15.MET October 2018 Q.6(B)

A d.c. motor takes an armature current of 110 A at 480 V. The resistance of the armature circuit is 0.2𝛀. The machine has six poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate - (i) The speed; (ii) The gross torque developed by the armature.

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Given

Ia = 110 V
V = 480 V
Ra = 0.2 ohm
No of poles , P = 6
No of conductors Z = 864
Lap connected, No of parallel path  A = P = 6
Flux per pole ø = 0.05 Wb

To find

i) Speed, N
ii) Gross torque , T

Solution

Terminal EMF , E = V - (Ia x Ra)
                              = 480 - (110 x 0.2)
                              = 458 V

i)                        E = ( ø x Z x N / 60) X (P / A)
                     
                       458 = (0.05 x 864 x N /60 ) x (6 / 6)

                        N   = 636 rpm

ii)    Gross torque  = 0.159 x ø x Z x Ia x (P / A) 

                          T  =  0.159 x 0.05 x 864 x 110 x (6 / 6)   

                         T   =  756.3 Nm