外文翻译--机电系统

机电系统 3.1 介绍 这章是处理有关电动机的数学模型和机电模型的。这些机电系统通过电磁感应实现电能和机械能之间的能量转换为基础的。机电系统的这个模型很重要 , 因为他们是大多数控制系统的重要组成。特别注意的是有持续直流场的电动机 , 是许多控制系统的基本的结构。这一个电动机被一个简单的模型描述 ,而且它可以直接地控制马达的扭矩。因为直流电动机的重要和简易，所以这章从介绍直流电动机的模型开始 , 而且呈现直流电动机典型的负载组态。然后根据机械能和电能转换的一般理论挑选主题并且强调在能功能上的一起呈现。提供给我们的 必需的背景来源于有关电动机的比较先进的模型。这包括一般的交流电动机和感应电动机的模型。 3.2 电动机 3.21 介绍 电动机在做旋转运动时不动的部分被称为定子。电动机做旋转运动的部分叫做转子。转子被固定在马达的负载轴上。转子的运动是由电磁洛仑兹力作用在转子上产生的马达扭矩决定的。洛仑兹力的产生有许多不同的方法，而电动机的特性是由洛仑兹力的产生方法决定的。电动机可分为直流电动机和交流电动机。电动机的扭矩能精确地被控制，所以直流电动机很适合应用软件控制。然而，最近在动力电子学方面的发展 , 已经使交流电动机的扭 矩也能被控制，从而使交流电动机现在也用于精确控制。关于电动机的基本参考是 (费兹杰罗， Kingsley 和 Umans 1983)，而一本包括控制方法的比较先进的教科书是 (Lenhard 1996) 。 3.2.2 基本的方程式 一个回转式的电动机有一个随着角速度旋转的电动机轴，而且有一些设备来设定电动机扭矩 T使得电动机轴有如下的等式： Jm m=T-TL TL是作用在轴上的负载扭矩。从轴传到电动机上的机械力是 Pm= T m 而传递到负载上的机械力是 PL= TL m 马达轴动力学可以被描述为一个带 有 T效应和输入端流量 m以及 TL效应和输出端流量 m的四端口。不同的电动机是以马达扭矩 T是如何产生的为特点的。在电动机中扭矩取决于电磁力在液压马达中受压液体的压力，而在涡轮中扭矩是取决于流动液体的流量 m变化所产生的力。马达的转速通常是用转每分钟来描述的。关系到 SI 单位 rad/ s 是 sra dsra drev 1 0 5.0602m in1 3.2.3 传动机构模型 一个电动机通常的转速是从 0到每分钟 3000转。专门设计的电动机可以达到每分钟 12000转。和这个相比教，汽车引擎一般是每分钟 800-6000转。 对于许多应用方 out,Tout in,Tin 面来讲，负载所要求的速度是低于马达的转速的，而且必须还要设置一个减速箱。这就使得负载可以有一个相对较低的转速，更重要的是，它提供了一个更高的扭矩。 带有传动定额 n(图 3.1) 的一个减速装置可以被描述为 out=n in Tin是作用在这个装置的输入端的轴的角速度， Tout是作用在装置输出端的轴的角速度。 对于 n1的减速装置，而且装置的额定配额是 10的话，则 n 1/10。输入扭矩和输出扭矩的关系可以通过比 较这个装置的输入动力和输出动力来得出。假设这是一个无损耗的装置，那么输入的动力就应该等于输出的动力 Tin in=Tout out 带入 out的表达式，我们得出 Tout=n1Tin 这也就是说一个减速装置可以使速度减少到原来的 1/10，而扭矩被放大到原扭矩的10倍。 一个带有比率 n的减速装置可以被描述为一个作用在输入端的变量 Tin和变量 in以及作用在输出端的变量 Tout和变量 out的二端对 out=n in Tout=n1Tin 3.2.4 电动机和传动机构 m,TL Jm TL JL 例如一个带有如下等式的电动机， Jm m=T-TL 这个电动机通过一个比率为 n的减速装置来驱动一个负载。可以得出负载轴的角速度为 L=n m,通过减速装置的输出扭矩 TL /n来驱动。负载的惯性量是 JL，假设作用在负载上的外部扭矩为 T,则计算负载的运动方式的等式为 JL L =n1TL- Te 如果负载方程式 (3.11) 被 n 乘并且加到电动机 (3.10) 的方程式 ,那么得出的就是马达系统的运动方式的等式。相应的，电动机的等式可以除以 n然后加到负载等式上。这将会得出负载系统的运动方式等式。 总结 : 电动机，传动机构和马达的负载的方程式是 (Jm +n2JL) m = T-nTe 电动机，传动机构和负载边的负载的方程式是 (1/n2)Jm+ JL L =n1T-Te 3.25 对平移的旋转装置的变换 m ,TL Fe m v 从旋转装置到平移的图 3.3 传动 一个轴的回转运动能被转换到平面移动，反之亦然，在如图 3.3 所显示的一个表面上装上一个转动的轮 .这种传动在平板 -齿轮传动、虚拟传动装置、滑轮以及在车轮和路面之间经常可见。假如轮子的半径为 r，轴的转速为 m，扭矩为 T，然后平移的速度将会是 v=r m 。记作用在传动部分的力为 F，则输入动力为 mTL，输出动力为 v。由于这个装置并不储存能量，然后可以得出 f= t/t 。这就表示 : 从转动到平移 的转换可以被描述为一个四端口。这个端口带有作用在输入端的变量TL和 m及作用在输出端的变量 F和 v。 v=r m 考虑一个通过半径为 r的轮子来驱动一个在做平移运动的物块。 假定负载的运动方程式为 mv =F-Fe f是作用在负载上面的一个外力。电动机的方程式为 Jm m m= T-TL 电动机和马达的负载的方程式是 (Jm +mr2) m =T-rFe 电动机和负载边的负载方程式是 (1/r2)Jm+ m v =r1T-Fe 3.26 扭矩特性 1 TL TL 2 m m 图 3.4: 左图所示为在 一个稳定的系统中由于增加的发动机速率 Wm而增加的负载扭矩 TL。右图所示为一个系统中的两个平衡点。平衡点 1是稳定的而平衡点 2是不稳定的，因为当发动机转速增加的时候负载扭矩比发动机扭矩降低的块。 在许多应用中负载扭矩将会取决于马达的速度。在图 3.4 的左边线图中显示了这样一个例子 ,负载扭矩随着速度的增加而增加的地方，就是当速度增加摩擦力也增加的地方，就好像汽车和自行车空气阻力一样。 而且，由于在发动机中的能量的损失的增加，发动机扭矩是马达轴转速降低的一个原因。则可以得出，如果发动机扭矩和负载扭矩都可以影响马达 的转速，那么 T=T( m)和 TL= TL( m)，则马达和负载的稳定性可以通过扭矩速度表来研究。通过马达模型的线性化研究 (3.1) , 则可以得出 : Jm m=k m 在这 mTk ( MMLT ) 是一个线性化常数。 从线性稳定性 理论我们可以得出当且仅当 k小于等于 0时，系统才稳定。这可以通过在如图 3.4 所显示的一个扭矩 - 速度的线图来研究。 摩擦力被定义为 TL( m)=Te+(Ts-Te)exp-(sm )2sgn( m)+B m 在这里 t代表库仑摩擦力而 Ts代表了静摩擦力 sgn( m)=11 ( m0,1) 常量 m是 Stribeck效应的特征速度而 B是黏性摩擦系数。对于这个摩擦特点的更细节的研究 ,见第 5 章 . 马达的扭矩可以直接控制，因此 t是一个常量。运动方式的等式是 Jm m=T-Te+(Ts-Te)exp-(sm )2sgn( m)- B m 为了简便假设 w0因此 sgn（ Wm） 1 Jm m=（ 2sm (Ts-Te)exp-(sm )2-B) m 这表示如果速度 Wm B2sm (Ts-Te)exp-(sm )2 系统对于常量马达扭矩 T是不稳定的。 electromechanical systems 3.1 introduction This chapter deals with mathematical models of electrical motors,and models of electromechanical.These electromechanical systems are based on energy conversion between electrical and mechanical energy due to the capacitive and inductive effects.This type of electromechanical systems are important,as they are vital component in most control systems.Special attention is given to the DC motor with constant field,which is a basic building block in many control systems.This motor is described by a simple model.,and it is possible to control the motor torque directly.Because of its importance and simplicity the chapter starts with the model of a DC motor,and presents typical load configuration for DC motor.Then selected topics from the general theory of electromechanical energy conversion is presented with emphasis on energy functions.This provides us with the necessary background to derive more advanced models of electrical motors.This includes the models of a general AC motor,and models for induction motors . 3.2 Electrical motors 3.21 Introduction An electrical motor with rotary motion has a stationary part called the stator .the rotary part of the motor is called the rotor.The rotor is fixed to the motor shaft which drives the load. The motion of the rotor is due to the motor torque which is set up by electromagnetic Lorentz forces acting on the rotor. There are many different ways of setting up an appropriate Lorentz force, and electrical motors are characterized depending on how this is done. Electrical motors are divided into DC motors and AC motors. DC motors are well suited for control applications, as the torque of the motor can be accurately controlled. The recent development in power electronics, however, has made it possible to control the torque also for AC motors,and, consequently, AC motors are now used for accurate control. A basic reference on electrical motors is (Fitzgerald, Kingsley and Umans 1983), while a more adnanced textbook including control methods is (Lenhard 1996). 3.2.2 Basic equations a rotary motor has a motor shaft that rotates with angular velocity ,and it has some device for setting up a motor torque T so that the motor shaft has the following equation of motion: Jm m=T-TL Here TL is the Load torque acting on the shaft. The mechanical power delivered from the motor to the shaft is Pm= T m while the mechanical power delivered to the load is PL= TL m The motor shaft dynamics can be described as a two-port with effort T and flow m at the input port, and effort T and flow m at the output port. Different types of motors are characterized according to how the motor torque T is generated. In electrical motors the torque is due to electromagnetic forces, in a hydraulic motor of the hydrostatic type it is due to the pressure force from a pressurized fluid, while in a turbine the torque is set up by the forces that result from the change of momentum in the flowing fluid. The speed of a motor is commonly given in revolutions per minute (rev/min). The relation to the SI unit rad/s is sra dsra drev 1 0 5.0602m in1 3.2.3 Gear model An electrical motor will typically have a speed range from zero up to about 3000 rev/min. Specially designed electrical motors may run up to 12000 rev/min. In comparison to this, car engines run from 800-6000 rev/min. For many applications the required speed range of the load is significantly less than the speed range of the motor, and a reduction gear must be used. This gives a lower speed of the load, and, more importantly, it gives a higher torque. A reduction gear with gear ration n (Figure 3.1) is described by out=n in where in is the angular velocity of the shaft on the input side of gear, and out is the angular velocity of the shaft on the output side of the gear. For a reduction gear n1, and a gear is said to have a gear ration of, say, 10 if n=1/10. the relation between the input torque Tin and the output Tout is found by comparing power in and power out for the gear. Suppose that the gear is lossless. Then power in is equal to power out, that is, out,Tout in,Tin Tin in=Tout out Inserting the expression for out we find that Tout=n1Tin This means that a reduction gear reduces the speed by a factor n, while it amplifies the torque by a factor 1/n. A gear with gear ratio n may be described as a two-port out=n in Tout=n1Tin with variables Tin and in at the input port, and variables Tout and out at the output port. 3.2.4 Motor and gear Consider a motor with equation of motion Jm m=T-TL that drives a load over a reduction gear with gear ratio n(Figure 3.2). Then the load has a shaft speed L=n mn ,and is driven by the output torque of the gear, which is TL /n. The inertia of the load is JL, and it is assumed m,TL Jm TL JL that an extemal torque T acts on the load. Then the equation of motion for the load is JL L =n1TL- Te If the load equation (3.11) is multiplied by n and added to the equation of the motor (3.10), then the result is the equation of motion for the system referred to the motor side. Alternatively, the motor equation (3.10) can be divided by n and added to the load equation (3.11). This will give the equation of motion of the system referred to the load side. To sum up: The equation of motor, gear and load referred to the motor side is (Jm +n2JL) m = T-nTe The equation of motion for motor, gear and load referred to the load side is (1/n2)Jm+ JL L =n1T-Te 3.25 transformation of rotation to translation m ,TL Fe m v Figure 3.3 Transmission from rotation to translation Rotational motion of a shaft can be transformed to translation motion and vice versa by mounting a wheel that rolls on a surface as shown in Figure 3.3.This type of transmission is commonly seen in rack-and pinion drivers, fiction gears, pulleys, and between car wheels and the road .Suppose that the wheel has radius r ,shaft speed m， and torque TL .Then the translational velocity will be v=r m.Denote the force acting form the wheel on the translating part by F .Then the input power will be mTL and the output power will be v .The gear dose not store energy ,and it follows that f=t/t .T his shows that : A motion to translation transmission can be described buy the two-port v=r m With variables TL and m at the input port , and variables F and v at the output port. Consider a motor which drives a mass in translational motion over a wheel with radius r. The load is assumed to have equation of motion mv =F-Fe Where f is an external force acting on the load .A motor described by Jm m m= T-TL The equation of motion for motor and load referred to the motor side is (Jm +mr2) m =T-rFe The equation of motion for motor and load referred to the load side is (1/r2)Jm+ m v =r1T-Fe 3.26 Torque characteristics 1 TL TL 2 m m Figure 3.4:To the left is shown a stable system where the load torque TL is increasing for increasing motor velocity m .To the right is shown a system with two equilibrium points. Equilibrium 1 is stable , while equilibrium 2 is unstable as the load torque TL decreases faster than the motor torque T when the motor velocity m increases. In many applications the load torque TL will depend on the motor speed. An example of this is shown in the left diagram of Figure 3.4, where the load torque increases with increasing speed. This will be the case for systems where the friction increases with the velocity, like the air resistance of a car or a bicycle. Moreover, the motor torque will typically be a decreasing function of the motor shaft speed m due to increasing energy loss in the motor. It tu