Download DC Motors Introduction D.C. motors are seldom used in ordinary

DC Motors
Introduction
D.C. motors are seldom used in ordinary application because all electric supply
caompaies furnish altenating current. However, for special applications such as in steel mills
mines and electric trains, it is advantageous to convert alternation current into direct current i
order to use d.c motors. The reason is that speed /torque characteristic of dc motors are
much more superior to that of a.c motrors there for it is not suprising to note that for insustrial
drives, d.c motors are also of three types viz. Series -wound shunt -wound and compound wound. The use of a particular motor depends upon the mechanical load it ahas to drive.
D.C. MOTOR PRINCIPLE
A mechine that converts d.c. Power into mechanical powert is known as a d.c motor. Its
operation is based on the principle that when a current carrying conductor is placed in a
magnetic field, the conductor experiences a mechanical force. The direction of this force is
given by Fleming's left had rule and magnitude i given by:
F=BIl newtorns
Basically, there is no constructional difference between a d.c motor and a d.c generator. The
same d.c. Machine can be run as a generator or motor.
WORKING OF D.C MOTOR
Consider a part of a moltipolar d.c. Motor as shown in Fig. 5.1 . when the terminals of the
motor are connected to an external source of d.c. Supply
(i) the field magnets are excited developing alternate N and S poles.
(ii) The armature conductors carry * current . All conductors under N pole carry current in
one direction while all the conductors under S- pole carry currents in the opposite
direction.
Suppose the conductorrs under N -pole carry currents into the plane of the paper and those
under S-pole carry currents out of the plane of the paper as shown in Fig. 5.1. Since each
amatre conductor is carrying current and is placed in the magnetic field, mechanical force
acts on it. Referring to Fig. 5.1 and applying Fleming's left hand rule, it is clear that force on
each conductor is tending to rotate the marmature in anticlockwise direction. All which sets
the armature rotationg. When the condutor moves from one side of a brush to the other the
current in that conductor is reversed and at the same time it comes under the influence of
next pole which is of opposite polarity. Consequently, the direction of force on the conductor
remains the same.
BACK OR COUNTER E.M.F.
When the armature of a d.c. Motor rotates under the influence of the driving torque, the
armature conductors move through the magnetic field and hence e.m.f. Is induced in them as
in a generator. The induced e.m.f acts in opposite direction to the applied voltage V (*Lenz's
law) and is known as back or counter e.m.f. Eb. The back e.m.f. Eb (=p φ Z N / 60 A) is always
less than the applied voltage V. although this differnence is small when the motor is running
under normal conditions.
Consider a shunt wound motor shown in Fig. 5.2. When d.c. Voltage V is appplied across the
motor terminals, the field magnets are excited and armature conductors are suppled with
current, Therefore, driving torque acts on the armature which begins to rotate. As the
aarmature rotates , back e.m.f. Eb is induced which opposes the applied voltage V. The
applied voltage V has to force current through the armature against the back e.m.f. Eb. The
electric work done in overcoming and causing the current to flow against Eb is converted into
mechanical energy developed in the armature. It follows, therefore, that energy conversion in
a d.c. Motor is only possible due to the production of back e.m.f.. Eb.
Net voltage across armature circuit = V-Eb.
If Ra is the armature circuit resistance then,
Since V and Ra are usually fixed, the value of Eb will determine the current drawn by the
motor. If the speed of the motor is high, then back e.m.f. Eb (=p φ Z N / 60 A) is large and
hence the motor will draw less armature curren and vice-versa.
SIGNIFICANCE OF BACK E.M.F.
The presence of back e.m.f. Makes the d.c. Motor a self -regulationg mechine i.e., it makes
the motor to draw as much armature current as is just sufficient to develop the torque required
by the load.
Armature current ,
(i) When the motor is running on no load, small torque is required to overcome the friction
and windage losses. Therefore, the armature current Ia is small and the back e.m.f. Is
nearly equal to the applied voltage.
(ii) If the motor is suddenly loaded, the firest effect is to cause the armature to slow down.
Therefore, the speed at which the armature conductors move throug the field is
reduced and hence the back e.m.f. Eb. Falls. The decreased back e.m.f. Allows a
larger current to flow through the armature and larger current means increased driving
torque. Thus the driving torque increases as the motor slows down. The motor will stop
slowing down when teh armature current is just sufficient to produce the increased
torque required by the load.
(iii) If the load on the motor is decreased, the driving torque is mimentarily in excess of the
requirement so that armature is accelerated. As the armature speed incteases, the
back emf Eb also increases and causes the armature current Ia to decrease. The motor
will stop accelerating when the armature current is just sufficient to produce the
reduced torque required by the load.
It follow, therefore , that back e.m.f in a d.c motor regulates the flow of armature current i.e, it
automatically changes the armature current to meet the load requirement.
VOLTAGE EQUATION OF D.C. MOTOR
Let in a d.c. Motor (See Fig. 5.3)
V = applied voltage
Eb= back e.m.f.
Ra = armatre resistance
Ia = armature current
Since back e.m.f. Eb acts in opposition to the applied voltage V, the net voltage across the
aarmature circuit is V-Eb. The armature current Ia is given by,
V=Eb+IaRa
or
This is known as voltage equation of the d.c. Motor.
POWER EQUUATION
If eq. (i) above is multiplied by Ia throughout, we get,
VIa = EbIa+I2aRa
This is know as power equation of the d.c. Motor.
VIa = electric power supplied to armature (armature input)
EbIa = * power developed by armature (armature output)
I2aRa = electric power wasted in armature (armature Cu loss)
Thus out of the armature input , a small portion (about 5%) is wasted as I2aRa and the
remaining portion EbIa is converted into mechanical power within the armature.
CONDITION FRO MAXIMUM POWER
The mechanical power developed by the motor is pm = EbIa
Pm = VIa - I2aRa
Now
Since , V and Ra are fixed, power developed by the motor depends upon armature current .
Tor maximum power, dPm/dIa should be zero.
Or IaRa = V/2
Now,
V= Eb + Ia Ra = Eb + V /2 [ IaRa = v/2]
Eb = V/2
Hence mechanical power devloped by the motor is maximum when back e.m.f. is equal
to hald the applied voltage.
Limitations : In practive, we never aim at achieving maximum power due to the following
reasons:
(i) The armature current under this condition is ver large – much excess of rated current
of the machine.
(ii) Half of the input power is waseted in the armature circuit. In fact, if we take into
account other losses (iron and mechanical), the efficiency will be well below 50%.
TYPES OF D.C. MOTORS
Like generators, there are there types of d.c. Motors characterised by the connections of field
winding in relation to the armature viz.
(i) Shunt -wound motor in which the field winding is connected in parallel with the
armature [see Fig 5.4]. The current through the shunt field winding is not the same as
thea armature current Shunt Field windings are designed to produce the necessary
m.m.f. By means of a relatively large number of turn of wire having high resistance.
Therefore, shunt field current is relatively small compared with the armature current.
(ii) Series – wound motor in which the firld winding is connected in series with the
armature [See Fig. 5.5]. There fore , series field winding carries the armature
current. Since the current passing through a series field winding is the same as the
armature current, series field windings must be designed with much fewer turns
than shunt field windings for the same m.m.f. Therefore, a series field winding has a
relatively small number of turns of thick wire and , therefore, will possess a low
resistance.
(iii) Compound – wound motor which has rwo field winding; one connected in parallel
with the armature and the other in series with it. There are two tyypes of compound
motor connections (like generators ). When the shunt field winding is directly
connected acress the armature terminals [See Fig. 5.6], it is called short – shunt
connections. When the shunt winding is so connected that it shunts the series
combination of armature and series field [see Fig. 5.7], it is called long shunt
connection.
The compound machines (generators or motors ) are always designed so that the flux
produced by shunt field winding is considerably larger than the flux produced by the series
field winding. Therefore, shunt field in compound machines is the basic dominant factor in the
production of the magnetic field in the machine.
ARMATURE TORQUE OF D.C. MOTOR
Torque is the turning moment of a force about an axis and is measured by the product of
force (F) and radius(r) at right angle to which the force acts i.e.
T= Fxr
In a d.c. Motor, each conductor is acted upon by acircumferential force F at a distance r, the
radius of the armature (Fig 5.11). Therefore each conductor exerts a torque, tending to rotate
the armature. The sum of the torques due to all armature conductors is known as gross or
armature torque (Ta).
Let in a d.c. Motor
r = average radius of armature in m
l = effective length of each conductor in m
Z = total number of armature conductors
A = number of parallel paths
i = current in each condutor = Ia/A
B = average flux desity in Wb/m2
Φ = flux per pole in Wb
P = number of poles
Force on each conductor, F = B i l newtons
Torque due to one conductor = F X r newtons – metre
Total armature torque, Ta = Zfr newtons – metre
= Zbilr
Now i = Ia/A, B = Φ/a where a is the x-sectional area of flux path per pole at radius r.
Clearly a= 2∏rl/P.
Ta = ZX( Φ/a)X(Ia/A)XlXr
or
Ta = 0.159 Z Φ Ia (P/A)N-m
Since Z, P and A are fixed for a given machine,
Ta α Φ Ia
Hence torque in a d.c. Motor is directly proportional to flux per pole and armature current.
(i) For a shunt motor , flux Φ is practically constant.
Ta α Ia
(ii) For a series motor, flux Φ is directly proportional to armature current Ia provided
magnetic saturation does not take place.
Ta α I2a
Alternative expression for Ta
Eb = p ΦZN/60A
p ΦZ/A = 60 x Eb/N
From equation (i) , we get the expression of Ta as :
Ta = 0.159x(60 X Eb/N)xIa
Ta = 9.55 x EbIa/N (N-m)
or Note that developed torque or gross torque means armature torque T a.
SHIFT TORQUE (Tsh)
The torque which is vailable at the motor shift for doing useful work is known as shaft torque.
It is represented by Tsh. Fig. 5.12 illustrates the concept of shaft torque. The total or gross
torque Ta developed in the armature of a motor is not available at the shaft because a part of
it is lost in overcoming the iron and frictional losses in the motor. Therefore, shaft torque Tsh is
somw what less than the armature torque Ta. The difference Ta-Tsh is called lost totque.
Clearly Ta – Tsh = 9.55 x Iron and frictional losses/N
For example, if the iron and frictional losses in a motor are 1600 W and the motor runs at 800
r.p.m, then,
Ta – Tsh = 9.55 x 1600/800 = 19.1(N-m)
As stated above, it is the shaft torque Tsh that produces the useful output. If the speed of the
motor is N r.p.m., then,
Outpurtin watts = 2∏NTsh/60
or Tsh = (output in watts /2∏N/60 )(N-m)
or Tsh = 9.55x Output in watts /N (N-m)
(60/2∏=9.55)
LOSSES IN A D.C. MOTOR
The losses occuring in a d.c. Motor are the same as ina d.c. Genetator [See Art 2.27]. These
are [See Fig. 5.21]:
(i) copper losses
(ii) iron losses or magnetic losses
(iii) mechnanical losses
As in a generator, these losses cause (a) an increase of machine temperature and (b)
reduction in the efficiency of d.c. Motor.
The following points may be noted:
(i) Apart form armature Cu loss, field Cu loss and brush contact loss, Cu losses also
occur in interpoles (commutationg poles ) and compensationg windings. Since these
windings carry armature current (Ia),
Loss in interpole winding = I2a x Resistance of interpole winding
Loss in compensation winding = I2a x Resistance oc compensating winding
(ii) Since d.c. Machines (generators or motors ) are generally operated at constant flux
density and constant speed, the iron losses are nearly constant.
(iii) The mechnanical losses (i.e. friction and windage) vary as the cube of the speed of
rotation of the d.c. Machine (generator or motor). Science d.c. Machines are generally
operated at constant speed, mechanical losses are considered to be constant.
EFFICIENCY OF A D.C. MOTOR
Like a d.c. Genetator , the efficiency of a d.c motor is the ratio of output power to the input
power i.e,
Efficiency, η = output/input x100
= (output/(output+losses ))x100
As for a generator (see Art 2.30), the efficiency of a d.c. Motor will be maximum when:
variable losses = Constant losses
Therefore , the efficiency curve of a d.c. Motor is similar in shape to that of a d.c. Generator
(Refer back to Fig. 5.57).
POWER STAGES
The power stages in a d.c. Motor are represented diagrammatically in Fig. 5.22.
A-B = copper losses
B- C – Iron and friction losses
Overall efficiency , ηc = C/A
Electrical efficiency , ηe = B/A
Mechanical efficiency , ηm = C/B
D.C. MOTOR CHARACTERISTICS
There are three principal types of d.c. Motors viz. Shunt motors, series motors and
compound motors. Both shunt and series types have only one field winding wound on the
core of each pole of the motor. Both shunt and series types have only one field winding
wound on the core of each pole of the motor. The compound type has two separate field
windings wound on the core of each pole. The perfoemance of a d.c. Motor can be judged
from its characteristic cureves known as mitor characteristics. Following are the three
important characteristics of a d.c. Motor:
(i) Torque and Armature current characteristic (Ta/Ia)
It is the cureve between armature torque Ta and armature current Ia of a d.c motor. It
is also known as electical characteristics of the motor.
(ii) Speed and armature current characteristiv (N/Ia)
It is the cureve between speed N and aramture curret Ia of a d.c. Motor. It is very
important characteristic as it is often the deciding factor in the selectionof the motor for a
particular application.
(iii) Speed and torque characteristic (N/Ta)
It is the curve between speed N and armature torque Ta of ad.c. Motor. It is also
known as mechanical characteristic.
CHARACTERISTICS OF SHUNT MOTORS
Fig. 5.30 shows the connection of a d.c shunt motor . The field current I sh is constant since
the field winding is directly connected to the supply voltage V which is assumed to be
constant. Hence, the flux in a shunt motor is * approximately constant.
(i) Ta/Ia Characteristic. We know that in a d.c. Motor,
Ta α φ Ia
Since the motor is perating from a constant supply voltage, flux φ is constant (neglecting
armature reaction).
Ta α Ia
Hence Ta/Ia characteristic is a straight line passing through the origin as shown in Fig. 5.31.
The shaft torque (Tsh) is less than Ta and is shown by a dotted line. It is clear from the
cureve that a very large current is required to start a heavy load. Therefore , a shunt motor
should not be started on heavy load.
(ii) N/Ta Characteristics. The speed N of a d.c. Motor is given by:
N α Eb/φ
The flux φ and back e.m.f. Eb in ashunt motor are almost constant under normal
conditions . Therefore, speed of a shunt motor will remain constant as the armature current
varies (dotted line AB in Fig. 5.32 ). Strictly speaking, when load is increased , Eb (= V-IaRa)
and φ decreases due to the armatur resistance drop and armature reaction respectively.
However, Eb decreases slightly more than φ so that the * speed of the motr decreases
slightly with load .(line AC in Fig. 5.32).
(iii) N/Ta Characteristic. The cureve is obtained by plotting the valued of N and Ta for
various armature current (See Fig. 5.33). It may be seen that speed falls somewhat as
the load torque increases.
Conclusion : Following two important conclusions are drawn from the slight above
characteristics.
(i) There is a slight change in the speed of a shunt motor form no-load to full-load. Hence
, it is essentially a costant -speed motor.
(ii) The starting towque is not high because Ta α Ia
CHARACTERISTICS OF SERIES MOTORS
Fig. 5.34 shows the connectiond of a series motor. Note that current passing through the field
winding is the same as that in the armature. If the mechnanical load on the motor increases,
the armature current also increases. Hence , the flux in a series motor increases with the
increase in armaure current and vice -versa.
(i) Ta/Ia Characteristics. We know that
Ta α φ Ia
Upto magnetic saturation. φ α Ia so that Ta α I2a
After magnetiv saturation φ is constant so that Ta α Ia
Thus upto magnetiv saturation, the armature torque is directly proportional to the
sequare of armature current. If Ia is doubled, Ta is almost quadrupled. Theredor , Ta/Ia curve
upto magnetiv saruration is a parabola( portion OA of the curve in Fig. 5.35). However, after
magnetic saturation, torque is directly proportional to the armature current. Therefore, Ta/Ia
curve after magnetic saturation is straight line (portion AB of the curve in Fig. 5.35).
It may be seen that in the initial portion of the curve (i.e. upto magnetiv saturation). Ta
α I2a. This means tht starting torque of a d.c. Series motor will be very high as compared to a
shunt motor (wher Ta α Ia).
(ii) N/Ia Characteristic. Thhe speed N of a series motor is given by ;
N α Eb/ φ where Eb = V- Ia(Ra+Rse)
When the armature current increases, the back e.m.f. Eb decreases dur to Ia(Ra+Rse) drop
while the flux φ increases. However, Ia(Ra+Rse) drop is quite small under normal conditions
and may be neglected.
N α I/ φ
α I / Ia upto magnetic saturation
Thus, upto magnetic saturation, the N/Ia curve follows the hyperbolic path as shown in Fig
5.36. After the flux become constant and so does the speed.
(iii) N/Ta characteristic. The N/Ta characteristic of a series motor is shown in Fig. 5.37. It is
clear that series motor develops high torque at low speed and vice-versa. It is because
and increase in torque requires an increase in armature current, which is also the field
current.
The result is that flux is strengthened and hence the speed drops ( N α I/ φ). Reverse
happens should the torque be low.
Conclusion : Follwing three imprtant conclusions are drawn from the above characteristics of
series motors:
(i) It has a high starting torque because initially Ta α I2a
(ii) It is a variable speed motor (see N/Ia curve in Fig. 5.36) i.e. it automatically
adjusts the speed as the load changes. Thus if the load decreases, its speed is
automatically raised and vice-versa.
(iii) At no-load, the armature current is very small ans so is the flux. Hence, the
speed rises to an excessive high value( N α I/ φ). This is dangerous for the
machine which may be destroyed dur to centrifugal forces set up in the rotationg
parts. Therefore, a series motor should never be started on no-load. However,
to start a series motor, mechanical load is first put and then the motor is started.
Note. The minimum load on a d.c. Motor should be grea enough to keep hte speed
within limits. If the speed becomes dangerously high, then motor must be disconnected from
the supply.
COMPOUND MOTORS
a compound motor has both series field and shunt field. The shunt field is always stronger
than the series field. Compound motors are of two types:
(i) Cumulative – compound motors in which series field aids the shunt field.
(ii) Differential -compound motors in which series field opposes the shunt field.
Differential compound motors are rarely used due to their poor torque characteristics at
heavy loads.
CHARACTERISTICS OF CUMULATIVE COMPOUNDS MOTORS
Fig. 5.38 shows the connections of a cumulative compound motor. Each pole carries a series
as well as shunt field winding: the series field aiding the shunt field.
(i) Ta/Ia charecteristic. As the load increases, the series field increases bbut shunt field
strangh remains constant. Consequently, total flux is increased and hence the
armature torque (Ta α φ Ia). It may be noted thtat torque of a cumulative – compound
motor is greate than that of shunt motr for a given armature current due to seires field
[See Fig. 5.39].
(ii) N/ Ia Characteristic. As explained above, as the load increases, the flux per pole also
increases. Consequently, the speed ( N α I/ φ) of the motor falls as the load increases
(see Fig. 5.40). It may be noted that as the load is added, the increased amount of flux
causes the speed to decrease more than does the speed of a shunt motor. Thus the
speed regulation of a cumulative compound motor is poorer than that of a shunt motor.
Note : Due to shunt field, the motor has a definite no load speed and can be operated
safely at
no – load.
(iii) N/Ta Characteristic. Fig 5.41 shows N/Ta charecteristic of a cumulative compound
motor for a given armature current, the torque of a cumulative compound motor is
more than that of a shunt motor but less than of a series motor.
Conclusions : a cumulative compound motor has characteristics intemediate between
series and shunt motors.
(i) Due to the presence of shunt field, the motor is pevented form running away at
no-load.
(ii) Due to the people of series field, the starting torque is increased.
COMPARISON OF THREE TYPES OF MOTORS
(i) The speed regulation of a shunt motor is better than that of a series motor.
However, speed regulation of a cumulative compound motor lies between shunt and
series motors (see Fig. 5.42)
(ii) For a given armature current, the starting torque of a series motor is more than that of
a shunt motor. However, the starting torque of a cumulative compound motor lies
between series and shunt motors (See Fig. 5.43).
(iii) Both shunt and cumulative compound motor have definite no-load speed. However, a
series motor has dangerously high speed at no-load.
APPLICATIONS OF D.C. MOTORS
1. Shunt motors. The characteristics of a shunt motor reveal that it is an approximately
constant speed motor. It os, therfore, used.
(i) where the speed is required to remain almost constant from no-load to full-load
(ii) where the load has to be driven at a number of speed and any one of which is required to
remain nearly constant
Insustrial use: Lathes, drills, boring mills, shapers, spinning and weaving machines etc.
2. Series motors. It is a variable speed motor i.e. speed is low at high torque and vice-
vers. However, at light or no-load, the motor tends to attain dengerusly high speed.
The motor has a high starting torque. It is, therfore, used
(i) wher large starting torque is required e.g. in elevatiors and electirc traction
(ii) where the load is subjected to heavy fluctuatuins and the speed is automatically
required to reduce at high torques and vice-versa.
Industrial use : Electric traction, cranes, elevators, air compressors, vacuumm cleaners
hair drier, sewing machines etc.
3. Compound motors. Diffenrential-compound motors are rarely used because of their
poor torque chareacteristics. However cumulative -compound motors are used where a
fairly constant speed is required with irregular loads or suddenly applied heavy loads.
Industrial use : Presses, shears, reciprocating machines etc.
TROUBLES IN D.C. MOTORS
Several toubles may arise in a d.c. Motor and afew of them are discussed below.:
1. Failure to start. Tis may be due to (i) ground fault (ii) open or short -circuit fault (iii)
wrong connections (iv) too low supply voltage (v) frozen bearing or (vi) ecessive load.
2. Sparking at brushes. This may be due to (i) troubles in brushes (i) troubles in
commutator (iii) troubles in armature or (iv) excessive load
(i) Brush troubles may arise due to insuffiecient contact surface, too short a brush ,
too little spring tension or wrong brush setting.
(ii) Commutator troubles may be dur to dirt on the commutator, high mica, rough
surface or eccentricity.
(iii) Armature troubles may be due to an pen armature coil. An open armature coil
will cause sparking each tim the open coil passes the brush. The location of this
open coil is noticeable by a burnt line between segments connection the coil.
3. Vibrations and pounding noises. These may be due to (i) worn bearing (ii) loose
parts (iii) rotating parts hitting stationary parts (iv) armature unbalanced (v)
misalignment of machine (vi) loose coupling etc.
4. Overheating. The overheating of motor may be due to (i) overloads (i) sparking
at the brushes (iii) short- circuited armature or field coils (iv) too frequent starts
or reversals (v) poor ventilations (vi) incorrect voltage.
Speed control of D.C. Motors
Introduction
Although a fro greter percentage of electric motors in service are a.c. Motors, the d.c. Motor is
of considerable industrila importance. The principal advantage of a d.c. Motor is that its speed
can be changed ove a wide range by a varietyof simple methods. Such a fine speed control is
generally not possible with a.c. Motors. In fact fine speed control is one of the reason for the
strong competitive position of d.c. Motors in the modern industrial application. In this chapete ,
we shall discuss the various methods of speed control of d.c. Motors.
SPEED CONTROL OF D.C. MOTORS
The speed of a d.c. Motor is given by :
N α I/ φ
N = K ((v-IaR)/Q) r.p.m
where R = Ra
= Ra + Rse
for shunt motor
From exp. (i) it is clear that there are three main methods of controlling the speed of a d.c.
Motor, namely:
(i) By varying the flux per pole (φ). This is known as flux control method.
(ii) By varying the resistance in the armature circuit. This is know as armature control
method.
(iii) By varying the applied voltage V. This is known as voltage control method.
SPEED CONTROL OF D.C. SHUNT MOTORS
The speed of shunt motor can be changed by (i) flux control method (ii) armature control
method (iii) voltage control method. The first method (i.e. flux control method) is frequently
used because it is simple and inexpensive.
1. Flux control method. It is based on the fact that by varying the flux φ, the motor speed
( N α I/ φ) can be changed and hence the name flux control method. In this method, a
variable resistance (known as shunt field rheostat) is placed in series with shunt field
winding as shown in Fig. 6.1.
The shunt field rheostat reduces the shunt field current Ish and hence the flux φ. Therefore ,
we can only * raise the speed of the motor above the normal speed (ssee Fig. 6.2). Generally,
the mehtod permits to increase the speed in the ratio 3:1. Wider speed ranges tend to
produce instablity and poor commutation.
Advantages
(i) This is an easy and convenient method.
(ii) It is an inexpensive method since very little power is wasted in th shunt field rheostat
due to relatively small value of Ish.
(iii) The speed contrl expercised by this method is independent of load on the machine.
Disadvantages
(i) Only speeds higher than normal speed can be obtained since the total field circuit
resistance cannot be reduced below Rsh – the shunt field winding resistance.
(ii) There is a limit to the maximum speed obtained since the total field circuit much
weakened, commutation becomes poorer.
Note. The field of a shunt motor in operation shuld never be opened because its speed will
increase to an extremely highvalue.
2. Armature control method. This method is based on the fact that by varying the voltage
available across the armature , the back e.m.f and hence the speed of the motor can
be changed. This is done by inserting a variable resistance Rc (known as controller
resistance) in series with the armature as shown in Fig. 6.3.
where N α V-Ia(Ra+Rc)
Rc = Controller resistance
Due to voltage droop in the controller resistance, the back e.m.f (Eb) is decreased . Since N α
Eb, the speed of the motor is reduced. The highest speed obtainable is that corresponding to
Rc =0 i.e., normal speed. Hence , this method can only provide speeds below the normal
speed (see fig. 6.4).
Disadvantages
(i) A large amount of power is wasted in the controller resistance since it carreis full
armature curren Ia.
(ii) The speed varies widely with load since the speed depends upon the voltage drop in
the controller resistance and hence on the armature current demanded by the load.
(iii) The output and efficiency of the motor are reduced.
(iv) This method results in poor speed regulation.
Due to above disadvantages, this method is seldom use dto control the speed o shunt
motors.
Note. The armature control method is a very common method for the speed control of d.c.
Series motors. The disadvantages of poor speed regulation is not important in a series motor
which is used only where varying speed service is required.
3. Voltage control method. In this method, the voltage sourve supplying the field current is
different from that which supplies the armature. This method avoids the disadvantages
of poor speed regulation and low efficiency as in armature control method. However, it
is quite expensive. Therefore this method of speed control is employed for large size
motors where efficiency is of grat importance.
(i) Multiple voltage control, In this method, the shunt field of the motor is connected
permanently across a fixed voltage source. The armature can be connected acreoss several
fiffernt voltage through a suitable switchgear. In this way, voltage applied across the armature
can be changed. The speed will be approximately proportional to the voltage applied across
the armature. Intermediate speeds can be obtained by means of a shunt field regulator.
(ii) Ward -Leonard system. In this method, the adjustable voltage for the armature is obtained
from and adjustable voltage generator while the field circuit is supplied from a separate
source. This is illustrated in Fig. 6.5. The armature of the shunt motor M (whose speed is to
be controlled ) is connected direclty to a d.c. geneg=rator G driven by a constant – speed a.c.
Motor A. The field of the shunt motor is supplied from the exciter E. The voltage of the
generator G can be varied by means of its field regulator. By reversing the field current of
generator G by controller FC, the voltage applied to the motor may be reversed. Sometimes,
a field regulator is included in the field circuit of shunt motor M for additional speed
adjustment. With this method , the motor may be operated at any speed upto its maximum
speed.
Advantages
(a) The speed of the motor can be adjusted through a wide range without resistance losses
which
results in high efficiency.
(b) The motor can be brought to a standstill quikly, simply by rapidly reducing the voltage of
generator G. When the generator voltage is reduced below the back e.m.f. Of the motor, this
back e.m.f. Sends current through the generator armature, establishing dynamic banking.
While this takes place the generator G operated as a motor driving motor A which returns
power to the line.
(c) This method is used for the speed control of large motors when a d.c. Supply is not
available.
The disadvantage of the method is that a special motor generator set is required for each
motor and the losses in this set are high if the motor is operating under light loads for long
periods.
Applications. The ward-Leonard system of speed control is expensive but is use where an
unusually wide and very sensitive speed control is desired. This arrangement offers an
excellent stepless speed control and is well suited for such applications as passenger
elevators, electric excavators etc.
SPEED CONTROL OF D.C. SERIES MOTORS
The speed control of d.c. Series motors can be obtained by (i) flux control method (ii)
armature – resistance control method. The latter method is mostly use
1. Flux control method. In this method, thhe fluux produced by the series motor is varied
and hence the speed. The variation of flux can be achieved in the following ways:
(i) Field diverters . In this method, a varoable resistance (called field diverter ) is connected in
parallel with series field winding as shown in Fig. 6.13. Its effect is to shunt some portion of
the line current from the series field winding , thus weakening the field and increasing the
speed ( N α I/ φ). The lowest speed obtainable is that corresponding to zero curent in the
diverter (i.e. diverte is open ). Obviously, the lowest speed obtainable is the normal speed of
the motor. Consequently, this method can only provide speed above the normal speed. The
series field diverter method is often employed in traction work.
(ii) Armature diverter. In order to obtain speed below th normal speed, avariable resistance
(called armature diverter) is connected in parallel with the armature as shown in Fig. 6.14.
The diverter shunts some of the line current, thus reducing the armature current. Now for a
giver load , if Ia is decreased, the flux φ must increase ( T α φ Ia). Since N α I/ φ , the motor
speed is decreased. By adjusting the armature diverter, any speed lower than the normal
speed can be obtained.
(iii) Tapped field control . In this method, flux is reduced (a nd hence speed is
increase) by decreasing the number of turns of hte series field winding as
shown in Fig. 6.15. The switch S can short circuit any part of hte field winding ,
the motor runs at normal speed and as the field turns are cut out, speed higher
than normal speed are achieved.
(iv) Paalleling field coils . This method is usually employed in the case of afan
motors. By regrouping the field coils as shown in Fig. 6.16. several ficed speed
can be obtained.
2. Armature – resistanc control. In this method, a variable resistance is directly connected
in series with the supply to the complete motr as hsown in Fig. 6.17. This reduces the
voltage availabe acreoss the armature and hence the speed falls . By changing the
value f variable resistance, any speed below the normal speed can be obtained. This is
the most common mtthod employed to control the speed of d.c seried motors. Although
this method has poor speed regulation, this has no significance for seris motors
because they are used in varying speed applications. The loss of power in the series
resistance for many applications of series motors is not too serious since in these
application, the control isutilised fro a large portion of th etime for reducing the speed
under light – load conditions and is only used internuttently when the motor is carrying
full load.
SERIES PARALLEL CONTROL
Another method used for the speed contol of d.c. Seris motors is teh series parallel method.
In this system which is widely used in traction system , two (or more ) similar d.c. Seris motors
are mehcanically coupled to the same load.
When the motor are connected in seris [See Fig. 6.23 (i)], each motor armature will recieve on
half the normal voltage. Therefor the speed will be low . When the notors are connected in
parallel , each motor armature receives the normal voltage and the speed is high [See Fig.
6.23 (ii)]. Thus we can obtain two speed. Note that forthe same load on the pair of motors, the
system would run appoximately four times the speed when the machines are in paralled as
when they are in series
NECESSITY OF D.C. MOTOR STARTER
At starting when the kkotor is stationary, there is no back e.m.f int hearmatue. Conswquetly if
the motor is directly switched on to the mains, the armature will draw a hevy current (Ia=V/Ra)
beecause of small armature resistance. As an example, 5 h.p., 220 V shunt motor has a ful –
load current of 20 A and an armature resistance o fabout 0.5 Ω . If this motor is directly
switched on to supply, it would take an armarue current of 220/0.5=440 A which is 22 times
the full – load curren . Thsi hight starting current may result in :
(i) burning of armature due to excessive heating effect,
(ii) damaging the commutator and brushes dur to heavy aparking,
(iii) excessive voltage drop in the line to which the motor is connected. The result is that
the operation of other appliance connected to the line may be impaired and in
particular cases, they may refuse to work.
In order to order to avaoid excessive current at starting , variable resistance ( know n as
startin resistance) is inserted in series with the armature circuit. This resistance is gradually
reduced as the motor gains speed ( and hence Eb increases) and eventually it is cut out
completely when the motor has attained full speed. The value of starting resistanc eis
generally such that starting current is limited to 1.25 to 2 times the full -load current.
TYPES OF D.C. MOTOR STARTERS
The starting opertion of a d.c. Motor consists in the insertion of external resistance into the
armature circuit to limit the starting current taken by the motor and the removal of this
resistance is steps as the motor accelertes. When the motor attains the normal speed, this
resistance is totally cut out of hte armature circuit. It is very important and desirable to provide
the starter with protective devices to enable the starter srm to return to OFF position.
(i) When the supply fails thus preventing the rmature being directly across themains when
this voltage is restored. For this purpose , we use no-volt releae coil.
(ii) When the motor becomes overlaoded or develops a faul causing the motor to take an
excessive current. For this purpose , we use overload release coil.
There ar two principal types of d.c motor starter viz. Thrree point starter and four point starter.
As we shall see, the two types of strters differ only in the manner in which the no – volt
release coi is connected.
THREE POINT STARTER
This type of starter is widely used for starting shunt snd compound motors.
Schemativ diagram. Fig.6.28 shows the schematic diagram of a three point starter for a shunt
motor which protective device. It is so called because it has three terminals L,Z and A. The
starter consist of strting resistance dicided into several sections and connected in series with
the armature. The a=tapping points of he starting resistance are brought out to a number of
studs. The three terminals L.Z and A of the starter are conected respectively to the postitive
line terminal, shunt field terminal and armature teminal. The other terminals of the armature
and shunt field windings are connected to the negative terminal of the supply. The no
voltrelease coil is connected in th shunt field circuit. One end of hte hadle is connected toeh
terminal L through the over load release coil. The other end of the handle moves against a
spiral spring and makes contact with each stud during starting operation, cutting out more and
more starting resistance as it passes over each stud in clockwise direction.
Operation
(i) To start with , the d.c. Supply is switched on with handle in the OFFposition.
(ii) The handle is now moved clockwise to the first stud. As soon as it comes in contact
with the first stud, the dhunt field winding tis dierelctly connected across the supply
while the whole starting resistance is inserted in series with the armature circuit.
(iii) 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 magnetivally by the no volt
release coil whichis energised by shunt field current.
(iv) If the supply voltage is suddenly interrupted or if the field excitation is accidently cut,t
he no volt release coil is demagnetise d and the handle goes back to the OFF position
under the pull of the spring. If no volt relwase 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.
(v) If the motor is over loade (or a fault occures , it wil draw excessive current formt he
supply. This current will increase the ampere turen of ht overload release coil and pull
te rmaature C, thus short circuiting the no- colt treleae coil. The no – volt coil is
demagnetise d and the handle is pulled to the OFFposition by the spring. Thus, the
motor is automatiallly disconnected rorm the supply.
Drawback. In a three point starter, the no volt release coil is connected in series tithe shunt
field circuit so that it carries the shunt field current. While excersing 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 may diconnected the
motor from th supply when it is not desired. This drewback is overcome in the four point
starter.
TESTING OF D.C. MACHINES
INTROCCTION
There are several tests that are conducated upon a d.c. Machine ( generator or motor ) to
judge its performance. One important test is performed to measure the efficiency of a d.c.
Machine . The efficiency of a d.c. Machine depends upon its losses. The smaller the losses,
the greater is the efficiency of the machine and cice versa.The considerationof losses in a d.c.
Machine is important for two princeipal reasons. First, losses deeine the efficicncy of the
mchine and appreciably influence its operationg cost, Secondly, losses determine the
heatingh of hte machine and hence the power output that may be obtained withou undue
deterioration of the indultation. In thi chapter, we shall focus or attentionon the various
methods for the determination of the efficiency of a d.c. Machine.
EFFICIENCY OF A D.C. MACHINE
The pwwer that a d.c. Machine receives is called the input ad the power it gives out is called
the outpuut. Therefore , the efficiency of a d.c. Machine like that of any energy tranferring
device, is given by:
Efficienvy = Output/Input
Output = Input – Loses and Input = Output Losses
Therefore, the efficiency of a d.c. Machine can also be expressed in the following forms:
Efficiency = Input – Losses /Input
Efficiency = Output / Output + Losses
The most obvious method of detemining the efficinecy of a d.c. Machine is to firectly load it
and measure the input power and putput powr . Then we can use eq (i) to detemine the
ddefficiency of hte machine . This method suffers form three main drawbacks. First , this
method requires tha application of load on the mahine. Secondlly for machine of large rating
the load sof htrquired sizes may not be availablle. Thirdly, even if it is possible to provide such
loads, large power will be dissipated, making it an expensive mathod.
The most common mehtod ofmeasuring the efficiency of a d.c. Machine is to deteminde its
losses ( instead of measuringthe input and output on load) . We can then use eq. (ii) or eq (iii)
to determine the efficiency of themachine . This method has the obvious advantage of
convenience and economy.
EFFICIENCY BY DIRECT LOADING
In this method , the d.c. Machine is loaded and output and input are measured to find the
efficiency. For this purpose two simple method can be used.
(i) Break test. In this mehtod a break is applied to a watr cooled pully mounted on the
motor shaft as hsown in Fig. 7.1. One end of hte rope is fixed to the floor via a spring
balance S and a known mass is sunspended at the other end. If the spring balance
relading is S kg -wt and th suspended mass has awheight to W kg-wt, then Net pull on
the rope= (W-S)kg-wt= (W-S)x9.81 newtons
If r is the radious of th pulley inmetrs , hten the shaft torque Tsh devleoped by the motor is
Tsh = (W-S)x9.81xr N-m
If the speed of th pully is N r.p.m. Then
Output power = 2лNtsh/60 =2лNx(W-S)xrx9.81/60 watts
Let V = supply voltage in volts
I current taken by the motr in amperes
Input to motor = VI watts
Efficiency = 2лNx(W-S)xrx9.81/60xVI
SWINBURNE'S METHOD FOR DETEMINING EFFICIENCY
In thsi method, the d.c. Machine (gentetor or motor ) is run as amotor at no load and losses
of the machine are detemined. Once the losses of ht emaahcine are known its efficiency at
any desires load machien are detemined. Once the losses of hte machineare known its
efficieny at any desired load can be determined in advance. It may noted taht this method is
applicable t thsoe machine in which fluex is practically constant at all loads eg. Shunt and
computn machines. Let us see how the efficiency of a d.c. Shunt machine( generator or
motor) is determined by this method . The test cosists of two steps:
(i) Detemination of hot resistance ofwindings. The aramture resistance and shunt field
resistance are measured using abattery vltmeter and ammetere. Since these
resistance are measured when te machine is cold, they must be converted to values
corresponding tothe temperture at which teh machine would work on full load.
Generally thse value ate mesured for a tempereture rise of 40 C above the room
temperature .Ltt the hot resistance of armaturse and shunt field be Ra and Rsh
respectively.
(ii) Determination of costant lossed . The machins is run as motor on noload with supply
voltage sdjustd to the reted voltage i.e. voltage stamped o the maeplate the speed
ofthe motor is adjusted to te rete dpeed with the help ofield regulator R as shown in Fig
7.2.
Let v = supply voltage
I0 = No – load currnetn read by mameter A1
Ish = Shunt field current read y ammetr A2
No load power input to armature = Via0 = V(I0-Ish)
Since the output of the motor is zero the no- load input power to the armature suppplies (a)
iron losses in the core (b) friction loss (c) windage loss (d) armature Culoss [I2a0 or (I0-Ish)2
Ra].
Constant losses Wc = Input to motor – Armature Cu loss
Wc = W I0-(I0-Ish)2Ra
Since constant lossssses are known the efficiency of th machine at any other load can be
determined suppose it is desired to determien the efficiency ofht machine at load current I
then
Armature current Ia = I – Ish
= I + Ish
Efficinecy when running as motor
Input powr to motor = VI
Armature Cu loss = I2aRa= (I-Ish)2Ra
Constant losses = Wc
Total losses = (I-Ish)2Ra + Wc
Motor efficiency Mm = Input -Losses/Input = VI(I-Ish)2Ra-Wc/VI
Efficicncy when running as a generator
Out put of generator = VI
Armauture Cu loss = (I+Ish)2 Ra
Constant losses = wc
Total losses = (I+Ish)2 Ra+Wc
Generator efficiency Mg = Output/Output+Losses = VI/VI+(I+Ish)2Ra+Wc
Advantages of swinburne's test
(i) The power rewuire to carry out hte test is small because it is a no -load test. Therefor
this method is quite economical.
(ii) The efficiency can be detrmined at any load becaude constant losses are known
(iii) This test is ver vonvenient
Disadvantages of Swinburne's tst
(i) It does not take into account the stray load losses that occure when the machine is
loaded.
(ii) This test does not aenable up the check the performane of the machine on full load.
For example it doed not indicte whether commutationon full load is satisfactory and
whether the temperature rise is within the specifice limits.
(iii) Thsi test does not give quite accurate efficency o fht machine. It is because iron
losses under actual load are greaer than those measured. This is manly due t armature
reaction distroting the field.
Transformer
INTRODUCTION
The transformer is probably one of the most useful electrical devices ever invented. It can
change the magnitude of alternating voltage or current from one valve to another. This useful
property of transformer is mainly responsible for the widespread use of alternating currents
rather than direct currents i.e., electric power is generated, transmitted and distributed in the
form of alternating current. Transformers have no moving parts, rugged and durable in
construction, thus requiring very little attention. Thy also have a very high efficiency-as high
as 99%. In this chapter, we shall study some of the basic properties of transformers.
8.1 TRANSFORMER
A transformer is a static piece of equipment used either for raising or lowering the voltage of
an a.c. supply with a corresponding decrease or increase in current. It essentially consists of
two windings, the primary and secondary, wound on a common laminated magnetic core as
shown in Fig.8.1. The winding connected to the a.c. sourse is called primary winding (or
primary) and the one connected to load is called secondary winding (or secondary). The
alternating voltage V1 whose magnitude is to be changed is applied to the primary. Depending
up on the number of turns of the primary (N1) amd secondary (N 2), an alternating e.m.f. E2 is
included in the secondary. This inuced e.m.f. E 2 in the secondary causes a secondary current
I2.
Consequently, terminal voltage V 2 will appear acrose the load. If V 2 > V 1, it is called a step
up-transformer. On the other hand, if V 2 < V 1, it is called a step-down transformer.
Working. When an alternating voltage V1 is applied to the primary, an alternating flux ф Is set
up in the core. This alternating flux links the windings and induces e.m.f.s
according to Faraday’s laws of electromagnetic induction. The e.m.f.
E1
E1 and E2
in them
is termed as primary
e.m.f. and e.m.f. E2 is termed as secondary e.m.f.
Clearly,
E1
Note that magnitudes of
= - N1dф
E1 and E2 *depend
up on the number of turns on the secondary and
primary respectively. If N 2 > N 1, then E2 > E1 (or V 2 > V 1) and we get a step-up transformer.
On the other hand, if N 2 < N 1, then E2 <E1 (or V 2 > V 1) and we get a step-down transformer. If
load is connected across the secondary winding, the secondary e.m.f. E 2 will cause a current
I2 to flow through the load. Thus, a transformer enables us to transfer a.c. power from one
circuit to another with a change in voltage level.
The following points may be noted carefully:
(i)
The transformer action is based on the laws of electromagnetic induction.
(ii)
There is no electrical connection between the primary and secondary. The a.c.
power is transferred from primary to secondary through magnetic flux.
(iii)
There is no change in frequency i.e., output power has same frequency as the input
power.
(iv)
The losses that occur in a transformer are:
(a) core losses – eddy current and hysteresis losses
(b) Copper losses – in the resistance of the windings
In practice, these losses are very small so that output power is nearly equal to the input
primary power. In other words, a transformer has very high efficiency.
PRACTICAL TRANSFORMER
A practical transformer differs from the ideal transformer in many respects. The practical
transformer has (i) iron losses (ii) winding resistances and (iii) magnetic leakage, giving rise to
leakage reactances.
(i) Iron losses. Since the iron core is subjected to alternating flux, there occurs eddy
current and hysteresis loss in it. These two losses together are known as iron losses or
core losses. The iron losses depend upon the supply frequency, maximum flux density
in the core, volume of the core etc . It may be noted that magnitude of iron losses in
quite small in a practical transformer.
(ii) Winding resistance . Since the windigs consists of copper conductors, it immediately
follows that both primary and secondary will have winding resistance. The primary
resistance R1 and secondary resistance R2 act in series with the respective windings
as shown in Fig. 8.6. When current flows through the windings, there will be power loss
as well as a loss in voltage due to IR drop . This will affect the power factor and E1 will
be less than V1 while V2 will be less than E2.
(iii) Leackage reactances. Both primary and secondary current produce flux. The flux φ
which links both the windings is the useful flux and is called mutal flux. However,
primary current would produce some flux φ1 which would not link the secondary
winding (See Fig. 8.7). Similarly, secondary current would produce some flux φ2 that
would not link the primary winding. The flux such as φ1 or φ2 which links only one
winding is called leakage flux. The leakage flux paths are mainly throughthe air. The
effect of these leakage fluxes would be the same as though inductive reactances were
connected in series with each winding of transformer that had no loakage flux as
ahown in Fig. 8.6. In other words , the effect of primar leakage flux φ1 is to introduce
an inductive reactance X1 in series with the primary winding as shown in Fig. 8.6.
Similarly, the secondary leakage flux φ2 introduces an dinductive reactance X2 in
series with the secondary winding. There will be no power loss due to leakage
reactane. However, the presence of lakage ractance in hte windings changes the
power factor as well as there is voltage loss due to IX drop.
Note : Although leakage flux in a transformer is quite small (about 5% of φ) compared
to the mutual flux φ yet it cannot be ignored. It is because leakage flux paths are through air
of high reluctance and hence require considerable e.m.f. It may be noted that energy is
conveyed from the primary winding to the secondary winding by mutual flux φ which links
both the windings.
PRACTICAL TRANSFORMER ON LOAD
We shall consider two cases (i) when such a transformer is assumed to hav eno einding
resisntance and leakage flux (ii) when the transformer has winding resistance and leakage
flux.
(i) No winding resisntace and leakage flux
Fig. 8.12 shows a practical transformer with the assumption that resistance and leakage
reactances of the windings are negligible. With this assumption, v2 = E2 and V1=E1. Let us
take the usual case of inductive load which causes the secondary current I2 to lag the
secondary voltage V2 by ф2. The total primary current I1 must meet two requirements viz.
(a) It must supply the no-load current l0 to meet the iron losses in the transformer and to
provide flux in the core.
(b) It must supply a current I2 to conteract the demagnetising effect to secondary current I2.
The magnitude of I2 will be such that N1I'2 = N2I2
or I'2 = (N2/N1) I2= KI2
The total primary current I1 is the phasor sumof I'2 and L0 ie,
I1 = I'2 + I0
where I'2 = -KI2
Note that I'2 is 180 0 out of phase with I2.
Phasor diagram Fig. 8.13 shows the phasor diagram for the usual caseof inductive load. Both
E1 and E2 lag behind the mutual flux ф by 90 0. The current I'2 represents the primary
current to nutralise the demagnetising effect of secondary current I2. Now I'2 = KI2 and is
antiphase with I2. I0 is the noload current of the transformer. The phasor sum of I'2 and I0
gives the total primary current I1. Note that in drawing the phasor diagram, the value of K is
assumed to be unity so that primary phasors are equal to secondary phasors.
Primar p.f = cos ф1
Secondary p.f. = cos ф2
Primary input power = V1I1 cos ф1
secondary output power = V2I2 cosф2
(ii) Transformer with resistance and leakage reactance
Fig. 8.17 shows a practical transformer having winding resistance and leakage reactances.
These are the actual conditions that exist in a transformer. There is voltage drop in R1 and X1
so that primary e.m.f. E1 is less that the applied voltage V1. Similarly, there is voltage drop in
R2 and X2 so that secondary terminal voltage V2 is less thatn the secondary e.m.f. E2. Let us
take the usual case of inductive load which causes the secondary current I2 to lag behind the
secondary voltage V2 by ф2. The total primary current I1 must meet two requirements viz.
(a) It must supply the no-load current I0 to meet the iron losses in the transformer and to
provide flux in the core.
(b) It must supply a current I'2 to counteract the demagnetising effect of secondary current I2.
The magnitude of I'2 will be such that :
N1I'2 = N2I2
or I'2 = (N2/N1)I2 = KI2
The total primary current I1 will be the phasor sum of I'2 an I0 id,
I1 = I'2 + I0 where I'2 = -KI2
V1 = =E1+I1(R1+jX1) where I1 = I0 + (-KI2)
= -E1 +I1+Z1
V2 = E2 – I2 (R2 + jX2)
= E2 – I2 Z2
Phasor diagram . Fig 8.18 shows the phasor diagram of the practical trasformer fot he usual
case of inductive load. Both E1 and E2 lag the mutual flux ф by 900. The current I'2
represents the primary current to neutralise the demagnetising effect of secondary current I2.
Now I'2 = KI2 and is opposite to I2. Also I0 is the no-load current of the transformer. The
phasor sum of I2 and I0 gives the total primary current I1.
Note that counter e.m.f that opposes the applied voltage V1 I1-E1. Therefore if we add I1R1 (
in phase with I1) and I1X1 (900 of I1) to -E1, we get the applied primary voltage V1. The
phasor E2 represents the induced e.m.f in the secondary by mutal flux ф. The secondary
terminal voltage V2 will be what is left over after subtracting I2R2 and I2X2 from E2.
Load power factor = cos ф2
primary power factor = cos ф1
Input power to transformer, P1 = V1I1cos ф1
Output power of transformer, P2 = v2 I2 cos ф2
Note : The reader may draw the phasor diagram of a loaded trnsformer for (i) unit p.f and (ii)
leading p.f. As an exercise.
IMPEDANCE RATIO
Consider a transformer having impedance Z2 in the secodary as shown in Fig. 8.19
Z2 = v2/I2
Z1= V1/I1
N2/Z1 = (V2/V1) x(I1/I2)
Z2/Z1 = K2
ie, impedance ration (Z2/Z1) is equal to the square of voltage transformation ration. In other
words, an impedance Z2 in secondry becomes Z2/K2 when transferred to primary. Likewise,
an impedance Z1 in the primary becomes K2Z1 when trensferred to the secondary.
Similarly, R2/R1 = K2 and X2/X1 = K2 s
Note the importance of above relations. We can transfer the parameters form one winding to
the other thus:
(i) A resistance R1 in the primary becomes K2R1 when transferred to the secondary.
(ii) A resistance R2 in the secondary becomes R2/ K2 when tranferred to the primary
(iii) A reactance X1 in the primary becomes K2 X1 when transferred to the secondary
(iv) A reactance X2 in the secondary becomes X2/ K2 when transferred to the primary.
Note : It is important to remember that :
(i)
When transferring resistance or reactance from primary to secondary, multiply it by K 2
(ii) when transferring resistance or reactance from sedondary to primary , divide it by K2.
(iii) When transferring voltage or current from one winding the other only K is used.
EXACT EQUIVALENT CIRCUIT OF A LOADED TRANSFORMER
Fig. 8.28 shows the exact equivalent circuit of a transformer on load. Here R1 is the primary
winding resistance and R2 is the secondary winding resistance. Similarly, X1 is the leakage
reactance of primary winding and X2 is the leakage rectance of the secondary winding. The
parallel circuit R0-X0 is the no-load equivalent circuit of the transformer. The resistance R0
represents the core lossess (hysteresis and eddy current losses) so thant current Iw which
supplies the core losses is shown passing through R0. The inductive reactance X0 represents
a loss-free coil which passes the magnetising current Im. The phasor sum of Iw and Im is the
no-load current I0 of the transformer.
Note that in the equivaleeeent circuit shown in Fig. 8.28, the imperfections of the transformer
have been taken into account by various circuit elements. Therefore, the transformer is now
the ideal one. Note that equuivalent circuit has created two normal electrical circuits
sseparated only by an ideal transformer whose function is to change values according to the
equation:
E2/E1 = N2/N1 = I'2/I1
The following points may be noted from the equivalent circuit:
(i) When the transformer is on no-load (ie, secondary terminals are open -circuited), there
is no current in the secondary winding. However, the primary draws a small no-load I0.
The no-load primary current I0 is composed of (a) magnetising current(Im) to create
magnetic flux in the core and (b) the current Iw required to supply the core losses.
(ii) When the secondary circuit of a transformer is closed through some external load ZL,
the voltage E2 induced in the secondary by mutual flux will produce a secondary
current I2. There will be I2R2 and I2 X2 drops in the secondary winding so that load
voltage V2 will be less than E2.
V2 = E2-I2 (R2+jX2) = E2 – I2Z
(iii) When the transformer is loaded to carry the secondary current I2, the primary current
consists of two components:
(a) The no-load current I0 to provide magnetising current and the current required to
supply the core losses .
(b) The primary current I'2 (-KI2) required to supply the load connected to the
secondary.
Total primary current I1 = I0 + ( - KI2)
(iv) Since the transformer in Fig. 8.28 is now ideal, the primary induced voltage E 1 can be
calculated from the relation :
E1/E2 = N1/N2
If we add I1R1 and I1X1 drops to E1 , We get the primary input voltage V1.
V1 = -E1 + I1 (R1 + j X1) = -E1 + I1Z1
V1 = -E1+I1Z1
SIMPLIFIED EQUIVALENT CIRCUIT OF A LOADED TRANSFORMER
The no-load current I0 of a transformer is small compared to the rated primary current.
Therefore, viltage drops in R1 and X1 due to I0 are negligible. The equivalent circuit shown in
Fig. 8.28 above can therefore , be simplified by transferring the shunt circuit R 0 – X0 to the
input terminals as shown in Fig. 8.29. This modificarion leads to only slight loss of accuracy.
(i) Equivalent circuit referred to primary. If all the secondary quantities are referred to the
primary, we get the equivalent circuit of te transformer referred to the primary as shown
in Fig. 8.30. (i) This further reduces to Fig. 8.30 (ii). Note that when secondar quantities
are referred to primary, resistances/reactances/impedances are divided by K 2, voltages
are divided by K and currents are multiplied by K.
R'2 = R2/K2 ; X'2 = X2/K2 ; Z'L = ZL/ K2 ; V'2 = V2/K ; I'2 = KI2
where R01 = R1 + R'2 ; X01 = X1 +X'2
(ii) Equivalent circuit referred to secondary. If all theprimary quantities are referred to
secondary, we get the equivalent circuit of the transformer referred to secondary as
shown in Fig. 8.32 (i) This further reduces to Fig. 8.32 (ii) Note that when primary
quantities are referred to secondary resistances/reactances/impedances are multiplied
by K2, voltages are divided by K and currents are divided by K.
APPROXIMATE EQUIVALENT CIRCUIT OF LOADED TRANSFORMER
The no-load current I0 in a transformer is only 1 – 3% of the rated primary current and may be
neglected without any serious error. The transformer can than be shown as in Fig. 8.36. This
is an approximate representation because no-load current has been neglected. Note that all
the circuit elements have been shown external so that the transformer is an ideal one.
As shown in Art. 8.11, if we refer all the quantities to one side (primary or secondary), the
ideal transformer stands removed and we ger the equivalent circuit.
(i) Equivalent circuit of transformer referred to primary. If all the secondary quantities are
referred to the primary, we get the secondary quatities are referred to primary,
resistance /resistance are divided by K2, voltages are divided by K and currents are
multiplied by K.
The equivalent circuit shown in Fig. 8.37 is an electrical circuit and can be solved for
various
currents and voltages. Thus if we find V'2 and I'2 , then actual secondary values can be
detrmined as under:
Actual secondary voltage V2 = K V'2
Actual secondary current I2 = I'2/K
(ii) Equivalent circuit of transformer referred to secondary. If all the primary quantities are
referred to secondary , we get the equivalent circuit of the transformer referred to
secondary as shown in Fig. 8.38. Note that when primary quantities are referred to
secondary, resistances/reactances are multiplied by K2, voltages are multiplied by K
and currents are divided by K.
The equivalent circuit shown in Fig. 8.38 is an electical circuit and can be solved for
various voltages and currents. Thus if we find V'1 and I'1, then actual primary values can be
determined as under:
Actual primary voltage , V1 = V'1/K
Actual primary current , I1 = KI'1
Note : The same final answers will be obtained where we use the equivalent circuit referred to
primary or secondary. The use of a particular equivalent circuit would depend upon the
conditions of the problem .
APPROXIMATE VOLTAGE DROP IN A TRANSFORMER
The approximate equivalent circuit of transformer referred to secondary is shown in Fig. 8.39.
At no-load , the secondary voltage is KV1. When a load having a lagging p.f. cos φ2 is applied,
the secondary carries a current I2 and voltage drops occur in (R2 + K2R1) and (X2 + K2X1).
Consequently, the secondary voltage falls from KV1 to V2. Referring to Fig. 8.39, we have,
V2 = K V1 -I2 [(R2 + K2 R1) + j(X2 +K2X1)]
= KV1 – I2 (R02 + j X02)
= KV1 – I Z02
Drop in secondary voltage = KV1 – V2 = I2X02
The phasor diagram is shown in Fig. 8.40. It is clear from the phasor diagram that drop in
secondary voltage is AC = I2Z02. It can be found as follows. With O as centre and OC as
radius , draw an arc cutting OA produced at M. then AC = AM = AN. From B, draw BD
perpendicular to OA produced. Draw CN perpendicular to OM and draw BL || OM.
Approcimately drop in secondary voltage = AN = AD+DN
= AD +BL
= I2 R02 cosφ2 + I2 X02 sin φ2
For a load having aleading p.f. cos φ2, we have,
Approximate voltage drop = I2R02cosφ2 – I2X02sinφ2
Note : If the circuit is referred to primary, then it can be easily established that:
Approximate voltage drop = I1R01cosφ2 ± I1X01 sinφ2
VOLTAGE REGULATION
The voltage regulation of a transformer is the arithmetic differnce (not phasor difference)
between the no-load secondary voltage (0V2) and the secondary voltage V2 on load expressed
as percentage of no-load voltage i.e,
% age volatage regulation = 0V2 -V2 /0V2 x 100
where 0V2 = No-load secondary voltage = KV1
V2 = Secondary voltage no load
As shown in Art 8.15, 0V2-V2 = I2 R02 cos φ2 ± I2 X02 sin φ2
The +ve sign is for lagging p.f. and -ve sign for lading p.f.
It may be noted that %age voltage regulation of athe transformer will be the same whether
primary or secondary side is considered.
TRANSFORMER TEST
The circuit constants , efficiency and voltge regultion ofa transformer can be determined by
two simple tests (i) open-circuit test and (ii) shor -circuit test. These tests are very convenient
as tehy provide the required information without actually loading the transformer . Further ,
the power required to carry out these tests is very small as compared with full-load output of
the transformer. These tests consist of measuring the input voltage, current and power to the
primary first with secondary open -circuited (open -circuit test ) and then with the secondary
short – circuited (short circuit test).
OPEN-CIRCUIT OR NO-LOAD TEST
This test is conducted to determine the iron losses (or core losses) and parameters R0 and X0
of the transformer. In this test, the rated voltage is applied to the primary (usually low-voltage
winding) while the secondary is left open -circuited. The applied primary voltage V1 is
measured by the voltmeter, the no-load current I0 by ameter and no -load input power W0 by
wattmeter as iron losses will occur in the transformer core. Hence wattmeter will record the
iron losses and small copper loss in the primary. Since no-load current I0 is very small
(usually 2-10% of rated current), Cu losses in the primary under no-load condition are
negligible as compared with iron losses. Hence wattmeter reading practically gives the iron
lossees in the transforme. It is reminded that iron losses are the same at all loads. Fig. 8.41
(ii) shows the equivalent circuit of transformer on no-load.
Iron losses Pi = Wattmeter reading = W0
No load current = Ammeter reading = I0
Applied voltage = Voltmeter reading = V1
Input power , W0 = V1 I0 cos φ0
N0-load p.f cos φ0 = W0/V1I0
Iw = I0cosφ0 ; Im = I0sin φ0
R0 = V1/Iw and X0 = V1/Lm
Thus open – circuit test enables us to detemine iron losses and parameters R0 and X0 of the
transformer.
SHORT CIRCUIT OR IMPEDANCE TEST
This test is conducted to determine R01 (orR02). X01 (or X02) and full-load copper loasses of
the transformer. In this test, the secondary (usually low-voltage winding) is short – circuited by
a thick conductor and variable low voltage is applied to the primary as shown in Fig.8.42 (i).
The low input voltage is gradually raised till at voltage Vsc, full-load current I1 flows in the
primary. Then I2 in the secondary also has full-load valllue since I1/I2 = N2/N1. Under such
conditions, the copper loss in the windings is the same as that on full load.
There is no ouput from the transforemer under short -circuit conditions. Therefore, input
power is all loss and this loss is almost entirely copper loss. It is because iron loss in the core
is negligibly small since the voltage Vsc is ver small. Hence the wattmeter will practically
register the full load copper losses in the transformer windings. Fig. 8.42 (ii) shows the
equivalent corcuit of a transformer on short circuit as referred to primary ; the no-load current
being neglected due to its smallness.
Full load Cu loss , Pc = Wttmeter winding = Ws
Applied voltage = voltmeter reading = Vsc
f.L primary current = Ammeter reading = I1
Pc = I21R1 + I21R'2 = I21R01
Where R01 is the total resistance of transformer referred to primary .
Total impedance referre to primary , Z01 = Vsc/I1
Total leakage rectance refrred to primary X01 = √Z201 -R201
short circuit p.f cost φs = Pc/VscI1
Thus short circuit test gives full load Cu loss, R01 and X01.
Note: The short circuit test will give full load Cu loss only if the applied voltage V sc is such so
as to circulate full-load currents in the windings. If in a short-circuit test, current value is other
than full-load value, the Cu loss will be corresponding to that current value.
ADVANTAGES OF TRNSFORMER TESTS
The above two simple transformer test offer the following advantages:
(i) The power required to carry out these tests is very small as cmplared to the full load
output of the transformer . In case of open circuit test , power required is equal to the
iron loss whereas for a short circuit test , power required is equal to full-load copper
loss.
(ii) These tests enable us to determine the efficiency of the transformer accurately at any
load and p.f. without actually loading the transformer.
(iii) The short- circuit enables us to determine R01 and X01 (or R02 and X02). We can thus
find the total voltage drop in the transformer as referred to primary or secondary. This
permits us to calculate voltage regulation of the transformer.