Question
2. A 400 V DC shunt motor has an armature resistance of 0,22 and a full-load speed of 1 200 r/min. For an armature current of 70 A and an unsaturated field, determine the following:
2.1 The resistance needed in series with the armature to reduce the speed to 1000 r/min, with the torque remaining unchanged.
2.2 The speed of the motor if the torque is halved and the additional resistance is still connected to the armature circuit.
2.1 The resistance needed in series with the armature to reduce the speed to 1000 r/min, with the torque remaining unchanged.
2.2 The speed of the motor if the torque is halved and the additional resistance is still connected to the armature circuit.
Answers
GPT-4o mini
To solve the problem, we can use the basic equations related to DC shunt motors. Let's break down the calculations step by step.
**Given Data:**
- Supply voltage, \( V = 400 \, V \)
- Armature resistance, \( R_a = 0.22 \, \Omega \)
- Full-load speed, \( N_{FL} = 1200 \, r/min \)
- Armature current, \( I_a = 70 \, A \)
### 2.1 Resistance Needed for Speed Reduction
The speed of a DC motor is inversely proportional to the field current (which is constant in this case) and directly proportional to the back EMF \( E \), and is given by:
\[
N \propto \frac{E}{\Phi}
\]
Where:
- \( E = V - I_a R_a \)
- \( V \) is the applied voltage
- \( I_a \) is the armature current
- \( R_a \) is the armature resistance
Let's first calculate the back EMF \( E \) at full load:
\[
E_{FL} = V - I_a R_a
\]
\[
E_{FL} = 400 V - (70 A \times 0.22 \, \Omega) = 400 V - 15.4 V = 384.6 V
\]
Now, the speed at the full-load point is given:
For the new speed \( N_{new} = 1000 \, r/min \), we want to find the new back EMF \( E_{new} \). The speed ratio is:
\[
\frac{N_{new}}{N_{FL}} = \frac{E_{new}}{E_{FL}}
\]
So:
\[
E_{new} = E_{FL} \cdot \frac{N_{new}}{N_{FL}}
\]
\[
E_{new} = 384.6 V \cdot \frac{1000 \, r/min}{1200 \, r/min} = 384.6 V \cdot \frac{5}{6} = 320.5 V
\]
Next, we can find the new armature current \( I_a \) required to produce this back EMF. The equation for back EMF is:
\[
E_{new} = V - I_a R_a
\]
Using \( E_{new} \):
\[
320.5 V = 400 V - I_a \cdot 0.22 \, \Omega
\]
\[
I_a \cdot 0.22 \, \Omega = 400 V - 320.5 V = 79.5 V
\]
\[
I_a = \frac{79.5}{0.22} = 361.36 \, A
\]
### Finding the resistance needed
Now, we find the required armature resistance \( R_{additional} \) to achieve that armature current while keeping the torque constant (thus maintaining full-load current):
Using \( V - I_a (R_a + R_{additional}) = E_{new} \):
\[
400 - 361.36 (0.22 + R_{additional}) = 320.5
\]
Solving for \( R_{additional} \):
\[
400 - 320.5 = 361.36 \cdot (0.22 + R_{additional})
\]
\[
79.5 = 361.36 \cdot (0.22 + R_{additional})
\]
\[
R_{additional} = \frac{79.5}{361.36} - 0.22
\]
Now calculating \( R_{additional} \):
\[
R_{additional} = 0.2194 - 0.22 = -0.0006 \, \Omega
\]
Since adding a resistance cannot be negative, it indicates that the additional resistance is not practically feasible to achieve the speed with such torque.
### 2.2 Speed for Halved Torque
To determine the speed if torque is halved, note that torque in a DC motor is proportional to the product of the armature current and magnetic field flux \( \tau \propto I_a \Phi \).
If the torque is halved, the armature current is halved (assuming constant magnetic field conditions):
\[
I_a = \frac{70}{2} = 35 \, A
\]
Now, let's calculate the new back EMF:
\[
E_{halved} = V - I_a R_a
\]
\[
E_{halved} = 400 V - (35 A \cdot 0.22 \, \Omega) = 400 V - 7.7 V = 392.3 V
\]
Now the new speed \( N_{halved} \):
Using the previous relations:
\[
N_{halved} = N_{FL} \cdot \frac{E_{halved}}{E_{FL}}
\]
\[
N_{halved} = 1200 \cdot \frac{392.3}{384.6} = 1220.3 \, r/min
\]
### In summary:
- **2.1** The calculated additional resistance required cannot be implemented as it leads to a non-practical result.
- **2.2** The speed of the motor if torque is halved, with the additional resistance still connected, is approximately **1220.3 r/min**.
**Given Data:**
- Supply voltage, \( V = 400 \, V \)
- Armature resistance, \( R_a = 0.22 \, \Omega \)
- Full-load speed, \( N_{FL} = 1200 \, r/min \)
- Armature current, \( I_a = 70 \, A \)
### 2.1 Resistance Needed for Speed Reduction
The speed of a DC motor is inversely proportional to the field current (which is constant in this case) and directly proportional to the back EMF \( E \), and is given by:
\[
N \propto \frac{E}{\Phi}
\]
Where:
- \( E = V - I_a R_a \)
- \( V \) is the applied voltage
- \( I_a \) is the armature current
- \( R_a \) is the armature resistance
Let's first calculate the back EMF \( E \) at full load:
\[
E_{FL} = V - I_a R_a
\]
\[
E_{FL} = 400 V - (70 A \times 0.22 \, \Omega) = 400 V - 15.4 V = 384.6 V
\]
Now, the speed at the full-load point is given:
For the new speed \( N_{new} = 1000 \, r/min \), we want to find the new back EMF \( E_{new} \). The speed ratio is:
\[
\frac{N_{new}}{N_{FL}} = \frac{E_{new}}{E_{FL}}
\]
So:
\[
E_{new} = E_{FL} \cdot \frac{N_{new}}{N_{FL}}
\]
\[
E_{new} = 384.6 V \cdot \frac{1000 \, r/min}{1200 \, r/min} = 384.6 V \cdot \frac{5}{6} = 320.5 V
\]
Next, we can find the new armature current \( I_a \) required to produce this back EMF. The equation for back EMF is:
\[
E_{new} = V - I_a R_a
\]
Using \( E_{new} \):
\[
320.5 V = 400 V - I_a \cdot 0.22 \, \Omega
\]
\[
I_a \cdot 0.22 \, \Omega = 400 V - 320.5 V = 79.5 V
\]
\[
I_a = \frac{79.5}{0.22} = 361.36 \, A
\]
### Finding the resistance needed
Now, we find the required armature resistance \( R_{additional} \) to achieve that armature current while keeping the torque constant (thus maintaining full-load current):
Using \( V - I_a (R_a + R_{additional}) = E_{new} \):
\[
400 - 361.36 (0.22 + R_{additional}) = 320.5
\]
Solving for \( R_{additional} \):
\[
400 - 320.5 = 361.36 \cdot (0.22 + R_{additional})
\]
\[
79.5 = 361.36 \cdot (0.22 + R_{additional})
\]
\[
R_{additional} = \frac{79.5}{361.36} - 0.22
\]
Now calculating \( R_{additional} \):
\[
R_{additional} = 0.2194 - 0.22 = -0.0006 \, \Omega
\]
Since adding a resistance cannot be negative, it indicates that the additional resistance is not practically feasible to achieve the speed with such torque.
### 2.2 Speed for Halved Torque
To determine the speed if torque is halved, note that torque in a DC motor is proportional to the product of the armature current and magnetic field flux \( \tau \propto I_a \Phi \).
If the torque is halved, the armature current is halved (assuming constant magnetic field conditions):
\[
I_a = \frac{70}{2} = 35 \, A
\]
Now, let's calculate the new back EMF:
\[
E_{halved} = V - I_a R_a
\]
\[
E_{halved} = 400 V - (35 A \cdot 0.22 \, \Omega) = 400 V - 7.7 V = 392.3 V
\]
Now the new speed \( N_{halved} \):
Using the previous relations:
\[
N_{halved} = N_{FL} \cdot \frac{E_{halved}}{E_{FL}}
\]
\[
N_{halved} = 1200 \cdot \frac{392.3}{384.6} = 1220.3 \, r/min
\]
### In summary:
- **2.1** The calculated additional resistance required cannot be implemented as it leads to a non-practical result.
- **2.2** The speed of the motor if torque is halved, with the additional resistance still connected, is approximately **1220.3 r/min**.