Question

The power input to a six pole, 3 phase, 50 Hz induction motor is 42 kW, the speed being 970 rpm. The stator losses are 1.2 kW and friction and windage losses are 1.8 kW. Find i) slip, ii) Rotor Output iii) Rotor copper loss and iv) Efficiency.

Answer 1

To solve this problem involving a three-phase induction motor, we can follow these steps:

### Given Data:
- Power input, \( P_{in} = 42 \, \text{kW} \)
- Stator losses, \( P_{stator} = 1.2 \, \text{kW} \)
- Friction and windage losses, \( P_{fw} = 1.8 \, \text{kW} \)
- Total losses, \( P_{loss} = P_{stator} + P_{fw} = 1.2 + 1.8 = 3.0 \, \text{kW} \)
- Speed, \( N = 970 \, \text{rpm} \)
- Number of poles, \( P = 6 \)
- Frequency, \( f = 50 \, \text{Hz} \)

### Step 1: Calculate Synchronous Speed
The synchronous speed \( N_s \) of a motor can be calculated using the formula:

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

Substituting the values:

\[
N_s = \frac{120 \cdot 50}{6} = 1000 \, \text{rpm}
\]

### Step 2: Calculate Slip
Slip \( s \) is calculated using the formula:

\[
s = \frac{N_s - N}{N_s}
\]

Substituting the values:

\[
s = \frac{1000 - 970}{1000} = \frac{30}{1000} = 0.03 \, \text{or} \, 3\%
\]

### Step 3: Calculate Rotor Output
The rotor output power \( P_{rotor} \) can be calculated as:

\[
P_{rotor} = P_{in} - P_{loss}
\]

Substituting the values:

\[
P_{rotor} = 42 \, \text{kW} - 3.0 \, \text{kW} = 39.0 \, \text{kW}
\]

### Step 4: Calculate Rotor Copper Loss
The rotor copper loss can be found using the slip and rotor output power. The rotor copper loss \( P_{copper} \) is given by:

\[
P_{copper} = s \cdot P_{rotor}
\]

Substituting the values:

\[
P_{copper} = 0.03 \cdot 39.0 \, \text{kW} = 1.17 \, \text{kW}
\]

### Step 5: Calculate Efficiency
Efficiency \( \eta \) can be calculated using the formula:

\[
\eta = \frac{P_{rotor}}{P_{in}} \times 100
\]

Substituting the values:

\[
\eta = \frac{39.0}{42.0} \times 100 = 92.86\%
\]

### Summary of Results:
i) Slip: \( 3\% \)

ii) Rotor Output: \( 39.0 \, \text{kW} \)

iii) Rotor Copper Loss: \( 1.17 \, \text{kW} \)

iv) Efficiency: \( 92.86\% \)

These calculations provide a comprehensive understanding of the motor's performance based on the given data.

Answer 2

To analyze the given three-phase induction motor, we need to determine several parameters including slip, rotor output, rotor copper loss, and efficiency. Let's break down each part of the problem.

### Given Data

1. **Power Input (P_in)**: 42 kW
2. **Speed (N)**: 970 RPM
3. **Stator Losses (P_st)**: 1.2 kW
4. **Friction and Windage Losses (P_fw)**: 1.8 kW
5. **Frequency (f)**: 50 Hz
6. **Number of Poles (p)**: 6

### i) Slip (s)

The slip \( s \) of the motor can be calculated using the formula:

\[ s = \frac{N_s - N_r}{N_s} \]

where:
- \( N_s \) is the synchronous speed.
- \( N_r \) is the rotor speed (which is given as 970 RPM).

First, calculate the synchronous speed \( N_s \). The synchronous speed can be calculated using the formula:

\[ N_s = \frac{120 \cdot f}{p} \]

where:
- \( f \) is the frequency (50 Hz)
- \( p \) is the number of poles (6)

Substitute the values:

\[ N_s = \frac{120 \cdot 50}{6} = 1000 \text{ RPM} \]

Now, calculate the slip \( s \):

\[ s = \frac{N_s - N_r}{N_s} = \frac{1000 - 970}{1000} = \frac{30}{1000} = 0.03 \]

So, the slip \( s \) is 0.03 or 3%.

### ii) Rotor Output (P_rotor_output)

To find the rotor output, we need to first find the total losses and subtract them from the input power:

1. **Total Losses**:

   \[
   \text{Total Losses} = \text{Stator Losses} + \text{Friction and Windage Losses} + \text{Rotor Copper Losses}
   \]

2. **Rotor Copper Losses**:

   The power input to the motor is used to overcome stator losses, friction and windage losses, rotor copper losses, and provide mechanical power. So:

   \[
   P_{\text{rotor output}} = P_{\text{input}} - \text{Stator Losses} - \text{Friction and Windage Losses} - \text{Rotor Copper Losses}
   \]

   The total power lost in the rotor (rotor copper losses) is given by:

   \[
   P_{\text{rotor copper losses}} = s \times P_{\text{rotor output}}
   \]

   Substitute this into the equation:

   \[
   P_{\text{rotor output}} = 42 \text{ kW} - 1.2 \text{ kW} - 1.8 \text{ kW} - (s \times P_{\text{rotor output}})
   \]

   \[
   P_{\text{rotor output}} = 42 - 1.2 - 1.8 - (0.03 \times P_{\text{rotor output}})
   \]

   \[
   P_{\text{rotor output}} = 39 - 0.03 \times P_{\text{rotor output}}
   \]

   Rearranging and solving for \( P_{\text{rotor output}} \):

   \[
   P_{\text{rotor output}} + 0.03 \times P_{\text{rotor output}} = 39
   \]

   \[
   1.03 \times P_{\text{rotor output}} = 39
   \]

   \[
   P_{\text{rotor output}} = \frac{39}{1.03} \approx 37.86 \text{ kW}
   \]

### iii) Rotor Copper Losses

Using the slip and rotor output, calculate the rotor copper losses:

\[
P_{\text{rotor copper losses}} = s \times P_{\text{rotor output}} = 0.03 \times 37.86 \approx 1.14 \text{ kW}
\]

### iv) Efficiency (η)

The efficiency can be calculated using:

\[
\eta = \frac{\text{Rotor Output}}{\text{Power Input}} \times 100\%
\]

\[
\eta = \frac{37.86}{42} \times 100\% \approx 90.1\%
\]

### Summary

1. **Slip**: 3%
2. **Rotor Output**: 37.86 kW
3. **Rotor Copper Losses**: 1.14 kW
4. **Efficiency**: 90.1%