教学目标:
1、 了解直流电机的基本结构、工作原理
2、 掌握直流电动机拖动系统的机械特性和动态特性
3、 掌握两种直流电动机起动方式;
4、 掌握直流电动机三种调速方式;
5、 他励直流电动机的制动方法有三种:能耗制动、反接制动和回馈制动;
6、 运用机械特性和运动方程式分析各种运转状态的过渡过程。
2.1 Elementary dc machine and Its Applications
1. Basic Principle
A dc machine is consisted of several arts:
carbon brushes air gap
copper commutator segments
armature windings ( coils)
armature iron ( rotor)
main pole (stator is salient pole)
2. Applications
hoists, fans, pumps, calendars
punch-presses, and cars
special applications:
steel mills, mines, electric trains
2.2 他励直流电动机的机械特性
机械特性的一般表达式
他励直流电动机机械特性是指电动机加上一定的电压 U 和一定的励磁电流 I f 时,电磁转矩与转速之间的关系,即 n=f(T) 。
固有机械特性
他励直流电动机固有机械特性是一条斜直线,跨三个象限,特性较硬。机械特性只表征电动机电磁转矩和转速之间的函数关系,是电动机本身的能力,至于电动机具体运行状态,还要看拖动什么样的负载。固有机械特性是电动机最重要的特性,在它的基础上,很容易得到电动机的人为机械特性。
人为机械特性
他励直流电动机的参数如电压、励磁电流、电枢回路电阻大小等改变后,其机械特性称为人为机械特性。
o 电枢回路串电阻的人为机械特性
o 改变电枢电压的人为机械特性
o 减少气隙磁通量的人为机械特性
减小气隙每极磁通的方法是减小励磁电流来实现的。电机磁路接近于饱和,增大每极磁通是难以做到的,改变磁通,都是减少磁通。
2.3 串励和复励直流电动机
串励直流电动机的机械特性
把励磁绕组串联在电枢回路就是串励直流电机,其正方向仍用电动机惯例
特点 :
是一条非线性的软特性;
当电磁转矩很小时,转速 n 很高;
电磁转矩 T 与电枢电流 I a 的平方成正比,因此起动转矩大,过载能力强。
复励直流电动机的机械特性
如果并励与串励两个励磁绕组的极性相同,叫积复励;极性相反,叫差复励。
积复励具有串励电动机的起动转矩大,过载能力强的优点,而没有空载转速很高的缺点。应用广泛,如无轨电车的拖动。
2.4 tarting of D.C. motor
1. Referring first to the shunt motor ,eqn shows that the downwards slop of the speed/torque curve is directly proportional to the armature resistance with constant flux and voltage.
To bring a motor rest up to speed from a constant supply voltage is a special case of speed control in which the armature-circuit resistance is varied. At standstill the e.m.f. is zero so the armature resistance alone limits the stating current and torque. With a value of say 0.05 p.u., the starting current at full voltage would be 1/0.05=20 p.u. Extra resistance is therefore required, and is connected in series with the armature.
2. By increasing the resistance, the characteristic will cut zero-speed axis to give a lower , though adequate starting torque and a reasonable starting current.
3. For a series motor, the effects of the additional circuit resistance can easily be calculated using the tabular method and a typical set of curves is shown on Fig. Rather fewer steps of starting resistance are necessary due to the shape of the characteristics.
A shunt motor rotating at 1500r/min is fed by a 120V source. The line current is 51A and the shunt-field resistance is 120 Ω . If the armature resistance is 0.1 Ω ,calculate the following:
a. The current in the armature
b. The counter-emf
c. The mechanical power developed by the motor
Solution:
a. The field current is
Im=120V/120 Ω =1A
The armature current is
I a= 51-1 = 50A
b.
本文关键字:直流电机 变频器基础,变频技术 - 变频器基础