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

Concept

The synchronous speed of an induction motor is given by:

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

where, N_s = Synchronous speed

f = Frequency

P = No. of poles

Calculation

Given, P = 6

f = 60 Hz

\(N_s={120\times 60\over 6}\)

Ns = 1200 rpm

Explanation:

Slip in Three-Phase Induction Motors

Definition: Slip in a three-phase induction motor is defined as the difference between the synchronous speed and the rotor speed, expressed as a percentage of the synchronous speed. Mathematically, slip (S) can be calculated using the formula:

S = [(Ns - Nr)/Ns] × 100

Where:

• Ns: Synchronous speed of the motor (in RPM).
• Nr: Rotor speed of the motor (in RPM).

The synchronous speed is determined by the supply frequency and the number of poles in the motor, using the formula:

Ns = (120 × f)/P

Where:

• f: Supply frequency (in Hz).
• P: Number of poles in the motor.

Slip is an essential parameter in induction motors as it determines the relative speed between the stator magnetic field and the rotor. A small slip ensures efficient operation of the motor while allowing torque generation.

Typical Value of Slip:

For a three-phase induction motor operating under full load conditions, the slip is typically very small, ranging between 0.05 (5%) and 0.1 (10%). This ensures the motor operates close to its synchronous speed while maintaining sufficient torque to drive the load.

Correct Option Analysis:

The correct option is:

Option 1: 0.05

The usual value of slip for a three-phase induction motor at full load is about 0.05 (5%). This small slip value indicates that the rotor speed is very close to the synchronous speed, which is desirable for efficient motor operation. At full load, the rotor speed slightly lags the synchronous speed, allowing current induction in the rotor, which generates torque to drive the load.

Why Slip Occurs:

Slip occurs in induction motors because the rotor must move slower than the synchronous speed to allow relative motion between the rotating magnetic field of the stator and the rotor conductors. This relative motion induces current in the rotor conductors, which interacts with the stator field to produce torque. Without slip, there would be no induced current, and thus no torque generation.

Factors Affecting Slip:

• Load on the motor: As the load increases, the slip increases because more torque is required to overcome the load.
• Rotor resistance: Higher rotor resistance generally results in higher slip values.
• Supply frequency and voltage: Variations in supply frequency and voltage can impact the synchronous speed and slip.

Advantages of Small Slip:

• Ensures high efficiency as the rotor operates close to the synchronous speed.
• Provides stable and consistent torque output under varying load conditions.
• Minimizes losses due to relative motion between the stator and rotor fields.

Important Information

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

Option 2: 0.1

A slip value of 0.1 (10%) is higher than the typical range for full-load operation in standard three-phase induction motors. While some motors may operate with a slip value of 10% under specific conditions, such as high rotor resistance or low efficiency designs, it is not representative of the usual value for standard motors operating at full load. Generally, efficient motors have slip values closer to 5%.

Option 3: 0.3

A slip value of 0.3 (30%) is significantly higher than the typical range for three-phase induction motors under full load. Such high slip values are usually indicative of either a malfunctioning motor or a motor designed for specific applications where high slip is acceptable, such as certain types of wound rotor motors or applications requiring variable speed. In standard induction motors, a slip of 30% would lead to substantial losses and inefficiency.

Option 4: 0.8

A slip value of 0.8 (80%) is far outside the normal operating range for three-phase induction motors. At such high slip values, the rotor speed would be extremely low compared to the synchronous speed, resulting in poor efficiency and substantial energy losses. This condition may occur in cases of rotor stalling or severe overload, but it is not representative of typical motor operation under full load conditions.

Conclusion:

Slip is a critical parameter in the operation of three-phase induction motors, allowing torque generation while ensuring efficient operation. Under full load conditions, the usual value of slip for standard motors is about 0.05 (5%), as described in option 1. This small slip ensures the rotor operates close to the synchronous speed, minimizing losses and providing stable torque output. Understanding the factors affecting slip and its implications is essential for selecting and operating induction motors effectively.

Concept

The slip of an induction motor is given by:

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

where, s = Slip

N_s = Synchronous speed

N_r = Rotor speed

Calculation

When the phase sequence is reversed, the synchronous speed reverses direction instantaneously, while the rotor continues to rotate in the same direction momentarily due to inertia.

Before reversal:

s = 0.05

N_r = (1-0.05)N_s

After reversal:

The new synchronous speed becomes -N_s (opposite direction). The rotor continues at 0.95N_s (same direction as before)

The slip in the reversed case is:

\(s'={N_s-(-0.95N_s)\over N_s}\)

\(s'=1.95\)

Concept

For a constant load torque, the slip varies inversely with the square of the voltage.

s α V

\({s_2\over s_1}=({V_1\over V_2})^2\)

Calculation

Given, s₁ = 0.06

V₁ = 400 V

V₂ = 200 V

\({s_2\over 0.06}=({400\over 200})^2\)

s₂ = 0.06 × 4

s2 = 0.24

Concept

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

\(f_r=sf_s\)

where, fr = Rotor induced frequency

fs = Supply frequency

s = Slip

The slip 's' is given by:

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

Ns = Synchronous speed

Nr = Rotor speed

Calculation

Given, fr = 90 oscillations per minute = 90 oscillations per 60 sec = 1.5 Hz

fs = 50 Hz

\(1.5=s\times 50\)

s = 0.03

\(0.03={1000-N_r\over 1000}\)

N_r = 970 rpm

Concept:

Slip in Induction Motor (s)

\(s=\frac{N_s\ -\ N_r}{N_s}\)   ---(1)

Where,

N_s is the stator frequency

N_r is the rotor frequency

Also, fr = s fs     ---(2)

Where,

fr is the rotor frequency

fs is the supply frequency

Explanation:

The rotor is locked means N_r = 0

So, from equation (1), s = 1

From equation (2),

fr = fs

Therefore, the rotor frequency of the induction motor = supply frequency.

• Induction motor is a type of electric motor in which alternating current from a power source is fed through a primary winding and induces a current in a secondary winding, with the parts arranged so that the resulting magnetic field causes a movable rotor to rotate with respect to a fixed stator.
• The torque-slip characteristics is represented by a rectangular hyperbola.
• For the immediate value of slip, the graph changes from one form to another.
• The torque equation of the induction motor is:

\(T = \frac{{Ks{R_2}E_{20}^2}}{{R_2^2 + {{\left( {s{X_{20}}} \right)}^2}}}\)

The torque slip characteristic curve is divided into three regions:

• Low slip region
• Medium slip region
• High slip region

Low slip region: (Near full load)

At synchronous speed, slip = 0, therefore the torque is zero.

When there is a light load, the speed is very near to synchronous speed.

The slip is very low and (sX₂₀)² is negligible in comparison with R₂. Therefore

\(T = \frac{{{K_1}s}}{{{R_2}}}\)

i.e. T ∝ S

High slip region: (After starting to maximum load)

As the slip increases, the speed of the motor decreases with the increase in load.

The term (sX₂₀)² becomes large.

The term R²₂ may be neglected in comparison with the term (sx₂₀)² and the torque equation becomes

\(T = \frac{{{K_3}{R_2}}}{{sX_{20}^2}}\)

i.e. \(T \propto \frac{1}{s}\) ​

Hence for heavy loads, torque is inversely proportional to slip.

Concept:

Rotor Resistance Control:

For induction motor, the torque in rotor resistance control methods is given as

\({\rm{{\rm T}}} = \frac{{{K_t}s}}{{{R_2}}}\)

where

s = slip

R₂ = Rotor Resistance

R₂¹ = Rotor resistance become three times = 3R₂

For full load torque \(s\; \propto \;{R_2}\)

Calculation:

Given

s₁ = 0.04

\(\begin{array}{l} {s_2} = \frac{{{R_2}{^1} }}{{{R_2}}} \times {s_1}\\ = \frac{3R_2}{R_2} \times 0.04 \end{array}\)

Slip (s):

In the induction motor, the rotor rotates with a speed of N slightly less than the synchronous speed (Ns). It slips the speed by a value is called slip speed.

Formula:

The slip speed can be calculated as

s = Ns - N

The percentage of the slip is calculated as 

\(s = \frac{N_s - N}{N_s}\times100\)

The rotor frequency is f' = s f.

f = Supply frequency

Calculation:

Given that,

Supply frequency f = 50 Hz

100 pulsations per minute i.e., (100 / 60) = 1.667 cycles /sec.

∴ The frequency of the rotor is

f' = 1.667 cycle per second = 1.667 Hz

The rotor frequency is f' = s f

⇒ f' = s x 50

⇒ s = 1.667 / 50 

∴ s = 0.0333

The slip of the motor is 3.33%.

Slip (s):

In the induction motor, the rotor rotates with a speed of N slightly less than the synchronous speed (N_s). It slips the speed by a value is called slip speed.

Formula:

The slip speed can be calculated as

s = N_s - N

The percentage of the slip is calculated as 

\(s = \frac{N_s - N}{N_s}\times100\)

The rotor frequency is f' = s f.

f = Supply frequency

Calculation:

Given that,

Supply frequency f = 50 Hz

At no-load, a voltmeter is connected on the rotor side. 

120 pulsations per minute i.e., (120 / 60) = 2 cycles /sec.

∴ The frequency of the rotor is

f' = 2 cycle per second = 2 Hz

The rotor frequency is f' = s f

⇒ f' = s x 50

⇒ s = 2 / 50 

∴ s = 0.04

The slip of the motor is 4%.

Reasons for synchronous motor is a better power factor as compared to that of an equivalent induction motor:

1. Synchronous motors can achieve efficiencies of >90% in some cases and are generally more energy-efficient than induction motors.

2. Synchronous machines has separate DC excitation which reduces machine's excitation dependency on main supply, hence it operates with better power factor

3. In the case of induction motors, a rotor is not supplied by dc supply, so total flux can be gained by stator itself, therefore, stator absorbs reactive power constantly and it has a low power factor

4. Higher PF means low requirement of MMF for energy transfer, hence low magnetizing current requirement.
5. Synchronous machines have separate DC excitation which reduces the machine's excitation dependency on the main supply, hence better PF.
whereas IM has no such provisions, hence low PF.

6. Power-factor is the ratio of working power to apparent power and is given as a percentage to show the efficiency of power distribution and its associated losses.

Additional Information

Difference between Three Phase Induction Motor and Synchronous Motor:

1. A three-phase synchronous motor is a doubly excited machine, whereas an induction motor is a single excited machine.
2. The armature winding of the synchronous motor is energized from an AC source and its field winding from a DC source. The stator winding of Induction Motor is energized from an AC source.
3. Synchronous Motor always runs at synchronous speed, and the speed of the motor is independent of load, but an induction motor always runs less than the synchronous speed. If the load increased the speed of the induction motor decreases.
4. The induction motor has self-starting torque whereas the synchronous motor is not self-starting. It has to be run up to synchronous speed by any means before it can be synchronized to AC supply.
5. A synchronous motor can be operated with lagging and leading power by changing its excitation. An induction motor operates only at a lagging power factor. At high loads, the power factor of the induction motor becomes very poor.
6. The Synchronous Motor can be used for power factor correction in addition to supplying torque to drive mechanical loads whereas an induction motor is used for driving mechanical loads only.
7. The synchronous motor is more efficient than an induction motor of the same output and voltage rating.
8. A synchronous motor is costlier than an induction motor of the same output and voltage rating.

The torque slip characteristics are shown in the figure below.

From the given diagram in the question,

• Point A represents the operation at a slip greater than 1.
• Points B and C represent the operation at a slip between 0 to 1.
• Point D represents the operation at a negative slip.

Concept:

The torque equation of the three-phase induction motor is given as

\(T =\frac{3\times E_{2}\times I_{2}\times Cos\phi}{\omega } \)

\(T = \frac{3}{{2\pi {N_s}}}\frac{{sE_2^2}}{{R_2^2 + {{(s{X_2})}^2}}}\)

Rotor current \({I_2} = \frac{{s{E_2}}}{{{Z_2}}}\)

Ns = speed of the motor

s = slip of the motor

R2 = rotor resistance

X2 = rotor reactance

E2 = rotor-induced emf

\(Cos\phi = Power factor\)

Hence it's independent of the type of machine.

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,

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

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

Concept:

Rotor resistance method of speed control:

Under load condition torque approximately

T ∝ (sV12) / (R2 + Re)

• In this method, some external resistance is inserted under the load conditions.
• Then the slip of the induction motor increases to maintain the load torque constant.
• As slip is increased, the speed of the motor will be reduced to below the rated speed.
• In this method, the motor acts as a constant torque variable power drive.

For torque constant(T = k)

\(\frac{{{{\bf{s}}_1}}}{{{{\bf{R}}_2}}} = \frac{{{{\bf{s}}_2}}}{{{{\bf{R}}_2} + {{\bf{R}}_{\bf{e}}}}}\)

Calculation:

For induction motor, the torque in rotor resistance control methods is given as

\({\rm{{\rm T}}} = \frac{{{K_t}s}}{{{R_2}}}\)

For full load torque \(s\; \propto \;{R_2}\)

\(\begin{array}{l} {s_2} = \frac{{{R_2} + {R_{ext}}}}{{{R_2}}} \times {s_1}\\ = \frac{{0.45 + 0.5}}{{0.45}} \times 0.05 \end{array}\)