A 500V dc shunt motor takes a total current of 125A and runs at 1400 rev/min. If theshunt field resistance is 75 Ω, the armature resistance is 0.32 Ω and the iron, friction andwindage losses amount to 1.9 kW, determine the overall efficiency of the motor

To determine the overall efficiency of the motor, we need to calculate the input power, output power, and then use the formula for efficiency.

Step 1: Calculate the Input Power The input power (Pin) to the motor is given by the product of the voltage (V) and the total current (I). Pin = V * I Pin = 500V * 125A = 62500W

Step 2: Calculate the Shunt Field Power The shunt field power (Psh) is given by the square of the shunt field current (Ish) times the shunt field resistance (Rsh). Ish = V / Rsh Ish = 500V / 75Ω = 6.67A Psh = Ish^2 * Rsh Psh = (6.67A)^2 * 75Ω = 3333.5W

Step 3: Calculate the Armature Power The armature power (Pa) is given by the product of the armature current (Ia) and the armature resistance (Ra). The armature current is the difference between the total current and the shunt field current. Ia = I - Ish Ia = 125A - 6.67A = 118.33A Pa = Ia^2 * Ra Pa = (118.33A)^2 * 0.32Ω = 4487.9W

Step 4: Calculate the Mechanical Power Developed The mechanical power developed (Pm) is the difference between the input power and the sum of the shunt field power and the armature power. Pm = Pin - Psh - Pa Pm = 62500W - 3333.5W - 4487.9W = 54678.6W

Step 5: Calculate the Output Power The output power (Pout) is the difference between the mechanical power developed and the iron, friction and windage losses (Pifw). Pifw = 1.9kW = 1900W Pout = Pm - Pifw Pout = 54678.6W - 1900W = 52778.6W

Step 6: Calculate the Efficiency The efficiency (η) of the motor is given by the ratio of the output power to the input power, multiplied by 100 to get a percentage. η = (Pout / Pin) * 100 η = (52778.6W / 62500W) * 100 = 84.44%

So, the overall efficiency of the motor is 84.44%.