\[TSFC = \frac{W_{f}}{Thrust\times time}\]\[W_{f} = 1000 \times 6.7 = 6700\,lb\]

\[time = \frac{W_{f}}{TSFC \times Thrust}\]

\[\Rightarrow time = \frac{6700}{0.5\times 60000} = \frac{6700}{30000} = 0.223\,hrs\]

\[ T = \left ( \dot{m}_{air} + \dot{m}_{fuel} \right )V_{e}-\dot{m}_{air}V_{\infty }\]\[\dot{m}_{fuel} = 0.03\dot{m}_{air}\]

\[\Rightarrow T = \left ( \dot{m}_{air} + 0.03\dot{m}_{air} \right )V_{e}-\dot{m}_{air}V_{\infty }\]

\[\Rightarrow T = \dot{m}_{air}\left ( 1.03V_{e}-V\infty  \right )\]

\[\Rightarrow \dot{m}_{air} = \frac{T}{1.03V_{e} – V_{\infty}} = \frac{25000}{1.03\left ( 1700 \right )-220}=16.33\,slug/s\]

\[\dot{m}_{air} = P_{\infty }V_{\infty}A_{inlet}\]

\[A_{inlet} = \frac{\dot{m}_{air}}{ P_{\infty }V_{\infty}} = \frac{16.33}{0.002377\left ( 220 \right )} = 31.23\,ft^{2}\]

(a) \[\dot{m} = \rho AV = 0.0023769\left ( \frac{\pi}{4}\times3.19^{2}\right )\times 1000 = 18.996\, Slug/S = 611.178\, lbm/S\]

\[F_{Thrust} = \dot{m}V_{relative}\]

\[\Rightarrow 10000 = 611.178\times V_{relative}\]

\[V_{relative} = 16.362\, ft/S\]

(b) \[\eta _{p} = \frac{2}{1 + \frac{V_{e}}{V_{a}}}\]
\[V_{e} = V_{rel} + V_{a} = 16.362 + 1000 = 1016.362\, ft/S\]
\[\frac{V_{e}}{V_{a}} = \frac{1016.362}{1000} = 1.016\]
\[\Rightarrow \eta _{p} = \frac{2}{1 + 1.016} = 0.992\]

Low-speed lift coefficient, \(C_{L} = C_{N}\cos\alpha\) 

\[\alpha = 25^{\circ} = 0.4363\,rad\]

\[\frac{C_{N}}{\left ( S/d \right )^{2}} = 2\pi \left ( \frac{\alpha }{S/d} \right )+4.9\left ( \frac{\alpha }{S/d} \right )^{1.7}\]

\[\frac{\alpha }{S/d} = \frac{0.4363}{0.425} = 1.0266\]

\[\frac{C_{N}}{\left (0.425 \right )^{2}} = 2\pi \left (1.0266 \right )+4.9\left (1.0266 \right )^{1.7}\]

\[\Rightarrow C_N=2.09\]

\[\Rightarrow C_{L} = 2.09\left ( \cos 25^{\circ} \right )=1.89\]

(a) \[M_{\infty}=2; \alpha = 1.5^{\circ} = 0.02618\,rad\]

\[C_{L}=\frac{4\alpha }{\sqrt{M_{\infty}^{2}}-1}= \frac{4\left (0.02618  \right )}{2^{2}-1}=0.0605\]

(b) \[C_{L} = \frac{C_{l}}{\sqrt{1+\left ( \frac{a_{0}}{\pi\,AR} \right )^{2}}+\frac{a_{0}}{\pi AR}}=\frac{0.0605}{\sqrt{1+\left ( \frac{2\pi}{\pi 2.56} \right )^{2}}+\frac{2\pi}{\pi 2.56}}=0.02951\]

(c) \[AR = 2.6; TR = 1; \wedge = 60^{\circ}\]

      \[ \beta =\sqrt{M_{\infty}^{2}-1}=\sqrt{2^{2}-1}=1.732\]

     \[Tan \wedge = Tan 60^{\circ}=1.732\]

    \[AR\,Tan \wedge = 2.6\times 1.732 = 4.503\]

    \[C_{Na} = \frac{4.1}{1.732}=2.367\, per\,rad\,= 0.0413\,per\,deg\]

    \[\Rightarrow C_{L}  = C_{Na}\alpha= 0.0413\left ( 1.5 \right )=0.062\]

Using Helmold’s relation \(a_{0}=0.105/deg = 6.016/radian;\alpha _{L=0} = -2.2^{\circ}\)

Therefore,

\[a = \frac{a_{0}}{\sqrt{1+\left ( \frac{a_{0}}{\pi AR} \right )^{2}}+\left ( \frac{a_{0}}{\pi AR} \right )}=\frac{6.016}{\sqrt{1+\left ( \frac{6.016}{15 \pi} \right )^{2}}+\frac{6.016}{15\pi}} = 2.0757/ radian = 0.0362 /deg\]

Therefore,

\[C_{L} = 0.0362[5-(-2.2)]=0.261\]

On using Prandtl’s lifting line theory,

\[a =\frac{a_{0}}{1+\frac{a_{0}}{\pi AR}}= \frac{6.016}{1+1.2766}=2.643 /rad = 0.0461/deg\]

\[C_{L}=0.0461(7.2)=0.332\]

Prandtl’s lifting line theory gives a higher result. 

(a) Lift coefficient at low speeds(incompressible flow), \(a_{0} = 6.016/radian\)

\[ a = \frac{a_{0}}{1 + \frac{a_{0}}{\pi e_{1} AR}}\]\[a = \frac{6.016}{1+\frac{6.016}{\pi\left ( 0.90 \right )4}}= 3.927/radian = 0.0685/degree\]\[C_{L0} = a\left ( \alpha -\alpha _{L=0} \right )=0.0685\left ( 5-\left ( -2.2^{\circ} \right ) \right )=0.493\](b) Lift coefficient for \(M_{\infty} = 0.7\). \[C_{L} = \frac{C_{L0}}{\sqrt{1-M_{\infty}^{2}}}=\frac{0.493}{1-0.7^{2}}=0.6903\]

For NACA 2412 angle of attack at 4°, coefficient of lift is  nearly 0.65 and coefficient of drag is nearly 0.04. 

Lift per unit span is 

\[{L}’ = \frac{1}{2}\rho_{\infty}V^{2}Cc_{l}\]

\[{L}’ = \frac{1}{2}\left ( 1.225 \right )\left ( 30 \right )^{2}1.5\left ( 0.65 \right ) = 537.5N/m\]

Drag per unit span is 

\[{D}’ = \frac{1}{2}\rho_{\infty}V^{2}Cc_{d}\]

\[{D}’ = \frac{1}{2}\left ( 1.225 \right )\left ( 30 \right )^{2}1.5\left ( 0.04 \right ) = 33.075N/m\]

 For steady, level flight, the weight is equal to the lift. 

\[C_{L} = \frac{L}{q_{\infty}S} = \frac{W}{q_{\infty}S}\]

\[  V_{\infty} = 160\left ( \frac{88}{60} \right )=234.7\,ft/s\]

\[ q_{\infty} = \frac{1}{2}\rho_{\infty} V_{\infty}^{2} = \frac{1}{2}\left ( 0.002377 \right )\left ( 234.7 \right )^{2} = 65.45\,lb/ft^{2}\]Therefore, 

\[ C_{L} = \frac{W}{q_{\infty}S}= \frac{506000}{\left ( 65.45 \right )\left (4605  \right )} = 0.362\]

Moment arm =\( 0.3c – 0.25c = 0.05c\)

\[M_{c/4} = -(0.05c)(200) = -10c = -10(5) = -50 ft-lb\, \textrm {per unit span}\]

The moment about the leading edge point is given by the lift acting through the center of pressure, with a moment arm \(0.3c\),

\[M_{LE} = -(0.3)(200) = -60c = -60(5) = -300 ft-lb\, \textrm{per unit span}\]