THREE PHASE INDUCTION MOTOR

1. If the frequency of 3-phase supply to the stator of a 3-phase induction motor is increased, then synchronous speed

1.  Increased✓

2.  Decreased

3.  Remain unchanged

4.  None of the above

 

 

Synchronous speed of the 3 phase induction motor is

Ns = 120f/p

From the above equation, it is clear that the synchronous speed is directly proportional to the applied frequency. I.e

Ns ∝ f

Hence by increasing the frequency, the synchronous speed of the 3 phase induction motor can be increased.

 

2.  A 3 phase induction motor is essentially a

1.  Constant speed motor✓

2.  Variable speed motor

3.  Very costly

4.  Easily maintainable

 

The induction motor is not a perfectly constant speed motor. At no load, the rotor lags behind the stator flux only a small amount since the only torque required is that needed to overcome small no-load losses. Hence it rotates at almost the synchronous speed (120 f/p rpm).

As load increases, it needs to generate torque to drive the load.as the mechanical load is added the rotor speed decreases. A decrease in rotor speed allows the constant speed rotating field to sweep across the rotor conductors at a faster rate, thereby inducing larger rotor current (since rotor impedance is low). This results in a large increase in torque which tends to bring the speed to the original value.

Although the motor speed does decrees slightly with increased load. Normally, maximum torque is produced at 5% slip, i.e. 95% of synchronous speed.the speed regulation of the induction motor is very close to the synchronous speed thus we can say that an induction motor is a nearly constant speed motor.

3.   If there are no copper losses in the rotor, then

1.  Rotor will not run

2.  Rotor will run at a very low speed

3.  Rotor will run at a very high speed

4.  Rotor will run at synchronous speed✓

 

Rotor gross output = rotor input − rotor copper loss ————- 1

Where rotor output = tg × 2πn/60

Rotor input = tg × 2πn_s/60

Where tg = gross torque

Ns = synchronous speed

N = rotor speed

When there is no copper loss then from equation 1

Rotor output = rotor input

Tg × 2πn/60 = tg × 2πn_s/60

N = ns

Hence if there are no copper losses in the rotor, then rotor output will become equal to rotor input and the rotor will run at synchronous speed.

4.  When the rotor of  three phase induction motor is blocked then the slip is

1.  0

2.  0.5

3.  0.1

4.  1✓

 

Since the rotor is blocked therefore it can’t rotate hence rotating speed nr = 0

Slip of an induction motor is given as

S = (ns – nr)/ns

S = (ns – 0)/ ns = ns/ns

S = 1

 

5.  The full load slip of an induction motor varies from

1.  10 – 20%

2.  100%

3.  50%

4.  2 – 7%✓

 

 “ an induction motor can be termed as a constant speed motor and the constant speed can only be achieved if the rotor speed is closed to the synchronous speed. Therefore a general-purpose induction motor is designed to operate in the low slip at full load in order to have good running performance. Depending on the rating, full load slip varies from 2 to 1%. Such a motor has high starting current (5-8 times) and low starting torque whereas for high hp motor, the slip varies from 3 – 7%.

 

6.  An 8 pole, three phase induction motor is supplied from 50 hz, a.c. supply. On full load, the frequency of induced  emf in the rotor is 2 hz. Then the full load slip and the corresponding speed.

1.  4% & 750

2.  4% & 720✓

3.  5% & 1000

4.  5% & 1500

 

Given value
p = 8 pole
f = 50 hz
f_r = 2 hz

Rotor frequency is given as
fr = sf
s = 2/50
s = 0.04
%s = 0.04 x 100 = 4% ——-full load slip

Now the corresponding speed i.e rotor speed

N = ns(1 – s)

Where ns = 120f/p = 120 x 50/8 = 750 r.p.m

N = 750(1 – 0.04)

N = 720 r.p.m

 

7.   A 4 pole, 3 phase, 50 hz induction motor runs at a speed of 1470 r.p.m. speed. Then the frequency of the induced e.m.f. in the rotor under this condition.

1.  1 hz✓

2.  2 hz

3.  5 hz

4.  6 hz

 

Given
p = 4
f = 50 hz
n = 1470 rpm

Ns = 120f/p = 120 x 50/4 = 1500 rpm

Slip of an induction motor

S = (ns – nr)/ns

= (1500 – 1470)/1500 = 0.02

Now rotor frequency f_r

F_r = sf = 0.02 x 50 = 1 hz

 

8.  For a 4 pole, 3 phase, 50 hz induction motor ratio of staler to rotor turns is 2. On a certain load, its speed is observed to be 1455 r.p.m. when connected to 415 v supply. Then the value of

I) the frequency of rotor e.m.f. in running condition.
Ii) the magnitude of induced e.m.f. in the rotor at standstill.
Iii) the magnitude of induced e.m.f. in the rotor under running condition. Assume star-connected stator.

1.  (i) → 2 hz (ii) →149.8 v (iii) → 11.5 v

2.  (i) → 1.5 hz (ii) →119.8 v (iii) → 3.594 v✓

3.  (i) → 1.5 hz (ii) → 3.594 v (iii) → 119.8 v

4.  (i) → 1.5 hz (ii) →20 v (iii) → 3.594 v

Given value
p = 4
f = 50 hz
n = 1455 rpm
e_1line = 415 v
turn ratio k =  rotor turns/stator turns = 1/2 = 0.5

Ns = 120f/p = 120 x 50/4 = 1500 rpm

Slip of an induction motor is given as

S = (ns – nr)/ns

= (1500 – 1455)/1500 = 0.03

(i)  fr = sf = 0.03 x50 = 1.5 hz

(ii) at standstill indcution motor acts as a transformer so

E2phe1ph=rotor turnstator turn =k

Where e_1ph = stator emf per phase in volts

E_2ph = rotor induced e.m.f per phase in volts at the start when the motor is at standstill.

But the ratio of the stator to rotor turns is given as 2, i.e

E2phe1ph=n2n1=12=k

Now the value of line voltage e_line = 415 v

In star connected the line voltage e_line = √3 × e_ph

E1ph=eline3=4153=239.6 volte2phe1ph=12e2ph=12×239.6e2ph=119.8 volt

(iii) in running condition

E_2r = se₂

Where e_2r = rotor induced emf per phase in running condition

E₂ = rotor induced emf per phase on standstill condition

E_2r = se₂ = 0.03 x 119.8 = 3.594 v

 

9.  The direction of rotation of 3-φ induction motor is clockwise when it is supplied with the 3-φ sinusoidal voltage having phase sequence of a-b-c. For counterclockwise rotation, the phase sequence of the power supply should be

1.  B-c-a

2.  A-c-b✓

3.  C-a-b

4.  Any of the above

The direction of rotation of a 3 phase induction motor can be reversed by interchanging any two of the three motor supply lines.

The phase sequence of the three-phase voltage applied to the stator winding is a-b-c. If this sequence is changed to a-c-b, it is observed that direction of rotation of the field is reversed i.e., the field rotates counterclockwise rather than clockwise. However, the number of poles and the speed at which the magnetic field rotates remain unchanged.

 

10.  The rotor winding of a 3-phase wound rotor induction motor is generally …………. Connected.

1.  Delta

2.  Star✓

3.  Partially delta & partially star

4.  None of the above

 

The stator can be either star or delta. Many motors have a winding scheme and coils with six connections that allow the windings to be connected in either star or delta. The rotor of a squirrel-cage induction motor has the number of rotor bars connected at both ends. That is essentially a delta connection because it is the only way the rotor bars can be connected to form a closed circuit. On the other hand, a wound rotor always has a star connection because that allows the external connection of resistors in series with the windings.

The advantages of star connection is as follows

1.  The phase voltages in star connection are 57.7 % of the line voltages, i.e. the armature winding in star connection is less exposed to voltage as compared to the delta connection which in turn prove more economic if we consider insulation, breakdown strength, the requirement of conductor material etc.

2.  Availability of neutral in the star connection, if the neutral is grounded then it also provides a path for the zero-sequence currents during faults, whereas in the delta connection the zero sequence currents flow within the delta circuit and hence increasing the load on the winding.

3.  Star connection reduces the number of slip rings required to connect the external resistance to the rotor of the induction motor.with star connection, only 3 slip rings are required.whereas in delta connected rotor 6 slip rings are required (2 for each phase).

4.  Vline= √3 vph or vph=vline/√3 in star connected

I.e. vph is reduced to √3 times the line voltage in a star connected system

Now voltage is related to insulation, and insulation is related to cost i.e higher the voltage, higher the insulation and higher the cost.

Since as told vph in a star is less, so the insulation in the conductor is less and hence the cost is reduced.

5.  In star connection, additional external resistance may be inserted in the rotor circuit at starting to increase the starting torque and decrease the starting current. As the motor gains speed, these external resistances are cut out of the rotor circuit.

 

11.  _________ are employed for the operation of jaw crushers

1.  Dc shunt wound motor

2.  Squirrel cage induction motor

3.  Belted slip ring induction motor✓

4.  Any dc motor

• Starting torque is directly proportional to the rotor resistance of an induction motor.
• In slip ring induction motor the ends of the rotor windings are externally connected by a variable rheostat (resistance is varied in order to give it proper starting and running current).
• So more the resistance, more the torque. When we add resistance to the rotor the torque is high, the slip is high and the current is reduced.
• Therefore slip ring induction motor is used for providing high starting torque.
• Belted slip ring induction motor is almost invariably used as, very often, the motor has to start against the heavy load or a stuck crusher.

 

12.. If ns and n are the speeds of rotating field and rotor respectively, then ratio rotor input/rotor output is equal to.

1.  N/ns✓

2.  Ns/n

3.  Ns − n

4.  N − ns

 

13. . At no-load, the iron loss of a 3-phase induction motor is

1.  Practically zero✓

2.  Large

3.  Small

4.  None of the above

 

As there is no load on induction motor, the rotor runs at a speed close to synchronous speed. Synchronous speed is the speed of the rotating magnetic field. Remember that the rotor speed is close to synchronous speed but never equal to it according to lenz law. So when the rotor runs very close to synchronous speed, the slip is very small approximately equal to zero. The frequency of rotor in running condition is given as sf. So rotor frequency is very very small. Since iron losses depend on the square of the frequency so they are almost negligible and therefore not considered.

14. . For higher efficiency of 3-phase induction motor, the slip should be

1.  Unity

2.  Large

3.  As small as possible✓

4.  Very large

 

An induction motor can be termed as a constant speed motor and the constant speed can only be achieved if the rotor speed is closed to the synchronous speed. Therefore a general-purpose induction motor is designed to operate in the low slip at full load in order to have good running performance. Depending on the rating, full load slip varies from 2 to 1%. Such a motor has high starting current (5-8 times) and low starting torque whereas, for high hp motor, the slip varies from 3-7%

Approximate efficiency of induction motor = (1 − s)

So it is clear that for the higher efficiency of 3 phase induction motor the slip should be as small as possible.

 

15.  A 3-phase induction motor is running at 2% slip. If the input to rotor is 1000 w, then mechanical power developed by the motor is

1.  500 w

2.  200 w

3.  20 w

4.  980 w✓

 

Mechanical power developed in 3-phase motor = (1 − s) × power input to rotor

= (1 – 0.02) x 1000 = 980 w

16.  The approximate efficiency of a 3-phase, 50 hz, 4-pole induction motor running at 1350 r.p.m. is

1.  90%✓

2.  60%

3.  45%

4.  100%

 

Given
p = 4
f = 50 hz
n = 1350 rpm

Ns = 120f/p = 120 × 50/4 = 1500 rpm

Slip s = (ns – n)/ns

= (1500 – 1350)/1500 = 0.1

Approximate efficiency of induction motor = (1 – s)

1 – 0.1 = 0.9 = 90%

17.  If the air gap between the rotor and stator of a 3-phase induction motor is increased, then

1.  No-load current is increased

2.  Leakage reactance is decreased

3.  Leakage reactance is increased✓

4.  Any of the above

 

If the air gap of an induction motor is increased, the following will happen:

• The permeability of the magnetic circuit rotor-to-stator will decrease.
• The magnetizing inductance of the motor thus decreases.
• The magnetizing current will increase. This will cause a poorer power factor at all loads.
• The magnetic flux in the air gap will decrease and leakage fluxes will increase. This will cause a reduction in the maximum available torque.

In summary, the maximum available torque will decrease, the power factor will worsen and the motor will run with increased slip.

So it is always good performance-wise to run with as small an air gap as possible, which will reverse all of these effects. But if the air gap is too small, rotor cooling is compromised, and if the rotor expands through overheating (e.g. by exceeding the recommended maximum number of starts per hour) the rotor can “pole” by rubbing or jamming with the stator.

18.  The conditions of an induction motor at no-load resemble those of a transformer whose secondary is

1.  Short-circuited

2.  Open-circuited✓

3.  Supplying a variable resistive load

4.  None of the above

 

An induction motor can resemble as a transformer.

• In a transformer, the ac voltage is supplied to the primary winding. This voltage produces current in the primary winding which sets up flux φ in the iron core. This flux links both the primary and the secondary winding. As the flux is alternating in nature, an emf is induced in the secondary winding. Hence, there is no electrical connection between the primary and secondary windings of a transformer. The secondary windings receive energy from the primary by electromagnetic induction.
• Similarly, in an induction motor, the three-phase ac supply is given to the stator winding which sets up a rotating magnetic field. This rotating magnetic field induces emf in the rotor conductors. Hence, the rotor receives energy from the stator by electromagnetic induction.
• The primary and secondary winding of a transformer resembles the stator and rotor winding of an induction motor. The resistance and leakage reactance of the stator and rotor winding of an induction motor is similar to the primary and secondary winding of a transformer.
• The operation of an induction motor under the no-load condition is similar to a transformer under open circuit condition. Similarly, a three-phase induction motor with its rotor blocked behaves similarly to a transformer under short circuit conditions.
• The equation for the emfs induced in the primary and secondary windings of a transformer is similar to that of the emfs induced in the stator and rotor winding of an induction motor.

 

19.  The conditions of induction motor on load resemble those of a transformer whose secondary is

1.  Short-circuited

2.  Open-circuited

3.  Supplying a variable resistive load✓

4.  None of the above

 

When a 3-phase induction motor is running at slips, the electrical equivalent of mechanical load is given by

Rl=r2(1–ss)

Where r₂ = rotor resistance

R_l = electrical equivalent of mechanical load on the motor

Thus when the induction motor is running on load, the condition is similar to that of a transformer whose secondary is supplying a purely variable resistance load.

20.  The speed of a squirrel cage induction motor is changed by

1.  Cascade connection

2.  Rheostatic control

3.  Pole changing method✓

4.  Any of the above

•  
• The pole changing method is used in squirrel cage im because like slip ring im we can’t add external resistance in the rotor to reduce speed.
• In this method, it is possible to have one or two speeds by changing the number of poles. This is possible by changing the connection of the stator winding with the help of simple switching.

 

21.  One of the speeds of a 2-speed squirrel cage induction motor is 800 r.p.m. (lower speed). The other speed will be

1.  400 rpm

2.  2400 rpm

3.  1200 rpm

4.  1600 rpm✓

 

Multispeed squirrel cage motors are provided with stator windings that may be reconnected to form the different number of poles. Two-speed motors usually have one stator winding that may be switched through suitable control equipment to provide two speeds; one of which is half of the other. Hence the speed of higher speed motor will be 1600 rpm

 

22.  In a three-phase induction motor, the number of poles in the rotor winding is always

1.  Equal to the number of poles in the stator

2.  Zero

3.  More than the number of poles in the stator

4.  Less than the number of poles in the stator

 

Answer 1. Equal to the number of poles in the stator

Explanation:

• If the motor is squirrel cage induction motor, the number of rotor poles automatically gets adjusted to the number of stator poles
• But if the motor is wound rotor induction motor and number of poles on rotor and stator don’t match then the resultant torque will be zero. And the motor will not run.