Question

A 250V, shunt motor on no load runs at 1000 rpm and takes 5A. The total armature and shunt field resistance are respectively 0.2Ω and 250Ω. Calculate the speed when loaded and taking a current of 50A, if armature reaction weakens the field by 3%.

Answer 1

To calculate the speed of a shunt motor when loaded, we can follow these steps:

### Step 1: Determine the Back EMF (E) at No Load

1. **Voltage (V)**: 250 V
2. **Shunt Field Current (I_f)**:
   \[
   I_f = \frac{V}{R_f} = \frac{250 \text{ V}}{250 \, \Omega} = 1 \text{ A}
   \]
   
3. **Total Current (I)** at no load:
   \[
   I = 5 \text{ A}
   \]
   
4. **Armature Current (I_a)**:
   \[
   I_a = I - I_f = 5 \text{ A} - 1 \text{ A} = 4 \text{ A}
   \]

5. **Armature Resistance (R_a)**:
   \[
   R_a = 0.2 \, \Omega
   \]

6. **Voltage Drop across Armature Resistance (V_a)**:
   \[
   V_a = I_a \cdot R_a = 4 \text{ A} \cdot 0.2 \, \Omega = 0.8 \text{ V}
   \]

7. **Back EMF (E)**:
   \[
   E = V - V_a = 250 \text{ V} - 0.8 \text{ V} = 249.2 \text{ V}
   \]

### Step 2: Calculate the Flux Change Due to Armature Reaction

1. **Armature Reaction Effect**: It weakens the field by 3%. Hence, the new field current under load can be calculated as follows:
   \[
   \text{New Field Current} = I_f' = I_f \cdot (1 - 0.03) = 1 \text{ A} \cdot 0.97 = 0.97 \text{ A}
   \]

2. **New Field Resistance Calculation**:
   \[
   R_f = 250 \, \Omega
   \]

3. **New Field Voltage (V_f)**:
   \[
   V_f = I_f' \cdot R_f = 0.97 \text{ A} \cdot 250 \, \Omega = 242.5 \text{ V}
   \]

### Step 3: Calculate the New Back EMF (E') When Loaded

1. **Load Current (I_load)**: 50 A
2. **New Armature Current (I_a')**:
   \[
   I_a' = I_load + I_f' = 50 \text{ A} + 0.97 \text{ A} \approx 50.97 \text{ A}
   \]

3. **Voltage Drop across Armature Resistance (V_a')**:
   \[
   V_a' = I_a' \cdot R_a = 50.97 \text{ A} \cdot 0.2 \, \Omega = 10.194 \text{ V}
   \]

4. **New Back EMF (E')**:
   \[
   E' = V - V_a' = 250 \text{ V} - 10.194 \text{ V} \approx 239.806 \text{ V}
   \]

### Step 4: Determine the New Speed (N')

1. **Relationship Between Speed and Back EMF**:
   Since speed is proportional to back EMF:
   \[
   \frac{N'}{N_0} = \frac{E'}{E}
   \]

2. **Substituting Known Values**:
   - \( N_0 \) (no-load speed) = 1000 rpm
   - \( E' \) = 239.806 V
   - \( E \) = 249.2 V

   \[
   \frac{N'}{1000} = \frac{239.806}{249.2}
   \]

3. **Calculate the New Speed (N')**:
   \[
   N' = 1000 \cdot \frac{239.806}{249.2} \approx 960.3 \text{ rpm}
   \]

### Final Result
The speed of the shunt motor when loaded with a current of 50 A, taking into account the armature reaction that weakens the field by 3%, is approximately **960.3 rpm**.

Answer 2

To determine the speed of the shunt motor under load, follow these steps:

1. **Calculate the no-load conditions:**
   - **Armature current (I_a) on no load**: \( I_a = 5 \text{ A} \)
   - **Shunt field current (I_f) on no load**:
     \[
     I_f = \frac{V}{R_f} = \frac{250}{250} = 1 \text{ A}
     \]
   - **Armature voltage (V_a) on no load**:
     \[
     V_a = V - I_f \cdot R_f = 250 - 1 \cdot 250 = 0 \text{ V} \text{ (which is not possible, thus }\ V_a = 250 \text{ V, since no load is ideal)}
     \]
   - **No-load voltage across the armature**:
     \[
     V_a = 250 - I_a \cdot R_a = 250 - 5 \cdot 0.2 = 249 \text{ V}
     \]

2. **Calculate the armature voltage under load:**
   - **Total load current**: \( I_{\text{total}} = 50 \text{ A} \)
   - **Shunt field current**: \( I_f = \frac{250}{250} = 1 \text{ A} \)
   - **Armature current under load**:
     \[
     I_a = I_{\text{total}} - I_f = 50 - 1 = 49 \text{ A}
     \]
   - **Voltage drop in the armature**:
     \[
     V_a = V - I_a \cdot R_a = 250 - 49 \cdot 0.2 = 250 - 9.8 = 240.2 \text{ V}
     \]

3. **Calculate the field weakening due to armature reaction:**
   - **Field weakening percentage**: 3%
   - **Effective field voltage**:
     \[
     V_f' = V_f \cdot (1 - 0.03) = 250 \cdot 0.97 = 242.5 \text{ V}
     \]

4. **Determine the speed ratio:**
   - The speed of a shunt motor is inversely proportional to the armature voltage and directly proportional to the field voltage:
     \[
     \frac{N_{\text{loaded}}}{N_{\text{no load}}} = \frac{V_a'}{V_a} \cdot \frac{V_f}{V_f'}
     \]
   - Where \( V_a' = 240.2 \text{ V} \) and \( V_a = 249 \text{ V} \), and \( V_f = 250 \text{ V} \) and \( V_f' = 242.5 \text{ V} \):
     \[
     \frac{N_{\text{loaded}}}{N_{\text{no load}}} = \frac{240.2}{249} \cdot \frac{250}{242.5}
     \]
   - Simplify to:
     \[
     \frac{N_{\text{loaded}}}{1000} = \frac{240.2}{249} \cdot \frac{250}{242.5} \approx 0.966 \cdot 1.031 = 0.996
     \]
   - **Speed under load**:
     \[
     N_{\text{loaded}} = 1000 \times 0.996 \approx 996 \text{ RPM}
     \]

Thus, the speed of the motor when loaded and drawing 50A, with armature reaction weakening the field by 3%, is approximately **996 RPM**.