A dc shunt motor runs at 1200 rpm on no-load drawing 5 A from 220 V mains. Its armature and field resistances are 0.25 \Omega and 110 \Omega respectively. When loaded (at motor shaft), the motor draws 62 A from the mains. What would be its speed? Assume that the armature reaction demagnetizes the field to the extent of 5%.
Also calculate the internal torque developed at no-load and on load. What is the motor shaft torque at load (this torque drives the mechanical load).
In a dc shunt motor
I_{L} = I_{a} + I_{f}Shunt field current, I_{f} = \frac{220}{110} = 2 A (constant)
At no load
I_{a0} = 5 – 2 = 3 A
E_{a0}= 220 – 0.25 \times 3 = 219.25 V
n_{0} = 1200 rpm
E_{a0} = \acute{K_{a}} \Phi n_{0}or 219.25 = \acute{K_{a}} \Phi \times 1200
T_{0} \omega _{0}=E_{a0}I_{a0}\left(\frac{2\pi \times 1200}{60} \right) T_{0} = 219.25 \times 3
or T_{0} = 5.23 Nm
This torque is absorbed in iron loss, and windage and friction losses of the motor (the shaft being at no load).
On load
I_{a1} = 62 – 2 = 60 A
E_{a1} = 220 – 0.25 \times 60 = 205 Vflux/pole \Phi _{1} = 0.95 \Phi
E_{a1} = \acute{K_{a}} \Phi_{1} n_{1}
205 = \acute{K_{a}} \times 0.95 \Phi n_{1}
Dividing Eq. (ii) by Eq. (i)
\frac{0.95 n_{1}}{1200}=\frac{205}{219.25}
or n_{1} = 1181 rpm
Drop in motor speed is only 1.6% on being loaded. This because the reduction in flux/pole due to armature reaction counters the drop in speed caused by armature resistance drop.
T_{1} \omega_{1} = E_{a1} I_{a1}
or T_{1} = 99.45 Nm
T_{1} (shaft) = 99.45 – 5.23
= 94.22 Nm