Rotor Cu Losses MCQ Quiz - Objective Question with Answer for Rotor Cu Losses - Download Free PDF

Explanation:

Slip Ring Induction Motor

Definition: A slip ring induction motor, also known as a wound rotor induction motor, is a type of induction motor where the rotor is connected to external resistances through slip rings. This arrangement allows for the addition of resistance to the rotor circuit, which can be varied to control the motor's speed and torque characteristics.

Working Principle: In a slip ring induction motor, the rotor winding is connected to external resistors via slip rings and brushes. When the motor is started, the external resistors are initially included in the rotor circuit. As the motor speeds up, the resistance can be gradually reduced to achieve the desired operating characteristics. This method provides better control over the starting current and torque compared to squirrel cage motors, where such adjustments are not possible.

Advantages:

• Improved starting torque: The external resistances allow for a higher starting torque, which is beneficial for applications requiring a high initial torque.
• Reduced starting current: The resistors limit the inrush current during startup, reducing the mechanical and electrical stress on the motor.
• Speed control: By varying the external resistance, the speed of the motor can be controlled, making it suitable for applications where variable speed operation is required.

Disadvantages:

• Complexity: The presence of slip rings and brushes makes the motor more complex and requires additional maintenance compared to squirrel cage induction motors.
• Higher cost: The additional components and complexity result in a higher initial cost and maintenance requirements.

Applications: Slip ring induction motors are commonly used in applications requiring high starting torque and variable speed operation. Examples include cranes, hoists, elevators, and conveyors.

Correct Option Analysis:

The correct option is:

Option 2: Slip ring induction motor

This option correctly identifies the type of motor where resistance is added to the rotor circuit to change its speed. The slip ring induction motor is designed to allow external resistances to be connected to the rotor, providing control over the motor's speed and torque characteristics.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Squirrel cage induction motor

This type of induction motor has a rotor constructed with bars short-circuited by end rings, forming a cage-like structure. The squirrel cage rotor does not allow for external resistance to be added to the rotor circuit, making it impossible to control the speed in the same manner as a slip ring induction motor. Therefore, this option is incorrect.

Option 3: Single phase induction motor

Single phase induction motors are typically used for smaller loads and do not have provisions for adding external resistance to the rotor circuit. The speed control methods for single phase motors are different and do not involve changing the rotor resistance. Hence, this option is incorrect.

Option 4: Synchronous motor

Synchronous motors operate at a constant speed determined by the supply frequency and the number of poles. They do not use rotor resistance for speed control, as their speed is not variable once synchronized with the supply frequency. This makes this option incorrect for the given statement.

Conclusion:

Understanding the characteristics and operational principles of different types of motors is crucial for correctly identifying the appropriate motor for specific applications. The slip ring induction motor stands out in its ability to use external resistors in the rotor circuit for speed and torque control, making it suitable for applications requiring high starting torque and variable speed operation. This feature sets it apart from other types of motors, such as squirrel cage induction motors, single phase induction motors, and synchronous motors.

Power flow in Induction Motor

\(P_g:P_c:P_d=1:s:1-s\)

where, Pg = Air gap Power

Pc = Rotor copper loss

Pd = Developed Power 

Power flow in Induction Motor

\(P_g:P_c:P_d=1:s:1-s\)

where, Pg = Air gap Power

Pc = Rotor copper loss

Pd = Developed Power 

Calculation

Given, Pd = 15 kW

s = 4%

The relationship between developed power and rotor copper loss is:

\({P_c\over P_d}={s\over 1-s}\)

\({P_c}=P_d\times{s\over 1-s}\)

\({P_c}=15\times{0.04\over 1-0.04}\)

P_c = 625 W

Slip power recovery scheme of a 3ϕ induction motor

Slip power recovery is one of the methods of controlling the speed of an Induction motor. 

In the rotor resistance control method, the slip power in the rotor circuit is wasted as I²R losses during the low-speed operation. The efficiency is also reduced.

In this method, the slip power from the rotor circuit can be recovered and fed back to the AC source so as to utilize it outside the motor. This increases the efficiency of the system.

When the power feeds into the rotor from the drive circuit, the slip is negative.

The significance of negative slip is that the rotor speed is greater than the synchronous speed.

Therefore, in the slip-power recover scheme motor speed is always above the synchronous speed i.e. N_R > N_S

The correct answer is option 4):(s/(1 - s))

Concept:

• The difference between the synchronous speed of the magnetic field, and the shaft rotating speed is slip
• The copper loss of the rotor is due to slip
• Rotor copper losses are equal to slip times rotor input. Hence the rotor copper losses are directly proportional to slip.

P_r:P_c = 1:s

where s= Slip of the induction motor.

P_r : P_c : P_m = 1 : s : 1-s.

Where,

P_r = Electrical power input in rotor;

P_m = Electrical Power converted to mechanical

P_c = I²R = Copper loss

S= slip

Hence The ratio of rotor copper loss with rotor output is:(s/(1 - s))

The correct answer is option 4):(s/(1 - s))

Concept:

• The difference between the synchronous speed of the magnetic field, and the shaft rotating speed is slip
• The copper loss of the rotor is due to slip
• Rotor copper losses are equal to slip times rotor input. Hence the rotor copper losses are directly proportional to slip.

P_r:P_c = 1:s

where s= Slip of the induction motor.

P_r : P_c : P_m = 1 : s : 1-s.

Where,

P_r = Electrical power input in rotor;

P_m = Electrical Power converted to mechanical

P_c = I²R = Copper loss

S= slip

Hence The ratio of rotor copper loss with rotor output is:(s/(1 - s))

Blocked Rotor Test: It is also known as the locked-rotor or short-circuit test.

This test is used to find:

• Short-circuit current with normal voltage applied to the stator
• Power factor on short-circuit
• Total leakage reactance X₀₁ of the motor is referred to as primary (i.e. stator) and the total resistance of the motor R₀₁ is referred to as primary.
   

Procedure:

• In this test, the rotor is locked (or allowed very slow rotation) and the rotor windings are short-circuited at slip-rings if the motor has a wound rotor.
• Just as in the case of a short-circuit test on a transformer, a reduced voltage (up to 15 or 20 percent of normal value) is applied to the stator terminals and is so adjusted that full-load current flows in the stator
• As in this case s = 1 (Motor rotor under rest), the equivalent circuit of the motor is exactly like a transformer, having a short-circuited secondary.
• The values of current, voltage, and power input on short-circuit are measured by the ammeter, voltmeter, and wattmeter connected in the circuits as before.
• Curves connecting the above quantities may also be drawn by taking two or three additional sets of readings at progressively reduced voltages of the stator.

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Rotor power output = 20 kW

s = 0.05

Airgap power \({P_g} = \frac{{20}}{{1 - 0.05}} = 21\;kW\)

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Given that, slip (s) = 4% = 0.04

Rotor power output (Pg) = 15 kW

Rotor copper losses, \({P_{cu}} = \frac{s}{{\left( {1 - s} \right)}} \times {P_g}\)

\(P_{cu} = \frac{{0.04}}{{1 - 0.04}} \times 15 \times {10^3} = 625\;W\)

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Given that, slip (s) = 4% = 0.04

Rotor power output (Pg) = 15 kW

Rotor copper losses, \({P_{cu}} = \frac{s}{{\left( {1 - s} \right)}} \times {P_g}\)

\(P_{cu} = \frac{{0.04}}{{1 - 0.04}} \times 15 \times {10^3} = 625\;W\)

Explanation:

Slip Ring Induction Motor

Definition: A slip ring induction motor, also known as a wound rotor induction motor, is a type of induction motor where the rotor is connected to external resistances through slip rings. This arrangement allows for the addition of resistance to the rotor circuit, which can be varied to control the motor's speed and torque characteristics.

Working Principle: In a slip ring induction motor, the rotor winding is connected to external resistors via slip rings and brushes. When the motor is started, the external resistors are initially included in the rotor circuit. As the motor speeds up, the resistance can be gradually reduced to achieve the desired operating characteristics. This method provides better control over the starting current and torque compared to squirrel cage motors, where such adjustments are not possible.

Advantages:

• Improved starting torque: The external resistances allow for a higher starting torque, which is beneficial for applications requiring a high initial torque.
• Reduced starting current: The resistors limit the inrush current during startup, reducing the mechanical and electrical stress on the motor.
• Speed control: By varying the external resistance, the speed of the motor can be controlled, making it suitable for applications where variable speed operation is required.

Disadvantages:

• Complexity: The presence of slip rings and brushes makes the motor more complex and requires additional maintenance compared to squirrel cage induction motors.
• Higher cost: The additional components and complexity result in a higher initial cost and maintenance requirements.

Applications: Slip ring induction motors are commonly used in applications requiring high starting torque and variable speed operation. Examples include cranes, hoists, elevators, and conveyors.

Correct Option Analysis:

The correct option is:

Option 2: Slip ring induction motor

This option correctly identifies the type of motor where resistance is added to the rotor circuit to change its speed. The slip ring induction motor is designed to allow external resistances to be connected to the rotor, providing control over the motor's speed and torque characteristics.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Squirrel cage induction motor

This type of induction motor has a rotor constructed with bars short-circuited by end rings, forming a cage-like structure. The squirrel cage rotor does not allow for external resistance to be added to the rotor circuit, making it impossible to control the speed in the same manner as a slip ring induction motor. Therefore, this option is incorrect.

Option 3: Single phase induction motor

Single phase induction motors are typically used for smaller loads and do not have provisions for adding external resistance to the rotor circuit. The speed control methods for single phase motors are different and do not involve changing the rotor resistance. Hence, this option is incorrect.

Option 4: Synchronous motor

Synchronous motors operate at a constant speed determined by the supply frequency and the number of poles. They do not use rotor resistance for speed control, as their speed is not variable once synchronized with the supply frequency. This makes this option incorrect for the given statement.

Conclusion:

Understanding the characteristics and operational principles of different types of motors is crucial for correctly identifying the appropriate motor for specific applications. The slip ring induction motor stands out in its ability to use external resistors in the rotor circuit for speed and torque control, making it suitable for applications requiring high starting torque and variable speed operation. This feature sets it apart from other types of motors, such as squirrel cage induction motors, single phase induction motors, and synchronous motors.

Power flow in Induction Motor

\(P_g:P_c:P_d=1:s:1-s\)

where, Pg = Air gap Power

Pc = Rotor copper loss

Pd = Developed Power 

The correct answer is option 4):(s/(1 - s))

Concept:

• The difference between the synchronous speed of the magnetic field, and the shaft rotating speed is slip
• The copper loss of the rotor is due to slip
• Rotor copper losses are equal to slip times rotor input. Hence the rotor copper losses are directly proportional to slip.

P_r:P_c = 1:s

where s= Slip of the induction motor.

P_r : P_c : P_m = 1 : s : 1-s.

Where,

P_r = Electrical power input in rotor;

P_m = Electrical Power converted to mechanical

P_c = I²R = Copper loss

S= slip

Hence The ratio of rotor copper loss with rotor output is:(s/(1 - s))

Blocked Rotor Test: It is also known as the locked-rotor or short-circuit test.

This test is used to find:

• Short-circuit current with normal voltage applied to the stator
• Power factor on short-circuit
• Total leakage reactance X₀₁ of the motor is referred to as primary (i.e. stator) and the total resistance of the motor R₀₁ is referred to as primary.
   

Procedure:

• In this test, the rotor is locked (or allowed very slow rotation) and the rotor windings are short-circuited at slip-rings if the motor has a wound rotor.
• Just as in the case of a short-circuit test on a transformer, a reduced voltage (up to 15 or 20 percent of normal value) is applied to the stator terminals and is so adjusted that full-load current flows in the stator
• As in this case s = 1 (Motor rotor under rest), the equivalent circuit of the motor is exactly like a transformer, having a short-circuited secondary.
• The values of current, voltage, and power input on short-circuit are measured by the ammeter, voltmeter, and wattmeter connected in the circuits as before.
• Curves connecting the above quantities may also be drawn by taking two or three additional sets of readings at progressively reduced voltages of the stator.

Concept:

Power flow in the Induction motor is as shown below.

Stator input power = P

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\) = P - stator loss

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Where, s = slip

Calculation:

Given-

Stator input power, P = 10.5 kW

Stator loss = 500 W

Slip, s = 6% = 0.06

Rotor input or air gap power, Pin = 10500 - 500 = 10000 W

Rotor copper loss, Pcu = s Pin = 0.06 x 10000

Pcu = 600 W

Now rotor copper loss per phase, P'cu = Pcu / 3

Pcu = 200 W