To solve the problem of finding the speed of a three-phase, 4-pole, 50 Hz induction motor when delivering maximum torque, and determining the resistance to be added to achieve 3/4th of the maximum torque at startup, follow these steps:
### 1. Speed of the Motor when Delivering Maximum Torque
**a. Find the Synchronous Speed:**
The synchronous speed \( N_s \) of the induction motor can be calculated using the formula:
\[ N_s = \frac{120 \times f}{P} \]
where:
- \( f \) is the frequency (50 Hz),
- \( P \) is the number of poles (4).
Substitute the given values:
\[ N_s = \frac{120 \times 50}{4} = 1500 \text{ RPM} \]
**b. Calculate the Slip at Maximum Torque:**
The slip \( s \) at which maximum torque occurs is given by:
\[ s_{max} = \frac{R_2}{\sqrt{R_2^2 + X_2^2}} \]
where:
- \( R_2 \) is the rotor resistance per phase,
- \( X_2 \) is the rotor reactance per phase.
From the problem, the rotor impedance is given as \( Z_2 = 0.03 + j0.15 \) ohms.
Thus:
- \( R_2 = 0.03 \) ohms,
- \( X_2 = 0.15 \) ohms.
So:
\[ s_{max} = \frac{0.03}{\sqrt{0.03^2 + 0.15^2}} \]
\[ s_{max} = \frac{0.03}{\sqrt{0.0009 + 0.0225}} \]
\[ s_{max} = \frac{0.03}{\sqrt{0.0234}} \]
\[ s_{max} = \frac{0.03}{0.153} \approx 0.196 \]
**c. Calculate the Rotor Speed:**
The actual rotor speed \( N_r \) is given by:
\[ N_r = N_s \times (1 - s) \]
Substitute \( s = 0.196 \):
\[ N_r = 1500 \times (1 - 0.196) \]
\[ N_r = 1500 \times 0.804 \]
\[ N_r \approx 1206 \text{ RPM} \]
### 2. Resistance to be Added for 3/4th of Maximum Torque at Startup
**a. Maximum Torque and Starting Torque Relationship:**
The torque \( T \) in an induction motor is proportional to the square of the slip \( s \) and inversely proportional to the rotor resistance \( R_2 \) at startup.
To achieve \( \frac{3}{4} \) of the maximum torque, we need to adjust the resistance such that the new torque is:
\[ T_{new} = \frac{3}{4} \times T_{max} \]
The resistance needed \( R_{2,new} \) for achieving this torque can be calculated by noting that the torque is directly proportional to \( \frac{R_2^2}{(R_2^2 + X_2^2)} \).
At startup (slip \( s = 1 \)), the torque is:
\[ T_{startup} \propto \frac{R_{2,new}^2}{R_{2,new}^2 + X_2^2} \]
Setting this to \( \frac{3}{4} \) of the maximum torque, we get:
\[ \frac{T_{new}}{T_{max}} = \frac{R_{2,new}^2}{R_{2,new}^2 + X_2^2} \times \frac{R_2^2 + X_2^2}{R_2^2} \]
\[ \frac{3}{4} = \frac{R_{2,new}^2}{R_{2,new}^2 + X_2^2} \]
Rearrange to solve for \( R_{2,new} \):
\[ \frac{3}{4} \times (R_{2,new}^2 + X_2^2) = R_{2,new}^2 \]
\[ \frac{3}{4} \times X_2^2 = \frac{1}{4} \times R_{2,new}^2 \]
\[ R_{2,new}^2 = 3 \times X_2^2 \]
\[ R_{2,new} = \sqrt{3} \times X_2 \]
Substitute \( X_2 = 0.15 \):
\[ R_{2,new} = \sqrt{3} \times 0.15 \approx 0.259 \text{ ohms} \]
**b. Resistance to be Added:**
The additional resistance \( R_{added} \) required is:
\[ R_{added} = R_{2,new} - R_2 \]
\[ R_{added} = 0.259 - 0.03 \]
\[ R_{added} \approx 0.229 \text{ ohms} \]
### Summary
1. **Speed of the Motor when Delivering Maximum Torque:** Approximately 1206 RPM.
2. **Resistance to be Added to Achieve 3/4th of Maximum Torque at Startup:** Approximately 0.229 ohms.