604
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72
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114
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Asked on 14th December 2023 in Aeronautics.
\[\frac{dh}{dt} = 3,000 ft/min = \frac{3000}{60} = 50\,ft/s\]\[\frac{dh}{dt} = P_{s} – \frac{V_{\infty }}{g}A\]\[\Rightarrow A = -\frac{\left ( 50 – 120 \right )32.2}{500} = 4.508 \,ft/s^{2}\]
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Asked on 14th December 2023 in Aeronautics.
\(v = 620\left ( \frac{5280}{3600} \right ) = 909.333\,ft/s; g = 32.2\,ft/s^{2}\)
Energy height =
\[h+\frac{v^{2}}{2g}=35000 + \frac{909.333^{2}}{2\left ( 32.2 \right )}=47839.853\,ft\]
- 367 views
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Asked on 14th December 2023 in Aeronautics.
\[\frac{T_{A}}{\left ( T_{A} \right )_{0}} = \frac{\rho }{\rho _{0}}\]
\(\left(T_{A} \right )_{0}\) is the thrust available at sea level. \(\rho _{0}\) is the standard sea level density.
From the equation we can see that the thrust produced by the turbojet engine is proportional to density of air. At sea level density of air is maximum so the amount of thrust produced by the turbojet engine is maximum. As the altitude increases, density of air decreases, so thrust produced by the turbojet engine also decreases.
Since thrust produced is proportional to velocity of aircraft, so as altitude increases thrust decreases and so velocity also decreases. Maximum thrust occurs at sea level, so maximum velocity is also at sea level.
- 385 views
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Asked on 14th December 2023 in Aeronautics.
\[\left ( C_{L} \right )_{max} = 1.55\,\rightarrow \mathrm{NACA}\,64-212\]\[\left ( C_{L} \right )_{max} = 1.52\,\rightarrow \mathrm{NACA}\,64-218\]
With plain flap deflection, \[\left ( C_{L} \right )_{max} = 1.72005\]
\(W = 960\,lb; S = 37.8\,ft^{2}\) \[ V_{stall} = \sqrt{\frac{2}{\rho _{\infty }}\left ( \frac{W}{S} \right )\frac{1}{\left ( C_{L} \right )_{max}}}=\sqrt{\frac{2}{0.002377}\left ( \frac{960}{37.8} \right )\frac{1}{1.72005}}=111.4601\,ft/s\]
\[\Rightarrow V_{stall} = 75.99552\,mph\]
- 356 views
- 1 answers
- 0 votes
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Asked on 13th December 2023 in Aeronautics.
(a) \[\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \frac{1}{2\sqrt{k\times C_{D,0}}} = \frac{1}{\sqrt{0.04\times 0.016}} = 19.7642\]
(b)\[\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max} = \frac{\frac{C_{D,0}}{3K}}{\frac{4C_{D,0}}{3}}=\frac{\sqrt{\frac{0.016}{3\left ( 0.04 \right )}}}{\frac{4\left ( 0.016 \right )}{3}} = \frac{0.36515}{0.02133}=17.12\]
(c)For \(\left ( \frac{C_{L}}{C_{D}} \right )_{max}\)\[C_{L} = \sqrt{\frac{C_{D,0}}{K}}=\sqrt{\frac{0.016}{0.04}} = 0.63246\]
\[S = 4.4\,m^{2};\rho = 1.225\,kg/m^{3},W = 3700\,N\]
\[V\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \sqrt{\frac{2W}{SC_{L}\rho }} = \sqrt{\frac{2\left ( 3700 \right )}{4.4\left ( 1.225 \right )\left ( 0.63246 \right )}}= 46.59\,m/s\]
For \(\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}\)\[C_{L} = 0.36515 V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}=V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\sqrt{\frac{0.63246}{0.36515}}= 61.316\,m/s\]
(d) At 10,000 ft, \[\frac{{\rho}’}{\rho} = \sigma = 0.73833\]
\[{V}'{\left(\frac{C_{L}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 46.59\times \frac{1}{0.73833}\]
\[\Rightarrow {V}’\left ( \frac{C_{L}}{C_{D}} \right )_{max} = 54.22\,m/s\]
\[{V}'{\left(\frac{C_{L}^{1/2}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 61.316 \times \frac{1}{\sqrt{0.73833}} = 71.36\,m/s\]
- 365 views
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Asked on 13th December 2023 in Aeronautics.
\[1\,mile = 5280\,ft;1\,bhp = 550\,ft.lb/s\]\[c_{t} = \frac{c\left ( v_{\infty} \right )}{\eta _{p}} = \frac{0.375\,\frac{lb}{bhp.h}\left ( 200\,mi/h \right )}{0.85} = \frac{0.375\frac{lb}{\frac{550\,ft\,.lb\times 3600\,s}{s}}\left ( 200\,\times \frac{5280\,ft}{h} \right )}{0.85}\]
\[\Rightarrow c_{t} = 0.235/h\]
- 140 views
- 1 answers
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Asked on 13th December 2023 in Aeronautics.
\[P_{eq,shaft} = P_{shaft} + P_{Thrust}\]\[P_{shaft} = 4910\]
\[P_{Thrust} = \frac{T_{j}V_{\infty}}{\eta _{p}} = \frac{250\left ( 500 \right )}{ 0.9} = 138888.89\]
\[\Rightarrow P_{eq,shaft} = 4910 + 138888.89 = 143798.89\]
- 150 views
- 1 answers
- 0 votes
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Asked on 12th December 2023 in Aeronautics.
\[\frac{T}{T_{V=0}} = 0.369M_{\infty}^{-0.305}\]\[\Rightarrow T_{3\,km} = 50000\left \{ 0.369\left ( 0.6 \right )^{-0.305} \right \} = 21560.57\,lb\]
- 135 views
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Asked on 11th December 2023 in Aeronautics.
\[TSFC = \frac{W_{f}}{Thrust\times time}\]\[W_{f} = 1000 \times 6.7 = 6700\,lb\]
\[time = \frac{W_{f}}{TSFC \times Thrust}\]
\[\Rightarrow time = \frac{6700}{0.5\times 60000} = \frac{6700}{30000} = 0.223\,hrs\]
- 136 views
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Asked on 8th December 2023 in Aeronautics.
\[ T = \left ( \dot{m}_{air} + \dot{m}_{fuel} \right )V_{e}-\dot{m}_{air}V_{\infty }\]\[\dot{m}_{fuel} = 0.03\dot{m}_{air}\]
\[\Rightarrow T = \left ( \dot{m}_{air} + 0.03\dot{m}_{air} \right )V_{e}-\dot{m}_{air}V_{\infty }\]
\[\Rightarrow T = \dot{m}_{air}\left ( 1.03V_{e}-V\infty \right )\]
\[\Rightarrow \dot{m}_{air} = \frac{T}{1.03V_{e} – V_{\infty}} = \frac{25000}{1.03\left ( 1700 \right )-220}=16.33\,slug/s\]
\[\dot{m}_{air} = P_{\infty }V_{\infty}A_{inlet}\]
\[A_{inlet} = \frac{\dot{m}_{air}}{ P_{\infty }V_{\infty}} = \frac{16.33}{0.002377\left ( 220 \right )} = 31.23\,ft^{2}\]
- 139 views
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