Induction Motor Torque MCQ Quiz - Objective Question with Answer for Induction Motor Torque - Download Free PDF

Concept

The EMF induced in the rotor of an induction motor is given by:

\(E_r=sE_o\)

where, Er = Rotor induced EMF

E_o = Standstill EMF

s = Slip

The slip 's' is given by:

\(s={N_s-N_r\over N_s}\)

Ns = Synchronous speed

Nr = Rotor speed

Explanation:

Three-Phase Squirrel-Cage Induction Motor

Definition: A three-phase squirrel-cage induction motor is a type of induction motor which is widely used in industrial and commercial applications. It is named "squirrel-cage" because the rotor structure resembles a squirrel cage. The rotor consists of short-circuited bars, typically made of aluminum or copper, which are connected by end rings. This type of motor is known for its rugged construction, reliability, and low maintenance requirements.

Increasing Starting Torque: In a three-phase squirrel-cage induction motor, various methods can be employed to increase the starting torque. Starting torque is crucial for applications where the motor needs to start under load. Here are some ways to achieve higher starting torque:

• Using Deep-Bar Rotors: Deep-bar rotors have rotor bars that are deeper and have a non-uniform cross-section. This design creates higher resistance during startup, which increases the starting torque. As the motor speeds up, the skin effect reduces the resistance, allowing the motor to operate efficiently at full speed. This method is effective because it provides high starting torque without compromising the motor's performance at normal operating speeds.
• Increasing the Number of Poles: By increasing the number of poles in the motor, the synchronous speed decreases. This lower synchronous speed results in higher slip at startup, which increases the starting torque. However, this method also reduces the motor's rated speed, which may not be desirable for all applications. It is a trade-off between starting torque and operating speed.
• Using a Double-Cage Rotor: Double-cage rotors have two sets of rotor bars with different electrical properties. The outer cage has higher resistance and lower reactance, providing high starting torque. The inner cage has lower resistance and higher reactance, which improves the motor's performance at normal operating speeds. This design allows for a good balance between high starting torque and efficient operation.

Correct Option Analysis:

The correct option is:

Option 1: A) Using deep-bar rotors, C) Increasing the number of poles, and D) Using a double-cage rotor.

This option correctly identifies the methods that can be used to increase the starting torque in a three-phase squirrel-cage induction motor. These methods are effective in improving the motor's starting performance while maintaining efficient operation at normal speeds.

Additional Information

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

Option 2: B) Increasing the frequency of operation, C) Increasing the number of poles, and D) Using a double-cage rotor.

Increasing the frequency of operation does not directly increase the starting torque. Instead, it affects the motor's speed and performance characteristics. Higher frequency operation typically increases the synchronous speed, which may not be desirable for applications requiring high starting torque. Therefore, this option is not suitable for increasing the starting torque.

Option 3: A) Using deep-bar rotors, B) Increasing the frequency of operation, and C) Increasing the number of poles.

As mentioned earlier, increasing the frequency of operation does not contribute to higher starting torque. While using deep-bar rotors and increasing the number of poles are effective methods, the inclusion of frequency increase makes this option incorrect for the specific goal of increasing starting torque.

Option 4: A) Using deep-bar rotors, B) Increasing the frequency of operation, and D) Using a double-cage rotor.

Similar to option 3, this option includes increasing the frequency of operation, which does not directly enhance the starting torque. While using deep-bar rotors and a double-cage rotor are valid methods, the inclusion of frequency increase renders this option incorrect for increasing starting torque.

Conclusion:

Understanding the methods to increase the starting torque in a three-phase squirrel-cage induction motor is crucial for selecting the appropriate approach based on the application requirements. Using deep-bar rotors, increasing the number of poles, and using a double-cage rotor are effective methods to achieve higher starting torque without compromising the motor's performance at normal operating speeds. These methods provide a balance between starting performance and operational efficiency, making them suitable for various industrial and commercial applications.

Maximum Starting Torque in Induction Motors:

In an induction motor, the maximum starting torque (also known as the "breakdown torque" or "pull-out torque") is the highest torque that the motor can produce under starting conditions. This occurs when the slip (s) is such that the rotor resistance is equal to the rotor reactance. At this point, the power factor of the rotor circuit is 0.707 lagging.

Here's a detailed explanation of the concept:

• In an induction motor, torque (T) is a function of slip (s). The torque produced by the induction motor is given by the formula:

  

  where \( R_2 \) is the rotor resistance and \( X_2 \) is the rotor reactance.
• As the slip increases from zero, the torque increases until it reaches a maximum value (maximum torque) and then starts to decrease.
• The condition for maximum torque can be derived by differentiating the torque equation with respect to slip (s) and setting the derivative to zero. This gives the condition for maximum torque as:

  

• At this point, the rotor impedance is minimum and the rotor current is at its maximum value. The rotor power factor at this condition is:

  

• Therefore, at the maximum starting torque, the rotor power factor is 0.707 lagging.

Hence, the correct answer is option 3 (0.707 Lagging).

Maximum Torque in Induction Motors:

The maximum torque in an induction motor, also known as pull-out torque, is the highest torque the motor can develop without losing synchronous speed. Understanding the factors that influence this torque is vital for motor design and application.

Factors Affecting Maximum Torque:

\(\tau _{max}=\frac{3V^{2}R_{2}}{2w_{s}\left(\frac{{R_{2}}^{2}}{S}+X_{2}\right)}\)

The maximum torque in an induction motor is given by the following formula:

• Where:
  • V = Supply voltage
  • R_2 = Rotor resistance
  • X_2 = Rotor reactance
  • ω_s = Synchronous speed
  • s = Slip
• The formula indicates that both rotor resistance (R_2) and rotor reactance (X_2) play a crucial role in determining the maximum torque. By varying either R_2 or X_2, we can influence the maximum torque produced by the motor.

Role of Rotor Reactance (X_2):

• Rotor reactance (X_2) is a function of the rotor inductance and the frequency of the rotor current. It impacts the impedance of the rotor circuit.
• An increase in rotor reactance leads to a higher impedance, which reduces the rotor current and, consequently, the torque.
• Conversely, a decrease in rotor reactance reduces the impedance, leading to an increase in rotor current and torque.

Role of Rotor Resistance (R_2):

• Rotor resistance (R_2) directly influences the torque produced by the motor.
• Increasing the rotor resistance improves the starting torque by making the rotor impedance more resistive. However, it can also lead to higher losses and lower efficiency.
• Reducing the rotor resistance decreases the starting torque but improves the overall efficiency at running conditions.

Combined Effect:

• Both rotor resistance and reactance affect the maximum torque. By adjusting these parameters, designers can optimize motor performance for specific applications.

Torque equation:

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sV^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

Where Ns is the synchronous speed

V = supply voltage

R2 = rotor resistance

X2 = rotor reactance

s is the slip

By the above expression, we can say that the torque of an induction motor depends on rotor resistance and slip.

Condition for maximum torque:

The condition to get the maximum torque at starting is,

\({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

Xm = R­m

Where,

Rm = Motor resistance per phase

Xm = Motor reactance per phase

At the starting of the three-phase slip ring induction motor

Slip (s) = 1 (At the starting N_r = 0)

Therefore, 

⇒ X_m = R_m

Hence the starting torque of an induction motor is maximum when rotor resistance equals rotor reactance.

Concept:

In an induction motor

Full load torque (T_FL) is given by

​\({T_{FL}} = \frac{{{P_{out}\times60}}}{{2\pi {N_r}}}\)

No-load torque is given by

\({T_{NL}} = \frac{{{P_{in}\times60}}}{{2\pi {N_s}}}\)

Synchronous speed (N_S) is given by

\({N_s} = \frac{{120\;f}}{P}\)

Slip is given by

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)

Where,

P_out = Output power

P_in = Input power

N_r = Rotor speed in rpm

s = slip

f = supply frequency in Hz

P = Number of poles

Calculation:

Given –

P = 4, f = 50 Hz,

 s = 4%, P­_out = 15 kW

\({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

\(0.04 = \frac{{1500 - {N_r}}}{{1500}}\)

N_r = 1440 rpm

Now the output torque of the machine at full load can be calculated as

\({T_{FL}} = \frac{{15 \times {{10}^3\times 60}}}{{2\pi \times 1440}}\)

T_FL = 99.4718 Nm

Concept:

Torque developed in Induction Motor:

• In the Induction motor, the torque is proportional to the product of flux per stator pole and rotor current.
• In addition to current and flux, the power factor also comes into existence.
   

 And we know that, T ∝ ϕ₂I₂cosϕ₂

Or, T = kϕ₂I₂cosϕ₂

Where I₂ is rotor current at standstill

Φ₂ is the angle between rotor EMF and rotor current

k is a constant

We have, E₂ ∝ ϕ₂

⇒ T = k₁E₂I₂cosϕ₂ …. (1)

Where E₂ is rotor EMF at a standstill and k₁ is another constant

If R₂ is rotor resistance per phase and X₂ is rotor reactance per phase at standstill,

Therefore rotor impedance at standstill per phase is given as,

\({Z_2} = \sqrt {R_2^2 + X_2^2} \)

\({I_2} = {\frac{{{E_2}}}{Z_2}} = \frac{{{E_2}}}{{\sqrt {R_2^2 + X_2^2} }}\)

\(cos{\phi _2} = \frac{{{R_2}}}{{{Z_2}}} = \frac{{{R_2}}}{{\sqrt {R_2^2 + X_2^2} }}\)

From equation (1)

\(T = {T_{st}} = {k_1}\frac{{E_2^2{R_2}}}{{R_2^2 + X_2^2}}\)

If the Induction motor is under the running condition and run with slip s,

E_r = sE₂

And, X_r = sX₂

Under this condition torque given as,

\(T = {T_r} = {k_1}\frac{{sE_2^2{R_2}}}{{{{\left( {{R_2}} \right)}^2} + {{\left( {s{X_2}} \right)}^2}}}\)

For developing maximum running torque,

\(\frac{{dT}}{{ds}} = 0\)

s = s_m

The torque at any load in an induction motor is the function of slip.

After solving it,

R₂ = s_m X

And, value of current at maximum starting torque is given as,
\({I_m} = \frac{{{E_r}}}{{\sqrt {R_2^2 + {{\left( {s_m{X_2}} \right)}^2}} }} = \frac{{E_r}}{{\sqrt { {{2R_2}^2}} }} \)

Calculation:

Given,

R₂ = 0.1 Ω

X₂ = 0.5 Ω

\(s_m = \frac{{{R_2}}}{{{X_2}}} = \frac{{0.1}}{{0.5}} = 0.2\)

E₂ = 150 volt

E_r = 0.2 × 150 = 30 volt

For maximum torque,

R₂ = s_mX₂ = 0.1 Ω

\({I_m} = \frac{{{E_r}}}{{\sqrt {R_2^2 + {{\left( {s_m{X_2}} \right)}^2}} }} = \frac{{E_r}}{{\sqrt { {{2R_2}^2}} }} = \frac{{30}}{{\sqrt { {{2(0.1)}^2}} }} =150\sqrt 2 \;A\)

Torque equation:

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sV^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

Where Ns is the synchronous speed

V = supply voltage

R2 = rotor resistance

X2 = rotor reactance

s is the slip

By the above expression, we can say that the torque of an induction motor depends on rotor resistance and slip.

Condition for maximum torque:

The condition to get the maximum torque at starting is,

\({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

Xm = R­m

Where,

Rm = Motor resistance per phase

Xm = Motor reactance per phase

At the starting of the three-phase slip ring induction motor

Slip (s) = 1 (At the starting N_r = 0)

Therefore, \({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

⇒ X_m = R_m

Hence the starting torque of an induction motor is maximum when rotor resistance equals rotor reactance.

From the below diagram we can observe that the maximum torque of an induction motor is independent of rotor resistance but slip at which maximum torque occurs depends on rotor resistance and they change on adding the additional resistance to the rotor circuit.

The maximum torque of an induction motor is given by

\({T_{max}} = \frac{{kE_{20}^2}}{{2{X_{20}}}}\)

∴ Maximum torque is directly proportional to supply voltage & maximum torque is inversely proportional to rotor reactance.

Hence, the maximum torque is dependent on the supply voltage & reactance of the rotor and is independent of the rotor resistance.

Sm the value of slip corresponding to the maximum torque is

\({S_m} = \frac{{{R_2}}}{{{X_{20}}}}\)

Therefore, slip at which maximum torque occurs is directly proportional to rotor resistance R2.

Starting torque with auto-transformer is x2 times the starting torque with the direct switching.

Current drawn from the supply = x2 Isc

x is the tapping of auto-transformer

Concept

The ratio of maximum to full load torque is given by:

\({T_{fl}\over T_{max}}={2S_{max}S_{fl}\over S^2_{max}+S^2_{fl}}\)

The maximum slip is given by:

\(S_{max}={R_2\over X_2}\)

where, S_max = Maximum slip

R₂ = Rotor resistance

X₂ = Rotor reactance

The full load slip is given by:

\(S_{fl}={N_S-N_R\over N_S}\)

S_fl = Full load slip

N_s = Synchronous speed

N_r = Rotor speed at full load

Calculation

Given, R2 = 0.04 Ω and X2 = 0.3 Ω

f = 50 Hz, P = 8

\(N_S={120f\over P}\)

\(N_S={120\times 50\over 8}=750\space rpm\)

\(S_{fl}={750-720\over 750}=0.04\)

\(S_{max}={0.04\over 0.3}=0.133\)

The ratio of maximum to full load torque is given by:

\({T_{fl}\over T_{max}}={2(0.133)(0.04)\over (0.133)^2+(0.04)^2}\)

\({T_{fl}\over T_{max}}=0.55\)

\({T_{max}\over T_{fl}}=1.8\)

Concept:

The ratio between the starting torque to full load torque is given by:

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={{s}_{fl}}\times {{\left( \frac{{{I}_{st}}}{{{I}_{fl}}} \right)}^{2}}\)

Where Tst is the starting torque

Tfl is the full load torque

sfl is the slip at full load

Ist is the starting current

Ifl is the rated current

Calculation:

The starting current of a 3-ϕ induction motor is 3 times the rated current. i.e.

Ist = 5 Ifl

Full load slip (sfl) = 4 %

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={0.04}\times {{\left( \frac{{5}}{{1}} \right)}^{2}}\)

\(= 0.04 \times {\left( {\frac{5}{1}} \right)^2} = 1.0\)

The correct answer is option 3):(325 rpm)

Concept:

Form the below diagram we can observe that the maximum torque of an induction motor is independent of rotor resistance but slip at which maximum torque occurs depends on rotor resistance and they change on adding the additional resistance to the rotor circuit

Sm the value of slip corresponding to the maximum torque is Sm=R2X20 Therefore, slip at which maximum torque occurs is directly proportional to rotor resistance R2.

Slip at maximum torque S_m=R2/X2, 

The speed at which maximum torque occurs is speed corresponding to, N = N_s (1 – s_m ) 

Calculation:

Given 

f = 50 Hz

P = 16

N_s = \(120 \times f \over P\)

= 375

Z = (0.02 +j 0.15) Ω 

S_m = ( \(0.02 \over 0.15\))

= 0.1333

N = Ns (1 – sm ) 

= 375(1- 0.133)

= 375 × 0.867

= 325 RPM

Concept:

Torque of induction motor is

\(\rm T=\frac{3}{\omega_s}.\frac{V^2}{\left(\frac{r_2}{s}\right)^2+x^2_2}×\frac{r_2}{s}\)

i.e, Torque is directly proportional to square of the voltage

∴ T ∝ V²

Calculation:

Case 1:

Torque = T₁

Voltage = V₁

Case 2:

Reduction in supply voltage by 10%

Torque = T₂

Voltage = V₂ = 0.9 V₁

\(\frac{T_2}{T_1}=\frac{V_2^2}{V_1^2}\)

\(\frac{T_2}{T_1}=\frac{{(0.9V_1)}^2}{{V_1}^2} = 0.81\)

T₂ = 0.81 T₁

% Change in T =  \(\frac{T_1-0.81T_1}{T_1}\times 100 \%= 19\%\)

Therefore, the correct answer is an option (2)

Torque in the 3-phase induction motor

The torque of a 3-phase induction motor is given by:

\(T={3\over ω_s}{V^2\over ({R_2\over s}^2)+X_2^2}{R_2\over s}\)

where, V = Supply

ω_s = Synchronous speed in radian/sec

R₂ = Rotor resistance

X₂ = Rotor reactance

s = Slip

Condition for maximum torque

The maximum torque of a 3-phase induction motor occurs at:

\(s={R_2\over X_2}\)

If the maximum torque has to occur in starting, the value of the slip (s)=1

\(1={R_2\over X_2}\)

\(R_2=X_2\)

∴ The maximum starting torque of a 3-phase induction motor occurs when the rotor resistance is equal to the rotor reactance.

The maximum torque at the start is given by:

\(T={3\over 2ω_s}{V^2\over X_2}\)

From the above expression, the maximum torque is independent of the rotor resistance but the slip at which maximum torque occurs depends upon the rotor resistance.