A 440-V shunt motor has armature resistance of 0.8 Ω and field resistant 200 Ω. What will be the back e.m.f when giving an output of 7.46 kW at 85 percent efficiency.
A 440-V shunt motor has armature resistance of 0.8 Ω and field resistant 200 Ω. What will be the back e.m.f when giving an output of 7.46 kW at 85 percent efficiency.
Right Answer is:
425.8 V
SOLUTION
Given data
Supply Voltage V = 440 V
Armature Resistance Ra = 0.8Ω
Field Resistance Rsh = 200Ω
Motor Output ηout = 7.46 kW = 7460 Watt
Motor Efficiency η = 85% = 0.85
The efficiency of the motor is given as
η = Motor Output power ⁄ Motor Input power
Motor Input Power = Motor Output Power ⁄ Efficiency
= 7460 ⁄ 0.85 = 8776.4 Watt
Now the Motor Line current or the Input current IL = P ⁄ V
IL = 8776.4 ⁄ 440 = 19.95 A
The Line current of DC shunt motor is the sum of Armature current and Shunt field current
IL = Ia + Ish
And Shunt Field Resistance is given as
Ish = V/Rsh = 440/200 = 2.2 A
∴ Ia = IL − Ish
Ia = 19.95 − 2.2 = 17.75 A
The Back EMF of DC Motor is given by
Eb = V − IaRa
= 440 − 17.75 × 0.8
= 440 − 14.2
= 425.8 V