外文翻譯--機(jī)電系統(tǒng)

機(jī)電系統(tǒng) 3.1 介紹 這章是處理有關(guān)電動(dòng)機(jī)的數(shù)學(xué)模型和機(jī)電模型的。這些機(jī)電系統(tǒng)通過(guò)電磁感應(yīng)實(shí)現(xiàn)電能和機(jī)械能之間的能量轉(zhuǎn)換為基礎(chǔ)的。機(jī)電系統(tǒng)的這個(gè)模型很重要 , 因?yàn)樗麄兪谴蠖鄶?shù)控制系統(tǒng)的重要組成。特別注意的是有持續(xù)直流場(chǎng)的電動(dòng)機(jī) , 是許多控制系統(tǒng)的基本的結(jié)構(gòu)。這一個(gè)電動(dòng)機(jī)被一個(gè)簡(jiǎn)單的模型描述 ,而且它可以直接地控制馬達(dá)的扭矩。因?yàn)橹绷麟妱?dòng)機(jī)的重要和簡(jiǎn)易，所以這章從介紹直流電動(dòng)機(jī)的模型開(kāi)始 , 而且呈現(xiàn)直流電動(dòng)機(jī)典型的負(fù)載組態(tài)。然后根據(jù)機(jī)械能和電能轉(zhuǎn)換的一般理論挑選主題并且強(qiáng)調(diào)在能功能上的一起呈現(xiàn)。提供給我們的 必需的背景來(lái)源于有關(guān)電動(dòng)機(jī)的比較先進(jìn)的模型。這包括一般的交流電動(dòng)機(jī)和感應(yīng)電動(dòng)機(jī)的模型。 3.2 電動(dòng)機(jī) 3.21 介紹 電動(dòng)機(jī)在做旋轉(zhuǎn)運(yùn)動(dòng)時(shí)不動(dòng)的部分被稱為定子。電動(dòng)機(jī)做旋轉(zhuǎn)運(yùn)動(dòng)的部分叫做轉(zhuǎn)子。轉(zhuǎn)子被固定在馬達(dá)的負(fù)載軸上。轉(zhuǎn)子的運(yùn)動(dòng)是由電磁洛侖茲力作用在轉(zhuǎn)子上產(chǎn)生的馬達(dá)扭矩決定的。洛侖茲力的產(chǎn)生有許多不同的方法，而電動(dòng)機(jī)的特性是由洛侖茲力的產(chǎn)生方法決定的。電動(dòng)機(jī)可分為直流電動(dòng)機(jī)和交流電動(dòng)機(jī)。電動(dòng)機(jī)的扭矩能精確地被控制，所以直流電動(dòng)機(jī)很適合應(yīng)用軟件控制。然而，最近在動(dòng)力電子學(xué)方面的發(fā)展 , 已經(jīng)使交流電動(dòng)機(jī)的扭 矩也能被控制，從而使交流電動(dòng)機(jī)現(xiàn)在也用于精確控制。關(guān)于電動(dòng)機(jī)的基本參考是 (費(fèi)茲杰羅， Kingsley 和 Umans 1983)，而一本包括控制方法的比較先進(jìn)的教科書(shū)是 (Lenhard 1996) 。 3.2.2 基本的方程式 一個(gè)回轉(zhuǎn)式的電動(dòng)機(jī)有一個(gè)隨著角速度旋轉(zhuǎn)的電動(dòng)機(jī)軸，而且有一些設(shè)備來(lái)設(shè)定電動(dòng)機(jī)扭矩 T使得電動(dòng)機(jī)軸有如下的等式： Jm m=T-TL TL是作用在軸上的負(fù)載扭矩。從軸傳到電動(dòng)機(jī)上的機(jī)械力是 Pm= T m 而傳遞到負(fù)載上的機(jī)械力是 PL= TL m 馬達(dá)軸動(dòng)力學(xué)可以被描述為一個(gè)帶 有 T效應(yīng)和輸入端流量 m以及 TL效應(yīng)和輸出端流量 m的四端口。不同的電動(dòng)機(jī)是以馬達(dá)扭矩 T是如何產(chǎn)生的為特點(diǎn)的。在電動(dòng)機(jī)中扭矩取決于電磁力在液壓馬達(dá)中受壓液體的壓力，而在渦輪中扭矩是取決于流動(dòng)液體的流量 m變化所產(chǎn)生的力。馬達(dá)的轉(zhuǎn)速通常是用轉(zhuǎn)每分鐘來(lái)描述的。關(guān)系到 SI 單位 rad/ s 是 sra dsra drev 1 0 5.0602m in1 3.2.3 傳動(dòng)機(jī)構(gòu)模型 一個(gè)電動(dòng)機(jī)通常的轉(zhuǎn)速是從 0到每分鐘 3000轉(zhuǎn)。專門設(shè)計(jì)的電動(dòng)機(jī)可以達(dá)到每分鐘 12000轉(zhuǎn)。和這個(gè)相比教，汽車引擎一般是每分鐘 800-6000轉(zhuǎn)。 對(duì)于許多應(yīng)用方 out,Tout in,Tin 面來(lái)講，負(fù)載所要求的速度是低于馬達(dá)的轉(zhuǎn)速的，而且必須還要設(shè)置一個(gè)減速箱。這就使得負(fù)載可以有一個(gè)相對(duì)較低的轉(zhuǎn)速，更重要的是，它提供了一個(gè)更高的扭矩。 帶有傳動(dòng)定額 n(圖 3.1) 的一個(gè)減速裝置可以被描述為 out=n in Tin是作用在這個(gè)裝置的輸入端的軸的角速度， Tout是作用在裝置輸出端的軸的角速度。 對(duì)于 n1的減速裝置，而且裝置的額定配額是 10的話，則 n 1/10。輸入扭矩和輸出扭矩的關(guān)系可以通過(guò)比 較這個(gè)裝置的輸入動(dòng)力和輸出動(dòng)力來(lái)得出。假設(shè)這是一個(gè)無(wú)損耗的裝置，那么輸入的動(dòng)力就應(yīng)該等于輸出的動(dòng)力 Tin in=Tout out 帶入 out的表達(dá)式，我們得出 Tout=n1Tin 這也就是說(shuō)一個(gè)減速裝置可以使速度減少到原來(lái)的 1/10，而扭矩被放大到原扭矩的10倍。 一個(gè)帶有比率 n的減速裝置可以被描述為一個(gè)作用在輸入端的變量 Tin和變量 in以及作用在輸出端的變量 Tout和變量 out的二端對(duì) out=n in Tout=n1Tin 3.2.4 電動(dòng)機(jī)和傳動(dòng)機(jī)構(gòu) m,TL Jm TL JL 例如一個(gè)帶有如下等式的電動(dòng)機(jī)， Jm m=T-TL 這個(gè)電動(dòng)機(jī)通過(guò)一個(gè)比率為 n的減速裝置來(lái)驅(qū)動(dòng)一個(gè)負(fù)載?？梢缘贸鲐?fù)載軸的角速度為 L=n m,通過(guò)減速裝置的輸出扭矩 TL /n來(lái)驅(qū)動(dòng)。負(fù)載的慣性量是 JL，假設(shè)作用在負(fù)載上的外部扭矩為 T,則計(jì)算負(fù)載的運(yùn)動(dòng)方式的等式為 JL L =n1TL- Te 如果負(fù)載方程式 (3.11) 被 n 乘并且加到電動(dòng)機(jī) (3.10) 的方程式 ,那么得出的就是馬達(dá)系統(tǒng)的運(yùn)動(dòng)方式的等式。相應(yīng)的，電動(dòng)機(jī)的等式可以除以 n然后加到負(fù)載等式上。這將會(huì)得出負(fù)載系統(tǒng)的運(yùn)動(dòng)方式等式。 總結(jié) : 電動(dòng)機(jī)，傳動(dòng)機(jī)構(gòu)和馬達(dá)的負(fù)載的方程式是 (Jm +n2JL) m = T-nTe 電動(dòng)機(jī)，傳動(dòng)機(jī)構(gòu)和負(fù)載邊的負(fù)載的方程式是 (1/n2)Jm+ JL L =n1T-Te 3.25 對(duì)平移的旋轉(zhuǎn)裝置的變換 m ,TL Fe m v 從旋轉(zhuǎn)裝置到平移的圖 3.3 傳動(dòng) 一個(gè)軸的回轉(zhuǎn)運(yùn)動(dòng)能被轉(zhuǎn)換到平面移動(dòng)，反之亦然，在如圖 3.3 所顯示的一個(gè)表面上裝上一個(gè)轉(zhuǎn)動(dòng)的輪 .這種傳動(dòng)在平板 -齒輪傳動(dòng)、虛擬傳動(dòng)裝置、滑輪以及在車輪和路面之間經(jīng)常可見(jiàn)。假如輪子的半徑為 r，軸的轉(zhuǎn)速為 m，扭矩為 T，然后平移的速度將會(huì)是 v=r m 。記作用在傳動(dòng)部分的力為 F，則輸入動(dòng)力為 mTL，輸出動(dòng)力為 v。由于這個(gè)裝置并不儲(chǔ)存能量，然后可以得出 f= t/t 。這就表示 : 從轉(zhuǎn)動(dòng)到平移 的轉(zhuǎn)換可以被描述為一個(gè)四端口。這個(gè)端口帶有作用在輸入端的變量TL和 m及作用在輸出端的變量 F和 v。 v=r m 考慮一個(gè)通過(guò)半徑為 r的輪子來(lái)驅(qū)動(dòng)一個(gè)在做平移運(yùn)動(dòng)的物塊。 假定負(fù)載的運(yùn)動(dòng)方程式為 mv =F-Fe f是作用在負(fù)載上面的一個(gè)外力。電動(dòng)機(jī)的方程式為 Jm m m= T-TL 電動(dòng)機(jī)和馬達(dá)的負(fù)載的方程式是 (Jm +mr2) m =T-rFe 電動(dòng)機(jī)和負(fù)載邊的負(fù)載方程式是 (1/r2)Jm+ m v =r1T-Fe 3.26 扭矩特性 1 TL TL 2 m m 圖 3.4: 左圖所示為在 一個(gè)穩(wěn)定的系統(tǒng)中由于增加的發(fā)動(dòng)機(jī)速率 Wm而增加的負(fù)載扭矩 TL。右圖所示為一個(gè)系統(tǒng)中的兩個(gè)平衡點(diǎn)。平衡點(diǎn) 1是穩(wěn)定的而平衡點(diǎn) 2是不穩(wěn)定的，因?yàn)楫?dāng)發(fā)動(dòng)機(jī)轉(zhuǎn)速增加的時(shí)候負(fù)載扭矩比發(fā)動(dòng)機(jī)扭矩降低的塊。 在許多應(yīng)用中負(fù)載扭矩將會(huì)取決于馬達(dá)的速度。在圖 3.4 的左邊線圖中顯示了這樣一個(gè)例子 ,負(fù)載扭矩隨著速度的增加而增加的地方，就是當(dāng)速度增加摩擦力也增加的地方，就好像汽車和自行車空氣阻力一樣。 而且，由于在發(fā)動(dòng)機(jī)中的能量的損失的增加，發(fā)動(dòng)機(jī)扭矩是馬達(dá)軸轉(zhuǎn)速降低的一個(gè)原因。則可以得出，如果發(fā)動(dòng)機(jī)扭矩和負(fù)載扭矩都可以影響馬達(dá) 的轉(zhuǎn)速，那么 T=T( m)和 TL= TL( m)，則馬達(dá)和負(fù)載的穩(wěn)定性可以通過(guò)扭矩速度表來(lái)研究。通過(guò)馬達(dá)模型的線性化研究 (3.1) , 則可以得出 : Jm m=k m 在這 mTk ( MMLT ) 是一個(gè)線性化常數(shù)。 從線性穩(wěn)定性 理論我們可以得出當(dāng)且僅當(dāng) k小于等于 0時(shí)，系統(tǒng)才穩(wěn)定。這可以通過(guò)在如圖 3.4 所顯示的一個(gè)扭矩 - 速度的線圖來(lái)研究。 摩擦力被定義為 TL( m)=Te+(Ts-Te)exp-(sm )2sgn( m)+B m 在這里 t代表庫(kù)侖摩擦力而 Ts代表了靜摩擦力 sgn( m)=11 ( m0,1) 常量 m是 Stribeck效應(yīng)的特征速度而 B是黏性摩擦系數(shù)。對(duì)于這個(gè)摩擦特點(diǎn)的更細(xì)節(jié)的研究 ,見(jiàn)第 5 章 . 馬達(dá)的扭矩可以直接控制，因此 t是一個(gè)常量。運(yùn)動(dòng)方式的等式是 Jm m=T-Te+(Ts-Te)exp-(sm )2sgn( m)- B m 為了簡(jiǎn)便假設(shè) w0因此 sgn（ Wm） 1 Jm m=（ 2sm (Ts-Te)exp-(sm )2-B) m 這表示如果速度 Wm B2sm (Ts-Te)exp-(sm )2 系統(tǒng)對(duì)于常量馬達(dá)扭矩 T是不穩(wěn)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