Question

A dc shunt motor is required to drive a constant power load at rated speed while drawing rated armature current. Neglecting saturation and all machine losses, if both the terminal voltage and the field current of the machine are halved then:
1. The speed becomes 2 pu but armature current remains at 1 pu
2. The speed remains at 1 pu but armature current becomes 2 pu
3. Both speed and armature current become 2 pu
4. Both speed and armature current remain at 1 pu

Answer 1

Correct Answer - Option 2 : The speed remains at 1 pu but armature current becomes 2 pu

Concept:

DC shunt motor

For constant power load P = constant

V₁ I_a1 = V₂ I_a2 and T₁ N₁ = T₂ N₂

V = terminal voltage

T = torque

N = speed of motor

I_a = armature current

I_f = field current

Calculation:

Given that

The terminal voltage and the field current of the machine are halved

V₂ = V₁ / 2  and I_f2 = I_f1 / 2

1) Armature current

V₁ I_a1 = V₂ I_a2

\({{\rm{I}}_{{\rm{a}}2}} = \frac{{{{\rm{V}}_1}{{\rm{I}}_{{\rm{a}}1}}}}{{{{\rm{V}}_2}}} = \frac{{{{\rm{V}}_1}{{\rm{I}}_{{\rm{a}}1}}}}{{{{\rm{V}}_1}/2}}\)

I_a2 = 2I_a1 = 2 pu

2) Speed of motor

T₁ N₁ = T₂ N₂  {T ∝ ϕ I_a}

\({{\rm{N}}_2} = \frac{{{{\rm{T}}_1}{{\rm{N}}_1}}}{{{{\rm{T}}_2}}} = \frac{{{{\rm{I}}_{{\rm{f}}1}}{{\rm{I}}_{{\rm{a}}1}}{{\rm{N}}_1}}}{{{{\rm{I}}_{{\rm{f}}2}}{{\rm{I}}_{{\rm{a}}2}}}}\)

\({{\rm{N}}_2} = \frac{{{{\rm{I}}_{{\rm{f}}1}}{{\rm{I}}_{{\rm{a}}1}}{{\rm{N}}_1}}}{{\left( {\frac{{{{\rm{I}}_{{\rm{f}}1}}}}{2}} \right)\left( {2{{\rm{I}}_{{\rm{a}}1}}} \right)}} = {{\rm{N}}_1}\)

Speed N₂ = 1 pu