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

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

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 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。