Homework 8

Not graded. Solutions. Concepts covered in this homework will be included in the final.

Note: all voltages are line-to-line unless specified otherwise.

Problem 1

The synchronous speed of a \(60 \ \text{Hz}\), three-phase induction motor is \(1200 \ \text{rpm}\).

a) How many poles does the motor have?

b) What the synchronous speed of the motor in a \(50 Hz \ \text{Hz}\) system?

Solutions.

a) Using the synchronous speed formula:

\[n_s = \frac{120f}{P}\]

Solving for \(P\):

\[P = \frac{120f}{n_s} = \frac{120 \times 60}{1200} = \frac{7200}{1200} = 6 \ \text{poles}\]

b) Repeating the calculation with \(f = 50 \ \text{Hz}\) and \(P = 6\):

\[n_s = \frac{120f}{P} = \frac{120 \times 50}{6} = \frac{6000}{6} = 1000 \ \text{rpm}\]

Problem 2

The output power of a four-pole, \(60 \text{ Hz}\), three-phase induction motor is \(100 \text{ hp}\) (\(1 \text{ hp}\) is about \(0.746 \text{ kW}\)). The motor operates at a slip of \(0.02\) and efficiency of \(90\%\). Compute:

a) Input power of the motor.

b) The output torque of the motor.

Solutions.

Solution

a) \(P_{in} = \frac{P_{out}}{\eta} = \frac{74.6 \ \text{kW}}{0.90} = 82.89 \ \text{kW}\)

b) The synchronous speed:

\[n_s = \frac{120f}{P} = \frac{120 \times 60}{4} = 1800 \ \text{rpm}\]

The rotor speed:

\[n_r = n_s(1 - s) = 1800(1 - 0.02) = 1764 \ \text{rpm}\]

Convert to rad/s:

\[\omega_r = \frac{2\pi n_r}{60} = \frac{2\pi \times 1764}{60} = 184.73 \ \text{rad/s}\]

Output torque:

\[T = \frac{P_{out}}{\omega_r} = 403.8 \ \text{N·m}\]

Problem 3

Given a \(60 \text{ Hz}\), \(12\) pole, \(3\)-phase Induction Motor with the following per phase equivalent circuit parameters (ignore the shunt elements):

\[R_1 = 0.8 \ \Omega \qquad R'_2 = 1.2 \ \Omega \qquad X_1 = 0.4 \ \Omega \qquad X'_2 = 0.3 \ \Omega \qquad V_1 = 240 \text{ V}\]

The slip is \(s = 0.06\). Find the following:

a) The speed of the motor.

b) The equivalent impedance of the circuit as seen from the source.

c) The stator current and the rotor current referred to the stator side \((I'_2)\).

d) The developed power and torque.

e) The copper loss of the motor.

Solutions.

a) The synchronous speed:

\[n_s = \frac{120f}{P} = \frac{120 \times 60}{12} = 600 \ \text{rpm}\]

The rotor speed:

\[n_r = n_s(1 - s) = 600(1 - 0.06) = 564 \ \text{rpm}\]

b)

The total series impedance per phase is:

\[Z_{eq} = \left(R_1 + \frac{R'_2}{s}\right) + j(X_1 + X'_2)=20.8 + j0.7 \ \Omega\]

c) Since shunt elements are ignored, \(I_1 = I'_2\):

\[I_1 = I'_2 = \frac{V_1}{Z_{eq}} = 11.53 \angle{-1.93°} \ \text{A}\]

d) The air-gap power (three-phase):

\[P_g = 3 \vert I'_2 \vert^2 \cdot \frac{R'_2}{s} = 7.98 \ \text{kW}\]

The developed mechanical power:

\[P_{dev} = 7.50 \ \text{kW}\]

The synchronous angular speed:

\[\omega_s = \frac{2\pi n_s}{60} = 62.83 \ \text{rad/s}\]

The developed torque:

\[T_d = \frac{P_g}{\omega_s} = 126.98 \ \text{N·m}\]

e) Stator copper loss:

\[P_{cu1} = 3 \vert I_1 \vert^2 R_1 = 319.2 \ \text{W}\]

Rotor copper loss:

\[P_{cu2} = s \cdot P_g = 478.7 \ \text{W}\]

Total copper loss:

\[P_{cu} = P_{cu1} + P_{cu2} = 797.9 \ \text{W}\]