Question

A 4-pole, 220V shunt motor has 540 lap wound conductors. It takes 32A from the supply mains and develops an output power of 5.595 KW. The field winding takes 1A. The armature resistance is 0.09 Ω and the flux per pole is 30 mwb. Calculate: (i) the speed and (ii) the torque developed in N-M.

Answer 1

Let's analyze the given information and calculate the required parameters step by step:

### Given Data:

- Supply Voltage (\( V \)) = 220 V
- Number of Poles (\( P \)) = 4
- Number of conductors (\( Z \)) = 540
- Total Current (\( I_{\text{total}} \)) = 32 A
- Field Current (\( I_{\text{field}} \)) = 1 A
- Armature Resistance (\( R_a \)) = 0.09 Ω
- Flux per pole (\( \phi \)) = 30 mWb = \( 30 \times 10^{-3} \) Wb
- Output Power (\( P_{\text{out}} \)) = 5.595 kW = 5595 W
- Type of winding: Lap wound

Now, we can proceed to solve the problem step by step.

### Step 1: Calculate the Armature Current

The total current supplied to the motor is split into two parts: field current and armature current.

\[
I_a = I_{\text{total}} - I_{\text{field}}
\]

\[
I_a = 32 \, A - 1 \, A = 31 \, A
\]

### Step 2: Calculate the Back EMF ( \( E_b \) )

The back EMF (\( E_b \)) is given by:

\[
E_b = V - I_a R_a
\]

Substitute the given values:

\[
E_b = 220 \, V - (31 \, A \times 0.09 \, \Omega) = 220 \, V - 2.79 \, V = 217.21 \, V
\]

### Step 3: Calculate the Speed ( \( N \) )

The back EMF \( E_b \) for a DC motor is also related to speed by the equation:

\[
E_b = \frac{P \phi Z N}{60 A}
\]

Where:
- \( P \) = number of poles = 4
- \( \phi \) = flux per pole = 30 mWb = \( 30 \times 10^{-3} \, \text{Wb} \)
- \( Z \) = number of armature conductors = 540
- \( N \) = speed in rpm
- \( A \) = number of parallel paths = number of poles (for lap-wound motor) = 4

Rearranging the equation to solve for speed \( N \):

\[
N = \frac{60 A E_b}{P \phi Z}
\]

Substitute the known values:

\[
N = \frac{60 \times 4 \times 217.21}{4 \times (30 \times 10^{-3}) \times 540}
\]

\[
N = \frac{52130.4}{64.8} = 804 \, \text{rpm}
\]

Thus, the speed of the motor is approximately **804 rpm**.

### Step 4: Calculate the Torque ( \( T \) )

The torque \( T \) developed by a DC motor can be found using the mechanical power output equation:

\[
P_{\text{out}} = \frac{2 \pi N T}{60}
\]

Rearranging for torque \( T \):

\[
T = \frac{P_{\text{out}} \times 60}{2 \pi N}
\]

Substitute the given values:

\[
T = \frac{5595 \, W \times 60}{2 \pi \times 804}
\]

\[
T = \frac{335700}{5049.6} = 66.47 \, \text{N-m}
\]

Thus, the developed torque is approximately **66.47 N-m**.

### Final Answer:
1. **Speed \( N \) = 804 rpm**
2. **Torque \( T \) = 66.47 N-m**

Answer 2

To solve for the speed and torque developed in a 4-pole, 220V shunt motor, we'll need to use the given parameters and some fundamental equations related to DC motors. Here's a step-by-step approach:

### Given Data

1. **Voltage (V)**: 220 V
2. **Number of poles (P)**: 4
3. **Total armature current (Iₐ)**: 32 A
4. **Field current (Iₓ)**: 1 A
5. **Output power (P_out)**: 5.595 kW
6. **Armature resistance (Rₐ)**: 0.09 Ω
7. **Flux per pole (Φ)**: 30 mWb = 30 × 10⁻³ Wb
8. **Number of lap wound conductors (Z)**: 540

### (i) Speed of the Motor

To find the speed of the motor, we first need to calculate the back EMF (Eₐ). The back EMF can be calculated using the formula:

\[ Eₐ = V - Iₐ \cdot Rₐ \]

where:
- \( Iₐ \) is the armature current.
- \( Rₐ \) is the armature resistance.

Substituting the given values:

\[ Eₐ = 220 - 32 \times 0.09 \]
\[ Eₐ = 220 - 2.88 \]
\[ Eₐ = 217.12 \text{ V} \]

The speed of the motor can be found using the formula for the back EMF:

\[ Eₐ = \frac{P \cdot Φ \cdot N}{60 \cdot A} \]

where:
- \( P \) is the number of poles.
- \( Φ \) is the flux per pole.
- \( N \) is the speed of the motor (in RPM).
- \( A \) is the number of parallel paths in the armature.

For a lap-wound armature, \( A = P \). In this case, \( A = 4 \).

Rearrange the formula to solve for \( N \):

\[ N = \frac{Eₐ \cdot 60 \cdot A}{P \cdot Φ} \]

Substitute the values:

\[ N = \frac{217.12 \cdot 60 \cdot 4}{4 \cdot 30 \times 10^{-3}} \]
\[ N = \frac{217.12 \cdot 60}{30 \times 10^{-3}} \]
\[ N = \frac{13027.2}{0.03} \]
\[ N = 434,240 \text{ RPM} \]

It looks like there was an error. Let’s recheck the calculation:

The formula should use the correct \( A \) and appropriate units. Recalculate carefully:

\[ N = \frac{217.12 \cdot 60}{30 \times 10^{-3} \cdot 4} \]
\[ N = \frac{217.12 \cdot 60}{0.12} \]
\[ N = \frac{13027.2}{0.12} \]
\[ N = 108,560 \text{ RPM} \]

### (ii) Torque Developed

The output power \( P_{out} \) is related to the torque \( T \) and speed \( N \) by:

\[ P_{out} = \frac{T \cdot N \cdot 2 \pi}{60} \]

Rearrange this formula to solve for \( T \):

\[ T = \frac{P_{out} \cdot 60}{N \cdot 2 \pi} \]

Substitute the given values:

\[ T = \frac{5595 \cdot 60}{434,240 \cdot 2 \pi} \]
\[ T = \frac{335,700}{2 \cdot 3.1416 \cdot 434,240} \]
\[ T = \frac{335,700}{2,727,788} \]
\[ T \approx 0.123 \text{ N-m} \]

So, the correct speed and torque are approximately:

- **Speed**: 108,560 RPM (after recalculating correctly)
- **Torque**: 0.123 N-m

If these results seem too high or unrealistic, verify the calculation steps or check the given parameters for consistency with real-world values.