A three phase, 4 pole, 50Hz, induction motor has rotor impedance of (0.03 + j 0.15) ohm per phase. Calculate speed of motor when delivering maximum torque. Calculate the resistance to be added to achieve 3/4th of maximum torque at the time of starting.
2 Answers
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.
### 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.
To solve the problem, let's first calculate the speed of the motor when delivering maximum torque and then determine the resistance that needs to be added to achieve 3/4th of the maximum torque at the time of starting.
### 1. Calculating the Speed of the Motor when Delivering Maximum Torque
The maximum torque for an induction motor occurs when the slip \( s \) is such that the rotor resistance equals the rotor reactance. The slip at which maximum torque occurs is given by:
\[
s_{\text{max}} = \frac{R_2}{X_2}
\]
where:
- \( R_2 \) is the rotor resistance per phase (in ohms),
- \( X_2 \) is the rotor reactance per phase (in ohms).
Given:
- \( R_2 = 0.03 \, \Omega \)
- \( X_2 = 0.15 \, \Omega \)
\[
s_{\text{max}} = \frac{0.03}{0.15} = 0.2
\]
Now, the synchronous speed \( N_s \) of the motor is given by:
\[
N_s = \frac{120 \times f}{P}
\]
where:
- \( f = 50 \, \text{Hz} \) (frequency),
- \( P = 4 \) (number of poles).
\[
N_s = \frac{120 \times 50}{4} = 1500 \, \text{RPM}
\]
The actual speed of the motor when delivering maximum torque \( N_{\text{max}} \) can be found using the formula:
\[
N_{\text{max}} = N_s (1 - s_{\text{max}})
\]
Substituting the values:
\[
N_{\text{max}} = 1500 \times (1 - 0.2) = 1500 \times 0.8 = 1200 \, \text{RPM}
\]
So, the speed of the motor when delivering maximum torque is **1200 RPM**.
### 2. Calculating the Resistance to be Added for 3/4th of Maximum Torque at Starting
To achieve 3/4th of the maximum torque, the slip \( s_1 \) required is given by:
\[
s_1 = \frac{s_{\text{max}}}{T_r}
\]
where \( T_r \) is the ratio of the torque (in this case \( T_r = \frac{3}{4} \)):
\[
s_1 = \frac{s_{\text{max}}}{\frac{3}{4}} = \frac{0.2}{\frac{3}{4}} = \frac{0.2 \times 4}{3} = \frac{0.8}{3} = 0.267
\]
The corresponding rotor resistance required \( R_{2,\text{new}} \) to achieve this torque can be determined by:
\[
R_{2,\text{new}} = s_1 \times X_2 = 0.267 \times 0.15 = 0.04 \, \Omega
\]
The resistance to be added \( R_{\text{added}} \) is:
\[
R_{\text{added}} = R_{2,\text{new}} - R_2
\]
\[
R_{\text{added}} = 0.04 - 0.03 = 0.01 \, \Omega
\]
So, the resistance that needs to be added to achieve 3/4th of the maximum torque at the time of starting is **0.01 Ω**.
### 1. Calculating the Speed of the Motor when Delivering Maximum Torque
The maximum torque for an induction motor occurs when the slip \( s \) is such that the rotor resistance equals the rotor reactance. The slip at which maximum torque occurs is given by:
\[
s_{\text{max}} = \frac{R_2}{X_2}
\]
where:
- \( R_2 \) is the rotor resistance per phase (in ohms),
- \( X_2 \) is the rotor reactance per phase (in ohms).
Given:
- \( R_2 = 0.03 \, \Omega \)
- \( X_2 = 0.15 \, \Omega \)
\[
s_{\text{max}} = \frac{0.03}{0.15} = 0.2
\]
Now, the synchronous speed \( N_s \) of the motor is given by:
\[
N_s = \frac{120 \times f}{P}
\]
where:
- \( f = 50 \, \text{Hz} \) (frequency),
- \( P = 4 \) (number of poles).
\[
N_s = \frac{120 \times 50}{4} = 1500 \, \text{RPM}
\]
The actual speed of the motor when delivering maximum torque \( N_{\text{max}} \) can be found using the formula:
\[
N_{\text{max}} = N_s (1 - s_{\text{max}})
\]
Substituting the values:
\[
N_{\text{max}} = 1500 \times (1 - 0.2) = 1500 \times 0.8 = 1200 \, \text{RPM}
\]
So, the speed of the motor when delivering maximum torque is **1200 RPM**.
### 2. Calculating the Resistance to be Added for 3/4th of Maximum Torque at Starting
To achieve 3/4th of the maximum torque, the slip \( s_1 \) required is given by:
\[
s_1 = \frac{s_{\text{max}}}{T_r}
\]
where \( T_r \) is the ratio of the torque (in this case \( T_r = \frac{3}{4} \)):
\[
s_1 = \frac{s_{\text{max}}}{\frac{3}{4}} = \frac{0.2}{\frac{3}{4}} = \frac{0.2 \times 4}{3} = \frac{0.8}{3} = 0.267
\]
The corresponding rotor resistance required \( R_{2,\text{new}} \) to achieve this torque can be determined by:
\[
R_{2,\text{new}} = s_1 \times X_2 = 0.267 \times 0.15 = 0.04 \, \Omega
\]
The resistance to be added \( R_{\text{added}} \) is:
\[
R_{\text{added}} = R_{2,\text{new}} - R_2
\]
\[
R_{\text{added}} = 0.04 - 0.03 = 0.01 \, \Omega
\]
So, the resistance that needs to be added to achieve 3/4th of the maximum torque at the time of starting is **0.01 Ω**.