## MAE3230 Fluid Mechanics课程简介

Physical properties of fluids:

Fluids are substances that can flow and take the shape of the container they occupy. The physical properties of fluids include density, viscosity, surface tension, and compressibility.

Density is defined as the mass per unit volume of a fluid. It is denoted by the symbol ρ and is measured in kg/m³.

Viscosity is a measure of a fluid’s resistance to deformation or flow. It is denoted by the symbol μ and is measured in Pa·s or N·s/m².

Surface tension is the force per unit length acting perpendicular to any line drawn on the surface of a liquid. It is denoted by the symbol γ and is measured in N/m.

Compressibility is a measure of how much a fluid’s volume changes when subjected to a pressure change. It is denoted by the symbol β and is measured in 1/Pa.

## PREREQUISITES

Outcome 1: Students will be able to use the fundamental principles and mathematical basis underlying the conservation equations.
Outcome 2: Be able to identify the guiding principles in a given fluid problem, to formulate the governing equations, and so to solve basic engineering problems.
Outcome 3: Recognize the difference between an ideal fluid and a viscous fluid, and to understand the limitations of the solutions for real practical fluid flows. Understand the difference between a simple solution and a real practical problem.

Outcome 4: Understand where their analysis might involve approximations and empirical approaches; for example, pipe flows and boundary layer flows.
Outcome 5: Have improved their ability to formulate an ordered approach to problem solving, using words of explanation in derivations, and algebra before substituting numerical values that allows neat analytical solutions and dimensional analysis.

## MAE3230 Fluid Mechanics HELP（EXAM HELP， ONLINE TUTOR）

1. Hydrostatic equilibrium: A plane-parallel atmosphere, composed of a perfect gas, is in hydrostatic equilibrium in an external field, $-\hat{z} g$.
(a) Derive an expression for the entropy gradient if the atmosphere is isothermal.
(b) Obtain expressions for $p(z)$ and $\rho(z)$ if the atmosphere is isentropic.
(c) Earth’s atmosphere: In the lower stratosphere, the air is isothermal. Use the condition of hydrostatic equilibrium to show that:
$$p(z) \propto \exp (-z / H)$$
where the scale height, $H=k T /\left(\mu m_p g\right)$. Estimate the scale height. (Use mean molecular weight $\mu=29$ and $T=300 \mathrm{~K}$ ). Now assuming that the air is isentropic, show that:
$$\frac{d T}{d z}=-\left(\frac{\gamma-1}{\gamma}\right) \frac{g \mu m_p}{k}$$
Here $\gamma \simeq 1.4$ is the ratio of specific heats for gases like Nitrogen and Oxygen. Why does the above expression vanish for $\gamma=1$ ?

(a) If the atmosphere is isothermal, then the temperature is constant throughout the atmosphere. The equation of hydrostatic equilibrium for a perfect gas is given by:

\frac{d p}{d z}=-\rho gdzdp​=−ρg

where $p$ is the pressure, $\rho$ is the density, $z$ is the height above a reference level, and $g$ is the acceleration due to gravity. Using the ideal gas law, $p=\rho k T / m$, where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of a molecule, we can write:

\frac{d}{d z}\left(\frac{p}{\rho}\right)=-g \frac{k T}{m}dzd​(ρp​)=−gmkT​

Since the temperature is constant, the right-hand side of the equation is a constant, and we can integrate both sides to obtain:

\frac{p}{\rho}=\mathrm{constant}-g \frac{k T}{m} zρp​=constant−gmkT​z

Taking the derivative of the logarithm of both sides with respect to $z$, we get:

\frac{d \ln \rho}{d z}=-\frac{m g}{k T}dzdlnρ​=−kTmg​

which is the expression for the entropy gradient in an isothermal atmosphere.

(b) If the atmosphere is isentropic, then the entropy is constant throughout the atmosphere. The equation of hydrostatic equilibrium can be written as:

\frac{d p}{d z}=-\frac{m g}{k T} pdzdp​=−kTmg​p

This is a first-order linear differential equation that can be solved by separation of variables:

\frac{d p}{p}=-\frac{m g}{k T} dzpdp​=−kTmg​dz

Integrating both sides, we get:

\ln p=-\frac{m g}{k T} z + \mathrm{constant}lnp=−kTmg​z+constant

or

p(z)=p_{0} \exp \left(-\frac{m g}{k T} z\right)p(z)=p0​exp(−kTmg​z)

where $p_{0}$ is the pressure at the reference level. Using the ideal gas law, we can write the density as:

\rho(z)=\frac{p(z) m}{k T}=\frac{p_{0} m}{k T} \exp \left(-\frac{m g}{k T} z\right)ρ(z)=kTp(z)m​=kTp0​m​exp(−kTmg​z)

(c) In the lower stratosphere, the air is isothermal, so we can use the expression for the entropy gradient obtained in part (a). The equation of hydrostatic equilibrium can be written as:

\frac{d p}{d z}=-\rho g=-\frac{p m g}{k T}dzdp​=−ρg=−kTpmg​

which is a first-order linear differential equation that can be solved by separation of variables:

\frac{d p}{p}=-\frac{m g}{k T} dzpdp​=−kTmg​dz

Integrating both sides, we get:

\ln p=-\frac{m g}{k T} z + \mathrm{constant}lnp=−kTmg​z+constant

or

p(z)=p_{0} \exp \left(-\frac{m g}{k T} z\right)p(z)=p0​exp(−kTmg​z)

where $p_{0}$ is the pressure at the reference level. The scale height is defined as $H=k T /(\mu m_p g)$, where $\mu$ is the mean molecular weight and $m_p$ is the mass of a proton. Using $\mu=29$ and $T=300 \mathrm{~K}$, we have:

Archimedes’ principle states that, when a solid body is totally or partially immersed in a fluid, the total buoyant upward force of the liquid on the body is equal to the weight of the displaced fluid. Prove the law assuming conditions of hydrostatic equilibrium. Using this result, estimate how much more would one weigh in vacuum.

To prove Archimedes’ principle, we start with the equation of hydrostatic equilibrium:

\frac{d P}{d z}=-\rho gdzdP​=−ρg

where $P$ is the pressure, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $z$ is the height above a reference level. We consider a solid body of volume $V$ and density $\rho_{s}$ that is partially or totally immersed in the fluid. The buoyant force on the solid body is equal to the weight of the fluid that is displaced by the solid body. Let $V_{d}$ be the volume of the fluid displaced by the solid body.

If the solid body is partially immersed in the fluid, then the displaced volume $V_{d}$ is equal to the volume of the part of the solid body that is immersed. If the solid body is totally immersed in the fluid, then $V_{d}$ is equal to the volume of the entire solid body. In either case, we have:

V_{d}=V_{s}Vd​=Vs​

where $V_{s}$ is the volume of the solid body.

The weight of the fluid displaced by the solid body is given by:

W_{d}=\rho V_{d} g=\rho_{s} V_{s} gWd​=ρVd​g=ρs​Vs​g

where $\rho$ is the density of the fluid.

The buoyant force on the solid body is given by:

F_{B}=\rho V_{d} g=\rho_{s} V_{s} gFB​=ρVd​g=ρs​Vs​g

Since the solid body is in hydrostatic equilibrium, the net force on the solid body must be zero. Therefore, the weight of the solid body must be equal to the buoyant force:

W=\rho_{s} V_{s} g=F_{B}W=ρs​Vs​g=FB​

This is Archimedes’ principle.

To estimate how much more one would weigh in vacuum, we consider the buoyant force on a human body due to air. The density of air at sea level is approximately $\rho=1.2 \mathrm{~kg/m^{3}}$. The weight of the air displaced by a human body of volume $V_{h}$ can be estimated as:

W_{d}=\rho V_{h} g=1.2 V_{h} gWd​=ρVh​g=1.2Vh​g

For a person with a volume of $V_{h}=0.1 \mathrm{~m^{3}}$, the weight of the displaced air is:

W_{d}=1.2 \times 0.1 \times 9.81 \mathrm{~m/s^{2}}=1.176 \mathrm{~N}Wd​=1.2×0.1×9.81 m/s2=1.176 N

This is the buoyant force due to air acting on the person. In vacuum, there would be no buoyant force, and the weight of the person would be:

W_{vacuum}=m gWvacuum​=mg

where $m$ is the mass of the person. The difference between the weight in air and the weight in vacuum is:

\Delta W=W_{vacuum}-W_{d}=m g-1.176 \mathrm{~N}ΔW=Wvacuum​−Wd​=mg−1.176 N

For a person with a mass of $m=70 \mathrm{~kg}$, the difference in weight is:

\Delta W=70 \times 9.81-1.176 \approx 682 \mathrm{~N}ΔW=70×9.81−1.176≈682 N

Therefore, a person would weigh approximately 682 N more in vacuum than in air.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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