Question

Power input to a 400V, 60Hz, 6 pole 3 φ I.M. running at 1140 rpm. is 40Kw. at 0.8 p.f. lag. Stator losses are 1000W. and friction windage losses are 2000W. Calculate - i) Slip ii) Rotor copper loss

Answer 1

To solve this problem, we'll follow these steps:

1. **Determine the synchronous speed (\(N_s\)) of the induction motor.**

2. **Calculate the slip (\(s\)) of the induction motor.**

3. **Determine the rotor copper losses.**

Here’s a step-by-step breakdown:

### 1. Determine the Synchronous Speed (\(N_s\))

The synchronous speed of an induction motor is given by:

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

where:
- \( f \) = frequency of the supply (60 Hz)
- \( P \) = number of poles (6)

Substitute the values:

\[ N_s = \frac{120 \times 60}{6} \]
\[ N_s = \frac{7200}{6} \]
\[ N_s = 1200 \text{ rpm} \]

### 2. Calculate the Slip (\(s\))

The slip \( s \) is given by:

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

where:
- \( N_s \) = synchronous speed (1200 rpm)
- \( N_r \) = rotor speed (1140 rpm)

Substitute the values:

\[ s = \frac{1200 - 1140}{1200} \]
\[ s = \frac{60}{1200} \]
\[ s = 0.05 \text{ or } 5\% \]

### 3. Determine the Rotor Copper Losses

To find the rotor copper losses, we need to first determine the total losses and then subtract the stator and friction/windage losses.

**Total Power Input (\(P_{in}\))**: 40 kW

**Stator Losses**: 1000 W

**Friction and Windage Losses**: 2000 W

**Total Losses (excluding rotor copper losses)**:

\[ \text{Total Losses} = \text{Stator Losses} + \text{Friction and Windage Losses} \]
\[ \text{Total Losses} = 1000 + 2000 \]
\[ \text{Total Losses} = 3000 \text{ W} \]

**Output Power (\(P_{out}\))** is:

\[ P_{out} = P_{in} - \text{Total Losses} - \text{Rotor Copper Losses} \]

However, to find the rotor copper losses, we use:

\[ \text{Input Power} = P_{out} + \text{Total Losses} \]
\[ 40000 = P_{out} + 3000 + \text{Rotor Copper Losses} \]

Since \( P_{out} \) is the mechanical power output of the motor and can be expressed as:

\[ P_{out} = \text{Rotor Power} \]

The rotor power is:

\[ \text{Rotor Power} = \text{Input Power} - \text{Stator Losses} - \text{Friction and Windage Losses} - \text{Rotor Copper Losses} \]

To find the rotor copper losses, we know:

**Electrical Power Supplied to Rotor ( \(P_{r_{elec}}\) )** = \( P_{in} - \text{Stator Losses} \):

\[ P_{r_{elec}} = 40000 - 1000 \]
\[ P_{r_{elec}} = 39000 \text{ W} \]

**Power Converted to Mechanical Work**:

\[ \text{Mechanical Power} = \text{Rotor Power} = \text{Rotor Power} - \text{Rotor Copper Losses} \]

Now:

\[ \text{Mechanical Power} = \text{Electrical Power Supplied to Rotor} \times (1 - s) \]
\[ \text{Mechanical Power} = 39000 \times (1 - 0.05) \]
\[ \text{Mechanical Power} = 39000 \times 0.95 \]
\[ \text{Mechanical Power} = 37050 \text{ W} \]

Subtracting the mechanical power from the input power minus the stator and friction/windage losses:

\[ \text{Rotor Copper Losses} = \text{Electrical Power Supplied to Rotor} - \text{Mechanical Power} \]
\[ \text{Rotor Copper Losses} = 39000 - 37050 \]
\[ \text{Rotor Copper Losses} = 1950 \text{ W} \]

### Summary

- **Slip (\(s\))**: 0.05 or 5%
- **Rotor Copper Losses**: 1950 W

Answer 2

To solve the problem, let's break it down into parts.

### Given Data:
- **Rated Voltage (V)**: 400V
- **Frequency (f)**: 60Hz
- **Number of Poles (P)**: 6
- **Power Input (P\(_{in}\))**: 40 kW = 40,000 W
- **Power Factor (PF)**: 0.8 (lagging)
- **Stator Losses**: 1000 W
- **Friction and Windage Losses**: 2000 W
- **Rotor Speed (N\(_r\))**: 1140 RPM

### I) **Slip (s)**

First, we need to calculate the synchronous speed (N\(_s\)):

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

Substitute the given values:

\[
N_s = \frac{120 \times 60}{6} = 1200 \text{ RPM}
\]

Now, slip \( s \) is calculated using the formula:

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

Substitute the synchronous speed \(N_s\) and rotor speed \(N_r\):

\[
s = \frac{1200 - 1140}{1200} = \frac{60}{1200} = 0.05
\]

So, the slip \( s \) is **5%**.

### II) **Rotor Copper Loss (P\(_{rcl}\))**

The total input power to the motor is 40,000 W, and the stator losses are 1000 W. The power transferred to the rotor \( P_{gross} \) can be calculated as:

\[
P_{gross} = P_{in} - \text{Stator Losses}
\]

\[
P_{gross} = 40,000 - 1,000 = 39,000 \text{ W}
\]

The rotor copper loss is given by:

\[
P_{rcl} = s \times P_{gross}
\]

Substitute the values of \( s \) and \( P_{gross} \):

\[
P_{rcl} = 0.05 \times 39,000 = 1,950 \text{ W}
\]

### Summary:
- **Slip (s)**: 5%
- **Rotor Copper Loss (P\(_{rcl}\))**: 1950 W