \Gamma=\frac{5000}{1.22 \times 50}=81.967 \mathrm{~m}^2 / \mathrm{s}

Let the speed of rotation corresponding to \Gamma=81.967 \mathrm{~m}^2 / \mathrm{s} \text { be } \nu_\theta . From Eq. (15.22), we have

\nu_\theta=\frac{\Gamma}{2 \pi R}=\frac{81.967}{2 \pi \times 1}=13.045 \mathrm{~m} / \mathrm{s}

Rotational speed is given by

N=\frac{60 \nu_\theta}{\pi D}=\frac{60 \times 13.045}{\pi \times 2}=124.57 \mathrm{rpm}                       \left[\because \nu_\theta=\frac{\pi D N}{60}\right]

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

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

=-\frac{81.967}{4 \pi \times 50 \times 1}=-0.1304=-\sin 7.49^{\circ}

= sin (180° + 7.49°) and sin (360° — 7.49)
= sin 187.49° and sin 352.51°

or                θ = 187.49° and 352.51°