物理学1A (航空)|PHYS1149 Physics 1A (Aviation)代写 unsw

This is an introductory level course in physics for students from all disciplines. The course has both a laboratory and theoretical component. Topics covered include the description of motion; forces and momentum; the dynamics of particles; kinetic and potential energy; the conservation of energy; oscillations and simple harmonic motion

这是一份unsw新南威尔斯大学PHYS1149的成功案例

物理学1A (航空)|PHYS1149 Physics 1A (Aviation)代写 unsw


问题 1.

An aircraft is equipped with a wing of symmetrical airfoils. The lift curve slope of the total aircraft is estimated to be $\frac{\partial C_{L}}{\partial \alpha}=0.8 \cdot 2 \pi \frac{1}{\mathrm{rad}}$. The stall angle of attack (AOA) is $12^{\circ}$. Wing area is $16 \mathrm{~m}^{2}$. Use $g=g_{0}$ and $\rho=\rho_{0}$.
What is the aircraft’s mass during a flight on which a stall speed of $50 \mathrm{kt}$ was observed.

证明 .

$$
\begin{array}{ll}
\text { Given: } \quad & \frac{\partial C_{L}}{\partial \alpha}=5.0265 \frac{1}{\mathrm{rad}} \quad \alpha_{\max }=12^{\circ}=0.20944 \mathrm{rad} \
& V=50 \mathrm{kt}=92.6 \mathrm{~km} / \mathrm{h}=25.72 \mathrm{~m} / \mathrm{s} \
& S_{W}=16 \mathrm{~m}^{2}
\end{array}
$$
$$
C_{L, \max }=\frac{\partial C_{L}}{\partial \alpha} \cdot \alpha_{\max }=5.0265 \frac{1}{\mathrm{rad}} \cdot 0.20944 \mathrm{rad}=1.05275
$$
$$
L=\frac{1}{2} \rho V^{2} \cdot C_{L, \max } \cdot S_{W}=\frac{1}{2} \rho_{0} V^{2} \cdot C_{L, \max } \cdot S_{W}=W=m \cdot g=m \cdot g_{0}
$$
$$
m=\frac{\rho_{0} \cdot V^{2}}{2 g_{0}} \cdot C_{L, \max } \cdot S_{W}=\frac{1.225 \mathrm{~kg} \cdot \mathrm{s}^{2} \cdot 25.72^{2} \cdot \mathrm{m}^{2}}{\mathrm{~m}^{3} \cdot 2 \cdot 9.80665 \cdot \mathrm{m} \cdot \mathrm{s}^{2}} \cdot 1.05275 \cdot 16 \mathrm{~m}=696 \mathrm{~kg}
$$
Answer: The aircraft’s mass is $696 \mathrm{~kg}$.






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PHYS1131 COURSE NOTES :

The category of effect of a failure is judged to be hazardous. Following $A C J N o$. 1 . to $J A R$ $25.1309$
a) What is the largest permissible failure probability?
b) What is the mean time to failure MTTF?
Solution
$\begin{array}{ll}F(t) & \text { probability of failure, } \ \lambda & \text { failure rate } \ \text { MTTF } & \text { mean time to failure } \ \text { FH } & \text { flight hour }\end{array}$
a) hazardous: $F(t=1 \mathrm{FH}) \leq 10^{-7}$
b) For small probabilities of failure: $\lambda \approx F / t=10^{-7} \cdot \frac{1}{\mathrm{FH}}$
$$
\mathrm{MTTF}=1 / \lambda=\frac{1}{10^{-7}} \mathrm{FH}=10000000 \mathrm{FH}
$$
Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than $10000000 \mathrm{FH}$




















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