19th Ave New York, NY 95822, USA

物理代写|电磁学代写Electromagnetism代考|PHYSICS7536

my-assignmentexpert™提供最专业的一站式服务：Essay代写，Dissertation代写，Assignment代写，Paper代写，Proposal代写，Proposal代写，Literature Review代写，Online Course，Exam代考等等。my-assignmentexpert™专注为留学生提供Essay代写服务，拥有各个专业的博硕教师团队帮您代写，免费修改及辅导，保证成果完成的效率和质量。同时有多家检测平台帐号，包括Turnitin高级账户，检测论文不会留痕，写好后检测修改，放心可靠，经得起任何考验！

物理代写|电磁学代写Electromagnetism代考|Mechanical Power

The mechanical power developed per unit volume is
$$\mathcal{P}{E M}=u_y \cdot \mathcal{F}{y E}=(1-s) \cdot \frac{\omega}{\ell} \cdot \mathcal{F}{y E}$$ Thus, the mechanical power developed for the two regions in the rotor are $$\mathcal{R}{E M}=\frac{1}{2} \cdot \sigma_1 \cdot(1-s) \cdot s \cdot\left(\frac{\omega \alpha}{\left|\gamma_1\right|}\right)^2 \cdot\left|a \cdot \frac{\cosh \left{\gamma_1(z+d)\right}}{\sinh \left(\gamma_1 d\right)}-b \cdot \frac{\cosh \left(\gamma_1 z\right)}{\sinh \left(\gamma_1 d\right)}\right|^2$$
over $0>z>-d$.
And
$$\mathcal{P}_{2 E M}=\frac{|b|^2}{2} \cdot \sigma_2 \cdot(1-s) \cdot s \cdot\left(\omega \cdot \frac{\mu_2}{\left|\gamma_2\right|}\right)^2 \cdot e^{\left(\gamma_2+\gamma_2\right)-(z+d)}$$
over $-d>z>-\infty$.
From Equations $8.64 b, 8.64 c, 8.67 b$ and $8.67 c$, we get
$$\frac{\mathcal{P}{1 E M}}{\mathcal{P}{1 E L}}=\frac{\mathcal{P}{2 E M}}{\mathcal{P}{2 E L}}=\frac{1-s}{s}$$
Now, if $v_1$ and $v_2$ indicate volume of the hysteresis ring and that of the base on which the ring is mounted, respectively, then from Equation 8.68
$$\int_{v_1} \mathcal{P}{1 E M} d v+\int{v_2} \mathcal{P}{2 E M} d v=\frac{1-s}{s} \cdot\left[\int{v_1} \mathcal{P}{1 E L} d v+\int{v_2} \mathcal{P}{2 E L} d v\right]$$ Therefore, $$\frac{P{E M}}{P_{E L}}=\frac{1-s}{s}$$
where $P_{E M}$ indicates the total mechanical power developed due to induction machine action, and $P_{E L}$ indicates total eddy current loss in the rotor of the machine.

The treatment so far has been concentrated on the operation of an induction machine. The influence of hysteresis is reflected through the expression for $\gamma_1$ in Equation $8.58 \mathrm{c}$, where the term $\beta$ indicates the lag angle due to hysteresis.

物理代写|电磁学代写Electromagnetism代考|Hysteresis Machine Action

To investigate the occurrence of electromechanical energy conversion in the rotor ${ }^9$, we can proceed with the Poynting theorem. The complex Poynting vector, $\mathbb{P}$, at any point in the hysteresis ring is given as
$$\mathbb{P}=\frac{1}{2} \cdot\left(E_1-u \times B_1\right) \times H_1^*$$
In view of the vector identity
$$\nabla \cdot(\mathbf{A} \times \mathbf{B}) \equiv \mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B})$$
we get
$$-\nabla \cdot \mathbb{P} \equiv-\frac{1}{2} H_1^* \cdot\left[\nabla \times\left(E_1-u \times B_1\right)\right]+\frac{1}{2}\left(E_1-u \times B_1\right) \cdot\left[\nabla \times H_1^\right]$$ Now, from Maxwell’s equations for moving medium, if displacement currents are neglected $$\nabla \times\left(E_1-u \times B_1\right)=-\frac{\partial B_1}{\partial t}$$ and $$\nabla \times \boldsymbol{H}_1^=\boldsymbol{J}_1^*$$

Therefore, in a reference frame fixed on the stator
$$-\nabla \cdot \mathbb{P}=\frac{1}{2} \boldsymbol{H}_1^* \cdot \frac{\partial \boldsymbol{B}_1}{\partial t}+\frac{1}{2}\left(\boldsymbol{E}_1-u \times \boldsymbol{B}_1\right) \cdot \boldsymbol{J}_1^*$$
or
$$-\nabla \cdot \mathbb{P}=j \omega \frac{1}{2} \boldsymbol{H}_1^* \cdot \boldsymbol{B}_1+\frac{1}{2} \cdot \frac{1}{\sigma_1} J_1 \cdot J_1^+\frac{1}{2} u \cdot\left(J_1^ \times \boldsymbol{B}_1\right)$$

Since, in view of Equation 8.59a
$$\boldsymbol{B}1=\alpha \cdot e^{-j \beta} \boldsymbol{H}_1$$ One may get from Equation 8.73 \begin{aligned} -\nabla \cdot \mathbb{P}= & j \omega \frac{1}{2} \alpha \cdot e^{-j \beta}\left(H{1 y} \cdot H_{1 y}^+H_{1 z} \cdot H_{1 z}^\right)+\frac{1}{2} \cdot \frac{1}{\sigma_1} J_{1 x} \cdot J_{1 x}^* \ & -u_y \cdot \frac{1}{2} \alpha \cdot e^{-j \beta} J_{1 x}^* \cdot H_{1 z} \end{aligned}
Equating real parts, we find in view of Equations $8.64 \mathrm{~b}$ and $8.67 \mathrm{~b}$
$$\mathcal{R} e[-\nabla \cdot \mathbb{P}] \stackrel{\text { def }}{=} \mathcal{P}{H L}+\mathcal{P}{E L}+\mathcal{P}{E M}+\mathcal{P}{H M}$$

物理代写|电磁学代写Electromagnetism代考|Mechanical Power

$$\mathcal{P}{E M}=u_y \cdot \mathcal{F}{y E}=(1-s) \cdot \frac{\omega}{\ell} \cdot \mathcal{F}{y E}$$因此，为转子中两个区域开发的机械功率为$$\mathcal{R}{E M}=\frac{1}{2} \cdot \sigma_1 \cdot(1-s) \cdot s \cdot\left(\frac{\omega \alpha}{\left|\gamma_1\right|}\right)^2 \cdot\left|a \cdot \frac{\cosh \left{\gamma_1(z+d)\right}}{\sinh \left(\gamma_1 d\right)}-b \cdot \frac{\cosh \left(\gamma_1 z\right)}{\sinh \left(\gamma_1 d\right)}\right|^2$$

$$\mathcal{P}{2 E M}=\frac{|b|^2}{2} \cdot \sigma_2 \cdot(1-s) \cdot s \cdot\left(\omega \cdot \frac{\mu_2}{\left|\gamma_2\right|}\right)^2 \cdot e^{\left(\gamma_2+\gamma_2\right)-(z+d)}$$ 通过$-d>z>-\infty$。 从方程$8.64 b, 8.64 c, 8.67 b$和$8.67 c$，我们得到 $$\frac{\mathcal{P}{1 E M}}{\mathcal{P}{1 E L}}=\frac{\mathcal{P}{2 E M}}{\mathcal{P}{2 E L}}=\frac{1-s}{s}$$ 现在，如果$v_1$和$v_2$分别表示迟滞环的体积和安装环的底座的体积，则由式8.68 $$\int{v_1} \mathcal{P}{1 E M} d v+\int{v_2} \mathcal{P}{2 E M} d v=\frac{1-s}{s} \cdot\left[\int{v_1} \mathcal{P}{1 E L} d v+\int{v_2} \mathcal{P}{2 E L} d v\right]$$因此，$$\frac{P{E M}}{P_{E L}}=\frac{1-s}{s}$$

物理代写|电磁学代写Electromagnetism代考|Hysteresis Machine Action

$$\mathbb{P}=\frac{1}{2} \cdot\left(E_1-u \times B_1\right) \times H_1^*$$

$$\nabla \cdot(\mathbf{A} \times \mathbf{B}) \equiv \mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B})$$

$$-\nabla \cdot \mathbb{P} \equiv-\frac{1}{2} H_1^* \cdot\left[\nabla \times\left(E_1-u \times B_1\right)\right]+\frac{1}{2}\left(E_1-u \times B_1\right) \cdot\left[\nabla \times H_1^\right]$$现在，从运动介质的麦克斯韦方程中，如果位移电流被忽略$$\nabla \times\left(E_1-u \times B_1\right)=-\frac{\partial B_1}{\partial t}$$和 $$\nabla \times \boldsymbol{H}_1^=\boldsymbol{J}_1^*$$

$$-\nabla \cdot \mathbb{P}=\frac{1}{2} \boldsymbol{H}_1^* \cdot \frac{\partial \boldsymbol{B}_1}{\partial t}+\frac{1}{2}\left(\boldsymbol{E}_1-u \times \boldsymbol{B}_1\right) \cdot \boldsymbol{J}_1^*$$

$$-\nabla \cdot \mathbb{P}=j \omega \frac{1}{2} \boldsymbol{H}_1^* \cdot \boldsymbol{B}_1+\frac{1}{2} \cdot \frac{1}{\sigma_1} J_1 \cdot J_1^+\frac{1}{2} u \cdot\left(J_1^ \times \boldsymbol{B}_1\right)$$

$$\boldsymbol{B}1=\alpha \cdot e^{-j \beta} \boldsymbol{H}1$$可以从公式8.73中得到\begin{aligned} -\nabla \cdot \mathbb{P}= & j \omega \frac{1}{2} \alpha \cdot e^{-j \beta}\left(H{1 y} \cdot H{1 y}^+H_{1 z} \cdot H_{1 z}^\right)+\frac{1}{2} \cdot \frac{1}{\sigma_1} J_{1 x} \cdot J_{1 x}^* \ & -u_y \cdot \frac{1}{2} \alpha \cdot e^{-j \beta} J_{1 x}^* \cdot H_{1 z} \end{aligned}

$$\mathcal{R} e[-\nabla \cdot \mathbb{P}] \stackrel{\text { def }}{=} \mathcal{P}{H L}+\mathcal{P}{E L}+\mathcal{P}{E M}+\mathcal{P}{H M}$$

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。