Chapter 8, Problem 8.54P

To determine

The wind force parallel and normal to the ship centreline and to estimate the power required to drive the motors.

Answer to Problem 8.54P

The wind force parallel and normal to the ship centreline are $6700lbf$ and $2700lbf$ respectively. The power required to drive the motors is $560hp$

Explanation of Solution

Given Information:

Flettner rotor ship length, $L=100ft$

Diameter of the rotor, $D=9ft$

Height of the rotor, $h=50ft$

Drag coefficient of water, $C_{D,boat}=0.006$

Area (Wetted), $A_{wetted}=3500f{t}^2$

Spin of rotor, $N=750r/\min$

Velocity of cross wind, $U_{\infty }=25ft/s$

Lift coefficient, $C_L=10$

Drag coefficient, $C_D=4$

Force on lift is given by:

$F_L=C_L\frac{\rho }{2}{U}_{\infty }^{2}DL$

$=10\left(\frac{0.00238}{2}\right){\left(25\right)}^2\left(9ft\right)\left(50ft\right)\left(2rotors\right)$

$F_L\approx 6700lbf(parallel)$

Force on drag is given by:

$F_D=C_D\frac{\rho }{2}{U}_{\infty }^{2}DL$

$=4\left(\frac{0.00238}{2}\right){\left(25\right)}^2\left(9ft\right)\left(50ft\right)\left(2rotors\right)$

$F_D\approx 2700lbf(normal)$

Assume,

$2\Pi R=2\Pi \left(4.5\right)$

$=28.3ft$ (Simulation of cylinder)

Shear stress is calculated by the relation:

${\tau }_w=C_f\frac{\rho }{2}{U}^2$

${\tau }_w\approx \frac{0.027}{{\mathrm{Re}}_L^{1/2}}\frac{\rho }{2}{\left(\Omega R\right)}^2$

Here,

$\Omega R=750\left(\frac{2\Pi }{2}\right)\left(4.5\right)$

$\Omega R=353\frac{ft}{s}$

${\mathrm{Re}}_L=\frac{0.00238\left(353\right)\left(28.3\right)}{3.71E-7}$

${\mathrm{Re}}_L=6.42E7$

$\therefore {\tau }_w\approx 0.308\frac{lbf}{f{t}^2}$

Torque is given by the relation:

$T={\tau }_w\left(\Pi DL\right)R$

$T=0.308\Pi \left(9\right)\left(50\right)\left(4.5\right)$

$T=1958ft-lbf$

Power(total) is calculated by the relation:

$P=T\Omega$

$P=\left(1958\right)\left(750\frac{2\Pi }{60}\right)\left(2rotors\right)/\left(550\frac{hp}{ft-lbf/s}\right)$

$P=\left(1958\right)\left(750\frac{2\Pi }{60}\right)\left(2rotors\right)/\left(550\frac{hp}{ft-lbf/s}\right)$

$P=560hp$

Conclusion:

The wind force parallel and normal to the ship centreline are = $6700lbf$ and $2700lbf$ respectively. The power required to drive the motors is = $560hp$.