Chapter 8, Problem 8.87P

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To determine

(a)

To estimate: the chord length.

Answer to Problem 8.87P

Chord length is $C\approx 0.219m$

Explanation of Solution

Given:

For fresh water, take $\rho =998kg/m^3, \text{V}=7\text{m}/\text{s}$

$,\alpha =2.5°$

Calculation:

$Lift=C_L\frac{\rho }{2}{V}^{2}bC=\frac{2\pi \sin \left[\frac{2.5}{57.3}+2\left(0.02\right)\right]}{1+\left(\frac{2}{8}\right)}\left(\frac{\frac{998kg}{m^3}}{2}\right){(7m/s)}^2\left(8C\right)C$

$C^2=\left(400kg\right)\left(9.81m/s^2\right)$

$C\approx 0.219m$.

To determine

(b)

The power required if $C_{D\infty }=0.01$.

Answer to Problem 8.87P

$\text{Power required}\approx 1.5h_p$

Explanation of Solution

Calculation:

At 7 m/s the power required depend upon the drag, with C_L=0.420

$C_D=C_{D\infty }+\frac{C_L^2}{\pi \left(AR\right)}=0.01+\frac{{(0.420)}^2}{\pi \left(8\right)}=0.0170$

$F=C_D\frac{\rho }{2}{V}^{2}bC=\left(0.0170\right)\left(\frac{\frac{998kg}{m^3}}{2}\right){(7m/s)}^2\left(8\times 0.219m\right)\left(0.219m\right)$

$F=159N$

$Power=FV=\left(159 N\right)\left(\frac{7m}{s}\right)=1120W=1.5h_p$.

To determine

(c)

To find: the top speed.

Answer to Problem 8.87P

Top speed $V_{max}=19.8m/s\approx 44mi/h$

Explanation of Solution

Given:

$C_{D\infty }=0.01, C=0.219m, b=1.75m, AR=8, Weight=3924N,$

$power=20h_p=14914W,\alpha ZL=-2.29°,\rho =998kg/m^3$

Concept used:

Set up the power equation again with velocity unknown;

$power=FV=\left[\left(C_{D\infty }+\frac{C_L^2}{\pi \left(AR\right)}\right)\left(\frac{\rho }{2}\right)V^{2}bC\right]V$

$C_L=\frac{2\left(weight\right)}{\rho V^{2}bC}=\frac{2\pi \sin \left(\alpha -{\alpha }_{zL}\right)}{1+2/\left(AR\right)}$

Put the value in above equations then solve

$C_L=0.0526, C_D=0.0101, \alpha =-1.69°$

$V_{max}=19.8m/s\approx 44mi/h$.