Bird J. Electrical Circuit Theory and Technology

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366 Electrical Circuit Theory and Technology

and torque T D



2pnZ



2n

i.e.,

T =

p8ZI

newton metres

21.7

For a given machine, Z, c and p are ﬁxed values

Hence torque,

T ∝ 8I

21.8

Problem 15. An 8-pole d.c. motor has a wave-wound armature

with 900 conductors. The useful ﬂux per pole is 25 mWb. Deter-

mine the torque exerted when a current of 30 A ﬂows in each

armature conductor.

p D 4, c D 2 for a wave winding,  D 25 ð 10

3

Wb, Z D 900,

D 30 A

From equation (21.7), torque T D

pZI

c

425 ð 10

3

90030

2

D 429.7Nm

Problem 16. Determine the torque developed by a 350 V d.c.

motor having an armature resistance of 0.5  and running at

15 rev/s. The armature current is 60 A.

V D 350 V, R

D 0.5 , n D 15 rev/s, I

D 60 A

Back e.m.f. E D V  I

D 350  600.5 D 320 V

From equation (21.6), torque T D

2n

32060

215

D 203.7Nm

Problem 17. A six-pole lap-wound motor is connected to a 250 V

d.c. supply. The armature has 500 conductors and a resistance of

1 . The ﬂux per pole is 20 mWb. Calculate (a) the speed and

(b) the torque developed when the armature current is 40 A

V D 250 V, Z D 500, R

D 1 ,  D 20 ð10

3

Wb, I

D 40 A,

c D 2p for a lap winding

(a) Back e.m.f. E D V  I

D 250  401 D 210 V

D.c. machines 367

E.m.f. E D

2pnZ

i.e. 210 D

2p20 ð 10

3

n500

Hence speed n D

210

20 ð 10

3

500

D 21 rev/s

or 21 ð 60 D 1260 rev/min

(b) Torque T D

2n

21040

221

D 63.66 Nm

Problem 18. The shaft torque of a diesel motor driving a 100 V

d.c. shunt-wound generator is 25 Nm. The armature current of the

generator is 16 A at this value of torque. If the shunt ﬁeld regulator

is adjusted so that the ﬂux is reduced by 15%, the torque increases

to 35 Nm. Determine the armature current at this new value of

torque.

From equation (21.8), the shaft torque T of a generator is proportional to

I

, where  is the ﬂux and I

is the armature current. Thus, T D kI

where k is a constant.

The torque at ﬂux 

and armature current I

is T

D k

Similarly, T

D k

By division

k



Hence



ð 16

0.85

ð I

i.e. I

16 ð 35

0.85 ð 25

D 26.35 A

That is, the armature current at the new value of torque is 26.35 A

Problem 19. A 100 V d.c. generator supplies a current of 15 A

when running at 1500 rev/min. If the torque on the shaft driving the

generator is 12 Nm, determine (a) the efﬁciency of the generator

and (b) the power loss in the generator.

(a) From Section 21.9, the efﬁciency of a

generator D

output power

input power

ð 100%

368 Electrical Circuit Theory and Technology

The output power is the electrical output, i.e. VI watts. The input

power to a generator is the mechanical power in the shaft driving

the generator, i.e. Tω or T2n watts, where T is the torque in Nm

and n is speed of rotation in rev/s. Hence, for a generator

efﬁciency, h D

T2n

ð 100%

i.e. h D

10015100

122



1500



i.e. efﬁciency= 79.6%

(b) The input power D output power Closses

Hence,T2n D VI C losses

i.e. losses D T2n  VI



122



1500



 [10015]

i.e. power loss D 1885  1500 D 385 W

Further problems on losses, efﬁciency, and torque may be found in

Section 21.17, problems 16 to 21, page 383.

21.12 Types of d.c. motor

and their characteristics

(a) Shunt-wound motor

In the shunt wound motor the ﬁeld winding is in parallel with the armature

across the supply as shown in Figure 21.16.

For the circuit shown in Figure 21.16,

Supply voltage, V D E C I

or generated e.m.f., E D V  I

Supply current, I D I

C I

, from Kirchhoff’s current law.

Figure 21.16

Problem 20. A 240 V shunt motor takes a total current of 30 A.

If the ﬁeld winding resistance R

D 150  and the armature resis-

tance R

D 0.4  determine (a) the current in the armature, and

(b) the back e.m.f.

D.c. machines 369

(a) Field current I

240

150

D 1.6A

Supply current I D I

C I

Hence armature current, I

D I  I

D 30  1.6 D 28.4A

(b) Back e.m.f. E D V  I

D 240  28.40.4 D 228.64 volts

Characteristics

The two principal characteristics are the torque/armature current and

speed/armature current relationships. From these, the torque/speed rela-

tionship can be derived.

(i) The theoretical torque/armature current characteristic can be derived

from the expression T / I

, (see Section 21.11). For a shunt-

wound motor, the ﬁeld winding is connected in parallel with the

armature circuit and thus the applied voltage gives a constant ﬁeld

current, i.e. a shunt-wound motor is a constant ﬂux machine. Since

 is constant, it follows that T / I

, and the characteristic is as

shown in Figure 21.17.

(ii) The armature circuit of a d.c. motor has resistance due to the arma-

ture winding and brushes, R

ohms, and when armature current

is ﬂowing through it, there is a voltage drop of I

volts. In

Figure 21.16 the armature resistance is shown as a separate resistor

in the armature circuit to help understanding. Also, even though the

machine is a motor, because conductors are rotating in a magnetic

ﬁeld, a voltage, E / ω, is generated by the armature conductors.

From equation (21.5) V D E C I

or E D V  I

Figure 21.17

However, from Section 21.5, E / n, hence n / E/,i.e.

speed of rotation, n /



V  I



21.9

For a shunt motor, V,  and R

are constants, hence as armature

current I

increases, I

increases and V  I

decreases, and

the speed is proportional to a quantity which is decreasing and is

as shown in Figure 21.18. As the load on the shaft of the motor

increases, I

increases and the speed drops slightly. In practice,

the speed falls by about 10% between no-load and full-load on

many d.c. shunt-wound motors. Due to this relatively small drop

in speed, the d.c. shunt-wound motor is taken as basically being a

constant-speed machine and may be used for driving lathes, lines of

shafts, fans, conveyor belts, pumps, compressors, drilling machines

and so on.

Figure 21.18

370 Electrical Circuit Theory and Technology

Figure 21.19

(iii) Since torque is proportional to armature current, (see (i) above), the

theoretical speed/torque characteristic is as shown in Figure 21.19.

Problem 21. A 200 V, d.c. shunt-wound motor has an armature

resistance of 0.4  and at a certain load has an armature current of

30 A and runs at 1350 rev/min. If the load on the shaft of the motor

is increased so that the armature current increases to 45 A, deter-

mine the speed of the motor, assuming the ﬂux remains constant.

The relationship E / n applies to both generators and motors. For a

motor,

E D V  I

,see equation 21.5

Hence E

D 200  30 ð 0.4 D 188 V,

and E

D 200  45 ð 0.4 D 182 V.

The relationship,



applies to both generators and motors. Since the ﬂux is constant, 

D 

Hence

188

182





1350





ð n

, i.e., n

22.5 ð 182

188

D 21.78 rev/s

Thus the speed of the motor when the armature current is 45 A is

21.78 ð 60 rev/min, i.e. 1307 rev/min

Problem 22. A 220 V, d.c. shunt-wound motor runs at

800 rev/min and the armature current is 30 A. The armature circuit

resistance is 0.4 . Determine (a) the maximum value of armature

current if the ﬂux is suddenly reduced by 10% and (b) the steady

state value of the armature current at the new value of ﬂux,

assuming the shaft torque of the motor remains constant.

(a) For a d.c. shunt-wound motor, E D V  I

. Hence initial gener-

ated e.m.f., E

D 220  30 ð 0.4 D 208 V. The generated e.m.f. is

also such that E / n, so at the instant the ﬂux is reduced, the speed

has not had time to change, and E D 208 ð 90/100 D 187.2V.

Hence, the voltage drop due to the armature resistance is 220 

187.2, i.e., 32.8 V. The instantaneous value of the current is

32.8/0.4, i.e., 82 A. This increase in current is about three times

the initial value and causes an increase in torque, (T / I

). The

motor accelerates because of the larger torque value until steady

state conditions are reached.

(b) T / I

and since the torque is constant,



D 

. The ﬂux  is reduced by 10%, hence

D.c. machines 371



D 0.9

Thus, 

ð 30 D 0.9

ð I

i.e. the steady state value of armature current, I

0.9

D 33

(b) Series-wound motor

In the series-wound motor the ﬁeld winding is in series with the armature

across the supply as shown in Figure 21.20.

Figure 21.20

For the series motor shown in Figure 21.20,

Supply voltage V D E C IR

C R



or generated e.m.f. E D V  IR

C R



Characteristics

In a series motor, the armature current ﬂows in the ﬁeld winding and is

equal to the supply current, I.

(i) The torque/current characteristic

It is shown in Section 21.11 that torque T / I

. Since the arma-

ture and ﬁeld currents are the same current, I, in a series machine,

then T / I over a limited range, before magnetic saturation of the

magnetic circuit of the motor is reached, (i.e., the linear portion of

the B–H curve for the yoke, poles, air gap, brushes and arma-

ture in series). Thus  / I and T / I

. After magnetic satura-

tion,  almost becomes a constant and T / I. Thus the theoretical

torque/current characteristic is as shown in Figure 21.21.

(ii) The speed/current characteristic

It is shown in equation (21.9) that n / V  I

/.Inaseries

motor, I

D I and below the magnetic saturation level,  / I. Thus

n / V  IR/I where R is the combined resistance of the series

ﬁeld and armature circuit. Since IR is small compared with V, then

an approximate relationship for the speed is n / V/I / 1/I since

V is constant. Hence the theoretical speed/current characteristic is

as shown in Figure 21.22. The high speed at small values of current

indicate that this type of motor must not be run on very light loads

and invariably, such motors are permanently coupled to their loads.

Figure 21.21

(iii) The theoretical speed/torque characteristic may be derived from

(i) and (ii) above by obtaining the torque and speed for various

values of current and plotting the co-ordinates on the speed/torque

characteristics. A typical speed/torque characteristic is shown in

Figure 21.23.

Figure 21.22

372 Electrical Circuit Theory and Technology

Figure 21.23

A d.c. series motor takes a large current on starting and the

characteristic shown in Figure 21.21 shows that the series-wound

motor has a large torque when the current is large. Hence these

motors are used for traction (such as trains, milk delivery vehicles,

etc.), driving fans and for cranes and hoists, where a large initial

torque is required.

Problem 23. A series motor has an armature resistance of 0.2 

and a series ﬁeld resistance of 0.3 . It is connected to a 240 V

supply and at a particular load runs at 24 rev/s when drawing 15 A

from the supply.

(a) Determine the generated e.m.f. at this load.

(b) Calculate the speed of the motor when the load is changed

such that the current is increased to 30 A. Assume that this

causes a doubling of the ﬂux.

(a) With reference to Figure 21.20, generated e.m.f., E, at initial load,

is given by

D V  I

R

C R



D 240  150.2 C0.3 D 240  7.5 D 232.5 volts

(b) When the current is increased to 30 A, the generated e.m.f. is

given by:

D V  I

R

C R



D 240  300.2 C0.3 D 240  15 D 225 volts

Now e.m.f. E / n

thus



i.e.,

232.5

225



24

2

n



since 

D 2

Hence speed of motor, n

24225

232.52

D 11.6 rev/s

As the current has been increased from 15 A to 30 A, the speed has

decreased from 24 rev/s to 11.6 rev/s. Its speed/current characteristic

is similar to Figure 21.22.

There are two types of compound wound motor:

(i) Cumulative compound, in which the series winding is so connected

that the ﬁeld due to it assists that due to the shunt winding.

Figure 21.24

D.c. machines 373

Figure 21.25

(ii) Differential compound, in which the series winding is so connected

that the ﬁeld due to it opposes that due to the shunt winding.

Figure 21.24(a) shows a long-shunt compound motor and Figure 21.24(b)

a short-shunt compound motor.

Characteristics

A compound-wound motor has both a series and a shunt ﬁeld winding,

(i.e. one winding in series and one in parallel with the armature), and is

usually wound to have a characteristic similar in shape to a series wound

motor (see Figures 21.21–21.23). A limited amount of shunt winding

is present to restrict the no-load speed to a safe value. However, by

varying the number of turns on the series and shunt windings and the

directions of the magnetic ﬁelds produced by these windings (assisting

or opposing), families of characteristics may be obtained to suit almost

all applications. Generally, compound-wound motors are used for heavy

duties, particularly in applications where sudden heavy load may occur

such as for driving plunger pumps, presses, geared lifts, conveyors, hoists

andsoon.

Typical compound motor torque and speed characteristics are shown in

Figure 21.25.

21.13 The efﬁciency of a

d.c. motor

It was stated in Section 21.9, that the efﬁciency of a d.c. machine is

given by:

efﬁciency,  D

output power

input power

ð 100%

Also, the total losses D I

C I

V C C (for a shunt motor) where C is

the sum of the iron, friction and windage losses.

For a motor, the input power D VI

and the output power D VI  losses

D VI  I

 I

V  C

Hence

efﬁciency h =



VI − I

− I

V − C



× 100% 21.10

The efﬁciency of a motor is a maximum when the load is such that:

= I

V Y C

374 Electrical Circuit Theory and Technology

Problem 24. A 320 V shunt motor takes a total current of 80 A

and runs at 1000 rev/min. If the iron, friction and windage losses

amount to 1.5 kW, the shunt ﬁeld resistance is 40  and the arma-

ture resistance is 0.2 , determine the overall efﬁciency of the

motor.

The circuit is shown in Figure 21.26.

Figure 21.26

Field current, I

320

D 8A

Armature current I

D I  I

D 80  8 D 72 A

C D iron, friction and windage losses D 1500 W

Efﬁciency,  D



VI  I

 I

V  C



ð 100%



32080  72

0.2  8320  1500

32080



ð 100%



25 600  1036.8 2560  1500

25 600



ð 100%



20 503.2

25 600



ð 100% D 80.1%

Problem 25. A 250 V series motor draws a current of 40 A. The

armature resistance is 0.15  and the ﬁeld resistance is 0.05 .

Determine the maximum efﬁciency of the motor.

The circuit is as shown in Figure 21.27.

D.c. machines 375

Figure 21.27

From equation (21.10), efﬁciency,

 D



VI  I

 I

V  C



ð 100%

However for a series motor, I

D 0 and the I

loss needs to be

R

C R



Hence efﬁciency,  D



VI  I

R

C R

  C



ð 100%

For maximum efﬁciency I

R

C R

 D C

Hence efﬁciency,  D



VI  2I

R

C R





ð 100%



25040  240

0.15 C 0.05

25040



ð 100%



10000  640

10000



ð 100%



9360

10000



ð 100% D 93.6%

Problem 26. A 200 V d.c. motor develops a shaft torque of

15 Nm at 1200 rev/min. If the efﬁciency is 80%, determine the

current supplied to the motor.

The efﬁciency of a motor D

output power

input power

ð 100%

The output power of a motor is the power available to do work at its shaft

and is given by Tω or T2n watts, where T is the torque in Nm and n