EMS+Lecture+16

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Transcript EMS+Lecture+16

Motor Operation
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
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Counter EMF
 Rotating armature cuts
through the magnetic field
 Voltage is induced in the
armature windings E = B l v
Zn
EO 
60
 This induced voltage is
called counter-electromotive
force (cemf), its polarity acts
against source voltage ES
 Power is taken from the
electrical system
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Electro Mechanical System
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Acceleration of the Motor
 The net voltage acting on the
armature circuit is: ES – EO
 The resulting armature current I
is limited only by the armature
resistance: I = (ES – EO) / R
 At rest, the induced voltage is
zero: EO,rest = 0 V
 Starting Current is 20 to 30 times greater
 The large current produces a large torque
I = ES / R
 As speed increases, the counter emf increases and
the voltage difference diminishes
 Resulting in a reduced current
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Example
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 EO,rest = 0 V so starting current is:
I = ES / 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 = (ES – EO) / R = (150 – 100)/1 = 50A
Armature current at 1460 rpm is
I = (ES – EO) / R = (150 – 146)/1 = 4A
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Machine Power and Torque
 Power and torque characteristics can be determined
over various shaft speeds
 Counter emf: EO = Zn/60
 Power supplied: Pin = Pa = ESI
 Voltage drop (IR losses): ES = EO + IR
 Separating power and losses
Pa = ESI = (EO + IR)I = EO I+ I2R
 The mechanical power : Pm = P = EOI
 The developed torque:
nT
Zn
Pm  P 
 EO I since EO 
so
9.55
60
nT  Zn 
ZI

I 
T

9.55  60 
6.28
Home Work Page 99 Example 5-2
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Electro Mechanical System
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Speed of Rotation
 We know that
EO = 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 fullload
 EO is approximately equal to ES
EO = Zn/60
n = 60EO /Z ≈ 60ES /Z
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Armature speed control
 ES 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 IX varies from zero to maximum or reverse
which in turn vary the ES and motor speed
 This method is known as Ward-Leonard system and is found in
steel mills, high rise elevators and paper mills etc.
 ES is adjusted slightly higher than EO, the armature will absorb
power because I flows into the positive terminal
 Let us reduce ES, now EO current I reverses as a result motor
torque reverses and dc motor suddenly becomes generator
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Armature Speed Control
 Speed is controlled by varying
the armature voltage ES
60 ES
n
Z
 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
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Electro Mechanical System
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