Lecture 16Electro Mechanical System1 DC motors are built the same way as generators  Armature of a motor connected to a dc power supply  When switch is closed a large current flows through the armature winding due to its low resistance  Armature is within a magnetic field  A force is exerted on the windings  The force causes a torque on the shaft  The shaft rotates Motor Operation

Lecture 16Electro Mechanical System2  Rotating armature cuts through the magnetic field  Voltage is induced in the armature windings E = B l v  This induced voltage is called counter-electromotive force (cemf), its polarity acts against source voltage E S  Power is taken from the electrical system Counter EMF

Lecture 16Electro Mechanical System3  The net voltage acting on the armature circuit is: E S – E O  The resulting armature current I is limited only by the armature resistance: I = (E S – E O ) / R  At rest, the induced voltage is zero: E O,rest = 0 V Acceleration of the Motor  Starting Current is 20 to 30 times greater  The large current produces a large torque I = E S / R  As speed increases, the counter emf increases and the voltage difference diminishes  Resulting in a reduced current

Lecture 16Electro Mechanical System4 Armature of a permanent magnet dc generator has a resistance of 1 ohm and generates 50V at a speed of 500 rpm. If the armature is connected to a 150V supply, find: a)The starting current b)The counter-emf when the motor runs at 1000 rpm. At 1460 rpm c)The armature current at 1000 rpm. At 1460 rpm a)At start-up E O,rest = 0 V so starting current is: I = E S / R = 150V / 1Ω = 150A b)Generator voltage at 500 rpm is 50 V, so counter-emf of the motor at 1000 rpm will be 100V and at 1460 rpm will be 146V c)Armature current at 1000 rpm is I = (E S – E O ) / R = (150 – 100)/1 = 50A Armature current at 1460 rpm is I = (E S – E O ) / R = (150 – 146)/1 = 4A Example

Lecture 16Electro Mechanical System5  Power and torque characteristics can be determined over various shaft speeds  Counter emf: E O = Zn  /60  Power supplied: P in = P a = E S I  Voltage drop (IR losses): E S = E O + IR  Separating power and losses P a = E S I = (E O + IR)I = E O I+ I 2 R  The mechanical power : P m = P = E O I  The developed torque: Machine Power and Torque Home Work Page 99 Example 5-2

Lecture 16Electro Mechanical System6  We know that E O = Zn  /60  The voltage drop across the armature resistance is always small compared to the supply voltage  Even as the load varies from no-load to full- load  E O is approximately equal to E S E O = Zn  /60 n = 60E O /Z  ≈ 60E S /Z  Speed of Rotation

Lecture 16Electro Mechanical System7  E S can be varied by connecting motor armature to a separately excited variable voltage dc generator G  Field excitation of the motor is kept constant, but the generator excitation current I X varies from zero to maximum or reverse which in turn vary the E S and motor speed  This method is known as Ward-Leonard system and is found in steel mills, high rise elevators and paper mills etc.  E S is adjusted slightly higher than E O, the armature will absorb power because I flows into the positive terminal  Let us reduce E S, now E O current I reverses as a result motor torque reverses and dc motor suddenly becomes generator Armature speed control

Lecture 16Electro Mechanical System8  Speed is controlled by varying the armature voltage E S  Motor speed changes proportionally to the armature voltage  The armature voltage is controlled by an external variable power supply  The field winding is separately excited by a constant voltage source Armature Speed Control