Question 15.20: A circular cylinder of 50 cm diameter is rotated about its a......

Given data:

Diameter of cylinder                      D = 50 cm = 0.5 m

Radius of cylinder                          R = 0.25 m

Length of cylinder                          L = 10 m

Velocity of water                          U_{\infty}=6 \mathrm{~m} / \mathrm{s}

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

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

When both the stagnation points coincide, the two angles are equal and is θ = -90°. Thus,

\sin \left(-90^{\circ}\right)=-\frac{\Gamma}{4 \pi U_{\infty} R}

or                    \frac{\Gamma}{4 \pi U_{\infty} R}=1

or                    \Gamma=4 \pi U_{\infty} R=4 \pi \times 6 \times 0.25=18.85 \mathrm{~m}^2 / \mathrm{s}

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

F_{L}=\rho L U_{\infty}\Gamma  (15.27)

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

= 1000 x 1 x 6 x 18.85 = 113100 N/m= 113.1 kN/m