DC MOTORS Part III
Transcript of DC MOTORS Part III
2
Starting of DC Motor
• A starter is a device to start and accelerate a motor.
• A controller is a device to start the motor, control and reverse the
speed of the DC motor and stop the motor.
• While starting the DC motor, it draws the heavy current which
damages the motor.
• The starter reduces the heavy current and protects the system from
damage.
3
Need of Starters for DC Motor
• The dc motor has no back emf.
• At the starting of the motor, the armature current is controlled by the
resistance of the circuit.
• The resistance of the armature is low, and when the full voltage is
applied at the standstill condition of the motor, the armature current
becomes very high which damage the parts of the motor.
• Because of the high armature current, the additional resistance is
placed in the armature circuit at starting.
• The starting resistance of the machine is cut out of the circuit when
the machine gains it speeds.
• The armature current of a motor is given by
4
• Thus, Ia depends upon E and Ra, if V is kept constant.
• When the motor is first switched ON, the armature is stationary.
• Hence, the back EMF Eb is also zero.
• The initial starting armature current Ias is given by the equation shown
below.
• Since, the armature resistance of a motor is very small, generally less
than one ohm.
• Therefore, the starting armature current Ias would be very large.
• For example – if a motor with the armature resistance of 0.5 ohms is
connected directly to a 230 V supply, then by putting the values in the
equation (2) we will get.
• This large current would damage the brushes, commutator and
windings.
5
• As the motor speed increases, the back EMF increases and the difference (V
– E) go on decreasing.
• This results in a gradual decrease of armature current until the motor attains
its stable speed and the corresponding back EMF.
• Under this condition, the armature current reaches its desired value.
• Thus, it is found that the back EMF helps the armature resistance in limiting
the current through the armature.
• Since at the time of starting the DC Motor, the starting current is very large.
• At the time of starting of all DC Motors, except for very small motors, an
extra resistance must be connected in series with the armature.
• This extra resistance is added so that a safe value of the motor is maintained
and to limit the starting current until the motor has attained its stable speed.
• The series resistance is divided into sections which are cut out one by one,
as the speed of the motor rises and the back EMF builds up.
• The extra resistance is cut out when the speed of the motor builds up to its
normal value.
Diagram
It has three terminals L, F and A
The starter consists of starting resistance divided into several sections and
connected in series with the armature.
The three terminals L, F and A of the starter are connected respectively to the
positive line terminal, shunt field terminal and armature terminal.
The other terminals of the armature and shunt field windings are connected to the
negative terminal of the supply.
Operation To start with, the DC supply is switched on with handle in the
OFF position.
The handle is now moved clockwise to the first stud. As soon as
it comes in contact with the first stud, the shunt field winding is
directly connected across the supply, while the whole starting
resistance is inserted in series with the armature circuit
As the handle is gradually moved over to the final stud, the
starting resistance is cut out of the armature circuit in steps. The
handle is now held magnetically by the no-volt release coil
(NVR) which is energized by shunt field current.
• If the supply voltage is suddenly interrupted the no-volt
release coil (NVC) is demagnetized and the handle goes back
to the OFF position under the pull of the spring.
• If no-volt release coil were not used, then in case of failure
of supply, the handle would remain on the final stud.
• If then supply is restored, the motor will be directly
connected across the supply, resulting in an excessive
armature current.
• If the motor is over-loaded (or a fault occurs), it will draw
excessive current from the supply.
• This current will increase the ampere-turns of the over-load
release coil and pull the armature C, thus short-circuiting the
no-volt release coil.
• The no-volt coil is demagnetized and the handle is pulled to
the OFF position by the spring.
• Thus, the motor is automatically disconnected from the
supply.
Drawbacks of 3 point starter
In a three-point starter, the no-volt release coil is connected in
series with the shunt field circuit so that it carries the shunt field
current.
While exercising speed control through field regulator, the field
current may be weakened to such an extent that the no-volt
release coil may not be able to keep the starter arm in the ON
position.
This drawback is overcome in the four point starter.
only difference between a three-point starter and a four-point
starter is the manner in which no-volt release coil is connected.
However, the working of the two starters is the same.
It may be noted that the three-point starter also provides
protection against an open-field circuit.
This protection is not provided by the four-point starter.
APPLICATIONS OF DC MOTORS
MOTORS.. APPLICATIONS…
DC SHUNT MOTOR
LATHES , FANS, PUMPS DISC AND
BAND SAW DRIVE REQUIRING
MODERATE TORQUES.
DC SERIES MOTOR ELECTRIC TRACTION, HIGH SPEED
TOOLS
DC COMPOUND MOTOR
ROLLING MILLS AND OTHER LOADS
REQUIRING LARGE MOMENTARY
TORQUES.
13
Numericals on DC Motor
14
Q.1: A DC shunt machine is connected to 250 V supply and has an
armature resistance of 0.12 ohm and resistance of field circuit is
100 ohm. Find the ratio of the speed as a generator to the speed as a
motor, line current in each case is 80 A.
Sol.: Given: V = 250 V, IL = 80 A, Ra = 0.12 ohm, Rsh = 100 ohm
Ish = V/Rsh = 250/100 = 2.5 A
For generator:
Iag = IL + Ish = 80+2.5 = 82.5 A
Eg = V + IagRa = 250 + 82.5 x 0.12 = 259.9 V
15
For Motor:
Iam = IL – Ish = 80 – 2.5 = 77.5 A
Em = V – IamRa = 250 – 77.5 x 0.12 = 240.7 V
N α E/Φ
Ng/Nm = (Eg x Φm)/(Em x Φg)
Since field current is same
Φg = Φm
Ng/Nm = Eg/Em = 260/240.7 =1.0798 = 1.08
16
Q. 2: The armature resistance of a 200 V shunt motor is 0.4 ohm and no load current is 2 A. When loaded and taken an armature current of 50 A, the speed is 1200 rpm. Find approximately the no load speed.
Sol.: Given: Ra = 0.4 ohm, V = 200 V, N1 = 1200 rpm,
Ia0 = 2A, Ia1 = 50 A
E0 = V – Ia0Ra = 200 – 2 x 0.4 = 199.2 V
E1 = V- Ia1Ra = 200 – 50 x 0.4 = 180 V
N α E/Φ
For shunt motor Φ is constant
N1/N0 = E1/E0 = 180/199.2 = 0.9036
N0= N1/0.9009= 1200/0.9036 = 1328.0212 rpm
17
Q. 3:A 250 V shunt motor on no load runs at 1000 rpm and takes 5 A. The total armature and shunt field resistance are 0.2 ohm and 250 ohm respectively. Calculate the speed when loaded and taking a current of 50 A. If the armature reaction weakens by 3%.
Sol.: Given: V = 250 V, N0 = 1000 rpm, IL = 5 A, Ra =0.2 ohm, Rsh = 250 ohm,
Ish = V/Rsh = 250/250 = 1 A
Ia0 = IL – Ish = 5-1 = 4 A
Ia1 = 50 – 1 = 49 A
E0 = V – Ia0Ra = 250 – 4 x 0.2 = 249.2 V
E1 = V – Ia1Ra = 250 – 49 x 0.2 = 240.2 V
Φ1 = 0.97 Φ0
18
N α E/Φ
N1/N0 = (E1 x Φ0)/(E0 x Φ1) = 240.2 x Φ0/(249.2 x 0.97 Φ0)
N1/N0 = 240.2/(249.2 x 0.97) = 0.9937
N1 = 1000 x 0.9937 = 993.7 rpm
19
Q. 4: A shunt generator delivers 50 kW at 250 V when running at 400 rpm. The armature and field resistance are 0.02 ohm and 50 ohm respectively. Calculate the speed of machine when running as a shunt motor and taking 50 kW input at 250 V. Allowed contact drop is 1 V per brush.
Sol.: Given: P = 50 kW, V = 250 V, Ng = 400 rpm, Ra = 0.02 ohm, Rsh = 50 ohm, Vb = 1 V per brush
IL = P/V = 50,000/250 = 200 A
Ish = V/Rsh = 250/50 =5 A
Iag = IL + Ish = 200+5 = 205 A
Eg = V + IagRa+ Vb = 250 + 205 x 0.02 +2 = 256.1 V
ILm = 200 A
Iam = IL – Ish = 200-5 = 195 A
20
Em = V – IamRa –Vb = 250 – 195 x 0.02 – 2 = 244.1 V
N α E [for constant Φ in shunt machine]
Ng/Nm = Eg/Em = 256.1/244.1
Nm = 400 x 0.9531 = 381.2573 rpm
21
Q. 5: A 4 pole 250 V wave connected shunt motor gives 10
kW when running at 1000 rpm and drawing Ia and Ish of 60
A and 1 A respectively. It has 560 conductors. Its Ra is 0.2
ohm. Assume a drop of 2 V per brush. Determine
(a) Total Torque, (b) Useful Torque, (c) useful Flux per pole,
(d) Rotational losses (e) Efficiency
Sol.: Given: P = 4, A= 2, V = 250 V, Puseful = 10 kW, N =
1000 rpm, Ia = 60 A, Ish = 1 A, Z = 560, Vb = 2 V
Eb = V – IaRa –Vb = 250 – 60 x 0.2 – 2 = 236 V
(a) Total Torque
22
T = (236 x 60 x 60)/(2 x 3.14 x 1000) = 135.3 Nm
(b) Useful torque (Tuseful) = Puseful/w = (P x 60)/(2 x 3.14 x n)
Tuseful = (10000 x 60)/(2 x 3.14 x 1000) = 95.5 Nm
(c) Eb = (PφNZ)/(60A)
Φ = (236 x 60 x 2)/ (4 x 1000 X 560) = 0.0126 Wb
Φ = 12.6 mWb
23
Q. 6: A 230 V shunt motor delivers 30 hp at the shaft at 1120
rpm. If the motor has an
efficiency of 87% at this load, determine:
a) The total input power.
b) The line current.
Sol.:
24
Q. 7: A separately excited DC motor has the following specifications:
Terminal voltage = 250 V, field voltage = 250 V, armature resistance
= 0.03 Ω, field resistance = 250 Ω. Initially the motor was running
at speed = 1103 rpm while supplied by the rated terminal voltage
and the armature current = 120 A. While supplying constant torque,
what is the speed of the motor if the terminal voltage is reduced to
200 V?
Sol.: Torque is constant and Vf is not changed (the field flux will be
constant)