$$1\,mile = 5280\,ft;1\,bhp = 550\,ft.lb/s$$ $$c_{t} = \frac{c\left ( v_{\infty} \right )}{\eta _{p}} = \frac{0.375\,\frac{lb}{bhp.h}\left ( 200\,mi/h \right )}{0.85} = \frac{0.375\frac{lb}{\frac{550\,ft\,.lb\times 3600\,s}{s}}\left ( 200\,\times \frac{5280\,ft}{h} \right )}{0.85}$$

$$\Rightarrow c_{t} = 0.235/h$$

$$P_{eq,shaft} = P_{shaft} + P_{Thrust}$$ $$P_{shaft} = 4910$$

$$P_{Thrust} = \frac{T_{j}V_{\infty}}{\eta _{p}} = \frac{250\left ( 500 \right )}{ 0.9} = 138888.89$$

$$\Rightarrow P_{eq,shaft} = 4910 + 138888.89 = 143798.89$$

$$\frac{T}{T_{V=0}} = 0.369M_{\infty}^{-0.305}$$ $$\Rightarrow T_{3\,km} = 50000\left \{ 0.369\left ( 0.6 \right )^{-0.305} \right \} = 21560.57\,lb$$

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

$$time = \frac{W_{f}}{TSFC \times Thrust}$$

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

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

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

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

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

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

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

(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 $M_{\infty} = 0.7$.  $$C_{L} = \frac{C_{L0}}{\sqrt{1-M_{\infty}^{2}}}=\frac{0.493}{1-0.7^{2}}=0.6903$$