外文翻譯--機電系統(tǒng)

機電系統(tǒng) 3.1 介紹 這章是處理有關電動機的數(shù)學模型和機電模型的。這些機電系統(tǒng)通過電磁感應實現(xiàn)電能和機械能之間的能量轉換為基礎的。機電系統(tǒng)的這個模型很重要 , 因為他們是大多數(shù)控制系統(tǒng)的重要組成。特別注意的是有持續(xù)直流場的電動機 , 是許多控制系統(tǒng)的基本的結構。這一個電動機被一個簡單的模型描述 ,而且它可以直接地控制馬達的扭矩。因為直流電動機的重要和簡易，所以這章從介紹直流電動機的模型開始 , 而且呈現(xiàn)直流電動機典型的負載組態(tài)。然后根據(jù)機械能和電能轉換的一般理論挑選主題并且強調在能功能上的一起呈現(xiàn)。提供給我們的 必需的背景來源于有關電動機的比較先進的模型。這包括一般的交流電動機和感應電動機的模型。 3.2 電動機 3.21 介紹 電動機在做旋轉運動時不動的部分被稱為定子。電動機做旋轉運動的部分叫做轉子。轉子被固定在馬達的負載軸上。轉子的運動是由電磁洛侖茲力作用在轉子上產(chǎn)生的馬達扭矩決定的。洛侖茲力的產(chǎn)生有許多不同的方法，而電動機的特性是由洛侖茲力的產(chǎn)生方法決定的。電動機可分為直流電動機和交流電動機。電動機的扭矩能精確地被控制，所以直流電動機很適合應用軟件控制。然而，最近在動力電子學方面的發(fā)展 , 已經(jīng)使交流電動機的扭 矩也能被控制，從而使交流電動機現(xiàn)在也用于精確控制。關于電動機的基本參考是 (費茲杰羅， Kingsley 和 Umans 1983)，而一本包括控制方法的比較先進的教科書是 (Lenhard 1996) 。 3.2.2 基本的方程式 一個回轉式的電動機有一個隨著角速度旋轉的電動機軸，而且有一些設備來設定電動機扭矩 T使得電動機軸有如下的等式： Jm m=T-TL TL是作用在軸上的負載扭矩。從軸傳到電動機上的機械力是 Pm= T m 而傳遞到負載上的機械力是 PL= TL m 馬達軸動力學可以被描述為一個帶 有 T效應和輸入端流量 m以及 TL效應和輸出端流量 m的四端口。不同的電動機是以馬達扭矩 T是如何產(chǎn)生的為特點的。在電動機中扭矩取決于電磁力在液壓馬達中受壓液體的壓力，而在渦輪中扭矩是取決于流動液體的流量 m變化所產(chǎn)生的力。馬達的轉速通常是用轉每分鐘來描述的。關系到 SI 單位 rad/ s 是 sra dsra drev 1 0 5.0602m in1 3.2.3 傳動機構模型 一個電動機通常的轉速是從 0到每分鐘 3000轉。專門設計的電動機可以達到每分鐘 12000轉。和這個相比教，汽車引擎一般是每分鐘 800-6000轉。 對于許多應用方 out,Tout in,Tin 面來講，負載所要求的速度是低于馬達的轉速的，而且必須還要設置一個減速箱。這就使得負載可以有一個相對較低的轉速，更重要的是，它提供了一個更高的扭矩。 帶有傳動定額 n(圖 3.1) 的一個減速裝置可以被描述為 out=n in Tin是作用在這個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系可以通過比 較這個裝置的輸入動力和輸出動力來得出。假設這是一個無損耗的裝置，那么輸入的動力就應該等于輸出的動力 Tin in=Tout out 帶入 out的表達式，我們得出 Tout=n1Tin 這也就是說一個減速裝置可以使速度減少到原來的 1/10，而扭矩被放大到原扭矩的10倍。 一個帶有比率 n的減速裝置可以被描述為一個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 out=n in Tout=n1Tin 3.2.4 電動機和傳動機構 m,TL Jm TL JL 例如一個帶有如下等式的電動機， Jm m=T-TL 這個電動機通過一個比率為 n的減速裝置來驅動一個負載?？梢缘贸鲐撦d軸的角速度為 L=n m,通過減速裝置的輸出扭矩 TL /n來驅動。負載的慣性量是 JL，假設作用在負載上的外部扭矩為 T,則計算負載的運動方式的等式為 JL L =n1TL- Te 如果負載方程式 (3.11) 被 n 乘并且加到電動機 (3.10) 的方程式 ,那么得出的就是馬達系統(tǒng)的運動方式的等式。相應的，電動機的等式可以除以 n然后加到負載等式上。這將會得出負載系統(tǒng)的運動方式等式。 總結 : 電動機，傳動機構和馬達的負載的方程式是 (Jm +n2JL) m = T-nTe 電動機，傳動機構和負載邊的負載的方程式是 (1/n2)Jm+ JL L =n1T-Te 3.25 對平移的旋轉裝置的變換 m ,TL Fe m v 從旋轉裝置到平移的圖 3.3 傳動 一個軸的回轉運動能被轉換到平面移動，反之亦然，在如圖 3.3 所顯示的一個表面上裝上一個轉動的輪 .這種傳動在平板 -齒輪傳動、虛擬傳動裝置、滑輪以及在車輪和路面之間經(jīng)常可見。假如輪子的半徑為 r，軸的轉速為 m，扭矩為 T，然后平移的速度將會是 v=r m 。記作用在傳動部分的力為 F，則輸入動力為 mTL，輸出動力為 v。由于這個裝置并不儲存能量，然后可以得出 f= t/t 。這就表示 : 從轉動到平移 的轉換可以被描述為一個四端口。這個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 考慮一個通過半徑為 r的輪子來驅動一個在做平移運動的物塊。 假定負載的運動方程式為 mv =F-Fe f是作用在負載上面的一個外力。電動機的方程式為 Jm m m= T-TL 電動機和馬達的負載的方程式是 (Jm +mr2) m =T-rFe 電動機和負載邊的負載方程式是 (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 一個穩(wěn)定的系統(tǒng)中由于增加的發(fā)動機速率 Wm而增加的負載扭矩 TL。右圖所示為一個系統(tǒng)中的兩個平衡點。平衡點 1是穩(wěn)定的而平衡點 2是不穩(wěn)定的，因為當發(fā)動機轉速增加的時候負載扭矩比發(fā)動機扭矩降低的塊。 在許多應用中負載扭矩將會取決于馬達的速度。在圖 3.4 的左邊線圖中顯示了這樣一個例子 ,負載扭矩隨著速度的增加而增加的地方，就是當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)動機中的能量的損失的增加，發(fā)動機扭矩是馬達軸轉速降低的一個原因。則可以得出，如果發(fā)動機扭矩和負載扭矩都可以影響馬達 的轉速，那么 T=T( m)和 TL= TL( m)，則馬達和負載的穩(wěn)定性可以通過扭矩速度表來研究。通過馬達模型的線性化研究 (3.1) , 則可以得出 : Jm m=k m 在這 mTk ( MMLT ) 是一個線性化常數(shù)。 從線性穩(wěn)定性 理論我們可以得出當且僅當 k小于等于 0時，系統(tǒng)才穩(wěn)定。這可以通過在如圖 3.4 所顯示的一個扭矩 - 速度的線圖來研究。 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效應的特征速度而 B是黏性摩擦系數(shù)。對于這個摩擦特點的更細節(jié)的研究 ,見第 5 章 . 馬達的扭矩可以直接控制，因此 t是一個常量。運動方式的等式是 Jm m=T-Te+(Ts-Te)exp-(sm )2sgn( m)- B m 為了簡便假設 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)對于常量馬達扭矩 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