6.MET August 2018 Q.6(B)

A 220 V, d.c. shunt motor has an armature resistance of 0.5 ohm and an armature current of 40 A on full load. Determine the reduction in flux necessary for a 50 per cent reduction in speed. The torque for both conditions can be assumed to remain constant.

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Given

V = 220V
Ra = O.5 ohm
Ia1 = 40 A

To find

Reduction in flux

Solution

Back emf Eb1 = V - (Ia1 Ra)
                        = 220 - (40 x 0.5)
                        = 200 V

Back emf Eb2 = 220 - (Ia2 x 0.5)

Eb2 / Eb1 = (220 - (0.5 Ia2)) / 200 --------------------------------------------------- (1)


Speed reduced by 50% , So N2 = 0.5 N1

Eb2 /Eb1 = (k x ø2 x N2) / (k x ø1 x N1)

                = (k x ø2 x 0.5 N1) / (k x ø1 x N1)

                = 0.5 ø2 / ø1 -------------------------------------------------(2)

For motor Torque is directly proportional to flux and armature current
As given in the problem torque at both condition is constant

T1 = T2
Ia1 ø1   = Ia2 ø2

ø1 / ø2 = Ia2 / Ia1 --------------------------------------------------(3)

Comparing (1) and (2)

0.5 ø2 / ø1 =  (220 - (0.5 Ia2)) / 200

Substituting (3) in above equation

0.5 Ia1 / Ia2 = (220 - (0.5 Ia2)) / 200

200 x 0.5 x 40 = 220 Ia2 - 0.5 Ia2

Ia22 - 440 Ia2 + 8000 = 0

Ia2 = 19 A or 421 A

ø2 / ø1  =     Ia1 /  Ia2

 ø2           = (Ia1 /  Ia2) x ø1

                = (40 / 421) x ø1
                
                = 9.5 % of ø1