38.MET March 2019 Q.6(B)

A 440V shunt motor takes an armature current of 30 A at 700 rev/min. The armature resistance is 0.7 ohm. If the flux is suddenly reduced 20 per cent, to what value will the armature current rise momentarily? Assuming unchanged resisting torque to motion, what will be the new steady values of speed and armature current? Sketch graphs showing armature current and speed as functions of time during the transition from initial to final, steady state conditions.

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Given

V = 400 V
Ia = 30 A
N1 = 700 
Ra = 0.7 Ω
Flux reduced to 20 %. Therefore Φ2 = 0.8 Φ1

To find

1.Momentarily increase in armature current
2.Speed & armature current
3.Graph

Solution

As we know that the voltage equation of motor is 
V = Eb + Ia Ra
440 = Eb + (30 x 0.7)
 Eb = 440 - 21
      = 419 V

We know that Eb ∝ Φ N

Consider emf Eb1 and flux Φ1 as initial values and emf Eb2 and flux Φ2 as after 20% flux reduction.
Taking rpm as constant.
Therefore Eb ∝ Φ 

Eb1 / Eb2 = Φ1 / Φ2
419 / Eb2 = Φ1 / 0.8 Φ1
         Eb2 = 335.2 V

1. Again V = Eb2 + Ia2 Ra
           440 = 335.2 + Ia2 x 0.7
   0.7 x Ia2 = 104.8
   
    Momentarily increase in armature current Ia2 = 149.71 A 

2. As we know that torque T ∝ Φ Ia
   
    It is given that after flux reduction torque remains constant.
    
    Therefore T1 = T2
              Φ1 Ia1 =  Φ2 Ia2
            Φ1 x 30 = 0.8  Φ1 x Ia2
                    Ia2 = 30 / 0.8

    Armature current Ia2 = 37.5 A

    V = Ebnew + Ia2new  Ra
    440 = Ebnew + (37.5 x 0.7)
    Ebnew = 413.75 V

    We know that Eb ∝ Φ N

     Eb1 / Ebnew = ( Φ1 / Φ2 ) x ( N1 / N2 )
     419 / 413.75 = (Φ1 / 0.8 Φ1) x (700 / N2)
           1.012 N2 = 1.25 x 700
                     N2 = 875 / 1.012

          Speed N2 = 864.6 rpm