A 20 kW, 250 V dc shunt motor has a full-load armature current of 85 A at 1100 rpm. The armature resistance is 0.18 \Omega .
Determine:
(a) the internal electromagnetic torque developed;
(b) the internal torque if the field current is suddenly reduced to 80% of its original value;
(c) The steady motor speed in part
(b) assuming the load torque to have remained constant.
Assume: magnetic circuit to be linear.
(a) E_{a} = 250 – 0.18 \times 85 = 234.7 V
E_{a}I_{a} = T \omega
234.7 \times 85 = T \times \frac{2\pi \times 1100}{60}
or T = 173.2 Nm
(b) Magnetic circuit linearity is assumed i.e. \Phi \propto I_{f} . Field suddenly reduced to 0.8 of original value. I_{a} is assumed to remain constant for that instant. Then
T_{1} = 0.8 \times 173.2 = 138.6 Nm(c) Under steady condition for the motor internal torque to build to the original value, new values of armature current and speed are established. These are obtained below:
T = \acute{K_{a}} K_{f} I_{f} \times 85 = \acute{K_{a}} K_{f} \times 0.8 I_{f} I_{a1} = 173.2 Nmor I_{a1} = 106.25 A
E_{a1} = 250 – 0.18 \times 106.25 = 230.9 V
and 234.7 = \acute{K_{a}} K_{f} I_{f} \times 1100
230.9 = \acute{K_{a}} \acute{K_{f}} 0.8 I_{f} n_{1}
Dividing we get \frac{230.9}{234.7} =\frac{0.8 n_{1}}{1100}
or n_{1} = 1353 rpm
Observe that both motor speed and armature current increase by reducing field current with constant load torque.