lecture 3 - aastmt
TRANSCRIPT
Lecture 3
DC Motors
A same DC machine can be used as a motor or generator. Construction of a DC motor is same as that of a DC
generator, however, the former converts electrical energy into mechanical energy.
The principle of working of a DC motor is that "whenever a current carrying conductor is placed in a magnetic
field, it experiences a mechanical force". The direction of this force is given by Fleming's left hand rule and it's
magnitude is given by F = BIL. When armature windings are connected to a DC supply, current flows in the
winding. Magnetic field is provided by field winding excitation. In this case, current carrying armature conductors
experience force due to the magnetic field, and this force will produce a torque to rotate the armature, thus rotating
the machine shaft. π»π¨ =ππ·
ππ π¨β π°π¨
I. Operation Principle of DC motor
When the armature of the motor is rotating, the conductors
are also cutting the magnetic flux lines and hence
according to the Faraday's law of electromagnetic
induction, an emf induces in the armature conductors. The
direction of this induced emf opposes the supplied armature
current (Ia), hence itβs called Back emf and given by
the emf equation of DC generator;
EA = πΎβ ππ, where K=ππ
2ππ΄
Torque per conductor;
πππππ = ππΉ = ππΌπππππΏπ΅where, π: the distance from the conductor axis to the force
point, πΉ: force applied on conductor, πΌππππ : the conductor
current, πΏ: Conductor length, π΅: magnetic field density
If there are A current paths in the machine, then the total
armature current IA is split among these paths, so the current in
a single conductor is;
πΌππππ =πΌπ΄
π΄
Then the torque per conductor is given by;
πππππ =ππΌπ΄πΏπ΅
π΄
The total flux (Pβ ) acting on the armature conductor is given
by;
πβ = π΅π΄π=π΅2πππΏWhere β is the flux per pole, P is total no of poles, and
π΄πis the conductor area facing the pole
Hence, πππππ =πβ πΌπ΄
2ππ΄
Total torque developed in armature;
π»π¨ =ππ·
ππ π¨β π°π¨= Kβ π°π¨.
Where Z: total number of condcutors
EA = πΎβ ππ EA πΌπ΄ = πΎβ ππ πΌπ΄
Hence, EA πΌπ΄ = ππ΄ππ = Pdev
In the DC motor, the commutator applies electric
current to the windings. By reversing the current
direction in the rotating conductors each half turn, a
steady unidirectional torque is produced. Hence, the
coil will rotate continually in the same direction.
Need of Commutator
III. Power flow diagramDC motors take in electric power and produce mechanical power. The efficiency of a DC machine is defined by
πΌπ΄2π π΄ +πΌπΉ
2π πΉ
DC motors are usually classified of the basis of their excitation configuration, as follows -
β’Separately excited (field winding is fed by external source)
β’Self excited -
β’ Series wound (field winding is connected in series with the armature)
β’ Shunt wound (field winding is connected in parallel with the armature)
IV. DC motor types
Separately excited Shunt Series
ππ = πΈπ΄ +πΌπ΄π π΄πΌπ = πΌπ΄
πΌπΉ =ππΉπ πΉ
ππ = πΈπ΄ +πΌπ΄π π΄πΌπ = πΌπ΄ + πΌπΉ
πΌπΉ =ππ π πΉ
Speed Regulation, (SR)
ππ = πΈπ΄ +πΌπ΄(π π΄+π π)πΌπ΄ = πΌπ
Generally, three characteristic curves are considered for DC motors which are,
(i) Developed Torque versus armature current
(ii) Speed versus armature current
(iii) Terminal characteristics (Speed versus developed torque)
These characteristics are determined by keeping following two relations in mind.
V. DC motor characteristics
πΈπ΄ = πΎβ ππ ππππ£ = πΎβ πΌπ΄
A separately excited dc motor is a motor whose field circuit is supplied from a separate constant-voltage power
supply, while a shunt dc motor is a motor whose field circuit gets its power directly across the armature terminals
of the motor. When the supply voltage to a motor is assumed constant, there is no practical difference in behavior
between these two machines.
When the load increases, the output torque required to drive the load will increase. Hence, the motor speed will
slow down. Consequently the internal generated voltage drops (πΈπ΄ = πΎβ ππ β) , increasing the armature current in
motor πΌπ΄ = (ππ βπΈπ΄β)/π π΄. As the armature current increases, the developed torque increase (ππππ£ = πΎβ πΌπ΄ β) and
finally the developed torque will be equal the load torque at a lower mechanical speed of rotation ππ.
Mechanical Load β ππ β, πΌπ΄ β , ππππ£ β
I. Separately excited/ Shunt DC motor
Torque vs. armature current
Generally, the developed torque is directly proportional to armature current and the relationship is in the form of a
straight line, assuming the field flux Ξ¦ to be constant as the supply voltage is constant.
Since, heavy starting load needs high starting current, shunt motor should never be started on a heavy load.
Speed vs. armature current
ππ = πΈπ΄ +πΌπ΄π π΄ πππ πΈπ΄ = πΎβ ππ »» ππ = πΎβ ππ + πΌπ΄ π π΄ »» ππ =(ππ β πΌπ΄π π΄)
πΎβ
As flux Ξ¦ is assumed constant, the speed decreases with armature current increase. But practically, due to armature reaction, Ξ¦
decreases with increase in armature current, and hence the speed decrease slightly. Hence, a shunt motor can be assumed as a
constant speed motor.
Torque vs. speed
ππ =(ππ β ππππ£ π π΄/πΎβ )
πΎβ As flux Ξ¦ is assumed constant, , the speed decreases with developed torque increase. But practically, due to armature reaction, Ξ¦
decreases with increase in armature current, and hence the speed decrease slightly. Thus, at heavy loads, the motor speed is
almost constant.
(ππππ£ = πΎβ πΌπ΄)
2. Adjusting the field resistance πΌπΉ =ππ
π πΉ(and thus the field flux). This can be applied to separately excited and shunt
motors
Hence, for a constant supply voltage, at a certain load, increasing the flux decreases the motor speed.
3. Inserting a resistor in series with the armature circuit. This can be applied to separately excited and shunt motors
Hence, for a constant supply voltage and fixed flux, at a certain load, increasing π π΄ decreases the motor speed
Speed Control of Separately excited and shunt DC Motors
1. Adjusting the supply voltage applied to the armature without changing the voltage applied to the field. Hence,
the flux is kept constant. This can be applied to separately excited motors only.
Hence, at a certain load, since the flux is fixed, increasing the armature voltage , increases the motor speed
ππ =(ππ β πΌπ΄ π π΄)
πΎβ
II. Series motor
In the DC series motor, the flux is directly proportional to the armature current. As the motor load increases, the armature
current increases hence the flux increases β = ππΌπ = ππΌπ΄
Torque vs. armature current
The developed torque is directly proportional to the square of the armature current and the Tdev-IA curve is parabola for smaller
values of IA.
Speed vs. armature current
ππ = πΈπ΄ +πΌπ΄(π π΄+π π ) πππ πΈπ΄ = πΎππΌπ΄ππ »» ππ = πΎππΌπ΄ππ + πΌπ΄ (π π΄+π π ) »» ππ =ππ
πΎππΌπ΄β
(π π΄+π π )
πΎπ
Hence, for series motor, the speed is inversely proportional to the armature current as shown in the speed-armature current curve.
When armature current is very small the speed becomes dangerously high. That is why a series motor should never be started
without some mechanical load
Torque vs. speed
ππππ£ = πΎππΌπ΄2 β«β« πΌπ΄ =
ππππ£
πΎπ»» ππ =
ππ
πΎπ ππππ£β
(π π΄+π π )
πΎπ
ππππ£ = πΎβ πΌπ΄ β«β«β«ππππ£ = πΎππΌπ΄2
For series motor, the speed is inversely proportional to the square root of the torque
Speed Control of series Motors
ππ =ππ πΎππΌπ΄
β(π π΄+π π )
πΎπ1. Adjusting the supply voltage applied
At a certain load, increasing the supply voltage , increases the motor speed
2. Inserting a series resistor into the motor circuit
At a certain load and certain supply voltage, increasing the circuit resistance, decreases the motor speed
Applications
β’ Separately excited DC motors are often used as actuators in trains and automotive traction applications.
β’ For their constant speed characteristics, shunt DC motors are used in fixed speed applications such as fans.
β’ Since the series motors can give high torque per ampere (since their toque is directly proportional to the
square of armature current), they can be used in applications that require high starting torque. Examples of
these applications include; starter motors in cars, and elevator motors.
β’ Discuss the theory of operation of DC motor
β’ What is the need of commutator for DC motor ?
β’ What are different DC motor types?
Draw their equivalent circuits
Draw their torque-armature current characteristics
Draw their speed-armature current characteristics
Draw their speed-torque characteristics
How to control their speed?
State their applications