A 250 V shunt motor on no load runs at 1000 rpm and takes 5 A. The total
armature and shunt field resistances are respectively 0.2 ohm and 250ohm.
Determine the speed when loaded taking a current of 50A, if armature reaction
weakens the field by 3%.
994 rpm
1523 rpm
O 306 rpm
01254 rpm
Answers
Answer:
hope it help
Explanation:
For initial operating point: IL1 = 10 A, ra = 0.2 Ω and supply voltage V = 220 V.
Field current If1 = 220/220 A = 1A
Armature current Ia1 = 10A – 1A = 9A
Now we write down the expressions for the torque and back emf.
Te1 = kt If1 Ia1 = kt × 1 × 9 = TL
Eb1 = kg If1 n = kg × 1 × 1000 = V – Ia1 ra = 220 – 9 × 0.2 = 218.2V
kg × 1 × 1000 = 218.2V
Since field resistance remains unchanged If2 = If1 = 1 A. Let the new steady armature current be
Ia2 and the new speed be n2. In this new condition the torque and back emf equations are
Te2 = kt ×1 × Ia2 = TL
Eb2 = kg × 1 × n2
= V – Ia2(ra + Rext)
∴kg × 1 × n2 = 220 – Ia2 × 5.2 V
Taking the ratios of Te2 and Te1 we get,
2
1
e
e
T
T
= 2 1
1 9
t
t
k I
k
× × a
× ×
Thus, Ia2 = 9 A
Now taking the ratio emfs 2
1
b
b
E
E , we get,
2 1
1 1000
g
g
k n
k
× ×
× ×
= 2 220 5.2
218
a − I ×
n rps -
V (supply)
If
TL
Figure 41.1: D.C shunt motor.
M Te
V (supply)
ra
Ia IL +
-
If
Eb
+
-
Rf
Version 2 EE IIT, Kharagpur
2
1000
n
=
220 9 5.2
218.2
− ×
or, 2
1000
n
=
173.2
218.2
or, n2 =
173.2 1000
218.2
×
∴ n2 = 793.76 rpm
It may be noted that, for constant load torque the steady state armature current does not
change with change in the value of the armature resistance.
Let us consider the next problem whose data are similar to the first problem except the fact
that load torque is a function of speed.
Answer:
The field current is \frac{250}{250}=1
250
250
=1 A
Under no load condition the armature current is 5-1=45−1=4 A. Hence back emf is
250-4\times 0.2=249.2250−4×0.2=249.2 volts
Since the armature current is 4 A which is very small as compared to 50 A as far as armature reaction effect in concerned, hence we neglect the effect.
The back emf when 50 A is drawn by the motor
250-(50-1)\times 0.2=240.2250−(50−1)×0.2=240.2 volts
Now 249.2=K\Phi \times 1000249.2=KΦ×1000
240.2=0.97 KN
Hence \frac{0.97\, KN\Phi }{K\Phi \times 1000}=\frac{240.2}{249.2}
KΦ×1000
0.97KNΦ
=
249.2
240.2
or N=\frac{240.2}{249.2}\times \frac{1000}{0.97}\simeq 993.7N=
249.2
240.2
×
0.97
1000
≃993.7
\simeq 994≃994 rpm.