To solve this problem, let's go step by step. We'll start by determining the required quantities: the ratio of maximum torque to full load torque and the speed at which the maximum torque occurs.
### 1. Given Data:
- Number of poles, \( P = 8 \)
- Supply frequency, \( f = 50 \) Hz
- Full load slip, \( s_{\text{FL}} = 2\% = 0.02 \)
- Rotor resistance per phase, \( R_2 = 0.001 \, \Omega \)
- Standstill rotor reactance per phase, \( X_2 = 0.005 \, \Omega \)
### 2. Synchronous Speed Calculation:
The synchronous speed \( N_s \) in RPM (revolutions per minute) for a 3-phase induction motor is given by:
\[
N_s = \frac{120 \times f}{P}
\]
Substituting the given values:
\[
N_s = \frac{120 \times 50}{8} = 750 \, \text{RPM}
\]
### 3. Full Load Speed:
The actual speed at full load \( N_{\text{FL}} \) is given by:
\[
N_{\text{FL}} = N_s \times (1 - s_{\text{FL}})
\]
Substituting the values:
\[
N_{\text{FL}} = 750 \times (1 - 0.02) = 750 \times 0.98 = 735 \, \text{RPM}
\]
### 4. Slip at Maximum Torque:
The slip at maximum torque \( s_{\text{max}} \) for an induction motor is given by:
\[
s_{\text{max}} = \frac{R_2}{X_2}
\]
Substituting the given values:
\[
s_{\text{max}} = \frac{0.001}{0.005} = 0.2
\]
### 5. Speed at Maximum Torque:
The speed at which maximum torque occurs \( N_{\text{max}} \) is given by:
\[
N_{\text{max}} = N_s \times (1 - s_{\text{max}})
\]
Substituting the values:
\[
N_{\text{max}} = 750 \times (1 - 0.2) = 750 \times 0.8 = 600 \, \text{RPM}
\]
### 6. Ratio of Maximum Torque to Full Load Torque:
The ratio of maximum torque \( T_{\text{max}} \) to full load torque \( T_{\text{FL}} \) is given by:
\[
\frac{T_{\text{max}}}{T_{\text{FL}}} = \frac{s_{\text{max}}}{s_{\text{FL}}}
\]
Substituting the values:
\[
\frac{T_{\text{max}}}{T_{\text{FL}}} = \frac{0.2}{0.02} = 10
\]
### Final Answers:
- **The ratio of maximum torque to full load torque** is \( 10 \).
- **The speed at which the maximum torque occurs** is \( 600 \, \text{RPM} \).