Chapter 5, Problem 5.72P

To determine

The drag of a full-scale aircraft.

Answer to Problem 5.72P

The aircraft (full scale) has a drag of $44.11kN$.

Explanation of Solution

Given Information:

Model chord length, $L=0.27m$

Wing Area, $A=0.63m^2$

Velocity of full-scale aircraft, $v=550mi/h$

$=245.87m/s$

V, mi/h	50	75	100	125
Drag, N	15	32	53	80

Drag coefficient for 1^st case

$C_{D,1}=\frac{2F_1}{\rho v_1^{2}A}$

Where $\rho =$ Density of air at $20°C$

$A$ = Area of the wing

$v_1=50mi/h=22.35m/s$

$F_1=15N$

$A=0.63m^2$

$\therefore C_{D,1}=\frac{2\times 15}{1.2\times {22.35}^2\times 0.63}=0.0794$

Similarly, drag coefficient for 2^nd case

$C_{D,2}=\frac{2F_2}{\rho v_2^{2}A}$

$v_2=75mi/h=33.53m/s$

$F_2=32N$

$C_{D,2}=\frac{2\times 32}{1.2\times {33.53}^2\times 0.63}=0.0753$

Drag coefficient for 3^rd case

$C_{D,3}=\frac{2F_3}{\rho v_3^{2}A}$

$v_3=100mi/h=44.7m/s$

$F_3=53N$

$C_{D,3}=\frac{2\times 53}{1.2\times {44.7}^2\times 0.63}=0.0702$

Drag coefficient for 4^th case

$C_{D,4}=\frac{2F_4}{\rho v_4^{2}A}$

$v_4=125mi/h=55.88m/s$

$F_4=80N$

$C_{D,4}=\frac{2\times 80}{1.2\times {55.88}^2\times 0.63}=0.0678$

Now, calculate Reynolds number for 1^st case

${\mathrm{Re}}_1=\frac{\rho v_{1}L}{\mu }$

Where L = Length of chord $=0.27m$

Viscosity, $\mu =1.8E-05kg/m.s$

${\mathrm{Re}}_1=\frac{1.2\times 22.35\times 0.27}{1.8\times {10}^{-5}}=402300$

Reynolds number for 2^nd case

${\mathrm{Re}}_2=\frac{\rho v_{2}L}{\mu }$

${\mathrm{Re}}_2=\frac{1.2\times 33.53\times 0.27}{1.8\times {10}^{-5}}=603540$

Reynolds number for 3^rd case

${\mathrm{Re}}_3=\frac{\rho v_{3}L}{\mu }$

${\mathrm{Re}}_3=\frac{1.2\times 44.7\times 0.27}{1.8\times {10}^{-5}}=804600$

Reynolds number for 4^th case

${\mathrm{Re}}_4=\frac{\rho v_{4}L}{\mu }$

${\mathrm{Re}}_4=\frac{1.2\times 55.88\times 0.27}{1.8\times {10}^{-5}}=1005840$

Aircraft is flying at 32800 ft at velocity $v=550mi/h$

$=245.87m/s$ . At this altitude,

Density, ${\rho }_1=0.4125kg/m^3$

Viscosity, ${\mu }_1=1.5E-05kg/m.s$

Length (full scale) $L_f=12\times 0.27=3.24m$

Reynolds number (full scale)

${\mathrm{Re}}_f=\frac{\rho v{L}_f}{\mu }$

${\mathrm{Re}}_f=\frac{0.4125\times 245.87\times 3.24}{1.5\times {10}^{-5}}$

$=21907195$

$\log ({\mathrm{Re}}_f)=\log (21907195)=7.34$

From the graph the best fit curve equation is

$y=-0.719x-0.094$

Let $y=\log (C_D)$

$x=\log (\mathrm{Re})$

$\log (C_D)=-0.719\log (\mathrm{Re})-0.094$

We know, $\log (\mathrm{Re})=7.34$

$\log (C_D)=-0.719\times 7.34-0.094$

$\log (C_D)=-1.408$

$C_D={10}^{-1.408}=0.039$

Drag (full scale) is calculated by the below formula

$F=\frac{C_D\rho v^2{A}_f}{2}$

Here, full scale area of the wing $A_f={12}^2\times 0.63=90.72m^2$

$\therefore F=\frac{0.039\times 0.4125\times {245.87}^2\times 90.72}{2}$

$F=44.113kN$

Conclusion:

The aircraft (full scale) has a drag of $44.11kN$.