Three Phase Induction Motor MCQ Quiz - Objective Question with Answer for Three Phase Induction Motor - 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

Crawling in an induction motor

• Crawling typically causes an induction motor to operate at a fraction (e.g., 1/3) of synchronous speed.
• Crawling is a phenomenon in squirrel-cage induction motors caused by harmonics in the supply voltage or due to the motor's design.
• This results in the motor getting "stuck" at around (1/3)^th of the synchronous speed instead of accelerating to near synchronous speed.
• It is commonly observed in motors with poor design or under heavy load during startup.
• Crawling is often accompanied by high starting currents and can produce noticeable noise, further impacting the motor's operation and potentially damaging the surrounding environment. 

Concept:

In an induction motor, the ratio of full load torque to maximum torque (pull-out torque) is influenced by rotor resistance. The torque equation for an induction motor is given by:

\( T \propto \frac{R_2}{s} \)

where:

• \(R_2\) = Rotor resistance
• \(s\) = Slip of the motor

Increasing the rotor resistance shifts the maximum torque point to a higher slip, closer to the full load operating point. This increases the ratio of full load torque to maximum torque, making the motor more efficient at handling varying loads.

Calculation:

- Increasing rotor resistance shifts the maximum torque to a higher slip region.
- This helps to bring the maximum torque point closer to the operating point of the motor.
- As a result, the ratio of full load torque to maximum torque improves.

Final Answer: Increasing rotor resistance increases the ratio of full load torque to maximum torque in an induction motor.

Equivalent circuit of a 3ϕ induction motor

The total resistance at the rotor is represented as \(\rm \frac{R_2}{s}\) where s is the slip.

Now, if we create an equivalent transformer circuit for an induction motor, the secondary resistance will be R₂. Thus, the load resistance will be \(\rm \frac{R_2(1-s)}{s}\).

Induction motor modelled as a transformer:

When all the rotor parameters are shifted to the stator side induction motor circuit is given by:

\(\rm \frac{R_2(1-s)}{s}\) is the resistance that represents mechanical load in the equivalent circuit.

Slip ring induction motors are designed to provide high starting torque with low starting current by adding external resistance in the rotor circuit during startup. This makes them ideal for applications requiring heavy starting loads like cranes, hoists, and elevators.

High efficiency at all loads → Not true; not as efficient as squirrel cage motors.

Low starting torque and high starting current → Opposite of the actual characteristic.

Operates only at synchronous speed → Incorrect; it operates below synchronous speed like all induction motors.

High starting torque and low starting current → Correct.

Rotor emf injection method:

For below-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is 180° out of phase with rotor emf.

E_2R is resultant emf in the rotor

E_2R = E₂ – E₁

\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)

R₂ is rotor resistance

T is the torque

s is the slip

Here, the value of rotor emf becomes less. To maintain constant torque, the value of slip will increase. Therefore, the speed will be decreased.

At this condition, effective rotor resistance increases.

For above-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is in phase with rotor emf.

E_2R is resultant emf in the rotor

E_2R = E₂ + E₁

\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)

R₂ is rotor resistance

T is the torque

s is the slip

Here, the value of rotor emf becomes more. To maintain constant torque, the value of slip will decrease. Therefore, the speed will be increased.

At this condition, effective rotor resistance decreases.

Induction generator always works with leading power factor since it will take large amount of reactive power to produce sufficient amount of working flux so that armature reaction is always magnetizing hence it will work always with leading pf.

Important Point

• induction generator is basically an induction motor, which runs above the synchronous speed
• when it acts as a generator it will supply the active power back to source, but for this supply of active power it needs reactive power as input to keep its winding excited
• in case if the induction motor is connected to the grid, it will draw the required reactive power for the excitation of windings, but if it is standalone system (ie. not connected to grid) then a capacitor bank will be always connected, and this will provide leading reactive power to keep the winding excited for the process of mechanical to electrical energy conversion
• since the reactive power is supplied by the capacitor, the induction generator is operating in leading power factor

Rheostat:

A rheostat is a type of variable resistor, whose resistance can be changed for varying the amount of electric current flowing through an electrical circuit. The word rheostat is made of two words (‘rheo’ meaning flow of current in Greek and ‘stat’ meaning stationary instrument). When placed in an electric circuit, the flow of electricity changed through two terminals: one terminal near the slider/adjustable contact and the other connected near the bottom.

A rheostat is internationally denoted by the following symbol:

The rheostat is generally used in applications where high voltage or current is required such as:

1. Changing the light intensity of a light bulb. An increase in the resistance of the rheostat decreases the flow of electric current, leading to the dimming of lights and vice-versa.
2. Generators
3. Motor speed
4. Heater and oven temperature control
5. Volume control

Additional Information

Some important electronic components symbols given below-

Magnetic reactance (X_m) is depends on airgap flux and the flux is depends on V/f.

X_m ∝ ϕ ∝ V/f

Hence, magnetizing reactance gets affected by reducing the rms value of the supply at the rated frequency.

Additional Information

The total resistance at rotor is represented as \(\frac{{{r_2}}}{s}\), where \(s\) is slip. Now, if we create an equivalent transformer circuit for induction motor, the secondary resistance will be \({r_2}\). Thus the load resistance will be \({r_2}\left( {\frac{1}{s} - 1} \right)\).

Induction motor modelled as a transformer

When all the rotor parameters are shifted to stator side induction motor circuit is given by

So, \(r_{2}^{'}\left( \frac{1}{s}-1 \right)\) is the resistance which shows the power which is converted to mechanical power output or useful power.

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.

Concept:

When 3-ϕ induction machine working as an induction generator, then 

Slip (s) = \(= \frac{{\left( { - {N_s} + {N_r}} \right)}}{{{N_s}}}\)

Where, 

N_s = Synchronous speed

N_r = Rotor speed

Frequency of rotor current = s × f

Where s is the slip

f is the supply frequency

Calculation:

Given that,

Supply frequency (f_s) = 60 Hz

Rotor current frequency (f_r) = 5 Hz

Number of poles = 4

Synchronous speed, \({{\rm{N}}_{\rm{s}}} = \frac{{120{{\rm{f}}_{\rm{s}}}}}{{\rm{P}}}\) 

\(= \frac{{120 × 60}}{4} = 1800\)

We know that,

f_r = (s) f_s

⇒ 5 = (s) (60)

\(\Rightarrow s = \frac{1}{{12}}\)

To work as an induction generator, rotor speed should be slip speed greater than synchronous speed, therefore 

\(\Rightarrow \frac{1}{{12}} = \frac{{ - 1800 + {N_r}}}{{1800}}\)

In a three phase induction motor,

a) The speed of stator is zero

b) The speed of stator magnetic field is synchronous speed

c) The speed of rotor magnetic field is synchronous speed

d) The speed of rotor is rotating speed

Power Stages in an Induction Motor:

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and Pcu is rotor copper loss.

Calculation:

Given,

s = 4% = 0.04

P₂ = 1000 W

From above concept,

P₂ : P_m = 1 : (1 - s)

1000 : P_m = 1 : (1 - 0.04)

1000 : P_m = 1 : 0.96

\(\frac{1000}{P_m}=\frac{1}{0.96}\)

P_m = 1000 × 0.96 = 960 W

• 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.

Slots in Induction motor:

• The speed of the induction motor is inversely proportional to the load torque. In semi-closed and closed slots, the air gap between the stator and rotor is small as compared to open slots. As the air gap is small, the requirement of magnetizing current to establish the flux in the air gap is less.
• It results in improved power factor In order to semi-closed slots or totally closed slots are used in induction motors.
• Among all the three types of slots, semi-closed type slots are preferred for induction machines as semi-closed slots having the partial advantages of open type and partial advantages of closed type slots.
• Open-type slots are generally preferred for synchronous and dc machines.
• In general, closed type slots are used in low hp motors, to control the starting current, as the leakage reactance offered by closed type slots is very high compared to other types of slots.
• Large size induction motors use open slots so that already prepared and properly insulated coils can be easily inserted in open slots.
• In order to improve the power factor semi-closed slots or totally closed slots are used in induction motors.