§AERE 261 Homework 4

Sep 25, 2025
Given:
Swing=22.9m2 S_{wing}=22.9m^2
h=7000ft h=7000ft
μ=1.7∗10−5kgm∗s \mu=1.7*10^{-5}\frac{kg}{m*s}
ρ=0.993kg/m3 \rho=0.993kg/m^3
V∞=89.5m/s V_\infty=89.5m/s
M=0.27 M=0.27

§1. (a)

Given:
S=36ft2=3.345m2 S=36ft^2=3.345m^2
t/c=0.08 t/c=0.08
Λ=20° \Lambda=20\degree
Q=1.05 Q=1.05
x/c=0.30 x/c=0.30
c=3.45ft=1.052m c=3.45ft=1.052m
Dependencies of CD,0C_{D,0}:
l=c=1.052m l=c=1.052m
R=ρV∞lμ=0.993kg/m3∗89.5m/s∗1.052m1.7∗10−5kgm∗s=5.5×106 R=\frac{\rho V_\infty l}{\mu}=\frac{0.993kg/m^3 * 89.5m/s * 1.052m}{1.7*10^{-5}\frac{kg}{m*s}}=5.5×10^6
Cf=0.455(log⁡10R)2.58(1+0.144M2)0.65=0.455(log⁡10(5.5×106))2.58(1+0.144(0.27)2)0.65=0.003289 C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(5.5×10^6))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.003289
FF=[1+0.6(x/c)m(tc)+100(tc)4][1.34M0.18(cosΛm)0.28] FF=\left[1+\frac{0.6}{(x/c)_m}\left(\frac{t}{c}\right)+100\left(\frac{t}{c}\right)^4\right]\left[1.34M^{0.18}(cos\Lambda_m)^{0.28}\right]
FF=[1+0.60.30(0.08)+100(0.08)4][1.34(0.27)0.18(cos(20°))0.28]=1.211 FF=\left[1+\frac{0.6}{0.30}\left(0.08\right)+100\left(0.08\right)^4\right]\left[1.34(0.27)^{0.18}(cos(20\degree))^{0.28}\right]=1.211
Swet=[1.977+0.52(t/c)]Sexposed=[1.977+0.52(0.08)]∗3.345m2=6.752m2 S_{wet}=[1.977+0.52(t/c)]S_{exposed}=[1.977+0.52(0.08)]*3.345m^2=6.752m^2
And finally CD,0C_{D,0}:
CD,0=Cf∗FF∗Q∗SwetSwing=0.003289∗1.211∗1.05∗6.752m222.9m2=0.001233 C_{D,0}=C_f*FF*Q*\frac{S_{wet}}{S_{wing}}=0.003289*1.211*1.05*\frac{6.752m^2}{22.9m^2}=\boxed{0.001233}

§1. (b)

Given:
Q=1.3 Q=1.3
Swet=0.88m2 S_{wet}=0.88m^2
Dependencies of CD,0C_{D,0}:
l=0.9m l=0.9m
R=ρV∞lμ=0.993kg/m3∗89.5m/s∗0.9m1.7∗10−5kgm∗s=4.705×106 R=\frac{\rho V_\infty l}{\mu}=\frac{0.993kg/m^3 * 89.5m/s * 0.9m}{1.7*10^{-5}\frac{kg}{m*s}}=4.705×10^6
Cf=0.455(log⁡10R)2.58(1+0.144M2)0.65=0.455(log⁡10(4.705×106))2.58(1+0.144(0.27)2)0.65=0.003376 C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(4.705×10^6))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.003376
Amax=π(0.5∗0.3m)2=0.071m2 A_{max}=\pi(0.5*0.3m)^2=0.071m^2
f=l(4/π)Amax=0.9m(4/π)∗0.071m2=2.993 f=\frac{l}{\sqrt{(4/\pi)A_{max}}}=\frac{0.9m}{\sqrt{(4/\pi)*0.071m^2}}=2.993
FF=1+0.35f=1+0.352.993=1.117 FF=1+\frac{0.35}{f}=1+\frac{0.35}{2.993}=1.117
And finally CD,0C_{D,0}:
CD,0=Cf∗FF∗Q∗SwetSwing=0.003376∗1.117∗1.3∗0.88m222.9m2=1.884×10−4 C_{D,0}=C_f*FF*Q*\frac{S_{wet}}{S_{wing}}=0.003376*1.117*1.3*\frac{0.88m^2}{22.9m^2}=\boxed{1.884×10^{-4}}

§1. (c)

I will be doing calculations for just a single landing gear and will introduce the coefficient fo 33 in the build up later summation.Given:
diameter=14in=0.356m \text{diameter}=14in=0.356m
width=7in=0.178m \text{width}=7in=0.178m
CD,wheel=0.25 C_{D,wheel}=0.25
Qwheel=1.2 Q_{wheel}=1.2
Dependency of CD,0C_{D,0}:
Sfrontal=diameter∗width=0.356m∗0.178m=0.0634m2 S_{frontal}=\text{diameter}*\text{width}=0.356m*0.178m=0.0634m^2
And finally CD,0C_{D,0}:
CD,0=CD∗SfrontalSwing∗Q=0.25∗0.0634m222.9m2∗1.2=8.306×10−4 C_{D,0}=C_D*\frac{S_{frontal}}{S_{wing}}*Q=0.25*\frac{0.0634m^2}{22.9m^2}*1.2=\boxed{8.306×10^{-4}}

§1. (d)

Given:
CD,0,vertical tail=0.0006 C_{D,0,\text{vertical tail}}=0.0006
CD,0,fuselage=0.0047 C_{D,0,\text{fuselage}}=0.0047
CD,0,wing=0.0076 C_{D,0,\text{wing}}=0.0076
The final CD,0:C_{D,0}:
CD,0=CD,0,vertical tail+CD,0,fuselage+CD,0,wing+CD,0,horizontal tail+CD,0,weather radar+3CD,0,landing gear C_{D,0}=C_{D,0,\text{vertical tail}}+C_{D,0,\text{fuselage}}+C_{D,0,\text{wing}}+C_{D,0,\text{horizontal tail}}+C_{D,0,\text{weather radar}}+3C_{D,0,\text{landing gear}}
CD,0=0.0006+0.0047+0.0076+0.001233+1.884×10−4+3∗8.306×10−4=0.0168 C_{D,0}=0.0006+0.0047+0.0076+0.001233+1.884×10^{-4}+3*8.306×10^{-4}=\boxed{0.0168}

§2.

Given:
m=1680kg m=1680kg
V∞=89.5m/s V_\infty=89.5m/s
h=7000ft h=7000ft
K=0.056 K=0.056
Thrust required in steady level flight:
q∞=12ρ∞V∞2=12∗0.993kg/m3∗(89.5m/s)2=3977Pa q_\infty=\frac{1}{2}\rho_\infty V_\infty^2=\frac{1}{2}*0.993kg/m^3 * (89.5m/s)^2=3977Pa
g=9.81m/s2 g=9.81m/s^2
W=mg=1680kg∗9.81m/s2=16480N W=mg=1680kg*9.81m/s^2=16480N
TR=q∞SCD,0+2KW2q∞S=3977Pa∗22.9m2∗0.0168+2∗0.056∗(16480N)23977Pa∗22.9m2=1864N T_R=q_\infty S C_{D,0}+\frac{2KW^2}{q_\infty S}=3977Pa * 22.9m^2 * 0.0168+\frac{2 * 0.056 * (16480N)^2}{3977Pa * 22.9m^2}=\boxed{1864N}