Question

A 3 φ I.M. has synchronous speed of 250rpm and 4% slip at full load. The rotor has a resistance of 0.02Ω/ph and stand still reactance of 0.15Ω/ph. Calculate i) The speed at which max torque is developed. ii) The ratio of maximum to F. L. torque. iii) The ratio of maximum to starting torque.

Answer 1

To solve the problem, we need to use some formulas related to induction motors. Here’s a step-by-step breakdown for each part of the question.

### Given Data:
- Synchronous Speed \( N_s = 250 \, \text{rpm} \)
- Slip at Full Load \( s_{FL} = 4\% = 0.04 \)
- Rotor Resistance \( R_r = 0.02 \, \Omega/\text{ph} \)
- Standstill Reactance \( X_r = 0.15 \, \Omega/\text{ph} \)

### i) Speed at which Maximum Torque is Developed

The speed at which maximum torque is developed occurs when the slip \( s \) is given by the formula:

\[
s_{max \, torque} = \frac{R_r}{\sqrt{R_r^2 + X_r^2}}
\]

1. **Calculate \( \sqrt{R_r^2 + X_r^2} \)**:

   \[
   \sqrt{R_r^2 + X_r^2} = \sqrt{(0.02)^2 + (0.15)^2}
   \]

2. **Find \( s_{max \, torque} \)**:

   \[
   s_{max \, torque} = \frac{0.02}{\sqrt{(0.02)^2 + (0.15)^2}}
   \]

3. **Calculate the speed**:

   The actual speed can be calculated using:

   \[
   N = N_s \times (1 - s_{max \, torque})
   \]

### ii) Ratio of Maximum to Full Load Torque

The ratio of maximum torque (\( T_{max} \)) to full load torque (\( T_{FL} \)) can be calculated using the formula:

\[
\frac{T_{max}}{T_{FL}} = \frac{s_{FL}}{s_{max \, torque}}
\]

### iii) Ratio of Maximum to Starting Torque

The ratio of maximum torque (\( T_{max} \)) to starting torque (\( T_{st} \)) can be calculated as:

\[
\frac{T_{max}}{T_{st}} = \frac{s_{st}}{s_{max \, torque}}
\]

Where \( s_{st} \) is the slip at starting:

\[
s_{st} = 1 \quad (\text{at starting})
\]

### Let's perform the calculations step by step.

Here are the calculated results:

### i) Speed at which Maximum Torque is Developed
The speed at which maximum torque is developed is approximately:
\[
N_{max \, torque} \approx 217.0 \, \text{rpm}
\]

### ii) Ratio of Maximum to Full Load Torque
The ratio of maximum torque to full load torque is:
\[
\frac{T_{max}}{T_{FL}} \approx 0.303
\]

### iii) Ratio of Maximum to Starting Torque
The ratio of maximum torque to starting torque is:
\[
\frac{T_{max}}{T_{st}} \approx 7.566
\]

These values provide insights into the performance characteristics of the induction motor under the given conditions. If you have any further questions or need additional calculations, feel free to ask!

Answer 2

To solve the given problem, we'll start by understanding the parameters and relationships involved.

### Given Data:
- Synchronous Speed, \( N_s \) = 250 rpm
- Slip at Full Load, \( s_{FL} \) = 4% = 0.04
- Rotor Resistance, \( R_2 \) = 0.02 Ω/phase
- Standstill Reactance, \( X_2 \) = 0.15 Ω/phase

### Formulae Needed:
1. **Rotor Speed \( N_r \) at Full Load:**
   \[
   N_r = N_s \times (1 - s_{FL})
   \]

2. **Speed at which Maximum Torque is Developed:**
   The slip at which maximum torque occurs (\( s_{max} \)) is given by:
   \[
   s_{max} = \frac{R_2}{X_2}
   \]
   The corresponding speed \( N_{max} \) is:
   \[
   N_{max} = N_s \times (1 - s_{max})
   \]

3. **Ratio of Maximum Torque to Full Load Torque:**
   \[
   \frac{T_{max}}{T_{FL}} = \frac{s_{FL}}{s_{max}} \times \frac{2}{1 + \sqrt{1 + \left(\frac{X_2}{R_2}\right)^2}}
   \]

4. **Ratio of Maximum Torque to Starting Torque:**
   \[
   \frac{T_{max}}{T_{start}} = \frac{1}{s_{max}} \times \frac{2}{1 + \sqrt{1 + \left(\frac{X_2}{R_2}\right)^2}}
   \]

### Step-by-Step Calculation:

#### 1) **Speed at which Maximum Torque is Developed:**
- First, calculate \( s_{max} \):
  \[
  s_{max} = \frac{R_2}{X_2} = \frac{0.02}{0.15} = 0.1333
  \]
- Now, calculate \( N_{max} \):
  \[
  N_{max} = N_s \times (1 - s_{max}) = 250 \times (1 - 0.1333) = 250 \times 0.8667 = 216.67 \text{ rpm}
  \]

#### 2) **Ratio of Maximum Torque to Full Load Torque:**
- First, calculate the ratio \( \frac{T_{max}}{T_{FL}} \):
  \[
  \frac{T_{max}}{T_{FL}} = \frac{0.04}{0.1333} \times \frac{2}{1 + \sqrt{1 + \left(\frac{X_2}{R_2}\right)^2}}
  \]
- Calculate \( \frac{X_2}{R_2} \):
  \[
  \frac{X_2}{R_2} = \frac{0.15}{0.02} = 7.5
  \]
- Then:
  \[
  \frac{T_{max}}{T_{FL}} = \frac{0.04}{0.1333} \times \frac{2}{1 + \sqrt{1 + (7.5)^2}}
  \]
  \[
  \frac{T_{max}}{T_{FL}} = 0.3 \times \frac{2}{1 + \sqrt{1 + 56.25}} = 0.3 \times \frac{2}{1 + \sqrt{57.25}}
  \]
  \[
  \sqrt{57.25} \approx 7.569
  \]
  \[
  \frac{T_{max}}{T_{FL}} = 0.3 \times \frac{2}{1 + 7.569} = 0.3 \times \frac{2}{8.569} = 0.3 \times 0.2334 = 0.070
  \]
  \[
  \frac{T_{max}}{T_{FL}} \approx 0.07
  \]

#### 3) **Ratio of Maximum Torque to Starting Torque:**
\[
\frac{T_{max}}{T_{start}} = \frac{1}{0.1333} \times \frac{2}{1 + \sqrt{1 + \left(\frac{X_2}{R_2}\right)^2}}
\]
We already calculated the part \( \frac{2}{1 + \sqrt{1 + (7.5)^2}} \):
\[
\frac{T_{max}}{T_{start}} = \frac{1}{0.1333} \times 0.2334 \approx 7.5 \times 0.2334 = 1.750
\]

### Final Answers:
1. **Speed at which Maximum Torque is Developed:** \( N_{max} \) = **216.67 rpm**
2. **Ratio of Maximum Torque to Full Load Torque:** \( \frac{T_{max}}{T_{FL}} \approx 0.07 \)
3. **Ratio of Maximum Torque to Starting Torque:** \( \frac{T_{max}}{T_{start}} \approx 1.75 \)