Question 4.7: A four-pole 220 V dc series motor has 240 slots in the armat...

In a series motor the armature winding and the field winding are connected in series across the supply voltage. Thus, the line current, field current, and the armature current are the same,

i.e.,

I_{a} = I_{f} = I_{L} = 80 A

The total member of armature conductors, Z = 240 × 6

= 1440

Armature is wave connected, and hence A = 2
No. of poles = 4

= 220 − 80 (0.1 + 0.05)

= 208 V

Again,                                                  E = \frac{ΦZNP}{60  A}

substituting values                           208 = \frac{1.75  ×  10^{-2}  ×  1440  ×  N  ×  4}{60  ×  2}

or,                                                         N = \frac{208  ×  60   ×  2  ×  10^{2}}{1.75  ×  1440  ×  4} = 248 rpm

Power developed by the armature = E × I_{a}

Power output = Power developed – Iron and Frictional losses

= 19.84 − 0.44
= 19.4 kW

If we want to convert in horse power, we use the relation 1 kW = 0.735 hp.
Thus, power output = 19.4 × 0.735 = 14.26 hp.