Posted on Categories:Thermodynamics, 热力学, 物理代写

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## 物理代写|热力学代写Thermodynamics代考|Transferring availability with heat transfer processes

You can move availability into or out of a system using heat transfer. In a heat transfer process, only a portion of the energy in a heat transfer process at a system boundary temperature $(T)$ can be transferred as availability. The amount of energy that doesn’t transfer in as avallability becomes an irreversibility in the system. This irreversibility occurs because not all heat transfer to a system can be converted into useful work. Some heat must be rejected to the ambient environment surrounding the system.

In theory, it’s possible to use a heat engine to produce useful work from the heat that’s rejected by a system to the environment. But even a heat engine has to reject some heat, so you can never completely get out of rejecting heat by producing useful work. In practice, replacing a heat rejection process with a heat engine usually isn’t cost-effective if the availability is relatively low compared to the dead state.

The portion of availability transferred in a system by heat, $A_{\text {heat }}$ at a boundary temperature $T$ in an environment at $T_0$ is calculated with this equation:
$$A_{\text {heat }}=\left(1-\frac{T_0}{T}\right) Q$$
Availability and heat move in or out of a system in the same direction as long as the temperature of the boundary is above the dead state temperature. If the system boundary temperature is below the dead state temperature, the availability transfers in the opposite direction of heat transfer.

Transferring availability with mass flow Mass flows through an open system, so it can move energy, entropy, and availability into or out of the system. The rate of total availability $\left(\dot{A}_{\max }\right)$ that’s transferred into or out of a system is calculated by the mass flow rate $(m)$ and the flow availability $\left(a_l\right)$, as shown in this equation: $\dot{A}{\text {mass }}=\dot{m} a{f .}$

For a closed system, there is no mass flow so there’s no transfer of availability by mass transfer, but the total availability $(A)$ in the system is calculated using the mass of the system $(m)$ and the specific availability $(a)$, as shown by the equation $A=m a$.

## 物理代写|热力学代写Thermodynamics代考|Understanding the Decrease in Availability Principle

The first law of thermodynamics states that energy cannot be created or destroyed but can only change form. This means that energy is a conserved property. In any system, the energy that goes into a system equals the energy that leaves and accumulates in the system. Energy may enter as a form of heat and leave as a form of work, as it does in a heat engine. Mass has similar qualities: The mass entering a system equals the mass leaving and accumulating in the system, even if it changes phase from a liquid to a gas.

Entropy and availability are not conserved properties like energy and mass. In Chapter 8 , I discuss the increase in entropy principle, whereby every process in a system causes entropy to increase from the perspective of the universe. Locally, you may cause entropy to decrease in a system by lowering its temperature or raising its pressure. But the result of entropy decreasing within a system is a proportionately greater increase in the entropy of the system’s surroundings. The net effect of entropy changes to the system and its surroundings is that entropy always increases.
The nature of availability is that it diminishes as it’s used to do work or provide heat. The inlet conditions of a thermodynamic process require highquality energy relative to the avallability at the outlet conditions, meaning the inlet conditions must have sufficient availability to perform the process. At the outlet of a work process, the availability has decreased such that it can’t do as much work as it could before the process.

You may locally increase the availability of energy by adding heat, work, or mass into a process, but the avallability of the source of heat, work, or mass decreases by supplying energy to the process. Globally, the decrease in availability is similar to the global increase in entropy. Eventually, the universe will have no availability for performing useful work. Irreversibility caused by heat transfer, friction, mixing, and so forth destroys availability and generates entropy. The destruction of availability, $A_{\text {destroyed }}$ is proportional to entropy generation, $S_{\mathrm{gen}}$, as shown in this equation: $A_{\text {destroyed }}=T_0 S_{\mathrm{gen}}$.

## 物理代写|热力学代写Thermodynamics代考|Transferring availability with heat transfer processes

$$A_{\text {heat }}=\left(1-\frac{T_0}{T}\right) Q$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Thermodynamics, 热力学, 物理代写

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## 物理代写|热力学代写Thermodynamics代考|Analyzing Isentropic Processes

You may think that every ideal process creates a change in entropy. Believe it or not, entropy doesn’t change in some ideal processes. An ideal compression or expansion process involving an ideal gas is reversible and adiabatic, has no change in entropy, and is called an isentropic process.
You can use either the constant specific heat assumption or the variable specific heat assumption when analyzing isentropic processes, as shown in the previous section “Calculating Entropy Change.” The constant specific heat method gives satisfactory results when the temperature change isn’t large, as in simple compression/expansion processes. The variable specific heat method gives the most accurate results, especially for large temperature changes in a process. I discuss analysis of isentropic processes using both methods in the following sections.

Using constant specific heat
For an isentropic process, the change in entropy equations shown in the previous section are set equal to zero. This gives three mathematical equations to relate temperature, pressure, and specific volume to each other. One equation relates temperature to pressure, the second relates temperature to specific volume, and the third relates pressure to specific volume. These equations are as follows:
\begin{aligned} & \left(\frac{T_2}{T_1}\right)=\left(\frac{v_1}{v_2}\right)^{k-1} \ & \left(\frac{T_2}{T_1}\right)=\left(\frac{P_2}{P_1}\right)^{(k-1) / k} \ & \left(\frac{P_2}{P_1}\right)=\left(\frac{v_1}{v_2}\right)^k \end{aligned}
Because temperature appears as a ratio in these equations, you must use absolute temperatures. The variable $k$ in these equations is called the ratio of specific heats. You can calculate $k$ for an ideal gas using the following equation:
$$k=\frac{c_p}{c_v}$$
For air, $k$ equals 1.4 for processes that are within a few hundred degrees Celsius of room temperature. Because specific heat varies with temperature, you should calculate $k$ using the specific heats at the average process temperature. No units are associated with $k$ because it’s a ratio.

## 物理代写|热力学代写Thermodynamics代考|Balancing Entropy in a System

In a thermodynamic system, energy can enter, leave, or be stored within the system by heat transfer, work, and mass flow. You use an energy balance equation to keep track of the energy flow in a system. Entropy can only enter or leave a system by mass flow and heat transfer. Entropy can be generated within the system by irreversibilities. You can write an entropy balance on a system to keep track of entropy flow, as follows:
$$\Delta S_{\text {systeme }}=S_2-S_1=S_{\text {in }}-S_{\text {cout }}+S_{\text {gen }}$$

The entropy balance means the change in entropy of a system ( $\Delta S_{\text {sustem }}$ ) during a process equals the difference between the final $\left(S_2\right)$ and initial $\left(S_1\right)$ entropy of the system.

Entropy generation $\left(S_{\text {gen }}\right)$ includes only the entropy generated within the system; it doesn’t include entropy generated in the surroundings. If the process within the system is internally reversible, the entropy generation is zero.
Heat is a disorganized form of energy, so entropy flows with it. Entropy enters the system $\left(S_{i n}\right)$ as heat is transferred to the system. Entropy is removed from the system $\left(S_{\text {out }}\right)$ as heat is transferred from the system. You can calculate the entropy transfer by heat $\left(S_{\text {hinat }}\right)$ in a system by dividing the heat transfer through a system boundary $\left(Q_k\right)$ by the absolute temperature $(T)$ of the boundary for each heat transfer process, as shown in the following equation:
$S_{\text {lhan }} \cong \sum \frac{Q_k}{T_k}$ where $k$ is the number of boundaries
A system may have more than one heat transfer process; in fact, many systems have a heat addition and a heat rejection process.

The quality of energy decreases in every thermodynamic process as it is expended to do work. In this section, I discuss the analysis of the change in availability $\left(\Delta A_{\text {system }}\right)$ for closed systems and open systems. In a closed system, the mass of the system remains fixed, whereas in an open system, mass is allowed to flow through. The system refers to the fluid (either a liquid or a gas) that’s used in a thermodynamic process to move heat or produce work.
The decrease in avallability of a system between two states represents the maximum amount of useful work output that can be done by the system. If the availability of a system increases between the initial and final states, then it represents the minimum amount of work input required by the system. The availability between two states is independent of the type of system used, the type of process in the system, and the type of heat and work interactions with the surroundings.

## 物理代写|热力学代写Thermodynamics代考|Analyzing Isentropic Processes

\begin{aligned} & \left(\frac{T_2}{T_1}\right)=\left(\frac{v_1}{v_2}\right)^{k-1} \ & \left(\frac{T_2}{T_1}\right)=\left(\frac{P_2}{P_1}\right)^{(k-1) / k} \ & \left(\frac{P_2}{P_1}\right)=\left(\frac{v_1}{v_2}\right)^k \end{aligned}

$$k=\frac{c_p}{c_v}$$

## 物理代写|热力学代写Thermodynamics代考|Balancing Entropy in a System

$$\Delta S_{\text {systeme }}=S_2-S_1=S_{\text {in }}-S_{\text {cout }}+S_{\text {gen }}$$

$S_{\text {lhan }} \cong \sum \frac{Q_k}{T_k}$，其中$k$是边界的数目

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Thermodynamics, 热力学, 物理代写

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## 物理代写|热力学代写Thermodynamics代考|Finding the coefficient of performance

In a cycle, the working fluid returns to its starting condition after the heat rejection process is complete. The heat transfer and work processes in a refrigeration cycle are repeated continuously to move heat from the lowtemperature reservoir to the high-temperature one.
The performance of a refrigerator isn’t measured by efficiency like a heat engine. Instead, it’s determined by the coefficient of performance (COP). The coefficient of performance is a measure of how much heat is transferred by the amount of work put into the refrigerator. In general, any measure of performance is (what you want $) /($ what you provide). For a refrigerator, what you want is to transfer heat. What you provide is work. The coefficient of performance is calculated much like the efficiency. The coefficient of performance is expressed as follows:
$$C O P=\frac{\text { Desired output }}{\text { Required input }}$$
For a refrigerator, the desired output is the amount of heat absorbed from the low-temperature reservoir $\left(Q_1\right)$. The heat removed from the reservoir equals the heat input to the refrigerator $\left(Q_i\right)$. The work into a refrigerator $\left(W_{\text {in }}\right)$ is equal to the difference between the heat output $\left(Q_{\infty}\right)$ and the heat absorbed $\left(Q_{\mathrm{n}}\right)$ by the refrigerator. That is, $W_{\mathrm{in}}=Q_{\text {out }}-Q_{\mathrm{in} \text {. }}$. The coefficient of performance for a refrigerator is calculated using this equation:
$$C O P_{\mathrm{R}}=\frac{Q_{\mathrm{in}}}{W_{\text {net.in }}}=\frac{Q_L}{W_{\text {netin }}}=\frac{Q_L}{Q_H-Q_L}$$
For a heat pump, the desired output is the amount of heat rejected to the warm energy reservoir $\left(Q_\mu\right)$, the interior of a house. The heat added to the reservoir equals the heat output of the refrigerator $\left(Q_{0 u}\right)$. You calculate the coefficient of performance for a heat pump by using this equation:
$$C O P_{\mathrm{HP}}=\frac{Q_{\text {att }}}{W_{\text {ret, in }}}=\frac{Q_H}{W_{\text {ret, in }}}=\frac{Q_H}{Q_H-Q_L}$$
You can find the coefficient of performance for a refrigerator or a heat pump by working out the following example. Suppose you have a refrigerator that absorbs 1 kilowatt of heat from the cold reservoir and rejects 1.3 kilowatts of heat to the warm reservoir. You can find the actual coefficient of performance for the refrigerator with the following equation:
$$C O P_{\mathrm{R}}=\frac{1 \mathrm{~kW}}{(1.3-1) \mathrm{kW}}=3.3$$

## 物理代写|热力学代写Thermodynamics代考|What Is Entropy?

Remember when you first got a new desk? You arranged all your papers and knick-knacks on it. It looked nice and neat. But if you’re like most people, things started to pile up on your desk and before you knew it, books, candy wrappers, sticky notes, and empty coffee cups took over. Yes, one aspect of entropy is at work here. Things that start out neat and tidy naturally become disordered. You can picture the universe this way. In the beginning, it was much smaller than it is today; its energy was concentrated into a very small space. But as the universe ages and expands, it becomes more disordered. Making something ordered again takes effort; you have to do some work.
Entropy has many different interpretations. Its definition depends on who you’re talking to. In principle, entropy is used by physicists, theologians, engineers, philosophers, information specialists, and economists, among other professionals. Entropy is often thought of as a measure of disorder of a system. But how can you quantify order or disorder? Entropy is a thermodynamic property of a substance that needs to be quantified in order to be useful.
In thermodynamics, you find microscopic and macroscopic perspectives on entropy.
Taking a microscopic view of entropy
On a microscopic level, entropy starts with the third law of thermodynamics, which I discuss in Chapter 2. At absolute zero temperature, the molecules in a substance have no energy to move, vibrate, or rotate. The entropy of the material is zero. As energy is added to a material, the entropy of the molecules increases because they become more energetic and more disorganized – the way your desk gets more cluttered the more you use it.

The entropy of a material increases as its temperature increases. Solid materials have less entropy than liquids, and liquids have less entropy than gases. As molecules in a material increase in temperature, they like to spread out and take up more room; that is, they become more disordered.

Pressure has the opposite effect on entropy of a material. As the pressure of a gas, liquid, or solid increases, the entropy decreases. However, liquids and solids are considered nearly incompressible, so the entropy decrease is minimal. Pressure forces molecules closer together; they become more ordered, so entropy decreases.

Scan through the thermodynamic property tables in the appendix to see how entropy increases with temperature and decreases with pressure.

## 物理代写|热力学代写Thermodynamics代考|Finding the coefficient of performance

$$C O P=\frac{\text { Desired output }}{\text { Required input }}$$

$$C O P_{\mathrm{R}}=\frac{Q_{\mathrm{in}}}{W_{\text {net.in }}}=\frac{Q_L}{W_{\text {netin }}}=\frac{Q_L}{Q_H-Q_L}$$

$$C O P_{\mathrm{HP}}=\frac{Q_{\text {att }}}{W_{\text {ret, in }}}=\frac{Q_H}{W_{\text {ret, in }}}=\frac{Q_H}{Q_H-Q_L}$$

$$C O P_{\mathrm{R}}=\frac{1 \mathrm{~kW}}{(1.3-1) \mathrm{kW}}=3.3$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Thermodynamics, 热力学, 物理代写

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## 物理代写|热力学代写Thermodynamics代考|Working with pumps, compressors, and turbines

Pumps are used in many applications, ranging from circulating water through filters in a swimming pool to providing drinking water for a city. Pumps use a work input to increase the pressure in a liquid and make it circulate in a network of pipes. In thermodynamics, the most common pump application is the use of a Rankine cycle in a power plant (see Chapter 12) to pressurize and circulate water through a boiler.

A compressor is similar to a pump except it pressurizes and circulates a gas instead of a liquid. Figure $6-4$ shows a diagram of a compressor that’s found in gas turbine engines, jet engines, and industrial facilities (see Chapter 10). The shape of the diagram indicates that the specific volume of the gas decreases as the pressure of the gas increases in the compressor. A compressor has a large number of blades, like a fan, mounted on a shaft. It may have many rows of blades, called stages, that increase the pressure step by step from one stage to the next. The blades for each stage get progressively smaller because the specific volume decreases as the gas is compressed. It takes much more work per unit mass to compress a gas than a liquid.

A turbine extracts work from a gas, such as steam in a Rankine cycle power plant (see Chapter 12) or air in a Brayton cycle engine (see Chapter 10). Figure 6-4 shows a diagram of a turbine.
Making assumptions for pumps, compressors, and turbines When you apply the first law of thermodynamics to a pump, compressor, or turbine, you usually make the following assumptions:

$\sim$ Turbines are usually insulated because they have hot gas flowing through them. Minimizing heat loss provides more work output. Compressors often have cooling to reduce the work input required. Pumps usually aren’t insulated because a heat loss or gain doesn’t really change the work input.
$\sim$ No change in potential energy occurs between the inlet and the outlet of the machine.

No change in kinetic energy occurs between the inlet and the outlet. For a pump, a liquid is incompressible, so the inlet and outlet velocities are the same. In a turbine, the change in kinetic energy can be sizable, but the change in enthalpy is usually much greater, so any change in kinetic energy is ignored for simplicity.

## 物理代写|热力学代写Thermodynamics代考|Writing the energy balance for a compressor, turbine, or pump

Using these assumptions, you can write the energy balance for a compressor, turbine, or pump as follows:
$$\left(\dot{Q}{\mathrm{in}}-\dot{Q}{\mathrm{ous}}\right)+\left(\dot{W}{\text {in }}-\dot{W}{\text {ost }}\right)=\dot{m}\left(h_{\text {oun }}-h_{\mathrm{in}}\right)$$
A pump and a compressor use the work-in term. A turbine uses the work-out term. Usually, there isn’t any heat transter into a pump, compressor, or turbine, so the heat-in term is zero. A compressor uses the heat-out term if it’s cooled. A turbine doesn’t have a heat-out term if it’s well insulated.

Analyzing a compressor
Here’s an example that shows you how to use the conservation of energy equation to determine the work required to operate the compressor of the jet engine. Suppose the compressor inlet enthalpy is $h_{i n}=254.7 \mathrm{~kJ} / \mathrm{kg}$. The compressor exit temperature is 500 Kelvin, and the heat loss $(q)$ from the compressor to the ambient air is 50 kilojoules per kilogram. The air mass flow rate through the compressor is 60 kilograms per second. You can find the work of the compressor as follows:

Write out the energy equation to solve for the rate of compressor work.
$$\dot{W}{\mathrm{zt}}=\dot{m}\left[q{\mathrm{cat}}+\left(h_{\mathrm{cut}}-h_{\mathrm{in}}\right)\right]$$

Look up the enthalpy of air for the compressor exit $h_{\text {ewt }}$ at $500 \mathrm{Kelvin}$ in Table A-l of the appendix.
$$h_2=503.5 \mathrm{~kJ} / \mathrm{kg}$$

Calculate the rate of compressor work.
Use the mass flow rate, the heat loss, and the change in enthalpy of the air in the energy equation.
$$\dot{W}_{\mathrm{ht}}=(60 \mathrm{~kg} / \mathrm{s})[50+(503.5-254.7) \mathrm{kJ} / \mathrm{kg}]\left(\frac{1 \mathrm{MW}}{1,000 \mathrm{~kJ} / \mathrm{s}}\right)=17.9 \mathrm{MW}$$

## 物理代写|热力学代写Thermodynamics代考|Working with pumps, compressors, and turbines

$\sim$ 涡轮机通常是绝缘的，因为它们有热气流过。最大限度地减少热损失提供更多的工作输出。压缩机通常具有冷却功能，以减少所需的工作输入。泵通常不是绝缘的，因为热量的损失或增加并不会真正改变输入的功。
$\sim$在机器的入口和出口之间没有势能的变化。

## 物理代写|热力学代写Thermodynamics代考|Writing the energy balance for a compressor, turbine, or pump

$$\left(\dot{Q}{\mathrm{in}}-\dot{Q}{\mathrm{ous}}\right)+\left(\dot{W}{\text {in }}-\dot{W}{\text {ost }}\right)=\dot{m}\left(h_{\text {oun }}-h_{\mathrm{in}}\right)$$

$$\dot{W}{\mathrm{zt}}=\dot{m}\left[q{\mathrm{cat}}+\left(h_{\mathrm{cut}}-h_{\mathrm{in}}\right)\right]$$

$$h_2=503.5 \mathrm{~kJ} / \mathrm{kg}$$

$$\dot{W}_{\mathrm{ht}}=(60 \mathrm{~kg} / \mathrm{s})[50+(503.5-254.7) \mathrm{kJ} / \mathrm{kg}]\left(\frac{1 \mathrm{MW}}{1,000 \mathrm{~kJ} / \mathrm{s}}\right)=17.9 \mathrm{MW}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 物理代写|热力学代写Thermodynamics代考|Conserving Mass in an Open System

The best way to begin every thermodynamic analysis is by defining a system. A system describes a region enclosed by an imaginary boundary (which may be fixed or flexible) that contains a mass or volume to use for analysis. A system that doesn’t allow mass to enter or leave is called a closed system. The mass inside a closed system is often called the control mass. A system that allows mass to enter and leave is called an open system. The volume of an open system is often called the control volume.

This chapter focuses on thermodynamic analysis using the conservation of mass and conservation of energy for open systems. Consenvation of mass means that the mass flow rate of material entering a system minus the mass flow rate leaving equals the mass that may accumulate within the system, as described by this equation:
$$\dot{m}{\text {in }}-\dot{m}{\text {out }}=\frac{d m_{3 \mathrm{~s}}}{d t}$$
When you see a “dot” over a variable like mass (mi), the dot means that the variable is on a rate basis or per unit time. The units for mass flow rate are kilograms per second.

Defining mass and volumetric flow rates
The size of the inlets and outlets of some open systems, such as nozzles and diffusers in jet engines, is important because they’re sized to take advantage of changes in kinetic energy. The mass flow rate entering or leaving an open system is related to the area of the opening $(A)$, the average fluid velocity normal to the inlet (V), and the fluid density $(\rho)$, as shown in this equation:
$$\dot{m}=\rho \mathrm{VA}$$
I use bold font for velocity (V) and italicized font for total volume $(V)$ to distinguish between these two variables throughout this book.

In this equation, the units of area are in square meters, velocity is in meters per second, and density is in kilograms per cubic meter.

The volumetric flow rate is related to the mass flow rate and is calculated either by using the average fluid velocity $(V)$ and the area $(A)$ of the opening, or by dividing the mass flow rate by the fluid density, as shown here:
$$\dot{V}=V A=\frac{\dot{m}}{\rho}$$
The units for volumetric flow rate are cubic meters per second.

## 物理代写|热力学代写Thermodynamics代考|Applying conservation of mass to a system

Here’s an example that shows you how to use the conservation of mass principle for an open system. Figure 6-1 shows a jet engine mounted on an aircraft. The system is defined by the dashed line around the engine. The system has two inlets, one for air and the other for fuel. The system has one outlet for exhaust.

The aircraft is flying at 250 meters per second. The air temperature is -50 degrees Celsius, and the pressure is 30 kilopascals. The air mass flow rate into the engine is 60 kilograms per second, and the fuel mass flow rate is 1 kilogram per second. The exhaust is 300 degrees Celsius and has a velocity of 1,000 meters per second. You can analyze this system to determine the volumetric flow rates into and out of the engine by following these steps:

Write the conservation of mass equation for the system.
No mass accumulates within the system, so the mass flow in equals the mass flow out.
$$\dot{m}{\text {air }}+\dot{m}{\text {tual }}=\dot{m}_{\text {estuast }}=(60+1) \mathrm{kg} / \mathrm{s}=61 \mathrm{~kg} / \mathrm{s}$$
To determine the volumetric flow rates, you need to find the gas density at the inlet and the exhaust.

Find the density $\rho_{\mathrm{in}}$ of the air at the inlet, using the ideal-gas-law equation.
$$\rho_{\mathrm{m}}=\frac{P}{R T_{\mathrm{im}}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(223 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.469 \mathrm{~kg} / \mathrm{m}^3$$

Find the density $\rho_{\text {ont }}$ of the exhaust at the exit, using the ideal-gas-law equation. Assume the exhaust has the properties of air.
$$\rho_{\text {out }}=\frac{P}{R T_{\text {ost }}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(573 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.182 \mathrm{~kg} / \mathrm{m}^3$$

Calculate the volumetric flow rate of air at the inlet, using the mass flow rate of the incoming air: $60 \mathrm{~kg} / \mathrm{s}$.
$$\dot{V}{\mathrm{in}}=\frac{\dot{m}{\text {air }}}{\rho_{\text {in }}}=\frac{60 \mathrm{~kg} / \mathrm{s}}{0.469 \mathrm{~kg} / \mathrm{m}^3}=128 \mathrm{~m}^3 / \mathrm{sec}$$

Calculate the volumetric flow rate of the exhaust at the exit, using the total mass flow rate of fuel plus air.
$$\dot{V}{\text {cut }}=\frac{\tilde{m}{\text {eathauf }}}{\rho_{\text {out }}}=\frac{61 \mathrm{~kg} / \mathrm{s}}{0.182 \mathrm{~kg} / \mathrm{m}^3}=335 \mathrm{~m}^3 / \mathrm{sec}$$
The volumetric flow rate depends on the density and mass flow rate of the air.

## 物理代写|热力学代写Thermodynamics代考|Conserving Mass in an Open System

$$\dot{m}{\text {in }}-\dot{m}{\text {out }}=\frac{d m_{3 \mathrm{~s}}}{d t}$$

$$\dot{m}=\rho \mathrm{VA}$$

$$\dot{V}=V A=\frac{\dot{m}}{\rho}$$

## 物理代写|热力学代写Thermodynamics代考|Applying conservation of mass to a system

$$\dot{m}{\text {air }}+\dot{m}{\text {tual }}=\dot{m}_{\text {estuast }}=(60+1) \mathrm{kg} / \mathrm{s}=61 \mathrm{~kg} / \mathrm{s}$$

$$\rho_{\mathrm{m}}=\frac{P}{R T_{\mathrm{im}}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(223 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.469 \mathrm{~kg} / \mathrm{m}^3$$

$$\rho_{\text {out }}=\frac{P}{R T_{\text {ost }}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(573 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.182 \mathrm{~kg} / \mathrm{m}^3$$

$$\dot{V}{\mathrm{in}}=\frac{\dot{m}{\text {air }}}{\rho_{\text {in }}}=\frac{60 \mathrm{~kg} / \mathrm{s}}{0.469 \mathrm{~kg} / \mathrm{m}^3}=128 \mathrm{~m}^3 / \mathrm{sec}$$

$$\dot{V}{\text {cut }}=\frac{\tilde{m}{\text {eathauf }}}{\rho_{\text {out }}}=\frac{61 \mathrm{~kg} / \mathrm{s}}{0.182 \mathrm{~kg} / \mathrm{m}^3}=335 \mathrm{~m}^3 / \mathrm{sec}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 物理代写|热力学代写Thermodynamics代考|Chilling with evaporators

A refrigeration system absorbs heat from a cold environment using a heat exchanger called an evaporator. Figure 4-7 shows a liquid-vapor refrigerant mixture that is colder than the local ambient environment entering the evaporator and boiling to become a superheated vapor. Evaporators are only found in refrigeration (and heat pump) systems; they aren’t used in heat engines. A fan is used to draw air over the evaporator coils. The refrigerant flows through a tube that’s bonded to fins. The fins help improve heat transfer from the ambient air to the refrigerant. The heat transfer rate in an evaporator is found using the same equation used for a condenser:
$$\dot{Q}=\dot{m}\left(h_{\text {out }}-h_{\mathrm{n}}\right)$$
The only difference is that the result is positive because heat is absorbed by the fluid. The enthalpy of the fluid at the evaporator inlet is $h_{i n}$, and the enthalpy at the evaporator exit is $h_{\text {out }}$.

Consider this example, which shows you how to calculate the heat transfer rate from a refrigerant evaporator. A refrigerator evaporator has an $\mathrm{R}-134 \mathrm{a}$ liquid-vapor mixture entering at 250 kilopascals pressure with a 20-percent quality ( discuss quality in Chapter 3 ). Refrigerant leaves the evaporator as a superheated vapor at 0 degrees Celsius. The refrigerant mass flow rate is 0.005 kilogram per second. You can find the heat transfer rate of the evaporator with the following steps:

Find the liquid enthalpy, $h_f$ and the vapor enthalpy, $h_{k^{\prime}}$ at $250 \mathrm{kPa}$ for the refrigerant liquid-vapor mixture entering the evaporator.
By using Table A-7 in the appendix, you find that $h_f=194.3 \mathrm{~kJ} / \mathrm{kg}$ and $h_{\mathrm{g}}=396.1 \mathrm{~kJ} / \mathrm{kg}$.

Calculate the enthalpy of the liquid vapor mixture, $h_{\mathrm{in}}$, at 20-percent quality by using the following equations:
\begin{aligned} & h_{\mathrm{in}}=h_f+x\left(h_{\mathrm{g}}-h_f\right) \ & h_{\mathrm{in}}=194.3 \mathrm{~kJ} / \mathrm{kg}+0.2(396.1-194.3) \mathrm{kJ} / \mathrm{kg}=234.7 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

Find the enthalpy $h_{\text {awt }}$ of the superheated vapor leaving the evaporator, using Table $\mathrm{A}-8$ in the appendix:
$$h_{\text {out }}=399.8 \mathrm{~kJ} / \mathrm{kg}$$

Calculate the heat transfer rate as follows:
$$\dot{Q}=(0.005 \mathrm{~kg} / \mathrm{sec})[(399.8-234.7) \mathrm{kJ} / \mathrm{kg}]=0.83 \mathrm{~kW}$$
Heat transfer to a fluid is a positive quantity.

## 物理代写|热力学代写Thermodynamics代考|Conservina Mass in a Closed Sustem

Every thermodynamic analysis begins by defining a system. A system describes the mass or volume you use for analysis. For example, a system can define the amount of gas contained within a piston and cylinder, the amount of air inside a football, or the amount of iced tea in a glass.

There are two basic categories of systems in thermodynamics. In a closed system, mass neither enters nor leaves the system during a process. In an open system, mass can enter and/or leave the system. This chapter focuses on closed systems. Chapter 6 addresses open systems.

Understanding how to define a system for thermodynamic analysis is important. Say you’re defining a system for a glass of iced tea. If you specify only the tea and the ice as the system, a process for that system may involve melting the ice to cool the tea. If you define the system as the ice, the tea, and the glass, a process may include melting the ice to cool the tea and the glass.
Mass, like energy, can be neither created nor destroyed, but it can change form. A solid mass can melt into a liquid, and a liquid can evaporate into a gas. Even in chemical reactions, the mass of the reactants equals the mass of the products (see Chapter 16). In each process, the mass doesn’t change. The principle of conservation of mass can be summed up as this: The net mass transferred into or out of a system equais the change in mass of a system. Mathematically, this is written as follows:
$$m_{\mathrm{h}}-m_{\text {out }}=\Delta m_{\mathrm{sys}}$$
The units of mass in the SI system are kilograms or grams. The conservation of mass can also be written on a rate basis with this equation:
$$\dot{m}{\text {in }}-\dot{m}{\text {cut }}=\frac{d m_{\text {sws }}}{d t}$$
When you see a “dot” over a variable like mass ( $m$ ), it means the variable is on a rate basis or per unit time. The units for mass flow rate are kilograms per second.
Because no mass flows in or out of a closed system during a thermodynamic process, the conservation of mass equation simplifies to the following equation: $m_{s \mathrm{~s}}=$ constant.

## 物理代写|热力学代写Thermodynamics代考|Chilling with evaporators

$$\dot{Q}=\dot{m}\left(h_{\text {out }}-h_{\mathrm{n}}\right)$$

\begin{aligned} & h_{\mathrm{in}}=h_f+x\left(h_{\mathrm{g}}-h_f\right) \ & h_{\mathrm{in}}=194.3 \mathrm{~kJ} / \mathrm{kg}+0.2(396.1-194.3) \mathrm{kJ} / \mathrm{kg}=234.7 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

$$h_{\text {out }}=399.8 \mathrm{~kJ} / \mathrm{kg}$$

$$\dot{Q}=(0.005 \mathrm{~kg} / \mathrm{sec})[(399.8-234.7) \mathrm{kJ} / \mathrm{kg}]=0.83 \mathrm{~kW}$$

## 物理代写|热力学代写Thermodynamics代考|Conservina Mass in a Closed Sustem

$$m_{\mathrm{h}}-m_{\text {out }}=\Delta m_{\mathrm{sys}}$$

$$\dot{m}{\text {in }}-\dot{m}{\text {cut }}=\frac{d m_{\text {sws }}}{d t}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 物理代写|热力学代写Thermodynamics代考|Working with springs

Springs have been used to store energy and do work for a long time although they’re generally used for doing small bits of work like running a clock, a wristwatch, or a wind-up toy. Work is defined by applying a force over a distance. When you push on a linear spring or twist a spiral spring, you apply work to do work on the spring. The spring is able to store energy until it’s released for doing work.

You may have noticed that when you push against a spring, the force increases with the amount the spring compresses. A material property called the spring constant $(k)$ determines how much force $(F)$ is required to compress or stretch a spring by a distance $(x)$. You calculate the force with the following equation:
$$F=k \cdot x$$
The units for each term in this equation are force in kilonewtons (kN), spring constant in kilonewtons per meter $(\mathrm{kN} / \mathrm{m})$, and distance in meters $(\mathrm{m})$.
$$W_{\text {spring }}=\int F d x=\frac{1}{2} k\left(x_2^2-x_1^2\right)$$
Figure $4-2$ shows that $x_1$ and $x_2$ are the initial and final positions of the spring relative to the relaxed length of the spring. The spring is at the relaxed length when the force on it equals zero. If the work starts with the spring at the relaxed position, then $x_1$ is zero. The units of work are in kilojoules (kJ).

## 物理代写|热力学代写Thermodynamics代考|Turning a shaft

A shaft is the most common way of putting work into motion. An automobile engine has pistons that are connected to a crankshaft that converts the reciprocating motion of the pistons into rotary motion used by the tires. A jet engine has a shaft that connects a turbine to a compressor so the turbine can turn the compressor. An electric motor has a shaft that rotates to do all kinds of work, like operating an elevator or mixing cake batter.
A shaft does not create work; it only carries work with rotary motion. The work carried by a shaft still uses a force moving over a distance. Figure 4-3 shows a perpendicular force $(F)$ applied to a shaft at the radius $(r)$ from the center axis to create torque $(T)$.

The torque on a shaft is related to the force by this equation:
$$T=F, r$$
The units for torque are newton-meters, force is expressed in newtons, and the radius is expressed in meters. The radius of the applied force is not the distance used in determining the work carried by the shaft. The work disof rotations $(n)$ turned by the shaft. The work distance is calculated using this equation:
$$s=(2 \pi r) n$$

The work transmitted by a shaft ( $W_{\text {shaft }}$ ) is calculated using the torque and the distance with this equation:
$$W_{\text {shaft }}=F \cdot s=(T / r)(2 \pi r) n=2 \pi \cdot n \cdot T$$
The units of work are kilojoules, so you have to use the following unit conversion in this equation:
$$1 \mathrm{~kJ}=1,000 \mathrm{~N} \cdot \mathrm{m}$$

## 物理代写|热力学代写Thermodynamics代考|Working with springs

$$F=k \cdot x$$

$$W_{\text {spring }}=\int F d x=\frac{1}{2} k\left(x_2^2-x_1^2\right)$$

## 物理代写|热力学代写Thermodynamics代考|Turning a shaft

$$T=F, r$$

$$s=(2 \pi r) n$$

$$W_{\text {shaft }}=F \cdot s=(T / r)(2 \pi r) n=2 \pi \cdot n \cdot T$$

$$1 \mathrm{~kJ}=1,000 \mathrm{~N} \cdot \mathrm{m}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 物理代写|热力学代写Thermodynamics代考|Getting properties in the vapor dome using quality

You can use the quality to determine many thermodynamic properties, such as the specific volume, internal energy, enthalpy, and entropy of a liquid-vapor mixture. (I discuss the thermodynamic property of entropy in Chapter 8.) The following equation shows you how to calculate the enthalpy of a liquid-vapor mixture, using the quality and enthalpy values from the preceding example in this section:
$$h=x\left(h_{\mathrm{g}}-h_\rho\right)+h_f=0.12(2,675-417.4) \mathrm{kJ} / \mathrm{kg}+417.4 \mathrm{~kJ} / \mathrm{kg}=688.3 \mathrm{~kJ} / \mathrm{kg}$$

This equation can be written in similar ways to find the internal energy $(u)$, specific volume (v), or entropy (s) of a mixture using quality, as shown in these equations:
Internal energy: $u=x\left(u_g-u_q\right)+u_r$
Specific volume: $v=x\left(v_s-v_q\right)+v_f$
Entropy: $s=x\left(s_s-s_\rho\right)+s_f$

## 物理代写|热力学代写Thermodynamics代考|From saturated vapor to superheated vapor

Imagine you want to do something crazy like capture the steam (a saturated vapor) that’s coming from your pot of boiling water and heat it up even more. For liability purposes, I’m not giving you any ideas on how to do this. Adding heat to the steam causes its temperature to increase. Steam that’s hotter than the saturation temperature is called superheated steam or superheated vapor. Please be careful with superheated steam, because it will burn you, and you can’t see it! As the temperature of a superheated vapor increases, its specific volume also increases as long as the pressure remains constant.
Making superheated vapor isn’t really a crazy idea, except when it’s done in your kitchen. In fact, you’re no doubt breathing a superheated vapor right now. Yes, air is a superheated vapor because it’s hotter than liquid air at the same pressure. Superheated vapors don’t have to be hot, even though they sound hot. Superheated vapors run around inside your air-conditioning system at home and in your car (see Chapter 13). Power plants use superheated vapors all the time. A steam generator or a boiler creates superheated steam so it can be used to turn a turbine and make electricity. I discuss the thermodynamic analysis of a power plant in Chapter 12 .

You can find the energy in a superheated vapor from its temperature and pressure. I show you how to do this in the section “Interpolating with two variables.”

## 物理代写|热力学代写Thermodynamics代考|Getting properties in the vapor dome using quality

$$h=x\left(h_{\mathrm{g}}-h_\rho\right)+h_f=0.12(2,675-417.4) \mathrm{kJ} / \mathrm{kg}+417.4 \mathrm{~kJ} / \mathrm{kg}=688.3 \mathrm{~kJ} / \mathrm{kg}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Thermodynamics, 热力学, 物理代写

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## 物理代写|热力学代写Thermodynamics代考|Knowing How Phase Changes Occur

A phase change in a material is one of the most frequently encountered processes in thermodynamics. Melting and freezing between solid and liquid phases are found in thermal energy storage systems, which are used for energy conservation purposes. In an electric power plant, water is boiled in a steam generator and condensed back into a liquid in the condenser (see Chapter 12). The refrigerator in your kitchen has a refrigerant that boils in the evaporator and condenses from a vapor to a liquid in the condenser (see Chapter 13).
Figure 3-5 shows a phase change from liquid to vapor on a temperature-volume $(T-v)$ diagram. As you add heat to a liquid, both the temperature and volume of the liquid increase until the liquid reaches the boiling point. During boiling under constant pressure, the temperature remains constant until all the liquid is turned into vapor. The volume for a given mass of a fluid increases a lot as it becomes a vapor. Often the volume of a vapor is 1,000 times more than the volume of a liquid. Additional heating of the vapor continues to increase both temperature and volume. In this section, I focus on the liquid-to-vapor phase change and introduce you to thermodynamic lingo on phase changes so you can talk like an engineer.

Volume changes due to a temperature change are small in a liquid compared to a vapor.

## 物理代写|热力学代写Thermodynamics代考|From compressed liquid to saturated liquid

Have you ever made a boiled egg? To properly prepare a boiled egg, you start with just enough water to cover the egg. The water in the pot is considered a compressed liquid. Even though nothing appears to be compressing the water, atmospheric pressure does a great job of compressing it. Without atmospheric pressure, the water in the pot would boil immediately and continue boiling until it all evaporated (without being heated at all).
Whether the water is in a pot on your stove or in a pipe feeding the boiler in a power plant, it’s a compressed liquid. As the water in your pot warms up, the volume of the water increases slightly. You can try this at home to see that it’s true. If the cold water barely covers the eggs, you’ll see that the eggs are completely submerged when the water is really hot because the volume of the water increases.

When the water is boiling, it’s a saturated liquid. A saturated liquid is a liquid that’s able to generate vapor if you add more heat to it. So you see bubbles forming in the liquid. The pressure inside the vapor bubbles is equal to the pressure that’s compressing the liquid. This pressure is called the saturation pressure. The temperature of the water when vapor bubbles are formed is called the saturation temperature.
The difference between saturation pressure and saturation temperature can be a bit tricky. Saturation pressure corresponds to a particular temperature, and saturation temperature corresponds to a particular pressure. You can think of it this way: For water, the saturation temperature at 1 atmosphere pressure is 100 degrees Celsius. The saturation pressure at 100 degrees Celsius is 1 atmosphere. After the water starts to boil, set your egg timer for 10 minutes to get perfect hard-boiled eggs.

If you know the temperature and the pressure of the water at any instant in time while the water is warming up, you can find thermodynamic properties like specific volume, internal energy, and enthalpy by using property tables like those in the appendix. You can also use thermodynamic properties to figure out how much energy it takes to make the water reach the boiling point, as shown in this example.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。