流体力学|Fluid mechanics I 3A1代写2023

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Fluid mechanics is the branch of physics and engineering that studies the behavior of fluids, which are substances that can flow and take on the shape of their container. Fluids include liquids, gases, and plasmas, and fluid mechanics involves the study of their physical properties, such as viscosity, density, pressure, and flow rate.

Fluid mechanics has a wide range of applications in various fields, such as aerospace engineering, civil engineering, chemical engineering, mechanical engineering, and environmental engineering. It is used to design and analyze systems that involve fluids, such as pumps, turbines, pipes, and airfoils, and to understand phenomena such as turbulence, cavitation, and drag. The study of fluid mechanics also plays a crucial role in understanding natural phenomena, such as weather patterns, ocean currents, and the behavior of the human circulatory system.

流体力学|Fluid mechanics I 3A1代写2023

问题 1.

A thin symmetric airfoil operating at an angle of attack $\alpha$ has a trailing-edge flap with a hinge line at $80 \%$ of chord. (a) The flap is deflected downwards by an angle $\delta$. Find an expression for the additional lift coefficient due to the flap deflection. You may apply the usual small-angle assumptions.

证明 .

(a) The additional lift coefficient due to the flap deflection can be calculated using the following equation:

$\Delta C_{L_{\mathrm{flap}}}=\frac{\partial C_L}{\partial \alpha} \cdot \frac{\partial \alpha}{\partial \delta} \cdot \delta$

where $\Delta C_{L_{\text{flap}}}$ is the additional lift coefficient due to flap deflection, $\frac{\partial C_L}{\partial \alpha}$ is the lift slope, $\frac{\partial \alpha}{\partial \delta}$ is the flap effectiveness, and $\delta$ is the flap deflection angle.

For a thin symmetric airfoil, the lift slope is given by:

$\frac{\partial C_L}{\partial \alpha}=2 \pi$

The flap effectiveness can be approximated using the following equation:

$\frac{\partial \alpha}{\partial \delta}=\frac{2}{\pi}\left(\frac{c \text { flap }}{c}\right)$

where $c_{\text{flap}}$ is the chord length of the flap and $c$ is the chord length of the airfoil.

Since the flap hinge line is at $80 %$ of the chord, we can assume that the flap chord length is $20 %$ of the airfoil chord length. Therefore, we have:

$\frac{\partial \alpha}{\partial \delta}=\frac{2}{\pi}\left(\frac{0.2 c}{c}\right)=\frac{0.4}{\pi}$

Substituting these values into the first equation, we obtain:

$\Delta C_{L_{\mathrm{flap}}}=\frac{\partial C_L}{\partial \alpha} \cdot \frac{\partial \alpha}{\partial \delta} \cdot \delta=2 \pi \cdot \frac{0.4}{\pi} \cdot \delta=0.8 \cdot \delta$

Therefore, the additional lift coefficient due to flap deflection is proportional to the flap deflection angle.

问题 2.

(b) If the flap is deflected by $10^{\circ}$, what is the magnitude of the additional lift coefficient?

证明 .

(b) If the flap is deflected by $10^{\circ}$, the magnitude of the additional lift coefficient is:

$\Delta C_{L_{\mathrm{flap}}}=0.8 \cdot 10^{\circ}=0.139$

问题 3.

(c) If the flap is not deflected, what change in angle of attack $\alpha$ is required to achieve the same lift as that in (b)?

证明 .

(c) If the flap is not deflected, the lift coefficient is given by:

$C_L=2 \pi \alpha$

To achieve the same lift coefficient as in part (b), we need to find the angle of attack $\alpha$ that satisfies:

$C_L=2 \pi \alpha+\Delta C_{L_{\text {flap }}}=2 \pi \alpha+0.139$

Equating this expression to the lift coefficient without flap deflection, we have:

$2 \pi \alpha+0.139=2 \pi \alpha_0$

where $\alpha_0$ is the angle of attack without flap deflection. Solving for $\alpha$, we obtain:

$\alpha=\alpha_0+\frac{0.139}{2 \pi}$

Therefore, the change in angle of attack required to achieve the same lift as in part (b) is proportional to the additional lift coefficient due to flap deflection, and is independent of the flap deflection angle.

这是一份2023年的剑桥大学University of Cambridge Fluid mechanics I 3A1流体力学代写的成功案例