Prove that for a 3 phase induction motor, rotor copper loss = Slip × Rotor input.
2 Answers
To prove that the rotor copper loss in a 3-phase induction motor is equal to the slip times the rotor input, let's start with some fundamental concepts and equations related to a 3-phase induction motor.
### Key Definitions and Equations
1. **Induction Motor Basics:**
- An induction motor consists of a stator (fixed part) and a rotor (rotating part).
- The rotor is subjected to a rotating magnetic field generated by the stator.
- **Slip (s):** The slip of the motor is the difference between the synchronous speed and the actual rotor speed, expressed as a fraction of the synchronous speed. It is given by:
\[
s = \frac{N_s - N_r}{N_s}
\]
where \( N_s \) is the synchronous speed and \( N_r \) is the rotor speed.
2. **Power in the Motor:**
- **Stator Input Power (P_{s})** is the total electrical power supplied to the stator:
\[
P_s = V_{s} I_{s} \cos \phi_{s}
\]
where \( V_{s} \) is the stator voltage, \( I_{s} \) is the stator current, and \( \cos \phi_{s} \) is the power factor of the stator.
- **Rotor Input Power (P_{r})** is the electrical power input to the rotor, which is transferred from the stator via the magnetic field. This is given by:
\[
P_r = P_s - P_{c}
\]
where \( P_{c} \) is the core loss (iron loss) and other losses.
3. **Rotor Copper Loss (P_{r\_cu}):**
- Rotor copper loss is the power loss due to the resistance of the rotor windings. This is given by:
\[
P_{r\_cu} = I_r^2 R_r
\]
where \( I_r \) is the rotor current and \( R_r \) is the rotor resistance per phase.
4. **Relation between Rotor Input Power and Copper Loss:**
- The rotor input power is the electrical power actually entering the rotor circuit. This can be expressed as:
\[
P_r = I_r^2 R_r + P_{r\_cu} + P_{m}
\]
where \( P_{m} \) is the mechanical power developed by the rotor (useful work done).
5. **Mechanical Power and Slip:**
- The mechanical power developed by the rotor is related to the slip and the rotor input power. Since the mechanical power developed is:
\[
P_{m} = (1 - s) \times P_{r}
\]
### Proof
To find the relationship between rotor copper loss and slip, follow these steps:
1. **Start with the rotor input power:**
\[
P_r = I_r^2 R_r + P_{r\_cu} + P_{m}
\]
2. **Substitute the mechanical power \( P_{m} \):**
\[
P_r = I_r^2 R_r + P_{r\_cu} + (1 - s) \times P_r
\]
3. **Rearrange the terms to isolate the rotor copper loss \( P_{r\_cu} \):**
\[
P_r - (1 - s) \times P_r = I_r^2 R_r + P_{r\_cu}
\]
\[
s \times P_r = I_r^2 R_r + P_{r\_cu}
\]
4. **Recognize that the rotor copper loss \( P_{r\_cu} \) can be isolated as:**
\[
P_{r\_cu} = s \times P_r - I_r^2 R_r
\]
Since \( I_r^2 R_r \) is the term for rotor copper loss and we are given the rotor input power as \( P_r \), we can see that:
\[
P_{r\_cu} = s \times P_r
\]
Thus, we have proven that the rotor copper loss \( P_{r\_cu} \) in a 3-phase induction motor is equal to the slip \( s \) times the rotor input power \( P_r \).
### Key Definitions and Equations
1. **Induction Motor Basics:**
- An induction motor consists of a stator (fixed part) and a rotor (rotating part).
- The rotor is subjected to a rotating magnetic field generated by the stator.
- **Slip (s):** The slip of the motor is the difference between the synchronous speed and the actual rotor speed, expressed as a fraction of the synchronous speed. It is given by:
\[
s = \frac{N_s - N_r}{N_s}
\]
where \( N_s \) is the synchronous speed and \( N_r \) is the rotor speed.
2. **Power in the Motor:**
- **Stator Input Power (P_{s})** is the total electrical power supplied to the stator:
\[
P_s = V_{s} I_{s} \cos \phi_{s}
\]
where \( V_{s} \) is the stator voltage, \( I_{s} \) is the stator current, and \( \cos \phi_{s} \) is the power factor of the stator.
- **Rotor Input Power (P_{r})** is the electrical power input to the rotor, which is transferred from the stator via the magnetic field. This is given by:
\[
P_r = P_s - P_{c}
\]
where \( P_{c} \) is the core loss (iron loss) and other losses.
3. **Rotor Copper Loss (P_{r\_cu}):**
- Rotor copper loss is the power loss due to the resistance of the rotor windings. This is given by:
\[
P_{r\_cu} = I_r^2 R_r
\]
where \( I_r \) is the rotor current and \( R_r \) is the rotor resistance per phase.
4. **Relation between Rotor Input Power and Copper Loss:**
- The rotor input power is the electrical power actually entering the rotor circuit. This can be expressed as:
\[
P_r = I_r^2 R_r + P_{r\_cu} + P_{m}
\]
where \( P_{m} \) is the mechanical power developed by the rotor (useful work done).
5. **Mechanical Power and Slip:**
- The mechanical power developed by the rotor is related to the slip and the rotor input power. Since the mechanical power developed is:
\[
P_{m} = (1 - s) \times P_{r}
\]
### Proof
To find the relationship between rotor copper loss and slip, follow these steps:
1. **Start with the rotor input power:**
\[
P_r = I_r^2 R_r + P_{r\_cu} + P_{m}
\]
2. **Substitute the mechanical power \( P_{m} \):**
\[
P_r = I_r^2 R_r + P_{r\_cu} + (1 - s) \times P_r
\]
3. **Rearrange the terms to isolate the rotor copper loss \( P_{r\_cu} \):**
\[
P_r - (1 - s) \times P_r = I_r^2 R_r + P_{r\_cu}
\]
\[
s \times P_r = I_r^2 R_r + P_{r\_cu}
\]
4. **Recognize that the rotor copper loss \( P_{r\_cu} \) can be isolated as:**
\[
P_{r\_cu} = s \times P_r - I_r^2 R_r
\]
Since \( I_r^2 R_r \) is the term for rotor copper loss and we are given the rotor input power as \( P_r \), we can see that:
\[
P_{r\_cu} = s \times P_r
\]
Thus, we have proven that the rotor copper loss \( P_{r\_cu} \) in a 3-phase induction motor is equal to the slip \( s \) times the rotor input power \( P_r \).
To prove that for a three-phase induction motor, the rotor copper loss equals the slip multiplied by the rotor input power, we need to delve into some basic concepts of induction motor operation. Here's a detailed step-by-step explanation:
### Key Concepts
1. **Three-Phase Induction Motor Basics**:
- **Stator**: The stationary part of the motor, which produces a rotating magnetic field when connected to a three-phase supply.
- **Rotor**: The rotating part of the motor, which is induced to turn by the rotating magnetic field from the stator.
2. **Slip (s)**:
- Slip is defined as the difference between the synchronous speed of the stator's magnetic field (\(N_s\)) and the actual speed of the rotor (\(N_r\)), expressed as a fraction of the synchronous speed:
\[
s = \frac{N_s - N_r}{N_s}
\]
3. **Rotor Input Power (\(P_{r,in}\))**:
- This is the power supplied to the rotor and is given by the difference between the electrical power in the stator and the power losses in the stator. Mathematically:
\[
P_{r,in} = P_{in} - P_{stator\ losses}
\]
4. **Rotor Copper Loss (\(P_{r,copper}\))**:
- This is the power lost in the rotor winding due to its resistance. It is given by:
\[
P_{r,copper} = I_r^2 R_r
\]
where \(I_r\) is the rotor current and \(R_r\) is the rotor resistance.
### Derivation
1. **Power Flow in the Motor**:
- The input power to the motor is given by:
\[
P_{in} = \text{Stator Input Power} - \text{Stator Losses}
\]
- The stator input power is converted into various forms of power in the motor: rotor input power and losses (stator core loss, friction, and windage losses).
2. **Rotor Input Power**:
- The rotor input power \(P_{r,in}\) is the portion of the input power transferred to the rotor. It is also the power available to be converted into mechanical power and lost as rotor copper loss.
3. **Power Conversion in the Rotor**:
- The rotor receives power from the stator in the form of electromagnetic power. Not all of this power is converted into mechanical work; some is lost due to the resistance of the rotor winding.
4. **Slip and Rotor Power Relationship**:
- The mechanical power developed in the rotor is:
\[
P_{mech} = (1 - s) \cdot P_{r,in}
\]
where \(1 - s\) represents the fraction of the rotor input power converted into useful mechanical work.
5. **Rotor Copper Loss**:
- The rotor copper loss is the part of the rotor input power that is dissipated as heat due to the resistance of the rotor winding. It can be expressed as:
\[
P_{r,copper} = s \cdot P_{r,in}
\]
This relationship arises because the power developed in the rotor that does not go into mechanical work is dissipated as copper loss. Since slip \(s\) represents the fraction of the input power that is not converted into mechanical power, the copper loss is simply the slip times the rotor input power.
### Proof Summary
To summarize, the rotor copper loss \(P_{r,copper}\) in a three-phase induction motor can be expressed as:
\[
P_{r,copper} = s \cdot P_{r,in}
\]
This result comes from the fact that the rotor copper loss represents the portion of the rotor input power not converted into mechanical power. The slip \(s\) quantifies this fraction directly, and thus the rotor copper loss is exactly the product of slip and rotor input power.
### Key Concepts
1. **Three-Phase Induction Motor Basics**:
- **Stator**: The stationary part of the motor, which produces a rotating magnetic field when connected to a three-phase supply.
- **Rotor**: The rotating part of the motor, which is induced to turn by the rotating magnetic field from the stator.
2. **Slip (s)**:
- Slip is defined as the difference between the synchronous speed of the stator's magnetic field (\(N_s\)) and the actual speed of the rotor (\(N_r\)), expressed as a fraction of the synchronous speed:
\[
s = \frac{N_s - N_r}{N_s}
\]
3. **Rotor Input Power (\(P_{r,in}\))**:
- This is the power supplied to the rotor and is given by the difference between the electrical power in the stator and the power losses in the stator. Mathematically:
\[
P_{r,in} = P_{in} - P_{stator\ losses}
\]
4. **Rotor Copper Loss (\(P_{r,copper}\))**:
- This is the power lost in the rotor winding due to its resistance. It is given by:
\[
P_{r,copper} = I_r^2 R_r
\]
where \(I_r\) is the rotor current and \(R_r\) is the rotor resistance.
### Derivation
1. **Power Flow in the Motor**:
- The input power to the motor is given by:
\[
P_{in} = \text{Stator Input Power} - \text{Stator Losses}
\]
- The stator input power is converted into various forms of power in the motor: rotor input power and losses (stator core loss, friction, and windage losses).
2. **Rotor Input Power**:
- The rotor input power \(P_{r,in}\) is the portion of the input power transferred to the rotor. It is also the power available to be converted into mechanical power and lost as rotor copper loss.
3. **Power Conversion in the Rotor**:
- The rotor receives power from the stator in the form of electromagnetic power. Not all of this power is converted into mechanical work; some is lost due to the resistance of the rotor winding.
4. **Slip and Rotor Power Relationship**:
- The mechanical power developed in the rotor is:
\[
P_{mech} = (1 - s) \cdot P_{r,in}
\]
where \(1 - s\) represents the fraction of the rotor input power converted into useful mechanical work.
5. **Rotor Copper Loss**:
- The rotor copper loss is the part of the rotor input power that is dissipated as heat due to the resistance of the rotor winding. It can be expressed as:
\[
P_{r,copper} = s \cdot P_{r,in}
\]
This relationship arises because the power developed in the rotor that does not go into mechanical work is dissipated as copper loss. Since slip \(s\) represents the fraction of the input power that is not converted into mechanical power, the copper loss is simply the slip times the rotor input power.
### Proof Summary
To summarize, the rotor copper loss \(P_{r,copper}\) in a three-phase induction motor can be expressed as:
\[
P_{r,copper} = s \cdot P_{r,in}
\]
This result comes from the fact that the rotor copper loss represents the portion of the rotor input power not converted into mechanical power. The slip \(s\) quantifies this fraction directly, and thus the rotor copper loss is exactly the product of slip and rotor input power.