RGPV EX503 UNIT II NOTES FOR BE STUDENTS OF V SEM EX BRANCH

1. READING MATERIAL FOR B.E STUDENTS
OF RGPV AFFILIATED
ENGINEERING COLLEGES PERSUING IN
ELECTRICAL AND ELECTRONICS BRANCH
Professor MD Dutt
Addl General Manager (Retd)
BHARAT HEAVY ELECTRICALS LIMITED
Professor(retd) in EX Department
Bansal Institute of Science and Technology
KOKTA ANANAD NAGAR BHOPAL
Presently Head of The Department ( EX)
Shri Ram College Of Technology
Thuakheda BHOPAL
Sub Code Ex503 Subject Electrical Machine II
UNIT II DC Machines II

2. RGPV Syllabus
EX 503 ELECTRICAL MACHINES II
UNIT I I
DC MACHINES II
Basic operation of DC motor, Torque equation, Operating characteristics of DC
motors, Starting of DC motors 2 Point, 3 Point and 4 point starters. Speed control of
DC motors. Losses and efficiency of DC machines, testing of DC machines , direct
testing, Swinburne test, Hopkinson’s test, Application of DC machines.
INDEX
Topic Page No
1 Basic operation of DC motor 3
2 Torque equation 4,5
3 Operating characteristics of DC motors, 6,7,8,9
4 Starting of DC motors 2 Point 10,11
5 3 Point starters 12,11
6 4 point starters 13
7 Speed control of DC motors 14,15,16,17
8 Losses and efficiency of DC machines 17,18,19,20
9 Testing of DC machines 21,22
10 Direct testing 23
11 Swinburne test 23,24,25
13 Hopkinson’s test 26,27,28
14 Application of DC machines. 28
15 Formulas 29

3. Q1 Basic principle of operation of a D.C. motor?
When a conductor carrying current is put in a magnetic field, a force is produced in it.
Let us consider one such conductor placed in a slot of armature and it is acted upon
the magnetic field from the north pole of the motor. By applying L.H.S it is found
that the conductor has tendency to move to the L.H.S. Since the conductor is placed on
the slot at circumference of rotor , The force acts in a tangential direction of rotor,
Thus a torque is developed on the rotor Similar torques are produced on all the rotor
conductors .Since the rotor is free to move, It starts rotating.
BACK EMF
When the motor of armature rotates, its conductors cut the magnetic flux, Therefore the
e.m.f. rotation Er is induced in them. In case of a motor, the e.m.f is known as BACK
E.M.F or counter e.m.f. The back e.m.f opposes the applied voltage , Since the back
E.M.F is induced due to generating action it magnitude is given
Eb = PNZ/60A
TORQUE OF A DC MACHINE

4. When the machine is operating as a motor the torque is transferred to the shaft of
the rotor and drives the mechanical load, The expression for the torque is same
for the generator or motor.
The Voltage equation of DC machine is
V= E+IaRa
Multiplying with Ia
We get
V Ia = EIa +Ia²Ra
VIa = Electrical input to the armature
Ia²Ra = copper loss in the armature
We also know that
Input = output + losses
EIa = Electrical equivalent of gross mechanical power developed by armature
τav = average electromagnetic torque developed by armature in Newton Mtrs
(Nm)
Mechanical power developed by the armature
Pm = ώτav = 2Пņτav
Pm = EIa = ώτav = 2Пņτav
But
E = nP ΦZ/A
Therefore

5. (nP ΦZ/A) Ia =2Пņτav
τav = (PZ/2П A ) X ΦIa
This is called Torque equation
For agiven machine the value of PZ A are constant
Therefore PZ/2П A is also constant = k
τav = k ΦIa
EQUIVALENT DIAGRAM
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RIEV
RIV

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6. OPERATING CHARACTERISTICS OF DC MOTORS
In both cases of shunt or separately excited D.C moter’s . the field is supplied from a
constant voltage so that the field current is constant. It is graph between two dependent
quantities.
SPEED ARMATURE CURRENT CHARECTERISTICS:- In a shunt motor,
Ish =V÷Rsh, if V is constant Rsh will also be constant, Hence the flux is constant but
at no load the flux decreases slightly due to armature reaction. If the effect of armature
Reaction is neglected the flux will remain constant. The motor speed is given by
N= V-Ia RA
Φ
If Φ
Is constant
N is proportional to V-Ia RA
This is the equation of straight line with a negative slope. That is the speed N of motor
decreases linearly with increase of current. Since the IaRa at full load is very small
compare to V the drop in speed from noload to full load is very small. For all practical
purposes the shunt motor is taken a constant speed motor.
TORQUE / ARMATURE CURRENT CHARECTERISTICS:- IF the flux Φ is
constant in a shunt motor, the torque would increase linearly with armature current Ia,
However for larger Ia , the net flux decreases due to the demagnetizing effect of
armature reaction. In view of this , the torque current characteristics deviates from
straight line as illustrated here below

7. τav = k ΦIa
SPEED TORQUE CHARECTERISTICS::- The speed torque characteristics is also
called the mechanical characteristics and under steady state conditions it can be
obtained as follows
ωm = Vt – Ia Ra
KaΦ
It is seen from the characteristics that with increase in torque the speed drops.

8. DC SERIES MOTOR
SPEED CURRENT CHARECTERISTICS::- If the armature reaction and
saturation is neglected, the main flux is directly proportional to the armature current Ia
, therefore it can be written as
Φ K Ia where K is constant
ωm o = Vt/ Ka Φ = Vt/Ka K Ia
Since ωm o is inversely proportional to Ia, the no load speed of series motor becomes
dangerously due to small no load current . In view of this, the series motor must always
be started and operate under load mechanically coupled with it.
TORQUE CURRENT CHARECTERISTICS::- We know that Φ = KIa which
shows that the torque is proportional to the Ia and therefore Torque current
characteristics is PARABOLA
SPEED TORQUE CHARECTERISTICS::- The speed verses Torque curve is
HYPERBOLA. It is seen from the curve that the increase in Torque τe does not
change the speed drop drastically

9. DC COMPOUND MOTORS:-
SPEED CURRENT CHARECTERISTICS::- With increase in Ia and Φse
increases. With increase in Ia the speed drops at faster rate in a cumulative compound
motor.
TORQUE CURRENT CHARECTERISTICS At no load the value of Ia = 0, Φse=0
and τe=0. As the armature current rises with load the shunt field Φsh remains almost
constant but series field Φse rises, as a result motor torque τe
SPEED TORQUE CHARECTERISTICS With the increase in motor torque τe
, armature current rises, with this field flux Φse rises, Consequently speed drops in
cumulatively compound DC motor.

10. STARTING OF DC MOTORS:-
At the time of starting the motor speed is zero. Therefore back emf is also zero.
Vt = Ea +Ia Ra Ea = 0
Vt = 0 + IaRa for shunt motor
Vt = 0+ Ia (Ra +Rs) for series motor and compound motor
With the rated voltage is applied the starting armature current is
= Vt/Ra for shunt motor , = Vt/ Ra+Rs for series motor and compound motor
Since the value of Ra and Rs is very small the motor draws large starting armature
current from supply means.
A 10Kw ,250V shunt motor having Ώ armature resistance 0.2 Ώ wiil draw
250/0.2 =1250A where as the rate current is 40A. Such heavy in rush of currents may
result the following
1. Detrimental sparking at commutator
2. Damage of armature winding and detoriation of insulation due to overheating
3. High starting torque and quick acceleration which may damage the rotating parts
of the rotor
4. Large dips in supply voltage.
In view of this , the armature current must be limited to a value that can be
commutated safely, by inserting a suitable external resistance in the armature
circuit. As the motor accelerates the back EMF is generated in the armature and
reduces the armature current to a small value. The external resistance inserted
must be gradually reduced as the rotor accelerates.
SHUNT AND COMPOUND STARTER’s A Primary function of starter is to
limit starting current in the armature circuit during starting or accelerating time.
The simplest type of starter is however modified to include few protective
devices such as over current release, no volt release etc.

11. THREE POINT STARTERS FOR DC SHUNT MOTOR
The handle H is kept at OFF position by spring S . For starting the motor , the
handle H is moved manually and when makes contact with the resistance stud 1
it is in the START position. In position the field winding receives the full
supply voltage., but the armature current is limited by the graded resistance R = (
R1+R2+R3+R4). The starter handle is then gradually moved from stud to stud ,
allowing the speed of the motor to build up until it reaches the RUN position. In
this position (a) the motor attains the full speed , (b) The supply is directly
across the both winding of motor(c) The resistance R is completely cut out. The
handle H is held in RUN position by an electromagnet energized by a no volt
Coil NVC. The no no volt trip coil is connected in series with field winding of
the motor. In the event of switching off, or when the supply voltage falls below
a predetermined value, or complete failure of supply while the motor is running.
NVC is deenergized, This results in release of the handle. . which is then pulled
back to the OFF position. By the action of spring. The current to the motor is cut
off, and the motor is not started without resistance R in the armature circuit..
The NVC also provides protection against an open circuit in the field winding.
The NVC is called no-volt or under voltage protection of motor. Without this
protection, the supply voltage might be restored with the handle in the RUN
position, Consequently full line voltage may be applied directly to the armature
resulting in a very large current.
The other protective device incorporated in the starter is overload protection.
Overload protection is provide by overload trip coil OLC and the NVC. The
overload coil is a small electromagnet. It carries the armature and for normal
values of armature current the magnetic pull of OLC is insufficient to attract
the strip P When the armature current exceeds the normal rated value ( that the
motor is overloaded) P is attracted by the electromagnet of OLC and closes the

12. contact aa, Thus NVC is short circuited , This results in release of handle H
which returns to OFF position and the motor supply is cut off.
DRAW BACK OF THREE POINT STARTER The motor with large speed
variation with armature voltage control suffers . To increase the speed of the
motor the field is to be decreased, therefore the current through the shunt field is
reduced. The field current may become very low. The very low field current
may not develop sufficient electromagnetic hold to over come the force of the
spring.

13. FOUR POINT STARTER The schematic connection diagram is given
herewith. The basic difference in the circuit of Four Point starter compare to 3
point starter is that , The holding coil is removed from the shunt field circuit and
is connected directly across the line with a current limiting resistance r in series.
Such arrangements form three parallel circuits.
1) Armature, starting resistance and overload release.
2) A variable resistance and shunt field winding
3) Holding coil and current limiting resistance.
With this arrangement, a change in field current for variation of speed of the
motor, does not affect the current through the holding coil, because the two
circuits are independent of each other.
Now a days automatic push button starters are used. In such starters the ON
push button is preset at the time of starting limiting the armature current. The
resistors are gradually disconnected by an automatic controlling arrangement
until full line voltage is available to the armature circuit. By the pressing of
OFF button, the circuit is disconnected.

14. SPEED CONTROL OF DC MOTORS
The speed of DC motor depends on the following equation:-
N = ( V-IaRa)/ Φ
From the above equation it is clear that the speed is dependent on supply voltage V
armature resistance and field flux Φ which is produced by field current. These three
factors are used for controlling speed of a DC motor.
1 Variation of armature resistance in the armature circuit.
2 Variation of field flux Φ
3 Variation of armature voltage.
ARMATURE RESISTANCE CONTROL :-In this method a variable series resistor Re
is connected in the armature circuit. In the case of shunt motor the field is directly
connected to the supply and therefore flux Φ Is not affected by this. In the case of
series motor the flux gets affected due to change in Ia and RA since Re resistor
carries full load armature current it must be designed keeping into this consideration
Disadvantages:-
i) A large amount of power is wasted in external resistance Re
ii) Control is limited to give speeds below the normal rated speed.
iii) For a given value of Re the speed reduction is not constant but varies with
motor load.

15. VARIATION OF FIELD FLUX Φ
In a shunt motor this is done by connecting a variable resister Re in series with field
winding. The resister Re is called the field regulator.. the Re reduces the field current
by virtue of this the Φ reduces, decrease in flux causes increase in speed. Increase in
flux causes decrease in speed. By this method the speed can be attained above normal
rated speed.
The variation in field current is done by two methods
i) By connecting a variable resistor parallel to the field winding. This is called
divertor method
ii) The second method is tapped field control.

16. ADVANTAGES
I) This is very easy and convenient method
II) Since the field current Ish is very small , the power loss in shunt field is also
very small
ARMATURE VOLTAGE CONTROL Ward Leonard system of speed control is
based on tis method. In this system M is the motor whose speed is to be controlled and
G is the separately excite DC generator, The generator is driven by 3 Phase induction
motor or synchronous motor. The combination of ac motor and DC generator is called
MG Set.
By changing generator field current the generated voltage is changed this voltage
increases the speed of the motor. With the armature voltage control constant torque
variable speed is obtained.
ADVANTAGES OF WARD LEONARD DERIVES
1) Smooth control of DC motor on both directions is possible
2) It has inherent regenerative breaking capacity
3) When load is intermittent like rolling mills, IM is used with flywheel to take
care of intermittent loading, in case of WARD LEONARD this is possible
without flywheel.

17. 4) By using over excited Synchronous motor a drive the system power factor can
be improved
DISADVANTAGES OF WARD LEONARD DERIVES
1) Higher initial cost due to MG set
2) Larger size and weight, require more floor area
3) Frequent maintenance and produces more noise.
4) Lower efficiency and higher total losses.
SOLID STATE CONTROL
Rotating MG sets are now a days replaced with solid state convertors to control the
speed of DC motors, The convertors are controlled rectifiers or choppers.
In case of AC supply, controlled rectifiers are used to convert fixed ac supply voltage
into variable D voltage.
When the supply is D.C Choppers are used to obtain variable DC voltage from the
fixed DC voltage supply.

18. LOSSES IN A DC MACHINES and efficiency
Following are the losses in the DC machines
1) Electrical or copper loss ( I²R Losses)
2) Core losses or Iron losses
3) Brush Losses
4) Mechanical losses
5) Stray load losses
ELECTRICAL LOSSES:- Windings having resistance consumes certain losses,
these are termed as copper losses because mostly windings are made of copper.
i) Armature copper loss Ia²Ra ( Ia is armature current)
ii) Shunt field copper loss Ish²Rsh
iii) Copper loss in the series field Ise²Rse
iv) Copper loss in the interpole winding which are in series with armature
Ia²Ri
v) Compound machines both series field and shunt field copper losses are
also there
vi) Copper losses are there in the compensating winding
CORE LOSES:-
The core losses are the hysteresis losses and Eddy current losses. Since
the machine usually operates at constant flux density and speed, these
losses are almost constant. These losses are about 20% of Full load losses.
BRUSH LOSES:- There is a power loss at the brush contact with
commutator and the carbon brushes. This loss can me measured by the
voltage drop at the brush contact and armature current.
Pbd = Vbd Ia
The voltage drop is more or less remains constant over a wide range of Ia
and it is assumed 2V ( approx)
MECHANICAL LOSSES:- The losses associated with mechanical effect
are called mechanical losses. These consists of friction losses at bearing

19. and windage losses ( fan losses) . the fans are used to take away the heat
produced due to I²R losses and iron losses inside the machine.
STRAY LOAD LOSSES:- These are miscellaneous losses which are due
to the following reasons:-
1) Distortion of flux due to armature reaction
2) Short circuit currents in the coils due to commutation.
These losses are difficult to find out, However they are taken as 1 % of
full load power output.
EFFICIENCY
InputPower
Losses
InputPower
LossesInputPower
InputPower
OutputPower




1

20. Let us assume that the
R = Total resistance
I = Output current
Ish = Current through the shunt field
Ia armature current I +Ish
V is the terminal voltage
Power loss in the shunt field
= V Ish
Mechanical Losses = Friction losses at bearings+ friction losses at commutator +
windage losses
Stray losses Core losses mechanical losses and shunt field copper losses are
considered as combined fixed losses.
ή = Output / Input
= VI / ( VI + Ia²Rat +Pk =VbdIa
Ia = I + Ish
Since Ish compare to I is very small we can consider
Ia ≡ I
ή = VI/ ( VI+ I²Rat +Vbd I +Pk)

21. LOAD for Maximum efficiency
Ifl = Full load current at maximum efficiency
Im = current at maximum efficiency
Im ² Rat = Pk
Im ² = Pk/ Rat
Current at Maximum ή = F.L Current X( Pk/F.L Copper loss) ²
TESTING OF DC MACHINES:-
Testing of Dc machines can be done by following three methods:-
1) Direct testing
2) Swinburne’s Test
3) Hopkinson’s Test
DIRECT TESTING :- This method is suitable for small machines. In direct testing
method DC machine is subjected to rated load and the entire out put is wasted. The
ration of out put power to the input power gives the efficiency.
For DC generators the out put power is wasted in resistors, the out put voltage and
current gives the out put power.
For Dc motors break test is carried out . the belt tightening hand wheels H1 and H2
help in So adjusting the load on the pulley. S1 S2 are the spring balance. These spring
balances are calibrated in Kg. The motor out put is given by
=ω(S1 –S2)r X 9081 Watts

22. S1 S2 are the spring balance tight side and slack side readings, r is the effective radius
of the pulley in meters = (½ outside pulley diameter + ½ belt thickness)
ω is the ( 2Пņ) is the motor speed in rad/sec
If Vt is the motor terminal voltage and Il is the line current, then power input to the
motor VtIl watts
The efficiency
ή % = {ω ( S1-S2) 9.81 X100 }/ VtIl
For series motor the break should be tight enough before motor is switched on
Disadvantages
1) The spring balance are not steady
2) The friction torque , at particular setting of hand wheel H1 and H2 does not
remain constant.

23. SWINBERN’s TEST
The machine is run as a motor at rated voltage and speed. Connection diagram is as
follows
Let V = supply voltage
I0 = No load current
Ish = Shunt Field current
Therefore No load armature current Ia0 = Io- Ish
NO load input VIo
No load power input supplies the following losses

24. i) Iron loss in the core
ii) Friction losses at bearing and commutator
iii) Windage losses
iv) Armature cupper loss at no load
When the machine is loaded the resistance of the armature and field
resistance changes due to temperature rise and due to I²R losses. Let the
resistance at the room temperature tOc is made by passing current through
armature and field from a low voltage DC supply. Let the rise be 50 degree
the hot resistance
Rt1 = R0 + α0t1
R(t1=50) = R0{1 + α0 (t1+5050)}
α0 = temperature coefficient of resistance at zero degree temperature
Stray losses = iron loss +friction loss + windage loss = input at no load
=VI0 – Pf –Pa0 = Ps
Pc = no load input – no load armature copper loss
Pc = Ps +Pf
By knowing the constant losses of the machine its efficiency can be
calculated at any other load.
Let I be the load current
Motor input VI
Armature cupper loss = Ia²Ra +Pc
ή = {VI – (Ia²Ra +Pc)}/ VI
Efficiency when running as generator
Ia = I + Ish

25. Out put = VI
ή gen = VI/ { VI +( I+Ish) ²Ra +Pc}
ADVANTAGES
i) It is convenient and economical, since power requirement is very less
ii) The efficiency at any load can be determined as the constant losses are
known
DISADVANTAGES
i) This cannot be done on series machines.
ii) No account is taken for iron loss from no load to full load. At full load
due to the armature reaction iron losses increases.

26. HPKINSON’S TEST:- This test is also called
1) Regenerative test
2) Back to back test
3) Heat run test
For conducting this test two identical machines are required, which are coupled
mechanically and also connected electrically parallel. One of them runs as
motor and other one runs as generator. When the machines run on full load the
supply gives the losses of the both the machines.
The switch S is kept open. The field current of motor is so adjusted by Rm to
enable to run the system at rated speed. Now the Rg the field rheostat is so
adjusted that the voltage across the generator armature is slightly higher than the
bus bar voltage by 1 or 2V. When this is achieved the switch S is closed. Under
this condition the generator is said to be floating. Any load can be thrown on by
changing excitation of machines. Let
V = Supply Voltage
Il = Line current
Im = Motor input current
Ig = Output current of generator
Iam = Motor armature current
Iag = Generator armature current
Ishm = Motor shunt field current
Ishg = Generator Shunt field current
Ra = Armature resistance
Rshm = Motor shunt field resistance
Rshg = Generator shunt field resistance
Eg = Generated Induce voltage
Em = Motor induced voltage ( Back EMF)
Eg = V + IshgRa
Em = V –IshmRa
Therefore
Eg> em
Eg α ΦgN
Em α ΦmN

27. Φg > Φm or Ishg > Ishm
Thus the excitation current of generator shall always be greater than motor.
Power input = V Il Total losses of both machines
Armature copper loss of motor = Iam² Ra
Armature copper loss of generator = Iag ²Ra
Field copper loss of motor = Ishm² Rshm
Field copper loss of generator = Ishg ² Rshg
For constant losses
Pc ( iron , windage are assumed equal
Pc = Power drawn – armature and shunt copper losses of both machine.
Since both machines are identical
Total constant loss per machine =1⁄2 Pc
The efficiency of generator
Out put = V Iag
Constant losses = 1⁄2 Pc
Copper loss of generator = Iag ²Ra
Copper loss og generator field = Ishg ² Rshg
ή = VIag/ ( VIag + Iag ²Ra + Ishg ² Rshg + 1⁄2 Pc )
Efficiency as motor
ή =Out put / In put = (Input – Losses )/ input
Input V Im = ( V ( Iam + Ishm)
Copper loss+ constant losses = Iam² Ra + Ishm² Rshm + 1⁄2 Pc
ή = { V (Iam +Ishm) – (Iam² Ra + Ishm² Rshm + 1⁄2 Pc)}
V (Iam +Ishm)

28. APPLICATION OF DC MACHINES:-
Presently the DC generators are overtaken by AC to DC rectifiers
which are static equipments. Thus DC generator are superseded by
rectified ac supply.
The main application of DC machines are follows
SERIES MOTOR:-
These motors are required where high starting torque is required and
speed can vary like traction, cranes.
SHUNT MOTOR :-
These motors are used where constant speed is required and starting
condition are not severe for example, Lifts, Fans, Blowers, Lathe etc.
COMPOUND MOTORS
These motors are used where high starting torque and fairly constant
speed is required, presses, shears, conveyors, elevators, rolling
machines.
Small DC machines ( In fractional KW ) are used primarily as control
device like tacho generator for speed sensing and servomotor for
positioning and tracking.