Speed control of DC motors

speed
control of
d.c.
motors


? factors controlling motor speed
? speed control of shunt motors
? speed control of series motors
? merits and demerits of rheostatic
control method
? series-parallel control
? electric braking
? electric braking of shunt motor
? electric braking of series motors
? electronic speed control method
for d.c. motors
? uncontrolled rectifiers
? controlled rectifiers
? thyristor choppers
? thyristor inverters
? thyristor speed control of separately-
excited d.c. motor
? thyristor speed control of d.c.
series motor
? full-wave speed control of a
shunt motor
? thyristor control of a shunt motor
? thyristor speed control of a series
d.c. motor
? necessity of a starter
? shunt motor starter
? three-point starter
? four-point starter
? starting and speed control of
series motors
? grading of starting resistance
? shunt motors
? series motor starters
? thyristor controller starters

     
learning objectives














 





contents
contents
1032 electrical technology
30.1. factors controlling motor speed
it has been shown earlier that the speed of a motor is given by the relation
n = . ( ) r.p.s.
? ?
=
? ?
a a a a v i r a v i r
k
z p
where ra = armature circuit resistance.
it is obvious that the speed can be controlled by varying (i) flux/pole, ? (flux control)
(ii) resistance ra of armature circuit (rheostatic control) and (iii) applied voltage v (voltage control).
these methods as applied to shunt, compound and series motors will be discussed below.
30.2. speed control of shunt motors
(i) variation of flux or flux control method
it is seen from above that n ? 1/?. by decreasing the flux, the
speed can be increased and vice versa. hence, the name flux or
field control method. the flux of a d.c. motor can be changed by
changing ish with help of a shunt field rheostat (fig. 30.1). since
ish is relatively small, shunt field rheostat has to carry only a small
current, which means i2r loss is small, so that rheostat is small in
size. this method is, therefore, very efficient. in non-interpolar
machine, the speed can be increased by this method in the ratio
2 : 1. any further weakening of flux ? adversely affects the
communication and hence puts a limit to the maximum speed
obtainable with the method. in machines fitted with interpoles, a
ratio of maximum to minimum speed of 6 : 1 is fairly common.
example 30.1. a 500 v shunt motor runs at its normal speed of 250 r.p.m. when the armature
current is 200 a. the resistance of armature is 0.12 ohm. calculate the speed when a resistance is
inserted in the field reducing the shunt field to 80% of normal value and the armature current is 100
ampere. (elect. engg. a.m.a.e. s.i. june 1992)
solution. eb1 = 500 ? 200 × 0.12 = 476 v eb2 = 500 ? 100 × 0.12 = 488 v
?2 = 0.8 ?1 n1 = 250 rpm n2 = ?
now, 2
1
n
n
=
2 1 2 1
2
1 2 1
488
or ,
250 476 0.8
b
b
e n
n
e
? ?
× = ×
? ? = 320 r.p.m.
example 30.2. a 250 volt d.c. shunt motor has armature resistance of 0.25 ohm, on load it takes
an armature current of 50 a and runs at 750 r.p.m. if the flux of motor is reduced by 10% without
changing the load torque, find the new speed of the motor. (elect. eng-ii, pune univ. 1987)
solution. 2
1
n
n
= 2 1
1 2
?
×
?
b
b
e
e
now, ta ? ? ia. hence ta1 ? ?1 ial and ta2 ? ?2 ia2.
since ta1 = ta2 ? ?1 ia1 = ?2 ia2
now, ?2 = 0.9 ?1 ? 50 ?1 = 0.9 ?, ia2 = 55.6 a
? eb1 = 250 ? (50 × 0.25) = 237.5 v eb2 = 250 ? (55.6 × 0.25) = 231.1 v
? 2
750
n
= 1
1
231.1
237.5 0.9
?
×
?
n2 = 811 r.p.m.
example 30.3. describe briefly the method of speed control available for dc motors.
a 230 v d.c. shunt motor runs at 800 r.p.m. and takes armature current of 50 a. find resistance
fig. 30.1
v
field
rheostat
speed control of d.c. motors 1033
to be added to the field circuit to increase speed to 1000 r.p.m. at an armature current of 80 a.
assume flux proportional to field current. armature resistance = 0.15 ? and field winding resistance
= 250 ?. (elect. technology, hyderabad univ. 1991)
solution. 2
1
n
n
= 2 1 2 1
1 2 1 2
?
× = ×
?
b b sh
b b sh
e e i
e e i
since flux ? field current
eb1 = 230 ? (50 × 0.15) = 222.5 v eb2 = 230 ? (80 × 0.15) = 218v
let rt = total shunt resistance = (250 + r) where r is the additional resistance
ish1 = 230/250 = 0.92 a, ish2 = 230/rt n1 = 800 r.p.m. n = 1000 r.p.m.
? 1000
800
=
218 0.92
222.5 230 /
×
t r
rt = 319 ? ? r = 319 ? 250 = 69 ?,
ish2 = 230
319
= 0.721
ratio of torque in two cases = 2 2 2
1 1 1
0.721 80
0.92 50
sh a
sh a
t i i
t i i
×
= = =
×
1.254
example 30.4. a 250 v, d.c. shunt motor has shunt field resistance of 250 ? and an armature
resistance of 0.25 ?. for a given load torque and no additional resistance included in the shunt
field circuit, the motor runs at 1500 r.p.m. drawing an armature current of 20 a. if a resistance of
250 ? is inserted in series with the field, the load torque remaining the same, find out the new speed
and armature current. assume the magnetisation curve to be linear.
(electrical engineering-i, bombay univ. 1987)
solution. in this case, the motor speed is changed by changing the flux.
now, 2
1
n
n
= 2 1
1 2
?
×
?
b
b
e
e
since it is given that magnetisation curve is linear, it means that flux is directly proportional to
shunt current. hence 2 2 1
1 1 2
= b × sh
b sh
n e i
n e i
where eb2 = v ? ia2 ra and eb1 = v ? ia1 ra.
since load torque remains the same ? ta ? ?1 ia1 ? ?2 ia2 or ?1 ia1 = ?2 ia2
? ia2 = 1 1
1 1
2 2
?
× = ×
?
sh
a a
sh
i
i i
i
now, ish1 = 250/250 = 1 a ish2 = 250/(250 + 250) = 1/2 a
? ia2 = 20 × 1
1/ 2
= 40 a ? eb2 = 250 ? (40 × 0.25) = 240 v and
eb1 = 250 ? (20 × 0.25) = 245 v ? 2 240 1
1500 245 1/ 2
= × n
? n2 = 2, 930 r.p.m.
example 30.5. a 250 v, d.c. shunt motor has an armature resistance of 0.5 ? and a field
resistance of 250 ?. when driving a load of constant torque at 600 r.p.m., the armature current is
20 a. if it is desired to raise the speed from 600 to 800 r.p.m., what resistance should be inserted in
the shunt field circuit ? assume that the magnetic circuit is unsaturated.
(elect. engg. amiete, june 1992)
solution. 2
1
n
n
= 2 1
1 2
?
×
?
b
b
e
e
since the magnetic circuit is unsaturated, it means that flux is directly proportional to the shunt
current.
1034 electrical technology
? 2
1
n
n
= 2 1
1 2
b × sh
b sh
e i
e i
where eb2 = v ? ia2 ra and eb1 = v ? ia1 ra
since motor is driving at load of constant torque,
ta ? ?1 ia1 ? ?2 ia2 ? ?2 ia2 = ?1 ia1 or ia2 = ia1 × 1
2
?
?
= ia1 × 1
2
sh
sh
i
i
now, ish1 = 250/250 = 1 a ish2 = 250/rt
where rt is the total resistance of the shunt field circuit.
? ia2 =
1 2
20
250/ 25
× = t
t
r
r
eb1 = 250 ? (20 × 0.5) = 240 v
eb2 =
2 800 250 ( / 25) 1
250 0.5 250 ( / 25)
25 600 240 250 /
? ? ? ? ? × ? = ? ? = ×
? ?
t t
t
t
r r
r
r
0.04 rt
2 ? 250 rt + 80,000 = 0
or rt =
250 62, 500 12,800 27
337.5
0.08 0.08
± ?
= = ?
additional resistance required in the shunt field circuit = 337.5 ? 250 = 87.5 ?.
example 30.6. a 220 v shunt motor has an armature resistance of 0.5 ? and takes a current
of 40 a on full-load. by how much must the main flux be reduced to raise the speed by 50% if the
developed torque is constant ? (elect. machines, amie, sec b, 1991)
solution. formula used is 2 2 1
1 1 2
?
= ×
?
b
b
n e
n e
. since torque remains constant,
hence ?1 ia1 = ?2 ia2 ? ia2 = ia1 . 1
2
?
?
= 40 x where x = 1
2
?
?
eb1 = 220 ? (40 × 0.5) = 200 v eb2 = 220 ? (40x × 0.5) = (220 ? 20 x) v
2
1
n
n =
3
2
... given ? 3
2
=
(220 20 )
200
? × x
x ? x2 ? 11x + 15 = 0
or x =
11 121 60 11 7.81
2 2
± ? ± = = 9.4* or 1.6
? 1
2
?
?
= 1.6 or 2
1
?
?
= 1
1.6
? 1 2
1
? ??
?
=
1.6 1 3
1.6 8
? = ? percentage change in flux = 3
100
8
× = 37.5%
example 30.7. a 220-v, 10-kw, 2500 r.p.m. shunt motor draws 41 a when operating at rated
conditions. the resistances of the armature, compensating winding, interpole winding and shunt
field winding are respectively 0.2 ?, 0.05 ?, 0.1 ? and 110 ?. calculate the steady-state values of
armature current and motor speed if pole flux is reduced by 25%, a 1 ? resistance is placed in series
with the armature and the load torque is reduced by 50%.
solution. ish = 220/110 = 2 a ia1 = 41 ? 2 = 39 a (fig. 30.2)
t1 ? ?1 ia1 and t2 ? ?2 ia2
? 2
1
t
t
= 1 1
2 2
?
×
?
a
a
i
i
* this figure is rejected as it does not give the necessay increase in speed.
speed control of d.c. motors 1035
or
1
2
= 3 2
4 39
× a i
? ia2 = 26 a
eb1 = 220 ? 39 (0.2 + 0.1 + 0.05) = 206.35 v
eb2 = 220 ? 26 (1 + 0.35) = 184.9 v
now, 2
2500
n
=
184.9 4
206.35 3
× n2 = 2987 r.p.m.
example 30.8. a 220 v, 15 kw, 850 r.p.m. shunt motor
draws 72.2 a when operating at rated condition. the
resistances of the armature and shunt field are 0.25 ? and
100 ? respectively. determine the percentage reduction in
field flux in order to obtain a speed of 1650 r.p.m. when
armature current drawn is 40 a.
solution. ish = 220/100 = 2.2 a ia1 = 72.2 ? 2.2 = 70 a
eb1 = 220 ? 70 × 0.25 = 202.5 v, eb2 = 220 ? 40 × 0.25 = 210 v.
now, 2
1
n
n
= 2 1
1 2
?
×
?
b
b
e
e
or 1
2
1650 210
850 202.5
?
= ×
?
? ?2 = 0.534 ?1
? reduction in field flux = 1 1
1
? ? 0.534?
? × 100 = 46.6%.
example 30.9. a 220 v shunt motor has an armature resistance of 0.5 ohm and takes an
armature current of 40 a on a certain load. by how much must the main flux be reduced to raise the
speed by 50% if the developed torque is constant ? neglect saturation and armature reaction.
(elect. machines, amie, sec b, 1991)
solution. t1 ? ?1 × ?1, ? ?1ia1, and t2 ? ?2 ia2 since, t1 = t2
? ?2 ia2 = ?1 ia1 or ?2/?1 = ia1/ia2 = 40/ia2
eb1 = 220 ? 40 × 0.5 = 200 v eb2 = 220 ? 0.5 ia2
now, 2
1
b
b
e
e
= 2 2
1 1
?
?
n
n
or 2 220 0.5
200
? a i
= 1
1 2
1.5 × 40
a
n
n i
? ia2 ? 40 ia2 + 24,000 = 0 or ia2 = 63.8 a
? 2
1
?
? =
40
63.8
= 0.627 or ?2 = 0.627 ?1 = 62.7% of?1
example 30.10. a d.c. shunt motor takes an armature current of 20 a from a 220 v supply.
armature circuit resistance is 0.5 ohm. for reducing the speed by 50%, calculate the resistance
required in the series, with the armature, if
(a) the load torque is constant
(b) the load torque is proportional to the square of the speed. (sambalpur univ., 1998)
solution. eb1 = v ? ia ra = 220 ? 20 × 0.5 = 210 v
210 ? n1
(a) constant load torque
in a shunt motor, flux remains constant unless there is a change in terminal voltage or there is a
change in the field-circuit resistance.
if torque is constant, armature-current then must remain constant. ia = 20 amp
with an external armature-circuit resistor of r ohms, 20 × (r + 0.5) = 220 ? eb2
the speed required now is 0.5 n1.
fig. 30.2
220 v
0.05 
1 
0.1 
110 
r
1036 electrical technology
fig. 30.3
with constant flux, eb ? speed. hence, 210 ? n1
eb2 ? 0.5 n1, eb2 = 105
r + 0.5 = (220 ? 105)/20 = 5.75, giving r = 5.25 ohms
(b) load torque is proportional to the square of speed.
with constant flux, developed torque at n1 r.p.m. ? ia1
tm1 ? 20
from the load side, tl1 ? n2
1
since tm1 = tl1
20 ? n1
2 ...(a)
at 50% speed, load torque, tl2 ? (0.5 n1)2
for motor torque, tm2 ? ia2
since tl2 = tm2, ia2 ? (0.5 n1)2 ...(b)
from eqn. (a) and (b) above
2
1
a
a
i
i
= 0.25, ia2 = 5 amp
2
1
b
b
e
e
= 2
220 ( 0.5)
210
? + ai r
= 0.5 n1
220 ? ia2 (r + 0.5) = 0.5 × 210
r + 0.5 =
220 105
5
?
= 23, r = 22.5 ohms
check : with the concept of armature power output : (applied here for part (b) only as an
illustration).
armature power–output = eb × ia = t × ?
when t ? (speed)2, eb ia = k1 t? = k2 ?3
at n1 r.p.m. 210 × 20 = k3 n1
3 ...(c)
with constant flux, at half speed
eb2 = 105
105 × ia2 = k3 (0.5 n1)3 ...(d)
from eqs. (c) and (d),
2 105
210 20
×
×
a i
=
3
1
3
1
0.125 × n
n
, giving ia2 = 5 amp
this gives r = 22.5 ohms.
example 30.11. a 250 v shunt motor runs at 1000 r.p.m. at no-load and takes 8 a. the total
armature and shunt field resistances are respectively 0.2 ohm, 250 ohm. calculate the speed when
loaded and taking 50 a. assume the flux to be constant.
(nagpur univ. summer 2000)
solution. the current distribution is shown in fig.
30.3.
at no load, il = 8 amp, if = 1 amp,
hence, ia = 7 amp
eb0 = 250 ? 7 × 0.2 = 248.6 volts,
= k ? × 1000
? k ? = 0.2486
250 v
a
aa
ff
f
a
50 a 49 a
ia
if 1 amp d.c. supply
speed control of d.c. motors 1037
at load, ia = 49 amp
eb1 = 250 ? 49 × 0.2 = 240.2
n1 = 240.2
0.2486
= 966.2 r.p.m.
notes : (i) the assumption of constant flux has simplified the issue. generally, armature reaction tends to
weaken the flux and then the speed tends to increase slightly.
(ii) the no load armature current of 7 amp is required to overcome the mechanical losses of motor as well
as driven load, at about 1000 r.p.m.
example 30.12. a 240 v d.c. shunt motor has an armature-resistance of 0.25 ohm, and runs at
1000 r.p.m., taking an armature current 40 a. it is desired to reduce the speed to 800 r.p.m.
(i) if the armature current remains the same, find the additional resistance to be connected in
series with the armature-circuit.
(ii) if, with the above additional resistance in the circuit, armature current decreases to 20 a,
find the speed of the motor. (bhartiar univ., november 1997)
solution. eb = 240 ? 0.25 × 40 = 230 v, 230 ? 1000 ...(a)
(i) for 800 r.p.m., eb2 ? 800 ...(b)
from (a) and (b), eb2 = 800
1000
× 230 = 184
240 ? (r + 0.25) × 40 = 184, r = 1.15 ohm
(ii) eb3 = 240 ? 20 (1.40) = 212
eb3 = 3
1000
n
× 230 = 212, n3 =
212
230
× 1000 = 922 r.p.m.
example 30.13. a 7.48 kw, 220 v, 990 r.p.m. shunt motor has a full load efficiency of 88%, the
armature resistance is 0.08 ohm and shunt field current is 2 a. if the speed of this motor is reduced
to 450 r.p.m. by inserting a resistance in the armature circuit, find the motor output, the armature
current, external resistance to be inserted in the armature circuit and overall efficiency. assume the
load torque to remain constant. (nagpur univ., november 1998)
solution. with an output of 7.48 kw and an efficiency of 88%, the input power is 8.50 kw.
losses are 1.02 kw.
input current = 85000/220 = 38.64 a
armature-current = 38.64 ? 2.00 = 36.64 a
power loss in the shunt field circuit = 220 × 2 = 440 w
copper-loss in armature-circuit = 36.642 × 0.08 = 107.4 w
no-load-loss at 900 r.p.m. = 1020 ? 107.4 ? 440 = 473 w
at 900 r.p.m. back-emf = eb1 = 200 ? (36.64 × 0.08) = 217.3 v
motor will run at 450 r.p.m. with flux per pole kept constant, provided the back-emf = eb2
= (459/900) × 217.1v = 108.5 v
there are two simplifying assumptions in this case, which must be stated before further calculations
:
1. load-torque is constant,
2. no load losses are constant.
(these statements can be different which leads to variations in the next steps of calculations.)
for constant load-torque, the condition of constant flux per pole results into constant armature
current, which is 36.64 a.
with an armature current of 36.64 a, let the external resistance required for this purpose be r.
36.64 r = 217.1 ? 108.5 = 108.6 v, r = 2.964 ohms
1038 electrical technology
total losses
total i2r - loss in armature = 36.642 × (2.964 + 0.08) = 4086 w
field-copper-loss + no load loss = 440 + 473 = 913
total loss = 4999 w
total output = 8500 ? 4999 = 3501 w
hence, efficiency = (3501/8500) × 100 = 41.2%
(note : because of missing data and clarification while making the statements in the question, there can be
variations in the assumption and hence in the final solutions.)
example 30.14. a d.c. shunt motor supplied at 230 v runs at 990 r.p.m. calculate the resistance
required in series with the armature circuit to reduce the speed to 500 r.p.m. assuming that armature
current is 25 amp. (nagpur univ., november 1997)
solution. it is assumed that armature resistance is to
(a) at 900 r.p.m. : eb1 = 230 = k × 900
(b) at 500 r.p.m. : eb2 = k × 500
therefore, eb2 = eb1 × 500/900 = 127.8 volts
the difference between eb1 and eb2 must be the droping in the external resistance to be added to the
armature circuit for the purpose of reducing the speed to 500 r.p.m.
eb1 ? eb2 = 25 × r
r = (230 ? 127.8)/25 = 4.088 ohms.
example 30.15. a 220 v d.c. shunt motor has an armature resistance of 0.4 ohm and a field
circuit resistance of 200 ohms. when the motor is driving a constant-torque load, the armaturecurrent
is 20 a, the speed being 600 r.p.m. it is desired to run the motor at 900 r.p.m. by inserting a
resistance in the field circuit. find its value, assuming that the magnetic circuit is not saturated.
(nagpur univ., november 1996)
solution. (i) at 600 r.p.m. if 1= 220/200 = 1.1 amp
eb1 = 220 ? (20 × 0.4) = 212 volts
tl = k1 × 1.1 × 20
the back e.m.f. = 212 = k2 × 1.1 × 600
or k2 = 212/660 = 0.3212
(ii) at 900 r.p.m. : tl = k1 × if 2 × ia2
due to constant load torque,
if 2 × ia2 = 1.1 × 20 = 22
eb2 = 220 ? (0.4 ia2) = k2 × if 2 × 900 = 289 if 2
220 ? (0.4 × 22/if 2) = 289 if 2
guess for approximate value of if 2 : neglecting armature -resistance droping and saturation, 50%
rise in speed is obtained with proportional decrease in if 2 related by
600
900
? 2
1
f
f
i
i
giving if 2 ? 0 .73 amp
[in place of if 2, ia2 can be evaluated first. its guess-work will give ia2 ? 1.5 × 20 ? 30 amp]
continuing with the solution of the equation to evaluate a value of if 2, accepting that value which
is near 0.73 amp, we have if 2 = 0.71865. [note that other value of if 2, which is 0.04235, is not
acceptable.]. corresponding ia2 = 30.6 amp. previous shunt field current, if 1 = 1.1, rf 1 = 200 ?.
new shunt field current, if 2 = 0.71865, rf 2 = 220/0.71865 = 306 ?. final answer is that a resistor of
106 ohms is to be added to the field circuit to run the motor at 900 r.p.m. at constant torque.
speed control of d.c. motors 1039
example 30.16. a 220 v d.c. shunt motor has an armature resistance of 0.40 ohm and fieldresistance
of 200 ohms. it takes an armature current of 22 a and runs at 600 r.p.m. it drives a load
whose torque is constant. suggest a suitable method to raise the speed to 900 r.p.m. calculate the
value of the controllable parameter. (nagpur univ., april 1998)
solution. at 600 r.p.m. ia1 = 22 amp, n1 = 600 r.p.m., if 1 = 220/200 = 1.1 amp.
eb1 = 220 ? (22 × 0.40) = 211.2 volts.
let the load-torque be denoted by t2, k1 and k2 in the equations below represent machine constants
appearing in the usual emf-equation and torque-equation for the d.c. shunt motor.
eb1 = 211.2 = k1 × if 1 × n1 = k1 × 1.10 × 600 or k1 = 211.2/660 = 0.32
tl = k2 × if 1 × ia1 = k2 × 1.10 × 22
since the load torque will remain constant at 900 r.p.m. also, the corresponding field current
(= if 2) and armature current (= ia2) must satisfy the following relationships :
tl = k2 if 2 ia2 = k2 × 1.10 × 22
or if 2 × ia2 = 24.2
and eb2 = 220 ? (ia2 × 0.40) = k1 × if 2 × 900
220 ? (ia2 × 0.40) = 0.32 × (24.2 × ia2) × 900
(alternatively, the above equation can also lead to a quadratic in if 2.)
this leads to a quadratic equation in ia2.
guess for if 2 : approximately, speed of a d.c. shunt motor is inversely proportional to the field
current. comparing the two speeds of 600 and 900 r.p.m., the value of if 2 should be approximately
given by
if 2 ? if 1 × (600/900) = 0.733 amp
guess for ia2 : for approximate conclusions, armature-resistance droping can be ignored. with
constant load-torque, armature-power must be proportional to the speed.
armature-power at 900 r.p.m.
armature-power at 600 r.p.m.
=
900
600
= 1.5
eb2 ia2 = 1.5 × eb1 ia1
neglecting armature-resistance dropings, eb1 ? v and eb2 ? v.
this gives ia2 = 1.5 × 22 = 33 amps
thus, out of the two roots for if 2, that which is close to 0.733 is acceptable. if quadratic equation
for ia2 is being handled, that root which is near 33 amp is acceptable.
continuing with the solution to quadratic equation for ia2, we have
220 ? 0.40 ia2 = 0.32 × (24.2 × ia2) × 900
220 ? 0.40 ia2 = 6969/ia2
i2
a2 ? 550 ia2 + 17425 = 0
this gives ia2 as either 33.75 amp or 791.25 amp.
from the reasoning given above, acceptable root corresponds to ia2 = 33.75 amp.
corresponding field current,if 2 = 24.2/33.75 = 0.717 amp
previous field circuit resistance = 200 ohms
new field circuit resistance = 220/0.717 = 307 ohms
hence, additional resistance of 107 ohms must be added to the shunt field circuit to run the motor
at 900 r.p.m. under the stated condition of constant load torque.
additional check : exact calculations for proportions of armature-power in two cases will give
the necessary check.
eb2 = 220 – (33.75 × 0.40) = 206.5
1040 electrical technology
as mentioned above, while guessing the value of ia2, the proportion of armature-power should
be 1.5.
2 2
1 1
b a
b a
e i
e i
=
206.56 33.75
211.2 22
×
× = 1.50
thus, the results obtained are confirmed.
example 30.17. a 250 v, 25 kw d.c. shunt motor has an efficiency of 85% when running at
1000 r.p.m. on full load. the armature resistance is 0.1 ohm and field resistance is 125 ohms. find
the starting resistance required to limit the starting current to 150% of the rated current.
(amravati univ., 1999)
solution.
output power = 25 kw, at full-load.
input power =
25,000
0.85
= 29412 watts
at full load, input current = 29412/250 = 117.65 amp
field current = 250/125 = 2 amp
f.l. armature current = 117.65 ? 2 = 115.65 amp
limit of starting current = 1.50 × 115.65 = 173.5 amp
total resistance in armature circuit at starting
= 250
173.5
= 1.441 ohms
external resistance to be added to armature circuit
= 1.441 ? 0.1 = 1.341 ohm.
tutorial problems 30.1
1. a d.c. shunt motor runs at 900 r.p.m. from a 460 v supply when taking an armature current of 25 a.
calculate the speed at which it will run from a 230-v supply when taking an armature current of 15
a. the resistance of the armature circuit is 0.8 ?. assume the flux per pole at 230 v to have
decreased to 75% of its value at 460 v. [595 r.p.m.]
2. a 250 v shunt motor has an armature resistance of 0.5 ? and runs at 1200 r.p.m. when the armature
current is 80 a. if the torque remains unchanged, find the speed and armature current when the field
is strengthened by 25%. [998 r.p.m. 64 a]
3. when on normal full-load, a 500 v, d.c. shunt motor runs at 800 r.p.m. and takes an armature current
42 a. the flux per pole is reduced to 75% of its normal value by suitably increasing the field circuit
resistance. calculate the speed of the motor if the total torque exerted on the armature is (a) unchanged
(b) reduced by 20%.
the armature resistance is 0.6 ? and the total voltage loss at the brushes is 2 v.
[(a) 1,042 r.p.m. (b) 1,061 r.p.m.]
4. the following data apply to d.c. shunt motor.
supply voltage = 460 v armature current = 28 a speed = 1000 r.p.m. armature resistance = 0.72
?. calculate (i) the armature current and (ii) the speed when the flux per pole is increased to 120%
of the initial value, given that the total torque developed by the armature is unchanged.
[(i) 23.33 a (ii) 840 r.p.m.]
5. a 100-v shunt motor, with a field resistance of 50 ? and armature resistance of 0.5 ? runs at a
speed of 1,000 r.p.m. and takes a current of 10 a from the supply. if the total resistance of the field
circuit is reduced to three quarters of its original value, find the new speed and the current taken from
the supply. assume that flux is directly proportional to field current. [1,089 r.p.m. 8.33 a]
speed control of d.c. motors 1041
6. a 250 v d.c. shunt motor has armature circuit resistance of 0.5 ? and a field circuit resistance of
125 ?. it drives a load at 1000 r.p.m. and takes 30 a. the field circuit resistance is then slowly
increased to 150 ?. if the flux and field current can be assumed to be proportional and if the load
torque remains constant, calculate the final speed and armature current. [1186 r.p.m. 33.6 a]
7. a 250 v, shunt motor with an armature resistance of 0.5 ? and a shunt field resistance of 250 ?
drives a load the torque of which remains constant. the motor draws from the supply a line current
of 21 a when the speed is 600 r.p.m. if the speed is to be raised to 800 r.p.m., what change must be
affected in the shunt field resistance ? assume that the magnetization curve of the motor is a straight
line. [88 ?]
8. a 240 v, d.c. shunt motor runs at 800 r.p.m. with no extra resistance in the field or armature circuit,
on no-load. determine the resistance to be placed in series with the field so that the motor may run
at 950 r.p.m. when taking an armature current of 20 a. field resistance = 160 ?. armature resistance
= 0.4 ?. it may be assumed that flux per pole is proportional to field current. [33.6 ?]
9. a shunt-wound motor has a field resistance of 400 ? and an armature resistance of 0.1 ? and runs
off 240 v supply. the armature current is 60 a and the motor speed is 900 r.p.m. assuming a
straight line magnetization curve, calculate (a) the additional resistance in the field to increase the
speed to 1000 r.p.m. for the same armature current and (b) the speed with the original field current of
200 a. [(a) 44.4 ? (b) 842.5 r.p.m.]
10. a 230 v d.c. shunt motor has an armature resistance of 0.5 ? and a field resistance of 76 2/3 ?. the
motor draws a line current of 13 a while running light at 1000 r.p.m. at a certain load, the field
circuit resistance is increased by 38 1/3 ?. what is the new speed of the motor if the line current at
this load is 42 a ? [1400 r.p.m.] (electrical engg. grad i.e.t.e. dec. 1986)
11. a 250 v d.c. shunt motor runs at 1000 r.p.m. and takes an armature current of 25 amp. its armature
resistance is 0.40 ohm. calculate the speed with increased load with the armature current of 50 amp.
assume that the increased load results into flux-weakening by 3%, with respect to the flux in previous
loading condition. (nagpur univ., april 1996)
hint : (i) first loading condition :
eb1 = 250 ? 25 × 0.40 = k1 × 1000
(ii) second loading condition :
eb2 = 250 ? 50 × 0.40 = 230 k1 × (0.97 ?) × n2. this gives n2. [988 r.p.m.]
(ii) armature or rheostatic control method
this method is used when speeds below the no-load speed are required. as the supply voltage
is normally constant, the voltage across the armature is varied by inserting a variable rheostat or
resistance (called controller resistance) in series with the armature circuit as shown in fig. 30.4 (a).
as controller resistance is increased, p.d. across the armature is decreased, thereby decreasing the
armature speed. for a load constant torque, speed is approximately proportional to the p.d. across the
armature. from the speed/armature current characteristic [fig. 30.4 (b)], it is seen that greater the
resistance in the armature circuit, greater is the fall in the speed.
v
field
ia
speed, n
controller
resistance
no resistance
in armature
rated load
resistance
in armature
armature current ia
( a )( b )
fig. 30.4
1042 electrical technology
let ia1 = armature current in the first case
ia2 = armature current in the second case
(if ia1 = ia2, then the load is of constant torque.)
n1, n2 = corresponding speeds, v = supply voltage
then n1 ? v ? ia1 ra ? eb1
let some controller resistance of value r be added to the armature circuit resistance so that its
value becomes (r + ra) = rt.
then n2 ? v ? ia2 rt ? eb2 ? 2 2
1 1
= b
b
n e
n e
(in fact, it is a simplified form of relation given in art. 27.9 because here ?1 = ?2.)
considering no-load speed, we have
0 0
?
=
?
a t
a a
n v i r
n v i r
neglecting ia0 ra with respect to v, we get
n = 0 1
? ? ? ? ?
? ?
a t i r
n
v
it is seen that for a given resistance rt the speed is a linear
function of armature current ia as shown in fig. 30.5 (a).
the load current for which the speed would be zero is found
by putting n = 0 in the above relation.
? 0 = 0 1
? ? ? ? ?
? ?
a t i r
n
v
or ia =
t
v
r
this is the maximum current and is known as stalling
current.
as will be shown in art. 30.5 (a), this method is very
wasteful, expensive and unsuitable for rapidly changing loads
because for a given value of rt, speed will change with load.
a more stable operation can be obtained by using a divertor
across the armature in addition to armature control resistance
(fig. 30.5 (b)). now, the changes in armature current (due to
changes in the load torque) will not be so effective in changing
the p.d. across the armature (and hence the armature speed).
example 30.18. a 200 v d.c. shunt motor running at 1000 r.p.m. takes an armature current of
17.5 a. it is required to reduce the speed to 600 r.p.m. what must be the value of resistance to be
inserted in the armature circuit if the original armature resistance is 0.4 ? ? take armature current
to be constant during this process. (elect. engg. i nagpur univ. 1993)
solution. n1 = 1000 r.p.m. eb1 = 200 ? 17.5 × 0.4 = 193 v
rt = total arm. circuit resistance n2 = 600 r.p.m. eb2 = (200 ? 17.5 rt)
since ish remains constant ?1 = ?2
? 600
1000
=
(200 17.5 )
193
? t r
rt = 4.8 ?
? additional resistance reqd. r = rt ? ra = 4.8 ? 0.4 = 4.4 ?.
it may be noted that brush voltage droping has not been considered.
example 30.19. a 500 v d.c. shunt motor has armature and field resistances of 1.2 ? and
500 ? respectively. when running on no-load, the current taken is 4 a and the speed is 1000 r.p.m.
calculate the speed when motor is fully loaded and the total current drawn from the supply is 26 a.
estimate the speed at this load if (a) a resistance of 2.3 ? is connected in series with the armature
and (b) the shunt field current is reduced by 15%. (electrical engg. i, sd. patel univ. 1985)
fig. 30.5 (a)
aia
series
field
divertor
fig. 30.5 (b)
field
series
resistance
divertor
speed control of d.c. motors 1043
solution. ish = 500/500 = 1 a ia1 = 4 ? 1 = 3 a
eb1 = 500 ? (3 × 1.2) = 496.4 v ia2 = 26 ? 1 = 25 a
eb2 = 500 ? (25 × 1.2) = 470 v ? 2
1000
n
=
470
496.4
n2 = 947 r.p.m.
(a) in this case, total armature circuit resistance = 1.2 + 2.3 = 3.5 ?
? eb2 = 500 ? (25 × 3.5) = 412.5 v ? 2
1000
n
=
412.5
496.4
n2 = 831 r.p.m.
(b) when shunt field is reduced by 15%, ?2 = 0.85 ?1 assuming straight magnetisation curve.
2
1000
n
= 412.5 1
496.4 0.85
× n2 = 977.6 r.p.m.
example 30.20. a 250-v shunt motor (fig. 30.6) has an armature current of 20 a when running
at 1000 r.p.m. against full load torque. the armature resistance is 0.5 ?. what resistance must be
inserted in series with the armature to reduce the speed to 500 r.p.m. at the same torque and what will
be the speed if the load torque is halved with this resistance in the circuit ? assume the flux to remain
constant throughout and neglect brush contact droping.
(elect. machines amie sec. b summer 1991)
solution. eb1 = v ? ia1 ra = 250 ? 20 × 0.5 = 240 v
let rt to be total resistance in the armature circuit i.e. rt = ra + r, where r is the additional
resistance.
? eb2 = v ? ia2 rt = 250 ? 20 rt
it should be noted that ia1 = ia2 = 20 a because torque
remains the same and ?1 = ?2 in both cases.
? 2
1
n
n
= 2 1 2
1 2 1
?
× =
?
b b
b b
e e
e e
or
500 250 20
1000 240
?
= t r
? rt = 6.5 ? hence, r = 6.5 ? 0.5 = 6 ?
since the load is halved, armature current is also halved
because flux remains constant. hence, ia3 = 10 a.
? 3
1000
n
=
250 10 6.5
240
? ×
or n3 = 771 r.p.m.
example 30.21. a 250-v shunt motor with armature resistance of 0.5 ohm runs at 600 r.p.m. on
full-load and takes an armature current of 20 a. if resistance of 1.0 ohm is placed in the armature
circuit, find the speed at (i) full-load torque (ii) half full-load torque.
(electrical machines-ii, punjab univ. may 1991)
solution. since flux remains constant, the speed formula becomes 2 2
1 1
= b
b
n e
n e
.
(i) in the first case, full-load torque is developed.
n1 = 600 r.p.m. eb1 = v ? ia1 ra1 = 250 ? 20 × 0.5 = 240 v
now, t ? ? ia ? ia (? ? is constant)
? 2
1
t
t
= 2
1
a
a
i
i
since t2 = t1 ia2 = ia1 = 20 a
eb2 = v ? ia2 ra2 = 250 ? 20 × 1.5 = 220 v,
n2 = 2
600
n
=
220
240
n2 =
600 220
240
×
= 550 r.p.m.
(ii) in this case, the torque developed is half the full-load torque.
fig. 30.6
0.5
20 a
250 a
1044 electrical technology
2
1
t
t
= 2
1
a
a
i
i
or 1
1
t / 2
t
= 2
20
a i
ia2 = 10 a eb2 = 250 ? 10 × 1.5 = 235 v
2
600
n
=
235
240
n2 = 600 × 235
240
= 587.5 r.p.m.
example 30.22. a 220 v shunt motor with an armature resistance of 0.5 ohm is excited to give
constant main field. at full load the motor runs of 500 rev. per minute and takes an armature
current of 30 a. if a resistance of 1.0 ohm is placed in the armature circuit, find the speed at (a)
full-load torque (b) double full-load torque. (elect. machines-i, nagpur univ. 1993)
solution. since flux remains constant, the speed formula becomes n2/n1 = eb2/eb1.
(a) full-load torque
with no additional resistance in the armature circuit,
n1 = 500 r.p.m. ia1 = 30 a eb1 = 220 ? 30 × 0.5 = 205 v
now, t ? ia (since ? is constant.) ? 2 2
1 1
= a
a
t i
t i
since t2 = t1 ia2 = ia1 = 30 a
when additional resistance of 1 ? is introduced in the armature circuit,
eb2 = 220 ? 30 (1 + 0.5) = 175 v n2 = ? 2
500
n
=
175
205
n2 = 427 r.p.m.
(b) double full-load torque
2
1
t
t
= 2
1
a
a
i
i
or 1
1
2t
t
= 2
30
a i
ia2 = 60 a
? eb2 = 220 ? 60 (1 + 0.5) = 130 v
? 2
500
n
=
130
205
n2 = 317 r.p.m.
example 30.23. the speed of a 50 h.p (37.3 kw) series motor working on 500 v supply is 750
r.p.m. at full-load and 90 per cent efficiency. if the load torque is made 350 n-m and a 5 ohm
resistance is connected in series with the machine, calculate the speed at which the machine will
run. assume the magnetic circuit to be unsaturated and the armature and field resistance to be 0.5
ohm. (electrical machinery i, madras univ. 1986)
solution. load torque in the first case is given by
t1 = 37,300/2? (750/60) = 474.6 n-m
input current, ia1 = 37,300/0.9 × 500 = 82.9 a
now, t2 = 250 n-m ia2 = ?
in a series motor, before magnetic saturation,
t ? ? ia ? ia2 ? t1 ? ia1
2 and t2 ? ia2
2
?
2
2
1
? ?
? ?
? ?
a
a
i
i
= 2
1
t
t
? ia2 = 82.9 × 250/ 474.6 = 60.2 a
now, eb1 = 500 ? (82.9 × 0.5) = 458.5 v
eb2 = 500 ? 60.2 (5 + 0.5) = 168.9 v
using 2
1
n
n
= 2 2
1 1
b × a
b a
e i
e i
, we get 2 168.9 82.9
750 458.5 60.2
= × n ? n2 = 381 r.p.m.
example 30.24. a 7.46 kw, 220 v, 900 r.p.m. shunt motor has a full-load efficiency of 88 per
cent, an armature resistance of 0.08 ? and shunt field current of 2 a. if the speed of this motor is
speed control of d.c. motors 1045
reduced to 450 r.p.m. by inserting a resistance in the armature circuit, the load torque remaining
constant, find the motor output, the armature current, the external resistance and the overall efficiency.
(elect. machines, nagpur univ. 1993)
solution. full-load motor input current i = 7460/220 × 0.88 = 38.5 a
? ia1 = 38.5 ? 2 = 36.5 a
now, t ? ? ia. since flux remains constant.
? t ? ia ? ta1 ? ia1 and ta2 ? ia2 or 2 2
1 1
a a
a a
t i
t i
=
it is given that ta1 = ta2 hence ia1 = ia2 = 36.5 a
eb1 = 220 ? (36.5 × 0.08) = 217.1 ?
eb2 = 220 ? 36.5 rt n1 = 900 r.p.m. n2 = 450 r.p.m.
now, 2
1
n
n
= 2 1 2
1 2 1
?
× =
?
b b
b b
e e
e e
(? ?1 = ?2)
? 450
900
=
200 36.5
217.1
? t r
rt = 3.05 ?
? external resistance r = 3.05 ? 0.08 = 2.97 ?
for calculating the motor output, it will be assumed that all losses except copper losses vary
directly with speed.
since motor speed is halved, stray losses are also halved in the second case. let us find their
value.
in the first case, motor input = 200 × 38.5 = 8,470 w motor output = 7,460 w
total cu losses + stray losses = 8470 ? 7460 = 1010 w
arm. cu loss = ia1
2 ra = 36.52 × 0.08 = 107 w field cu loss = 220 × 2 = 440 w
total cu loss = 107 + 440 = 547 w ? stray losses in first case = 1010 ? 547 = 463 w
stray losses in the second case = 463 × 450/900 = 231 w
field cu loss = 440 w, as before arm. cu loss = 36.52 × 3.05 = 4,064 w
total losses in the 2nd case = 231 + 440 + 4,064 = 4,735 w
input = 8,470 w – as before
output in the second case = 8,470 ? 4,735 = 3,735 w
? overall ? = 3,735/8,470 = 0.441 or 44.1 per cent*
example 30.25. a 240 v shunt motor has an armature current of 15 a when running at 800
r.p.m. against f.l. torque. the arm. resistance is 0.6 ohms. what resistance must be inserted in
series with the armature to reduce the speed to 400 r.p.m., at the same torque ?
what will be the speed if the load torque is halved with this resistance in the circuit ? assume the
flux to remain constant throughout. (elect. machines-i nagpur univ. 1993)
solution. here, n1 = 800 r.p.m., eb1 = 240 ? 15 × 0.6 = 231 v
flux remaining constant, t ? ia. since torque is the same in both cases, ia2 = ia1 = 15 a. let
r be the additional resistance inserted in series with the armature. eb2 = 240 ? 15 (r + 0.6) n2
= 400 r.p.m.
? 400
800
=
240 15 ( 0.6)
231
? r +
r = 7.7 ?
* it may be noted that efficiency is reduced almost in the ratio of the two speeds.
1046 electrical technology
when load torque is halved :
with constant flux when load torque is halved, ia is also halved. hence, ia3 = ia1/2 = 15/2 = 7.5 a
? eb3 = 240 ? 7.5 (7.7 + 0.6) = 177.75 v n3 = ?
3
1
n
n
= 3
1
b
b
e
e
or 3
800
n
=
177.75
231
n3 = 614.7 r.p.m.
example 30.26. (a) a 400 v shunt connected d.c. motor takes a total current of 3.5 a on no
load and 59.5 a at full load. the field circuit resistance is 267 ohms and the armature circuit
resistance is 0.2 ohms (excluding brushes where the droping may be taken as 2 v). if the armature
reaction effect at ‘full-load’ weakens the flux per pole by 2 percentage change in speed from no-load
to full-load.
(b) what resistant must be placed in series with the armature in the machine of (a) if the fullload
speed is to be reduced by 50 per cent with the gross torque remaining constant ? assume no
change in the flux. (electrical machines, amie sec. b, 1989)
solution. (a) shunt current ish = 400/267 = 1.5 a. at no load, ia1 = 3.5 ? 1.5 = 2 a, eb1
= v ? ia1 rai ? brush droping = 400 ? 2 × 0.2 ? 2 = 397.6 v. on full-load, ia2 = 59.5 ? 1.5 = 58 a, eb2 =
400 ? 58 × 0.2 ? 2 = 386.4 v.
? 1
2
b
b
e
e
=
1 1
2 2
?
?
n
n
or 1 1 1
1 2 2
397.6
1.0084
386.4 0.98
n n
n n
?
= =
?
% change in speed = 1 2
1
100
?
× n n
n
= ( 1 )
1 100
1.0084
? × = 0.833
(b) since torque remains the same, ia remains the same, hence ia3 = ia2. let r be the resistance
connected in series with the armature.
eb3 = v ? ia2 (ra + r) ? brush droping
= 400 ? 58 (0.2 + r) ? 2 = 386.4 ? 58 r
? 2
3
b
b
e
e
= 2 2 2
3 3 3
?
=
?
n n
n n
(q ?2 = ?3)
386.4
386.4 ? 58r
=
1
0.5
r = 3.338 ?
example 30.27. a d.c. shunt drives a centrifugal pump whose torque varies as the square of the
speed. the motor is fed from a 200 v supply and takes 50 a when running at 1000 r.p.m. what
resistance must be inserted in the armature circuit in order to reduce the speed to 800 r.p.m. ? the
armature and field resistance of the motor are 0.1 ? and 100 ? respectively.
(elect. machines, allahabad univ. 1992)
solution. in general, t ? ? ia
for shunt motors whose excitation is constant,
t ? ia ? n2, as given.
? ia ? n2. now ish = 200/100 = 2 a ? ia1 = 50 ? 2 = 48 a
let ia2 = new armature current at 800 r.p.m.
then 48 ? n1
2 ? 10002 and ia2 ? n2
2 ? 8002
? 2
48
a i
= ( )2
800
1000
= 0.82 ? ia2 = 48 × 0.64 = 30.72 a
speed control of d.c. motors 1047
? eb1 = 200 ? (48 × 0.1) = 195.2 v eb2 = (200 ? 30.72 rt) v
now, 2
1
n
n
= 2
1
b
b
e
e
? 800
1000
=
200 30.72
195.2
? t r
, rt = 1.42 ?
additional resistance = 1.42 ? 0.1 = 1.32 ?
example 30.28. a 250 v, 50 h.p. (373 kw) d.c. shunt motor has an efficiency of 90% when
running at 1,000 r.p.m. on full-load. the armature and field resistances are 0.1 ? and 115 ?
respectively. find
(a) the net and developed torque on full-load.
(b) the starting resistance to have the line start current equal to 1.5 times the full-load current.
(c) the torque developed at starting. (elect. machinery-i, kerala univ. 1987)
solution. (a) tsh = 9.55 × 37,300/1000 = 356.2 n-m
input current =
37,300
250 × 0.9 = 165.8 a ish =
250
125
= 2 a
? ia = 165.8 ? 2 = 163.8 a eb = 250 ? (163.8 × 0.1) = 233.6 v
? ta = 9.55
233.6 163.8
1000
×
= 365.4 n-m
(b) f.l. input line i = 165.8 a permissible input i = 165.8 × 1.5 = 248.7 a
permissible armature current = 248.7 ? 2 = 246.7 a
total armature resistance = 250/246.7 = 1.014 ?
?starting resistance required = 1.014 ? 0.1 = 0.914 ?
(c) torque developed with 1.5 times the f.l. current would be practically 1.5 times the f.l.
torque.
i.e. 1.5 × 365.4 = 548.1 n-m.
example 30.29. a 200 v shunt motor with a shunt resistance of 40 ? and armature resistance
of 0.02 ? takes a current of 55 a and runs at 595 r.p.m. when there is a resistance of 0.58 ? in series
with armature. torque remaining the same, what change should be made in the armature circuit
resistance to raise the speed to 630 r.p.m. ? also find
(i) at what speed will the motor run if the load torque is reduced such that armature current is
15 a.
(ii) now, suppose that a divertor of resistance 5 ? is connected across the armature and series
resistance is so adjusted that motor speed is again 595 r.p.m., when armature current is
50 a. what is the value of this series resistance ? also, find the speed when motor current
falls of 15 a again.
solution. the circuit is shown in fig. 30.7.
200 v
ai
ir
i
(200 -ir)
r
5w
0.58 w
220 w 50 w
0.02 w ia
5 a 55 a
40 w
shunt
fig. 30.7 fig. 30.8
1048 electrical technology
ish = 200/40 = 5 a ? ia1 = 55 ? 5 = 50 a
armature circuit resistance = 0.58 + 0.02 = 0.6 ?
? eb1 = 200 ? (50 × 0.6) = 170 v
since torque is the same in both cases, ia1 ?1 = ia2 ?2
moreover, ?1 = ?2 ? ia1 = ia2 ? ia2 = 50 a
now eb1 = 170 v, n1 = 595 r.p.m., n2 = 630 r.p.m., eb2 = ?
using 2
1
n
n
= 2
1
b
b
e
e
(? ?1 = ?2)
we get eb2 = 170 × (630/595) = 180 v
let r2 be the new value of armature circuit resistance for raising the speed from 595 r.p.m. to 630
r.p.m.
? 180 = 200 ? 50 r2 ? r = 0.4 ?
hence, armature circuit resistance should be reduced by 0.6 ? 0.4 = 0.2 ?.
(i) we have seen above that
ia1 = 50 a, eb1 = 170 v, n1 = 595 r.p.m.
if ia2 = 15 a, eb2 = 200 ? (15 × 0.6) = 191 v
? 2
595
n
=
191
170
? n2 = 668.5 r.p.m.
(ii) when armature divertor is used (fig. 30.8).
let r be the new value of series resistance
? eb3 = 200 ? ir ? (50 × 0.02) = 199 ? ir
since speed is 595 r.p.m., eb3 must be equal to 170 v
? 170 = 199 ? ir ? ir = 29 v p.d. across divertor = 200 ? 29 = 171 v
current through divertor id = 171/5 = 34.2 a ? i = 50 + 34.2 = 84.2 a
as ir = 29 v ? r = 29/84.2 = 0.344 w
when ia = 15 a, then id = (i ? 15) a
p.d. across divertor = 5 (i ? 15) = 200 ? 0.344 i ? i = 51.46 a
eb4 = 200 ? 0.344 i ? (15 × 0.02)
= 200 ? (0.344 × 51.46) ? 0.3 = 182 v
? 4
1
n
n
= 4
1
b
b
e
e
or 4 182
595 170
= n ? n4 = 637 r.p.m.
the effect of armature divertor is obvious. the speed without divertor is 668.5 r.p.m. and with
armature divertor, it is 637 r.p.m.
(iii) voltage control method
(a) multiple voltage control
in this method, the shunt field of the motor is connected permanently to a fixed exciting voltage,
but the armature is supplied with different voltages by connecting it across one of the several different
voltages by means of suitable switchgear. the armature speed will be approximately proportional to
these different voltages. the intermediate speeds can be obtained by adjusting the shunt field regulator.
the method is not much used, however.
(b) ward-leonard system
this system is used where an unusually wide (upto 10 : 1) and very sensitive speed control is
required as for colliery winders, electric excavators, elevators and the main drives in steel mills and
blooming and paper mills. the arrangement is illustrated in fig. 30.9.
m1 is the main motor whose speed control is required. the field of this motor is permanently
connected across the d.c. supply lines. by applying a variable voltage across its armature, any desired
speed control of d.c. motors 1049
speed can be obtained. this variable voltage is supplied by a motor-generator set which consists of
either a d.c. or an a.c. motor m2 directly coupled to generator g.
fig. 30.9
the motor m2 runs at an approximately constant speed. the output voltage of g is directly fed to
the main motor m1. the voltage of the generator can be varied from zero up to its maximum value by
means of its field regulator. by reversing the direction of the field current of g by means of the
reversing switch rs, generated voltage can be reversed and hence the direction of rotation of m1. it
should be remembered that motor generator set always runs in the same direction.
despite the fact that capital outlay involved in this system is high because (i) a large output
machine must be used for the motor generator set and (ii) that two extra machines are employed, still
it is used extensively for elevators, hoist control and for main drive in steel mills where motor of
ratings 750 kw to 3750 kw are required. the reason for this is that the almost unlimited speed
control in either direction of rotation can be achieved entirely by field control of the generator and the
resultant economies in steel production outwiegh the extra expenditure on the motor generator set.
a modification of the ward-leonard system is known as ward-leonard-ilgner system which
uses a smaller motor-generator set with the addition of a flywheel whose function is to reduce fluctuations
in the power demand from the supply circuit. when main motor m1 becomes suddenly overloaded,
the driving motor m2 of the motor generator set slows down, thus allowing the inertia of the
flywheel to supply a part of the overload. however, when the load is suddenly thrown off the main
motor m1, then m2 speeds up, thereby again storing energy in the flywheel.
when the ilgner system is driven by means of an a.c. motor (whether induction or synchronous)
another refinement in the form of a ‘slip regulator’ can be usefully employed, thus giving an additional
control.
the chief disadvantage of this system is its low overall efficiency especially at light loads. but as
said earlier, it has the outstanding merit of giving wide speed control from maximum in one direction
through zero to the maximum in the opposite direction and of giving a smooth acceleration.
example 30.30. the o.c.c. of the generator of a ward-leonard set is
field amps : 1.4 2.2 3 4 5 6 7 8
armature volts : 212 320 397 472 522 560 586 609
the generator is connected to a shunt motor, the field of which is separately-excited at 550 v. if
the speed of motor is 300 r.p.m. at 550 v, when giving 485 kw at 95.5% efficiency, determine the
excitation of the generator to give a speed of 180 r.p.m. at the same torque. resistance of the motor
motor generator variable speed motor
d.c. supply line
shunt
field
r.s.
field
regulator
shunt
field
1050 electrical technology
armature circuit = 0.01 ?, resistance of the motor field = 60 ?, resistance of generator armature
circuit = 0.01 ?. ignore the effect of armature reaction and variation of the core factor and the
windage losses of the motor.
solution. motor input = 485 × 103/0.955 = 509,300 w
motor to motor field = 550/60 = 55/6 a
input to motor field = 550 × 55/6 = 5,040 w
? motor armature input = 509,300 ? 5,040 = 504,260 w
? armature current = 504,260/550 = 917 a
back e.m.f. eb1 at 300 r.p.m. = 550 ? (917 × 0.01) = 540.83 v
back e.m.f. eb2 at 180 r.p.m. = 540.83 × 180/300 = 324.5 v
since torque is the same, the armature current of the main motor is also the same i.e. 917 a
because its excitation is independent of its speed.
? v = 324.5 + (917 × 0.01) = 333.67 v
generated e.m.f. = v + ia ra ...for generator
333.67 + (917 × 0.011) = 343.77 v.
if o.c.c. is plotted from the above given data, then it
would be found that the excitation required to give 343.77
v is 2.42 a.
? generator exciting current = 2.42 a
30.3. speed control of series motors
1. flux control method
variations in the flux of a series motor can be brought
about in any one of the following ways :
(a) field divertors
the series winding are shunted by a variable resistance
known as field divertor (fig. 30.10). any desired amount
of current can be passed through the divertor by adjusting
its resistance. hence the flux can be decreased and
consequently, the speed of the motor increased.
(b) armature divertor
a divertor across the armature can be used for giving speeds lower than the normal speed (fig.
30.11). for a given constant load torque, if ia is reduced due to armature divertor, the ? must
increase.
(? ta ? ? ia). this results in an increase in current taken from the supply (which increases the flux
and a fall in speed (n ? i/?)). the variation in speed can be controlled by varying the divertor
resistance.
fig. 30.10
aia
series
field
divertor
series
field
armature
divertor
series field
fig. 30.11 fig. 30.12