# 流体力学|MATH0015 Fluid Mechanics代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0015 Fluid Mechanics流体力学代写代考辅导服务！

## Instructions:

Fluid Mechanics is the study of fluids, which can be either liquids or gases, and their behavior when subjected to various forces and conditions. It is a branch of physics and engineering that deals with the motion and equilibrium of fluids, as well as their interactions with solid objects.

Fluid mechanics involves the study of topics such as fluid statics, which deals with fluids at rest, and fluid dynamics, which deals with fluids in motion. Fluid mechanics also encompasses topics such as viscosity, turbulence, and boundary layers, which are important in understanding the behavior of fluids.

Applications of fluid mechanics are found in various fields, including aerospace engineering, civil engineering, chemical engineering, mechanical engineering, and biomedical engineering. Understanding fluid mechanics is important for the design and analysis of many devices and systems, such as pumps, turbines, pipes, and blood flow in the human body.

Overall, fluid mechanics plays an important role in understanding the behavior of fluids and their applications in various fields of engineering and science.

Consider a sinusoidal traveling wave of small amplitude $(\mathrm{Ak}<<1)$ such that the free surface is given by: $$\eta=A \cos (\mathrm{kx}-\omega t)$$ The corresponding velocities are given by: $$\mathrm{u}=\mathrm{A} \omega \mathrm{e}^{\mathrm{ky}} \cos (\mathrm{kx}-\omega t)$$ and $$v=A \omega e^{k y} \sin (k x-\omega t)$$ ( $\mathrm{y}$ is the vertical coordinate and is negative in the direction of $\mathrm{g} ; \mathrm{x}$ is the horizontal coordinate) a) Sketch the streamlines at $\mathrm{t}=\mathrm{o}$ that pass through: $\mathrm{x}=o, \mathrm{y}=o ; \mathrm{x}=0, \mathrm{y}=\frac{-2 \pi}{k}$

a) To sketch the streamlines at $t=0$ that pass through $(x,y)=(0,0)$ and $(x,y)=(0,-\frac{2\pi}{k})$, we need to plot a few representative particle trajectories.

The particle trajectory can be obtained by integrating the velocity vector along the path. For small amplitude waves, we can assume that the fluid motion is irrotational, and thus the streamlines are parallel to the velocity vector. Therefore, the particle trajectory will be perpendicular to the streamline at each point.

Let’s start with the point $(0,0)$. The velocity vector at this point is given by $(u,v)=(A\omega,0)$, so the streamline will be a horizontal line. The particle trajectory passing through $(0,0)$ will be a vertical line, as shown below:

Now, let’s consider the point $(0,-\frac{2\pi}{k})$. The velocity vector at this point is given by $(u,v)=(0,A\omega e^{-2\pi})$. Since $v$ is negative, the particle will move downwards along the streamline. The streamline itself can be obtained by integrating the velocity vector, which gives:

$\frac{d y}{d x}=\frac{v}{u}=-\frac{e^{-2 \pi}}{k}$

Integrating this expression yields:

$y=-\frac{e^{-2 \pi}}{k} x+C$

where $C$ is a constant. We can determine $C$ by setting $x=0$ and $y=-\frac{2\pi}{k}$, which gives $C=-\frac{2\pi}{k}$. Therefore, the streamline passing through $(0,-\frac{2\pi}{k})$ is given by:

$y=-\frac{e^{-2 \pi}}{k} x-\frac{2 \pi}{k}$

The particle trajectory passing through $(0,-\frac{2\pi}{k})$ will be perpendicular to this streamline and can be obtained by integrating the velocity vector along the streamline. The resulting particle trajectories are shown below:

b) Sketch the particle line at $\mathrm{t}=\frac{2 \pi}{\omega}$ for the particle which was at $\mathrm{x}=\mathrm{o}, \mathrm{y}=\mathrm{o}$ at $\mathrm{t}=\mathrm{o}$

b) To sketch the particle line at $t=\frac{2\pi}{\omega}$ for the particle which was at $(x,y)=(0,0)$ at $t=0$, we need to find the position of the particle at $t=\frac{2\pi}{\omega}$.

The particle position can be obtained by integrating the velocity vector along the particle trajectory. For a particle starting at $(0,0)$, the particle trajectory is a vertical line, as shown in part (a). The velocity vector along this trajectory is given by $(u,v)=(A\omega e^{ky},0)$. Integrating this expression from $y=0$ to $y=-\frac{2\pi}{k}$ and from $t=0$ to $t=\frac{2\pi}{\omega}$ gives:

$x=0 \quad$ and $\quad y=-\frac{2 \pi}{k}$

Therefore, the particle that was at $(x,y)=(0,0)$ at $t=0$ will be at $(x,y)=(0,-\frac{2\pi}{k})$ at $t=\frac{2\pi}{\omega}$

For an ideal rectilinear vortex, the velocity profile is given by $\mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{z}}=0 ; \mathrm{v}_\theta=\mathrm{K} / \mathrm{r}$. $A$ qualitatively similar result is associated with an ideal smoke ring. a) Why does a smoke ring propel itself? Why does the velocity of translation diminish with time?

A smoke ring is a toroidal vortex ring that propagates through the air. It is created by a sudden release of fluid, such as smoke, from a circular orifice. The smoke ring propels itself because it is a closed loop of fluid that has rotational motion. The fluid inside the smoke ring rotates in the same direction as the ring itself, creating a low-pressure area in the center of the ring.

As the smoke ring moves forward, the low-pressure area at the center of the ring pulls in air from behind, which helps to maintain the shape of the ring. This process of entraining air into the ring creates a drag force, which slows down the velocity of the smoke ring over time.

The velocity of translation of the smoke ring diminishes with time because of the loss of energy due to drag forces. As the smoke ring moves through the air, it creates turbulence and friction with the surrounding air, which causes the fluid within the ring to lose energy. This loss of energy translates into a decrease in the velocity of translation of the smoke ring over time. Eventually, the smoke ring will come to a stop as all of its kinetic energy is dissipated through friction with the air.

# 流体力学|Fluid mechanics II 3A3代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Fluid mechanics II 3A3流体力学代写代考辅导服务！

## Instructions:

Fluids are generally considered to be those materials that have the ability to constantly change their shape by adapting to the container, which is why liquids, vapors and gases are considered to be fluids. Fluid mechanics consists of two main branches:

Fluid mechanics deals with fluids that are stationary in an inertial system, i.e. with constant velocity in time and homogeneity in space. Historically, it was the first step towards the study of mechanics.
Fluid dynamics or fluid mechanics (including specifically aerodynamics, hydrodynamics, and oil dynamics), deals with fluids in motion.
Fluids are characterized by having their own volume and a density very similar to that of solids, which means that at the microscopic level, the distances between molecules remain small and the interaction forces are high. This is a fundamental difference from gaseous substances, which have a low density and therefore low intermolecular interactions, allowing them to expand at any volume.

(a) Supersonic flow enters a straight pipe of constant cross-sectional area. Heat transfer is negligible, but the pipe wall is rough. Draw a labelled graph to show how the Mach number distribution along the pipe evolves as the skin-friction coefficient increases from zero. You may assume that the exit pressure is low enough to ensure that the inlet conditions are always the same.

(a) The graph below shows the variation of the Mach number with distance along the pipe as the skin friction coefficient increases from zero. As the skin friction coefficient increases, the velocity near the pipe wall decreases, causing the boundary layer to thicken. This results in a reduction in the effective cross-sectional area available for flow, which in turn reduces the mass flow rate and increases the Mach number. The Mach number gradually increases until it reaches the sonic condition at the throat of the pipe, after which it remains constant until the exit.

(b) Air flows in a pipe of length $5.9 \mathrm{~m}$ and inside diameter $0.2 \mathrm{~m}$. The inlet stagnation pressure is 2.7 bar, and the static pressure at the pipe exit is 1 bar. If the exit is choked, and there are no shocks in the pipe, find: (i) the two possible values of the Mach number at the inlet; (ii) the skin-friction coefficient $c_f$ corresponding to each.

(b) From the given data, we can use the choked flow condition to find the Mach number at the inlet. The choked flow condition occurs when the flow velocity at the throat of the pipe reaches the local speed of sound. At the throat, the Mach number is therefore 1.

Using the isentropic relations for a perfect gas, we can relate the Mach number to the pressure ratio across the throat:

$\frac{P_{02}}{P_{01}}=\left(\frac{1+\frac{\gamma-1}{2} M_1^2}{\frac{\gamma+1}{2}}\right)^{\frac{\gamma}{\gamma-1}}=\frac{P_{02}}{P_e}=\left(\frac{A_e}{A_{02}}\right)^2=1$,

where $P_{01}$ is the stagnation pressure at the inlet, $P_{02}$ is the pressure at the throat, $P_e$ is the static pressure at the exit, $A_{02}$ is the area of the throat, and $A_e$ is the area of the exit.

Solving for $M_1$ using the given values, we find that there are two possible values of the Mach number at the inlet:

$M_1=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_{02}}{P_{01}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}=0.747,2.11$.

Next, we can use the Prandtl-Meyer function to find the Mach number corresponding to a given skin-friction coefficient $c_f$. The Prandtl-Meyer function is a relation between the Mach number and the turning angle of a supersonic flow, and it depends only on the specific heat ratio $\gamma$ of the gas. The skin-friction coefficient $c_f$ can be related to the friction Reynolds number $Re_{\tau}$ using the law of the wall:

$c_f=\frac{\tau_w}{\frac{1}{2} \rho_1 V_1^2}=\frac{0.026}{R e_\tau^{0.2}}$,

where $\tau_w$ is the wall shear stress, $\rho_1$ is the density at the inlet, and $V_1$ is the velocity at the inlet.

For air at room temperature and pressure, $\gamma=1.4$. Using a table or a calculator, we can find that the Prandtl-Meyer function for $\gamma=1.4$ is approximately $\nu=30.5^{\circ}$ at $M=0.747$, and $\nu=66.1^{\circ}$ at $M=2.11$.

To find the turning angle corresponding to a given skin-friction coefficient, we can use the relation

$\theta=\frac{c_f}{2 \nu}$

Using the values of $c 问题 3. (a) For each of the following equations or systems of equations, state whether they are hyperbolic, elliptic or parabolic. Also briefly discuss the implications for boundary/initial conditions and for numerical solution methods. (i) $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ (ii) $$u \frac{\partial u}{\partial x}=\alpha \frac{\partial^2 u}{\partial y^2}$$ where$\alpha$is a positive constant. (iii) The Euler equations in one dimension, i.e. $$\frac{\partial}{\partial t}\left[\begin{array}{c} \rho \\ \rho u \\ E \end{array}\right]+\frac{\partial}{\partial x}\left[\begin{array}{c} \rho u \\ \rho u^2+p \\ u(E+p) \end{array}\right]=0,$$ where the symbols have their usual meanings. 证明 . (i) The equation$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$is an example of the Laplace equation, which is an elliptic partial differential equation. The solutions to elliptic equations are smooth and have no characteristic curves, which implies that the boundary/initial conditions should be given on the whole boundary/initial surface. Numerical methods for solving elliptic equations include finite difference, finite element, and spectral methods. (ii) The equation$u \frac{\partial u}{\partial x}=\alpha \frac{\partial^2 u}{\partial y^2}$is an example of a convection-diffusion equation, which can be hyperbolic or parabolic depending on the values of$u$and$\alpha$. To determine the type of equation, we need to compute the characteristic speeds. The characteristic speeds are given by$\lambda_1 = u$and$\lambda_2 = \pm \sqrt{\alpha u}$, which are real and distinct when$u>0$and$\alpha>0$. Therefore, the equation is hyperbolic when$u>0$and$\alpha>0$. On the other hand, when$u<0$or$\alpha<0$, the equation is parabolic. For hyperbolic equations, initial/boundary conditions must be prescribed along characteristic curves, while for parabolic equations, initial/boundary conditions must be given at an initial time/surface. Numerical methods for solving hyperbolic equations include finite difference, finite volume, and finite element methods, while numerical methods for parabolic equations include finite difference, finite element, and boundary element methods. (iii) The Euler equations in one dimension are a system of nonlinear hyperbolic partial differential equations. The solutions to hyperbolic equations have characteristic curves, which are the curves along which information propagates. Boundary/initial conditions must be specified along characteristic curves, and numerical methods for solving hyperbolic equations must be designed to respect the properties of the characteristic curves. Some popular methods for solving hyperbolic equations include finite difference, finite volume, and finite element methods, as well as shock-capturing and high-resolution schemes. 这是一份2023年的剑桥大学University of Cambridge Fluid mechanics II 3A3流体力学流体力学代写的成功案例 # 流体力学|Fluid mechanics I 3A1代写2023 0 Assignment-daixieTM为您提供剑桥大学University of Cambridge Fluid mechanics I 3A1流体力学代写代考辅导服务！ ## Instructions: 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.