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To solve this example for a three-phase induction motor, we need to calculate the **speed**, **output torque and power**, **stator current**, **power factor**, and **efficiency** when the motor is operated at rated voltage and frequency, given a slip of 1.6%. 

The provided motor data are as follows:

- \( R_1 = 0.271 \, \Omega \)
- \( R_2 = 0.188 \, \Omega \)
- \( X_1 = 1.12 \, \Omega \)
- \( X_2 = 1.91 \, \Omega \)
- \( X_m = 23.10 \, \Omega \)
- \( P_{\text{core losses}} = 320 \, \text{W} \)
- Motor power: 20 kW (rated power)
- Rated voltage: 460 V (line-to-line)
- Frequency: 60 Hz
- Slip: 1.6% (\(s = 0.016\))
- Number of poles: 6

### 1. **Speed Calculation**:

The synchronous speed \( N_s \) of the motor is given by the formula:

\[
N_s = \frac{120 \times f}{P}
\]

where:
- \( f = 60 \, \text{Hz} \) (frequency)
- \( P = 6 \) (number of poles)

Substituting the values:

\[
N_s = \frac{120 \times 60}{6} = 1200 \, \text{rpm}
\]

The rotor speed \( N_r \) is related to the slip by:

\[
N_r = N_s (1 - s)
\]

where \( s = 0.016 \) is the slip. Thus:

\[
N_r = 1200 (1 - 0.016) = 1200 \times 0.984 = 1180.8 \, \text{rpm}
\]

### 2. **Output Power and Torque**:

The output power is given by:

\[
P_{\text{out}} = P_{\text{input}} - P_{\text{losses}} = 20,000 \, \text{W} - 320 \, \text{W} = 19,680 \, \text{W}
\]

To find the torque, use the formula:

\[
T = \frac{P_{\text{out}}}{\omega_r}
\]

where \( \omega_r \) is the rotor angular velocity in radians per second:

\[
\omega_r = \frac{2\pi N_r}{60} = \frac{2\pi \times 1180.8}{60} \approx 123.58 \, \text{rad/s}
\]

Thus, the torque is:

\[
T = \frac{19,680}{123.58} \approx 159.3 \, \text{Nm}
\]

### 3. **Stator Current**:

We can calculate the stator current using the power and voltage. For a three-phase system:

\[
P_{\text{input}} = \sqrt{3} V_{LL} I_1 \cos \phi
\]

However, we need additional information about the power factor (\( \cos \phi \)) and the impedance to find the exact stator current.

### 4. **Power Factor**:

The power factor is related to the impedance of the motor. Without more information on the phase angle, we can estimate using:

\[
\cos \phi = \frac{R_{\text{total}}}{Z_{\text{total}}}
\]

### 5. **Efficiency**:

The efficiency of the motor is given by:

\[
\eta = \frac{P_{\text{out}}}{P_{\text{input}}}
\]

With the input power at 20 kW and output power at 19.68 kW:

\[
\eta = \frac{19,680}{20,000} = 0.984 \, \text{or} \, 98.4\%
\]

### Summary:
- **Speed**: \( N_r = 1180.8 \, \text{rpm} \)
- **Torque**: \( T = 159.3 \, \text{Nm} \)
- **Efficiency**: \( 98.4\% \)
- **Power Factor** and **Stator Current** would need further impedance calculations.

Would you like further details on calculating the stator current or power factor?

Here are 5 related questions you might be interested in:
1. How can the equivalent circuit parameters of an induction motor be used to compute current?
2. What factors affect the slip of an induction motor?
3. How does the rotor resistance influence motor performance?
4. What is the significance of the core loss in motor efficiency?
5. How do changes in load affect the slip and speed of an induction motor?

**Tip**: Always ensure to compute synchronous speed correctly, as it's crucial for determining motor slip and performance.

For a slip of 1.6 percent, compute the speed, output torque and power, stator current, power factor, and efficiency when the motor is operated at rated voltage and frequency.

Learn how to compute the speed, torque, power, stator current, power factor, and efficiency of a 20 kW, six-pole, Y-connected induction motor at rated voltage and frequency with a slip of 1.6%. This example covers essential concepts such as induction motor theory, slip speed relation, and power equations.

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