§AERE 261 Homework 3

Sep 25, 2025
Hello, grader of my homework! In order to preserve my sanity, I am moving away from clunky Google Docs. I am now trying out Markdown and LaTeX\LaTeX with the GitHub flavor so my homeworks should be prettier for you and easier for me to write.

§Knowns

M∞=0.27M_{\infty}=0.27V∞=89.6m/sV_{\infty}=89.6m/sρ=0.993kg/m3\rho=0.993kg/m^3μ=1.7∗10−5kgm∗s\mu=1.7*10^{-5}\frac{kg}{m*s}b=12.4mb=12.4mS=22.9m2S=22.9m^2

§1. (a)

AR=b2S=(12.4m)222.9m2=6.71 AR=\frac{b^2}{S}=\frac{(12.4m)^2}{22.9m^2}=\boxed{6.71}

§1. (b)

More info that we got:αstall=17.5deg⁡\alpha_{\text{stall}}=17.5\degαL=0=−1.19deg⁡\alpha_{L=0}=-1.19\dega0=0.119deg⁡−1a_0=0.119\deg^{-1}e=0.92e=0.92I'm going to use the degree version of the formula to extend the 2D airfoil to a 3D wing.
a=a01+57.3deg⁡a0πeAR=0.119deg⁡−11+57.3deg⁡∗0.119deg⁡−1π∗0.92∗6.71=0.0880deg⁡−1 a=\frac{a_0}{1+\frac{57.3\deg a_0}{\pi e AR}}=\frac{0.119\deg^{-1}}{1+\frac{57.3\deg*0.119\deg^{-1}}{\pi*0.92*6.71}}=0.0880\deg^{-1}
Now for CLC_L
CL=a(α−αL=0)=0.0880deg⁡−1(α+1.19deg⁡) C_L=a(\alpha-\alpha_{L=0})=\boxed{0.0880\deg^{-1}(\alpha+1.19\deg)}

§1. (c)

α=αstall=17.5deg⁡ \alpha=\alpha_{\text{stall}}=17.5\deg
CL=0.0880deg⁡−1(17.5deg⁡+1.19deg⁡)=1.643 C_L=0.0880\deg^{-1}(17.5\deg+1.19\deg)=\boxed{1.643}

§1. (d)

Givenmfull=1680kgm_\text{full}=1680kgg=9.81m/s2g=9.81m/s^2Converting mass to force (or weight):
L=W=mfullg=1680kg∗9.81m/s2=16480N=16.48kN L=W=m_\text{full}g=1680kg*9.81m/s^2=16480N=16.48kN
Rewriting CLC_L in terms of LL
L=12ρ∞V∞2SCL L=\frac{1}{2}\rho_\infty V_\infty^2 S C_L
CL=2Lρ∞V∞2S=2∗16.48kN0.993kg/m3∗(89.6m/s)2∗22.9m2=0.1805 C_L=\frac{2L}{\rho_\infty V_\infty^2 S}=\frac{2*16.48kN}{0.993kg/m^3 * (89.6m/s)^2 * 22.9m^2}=\boxed{0.1805}

§1. (e)

Pulling this from up above:CL,stall=1.643C_{L,\text{stall}}=1.643L=16.48kNL=16.48kNSolving for V∞V_\infty with CLC_L equal to the stall value:
W=12ρ∞V∞2SCL,stall W=\frac{1}{2}\rho_\infty V_\infty^2 S C_{L,\text{stall}}
V∞=2Wρ∞SCL,stall=2∗16.48kN0.993kg/m3∗22.9m2∗1.643=29.7m/s V_\infty=\sqrt{\frac{2W}{\rho_\infty S C_{L,\text{stall}}}}=\sqrt{\frac{2*16.48kN}{0.993kg/m^3 * 22.9m^2 * 1.643}}=\boxed{29.7m/s}

§1. (f)

There is no wing sweep so Λ=0\Lambda=0And AR from above: AR=6.71AR=6.71
e0=1.78(1−0.045AR0.68)−0.64=1.78(1−0.045∗6.710.68)−0.64=0.8477 e_0=1.78(1-0.045AR^{0.68})-0.64=1.78(1-0.045*6.71^{0.68})-0.64=\boxed{0.8477}

§1. (g)

K=1piARe=1pi∗6.71∗0.8477=0.0560 K=\frac{1}{piARe}=\frac{1}{pi*6.71*0.8477}=0.0560
And now the drag coefficient:
CD=CD,0+KCL2 C_D=C_{D,0}+KC_L^2
From last homework: Cd,0=0.006C_{d,0}=0.006
CD=0.006+0.0560∗(0.1805)2=0.00782 C_D=0.006+0.0560*(0.1805)^2=\boxed{0.00782}

§2. (a)

Atop=12(103.5in)(53.9in)+(207.5in)(53.9in)=13970in2=9.013m2 A_{top}=\frac{1}{2}(103.5in)(53.9in)+(207.5in)(53.9in)=13970in^2=9.013m^2
Aside=(39.4in)(69.3in)+(138.2in)(66in)+12(103.5in)(66in)=15267in2=9.8497m2 A_{side}=(39.4in)(69.3in)+(138.2in)(66in)+\frac{1}{2}(103.5in)(66in)=15267in^2=9.8497m^2
Swet,fuselage=3.4(Atop+Aside2)=3.4(9.013m2+9.8497m22)=32.1m2 S_{\text{wet,fuselage}}=3.4\left(\frac{A_{top}+A_{side}}{2}\right)=3.4\left(\frac{9.013m^2+9.8497m^2}{2}\right)=\boxed{32.1 m^2}

§2. (b)

GivenAmax=2.22m2A_{\text{{max}}}=2.22m^2l=7.9ml=7.9mSwet=32.26m2S_\text{wet}=32.26 m^2 (value given in the question differs a negligible amount from my calculations so I am going with the one provided in the question)t/c=0.15t/c=0.15M=0.27M=0.27Swing=22.9m2S_\text{wing}=22.9m^2Swet=32.1m2S_{\text{wet}}=32.1 m^2This is from the estimation documentQc=1.0Q_c=1.0
R=ρVlμ=0.993kg/m3∗89.6m/s∗7.9m1.7∗10−5kgm∗s=4.135×107 R=\frac{\rho V l}{\mu}=\frac{0.993kg/m^3 * 89.6m/s * 7.9m}{1.7*10^{-5}\frac{kg}{m*s}}=4.135×10^7
Cf=0.455(log⁡10R)2.58(1+0.144M2)0.65=0.455(log⁡10(4.135×107))2.58(1+0.144(0.27)2)0.65=0.0023996 C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(4.135×10^7))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.0023996
f=ld=l(4/π)Amax=7.9m(4/π)∗2.22m2=4.699 f=\frac{l}{d}=\frac{l}{\sqrt{(4/\pi)A_{\text{{max}}}}}=\frac{7.9m}{\sqrt{(4/\pi)*2.22m^2}}=4.699
FFfuselage=0.9+5f1.5+f400=0.9+54.6991.5+4.699400=1.403 FF_{\text{fuselage}}=0.9+\frac{5}{f^{1.5}}+\frac{f}{400}=0.9+\frac{5}{4.699^{1.5}}+\frac{4.699}{400}=1.403
CD,fuselage=CfFFQcSwetSwing=0.0023996∗1.403∗1.0∗32.1m222.9m2=0.004719 C_{\text{D,fuselage}}=C_f FF Q_c \frac{S_\text{wet}}{S_\text{wing}}=0.0023996 * 1.403 * 1.0* \frac{32.1 m^2}{22.9m^2}=\boxed{0.004719}

§2. (c)

t/c=0.15t/c=0.15(x/c)m=0.3(x/c)_m=0.3M=0.27M=0.27Λm=0\Lambda_m=0Qwing=1.0Q_\text{wing}=1.0Swing=22.9m2S_\text{wing}=22.9m^2R=9.67∗106R=9.67*10^6Assuming turbulent flow
Cf=0.455(log⁡10R)2.58(1+0.144M2)0.65=0.455(log⁡10(9.67∗106))2.58(1+0.144(0.27)2)0.65=0.002999 C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(9.67*10^6))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.002999
FF=[1+0.6(x/c)m(tc)+100(tc)4][1.34M0.18(cos⁡Λm)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]
=[1+0.60.3(0.15)+100(0.15)4][1.34(0.27)0.18(cos⁡0)0.28]=1.430 =\left[1+\frac{0.6}{0.3}\left(0.15\right)+100\left(0.15\right)^4\right]\left[1.34(0.27)^{0.18}(\cos0)^{0.28}\right]=1.430
Swet=[1.977+0.52(t/c)]Sexposed=[1.977+0.52(0.15)](22.9m2)=47.06m2 S_\text{wet}=[1.977+0.52(t/c)]S_\text{exposed}=[1.977+0.52(0.15)](22.9m^2)=47.06 m^2
CD,wing=CfFFQwingSwetSwing=0.002999∗1.430∗1.0∗47.06m222.9m2=0.008813 C_{\text{D,wing}}=C_f FF Q_\text{wing} \frac{S_\text{wet}}{S_\text{wing}}=0.002999 * 1.430 * 1.0 * \frac{47.06 m^2}{22.9m^2}=\boxed{0.008813}
CDC_D from xflr5 was 0.004750.00475 but our approximation is 0.0088130.008813 which almost double. Most of these equations are "hand-wavy" and extremely imprecise so this is somewhat expected. I have triple checked my numbers so I am unsure why they're so different.

§2. (d)

CD=CD,0+KCL2=CD,fuselage+CD,wing+KCL,wing2 C_D=C_{D,0}+KC_L^2=C_{D,fuselage}+C_{D,wing}+KC_{L,wing}^2
=0.004719+0.008813+0.0560CL2=0.0135+0.0560CL2 =0.004719+0.008813+0.0560C_L^2=\boxed{0.0135 + 0.0560C_L^2}