A DC shunt motor draws a full load armature current 22,5 A from a 200volt supply. The armature and shunt field resistances are 0,2 ohms and 120ohms, respectively. Determine the efficiency of the motor if the constant losses are assumed to be 320watts

Answer 1

First, calculate the total input power to the motor:
Total input power = Voltage * Current
Total input power = 200V * 22.5A
Total input power = 4500W

Next, calculate the total armature copper losses:
Armature copper losses = (Armature current)^2 * Armature resistance
Armature copper losses = (22.5)^2 * 0.2
Armature copper losses = 101.25W

Next, calculate the total shunt field copper losses:
Shunt field copper losses = (Armature current)^2 * Shunt field resistance
Shunt field copper losses = (22.5)^2 * 120
Shunt field copper losses = 30375W

Then, calculate the total copper losses:
Total copper losses = Armature copper losses + Shunt field copper losses
Total copper losses = 101.25W + 30375W
Total copper losses = 30476.25W

Given that constant losses are 320W, we can calculate the mechanical output power:
Mechanical output power = Total input power - (Total copper losses + Constant losses)
Mechanical output power = 4500W - (30476.25W + 320W)
Mechanical output power = 4500W - 30796.25W
Mechanical output power = 703.75W

Finally, calculate the efficiency of the motor:
Efficiency = (Mechanical output power / Total input power) * 100
Efficiency = (703.75W / 4500W) * 100
Efficiency = 15.64%

Therefore, the efficiency of the motor is 15.64%.