Question

A four-pole induction motor, 460 V, 25 hp, 60 Hz, connected in Y, has the following impedances in ohm per phase referring to the stator circuit:

R1=0.641W, R2=0.332W, X1=1.106W, X2=0.464W, Xm=26.3W

The total turnover losses are 1,100 W and are supposed to be constant. The losses in the core are grouped with the losses by rotation. For a rotor slide of 2.2% at nominal voltage and frequency, find the following engine quantities:

A) Speed,
b) Stator current,
c) Power factor,
d) Pd and Psal.
e) td and tload,
f) Efficiency

Answer 1

The values obtained are: speed of 1800 RPM, stator current of 29.4 Amps, power factor of 0.502, Pd of 277.5 Watts and Psal of 116537 Watts, td of 4833.2625 Nm and tload of 0.025 Nm, and an efficiency of 99.9995%.

Step-by-step explanation:

To find the requested engine quantities, we need to calculate the values of each quantity based on the given information. Let's start with the calculations:

1. Speed (N): The speed of an induction motor can be calculated using the formula N = 120 * (f / p), where N is the speed in RPM, f is the frequency in Hz, and p is the number of poles. Substituting the given values, we have N = 120 * (60 / 4) = 1800 RPM.
2. Stator current (Is): The stator current can be calculated using the formula Is = (P / (sqrt(3) * V)), where Is is the stator current in Amps, P is the power in Watts, and V is the line voltage. Substituting the given values, we have Is = (25000 / (sqrt(3) * 460)) = 29.4 Amps.
3. Power factor (PF): The power factor can be calculated using the formula PF = P / (sqrt(3) * Is * V), where PF is the power factor, P is the power in Watts, Is is the stator current in Amps, and V is the line voltage. Substituting the given values, we have PF = (25000 / (sqrt(3) * 29.4 * 460)) = 0.502.
4. Pd and Psal: Pd represents the power dissipated in the rotor and Psal represents the power supplied to the stator. These values can be calculated using the formula Pd = (Is^2 * R2) and Psal = (3 * V * Is * PF), where R2 is the rotor resistance. Substituting the given values, we have Pd = (29.4^2 * 0.332) = 277.5 Watts and Psal = (3 * 460 * 29.4 * 0.502) = 116537 Watts.
5. td and tload: td represents the torque developed by the motor and tload represents the external load torque. These values can be calculated using the formula td = (750 * (Psal / N)) and tload = (Pd / (2 * pi * N)). Substituting the given values, we have td = (750 * (116537 / 1800)) = 4833.2625 Nm and tload = (277.5 / (2 * pi * 1800)) = 0.025 Nm.
6. Efficiency: The efficiency of the motor can be calculated using the formula Efficiency = ((td - tload) / td) * 100%. Substituting the calculated values, we have Efficiency = ((4833.2625 - 0.025) / 4833.2625) * 100% = 99.9995%.

So, the values of the requested engine quantities are as follows:
a) Speed is 1800 RPM.
b) Stator current is 29.4 Amps.
c) Power factor is 0.502.
d) Pd is 277.5 Watts and Psal is 116537 Watts.
e) td is 4833.2625 Nm and tload is 0.025 Nm.
f) Efficiency is 99.9995%.