Question

The power input to a 500V 50Hz 6Pole 3 phase induction motor running at 975 rpm is 40 kW. The stator losses are 1kW and friction and windage losses are 2kW. Calculate: (i) Slip (ii) Rotor copper loss (iii) Shaft power (iv) Efficiency

Answer 1

To solve this problem, we can break it down step by step.

### Given Data:
- **Voltage (V)**: 500 V
- **Frequency (f)**: 50 Hz
- **Poles (P)**: 6
- **Speed (N)**: 975 RPM
- **Input Power (P_in)**: 40 kW
- **Stator Losses (P_stator)**: 1 kW
- **Friction and Windage Losses (P_friction)**: 2 kW

### Step 1: Calculate the Synchronous Speed (N_s)
The synchronous speed of the 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 (S)
Slip (S) is given by the formula:

\[
S = \frac{N_s - N}{N_s} \cdot 100
\]

Substituting the values:

\[
S = \frac{1000 - 975}{1000} \cdot 100 = 2.5\%
\]

### Step 3: Calculate Rotor Copper Loss (P_rotor)
The rotor copper loss can be calculated as:

\[
P_{rotor} = P_{in} - P_{stator} - P_{friction}
\]

Substituting the values:

\[
P_{rotor} = 40 \text{ kW} - 1 \text{ kW} - 2 \text{ kW} = 37 \text{ kW}
\]

### Step 4: Calculate Shaft Power (P_shaft)
Shaft power is given by:

\[
P_{shaft} = P_{in} - P_{stator} - P_{friction} - P_{rotor}
\]

However, we already know that the rotor copper loss is included in the input power calculation, so we can consider:

\[
P_{shaft} = P_{in} - P_{stator} - P_{friction}
\]

Substituting the values:

\[
P_{shaft} = 40 \text{ kW} - 1 \text{ kW} - 2 \text{ kW} = 37 \text{ kW}
\]

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

\[
\eta = \frac{P_{shaft}}{P_{in}} \cdot 100
\]

Substituting the values:

\[
\eta = \frac{37 \text{ kW}}{40 \text{ kW}} \cdot 100 = 92.5\%
\]

### Summary of Results:
1. **Slip (S)**: \(2.5\%\)
2. **Rotor Copper Loss (P_rotor)**: \(37 \text{ kW}\)
3. **Shaft Power (P_shaft)**: \(37 \text{ kW}\)
4. **Efficiency (η)**: \(92.5\%\)

Answer 2

Let's break down the calculations step by step:

### (i) Slip Calculation

1. **Synchronous Speed** (\(N_s\)) of the motor:
   \[
   N_s = \frac{120 \times f}{P}
   \]
   where \( f = 50 \text{ Hz} \) and \( P = 6 \text{ poles} \).
   \[
   N_s = \frac{120 \times 50}{6} = 1000 \text{ RPM}
   \]

2. **Slip** (\(s\)) is given by:
   \[
   s = \frac{N_s - N_r}{N_s}
   \]
   where \( N_r = 975 \text{ RPM} \).
   \[
   s = \frac{1000 - 975}{1000} = \frac{25}{1000} = 0.025 \text{ or } 2.5\%
   \]

### (ii) Rotor Copper Loss

1. **Total Power Input** (\(P_{in}\)) is given as 40 kW.
2. **Stator Losses** and **Friction & Windage Losses** are provided, so:
   \[
   P_{out} = P_{in} - \text{Stator Losses} - \text{Friction and Windage Losses} - \text{Rotor Copper Loss}
   \]

   Rearranging to find Rotor Copper Loss:
   \[
   P_{out} = P_{in} - (\text{Stator Losses} + \text{Friction and Windage Losses} + \text{Rotor Copper Loss})
   \]

   Given that:
   \[
   \text{Stator Losses} = 1 \text{ kW}
   \]
   \[
   \text{Friction and Windage Losses} = 2 \text{ kW}
   \]
   Therefore:
   \[
   P_{out} = 40 \text{ kW} - 1 \text{ kW} - 2 \text{ kW} - \text{Rotor Copper Loss}
   \]
   \[
   \text{Rotor Copper Loss} = P_{out} - (40 - 1 - 2) = P_{out} - 37
   \]

   To calculate \(P_{out}\) (Shaft Power):
   \[
   P_{out} = \text{Power Input} - (\text{Stator Losses} + \text{Friction and Windage Losses})
   \]
   \[
   P_{out} = 40 \text{ kW} - 1 \text{ kW} - 2 \text{ kW} = 37 \text{ kW}
   \]

   Rotor Copper Loss:
   \[
   \text{Rotor Copper Loss} = P_{out} - 37 \text{ kW} = 0
   \]

### (iii) Shaft Power

Shaft Power is the output power:
\[
P_{out} = 37 \text{ kW}
\]

### (iv) Efficiency

1. **Efficiency** (\(\eta\)) is calculated as:
   \[
   \eta = \frac{\text{Shaft Power}}{\text{Power Input}} \times 100
   \]
   \[
   \eta = \frac{37 \text{ kW}}{40 \text{ kW}} \times 100 = 92.5\%
   \]

### Summary

1. **Slip**: 2.5%
2. **Rotor Copper Loss**: 0 kW
3. **Shaft Power**: 37 kW
4. **Efficiency**: 92.5%