MAILAM ENGINEERING COLLEGE · 16. Name the various methods for predetermining the voltage...

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Dr.A.Jeralidine Viji & M.Vijayaragavan / EEE / MEC Unit-01

MAILAM ENGINEERING COLLEGE Department of Electrical and Electronics Engineering

SUB CODE & NAME: EE6504 & ELECTRICAL MACHINES - II

UNIT – 01

SYNCHRONOUS GENERATOR

PART-A

1. What do you mean by the salient-pole type rotor?

Salient - pole type rotor means a low and moderate speed rotor having large diameter

and small axial length with projected poles coming out of the rotor frame the outer surface of

which almost follows the inner cylindrical surface of the stator frame.

2. Define voltage regulation of an alternator.Dec-2011, 2013,2015,May 2016,2017.

The voltage regulation of an alternator is defined as the increase in terminal voltage when

full load is thrown off, assuming field current and speed remaining the same.

% 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 = 𝑬𝟎 − 𝑽

𝑽 𝒙 𝟏𝟎𝟎

E0 = No load terminal voltage

V = Full load rated terminal voltage.

3. What are the advantages of having rotating field system?

Better insulation, Ease of current collection, More rigid construction, Reduced armature

leakage reactance, Lesser number of slip rings, Lesser rotor weight & inertia, improved ventilation &

heat dissipation.

4. Why EMF method is called Pessimistic method? May-2011

The value of voltage regulation obtained by EMF method is always more than the actual value,

therefore it is called Pessimistic method.

5. Why MMF method is called Optimistic method?

The value of voltage regulation obtained by MMF method is less than the actual value,

therefore it is called Optimistic method.

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6. Compare salient pole rotor & smooth cylindrical rotor. Dec-2011

Distinguish the use of salient pole and round rotor synchronous machines.

May-2015,May 2106

S.No Salient pole rotor Cylindrical rotor

1 Large diameter and short axial length Small diameter and long axial length

2 Used for low speed alternators Used for high - speed turbo alternators

3 Has projecting poles No projecting poles

4 Needs damper windings Does not need damper windings.

5 Windage loss is more Windage loss is less.

7. How is the armature winding in alternators different from those used in dc machines?

The armature winding of the alternator is placed in the stator, but the in case of dc machines,

armature winding is placed in rotor.

8. What are squirrel-cage windings of alternators? How and why are they used?

Damper windings are squirrel cage windings of the alternators. This winding is placed in rotor

pole shoes.

9. Write down the equation for frequency of emf induced in an altenator.

Frequency of emf induced in an Alternator, f expressed in cycles per second or Hz, is given by

the following equation

𝑓 = 𝑃 𝑁

120 𝐻𝑧

Where P- Number of poles

N-Speed in rpm.

10. Name the types of Alternator based on their rotor construction.

Smooth cylindrical type alternator

Salient pole alternator.

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11. What are the advantages of salient pole type construction used for Synchronous machines?

May-2012,APR-MAY 2018

They allow better ventilation

The pole faces are so shaped that the radial air gap length increases from the pole center

to the pole tips so that the flux distribution in the air-gap is sinusoidal in shape which

will help the machine to generate sinusoidal emf.

Due to the variable reluctance the machine develops additional reluctance

power which is independent of excitation.

12. Why is short pitch winding preferred over full-pitch winding? May-2012

Distorting harmonics can be reduced or totally eliminated.

Conductor material, copper, is saved in the back and front end connections due to less

coil-span.

Fractional slot winding with fractional number of lots/phase can be used which in turn

reduces the tooth ripples.

Mechanical strength of the coil is increased.

13. Define winding (distribution) factor. May-2011

The winding factor Kd is defined as the ratio of phasor addition of emf induced in all the coils

belonging to each phase winding to their arithmetic addition ( Kd= Vector sum/ Arithmetic sum)

14. Why are alternators rated in kVA and not in kW? Dec-2012

Apart from the constant loss incurred in Alternators is the copper loss, occurring in the 3 -

phase winding which depends on I2 R, the square of the current delivered by the generator.

As the current is directly related to apparent - power delivered by the generator, the Alternators

have only their apparent power in VA/ kVA / MVA as their power rating.

15. What is the necessity for predetermination of voltage regulation?

Most of the Alternators are manufactured with large power rating, hundreds of kW or MW, and

also with large voltage rating up to 33kV. For Alternators of such power and voltage ratings

conducting load test is not possible. Hence other indirect methods of testing are used and the

performance like voltage regulation then can be predetermined at any desired load currents and power

factors.

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16. Name the various methods for predetermining the voltage regulation of 3-phase Alternator.

Synchronous impedance / EMF method

Ampere-turn / MMF method

Potier / ZPF method.

ASA method

17. What are the advantages and disadvantages of estimating the voltage regulation of an

Alternator by EMF method?

Advantages:

Simple no load tests (for obtaining OCC & SCC) are to be conducted

Calculation procedure is much simpler

Disadvantages:

The value of voltage regulation is always higher than the actual value.

18. What are the tests data required for predetermining the voltage regulation of an Alternator

by MMF method?

Data required for MMF method are :

Effective resistance per phase of the 3-phase winding R

Open circuit characteristic (OCC) at rated speed/frequency

Short circuit characteristic (SCC) at rated speed/frequency

19. State the condition to be satisfied before connecting two alternators in parallel. Nov-Dec 2016

The terminal voltage magnitude of the incoming alternator must be made equal to the

existing alternator or the bus-bar voltage magnitude.

The phase sequence of the incoming Alternator voltage must be similar to the bus-bar

voltage.

The frequency of the incoming Alternator voltage must be the same as the bus-bar

voltage.

20. What are the two components of field current required for the predetermination of

regulation by MMF method?

Rated voltage

Rated current.

21. Define short circuit ratio of an alternator.

The ratio of field current required to produce rated voltage on open circuit to field current

required to produce rated current on 3 phase short circuit.

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22. What is synchronous reactance?

The sum of leakage reactance XL and fictitious reactance Xa is known as synchronous

reactance XS .

Xs = XL + Xa

23. Define Pitch (coil span) factor.

The ratio of vector sum of induced emfs per coil to the arithmetic sum of induced emfs per

coil.

24. What is armature reaction? Dec-2012, 2013,2015

The effect of armature flux on main flux is called armature reaction.

25. What is the use of potier triangle?

It is used to determine the voltage regulation of an alternator.

26. State the use of slip test of an alternator.

To determine the Xd and Xq.

27. What is infinite bus-bar?

The source or supply lines with non variable voltage and frequency are called infinite bus-bar.

28. Calculate the distribution factor for a 36- slot, 4 -pole, single layer three phase winding.

𝑛 = 36

4 = 9 ; 𝛽 =

180°

9= 20° ; 𝑚 =

36

3 𝑥 4 = 3

𝑘𝑑 = sin

𝑚𝛽2

𝑚 sin𝛽2

= sin

3 𝑥 202

3 sin202

= 0.96

29. Sketch salient pole and non salient pole rotors.

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30. How will you distinguish between the two types of synchronous generator from their

appearance? May-2014

Salient pole type – low core length and large diameter

Smooth cylindrical type – large core length and small diameter

31. Two reaction theory is applied only to salient pole machines? State the reason.

Dec-2014,APR-MAY 2018

A multipolar machine with cylindrical rotor has a uniform air-gap and therefore, its reactance

remains the same, irrespective of the spatial position of rotor. But in case of salient pole machines, the

airgap is not uniform and its reactance varies with rotor position. Because of non-uniformity of the

reluctance of the magnetic paths, the mmf of armature is divided into two components viz,

(i) one component is located along the axis of salient pole rotor known as direct-axis (d-axis)

component

(ii) Other component located perpendicular to the axis of salient pole rotor known as quadrature axis

(q-axis) component.

These facts form the basis of the two reaction theory applied to salient pole machines.

32. What is meant by alternator on infinite bus bars? May-2014

An infinite busbar means a source whose terminal voltage and frequency remains constant

irrespective of whether a new voltage source is connected to it or a load is put on it. Thus, alternators

on infinite bus bar means that their terminal voltage and frequency remains constant, equal to bus-bar

voltage and frequency.

33. Draw typical open circuit and short circuit characteristics of synchronous machine.

May-2015

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34. What are the essential elements for generating emf in alternators? Dec-2014

Prime mover, Stator, Rotor, Exciter

35. What is a capability curve?

The limits within which the synchronous machines operate safely is called capability curves

which are also known as operating charts or capability charts.

36. Distinguish between Full pitch coil and short pitch coil? Nov – Dec 2016

Full pitch coil Short Pitch Coil

Pole pitch equal is equal coil span Pole pitch is greater than coil span

It’s does not removing harmonics content It’s removing harmonics content so that pure

sinusoidal waveform can be obtained.

Voltage magnitude does not reduced Voltage magnitude reduced.

37.What do you mean by single layer and double layer winding? Apr-May 2017

Single Layer Winding:

In this type of winding, the complete slot is containing only one coil side of a coil. This type of winding is not

normally used for machines having commutators

Double layer Winding:

It consists of identical coils with one coil side of each coil in top half of the slot and the other coil side in

bottom half of another slot which is nearly one pole pitches away.

38. What is the necessity of chording in the armature winding of a synchronous machine? NOV-

DEC 2017

(1) Reduce the MMF harmonics produced by the armature winding and

(2) Reduce the EMF harmonics induced in the winding, without reducing the magnitude of the

fundamental EMF wave to a great extent.

39. Distinguish between transient and sub-transient reactances? NOV-DEC 2017

Sub transient Reactance, usually denoted as X''d (X double prime sub d), is the reactance used to

determine the current during the first cycle after the occurrence of the fault. In about 0.1 second this

reactance increases to the level known as Transient Reactance usually denoted as X'd,

and after 0.5 to 2 seconds it increases to the leven known as Synchronous Reactance and denoted as

Xd, and this determines the fault current after a steady condition is reached.

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Part – B

1. What are the advantages of stationary armature and rotating field type of

alternators compared to the other type?

What are the two types of rotors used in alternators of rotating field type?

Dec-2012, May-2012

Describe the salient constructional features of ac generators driven by (1)

diesel engines (2) steam engines. Dec-2014,Nov –Dec 2016

Describe the principle and construction of slow speed operation generation

with neat diagram? APR-MAY 2018

Advantages of stationary armature:

The field winding of an alternator is placed on the rotor and is connected to D.C. supply

through two slip rings. The 3-phase armature winding is placed on the stator. This arrangement

has the following advantages.

Better insulation-It is easier to insulate stationary winding for high voltages for which the

alternators are usually designed.

Easy of current collection-The stationary 3-phase armature can be directly connected to load

without going through large, unreliable slip rings and brushes.

Lesser no of slip rings-Only two slip rings are required for D.C. supply to the field winding on

the rotor. Since the exciting current is small, the slip rings and brush gear required are of light

construction.

Reduced armature leakage reactance.

Due to simple and robust construction of the rotor, higher speed of rotating D.C. field is

possible.

Improved ventilation and heat dissipation.

Construction and operation of Alternator:

A.C. system has a number of advantages over D.C. system. The machine which produces 3-

phase power from mechanical power is called an alternator or synchronous generator.

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An alternator operates on the same fundamental principle of electromagnetic induction as a

D.C. generator i.e., when the flux linking a conductor changes, an e.m.f. is induced in the

conductor.

Alternator also has an armature winding and a field winding. But there is one important

difference between the two.

In a D.C. generator, the armature winding is placed on the rotor in order to provide a way of

converting alternating voltage generated in the winding to a direct voltage at the terminals

through the use of a rotating commutator.

The field poles are placed on the stationary part of the machine. Since no commutator is

required in an alternator, it is usually more convenient and advantageous to place the field

winding on the rotating part (i.e., rotor) and armature winding on the stationary part (i.e.,

stator).

Construction:

An alternator has 3-phase winding on the stator and a DC field winding on the rotor.

1. Stator

It is the stationary part of the machine and is built up of sheet-steel laminations having

slots on its inner periphery. A 3-phase winding is placed in these slots and serves as the

armature winding of the alternator. The armature winding is always connected in star and the

neutral is connected to ground.

2. Rotor

The rotor carries a field winding which is supplied with direct current through two slip

rings by a separate D.C. source. This D.C. source (called exciter) is generally a small D.C.

shunt or compound generator mounted on the shaft of the alternator.

Rotor construction

Salient (or projecting) pole type

Non-salient (or cylindrical) pole type

Salient pole type

In this type, salient or projecting poles are mounted on a large circular steel frame

which is fixed to the shaft of the alternator.

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The individual field pole windings are connected in series in such a way that when the

field winding is energized by the D.C. exciter, adjacent poles have opposite polarities.

Low and medium-speed alternators (120-400 r.p.m.) such as those driven by diesel

engines or water turbines have salient pole type rotors due to the following reasons:

(i) Excessive windage loss if driven at high speeds and would tend to produce noise.

(ii) Salient-pole construction cannot be made strong enough to withstand the mechanical

stresses to which they may be subjected at higher speeds. Salient-pole type rotors have large diameters

and short axial lengths.

Non-salient pole type

In this type, the rotor is made of smooth solid forged-steel radial cylinder having a

number of slots along the outer periphery.

The field windings are embedded in these slots and are connected in series to the slip

rings through which they are energized by the DC exciter.

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The regions forming the poles are usually left unslotted. It is clear that the poles formed

are non-salient i.e., they do not project out from the rotor surface.

High-speed alternators (1500 or 3000 rpm) are driven by steam turbines and use non-

salient type rotors due to the following reasons:

(i) Mechanical robustness and gives noiseless operation at high speeds.

(ii) The flux distribution around the periphery is nearly a sine wave and hence a better emf

waveform is obtained than in the case of salient-pole type.

Since steam turbines run at high speed and a frequency of 50 Hz is required, we need a small

number of poles on the rotor of high-speed alternators (also called turbo alternators). turbo alternators

possess 2 or 4 poles and have small diameters and very long axial lengths.

Alternator Operation

The rotor winding is energized from the d.c. exciter and alternate N and S poles are developed

on the rotor.

When the rotor is rotated in anti-clockwise direction by a prime mover, the stator or armature

conductors are cut by the magnetic flux of rotor poles.

Consequently, emf is induced in the armature conductors due to electromagnetic induction.

The induced emf is alternating since N and S poles of rotor alternately pass the armature conductors.

The direction of induced e.m.f. can be found by Fleming's right hand rule and frequency is

given by,

𝑓 = 𝑁 𝑃

120 𝐻𝑧

Where, N - Speed of rotor in rpm

P - no. of poles

The magnitude of the voltage induced in each phase depends upon the rotor flux, the number

and position of the conductors in the phase and the speed of the rotor.

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Figure (i) shows star-connected armature winding and D.C. field winding. When the rotor is

rotated, a 3-phase voltage is induced in the armature winding. The magnitude of induced e.m.f.

depends upon the speed of rotation and the D.C. exciting current. The magnitude of e.m.f. in each

phase of the armature winding is the same. However, they differ in phase by 120° electrical as shown

in the phasor diagram.

Frequency

The frequency of induced emf in the armature conductors depends upon speed and the number

of poles.

Let N = rotor speed in rpm

P = number of rotor poles

f = frequency of emf in Hz

Consider a stator conductor that is successively swept by the N and S poles of the rotor.

If a positive voltage is induced when a N-pole sweeps across the conductor, a similar negative

voltage is induced when a S-pole sweeps by. This means that one complete cycle of emf is generated

in the conductor as a pair of poles passes it i.e., one N-pole and the adjacent following S-pole. The

same is true for every other armature conductor.

𝑁𝑜. 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒𝑠 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑁𝑜. 𝑜𝑓 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 = 𝑃

2

𝑁𝑜. 𝑜𝑓 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 = 𝑁

60

𝑁𝑜. 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 = 𝑃

2

𝑁

60 =

𝑃 𝑁

120

But number of cycles of emf per second is frequency.

𝑓 = 𝑃 𝑁

120

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It may be noted that N is the synchronous speed and is generally represented by Ns. For a

given alternator, the number of rotor poles is fixed and, therefore, the alternator must be run at

synchronous speed to give an output of desired frequency. For this reason, an alternator is sometimes

called synchronous generator.

Winding factors

Distribution factor (Kd):

In concentrated type winding, all coil sides are placed in one slot under a pole. So the resultant

emf / phase is equal to the arithmetic sum of individual coils emf in that phase.

In distributed winding, coil sides are distributed in different slots under a pole. So the emf /

phase are same in all coils but have some phase difference.

The distribution factor is defined as the ratio of resultant emf when coils are distributed to the

resultant emf when coils are concentrated. It is less than unity.

𝑘𝑑 = 𝑒𝑚𝑓 𝑤𝑖𝑡𝑕 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑤𝑖𝑛𝑑𝑖𝑛𝑔

𝑒𝑚𝑓 𝑤𝑖𝑡𝑕 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑛𝑑𝑖𝑛𝑔

𝑘𝑑 = sin

𝑚𝛽2

𝑚 sin𝛽2

Pitch factor (Kc):

The pitch factor is defined as the ratio of resultant emf when coil is short pitch to the resultant

emf when coil is full pitched.

𝐾𝑐 = 𝑒𝑚𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑠𝑕𝑜𝑟𝑡 𝑝𝑖𝑡𝑐𝑕 𝑐𝑜𝑖𝑙

𝑒𝑚𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑓𝑢𝑙𝑙 𝑝𝑖𝑡𝑐𝑕 𝑐𝑜𝑖𝑙

𝐾𝑐 = cos 𝛼

2

2. Derive the emf equation of an alternator.Dec-2012

Develop the formula for the induced emf in an alternator. Dec-2014,

2013,2015,Nov-Dec 2016,2017

Z-No. Of conductors or coil sides in series per phase ∅ =Flux per pole in webers

P=Number of rotor poles

N=Rotor speed in r.p.m

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In one revolution(i.e..60/N Second),each stator conductor is cut by P∅ webers

i.e..

d∅ =P∅ , dt=60N

∴ 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑒𝑚𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑜𝑛𝑒 𝑠𝑡𝑠𝑡𝑜𝑟 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟

=𝑑∅

𝑑𝑡=

𝑃∅

60𝑁

=𝑃∅𝑁

60 𝑉𝑜𝑙𝑡𝑠

Since there are Z conductors in series per phase

∴ 𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑒𝑚𝑓

𝑝𝑕𝑎𝑠𝑒=

𝑃∅𝑁

60× 𝑍

=𝑃∅𝑍

60×

120𝑓

𝑃 ∵ 𝑁 =

120𝑓

𝑃

R.M.S value of e.m.f/Phase= Average Value /phase×Form factor

= 2𝑓∅𝑍 × 1.11 = 2.22𝑓∅𝑍 𝑉𝑜𝑙𝑡𝑠

∴ 𝐸𝑟 .𝑚 .𝑠/𝑃𝑕𝑎𝑠𝑒 = 2.22𝑓∅𝑍 𝑉𝑜𝑙𝑡𝑠

If 𝐾𝑃 and 𝐾𝐷 are the pitch factor and distribution factor of the armature winding. then

𝐸𝑟 .𝑚 .𝑠/𝑃𝑕𝑎𝑠𝑒 = 2.22𝐾𝑃𝐾𝐷 𝑓∅𝑍 𝑉𝑜𝑙𝑡𝑠

𝐸𝑟 .𝑚 .𝑠/𝑃𝑕𝑎𝑠𝑒 = 4.44𝐾𝑃𝐾𝐷 𝑓∅𝑍 𝑉𝑜𝑙𝑡𝑠

The line voltage will depend upon whether the winding is star or delta connected.

3) Define Armature reaction and explain the effect of armature reaction on

different power factor loads of synchronous generator? DEC 2015,May 2016

When a load is connected to the synchronous Generator, a current will flow through the stator

winding, i.e. the armature winding of the synchronous generator. According to the power factor of the

load, the armature flux will be in different phase positions with respect to the main flux in the armature

winding. Below figure describes the armature reaction effect at the leading pole tips is weakened and

the flux at trailing pole tips is strengthened.(Fig. a).

As a result, the average field strength remains the same. Under the circumstance, the main field

is distorted. But at zero power factor lagging, the magnetic field due to armature current causes

demagnetizing(Fig b)of the main field whereas in case of zero power factor leading, the magnetic field

due to armature current causes magnetizing(fig c).

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Fig: Armature reaction Effect at different P.F Load

Synchronous reactance:

It may be observed that there is some effect of armature reaction, which will naturally, in turn,

produce an effect on the generated armature voltage of the synchronous machine.Hence, besides the

armature resistance and the leakage reactance drop effect, there will be an armature reaction effect on

the generated voltage in the stator of the synchronous machine. This effect of armature reaction is

equivalents that of an inductive reactance. This armature reaction effect is therefore considered to be

the fictitious reactance drop. So the combined leakage reactance is combined vector ally with

synchronous reactance to obtain synchronous impedance.

4. Explain the method to predetermine the voltage regulation of an alternator?

Dec-2012

Explain the EMF and MMF method of evaluating the synchronous

reactance. May-2015

Describe the method of determining the voltage regulation of an alternator

by synchronous impedance method. Dec-2014

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List the methods used to predetermine the voltage regulation of synchronous

machine and explain the MMF method. Dec-2013

Describe the potier method of determining the regulation of an alternator.

May-2013,2017, MAY-DEC 2018

Regulation is found by the following expression,

% 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 = 𝑬𝟎 − 𝑽

𝑽 𝒙 𝟏𝟎𝟎

Where, V – terminal voltage

E0 – induced voltage

E0 is determined by the following methods:

EMF method or synchronous impedance method

MMF method or Ampere turns method

ZPF method or Potier method

ASA method

1. EMF method or Synchronous Impedance method:

Conduct tests to find

OCC (upto 125% of rated voltage)

SCC (for rated current)

Armature resistance (per phase)

V - rated voltage

ISC - short circuit current corresponding to the field current producing

the rated voltage.

V - rated voltage

ISC - short circuit current corresponding to the field current producing the rated voltage.

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Synchronous Impedance per phase,

𝑍𝑠 = 𝑉

𝐼𝑠𝑐

𝑋𝑠 = 𝑍𝑠2 − 𝑅𝑎

2

For any load current I and phase angle Φ, find E0 as the vector sum of V, IRa and IXs.

For lagging power factor:

𝑬𝟎 = 𝑽𝐜𝐨𝐬𝜱 + 𝑰𝑹𝒂 𝟐 + 𝑽 𝐬𝐢𝐧 𝜱 + 𝑰𝑿𝒔

𝟐

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For unity power factor:

𝑬𝟎 = 𝑽 + 𝑰𝑹𝒂 𝟐 + 𝑰𝑿𝒔

𝟐

For leading power factor:

𝑬𝟎 = 𝑽𝐜𝐨𝐬𝜱 + 𝑰𝑹𝒂 𝟐 + 𝑽 𝐬𝐢𝐧 𝜱 − 𝑰𝑿𝒔

𝟐

2. MMF method or Ampere turns method: NOV-DEC 2017




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Steps:

By suitable tests plot OCC and SCC.

From the OCC find the field current Ifl to produce rated voltage V.

From SCC find the magnitude of field current If2 to produce the required armature

current.

Draw If2 at angle (90+Φ) from If1, where Φ is the phase angle of current from voltage. If

current is leading, take the angle of If2 as (90-Φ).

Find the resultant field current, If and mark its magnitude on the field current axis.

From OCC. find the voltage corresponding to If , which will be E0.

3. ZPF (Zero Power Factor) method or Potier method:




ZPF (for rated current and rated voltage)

Armature Resistance (if required)

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Steps:

By suitable tests plot OCC and SCC.

Draw tangent to OCC (air gap line).

Conduct ZPF test at full load for rated voltage and fix the point B.

Draw the line BH with length equal to field current required to produce full load current at

short circuit.

Draw HD parallel to the air gap line so as to touch the OCC.

Draw DE parallel to voltage axis. Now, DE represents voltage drop IXL and BE represents

the field current required to overcome the effect of armature reaction.

Triangle BDE is called Potier triangle and XL is the Potier reactance.

Find E from V, IXL and Φ. Consider Ra also if required. The expression to use is,

𝑬 = 𝑽 𝐜𝐨𝐬𝜱 + 𝑰𝑹𝒂 𝟐 + 𝑽 𝐬𝐢𝐧𝜱 + 𝑰𝑿𝑳

𝟐

Find field current corresponding to E.

Draw FG with magnitude equal to BE at angle (90+Ψ) from field current axis,

where Ψ - the phase angle of current from voltage vector E (internal phase angle).

The resultant field current is given by OG. Mark this length on field current axis.

From OCC find the corresponding E0.

21


4. ASA method:

Conduct tests to find,



ZPF (for rated current and rated voltage)


Steps:

Follow steps 1 to 7 as in ZPF method.

Find If1 corresponding to terminal voltage V using air gap line (OF1 in figure).

Draw If2 with length equal to field current required to circulate rated current during

short circuit condition at an angle (90+Φ) from If1. The resultant of If1 and If2 gives If

(OF2 in figure).

Extend OF2 upto F so that F2F accounts for the additional field current accounting for

the effect of saturation. F2F is found for voltage E as shown.

Project total field current OF to the field current axis and find corresponding voltage E0

using OCC.

22


5. Illustrate a method for determining the direct and quadrature axis reactances of

a salient pole synchronous generator. May-2015,NOV-DEC 2017

Describe a method of determining direct and quadrature axis reactance of

salient pole alternator. May-2014

Describe the slip test for finding Xd and Xq.May-2013

Andrew Blondel had proposed the two reaction theory which resolves the given armature mmfs

into two mutually perpendicular components follows.

One components is located along the axis of salient pole rotor known as direct axis(or-d

axis)component.

The other component is located perpendicular to the axis of salient pole rotor known as

quadrature axis (or q axis) component.

Airgap is uniform in the cycle cylindrical rotor synchronous machine whereas it is non uniform for a

salient pole synchronous mahine.In a salient pole machine, the reluctance of the magnetic circuit along

the direct axis is much less than the reluctance along the quadrature axis.

Because of non-uniformity of the reluctance of the magnetic paths, the mmf of the armature is divided

into two components namely

i) A direct component and

ii) Quadrature component

When armature current is in phase with the excitation voltage, the entire mmf of the armature acts at

right angles to the axis of the salient poles and therefore, all the armature mmf is in quadrature. On the

other hand, if the armature current is in quadrature with the excitation voltage E0,the entire mmf of the

armature acts directly upon the magnetic paths through the salient poles and all of the armature mmf is

direct acting, either directly opposing or directly aiding the mmf of the salient pole field winding.

When the phase difference between armature current and excitation voltage is in between 0 and

90˚,the armature will have both a direct acting and a quadrant components. the direct acting

components is proportional to the sin of the phase angle between the armature current and excitation

voltage whereas the quadrature component is proportional to the cosine of the phase angle between the

armature current and the excitation voltage.

The two reactance concept is similar to the synchronous impedance concept in that the effect of

armature reaction is taken into account by means of equivalent armature reactance voltage.

However, owing to the difference in the reluctance of the magnetic paths upon which the two

components of the armature mmf act,the value of the equivalent reactance for the direct component of

the armature mmf is greater than the value of the equivalent reactance for the quadrature component of

23


the armature mmf. Thus the two reactance concept of salient pole machine replace the effect of

armature reaction by two fictitious voltages.

These reactance voltages are respectively 𝐼𝑑𝑋𝑠𝑑 and 𝐼𝑞𝑋𝑠𝑞 ,where 𝐼𝑑 𝑎𝑛𝑑 𝐼𝑞 are the components of the

armature components of the armature current along direct and quadrature axis respectively. Each of

these components of armature current also, produces a leakage reactance voltage caused by the

armature leakage flux.

The armature leakage reactance is assumed to have the same value 𝑋𝐿 for both components of the

armature current.Therefore,there is synchronous reactance for each components of the armature mmf

as below

Direct axis synchronous reactance 𝑋𝑠𝑑 = 𝑋𝑠𝑑 + 𝑋𝐿 ……………………….. (1)

Quadrature axis synchronous reactance 𝑋𝑠𝑞 = 𝑋𝑠𝑞 + 𝑋𝐿…………………… (2)

The voltage equation foe each phase of the armature based on the two reactance concept is given

by𝑉 = 𝐸0 − 𝐼𝑅𝑎 − 𝐼𝑑𝑋𝑠𝑑 − 𝐼𝑞𝑋𝑠𝑞…………………………………. (3)

24


Measurement of Xd and Xq using slip test:

The direct axis and quadrature axis reactance’s can be measured by slip test. the machine is driven by

an auxiliary motor (preferably DC motor) at a speed slightly less or slightly more than the

synchronous speed.

Figure: Connection of Slip Speed

The field winding is kept open circuited and a low voltage 3-phase supply (about 20% of the rated

voltage)is applied to the armature terminals. The direction of rotation should be the same as the

direction of rotating field. If this condition is fulfilled a small AC voltage would be indicated by the

voltmeter across the field winding.

The relative velocity between armature mmf and field poles equal to slip speed,i.e.,difference between

synchronous speed and rotor speed.the stator mmf moves slowely past the field poles at slip speed.this

would cause the armature current to vary cyclically as shown in below figure.

When the peak of the armature mmf is in line with the field poles, the reluctance offered by the

magnetic circuit is minimum. the ratio of armature terminal voltage per phase to armature phase

current gives 𝑋𝑑 .

After one quarter of slip cycle, the peak of armature mmf is in line with q-axis and the reluctance

offered by the magnetic circuit is minimum. the ratio of armature terminal voltage per phase to

armature phase current gives 𝑋𝑞 .when the armature mmf is in line with field poles, the armature fux

linkage with field winding is maximum and rate of change of this flux linkage is zero.

25


On the other hand, when armature mmf is in line with q-axis, the flux linkage with field winding is

minimum and rate of change of this flux linkage is maximum. so that induced voltage across the field

winding is maximum.

This helps is locating the points of maximum and minimum on armature voltage and current

oscillograms.the slip should be very small. so that inertia of moving parts of instruments does not

cause errors in measurements. Greater accuracy can be achived by using recording oscillogram.

In view of the error involved in reading oscillograms.this test should be used only to find the ratio of

𝑋𝑑 /𝑋𝑞 .The value of 𝑋𝑑 can be found from the open circuit and short circuit tests, as in the case of

cylindrical rotor machine.

26



Xd and Xq


Steps:

Energise the alternator with field unexcited and driven close to synchronous speed by a prime

mover.

Measure the line voltage and line current of the alternator.

Find Xd and Xq by the following expressions,

𝑿𝒅 = 𝑽𝒎𝒂𝒙

𝟑 𝑰𝒎𝒊𝒏

𝑿𝒒 = 𝑽𝒎𝒊𝒏

𝟑 𝑰𝒎𝒂𝒙

Find Id as follows

𝜳 = 𝐭𝐚𝐧−𝟏𝑽 𝐬𝐢𝐧𝜱 + 𝑰 𝑿𝒒

𝑽𝐜𝐨𝐬𝜱 + 𝑰 𝑹𝒂 ; 𝑰𝒅 = 𝑰 𝐬𝐢𝐧𝜳

Then expression for E0 is,

𝑬𝟎 = [ 𝑽𝐜𝐨𝐬𝜱 + 𝑰 𝑹𝒂 𝟐 + 𝑽 𝐬𝐢𝐧 𝜱 + 𝑰 𝑿𝒒

𝟐 ]

𝟏𝟐 + 𝑰𝒅 (𝑿𝒅 − 𝑿𝒒)

27


6. State and explain the conditions for parallel operation of alternators.

Dec-2012,NOV-DEC 2017

Necessity for parallel operation of alternators:

Several small units in parallel are more reliable than one large unit operating singly. If there is

any failure of a particular unit, the others can maintain the system in order.

Overhauling and repairing of any small unit can be easily done.

For any further extension, another unit is parallel can be easily installed.

The cost of standby units is also small.

The most important point in the parallel operation of alternators is the synchronizing of

alternators.

Conditions for synchronization:

The terminal voltage of the alternator to be connected ti the bus, where all other

alternators are already connected, must be equal to the busbar voltage.

The frequency of the alternator voltage to be connected to the bus must be equal to that

of the busbar voltage.

The phase of the voltage of the machine to be connected, relative to the load, must be

same as that of the busbar.

28


Synchronization of alternator with busbar:

Synchronization if the alternator with the busbar can be made with the following methods:

Three dark lamps method

Two bright lamps and one dark lamp method

Synchroscope

Three dark lamps method:

From the diagram, it is clear that if the phases of the busbar are connected with the phases of

the alternator at the moment when all the conditions of parallel operation are satisfied, then three

lamps L1, L2, L3 should be more or less dark.

Two bright lamps and one dark lamp method:

Suppose we have to connect the alternator 2(R’Y’B’)in parallel with the alternator 1(RYB) or the

busbar.three lamps L1,L2,L3 are connected between RR’,YB’ and BY’ (in below Figure).If the

voltage waveforms of RYB and R’Y’B’ are identical, then L1 will be dark and L2 and L3 will be

bright since they get the full line voltage across them.

29


Suppose R’Y’B’(i.e., incoming alternator 2)is moving fast, then as per below shown in figure the

voltage YB’ decreases and Y’B increases. the voltage across RR’will not be zero. so the lamp L1 starts

glowing, lamp L2 will be less bright than lamp L3.on other hand if alternator 1 moves fast, then the

opposite scenario will be observed. Lamp L2 will be more bright than lamp L3 and the lamp L1 starts

glowing From this, we get an idea as to which system is moving fast and accordingly, we have to

adjust to make the bus bar voltage and the incoming alternator in synchronism.

Synchroscope:

The best method of synchronizing alternators is by means of a single phase device known as

synchroscope.which provides a more accurate indication of synchronism than do Lamps.

The Synchroscope is an instrument for indicating difference of phase and frequency between two

voltages. It is essentially a spilt-phase motor in which torque is developed if the frequencies of the two

voltages differ.

A pointer, which is attached to the rotating part of the instrument. Move over the dial face in either a

clockwise or counter clock wise direction. Depending on weather the incoming machine is fast or

slow.

The synchroscope has two pair of terminals. one pair marked as existing has to be connected to the

existing has to be connected to the existing alternator or bus bar terminals and the other pair with

marking ‘incoming’ has to be connected to the corresponding terminals of the alternators as shown in

below figure.

The Synchroscope has a circular dial in which a thick line is marked at the top.a clockwise symbol

with letter ‘F’ on one side and an anticlockwise symbol with letter’s’ marked on the other side. the

pointer is capable of rotating in both directions. After equal voltage magnitude condition is satisfied

30


the operator has to look into the synchroscope. The rate of rotation of the pointer indicates the amount

of frequency difference between the alternators. The direction of rotation indicates whether the

incoming alternator frequency is higher or lower than the existing alternator i.e.. Whether the

incoming alternator is fast or slow. Suitable correction is then made in the speed of the alternator and

the rate of rotation is reduced to the smallest possible value. The TPST switch is closed to synchronize

the incoming alternator when the pointer faces the top thick line marking.

Synchronizing Torque

Let, synchronizing torque be Ts in Nm.

3 𝑃𝑠 = 𝑇𝑠 𝑥 2 𝜋 𝑁𝑠

60

𝑻𝒔 = 𝟑 𝑷𝒔 𝟔𝟎

𝟐 𝝅 𝑵𝒔

31


7. Derive an expression for real and reactive power outputs of asynchronous

generator. May-2015

The flow of active and reactive power in a synchronous link will be studied.the approach will be

analytical and armature resistance will be considered for generality of results.

Below figure shows the schematic of a synchronous generator wherein 𝐸 𝑓 leads 𝑉 𝑡 by angle 𝛿 the

synchronous impedance * is

𝑍 𝑠 = 𝑅𝑎 + 𝑗𝑋𝑠 = 𝑍𝑠 < 𝜃 (1)

As shown by the impedance triangle is

𝜃 = 𝑡𝑎𝑛−1 𝑋𝑠

𝑅𝑎 (2)

And 𝛼 = 90° − 𝜃 = 𝑡𝑎𝑛−1 𝑅𝑎

𝑋𝑠 (3)

The armature current in fig shown in below can be expressed as

𝐼 𝑎 =𝐸𝑓 < 𝛿 − 𝑉𝑡 < 0°

𝑍𝑠 < 𝜃 (4)

The complex power output is

𝑆 𝑒 = 𝑃𝑒 + 𝑗𝑄𝑒 = 𝑉𝑡 < 0°𝐼 𝑎∗ 5

Substituting for 𝐼𝑎 from Eq (4), in Eq. (5)

32


𝑃𝑒 + 𝑗𝑄𝑒 = 𝑉𝑡 < 0 𝐸𝑓 < 𝛿 − 𝑉𝑡 < 0°

𝑍𝑠 < 𝜃

∗

=𝑉𝑡𝐸𝑓

𝑍𝑠< 𝜃 − 𝛿 −

𝑉𝑡2

𝑍𝑠< 𝜃 6

Equating the real and imaginary parts Eq. (6), the following expressions for real and reactive power

output are obtained as

𝑃𝑒 𝑜𝑢𝑡 = −𝑉𝑡

2

𝑍𝑠𝑐𝑜𝑠𝜃 +

𝑉𝑡𝐸𝑓

𝑍𝑠cos( 𝜃 − 𝛿) (7𝑎)

𝑄𝑒 𝑜𝑢𝑡 = −𝑉𝑡

2

𝑍𝑠𝑆𝑖𝑛𝜃 +

𝑉𝑡𝐸𝑓

𝑍𝑠𝑆𝑖𝑛( 𝜃 − 𝛿) (7𝑏)

Net mechanical power input to the machine is given by

𝑃𝑚 𝑖𝑛 = 𝑃𝑒, = 𝑅𝑒 𝑆𝑒

′ = 𝐸𝑓 < 𝛿 𝐸𝑓 < 𝛿 − 𝑉𝑡 < 0°

𝑍𝑠 < 𝜃

∗

= 𝐸𝑓2𝑐𝑜𝑠𝜃 −

𝑉𝑡𝐸𝑓

𝑍𝑠cos( 𝛿 + 𝜃) (8)

It is convenient to express the above results in terms of angle 𝛼 defined in the impedance triangle

figure. Equation 7a,7b and 8 then modify as below.

𝑃𝑒 𝑜𝑢𝑡 = −𝑉𝑡

2

𝑍𝑠𝑅𝑎 +

𝑉𝑡𝐸𝑓

𝑍𝑠sin(𝛿 + 𝛼) (8𝑎)

𝑄𝑒 𝑜𝑢𝑡 = −𝑉𝑡

2

𝑍𝑠𝑋𝑠 +

𝑉𝑡𝐸𝑓

𝑍𝑠𝑐𝑜𝑠 (𝛿 + 𝛼) (8𝑏)

𝑃𝑚 𝑖𝑛 =𝐸𝑓

2

𝑍𝑠2

𝑅𝑎 +𝑉𝑡𝐸𝑓

𝑍𝑠sin 𝛿 − 𝛼 9

The real electrical power output, 𝑃𝑒 as per Eq 8a is plotted in below fig from which it is observed that

its maximum value is

33


𝑃𝑒 𝑜𝑢𝑡 𝑚𝑎𝑥 = −𝑉𝑡

2𝑅𝑎

𝑍𝑠2 +

𝑉𝑡𝐸𝑓

𝑍𝑠 10

Occurring at 𝛿 = 𝜃,which defines the limits of steady state stability.The machine will fall out step for

angle 𝛿 > 𝜃. 𝑜𝑓 𝑐𝑜𝑢𝑟𝑠𝑒, 𝜃 will be 90°.if resistance is negligible in which case the stability limit will

be at 𝛿 = 90° in follows equation (9 )that

𝑃𝑚 𝑖𝑛 𝑚𝑎𝑥 =𝐸𝑓

2

𝑍𝑠2 𝑅𝑎 +

𝑉𝑡𝐸𝑓

𝑍𝑠; 𝑎𝑡 𝛿 = 90° + 𝛼 = 𝜃 + 2𝛼 (11)

Since the angle 𝛿 𝑖𝑛 𝐸𝑞 (11) is more than 𝜃,the maximum mechanical power input operation for a

generator lies in the unstable region.

Equation (8a),(8b),(9 )and (10) simplify as below when armature resistance is neglected.

𝑃𝑒 𝑜𝑢𝑡 =𝑉𝑡𝐸𝑓

𝑋𝑠sin 𝛿 (12)

𝑄𝑒 = 𝑜𝑢𝑡 = −𝑉𝑡

2

𝑋𝑠+

𝑉𝑡𝐸𝑓

𝑋𝑠cos𝛿 (13)

𝑃𝑚 𝑖𝑛 = 𝑃𝑒 𝑜𝑢𝑡 =𝑉𝑡𝐸𝑓

𝑋𝑠sin 𝛿 (14)

𝑃𝑒 𝑜𝑢𝑡 𝑚𝑎𝑥 = 𝑃𝑚 𝑖𝑛 𝑚𝑎𝑥 =𝑉𝑡𝐸𝑓

𝑋𝑠; 𝑎𝑡 𝛿 = 90° (15)

34


8. Discuss about the effect of change in excitation and mechanical input in

synchronous machine.

When an alternator is running in parallel with other alternator, the load taken up by it is totally

determined by driving torque or the power input of the prime mover. If any change of excitation is

carried out, it does not change its kW output but merely changes kVA or merely changes the power

factor at which the load is delivered.

Change of excitation:

Let us assume that the two alternators operating in parallel are identical, that is, they are

supplying half of the active load and reactive load or, equal to the power factor of the load. If the

excitation of the alternator 1 is increased E1 > E2 and it causes a circulating current (𝐼𝑐) which flows

through the armature and round the bus bars. From the shown in the below diagram, (𝐼𝑐) is added

vectorally to the load current of alternator 1 and subtracted from the load current of alternator 2 which

causes a change in load current. Therefore, alternator 1 and alternator 2 will deliver the load current at

power factor cos∅1 and cos∅2 respectively where cos∅1 > cos∅2 . Although the two machines

deliver the load currents at different power factors, it has no effect on kW output, but kVAR supplied

by alternator 1 is increased whereas kVAR supplied by alternator 2 is decreased as shown in the below

diagram.

FIG: Effect of change in excitations

Change in mechanical input or steam supply:

35


Let us assume that the excitation of the alternators remain unaltered during their operation in

parallel. Let the steam supply to alternator 1 be increased, so that input to its prime mover is increased.

Alternator 1 cannot over run alternator 2 because the speeds of the two alternators are tied by their

synchronous bond. E1 advances E2 by a small angle δ.

Fig: Effect of change of stream supply

Hence resultant voltage (Er) is produced and it acts on the local circuit resulting in a current Ir

which lags Er by an angle 90˚. Therefore, power per phase of alternator 1 is increased whereas power

per phase of alternator 2 is decreased. Since the increase in steam input has no effect on the division of

reactive power, the active power output of alternator 1 is increased whereas active power output of

alternator 2 is decreased.

9.Explain the power-angle characteristics of the salient pole synchronous machine.

Show in the one-line diagram of a salient-pole synchronous machine connected to infinite bus-bars of

voltage 𝑉𝑏 through a line series reactance 𝑋𝑒𝑥𝑡 (per phase). The total d- and q-axis reactance’s are then

𝑋𝑑 = 𝑋𝑑𝑔 + 𝑋𝑒𝑥𝑡

𝑋𝑞 = 𝑋𝑞𝑔 + 𝑋𝑒𝑥𝑡

36


The resistances of machine armature and line are assumed negligible. Shown in the below diagram

gives the pharos diagram when the machine is generating. It is easy to see from this diagram that the

real power delivered to bus-bars is

𝑃𝑒 = 𝐼𝑑𝑉𝑏 sin 𝛿 + 𝐼𝑑𝑉𝑏 cos𝛿 (1)

Now 𝐼𝑑 =𝐸𝑓−𝑉𝑏 cos 𝛿

𝑋𝑑 (2)

And 𝐼𝑑 =𝑉𝑏 sin 𝛿

𝑋𝑑 (3)

Substituting in Eq. (1),

𝑃𝑒 =𝐸𝑓𝑉𝑏

𝑋𝑑sin 𝛿 + 𝑉𝑏

2 𝑋𝑑−𝑋𝑞

2𝑋𝑑𝑋𝑞sin 2𝛿 (4)

37


Equation (4) gives the expression for the electrical power output of a salient-pole generator. The same

expression would give the electrical power input of a motoring machine wherein 𝐸𝑓 lags 𝑉𝑏 by angle𝛿.

The second term in Eq (4) compared to a cylindrical motor arises on account of saliency and is known

as the reluctance power (torque). The reluctance power varies as sin 2𝛿 with a maximum value

at𝛿 = 45°. It is to be further observed that this term is independent of field excitation and would be

present even if the field is unexcited *. A synchronous motor with salient poles but no field winding is

known as the reluctance motor. If is used for low-power, constant-speed applications where special

arrangements for dc excitation would be cumbersome.

The power angle plots of both the terms of Eq (4) along with the form of the resultant power-angle

curve are shown in the below diagram. It is immediately observed that 𝑃𝑒 ,𝑚𝑎𝑥 = 𝑃𝑝𝑢𝑙𝑙 −𝑜𝑢𝑡 now occurs

at𝛿 < 90°(usually at 𝛿 about 70°) and further its magnitude is larger than for a cylindrical machine

with same 𝑉𝑏 ,𝐸𝑓 and 𝑋𝑠=𝑋𝑑 on account of the reluctance power term.

38


10.Discuss in detail about the short circuit transients occurring in synchronous

machine.

The synchronous machine is mainly used in power generation. In a power system network,

there is probability of load fluctuations, faults and many other failures. Hence, there will occur the

transient condition of the system. Now, we have to see the impact on the synchronous reactance of

synchronous machine when the system goes to transient state from the steady-state condition.

Shown in the below diagram describes symmetrical short-circuit condition on synchronous machine.

Here three periods can be observed-sub transient, transient and steady state periods. When the machine

is short circuited, a large value of short-circuit current will flow inducing voltages and currents both in

the damper and the field windings.

Fig: Symmetrical short circuit condition for synchronous machines

As per Len’s law the induced voltages and currents will oppose the very cause and the short-

circuit will therefore be gradually reduced and finally it will attain a steady-state value. Since the

damper winding consists of a few thick bars and possesses a very low time constant, its induced

current will first vanish.

The initial period of decay of the short-circuit current is termed the sub transient period. The

reactance during this period is termed the sub transient reactance. After that the effect of the field

current will appear.

This period is termed transient period and the reactance involved during this period is termed

transient reactance. Hence the field winding will have comparatively a higher time constant.

So the transient reactance will be higher than the sub transient reactance. When the effect of

field winding has vanished, the steady-state reactance will appear. The steady-state reactance will be

39


higher than the transient reactance. Usually the sub transient reactance, transient reactance and steady-

state reactance are symbolized as 𝑋𝑑" , 𝑋𝑑

′ and 𝑋𝑑 respectively. As we have seen,

𝑋𝑑" < 𝑋𝑑

′ < 𝑋𝑑

11.Sketch the capability curves of a synchronous machine and explain it.

The capability curve of the synchronous generator defines the bounds within which it can operate

safely. Various bounds imposed on the machine are:

1. MVA-loading cannot exceed the generator rating. This limit is imposed by the stator heading.

2. MW-loading cannot exceed the turbine rating which is gives by MVA(rating) × pf*(rating).

3. The generator must operate a safe margin away from the steady-state stability limit (𝛿 = 90°).

this can be laid down as a maximum allowable value of 𝛿.

4. The maximum field current cannot exceed a specified value imposed by rotor heading.

To draw the capability curve of the synchronous generator, its pharos diagram is used which is redraw

in the below diagram armature resistance is neglected. After multiplying voltage magnitude of each

voltage pharos by (3𝑉𝑡 𝑋𝑠 ), the phasor diagram is redrawn in below diagram. It is immediately

recognized that OMN is the complex power triangle (in 3-phase values)

40


wherein obviously Q is positive for lagging power factor, ∅ being values to the angle of OM from the

P-axis. A mere scale change will convert these values to the units of MVA, MW and MVAR.

Constant S operation will lie on a circle centered at O and radius OM. Constant P operation will lie on

a line parallel to QO’-axis. Constant-excitation (𝐹𝑓) operation will lie on a circle centered at O’ of

radius OM (3𝑉𝑡 𝐸𝑓 𝑋𝑠 ). Constant-pf operation will lie on a radial line through O.

Now with specified upper limits of S, P and 𝐸𝑓 (field current), the boundaries of the capability curve

can be drawn as in the below diagram. The limit on the left side is specified by𝛿(max), the safe

operating value from the point of view of transient stability. Since the minimum excitation operation

corresponds to𝛿 = 90°, the machine operation is at a safe limit from 𝐸𝑓 (min) by specifying𝛿(max).

41


Fig: Capability curves of Synchronous machines.

42


Problems:

1. A 4 pole alternator has an armature with 25 slots and 8 conductors per slot and

rotates at 1500 rpm and the flux per pole is 0.05 wb. Calculate the emf generated,

if winding factor is 0.96 and all the conductors are in series. Dec-2012

Solution: P=4,𝑁𝑠 = 1500 𝑟. 𝑝. 𝑚, 25 𝑠𝑙𝑜𝑡𝑠, 8 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠/𝑠𝑙𝑜𝑡

𝑁𝑠 =120𝑓

𝑃 𝑖. 𝑒 𝑓 =

1500 × 4

120= 50𝐻𝑧, 𝜑 = 0.05𝑤𝑏

𝑧 = 25 × 8 = 200, 𝑍𝑝𝑕 =𝑍

3=

200

3, 𝑇𝑝𝑕 =

𝑍𝑝𝑕

2=

100

3

𝐾𝑐 = 1, 𝐾𝑑 = 0.96

𝐸𝑝𝑕 = 4.44𝐾𝑐𝐾𝑑 𝜑𝑓𝑇𝑝𝑕 = 355.2𝑉

Assuming Star Connection ,𝐸𝐿𝑖𝑛𝑒 = 3𝐸𝑝𝑕 = 615.224𝑉

2. A 220 V, 50 Hz, 6 pole star connected alternator with ohmic resistance of 0.06

ohm per phase, gave the following data for open circuit, short circuit and full load

zero power factor characteristics. Find the percentage voltage regulation at full

load current of 40 A at power factor of 0.8 lag by (1) emf method (2) mmf method

(3) ZPF method. Compare the results so obtained.

Field

current, A 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.80 2.20 2.60 3.00 3.40

Open

circuit

voltage, V

29.0 58.0 87.0 116 146 172 194 232 261.5 284 300 310

Short

circuit

current Isc

6.6 13.2 20.0 26.5 32.4 40.0 46.3 59.0

ZPF

terminal

voltage, V

0 29 88 140 177 208 230

May-2015,2016

43


Solution: Rated per phase voltage 𝑽𝒕 =𝟐𝟐𝟎

𝟑= 𝟏𝟐𝟕 𝑽

Per phase values for O.C.C and Z.P.f.C are tabulated below and O.C.C.,S.C.C.

Field

current, A 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.80 2.20 2.60 3.00 3.40

𝐸𝑓 𝑖𝑛 𝑉 16.3 33.5 50.2 67.0 84.3 99.3 112 134 151 164 173.2 179.0

ZPF

terminal

voltage, V

- - - - - 0 16.73 50.8 80.8 102 120 132.7

(a) E.m.f. method. The values of the synchronous impendence 𝑍𝑠 and synchronous reactance

𝑋𝑠,are tabulated below for different values of excitations(taking 𝐸𝑓 from O.C.C. and 𝐼𝑠𝑐 from

S.C.C. for the same field current):

𝐼𝑓𝑖𝑛 𝐴 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.80

𝐸𝑓 𝑖𝑛 𝑉 16.73 33.50 50.2 67 84.3 99.3 112 134

𝐼𝑠𝑐 𝑖𝑛 𝐴 6.6 13.2 20.0 26.5 32.4 40.0 46.3 59.0

𝑍𝑠𝑖𝑛 Ω 2.535 2.535 2.51 2.53 2.51 2.48 2.42 2.27

𝑋𝑠𝑖𝑛 Ω 2.53 2.53 2.51 2.53 2.51 2.48 2.42 2.27

Here 𝑋𝑠 ≅ 𝑍𝑠 , since 𝑟𝑎 is quite small.

For full load and power factor of 0.8 lagging, the pharos diagram is similar to that given in the below

diagram. With 𝑉𝑡 as the reference pharos,

𝑉 𝑡 = 127 + 𝑗0.00 𝐼 𝑎 = 40 0.8 − 𝑗0.6 = 32 − 𝑗24 .

𝐸 𝑓 = 𝑉 𝑡 + 𝐼 𝑎 𝑟𝑎 + 𝑗𝑋𝑠

=127+ (32-j24)(0.06+j2.27)=182.92+j70.16

𝑜𝑟 𝐸𝑓 = 182.92 2 + 70.16 2 = 195.5 𝑣𝑜𝑙𝑡𝑠.

∴ Percentage voltage regulation =195.5−127

127× 100 = 53.9%.

44


Excitation voltage 𝐸𝑓 can also be calculated by referring to.

Note that minimum value of 𝑍𝑠, corresponding to maximum short-circuit current, has been used here.

(a) M.M.F method. Voltage behind armature resistance 𝑟𝑎 ,

𝐸′ = 𝑉 𝑡 + 𝐼 𝑎𝑟𝑎 .

For convenience, take 𝐼𝑎 as the reference pharos.

∴ 𝐸′ = 127 0.8 + 𝑗0.6 + 40 0.06 = 104 + 𝑗76.2 (1)

𝑜𝑟𝐸′ = (104)2 + (76.2)2 = 129.0 𝑉.

For 𝐸′=129.0 V, the field excitation 𝐹𝑟1 from O.C.C. is equal to 1.69 A.

From S.C.C., 𝐹𝑎 + 𝐹𝑎𝑙 = the field current required to circulate full-load short circuit

current=1.20 A.

From (1), the angle 𝛼 can be find from

𝛼 = tan−176.2

104= 36.2°.

45


In phasor form, 𝐹 𝑓 = 1.69[cos 90 + 𝛼 + 𝑗 sin(90 + 𝛼)]

=1.69[-sin 𝛼 + 𝑗 cos 𝛼]

=1.69[-0.591+j0.807]=-1+j1.365.

𝐹 𝑓 = 𝐹 𝑟1 − 𝐹𝑎 + 𝐹𝑎𝑙 = −1 + 𝑗1.365 − 1.20 = −2.20 + 𝑗1.365

Or 𝐹𝑓 = 2.59A

Field m.m.f can also be computed by referring to fig: Where AB=𝐹𝑟1 = 1.69𝐴; 𝐵𝐶 = (𝐹𝑎 +

𝐹𝑎𝑙)=1.20 𝐴 𝑎𝑛𝑑 𝛼 = 36.2°

𝐹𝑓 = 1.69 + 1.20𝑠𝑖𝑛36.2 2 + 1.20𝑐𝑜𝑠36.2 2 =2.5868≅ 2.59 𝐴

Corresponding to 𝐹𝑓 = 2.59A,𝐸𝑓 𝑓𝑟𝑜𝑚 𝑂𝐶𝐶 𝑖𝑠 163.5 𝑉

∴ 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =163.5 − 127

127× 100 = 28.75%.

c)Zero power factor method: First of all, the potier triangle ABC is drawn as described before. Point A

corresponds to the rated voltage of 127 V on the z.p.f.the line AD is drawn parallel and equal to

F’O=1.2 A.Then DC is drawn parallel to the air gap line, meeting the O.C.C.at point C. Perpendicular

CB on AD gives 𝐼𝑎𝑥𝑎𝑙 drop equal to 30 volts

∴ 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑙𝑒𝑎𝑘𝑎𝑔𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑥𝑎𝑙 =30

40= 0.75Ω

The air gap voltage 𝐸𝑟 = 148.6 𝑉, 𝑓𝑟𝑜𝑚 𝐸𝑟 = 𝑉𝑡 + 𝐼𝑎 𝑟𝑎 + 𝑗𝑥𝑎𝑙

With Ia as the reference phasor, 𝐸𝑟 = 127 0.8 + 𝑗0.6 + 40 0.06 + 𝑗0.75 = 104 + 𝑗106.2

∴ 𝐸𝑟 = 10.4 2 + (106.2)2 =148.6 volts

Corresponding to

𝐸𝑟 = 148.6 𝑉, 𝑡𝑕𝑒 𝑓𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐹𝑟 𝑓𝑟𝑜𝑚 𝑜. 𝑐. 𝑐 𝑖𝑠 2.134 𝐴. 𝑡𝑕𝑒 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑚𝑚𝑓 𝐹𝑎 ,

𝑓𝑟𝑜𝑚 𝑝𝑜𝑡𝑖𝑒𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝐴𝐵 = 0.84A

Now the angle between 𝐸𝑟 𝑎𝑛𝑑 𝐼𝑎 𝑖𝑠, 𝑠𝑎𝑦 𝛽, 𝑡𝑕𝑒𝑛 𝐸𝑟 𝑣𝑎𝑙𝑢𝑒

𝛽 = tan−1 106.2

104.0 = 45.6°

46


∴ 𝐹𝑟 = 2.134 𝑐𝑜𝑠 90 + 45.6 + 𝑗𝑠𝑖𝑛 90 + 45.6

= (-1.524+j1.494) Amp

𝐹𝑎=0.84 A

∴ 𝐹𝑓 = 𝐹𝑟 − 𝐹𝑎 = −1.524 + 𝑗1.494 − (0.84) = −2.364 + 𝑗1.494

𝑜𝑟 𝐹𝑓 = 2.797𝐴

Field m.m.f 𝐹𝑓 can also be calculated by referring fig Where AB=2.134 ABC=𝐹𝑎 = 0.84 𝐴 𝑎𝑛𝑑 𝛽 =

45.6°

∴ 𝐹𝑓 = 2.134 + 0.84 sin 45.6° 2 + 0.84𝑐𝑜𝑠45.6° 2 = 2.797 𝐴

For 𝐹𝑓 = 2.797 𝐴, 𝑡𝑕𝑒 𝑒𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 𝑂. 𝐶. 𝐶 𝑖𝑠 169 𝑉

∴ 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑉𝑜𝑙𝑡𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =169 − 127

127× 100 = 33.1%

3. A 3-phase 16 pole alternator has stator connected winding with 144 slots and 10 conductors

per slot. The flux per pole is 0.04 wb and is distributed sinusoidally. The speed is 375 rpm. Find

the frequency, phase emf and line emf. The coil span is 120∘ electrical. Dec-2013

Given Data:

Phase-3, pole-16, slots-144, conductors-10 conductors per slot, flux-0.04 wb, Ns-375 rpm

Find: i) f ii)Phase voltage iii)Line voltage

Solutions:

Ns =120f

P

f =NsP

120=

375 × 16

120= 50Hz

EPh = 4.44 × ϕ × f × TPh × Kp × Kd

𝑇𝑃𝑕 =𝑧𝑃𝑕

2

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 = 144 × 10 = 1440 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟

zPh = Total conductor

3=

1440

3= 480

TPh =480

2= 240

𝐾𝑝 =𝑐𝑜𝑠𝛼

2

α = 180° − 150° = 30°

Kp =cos30°

2= 0.965

47


Kd =sin

mβ2

msinβ2

Pole pitch= 144

9= 9

β =180

pole pitch=

180

9= 20

m=3

Kd =sin

3 × 202

3sin202

=sin 30

3sin10= 0.959

𝐸𝑃𝑕 = 4.44 × 0.04 × 50 × 240 × 0.965 × 0.959

EPh = 1973.9V

𝐄𝐋 = 𝟑 × 𝐄𝐏𝐡 = 𝟑 × 𝟏𝟗𝟕𝟑. 𝟗𝐕 = 𝟑𝟒𝟏𝟖. 𝟗𝐕

4)The open and short circuit test readings for a 3-phases,star connected 1000 KVA,2000 V,50

Hz, Synchronous Generator are: APR-MAY 2018

Fields,amperes: 10 20 25 30 40 50

OC voltage,V: 800 1500 1760 2000 2350 2600

SC armature,,A - 200 250 300 - -

The armature effective resistance is 0.2𝛀per phase.Draw the characterstics curves and estimates

the full load percentage regulation at 0.8 p.f lagging and 0..8 p.f leading. NOV-DEC 2015,May

2017,

Solutions:

Figure shows open circuit characterstics and short circuit characterstics

The Phase voltage are:462,866,1016,1357,1502

48


Full load phase voltage=2000

3= 1155 𝑉

Full load current=1,000 ,000

2000 / 3= 288.7𝐴

Voltage /phase at full load at o.8 p.f

=𝑉 + 𝐼𝑅𝑎𝑐𝑜𝑠𝜙 = 1155 + 288.7 × 0.2 × 0.8 = 1200𝑉

From open circuit curve it is found that field current necessary to produce this voltage=32 A

From short circuit characterstics it is found that field current necessary to produce full load current of

288.7 A is=29 A

a) Cos∅ = 0.8, ∅ = 36.86° 𝑙𝑎𝑔

in fig (b),AB=32 A,BC=29 A and is at an angle (90+36.86°)

Total field current at full load 0.8 P.F lagging is AC=54.6A

Open circuit volt corresponding to a field current of 54,6 is=1555 V

% Regulation=1555 −1155

1155× 100 = 34.6%

b)In this case As PF is leading,BC is drawn with AB at an Angle of

90° − 36.86° = 53.14°

AC=27.4A

Open circuit voltage corresponding to 27.4 A of field excitation is 1080 V

% Regulation=1080 −1155

1155× 100 = −6.4%

1

Dr. A. Jeralidine Viji & M.Vijayaragavan / EEE / MEC Unit-02

MAILAM ENGINEERING COLLEGE Department of Electrical and Electronics Engineering

SUB CODE & NAME: EE6504 & ELECTRICAL MACHINES - II

UNIT - 02

SYNCHRONOUS MOTOR

PART-A

1. What is hunting? How is hunting minimized? Dec-2013, 2012, May-2012,Dec 2015,Nov-

Dec 2016

When a synchronous motor is used for driving a fluctuating load, the rotor starts oscillating

about its new position of equilibrium corresponding to the new load. This is called hunting or phase

winding. To preventing hunting damper windings are embedded in the face of the field poles of the

motor.

2. When is synchronous motor said to receive 100% excitation? Dec 2015

When Eb=V, synchronous motor a said to receive 100% excitation.

3. How does a change of excitation affect its power factor? APR-MAY 2018 or A 3-phase

synchronous motor driving a constant load torque draws power from infinite bus at

leading power factor. How power angle and power factor change if the excitation is increased?

NOV-DEC 2017

Under Excitation Lagging P.F Eb <V

Over Excitation Leading P.F Eb >V

Critical Excitation Unity P.F Eb =V

Normal Excitation Lagging Eb =V

.

4. When a synchronous motor is said to be under-excited? What will be the pf at this

condition?

Excitation emf Eb is less than the supply voltage Eb < V.

Lagging Power factor.

5. What are the inherent disadvantages of synchronous motor?May 2016

Higher cost

Necessity of a dc excitation motor

Greater initial cost

High maintenance cost

2


6. Mention four applications of synchronous motor.

Power Factor correction

Constant speed, constant load drives

Voltage regulation of transmission lines

As frequency changes.

7. What is the role of synchronous motor in transmission lines? How?

Synchronous motor acts as a voltage regulator in transmission lines.

When line voltage decreases due to inductive load, motor excitation is increased

thereby increasing its power factor which compensates for the line voltage drop.

When the line voltage increases due to the line capacitance effect, synchronous

motor excitation is decreased, thereby making its power factor lagging which

helps to maintain the transmission line voltage as its normal voltage.

8. List the advantages and disadvantages of synchronous motor.

Advantages of synchronous motor:

The speed is constant and independent of load.

These motors usually operate at higher frequencies.

Electromagnetic power varies linearly with the voltage.

These motors may be constructed with wider air gaps than the induction motors,

which make them better mechanically.

An over excited synchronous motor having a leading power factor can be operated in

parallel with induction motor.

Disadvantages of synchronous motor:

It cannot be started under load.

It requires dc excitation which must be supplied from the external source.

It has tendency to hunt.

It cannot be used for variable speed jobs as there is no possibility of speed

adjustment.

Collector rings and brushes are required.

9. Define pull out torque in synchronous motor?

The maximum torque in which the motor can develop without pulling out of step or

synchronism is called the pull out torque.

3


10. What is meant by synchronous condenser? Or How the synchronous motor can be used

as synchronous condenser ? Dec-2014, 2012,May 2017,APR-MAY 2018

Synchronous motor operating at an over excited condition is called synchronous condenser.

The synchronous condensers having leading power factor are widely used for improving power

factor of those power systems which employ a large number of induction motors and other lagging

power factor loads.

11. Define pull in torque in synchronous motor.

It pertains to the ability of the machine to pull into synchronism when changing from

induction to synchronous motor operation.

12. What are V curves?May-2013,May 2017.

The V curves show the relation that exists between the armature current and the field current

for different constant power input.

13. Give the expression for gross mechanical power developed by synchronous motor.

Pm =3 Eb V sin δ

Xs

Where Eb = Excitation emf, V=Supply voltage

Xs = Synchronous reactance δ=Load angle

14. In what operating condition is a synchronous motor referred to as synchronous condenser

or phase modifier?

On over excited conditions.

15. Name the important characteristics of a synchronous motor not found in an induction

motor? May-2014

The essential features of synchronous motor are,

The rotor speed is synchronous with stator rotating field.

The power factor can be easily varied by varying its field current.

It is used for constant speed operation.

16. What is the common starting method used for synchronous motor? May-2013, 2014

Starting with the help of damper winding.

Starting with the help of separate small induction motor.

Starting by using an ac motor coupled to the synchronous motor.

17. Why does the synchronous motor always run at synchronous speed? May-2012

A synchronous motor always runs at synchronous speed because of the magnetic locking

between the stator and rotor poles.

4


18. What is the function of the damper winding? Dec-2013,Nov-Dec 2016,2017

The uses of damping winding are

To develop necessary starting torque.

To prevent hunting or phase swinging.

19. Why is synchronous motor non self-starting?

Why can’t the synchronous motor self start? Explain. May-2015

If a three phase supply is given to the stator of a stationary synchronous machine with the

rotor excited, no steady starting torque will be developed instead, a sinusoidally time varying torque

is developed, the average of which is zero and what is why synchronous motor is not self starting.

20. What does synchronous phase modifier mean?

Synchronous condensers are sometimes operated at power factors ranging from lagging

through unity to leading for voltage control. When operated I this way, a synchronous condenser is

called a synchronous phase modifier.

21. How a synchronous machine is different form induction motor?

S. No Synchronous machines Induction motor

1 Machine speed is constant by varying

the load.

Machine speed is not constant by varying

the load.

2 DC excitation is required. DC excitation is not required.

3 Not self starting machine. Self starting machine.

4 High cost Low cost

22. What could be the reasons if a 3-phase synchronous motor fails to start? Dec-2014

It is usually due to the following reasons:

Voltage may be too low

Some faulty connections in auxillary apparatus

Too much starting load

Open circuit in one phase or short circuit

Field excitation may be excessive

23. Draw the typical torque angle characteristics of synchronous machine. May-2015

5


24) How we can change the speed of synchronous motor? May 2016

Normally synchronous motor is constant speed motor it always rotate synchronous speed. it either

run at synchronous speed or in standstill position. never run below or above synchronous speed.

PART-B

1. Explain briefly principle operation of synchronous motor?NOV-DEC 2017

Synchronous motor:

A synchronous motor is electrically identical with an alternator (or) AC generator. In fact a

given synchronous machine may be used at least theoretically as an alternator when driven

mechanically or as a motor, when driven mechanically just as in the case of DC machine. Most

synchronous motors are rated between 150KW and 15MW and run at speeds ranging from 150 to

1800 rpm.

Characteristics of synchronous motors:

It runs either at synchronous speed or not at all.

It’s not inherently self starting.

It’s operated under a wide range of pf; it cannot be used for power correction’s

purpose.

Purpose of working:

Synchronous motor works on the principle of the magnetic locking. When two unlike poles

are brought near each other. If the magnets are strong there exists a tremendous force of attraction

between those two poles. In such condition the tow magnets are said to be magnetically locked.

6


When a balanced 3 phase ac supply is given to a three phase stator winding of the

synchronous motor. It produces a rotating magnetic field. The speed of the magnetic field is known

as the synchronous peed. From fig. The two stator poles are marked as Ns and Ss. Assume they are

rotate in clockwise directions at synchronous speed.

From fig (a) The stator points X and Y like poles Ns and Nr of rotor repel each other.

Similarly Ss of stator and Sr of rotor also repel each other. Now the rotor will begin to rotate in the

anticlockwise direction.

Half a cycle latter the position of the stator poles are interchanged Ns is at the point Y and

Ss at point X. Now Ns attracts Sr and Ss attracts Nr. Hence the rotor tends to rotate clockwise since

the repulsion and attraction takes place in every half a cycle alternatively, the rotor is stationary.

Therefore synchronous motor is not a self starting motor.

It needs a two separate supplies-one a D.C source for excitation of the rotor and other, a

three phase supply for the stator. Because of the inter locking between the stator and rotor poles the

motor runs only at one speed, the synchronous peed.

2. Derive the equivalent circuit diagram of synchronous motors and derive an equivalent for

the power and torque output of a synchronous motor?

Dec-2013, May-2013, 2012,Dec 2015

Derive the torque equation for synchronous motor.

From the figure, the applied voltage V is the vector sum of reversed back emf (ie) –Eb and

the impedance drop Ia Zs. In other words, V = - Eb + Ia Zs. The angle δ between the phasors V and

Eb is called the load angle or power angle of the synchronous motor.

7


Except for very small machines,the armature resistance of a synchronous motor is negligible as

compared to its synchronous reactance.Hence the equivalent circuit for the motor become as shown

in below figure

From the phasor diagram

𝐴𝐵 = 𝐸𝑏𝑠𝑖𝑛𝛿 𝑎𝑛𝑑

𝑐𝑜𝑠𝜙 =𝐴𝐵

𝐼𝑎𝑋𝑠

So AB = 𝐼𝑎𝑋𝑠𝑐𝑜𝑠𝜙

∴ 𝐸𝑏𝑠𝑖𝑛𝛿 = 𝐼𝑎𝑋𝑠𝑐𝑜𝑠𝜙 = 𝐼𝑎𝑐𝑜𝑠𝜙 =𝐸𝑏𝑠𝑖𝑛𝛿

𝑋𝑠

𝑃 = 𝑉𝐼𝑎𝑐𝑜𝑠𝜙

𝑃 =𝑉𝐸𝑏𝑠𝑖𝑛𝛿

𝑋𝑠

8


𝑃𝑖𝑛 =3𝐸𝑏𝑉

𝑋𝑠𝑠𝑖𝑛𝛿 𝑓𝑜𝑟 3 𝑝𝑕𝑎𝑠𝑒𝑠

Since stator copper loss has been neglected, 𝑃𝑖𝑛 𝑎𝑙𝑠𝑜 representes the gross mechanical

power(𝑃𝑚 )developed by the motor

𝑃𝑚 =3𝐸𝑏𝑉

𝑋𝑠𝑠𝑖𝑛𝛿

Gross torque developed by the motor

𝑇 =𝑃𝑚𝜔𝑚

𝑇 =3𝐸𝑏𝑉

𝜔𝑚𝑋𝑠

∴ 𝜔𝑚 =2𝜋𝑁

60

𝑇 =9.55𝑃𝑚

𝑁𝑁𝑚

Operation on infinite bus bars

The synchronous motor connected to an infinite bus bar behaves similarly for the changes

in the load at constant excitation. As the load increases, the load angle increases, current increases

and power factor changes. The changes in the power factor depends on the excitation used for

synchronous motor (ie) whether it is normally excited (Ebph = Vph), over excited (Ebph > Vph), or

over excited (Ebph < Vph).

Thus the changes in load on synchronous motor connected to an infinite bus bar can be summarized

as,

1. Irrespective of excitation, as load increases, the load angle δ and armature current Ia increases.

2. When the motor is normally excited (Ebph = Vph), then as load increases, the change in current is

more significant than the change in power factor. The power factor tends to become more and more

lagging as the load increases.

3. When the motor is over excited or under excited, the power factor changes are more significant

than the changes in the current as load changes.

4. When the motor is over excited or under excited, the power factor tends to approach to unity as

the load increases.

9


3. Describe a laboratory method of obtaining V and inverted V curves of a synchronous

motor? Dec-2014, 2013, 2012,2015,May 2016,Nov-Dec 2016,May 2017.

We know that if excitation is varied from under excitation to over excitation. Varying the

current Ia decreases.

So it becomes min at unity PF and then again increases but initial lagging current becomes

unity and the becomes leading in nature.

This can be shown in fig.

Excitation table:





Excitation can be increased by increasing the field current passing through the field winding

of synchronous motor.

IF graphs plotted Ia vs If, then its shape looks like and English alphabet V. In such a graphs,

are obtained at various load conditions. Such curves are called V-Curves of Synchronous

motors.

10


If graphs plotted If vs cosφ then shape of the graph looks like and inverted V. Such curves

obtained by plotting P.F against If at various load conditions are called V-curves of

synchronous motors.

4. Derive the equation for the Power input and power developed by the synchronous motor.

Net input to the synchronous motor is the three phase input to the stator.

∴ 𝑃𝑖𝑛 = 3𝑉𝐿𝐼𝐿𝑐𝑜𝑠∅ 𝑊

Where 𝑉𝐿=Applied Line Voltage

𝐼𝐿 = 𝐿𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑟𝑎𝑤𝑛 𝑏𝑦 𝑡𝑕𝑒 𝑚𝑜𝑡𝑜𝑟

𝑐𝑜𝑠∅ = 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑝. 𝑓 𝑜𝑓 𝑠𝑦𝑛𝑐𝑕𝑟𝑜𝑛𝑜𝑢𝑠 𝑚𝑜𝑡𝑜𝑟

Or

𝑃𝑖𝑛 = 3 𝑝𝑒𝑟 𝑝𝑕𝑎𝑠𝑒 𝑝𝑜𝑤𝑒𝑟

= 3 × 𝑉𝑝𝑕 𝐼𝑝𝑕𝑐𝑜𝑠∅ 𝑊

Now in stator,due to its resistance 𝑅𝑎 per phase there are ststor copper losses.

Total stator copper losses=3 × 𝐼𝑎𝑝𝑕 2

× 𝑅𝑎

∴ the remaining power is converted to the mechanical power,called gross mechanical power

developed by the motor denoted as 𝑃𝑚

∴ 𝑃𝑚 = 𝑃𝑖𝑛 − 𝑆𝑡𝑎𝑡𝑜𝑟 𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠𝑒𝑠

Now 𝑃 = 𝑇 × 𝜔

∴ 𝑃𝑚 = 𝑇𝑔 ×2𝜋𝑁𝑠

60 𝐴𝑠 𝑠𝑝𝑒𝑒𝑑 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑁𝑠

∴ 𝑃𝑚 × 60

2𝜋𝑁𝑠 𝑁𝑚

This is the gross mechanical torque developed. In Dc motor ,electrical equivalent of gross

mechanical power developed is 𝐸𝑏 × 𝐼𝑎 , Similar in synchronous motor the electrical equivalent of

gross mechanical power developed is given by,

𝑃𝑚 = 3𝐸𝑏𝑝𝑕 × 𝐼𝑎𝑝𝑕 × 𝑐𝑜𝑠 𝐸𝑏𝑝𝑠 ∧ 𝐼𝑎𝑝𝑕

i) For Lagging P.F= 𝐸𝑏𝑝𝑠 ∧ 𝐼𝑎𝑝𝑕 = ∅ − 𝛿

ii) For Leading P.F= 𝐸𝑏𝑝𝑠 ∧ 𝐼𝑎𝑝𝑕 = ∅ + 𝛿

iii) For Unity P.F= 𝐸𝑏𝑝𝑠 ∧ 𝐼𝑎𝑝𝑕 = 𝛿

In general

𝑃𝑚 = 3𝐸𝑏𝑝𝑕 × 𝐼𝑎𝑝𝑕 × cos(∅ ± 𝛿

11


Positive sign for leading power factor and Negative sign for lagging P.F.

Net output of the motor then can be obtained by subtracting friction and windage i.e.

mechanical losses from gross mechanical power developed.

∴ 𝑃𝑜𝑢𝑡 = 𝑃𝑚 − 𝑀𝑒𝑐𝑕𝑎𝑛𝑖𝑐𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠

∴ 𝑇𝑠𝑕𝑎𝑓𝑡 =𝑃𝑜𝑢𝑡 × 60

2𝜋𝑁𝑠 𝑁𝑚

𝑊𝑕𝑒𝑟𝑒 𝑇𝑠𝑕𝑎𝑓𝑡 = 𝑠𝑕𝑎𝑓𝑡 𝑡𝑜𝑟𝑞𝑢𝑒 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑡𝑜 𝑙𝑜𝑎𝑑

𝑃𝑜𝑢𝑡 = 𝑃𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑡𝑜 𝑙𝑜𝑎𝑑

∴ 𝜂 =𝑃𝑜𝑢𝑡𝑃𝑖𝑛

× 100 ……… .𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

The power flow in synchronous motors can be summarized as shown in figure.

5. Discuss briefly why synchronous motor is not self starting. Explain the different methods

for starting a synchronous motor? May-2012, Dec-2014, 2012,May 2016,Nov-Dec 2016,2017

METHODS OF STARTING:

The rotor (which is as yet unexcited) is speeded upto synchronous (or) near synchronous

speed by some arrangement and then excited by the DC source.

The moment this synchronously rotating rotor is excited, it’s magnetically locked into

position with the rotor.

Because of this interlocking of stator and rotor poles that the motor has either to run

synchronously (or) not at all.

The various methods to start the synchronous motors are:

Using pony motors.

Using damper winding

12


As a slip ring induction motor,

Using small DC machine coupled to it.

(i) Using pony motors:

In this method, the rotor is brought to the synchronous speed with the help of some

external device like small induction motor. Such device is called pony motors.

Once magnetic locking achieved the pony motors is developed. Then motor rotates

at synchronous speed continuously.

(ii) Using damper winding:

The synchronous motor is made self starting by providing a special winding on the

rotor poles, known as damper winding or squirrel cage winding. The damper

winding consists of short circuited copper bars embedded in the face of the field

poles.

When 3 phase ac supply is give it runs at a speed near the synchronous speed. At this

staged excitation is given to the field winding.

The rotor will be pulled into synchronous speed. Thus the damper windings are used

to make the machines as self starting and to minimize hunting.

Hunting is nothing but when the load is suddenly increased (or) decreased the rotor

oscillates about its synchronous position. This action is called hunting.

(iii) As a slip ring induction motor:

The above method of starting synchronous motor as a squirrel cage I.M does not provide

high starting torque.

So to achieve this instead of damper winding its designed to form a 3 phase star (or) delta

connected winding.

The three ends of this winding are brought out through slip rings.

An external rheostat introduced in series with the rotor circuit. So when stator is excited the

motor starts as a slip ring I.M due to resistance added. In the rotor provide high starting

torque.

Now the resistance is gradually cut off. At rotor gathers speed when motor attains speed

near synchronous. The D.C excitation is provided to the rotor.

Then moltor gets pulled into synchronism and starts rotating at synchronous speed.

(iv) Using small dc machines:

The d.c motor coupled in synchronous machines. This is used to as a d.c motor to rotate the

synchronous motor at a synchronous speed, then the excitation to the rotor is provided. Once

motor starts running as a synchronous motor, the same dec machines acts as a D.C

generator called exciter.

The field of the synchronous motor is then excited by this exciter itself. The synchronous

motor then excited and synchronized with AC supply mains. At the moment of

synchronizing the synchronous motor is switched on with the ac mains and either the dc

13


motor is disconnected from the dc supply mains or the field of the dc machines is

strengthened until it begins to function as a generator.

Now the synchronous machine is operating as a motor, from AC supply mains and dc

machines acts as load on it. The synchronous motor can also be started by the exciter

mounted on an overhung synchronous motor bracket and shaft extension.

6. Explain the effect of changing excitation on armature current and power factor?

Dec-2012, May-2014, 2012

Illustrate through neat phasor diagram, the functioning of synchronous machine with

varying excitation under constant real power load. May-2015,2016,MAY-JUNE 2018

OPERATION OF SYNCHRONOUS MOTOR AT CONSTANT VARIABLE LOAD

EXCITATION:

Consider a synchronous motor operating at certain load. The corresponding load angle is δ.

At start consider normal behavior of the synchronous motor, where excitation is adjusted to get EB

=V. Such an excitation is called normal excitation of the motors.

(i)NORMAL EXCITATION

Motor is drawing certain current Ia from the supply and power I/P to the motor is say P in. So

the P.F of the motor is lagging in nature. Shown in fig (a)

When excitation is changed, Eb changes, but there is hardly any change I the losses of the

motor.

So the power I/P also remain same for constant load demanding same power o/p.

(ii)UNDER EXCIATION:

In this case (Eb < V) is called under excitation.

Due to this Er increase but keep la cos constant. It shows fig(b). So in this case current

drawn by the motor increase the P.F cos decreases and becomes more and more lagging in

nature.

(iii)OVER EXCITATION:

In this case (Eb > V) is called over excitations.

Due to this increased magnitude of Eb, Er also increases but phase of Er also changes.

Son in this condition P.F of the motor becomes leading in nature. So in this case motor

works on leading power factor.

14


(iv)CRICTICAL EXCITATION:

In this case both Eph and Vph are equal and Ia cosφ constant. So cosφ=1. The P.F is

unity.Hence





15


7. Write short notes on hunting and damper winding.

Illustrate the phenomenon of hunting and the use of damper winding with the help of

dynamic equations. May-2015

Hunting:

When synchronous motor is on no load, the stator and rotor pole axes almost coincide

with each other.

When motor is loaded, the rotor pole axis falls back with respect to stator. The angle by

which rotor retards is called load angle or angle of retardation, δ.

If the load connected to the motor is suddenly changed by a large amount, then rotor tries

to take its new equilibrium position.

But due to inertia of the rotor, it cannot achieve its final position instantaneously. While

achieving its new position due to inertia it passes beyond its final position corresponding to new

load. This will produce more torque than what is demanded. This will try to reduce the load angle

and rotor swings in other direction. So there is periodic swinging of the rotor on both sides of the

new equilibrium position, corresponding to the load.

Fig. Hunting in synchronous motor

Such oscillation of the rotor about its new equilibrium position, due to sudden application

or removal of load is called swinging or hunting in synchronous motor.

Due to such hunting, the load angle δ changes its value about its final value, As δ changes,

for same excitation, (ie) Ebph the current drawn by the motor also changes. Hence during hunting

there are changes in the current drawn by the motor which may cause problem to the other

appliances connected to the same line. The change in armature current due to hunting is shown in

figure.

16


Fig. Current variations during hunting

If such oscillations continue for longer period, there are large fluctuations in the current. If

such variations synchronise with the natural period of oscillation of the rotor, the amplitude of the

swing may become so great that motor may come out of synchronism. At this instant mechanical

stresses on the rotor are severe and current drawn by the motor is also very large. So motor gets

subjected to large mechanical and electrical stresses.

Damper winding:

The short circuited winding places in the slots provided in pole faces is called damper

winding.

When the rotor starts oscillating (ie) when hunting starts a relative motion between

damper winding and the rotating magnetic field is created. Due to this relative motion, emf gets

induced in the damper winding. According to lenz’s law, the direction of induced emf is always so

as to oppose the cause producing it. The cause is the hunting. So such induced emf opposes the

hunting. The induced emf tries to damp the oscillations as quickly as possible. Thus hunting is

minimized due to damper winding.

Fig. Effect of damper winding on hunting

17


The time required by the rotor to take its final equilibrium position after hunting is called

as setting time of the rotor. If the load angle δ is plotted against time, the schematic representation

of hunting can be obtained as shown in figure. It is shown that due to damper winding the setting

time of the rotor reduces considerably.

8. Explain how synchronous motor can be operated as a synchronous condenser?

May-2012, NOV-DEC 2017

When synchronous motor is over excited it takes leading P.F current. If synchronous motor

is on no load where load angle (δ) is very small and its over excited. The P.F angle increases

almost upto 90.

And motor runs almost zero leading P.F condition. Hence over excited synchronous .motor

operating on no load condition is called as synchronous condenser (or) Synchronous

capacitor.

DE-MERITS OF POOR PF:

Need more conductor size.

Fixed active power (p), large KVA rating need. Hence increases the cost.

More cu losses

Poor Effeciency.

USE OF SYNCHRONOUS CONDENSER IN PF IMPROVEMENT:

Now let Vph is the voltage applied and Iph is the current lagging Vph by angle. This

P.F is very low, lagging.

The synchronous motor acting as a synchronous. Condenser is now connected across

the same supply. This draws a leading current of Iph.

The total current drawn from the supply is now phasor of Iph and I2ph. This total

current I1 now lags Vph by small angle due to which effective power factor get

improved.

18


MERITS OF SYNCHRONOUS MOTOR:

Speed is constant and independent of load.

Usually operate at higher efficiencies.

Mechanically it’s better than I.M

An over excited synchronous motor having a leading P.F can be operated in parallel

with I.M

Electromagnetic power varies linearly with the voltage.

DE-MERITS:

Construction is complicated.

It’s not self start m/c.

Speed control is no possible.

It required separate DC sources.

Collector rings and brushes are required.

Higher cost.

APPLICATIONS:

Used as P.F correction devices.

Used as phase advances.

Used as a phase modifiers for voltage regulation of the transmission lines.

In typical application of high speed.

Blowers

DC generators

Line Shifts

Centrifugal pumps

Compressors

Rubber and Paper Mill.

9. Derive the expression for power developed in a synchronous motor. Also find the condition

for maximum power developed? May-2013,2017.

OL – Supply voltage / phase

I – Armature current

LM – Back emf at a load angle of δ

OM – Resultant voltage, ER

Er = IZs (or I Xs if Ra is negligible)

19


I lags/leads V by an angle ∅ and lags behind 𝐸𝑟 by an angle 𝜃.

∴ 𝜃 = 𝑡𝑎𝑛−1 𝑋𝑠

𝑅𝑎

Line NS is drawn at angle 𝜃 𝑡𝑜 LM

Line LN and QS are perpendicular to NS.

Mechanical power developed per phase in the rotor

𝑃𝑚𝑒𝑐 𝑕 = 𝐸𝑏 𝐼 𝑐𝑜𝑠𝜓

Δ𝑂𝑀𝑆, 𝑀𝑆 = 𝐼𝑍𝑠𝑐𝑜𝑠𝜓

𝑀𝑆 = 𝑁𝑆 − 𝑁𝑀 = 𝐿𝑄 − 𝑁𝑀

𝐼𝑍𝑠𝑐𝑜𝑠𝜓 = 𝑉𝑐𝑜𝑠 𝜃 − 𝛿 − 𝐸𝑏𝑐𝑜𝑠𝜃

Or 𝐼 𝑐𝑜𝑠𝜓 =𝑉

𝑍𝑠𝑐𝑜𝑠 𝜃 − 𝛿 −

𝐸𝑏

𝑍𝑠𝑐𝑜𝑠𝜃

𝑃𝑚𝑒𝑐 𝑕 𝑃𝑕𝑎𝑠𝑒 = 𝐸𝑏 𝑉

𝑍𝑠𝑐𝑜𝑠 𝜃 − 𝛿 −

𝐸𝑏


𝑃𝑚𝑒𝑐 𝑕 𝑃𝑕𝑎𝑠𝑒 =𝐸𝑏𝑉

𝑍𝑠cos 𝜃 − 𝛿 −

𝐸𝑏2


This is the expression for the mechanical power developed in terms of load angle (𝛼)and the

internal angle 𝜃 of the motor for a constant voltage V&𝐸𝑏 .

Condition for maximum power developed can be found by differentiating the above

expression with respect to load angle & the equating it to zero.

20


𝑑𝑃𝑚𝑒𝑐 𝑕

𝑑𝛼=

−𝐸𝑏𝑉

𝑍𝑠sin 𝜃 − 𝛿 = 0

∴ sin 𝜃 − 𝛿 = 0 𝑜𝑟 𝜃 = 𝛿

∴ 𝑉𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢 𝑝𝑜𝑤𝑒𝑟,

𝑃𝑚𝑒𝑐 𝑕 𝑚𝑎𝑥 =𝐸𝑏𝑉

𝑍𝑠−𝐸𝑏

2

𝑍𝑠𝑐𝑜𝑠 𝛿 =

𝐸𝑏𝑉

𝑍𝑠−

𝐸𝑏2

𝑍𝑠𝑐𝑜𝑠 𝜃

This shows that the maximum power and hence torque depends on V and 𝐸𝑏 that is

excitations.

Maximum value of 𝜃 𝑎𝑛𝑑 𝑕𝑒𝑛𝑐𝑒 𝛼 is the same but maximum torque will be proportional to

the maximum power developed.

If 𝑅𝑎 is neglected ,then 𝑍𝑠 = 𝑋𝑠 and 𝜃 = 90°

𝑐𝑜𝑠𝜃 = 0

𝑃𝑚𝑒𝑐 𝑕 =𝐸𝑏𝑉

𝑋𝑠cos(90° − 𝛿)


𝑋𝑠𝑠𝑖𝑛 𝛿

This gives the value of mechanical power developed in terms of 𝛿 the basics variable of a

synchronous machine.


𝑋𝑠,𝑤𝑕𝑒𝑛 (𝛿 = 90°)

𝑡𝑕𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝑡𝑕𝑒 𝑝𝑢𝑙𝑙𝑜𝑢𝑡, 𝑡𝑜𝑟𝑞𝑢𝑒.

To determine the value of excitation or induced e.m.f 𝐸𝑏 to give maximum power developed

possible, differentiate with respect to 𝐸𝑏 and equate to zero.

21


𝑑𝑃𝑚𝑒𝑐 𝑕

𝑑𝐸𝑏=

𝑉

𝑍𝑠−

2𝐸𝑏

𝑍𝑠𝑐𝑜𝑠𝜃 = 0 𝑜𝑟 𝐸𝑏 =

𝑉

2𝑐𝑜𝑠𝜃

Substituting 𝐸𝑏 =𝑉

2𝑐𝑜𝑠𝜃 𝑖𝑛 (𝑃𝑚𝑒𝑐 𝑕)𝑚𝑎𝑥

(𝑃𝑚𝑒𝑐 𝑕)𝑚𝑎𝑥 =𝑉2

2𝑍𝑠𝑐𝑜𝑠𝜃−

𝑉2

4𝑍𝑠𝑐𝑜𝑠𝜃=

𝑉2

4𝑍𝑠𝑐𝑜𝑠𝜃=

𝑉2

4𝑅𝑎

𝑤𝑕𝑒𝑟𝑒 𝑅𝑎 = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑚𝑜𝑡𝑜𝑟.

𝐻𝑒𝑛𝑐𝑒 (𝑃𝑚𝑒𝑐 𝑕)𝑚𝑎𝑥 =𝑉2

4𝑅𝑎

10. What are constant excitation circles and constant power circle for a synchronous motor?

How are they derived?Dec-2014 , NOV-DEC 2017

Constant excitation circle:

As 𝑬𝒃𝒑𝒉 depends on flux, for constant excitations Ebph is constant.for constant excitation,if load is

varied then 𝛿 keeps on charging,due to which 𝑉𝑝𝑕 − 𝐸𝑝𝑕 = 𝐸𝑅𝑝𝑕 = 𝐼𝑎𝑝𝑕𝑍𝑠 keeps on changing.the

locous of extremities of 𝐸𝑅𝑝𝑕 = 𝐼𝑎𝑝𝑕𝑍𝑠 is a circle and as 𝑍𝑠 is constant,represents current locus for

the synchronous motor under constant excitations and variable load conditions.As 𝛿

increases, 𝐼𝑎𝑝𝑕𝑍𝑠 increases and motor draws more current.As load decreases, 𝛿 decreases hence

𝐼𝑎𝑝𝑕𝑍𝑠 decreases and motor draws less current.such a current locus is shown in figure.

Constant power circle or Blondel diagram:

The Blondel diagram of a synchronous motor is an extension of ansimple phasor diagram of a

synchronous motor.

For a synchronous motor,the power input to the motor per phase is given by,

𝑃𝑖𝑛 = 𝑉𝑝𝑕 𝐼𝑎𝑝𝑕 𝑐𝑜𝑠∅

22


The gross mechanical power developed per phase will be equal to the difference between 𝑃𝑖𝑛 per

phase and the per phase copper losses of the winding.

Copper loss per phase= 𝐼𝑎𝑝𝑕 2𝑅𝑎

𝑃𝑚=𝑉𝑝𝑕 𝐼𝑎𝑝𝑕 𝑐𝑜𝑠∅ − 𝐼𝑎𝑝𝑕 2𝑅𝑎

For mathematical convenience let 𝑉𝑝𝑕 = 𝑉 𝑎𝑛𝑑 𝐼𝑎𝑝𝑕 = 𝐼,

∴ 𝑃𝑚 = 𝑉𝐼𝑐𝑜𝑠∅ − 𝐼2𝑅𝑎

𝐼2𝑅𝑎 − 𝑉𝐼𝑐𝑜𝑠∅ + 𝑃𝑚 = 0

𝐼2 − 𝑉𝐼𝑐𝑜𝑠 ∅

𝑅𝑎+

𝑃𝑚

𝑅𝑎= 0 …………………………………(1)

Now consider the phasor diagram as shown in figure.,

The above equation represents polar equations to a circle.to obtain this circle in a phasor

diagrams,draw a line OY at an angle 𝜃 𝑤𝑖𝑡𝑕 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑂𝐴.

∴ ∠𝑌𝑂𝐴 = 𝜃

∴ ∠𝑌𝑂𝐵 = 𝜑

23


The circle represented by equation (1) has a center at same point o’ on the line OY.the circle drawn

with center as o’ and radius as O’B represents circle of constant power.This is called Blondel

diagram,shown in the fig.

Thus if excitation is varied while the power is kept constant,then working point B will move along

the circle of constant power.

11) A synchronous motor having 40% reactance and a negligible resistance is to be operated

at rated load at 0.8 power factor lag and 0.8 power factor lead. What are the values of

induced emf?

Given Data:

V=100v,Impedance drop =𝐼𝑎𝑋𝑠 = 40𝑣

i) UPF:

𝜃 = 90°

𝐸𝑏 = 1002 + 402 = 108𝑉

ii)0.8 P.F lagging:

Here ∟𝐶𝐴𝐵 = 𝜃 − 𝜑 = 90° − 36.86° = 53.13°

𝐸𝑏2 = 1002 + 402 − 2 × 100 × 40 × 𝑐𝑜𝑠 53.13

𝐸𝑏 = 82.5 𝑉

iii)0.8 P.f leading:

24


Here (𝜃 + 𝜑) = 90 + 36.86 = 126.86°

𝐸𝑏2 = 1002 + 402 − 2 × 100 × 40 × 𝑐𝑜𝑠126.9

𝐸𝑏 = 128 𝑉

12) A 3000 V,3-phase synchronous motor running at 1500 rpm has its excitations kept

constant corresponding to no load terminal voltage of 3000V.Determine the power

input,power factor and torque developed for an armature current of 250 A.if the

synchronous reactance is 5 ohm per phase and armature resistance is neglected.

Solution:

Voltage/phase=3000

3= 1732 𝑉

Induced emf=1732V

Impedance drop=𝐼𝑎𝑋𝑠 = 750 𝑉

As shown in figure ,the armature current 𝐼𝑎 𝑖𝑠 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑡𝑜 𝑙𝑎𝑔 𝑏𝑦 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝜑.since 𝑅𝑎 is

negligible,𝜃 = 90°

∟ 𝐶𝐴𝐵 = 90 − 𝜑

Considering ∆𝐶𝐴𝐵,𝑤𝑒 𝑕𝑎𝑣𝑒

17322 = 17322 + 7502 − 2 × 1732 × 750 × cos(90 − 𝜑)

𝑠𝑖𝑛𝜑 = 0.2165

25


𝜑 = 12.5°

Cos𝜑 = 0.976 𝑙𝑎𝑔

Input power 𝑃𝑖𝑛 = 3 × 3000 × 250 × 0.976

𝑃𝑖𝑛 = 1267.86𝐾𝑤

Speed 𝑁𝑠 = 1500 𝑟𝑝𝑚.

Torque developed 𝑇𝑔 =9.55×𝑃𝑚

𝑁𝑠=

9.55×1267 .68×103

1500= 8072 𝑁 −𝑚

13) A 3-phase,11000 V,star connected synchronous motor takes a load of 100 A. The

effective synchronous reactance and resistance per phase are 30 𝛀 and 0.8 𝛀 respectively.

Find the power supplied to the motor and the induced emf for (1)0.8 pf lag, (2) 0.8 pf lead.

Given data:

Supply voltage 𝑉𝐿 = 11000 𝑉, Load current 𝐼𝐿 = 100 𝐴,

Synchronous reactance 𝑋𝑠 = 30 Ω, Resistance𝑅𝑎 =0.8 Ω.

To find:

Power supplied to the motor and induced emf for (1) 0.8 pf lag (2)0.8 pf lead.

Solutuion:

Input power 𝑃𝑖𝑛 = 3𝑉𝐿𝐿𝐿 cos∅ = 3 ∗ 11000 ∗ 100 ∗ 0.8 = 1524.2𝑘𝑊

Synchronous impedance 𝑍𝑠 = 𝑅𝑎2 + 𝑋𝑠

2= 0.82 + 302 = 30.01Ω

𝜑 = cos−1 0.8 = 36.87°

Intenal angle 𝜃 = tan−1 𝑋𝑠

𝑅𝑠 = tan−1

30

0.8 = 88.47°

𝐸𝑟 = 𝐼𝑍𝑠 = 100 × 30.01 = 3001𝑉.

Per phase voltage V=11000/ 3=6351 V

𝐸𝑏 = 𝑉2 + 𝐸𝑟2 − 2𝑉𝐸𝑅 cos(𝜃 − ∅)

𝐸𝑏 = 63512 + 30012 − 2 ∗ 6351 ∗ 3001 ∗ cos(88.47° − 36.87°)

𝐸𝑏 = 5066 𝑉

II) 0.8 power factor lead

𝑃𝑖𝑛 = 3𝑉𝐿𝐿𝐿 cos∅ = 3 ∗ 11000 ∗ 100 ∗ 0.8 = 1524.2𝑘𝑊

Back emf 𝐸𝑏 = 𝑉2 + 𝐸𝑟2 − 2𝑉𝐸𝑅 cos(𝜃 − ∅)

26


𝐸𝑏 = 63512 + 30012 − 2 ∗ 6351 ∗ 3001 ∗ cos(88.47° + 36.87°)

𝐸𝑏 = 8449.26

14) A synchronous motor absorbing 75kW is connected in parallel with a factor load of

300kW having a lagging power factor of 0.9. If the combined load has lagging power

factor of 0.95, what is the value of leading kVAR supplied by the motor and what power

factor is it working?

Solution:

Factory load 𝑃𝐿 = 300𝑘𝑊

Power factor cos∅𝐿 = 0.9 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

Load kVAR ∅𝐿 = 300 tan(cos−1 0.9) = 300 ∗ 0.484 = 145.29𝑘𝑉𝐴𝑅

Synchronous motor load 𝑃𝑚 = 75𝑘𝑊

Total load 𝑃 = 𝑃𝐿 + 𝑃𝑚 = 300 + 75 = 375𝑘𝑊

Combined power factor cos∅ = 0.95 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

∅ = cos−1(0.95) = 18.19°

Combined kVAR 𝑄 = 𝑃 tan ∅ = 375 ∗ tan 18.19 = 123.22𝑘𝑉𝐴𝑅(𝑙𝑎𝑔)

Leading kVAR supplied by the motor

𝑄𝑚 = 𝑄𝐿 −𝑄 = 145.29 − 123.22 = 22.07𝑘𝑉𝐴𝑅

kVA supplied by the motor 𝑆𝑚 = 𝑃𝑚2 + 𝑄𝑚2 = 752 + (22.07)2 = 78.18𝑘𝑉𝐴𝑅

Power factor of the motor cos∅ =𝑃𝑀

𝑆𝑀=

75

78.18= 0.959(𝑙𝑒𝑎𝑑)

15)A 75 KW ,three phase Y connected 50 Hz,440 V,Cylindrical rotor synchronous motor operates

at rated conditions with 0.8 p.f leading. the motor efficiency excluding field and stator losses is

95% and Xs=2.5 ohm. Calculate

i) Mechanical power developed

ii) Armature Current

iii) Back Emf

iv) Power Angle

v) Maximum or pull out torque of the motor. APR-MAY 2018

Soloutions:

𝑁𝑠 =120

50 × 4= 1500 𝑅𝑝𝑚 = 25 𝑟𝑝𝑠

i) Since power input is known

∴ 3 × 440 × 𝐼𝑎 × 0.8 = 78.950; 𝐼𝑎 = 129𝐴

27


iii)Applied Voltage/phase=440/ 3 =254 V.Let V=254<0°

now V=𝐸𝑏 + 𝑗𝑋𝑠=516,<-30°

iv)𝛼 = −30°

v)Pull out torque occurs when 𝛼 = 90°

𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑚 = 3𝐸𝑏𝑉

𝑋𝑠𝑠𝑖𝑛𝛿 = 3

256 × 516

2.5= sin 90° = 157.275𝑊

Pull out torque =9.55×157.275

1500= 1000𝑁 −𝑚

1

Dr.A.Jeraldine Viji & M.Vijayaragavan / EEE / MEC Unit-03

MAILAM ENGINEERING COLLEGE

Department of Electrical and Electronics Engineering

UNIT - 03

THREE PHASE INDUCTION MOTOR

PART-A

1. State the principle of 3-phase induction motor.

While starting, rotor conductors are stationary and they cut the revolving magnetic field and so an

emf is induced in them by electromagnetic induction. This induced emf produces a current if the circuit is

closed. This current opposes the cause by Lenz’s law and hence the rotor starts revolving in the same

direction as that of the magnetic field.

2. What are the two types of 3-phase induction motor? May-2012

Squirrel cage and slip ring and wound type induction motor.

3. How will you change the direction of rotation of three phase induction motor? May-2011 ,Nov-

Dec 2016

By changing the phase sequence of the 3-phase supply.

4. Why an induction motor is called as rotating transformer?

The rotor receives same electrical power in exactly the same way as the secondary of a two

winding transformer receiving its power from primary. That is why induction motor is called as rotating

transformer.

5. What is the function of slip ring in 3-phase induction motor? May-2013

Slip rings are used to connect external stationary circuit to internal rotating circuit.

6. Why an induction motor is called asynchronous motor?

The induction motor does not rotate synchronous speed. It always rotate below synchronous

speed that why it is called asynchronous motor.

7. What is slip? Dec-2013, 2012, 2011, 2009, May-2013,2017

Define slip of an induction motor. Dec-2014

The difference between the synchronous speed Ns and the actual speed N of the rotor is known as

slip.

2


% 𝑆𝑙𝑖𝑝 = 𝑁𝑠 − 𝑁

𝑁𝑠 𝑥 100

8. Why an induction motor never runs at its synchronous speed? MAY-JUNE 2018

If it runs at synchronous speed then there would be no relative speed between the two, hence no

rotor emf, so no rotor current, then no rotor torque to maintain rotation.

9. What are slip rings?

The slip rings are made of copper alloys and are fixed around the shaft insulating it. Through

these slip rings and brushes rotor winding can be connected to external circuit.

10. What is mean by end ring? May-2004

In squirrel cage rotor, the copper bars are placed in the sots. These bars are short circuited at each

end with the help of conducting copper ring is called end ring.

11. Compare squirrel cage rotor and slip ring rotor.

S.No Squirrel cage rotor Slip ring rotor

1 Simple construction Complicated

2 Rotor consists of copper bars. Three phase winding

3 Slip ring and brushes are not present Present

4 External resistance can’t be added Added

5 Moderate torque High starting torque

6 Speed control by rotor resistance is not

possible Possible

12. Give the condition for maximum torque for 3-phase induction motor, when it is running?

May-2013, 2009,2016,Dec 2015

The rotor resistance and reactance should be same for maximum torque i.e. R2 = S X2.

13. Under what condition the slip in an induction motor is? a)negative b)greater than 1

May-2010

When rotor is running at a speed above the synchronous speed slip is negative.

When motor is rotated in opposite direction to that of rotating field slip is greater than 1.

3


14. What are the two fundamental characteristics of a rotating magnetic field? Dec-2010

The resultant of three alternating fluxes separated from each other by 120 degree has constant

Amplitude of 1.5 φm .

The resultant always keeps on rotating with a certain speed in space (Ns).

15. What is induction generator? May-2011

When the slip of the induction motor is negative the induction motor that runs as a generator is called

induction generator.

16. What are the purposes that could be served by external resistors connected in the rotor Circuit

of phase wound IM? May-2006

a ) increasing starting torque. b) For speed control c) limiting starting current.

17. What are the merits of inner and outer cage of double cage induction motor?

Dec-2012, 2006

Merits of inner cage: Leakage reactance is high, Resistance is small.

Merits of outer cage: High starting torque, Resistance is high.

18. Define Synchronous speed in a 3-phase IM? May-2004

The speed at which the revolving flux rotates is called synchronous speed Ns and is given by

𝑁𝑠 = 120 𝑓

𝑃

Where f – Supply Frequency

P- Number of poles on the stator.

19. What are the losses in induction motor? May-2006

a) Constant losses b) Variable losses.

20. What is cogging? May-2008

When the number of rotor slots is equal to stator slots, precisely the same order harmonics are

strongly produced, all rotating at corresponding speeds in both stator and rotor. Thus harmonics of every

order would try to exert synchronous torque at their corresponding synchronous speeds and the motor

would refuse to start. That is magnetic locking between the stator and the rotor slots. This is known as

cogging or magnetic locking.

4


21. What is crawling in IM? May- 2011

The tendency of the motor to run stably at speeds as low as one seventh of its synchronous speed

with a low pitched howling sound is called crawling.

22. What are the applications of 3-phase IM? May-2012, 2004

Squirrel Cage type - Drilling machines, Grinders, fans and blowers, lathes &

Wound type - (for high starting torque) lifts, hoists cranes elevators and compressors

23. What are the characteristics of double squirrel cage motor, compared to a squirrel cage motor?

Dec- 2003

(i) High starting torque (ii) Excellent running performance

24. Name the tests to be conducted for predetermining the performance of 3-phase induction

machine. Dec-2006

No load test & Blocked rotor test

25. What is Circle diagram of an IM?

When an IM operates on constant voltage and constant frequency source, the loci of stator current

phasor is found to fall on a circle. This circle diagram is used to predict the performance of the machine

at different loading conditions as well as mode of operation.

26. Why the slots on the induction motors are usually skewed? May-2011,2016,2017,Dec-

2014,2015

o To make the motor run quietly by reducing the magnetic horn (noise).

o To reduce the locking tendency of the rotor

27. Why an induction motor, at no load, operates at very low power factor? APR-MAY 2018

An induction motor draws a large magnetizing current (Im) to produce the required flux in the air-

gap. ... Therefore, the induction motor takes a high no-load current lagging the applied voltage by a large

angle. Hence the power factor of an induction motor on no load is low

28. What is locked rotor torque?

It is a torque under blocked rotor condition.

5


29. What is known as a double cage induction motor? Draw the torque-slip characteristic of

double-cage induction motor. NOV-DEC 2017

Induction motor performance can be improved by two cages. That is upper cage and lower cage.

This type of induction motor is called double cage induction motor.

30. State the advantages of skewing? Dec-2011,Nov-Dec 2016.

It reduces humming and hence quite running of motor is achieved. It reduces magnetic locking of

the stator and rotor.

31. Mention the losses that occur in an induction motor? May-2012

Stator Losses

i)Stator core Loss

ii)Stator Copper Loss

Rotor Loss

i)Rotor Copper Loss

Mechanical losses

32. How do change in supply voltage and frequency affect the performance of a 3 phase induction

motor? May-2014

T α V2

33. The starting torque of a squirrel cage induction motor cannot be altered when the applied

voltage is constant why? May-2014

The torque is directly proportional to the square of the induced emf at standstill.

6


34. What are the merits and demerits of double squirrel cage induction motors? Dec-2014

Merits of inner cage: Leakage reactance is high, Resistance is small.

Merits of outer cage: High starting torque, Resistance is high.

35. How much is the developed torque in an induction motor at synchronous speed? Explain.

May-2015

T = 0

36. State a method by which starting torque of the induction motor can be increased.

May-2015

By increasing the rotor resistance, starting torque can be increased.

37. What measure can be taken for minimizing the effect of crawling in a 3-phase induction motor?

NOV-DEC 2017

By choosing proper combination of stator and rotor slots we can minimize crawling.

PART-B

1. Explain the operation and construction of 3-phase induction motor.

May-2011, Dec-2013, 2012,APR-MAY 2018

Explain the working principle of a 3-phase induction motor. NOV-DEC -2014,2015,2017

Introduction

The three-phase induction motors are the most widely used electric motors in industry. They run

at essentially constant speed from no-load to full-load.

It is simple, rugged, low-priced, easy to maintain

Principle of Operation

When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic field is set

up which rotates round the stator at synchronous speed Ns (= 120 f/P).

7


The rotating field passes through the air gap and cuts the rotor conductors, w h ic h a s ye t ,

a r e stationary. Due to the relative speed between the rotating flux and the stationary rotor, e.m.f.s is

induced in the rotor conductors. Since the rotor circuit is short-circuited, currents start flowing in

the rotor conductors.

The current-carrying rotor conductors are

placed in the magnetic field produced by the

stator. Consequently, mechanical force acts on

the rotor conductors. The sum of the mechanical

forces on all the rotor conductors produces a

torque which tends to move the rotor in the same

direction as the rotating field.

The fact that rotor is urged to follow the

stator field (i.e., rotor moves in the direction of stator field) can be explained by Lenz's law. According

to this law, the direction of rotor currents will be such that they tend to oppose the cause producing them.

Now, the cause producing the rotor currents is the relative speed between the rotating field and the

stationary rotor conductors.

Construction

A 3-phase induction motor has two main parts, stator and rotor.

Stator

The stator carries a 3-phase winding (called stator winding) while the rotor carries a short-circuited

winding (called rotor winding). Only the stator winding is fed from 3-phase supply. The rotor winding

derives its voltage and power from the externally energized stator winding through electromagnetic

induction and hence the name.

The rotor is separated from the stator by a small air-gap which ranges from 0.4 mm to 4 mm,

depending on the power of the motor. It consists of a steel frame which encloses a hollow, cylindrical

core made up of thin laminations of silicon steel to reduce hysteresis and eddy current losses.

A number of evenly spaced slots are provided on the inner periphery of the lamination. The

insulated connected to form a balanced 3-phase star or delta connected circuit. The 3-phase stator

8


winding is wound for a definite number of poles as per requirement of speed. Greater the number of

poles, lesser is the speed of the motor and vice-versa.

When 3-phase supply is given to the stator winding, rotating magnetic field of constant

magnitude is produced. This rotating field induces currents in the rotor by electromagnetic induction.

Rotor

The rotor, mounted on a shaft, is a hollow laminated core having slots on its outer periphery.

The winding placed in these slots (called rotor winding) may be one of the following two types:

1. Squirrel cage type 2. Wound type or slip ring induction type.

Squirrel Cage Rotor.

It consists of a laminated cylindrical core having parallel slots on its outer periphery. One

copper or aluminum bar is placed in each slot.

All these bars are joined at each end by metal rings called end rings. This forms a permanently

short-circuited winding which is indestructible. The entire construction (bars and end rings)

resembles a squirrel cage and hence the name.

The rotor is not connected electrically to the supply but has current induced in it by transformer

action from the stator.

Those induction motors which employ squirrel cage rotor are called squirrel cage induction

motors. Most of 3-phase induction motors use squirrel cage rotor as it has a remarkably simple and

robust construction.

9


However, it suffers from the disadvantage of a low starting torque. It is because the rotor bars

are permanently short-circuited and it is not possible to add any external resistance to the rotor circuit to

have a large starting torque.

Wound rotor.

It consists of a laminated cylindrical core and carries a 3-phase winding, similar to the one on the

stator. The rotor winding is uniformly distributed in the slots and is usually star-connected.

The open ends of the rotor winding are brought out and joined to three insulated slip rings

mounted on the rotor shaft with one brush resting on each slip ring.

At starting, the external resistances are included in the rotor circuit to give a large start ing

torque. These resistances are gradually reduced to zero as the motor runs up to speed.

The external resistances are used during starting period only. When the motor attains normal

speed, the three brushes are short-circuited so that the wound rotor runs like a squirrel cage rotor.

Advantages

(i) It has simple and rugged construction.

(ii) It is relatively cheap.

(iii) It requires little maintenance.

(iv) It has high efficiency and reasonably good power factor.

(v) It has self starting torque.

10


Disadvantages

(i) It is essentially a constant speed motor and its speed cannot be changed easily,

(ii) Its starting torque is inferior to D.C. shunt motor.

2. Explain how a revolving magnetic field is produced when 3-phase supply is given to 3-phase

induction motor. May-2011

Rotating Magnetic Field due to 3-Phase Currents

When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is

produced. This field is such that its poles do no remain in a fixed position on the stator but go on

shifting their positions around the stator. For this reason, it is called a rotating. It can be shown that

magnitude of this rotating field is constant and is equal to 1.5 φm where φm is the maximum flux due to

any phase.

The three phases X, Y and Z are energized from a 3-phase source and currents in these phases

are indicated as Ix, Iy and Iz

Øx = Øm sin wt

Øy = Øm sin (wt – 120°)

Øz = Øm sin (wt – 240°)

3-phase supply produces a rotating field of constant magnitude equal to 1.5 Øm

11


12


Thus the magnitude of φr once again remain same. The resultant of the three alternating fluxes, separated

from each other by 120o , has a constant magnitude of 1.5 Øm where Øm is the maximum amplitude of

an individual flux due to any phase.

Speed of rotating magnetic field

The speed at which the rotating magnetic field revolves is called the synchronous speed

(Ns). The time instant 4 represents the completion of one-quarter cycle of alternating current Ix from

the time instant 1. During this one quarter cycle, the field has rotated through 90°. At a time instant

represented by 13 or one complete cycle of current Ix from the origin, the field has completed one

revolution.

Therefore, for a 2-pole stator winding, the field makes one revolution in one cycle of current.

In a 4-pole stator winding, it can be shown that the rotating field makes one revolution in two cycles of

current. In general, fur P poles, the rotating field makes one revolution in P/2 cycles of current.

Cycles of current = (P/2) x revolutions of field

Since revolutions per second is equal to the revolutions per minute (Ns) divided by 60 and the number of

cycles per second is the frequency f,

𝑓 =𝑃

2×

𝑁𝑠

60×

𝑁𝑠𝑃

120

𝑁𝑠 =120𝑓

𝑃

The speed of the rotating magnetic field is the same as the speed of the alternator that is

supplying power to the motor if the two have the same number of poles. Hence the magnetic flux is said to

rotate at synchronous speed.

13


Direction of rotating magnetic field

The phase sequence of the three-phase voltage applied to the stator winding is X-Y-Z. If this

sequence is changed to X-Z-Y, 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. Thus it is necessary only to change the phase sequence

in order to change the direction of rotation of the magnetic field. For a three-phase supply, this can be

done by interchanging any two of the three lines.

3. Explain slip and maximum torque of a 3 phase induction motor. May-2011,2016,

Dec-2013,2015,2017

Slip

Rotor tries to catch the direction of rotating field. In practice, the rotor can never reach the

speed of stator flux. If it did, there would be no relative speed between the stator field and rotor

conductors, no induced rotor currents and, therefore, no torque to drive the rotor. The friction and

windage would immediately cause the rotor to slow down.

Hence, the rotor speed (N) is always less than the suitor field speed (Ns). This difference in speed

depends upon load on the motor.

The difference between the synchronous speed Ns of the rotating stator field and the actual rotor

speed N is called slip. It is usually expressed as a percentage of synchronous speed

𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑆𝑙𝑖𝑝 𝑆 =𝑁𝑠 − 𝑁

𝑁𝑠× 100

The quantity Ns – N is sometimes called slip speed,

When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %.

In an induction motor, the change in slip from no-load to full-load is hardly 0.1% to 3% so

that it is essentially a constant-speed motor.

Condition for Maximum Starting Torque

It can be proved that starting torque will be maximum when rotor resistance/phase is

equal to standstill rotor reactance/phase.

14


Now 𝑇𝑠 = 𝐾1𝑅2

𝑅22+𝑋2

2

Differentiating eq.(i) w.r.t 𝑅2 and equating the result to zero, we get

𝑑𝑇𝑠

𝑑𝑅2= 𝐾1

1

𝑅22 + 𝑋2

2 −𝑅2 2𝑅2

𝑅22 + 𝑋2

2 2 = 0

Or 𝑅22 + 𝑋2

2 = 2𝑅22

Or 𝑅2 = 𝑋2

As the rotor resistance is increased from a relatively low value, the starting torque increases until

it becomes maximum when R2 = X2. If the rotor resistance is increased beyond this optimum value, the

starting torque will decrease.

Torque Under Running Conditions (NOV-DEC 2017)

Let the rotor at standstill have per phase induced e.m.f. E2, reactance X2 and resistance R2.

Then under running conditions at slip s,

Rotor e.m.f/phase,𝐸2′ = 𝑠𝐸2

Rotor reactance /phase,𝑋2′ = 𝑠𝑋2

Rotor impedance/phase,𝑍2′ = 𝑅2

2 + 𝑠𝑋2 2

Rotor current/Phase, 𝐼2′ =

𝐸2′

𝑍2′ =

𝑠𝐸2

𝑅22+ 𝑠𝑋2 2

Rotor P.F., cos∅𝑚′ =

𝑅2

𝑅22+ 𝑠𝑋2

2

Running Torque,𝑇𝑟 ∝ 𝐸2′ 𝐼2

′ 𝑐𝑜𝑠∅2′

∝ ∅𝐼2′ 𝑐𝑜𝑠∅2

′ ∵ 𝐸2

′∝ ∅

15


∝ ∅ ×𝑠𝐸2

𝑅22

+ 𝑠𝑋2 2

×𝑅2

𝑅22 + 𝑠𝑋2

2

∝∅𝑠𝐸2𝑅2

𝑅22 + 𝑠𝑋2

2

∝𝐾∅𝑠𝐸2𝑅2

𝑅22 + 𝑠𝑋2

2

∝𝐾1𝑠𝐸2

2𝑅2

𝑅22+ 𝑠𝑋2 2

∵ 𝐸2 ∝ ∅

If the stator supplies voltage V is constant, then stator flux and hence E2 will be constant

𝑇𝑟 =𝐾2𝑠𝑅2

𝑅22 + 𝑠𝑋2 2

It may be seen that running torque is:

(ii) Directly proportional to slip i.e., if slip increases (i.e., motor speed decreases), the

torque will

increase and vice-versa,

(ii) Directly proportional to square of supply voltage (E2 α V).

It can be shown that value of 𝐾1 = 3/2𝜋𝑁𝑠where 𝑁𝑠 is in r.p.s

∴ 𝑇𝑟 =3

2𝜋𝑁𝑠.

𝑠𝐸22𝑅2

𝑅22 + 𝑠𝑋2 2

=3

2𝜋𝑁𝑠.𝑠𝐸2

2𝑅2

𝑍2, 2

At starting S=1 so that starting torque is

𝑇𝑟 =3

2𝜋𝑁𝑠

.𝐸2

2𝑅2

𝑅22 + 𝑋2

2

Maximum Torque under Running Conditions

In order to find the value of rotor resistance that gives maximum torque under running

conditions, differentiate below equation w.r.t. S and equate the result to zero i.e.,


𝑅22 + 𝑠𝑋2 2

𝑑𝑇𝑟

𝑑𝑠=

𝐾2 𝑅2 𝑅22 + 𝑆2𝑋2

2 − 2𝑠𝑋2

2(𝑠𝑅2)

𝑅22 + 𝑆2𝑋2

2

2

16


𝑅22 + 𝑆2𝑋2

2 − 2𝑠𝑋2

2 = 0

𝑅22 = 𝑠𝑋2

Slip corresponding to maximum torque, s = R2/X2

𝑇𝑚 =3

2𝜋𝑁𝑠.𝐸2

2

2𝑋2 𝑁 − 𝑚

4. Explain the Torque-Slip Characteristics of induction motor. May-2014, 2013, 2012, 2008, Dec-

2012, 2011,Nov-Dec 2016.May 2017

Draw and explain the slip-torque characteristics of a typical 3-phase induction motor. Mark the

starting and maximum torque regions on the diagram so drawn. Dec-2014

Draw the torque-slip characteristics of an induction motor for varying frequency, stator voltage

and rotor resistance. May-2015

Torque equation of 3 phase induction motor is given by,


𝑅22 + 𝑠𝑋2 2

Where k and E2 constant

When rotor rotate normal speed that is close to synchronous speed.

S = 0 , T = 0 Hence curve starts at point ‘0’.

This characteristics consist of two regions

1. Stable region

2. Unstable region.

Stable region:

Slip value very small that is ( s X2)2 is very small as compared to R2

2. Hence neglecting s

2X2

2

T α s R2/ R22

T α s ( R2 is constant)

Slip value is directly proportional to the torque. In this region as the ( T ↑ s ↑) load increases, speed

decreases or the slip increases. So the characteristic is approximately a straight line.

17


Unstable region:

When the slip increased from Sm the region is called unstable region. Here the slip value is high

that is the values between Sm and 1. The term R22 may be neglected as compared to s

2X2

2.

T α 1/ s

The torque is inversely proportional to slip. That is slip increases torque decreases ( S ↑ T ↓ )

5. Draw the Equivalent Circuit of 3-Phase Induction Motor at Any Slip.

May-2013, 2011, Dec-2012, 2016,Nov-Dec 2016.

Draw the equivalent circuit and derive expressions for maximum torque and

power of a three phase induction motor. May-2015 ,APR-MAY 2018

In a 3-phase induction motor, the stator winding is connected to 3-phase supply and the rotor

winding is short-circuited. The energy is transferred magnetically from the stator winding to the short-

circuited, rotor winding.

Therefore, an induction motor may be considered to be a transformer with a rotating secondary

(short-circuited). The stator winding corresponds to transformer primary and the rotor finding corresponds

to transformer secondary.

In view of the similarity of the flux and voltage conditions to those in a transformer, one can

expect that the equivalent circuit of an induction motor will be similar to that of a transformer.

18


Equivalent Circuit of the Rotor

𝑰𝟐′ =

𝑠𝐸2

𝑅22+ 𝑠𝑋2

2

Mathematically ,this value is unaltered by writing it as

𝑰𝟐′ =

𝑬𝟐

𝑹𝟐𝑺

𝟐

+ 𝑿𝟐 𝟐

𝑅2

𝑠= 𝑅2 + 𝑅2

1

𝑠− 1

𝑅𝐿 = 𝑅2 1

𝑠− 1

Transformer Equivalent Circuit of Induction Motor

The following points may be noted from the equivalent circuit of the induction motor:

(i) At no-load, the slip is practically zero and the load R’L is infinite. This condition resembles that in

a transformer whose secondary winding is open-circuited,

(ii) At standstill, the slip is unity and the load R’L is zero. This condition resembles that in a

transformer whose secondary winding is short-circuited,

19


When the motor is running under load, the value of R’L will depend upon the value of the slip s.

This condition resembles that in a transformer whose secondary is supplying variable and purely resistive

load.

(iii) The equivalent electrical resistance R’L related to mechanical load is slip or speed dependent. If the slip s

increases, the load R’L decreases and the rotor current increases and motor will develop more mechanical

power. This is expected because the slip of the motor increases with the increase of load on the motor

shaft.

Approximate Equivalent Circuit of Induction Motor

The above approximate circuit of induction motor is not so readily justified as With the transformer.

This is due to the following reasons:

i) Unlike that of a power transformer, the magnetic circuit of the induction motor has an air-gap.

Therefore, the exciting current of induction motor (30 to 40% of full-load current) is much higher

than that of the power transformer. Consequently, the exact equivalent circuit must be used for

accurate results,

(ii) The relative values of X1 and X2 in an induction motor are larger than the Corresponding ones to

be found in the transformer. This fact does not Justify the use of approximate equivalent circuit (iii)

in a transformer, the windings are concentrated whereas in an induction motor, the windings are

distributed. This affects the transformation ratio.

In spite of the above drawbacks of approximate equivalent circuit, it yields results that are satisfactory

for large motors. However, approximate equivalent circuit is not justified for small motors.

6. A 3-ph, 400-V induction motor gave the following test readings;

No-load : 400 V, 1250 W, 9 A,

Short-circuit: 150 V, 4 kW, 38 A.

Draw the circle diagram, if the normal rating is 14.9 kW. Find from the circle diagram, the full-

load value of current, pf and slip. May-2005

20


Solution: From No load test

From blocked rotor test:

Short-circuit current with normal voltage is ISN = 38 (400/150) = 101.3 A

Power taken would be = 4000 (400/150)2

= 28,440 W

OO′ represents I0 of 9 A. If current scale is 1 cm = 5 A then vector OO′ = 9/5 = 1.8 cm and is

drawn at an angle of φ0 = 78.5º with the vertical OV.

Similarly, OA represents ISN equal to 101.3 A. It measures 101.3/5 = 20.26 cm and is drawn at

an angle of 66.1º, with the vertical OV.

Line O′D is drawn parallel to OX. NC is the right-angle bisector of O′A. The semi-circle Ο ′AD

is drawn with C as the centre.

This semi-circle is the locus of the current vector for all load conditionsfrom no-load to short-

circuit. Now, AF represents 28,440 W and measures 8.1 cm.

Hence, power scale becomes : 1 cm = 28,440/8.1 = 3,510 W. Now, full-load motor output

= 14.9 × 103 = 14,900 W.

21


According to the above calculated power scale, the intercept between the semi-circle and output

line O′A should measure = 14,900/3510 = 4.25 cm.

For locating full-load point P, BA is extended. AS is made equal to 4.25 cm and SP is drawn

parallel to output line O′A. PL is perpendicular to OX.

Line current= OP = 6 cm = 6 × 5 = 30 A; φ = 30º (by measurement)

p.f.=cos 30º = 0.886 (or cos φ = PL/OP = 5.2/6 = 0.865)

slip= rotor Cu loss / rotor input

In Fig. EK represents rotor Cu loss and PK represents rotor input.

Slip = EK/PK = 0.3/4.5 = 0.067 or 6.7%.

7. Draw the circle diagram for a 3.73 kW, 200-V, 50-Hz, 4-pole, 3phase star-connected induction

motor from the following test data :

No-load : Line voltage 200 V, line current 5 A; total input 350 W

Blocked rotor : Line voltage 100 V, line current 26 A; total input 1700 W

Estimate from the diagram for full-load condition, the line current, power factor and also the

maximum torque in terms of the full-load torque. The rotor Cu loss at standstill is half the total

Cu loss.

Solution. No load test

Blocked-rotor test :

Short-circuit current with normal voltage, ISN = 26 × 200/100 = 52 A

Short-circuit/blocked rotor input with normal voltage = 1700(52/26)2 = 6,800 W

22


In the circle diagram voltage is represented along OV which is drawn perpendicular to OX.

Current scale is 1 cm = 2 A

Line OA is drawn at an angle of φ0 = 78º15′ with OV and 2.5 cm in length. Line AX′ is drawn

parallel to OX. Line OB represents short-circuit current with normal voltage i.e. 52 A and measures

52/2 = 26 cm. AB represents output line. Perpendicular bisector of AB is drawn to locate the centre

C of the circle. With C as centre and radius = CA, a circle is drawn which passes through points

A and B. From point B, a perpendicular is drawn to the base.

BD represents total input of 6,800 W for blocked rotor test. Out of this, ED represents no-load

loss of 350 W and BE represents 6,800 − 350 =6,450 W.

Now BD = 9.8 cm and represents 6,800 W.pow er scale=6,800/9.8 = 700 watt/cm or 1 cm = 700

W

BE which represents total copper loss in rotor and stator, is bisected at point T to separate the two

losses. AT represents torque line.

Now, motor output = 3,730 watt. It will be represented by a line = 3,730/700 = 5.33 cm

The output point P on the circle is located thus :

DB is extended and BR is cut = 5.33 cm. Line RP is drawn parallel to output line AB and cuts

the circle at point P. Perpendicular PS is drawn and P is joined to origin O.

Point M corresponding to maximum torque is obtained thus :

23


From centre C, a line CM is drawn such that it is perpendicular to torque line AT. It cuts the

circle at M which is the required point. Point M could also have been located by drawing a line parallel

to the torque line. MK is drawn vertical and it represents maximum torque.

Now, in the circle diagram, OP = line current on full-load = 7.6 cm.

Hence, OP represents 7.6 ×2 = 15.2 A.

Power factor on full-load = SP/OP

= 6.45/7.6

= 0.86

Max. torque/F.L. torque= MK/PG

=10/5.6 = 1.8

Max. torque = 180% of full-load torque.

8. Explain briefly construction of circle Diagram and also Draw the circle diagram from no-load

and short-circuit test of a 3-phase. 14.92 k W, 400-V, 6-pole induction motor from the following test

results (line values).

No-load : 400-V, 11 A, P.f. = 0.2

Short-circuit : 100-V, 25 A, P.f. = 0.4

Rotor Cu loss at standstill is half the total Cu loss.

From the diagram, find (a) line current, slip, efficiency and p.f. at full-load (b) the maximum

torque. May 2016,2017

Solution.

No-load p.f.= 0.2 ; φ0 = cos− 1 (0.2) = 78.5º

Short-circuit p.f.=0.4 : φs = cos− 1 (0.4) = 66.4º

S.C. current ISN if normal voltage were applied = 25 (400/100) = 100 A

S.C. power input with this current = × 400 × 100 × 0.4 = 27,710 W

Assume a current scale of 1 cm = 5 A.

No-load current vector OO′ represents 11 A. Hence, it measures 11/5 = 2.2 cm and is drawn

at an angle of 78.5º with OY.

Vector OA represents 100 A and measures 100/5 = 20 cm. It is drawn at an angle of 66.4º with OY.

O′D is drawn parallel to OX. NC is the right angle bisector of O′A.

With C as the centre and CO′ as radius, a semicircle is drawn as shown.

24


AF represents power input on short-circuit with normal voltage applied. It measures 8 cm and (as

calculated above) represents 27,710 W.

Hence, power scale becomes 1 cm=27,710/8 = 3,465 W

F.L motor output = 14,920 W. According to the above power scale, the intercept between the semicircle

and the output line O′A should measure = 14,920/3,465 = 4.31 cm.

Hence, vertical line PL is found which measures 4.31 cm. Point P represents the full-load operating point.

(a) Line current = OP = 6.5 cm which means that full-load line current

= 6.5 × 5 = 32.5 A. φ = 32.9º (by measurement)

∴ cos 32.9º= 0.84 (or cos φ = PL/OP = 5.4/6.5 = 0.84)

slip = EK/PK = 0.3/5.35= 0.056or 5.6% ;

η = PE/PL = 4.3/5.4 = 0.8 or 80%

(b) For finding maximum torque, line CM is drawn ⊥ to torque line O′H. MT is the vertical

intercept between the semicircle and the torque line and represents the maximum torque of the

motor in synchronous watts

Now, MT = 7.8 cm (by measurement)

∴ Tmax = 7.8 × 3465 = 27,030 synch. Watt

1. Construction of the Circle Diagram

Circle diagram of an induction motor can be drawn by using the data obtained from

(1) no-load (2) short-circuit test and (3) stator resistance test.

25


Step No. 1

From no-load test, I0 and φ0 can be calculated. Hence, I0 can be laid off lagging φ0 behind the

applied voltage V.

Step No. 2

Next, from blocked rotor test or short-circuit test, short- circuit current ISN corresponding to

normal voltage and φS are found. The vector OA represents ISN = (ISV/VS ) in magnitude and phase.

Vector O′A represents rotor current I2′ as referred to stator.

Clearly, the two points O′ and A lie on the required circle. For finding the centre C of this circle,

chord O′A is bisected at right angles–its bisector giving point C. The diameter O′D is drawn perpen-

dicular to the voltage vector.

As a matter of practical contingency,it is recommended that the scale of current vectors should be so

chosen that the diameter is more than 25 cm,in order that the performance data of the motor may be read

with reasonable accuracy from the circle diagram.

With centre C and radius = CO′, the circle can be drawn. The line O′A is known as out-put line.

It should be noted that as the voltage vector is drawn vertically, all vertical distances represent the active

or power or energy components of the currents.

For example, the vertical component O′P of no-load current OO′ represents the no-load input, which

supplies core loss, friction and windage loss and a negligibly small amount of stator I2 R loss. Similarly,

the vertical component AG of short-circuit current OA is proportional to the motor input on short-

circuit or if measured to a proper scale, may be said to equal power input

26


Step No. 3

Torque line. This is the line which separates the stator and the rotor copper losses. When the

rotor is locked, then all the power supplied to the motor goes to meet core losses and Cu losses in the

stator and rotor windings. The power input is proportional to AG. Out of this, FG (= O′P) represents

fixed losses i.e. stator core loss and friction and windage losses. AF is proportional to the sum of the

stator and rotor Cu losses. The point E is such that

As said earlier, line O′E is known as torque line.

How to locate point E?

i.. Squirrel-cage Rotor. Stator resistance/phase i.e. R1 is found from stator-resistance test.

Now, the short-circuit motor input Ws is approximately equal to motor Cu losses (neglecting iron losses).

(ii) Wound Rotor. In this case, rotor and stator resistances per phase r2 and r1 can be easily

computed. For any values of stator and rotor currents I1 and I2 respectively, we can write

Maximum Quantities

It will now be shown from the circle diagram , that the maximum values occur at the positions

stated below :

(i) Maximum Output

It occurs at point M where the tangent is parallel to output line O′A. Point M may be located by

27


drawing a line CM from point C such that it is perpendicular to the output line O′A. Maximum output is

represented by the vertical MP

(ii) Maximum Torque or Rotor Input

It occurs at point N where the tangent is parallel to torque line O′E. Again, point N may be found

by drawing CN perpendicular to the torque line. Its value is represented by NQ . Maximum torque is also

known as stalling or pull-out torque.

(iii) Maximum Input Power

It occurs at the highest point of the circle

i.e. at point R where the tangent to the circle is horizontal. It is proportional to RS. As the point R is

beyond the point of maximum torque, the induction motor will be unstable here. However, the maximum

input is a measure of the size of the circle and is an indication of the ability of the motor to carry short-

time over-loads. Generally, RS is twice or thrice the motor input at rated load.

9. Explain the Losses and efficiency of 3-phase induction motor

Losses

1. Constant losses

2. Variable losses.

1.Constant losses:

a. Core losses- This losses occurs in Stator and rotor core. This losses also called iron losses.

Iron losses- It includes eddy current losses and hysteresis losses. Eddy current losses are

minimized by using lamination .

Hysteresis losses- This losses are minimized by selecting proper material ( silicon materials).

Fr = sf

Where,

fr- rotor frequency

s- slip

f- supply frequency or stator frequency.

Rotor frequency is s times the supply frequency.

b.Mechanical loss- It include frictional and windage losses .

2.Variable losses:- This losses are also called copper losses.

28


- Copper losses in stator& rotor due to current flowing in winding.

- as load changes as current changes.

- rotor copper loss = 3 I22R2

Efficiency:

η = Pout/ Pin ˟ 100.

The maximum efficiency occurs when variable losses becomes equal to constant losses. When the motor is no load

, current drawn by the motor is small. Hence efficiency is low.

Load increases I increases so copper losses also increases - Variable losses achieve the same value as that of

constant losses, efficiency attains its maximum value.

If load increases more- variable losses ˃ constant losses. Hence deviating from condition for maximum,

efficiency starts decreases.

10. Write a brief note on

a. Double Squirrel-Cage Motors with speed torque charactersticsMay-2013, 2012,

Nov-Dec 2016

b. Induction generator Nov-Dec 2016, NOV-DEC 2017

c. Synchronous induction motor. Dec-2014, 2010, May-2012

a. Double Squirrel-Cage Motors

One of the advantages of the slip-ring motor is that resistance may be inserted in the rotor circuit

to obtain high starting torque (at low starting current) and then cut out to obtain optimum running

conditions. However, such a procedure cannot be adopted for a squirrel cage motor because its cage is

permanently short-circuited. In order to provide high starting torque at low starting current, double-cage

construction is used.

(i) The outer winding consists of bars of smaller cross-section short-circuited by end rings.

Therefore, the resistance of this winding is high. Since the outer winding has relatively open slots

and a poorer flux path around its bars, it has a low inductance. Thus the resistance of the outer

squirrel-cage winding is high and its inductance is low.

(ii) The inner winding consists of bars of greater cross-section short-circuited by end rings. Therefore,

the resistance of this winding is low. Since the bars of the inner winding are thoroughly buried in

iron, it has a high inductance. Thus the resistance of the inner squirrel-cage winding is low and its

inductance is high.

29


When a rotating magnetic field sweeps across the two windings, equal e.m.f.s are induced in each.

(i) At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz), making the reactance of the

lower winding much higher than that of the upper winding. Because of the high reactance of the lower

winding, nearly all the rotor current flows in the high-resistance outer cage winding. This provides the

good starting characteristics of a high-resistance cage winding. Thus the outer winding gives high starting

torque at low starting current.

(ii) As the motor accelerates, the rotor frequency decreases, thereby lowering the reactance of the

inner winding, allowing it to carry a larger proportion of the total rotor current At the normal operating

speed of the motor, the rotor frequency is so low (2 to 3 Hz) that nearly all the rotor current flows in the

low-resistance inner cage winding. This results in good operating efficiency and speed regulation.

Induction Generator

Induction Motor Operating as a Generator:

When run faster than its synchronous speed, an induction motor runs as a generator called an

Induction generator. It converts the mechanical energy it receives into electrical energy and this energy is

released by the stator. As soon as motor speed exceeds its synchronous speed, it starts delivering active

power P to the 3-phase line. However, for creating its own magnetic field, it absorbs reactive power Q from

the line to which it is connected. As seen. Q flows in the opposite direction to P.

The active power is directly proportional to the slip above the synchronous speed. The reactive

power required by the machine can also be supplied by a group of capacitors connected across its

terminals. This arrangement can be used to supply a 3-phase load without using an external source. The

frequency generated is slightly less than that corresponding to the speed of rotation. The terminal voltage

increases with capacitance. If capacitance is insufficient, the generator voltage will not build up. Hence,

capacitor bank must be large enough to supply the reactive power normally drawn by the motor.

30


Synchronous induction motor:

A three phase slip ring induction motors runs at a constant speed when its rotor winding is fed

from DC source. Such motors are then called synchronous induction motors. Consider a normal three

phase slip ring induction motors. Consider a normal three phase slip ring induction motor with a

conventional three phase rotor.

A DC supply is fed to the rotor windings. Due to this, an alternate n and s poles on the rotor as in

the case when the rotor carries alternating currents working as an induction motor. However, the essential

difference is that the DC excitation being fixed, the pole axis due to DC excitation is also fixed in space

and does not shift as in the case when the rotor carries alternating currents.

These fixed rotor poles get magnetically locked with the rotating magnetic field produced by the

three-phase stator winding carrying alternating currents and the motor runs at a constant speed equal to

the synchronous speed. Below fig shows schematic diagram of starting and synchronizing a synchronous

induction motor. The machines start as an ordinary slip-ring induction motor with additional resistance

inserted in the rotor circuit through the slip rings when the switch is in the start position.

31


When the additional resistance is completely cut-out and the motor runs with a small slip, the

switch is changed over to run position and there by connects the exciter with the rotor windings. And

provided with DC excitation in the rotor winding, pulls in to step as in the case of synchronous motor.

Since synchronous induction motor starts as a slip-ring induction motor, it is possible to start the motor

with heavy loads.

Hence the synchronous induction motor is essentially a motor, having the induction motor

features like high starting torque combined with the synchronous motor features like constant

speed and power factor control.

32


11. A 3-phase induction motor has a starting torque of 100% and a maximum torque of 200% of

the full load torque. Find slip at maximum torque. Dec-2012

Given Data:

Starting torque 𝑇𝑠𝑡 = 100% 𝑜𝑓 𝑇𝑓 ,𝑇𝑠𝑡 = 𝑇𝑓 ,𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑡𝑜𝑟𝑞𝑢𝑒 𝑇𝑚𝑎𝑥 = 200% 𝑜𝑟 𝑇𝑓

To find:

Slip at maximum Torque.

Soloution:

𝑇𝑠𝑡

𝑇𝑓= 1;

𝑇𝑚𝑎𝑥

𝑇𝑓= 2

𝑇𝑠𝑡

𝑇𝑚𝑎𝑥=

1

2= 0.5

We know

𝑇𝑠𝑡

𝑇𝑚𝑎𝑥=

2𝑎

1 + 𝑎2= 0.5

∴ 1 + 𝑎2 0.5 = 2𝑎 𝑜𝑟 0.5𝑎2 − 2𝑎 + 0.5 = 0

i.e., 𝑎2 − 4𝑎 + 1 = 0

𝑥 =4 ± 16 − 4

2=

4 ± 3.46

2= 0.27

𝑎 =𝑅2

𝑋2= 0.27 = 𝑆𝑚

𝑆𝑚 = 0.27

33


12. The active Power input to a 415 V,50 Hz,6 pole,3 phase induction motor running at 970 rpm is 41

Kw.The input P.F is 0.9.The stator losses amount to 1.1 Kw and the mechanical losses total 1.2

KW.Calculate line copper loss,mechanicl power output and Efficiency.

Given Data:

V=415 V, F=50 Hz,P=6, N=970 Rpm, 𝑃𝑚 = 41 𝐾𝑤, Input P.F=0.9,Stator Losses 𝑃𝑆𝐿 = 1.1 𝐾. 𝑊 ,Mechanical

Losses 𝑃𝑚𝐿 = 1.2 𝐾. 𝑊

To find:

Line current, Slip, Rotor Copper Losses, Mechanical power output and efficiency.

Soloution:

i) Line current(𝐼𝐿)

Input Power 𝑃𝑚 = 3𝑉𝐿𝐼𝐿𝑐𝑜𝑠∅

𝐼𝐿 =𝑃𝑚

3𝑉𝐿𝑐𝑜𝑠∅=

41 × 103

3 × 415 × 0.9

𝐼𝐿 = 63.37𝐴

ii) Slip (S)

𝑁𝑠 =120𝑓

𝑃=

120 × 50

6= 1000 𝑟𝑝𝑚

Slip S=𝑁𝑠−𝑁

𝑁𝑠=

1000 −970

1000= 0.03 𝑜𝑟 3%

iii) Rotor copper Loass (𝑃𝑐𝑢 )

Rotor input power 𝑃2 = 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 − 𝑆𝑡𝑎𝑡𝑜𝑟 𝐿𝑜𝑠𝑠𝑒𝑠

= 𝑃𝑖𝑛 − 𝑃𝑆𝐿 = 41 − 1.1

𝑃2 = 39.9 𝐾. 𝑊

Rotor copper Loss 𝑃𝑐𝑢 = 𝑠𝑃2 = 0.03 × 39.9

𝑃𝑐𝑢 = 1.197 𝐾. 𝑊

iv) Mechanical Power developed 𝑃𝑚 = 𝑃2 − 𝑃𝑐𝑢 = 39.9 − 1.197

𝑃𝑚 = 38.703 𝐾. 𝑊

Output power 𝑃𝑜𝑢𝑡 = 𝑃𝑚 − 𝑃𝑚𝑙 = 38.703 − 1.2

𝑃𝑜𝑢𝑡 = 37.503 𝐾. 𝑊

v) Efficiency (𝜂) 𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑚𝑎𝑥× 100 =

37.503

41× 100

𝜂 = 91.47%

34


13. A 6 pole, 50 Hz, 3-Phase, induction motor running on full load develops a useful torque of 160

N-m. When the rotor emf makes 120 complete cycle per minute. Calculate the shaft power input.

If the mechanical torque lost in friction and that for core loss in 10 N-m find

i)The copper loss in the rotor windings.

ii)The input the motor.

iii)The Efficiency. The total stator loss be 800 W. May-2015, Dec-2011

Solution:

𝑓2 = 𝑠𝑓 =120

60= 2 𝐻𝑧

∴ 𝑆 =2

50= 0.04 𝑜𝑟 4%

𝑛𝑠 = 1000 𝑟𝑝𝑚

𝑛 = 1 − 0.04 × 1000 = 960 𝑟𝑝𝑚

𝜔 =960 × 2𝜋

60= 100.53 𝑟𝑎𝑑

𝑠𝑒𝑐

Shaft power output=160× 100.53

= 16.085 𝐾𝑊

𝑀𝑒𝑐𝑕𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑃𝑚 = 160 + 10 × 100.53

𝑃𝑚 = 17.09 𝐾𝑊

Note that torque of rotational loss is added to shaft power

𝑃𝑚 = 3𝐼2′2𝑅2

′ 1

𝑠− 1

Rotor copper Loss =3𝐼2′2𝑅2 = 𝑃𝑚

𝑆

1−𝑆

= 17.090 ×0.04

1 − 0.04= 712 𝑊

𝑖𝑛𝑝𝑢𝑡 𝑡𝑜 𝑚𝑜𝑡𝑜𝑟 = 17.09 + 0.712 + 0.8 = 18.602 𝐾. 𝑊

𝜂 =16.084

18.602= 86.47%

35


14. 746 kW, three phase, 50 HZ, 16 pole induction motor has a rotor impedance of (0.02+j0.15)Ω at

obtained at 360rpm.Calculate the i) ratio of maximum to full-load torque ii) speed of maximum

torque and iii) rotor resistance to be added to get maximum starting torque ? May-2014

Given Data:

Supply Frequency f=50 Hz,Number of Poles P=16,Rotor impedance ((𝑅2 + 𝑗𝑋2) = (0.02 + 𝑗0.15)Ω

Speed of full load torque N=360 Rpm

To Find:

i)𝑇𝑚𝑎𝑥

𝑇𝑓𝑙

ii)Speed of maximum torque

iii)Rotor resistance to be added to get maximum starting torque.

Soloution:

Synchronous Speed 𝑁𝑠 =120𝑓

𝑝=

120×50

16=375 rpm

Full load Slip 𝑆𝑓 = 𝑁𝑠−𝑁𝑟

𝑁𝑠=

375−360

375= 0.04

𝑎 =𝑅2

𝑋2=

0.02

0.15= 0.133

𝑇𝑓𝑙

𝑇𝑚𝑎𝑥=

2𝑎𝑠𝑓𝑎2 + 𝑆𝑟

2=

2 × 0.133 × 0.04

0.133 2 + 0.044 2= 0.55

𝑇𝑚𝑎𝑥

𝑇𝑓𝑙=

1

0.55= 1.818

𝑇𝑚𝑎𝑥

𝑇𝑓𝑙= 1.818

Slip at which maximum torque occurs 𝑆𝑚 =𝑅2

𝑋2=

0.012

0.015= 0.133

Motor speed 𝑁 = 𝑁𝑠 1 − 𝑆𝑚 = 375 1 − 0.133

N=325 rpm

For maximum Starting Torque,𝑅2 = 𝑋2

Hence total rotor resistance per phase =0.15Ω

∴ External resistance required per phase = 0.15-0.02=0.13 Ω

36


15. Draw the circle diagram of a 15 HP, 230 V, 50 Hz, 3-phase slipring Induction motor with a star

connected stator and rotor. The winding ratio is unity. The following are the test readings

No load test: 230 V, 9 A, pf=0.2143,

Blocked rotor test: 115 V, 45 A, p.f = 0.454

Find

i) Starting torque ii) maximum torque iii) maximum power factor

iv) slip for maximum torque v) maximum power output Dec-2013

Solution: From No load Test

Cos ∅0 = 0.2143 ∅0 = 77.62°

𝐼0 = 9𝐴 , 𝑉𝑜 = 230 𝑉

From S.C Test:

Cos ∅𝑆𝐶 = 0.454 ∅𝑆𝐶 = 62.99°

𝐼𝑆𝐶 = 45𝐴 , 𝑉𝑆𝐶 = 115 𝑉

𝐼𝑆𝑁=𝐼𝑆𝐶 × 𝑉𝐿

𝑉𝑆𝐶

= 45 × 230

115

𝑰𝑺𝑵 = 𝟗𝟎 𝑨

And

𝑊𝑆𝑁 = 3 × 𝑉𝐿 × 𝐼𝐿 × 𝑐𝑜𝑠∅𝑆𝐶

= 3 × 230 × 90 × 0.454

𝑾𝑺𝑵 = 𝟏𝟔𝟐𝟕𝟕. 𝟒𝟔𝑾

Choose current scale say 1 CM=5A

1. Draw vector OO’=𝐼0 = 9 𝐴 𝑖. 𝑒 1.8 𝐶𝑚 𝑎𝑠 𝑝𝑒𝑟 current scale at an angle of 77.62° w.r.t voltage axis.

2. Draw horizontal line from O’ parallel to X axis.

3. Draw vector OA=𝐼𝑆𝑁 = 90 𝐴 i.e 18 cm as per scale at an angle of 62.99° w.r.t voltage axis.

4.Join O’A this is output Line.

5.draw perpendicular bisector of O’A to meet horizontal line drawn from O’at point C.This is center of

circle.

6.With C as center and CO’ as radius,draw a semicircle to meet horizointal line from O’ at B.

7.Draw perpendicular from A on the X-axis meeting it at D.

l(AD)=8 cm=𝑊𝑆𝑁

Power scale=𝑊𝑆𝑁

𝑙(AD )=

16277 .46

8

Power scale=2034.68 W/Cm

8.Now rotors cu loss are half the total copper loass

37


𝐴𝐸

𝐴𝐹=

1

2𝑖. 𝑒 𝐴𝐸 =

1

2𝐴𝐹

Join O’E this is the torque line.

9.To locate full load point,draw AA’such that AA’ represents full load output to the power scale.

AA’=11190

2034 .68= 5.49 𝑐𝑚

i) Maximum Output =l(MN)×Power scale

=5.49×Power scale

5.49× 2034.68

Maximum Output =11170.39 Watts

ii) Starting Torque =l(AE)×Power scale

=4× 2034.68

Starting Torque =8138.72 Synchronous Watts.

iii) Maximu Torque = l(JK)×Power scale

=7.1 × 2034.68

Maximu Torque = 14446.28 Synchronous Watts.

iv) Maximum Power Factor = cos 𝑯𝑰

𝑶𝑯

=cos 4.1

4.8

=cos[0.854]

P.F=0.99 lag

v) Slip=𝑸𝑹

𝑷𝑹=

𝟎.𝟔𝟓𝒄𝒎

𝟓𝒄𝒎

Slip = 0.13× 𝑾𝟎 = 𝟏𝟑%

38


39


16. A 15 KW, 400V, 50 HZ, three phase star connected i.m gave the following test results

No load test: 400 V, 9 A, 1310 W

Blocked Rotor test: 200 V, 50 A, 7100 W

Stator and rotor ohmic losses at stand still are equal. Draw the induction motor circle diagrams

and calculate

i) Line current ii) Power factor iii) slip

iv) Torque and efficiency at full load

May-2014,APR-MAY 2018

From No load Test

Cos ∅0 = 0.2140 ∅0 = 77.87°

𝐼0 = 9𝐴 , 𝑉𝑜 = 400 𝑉

From S.C Test:

Cos ∅𝑆𝐶 = 0.409 ∅𝑆𝐶 = 65.8°

𝐼𝑆𝐶 = 50𝐴 , 𝑉𝑆𝐶 = 200 𝑉

𝐼𝑆𝑁=𝐼𝑆𝐶 × 𝑉𝐿

𝑉𝑆𝐶

= 50 × 400

200

𝑰𝑺𝑵 = 𝟏𝟎𝟎 𝑨

And

𝑊𝑆𝑁 = 3 × 𝑉𝐿 × 𝐼𝑆𝑁 × 𝑐𝑜𝑠∅𝑆𝐶

= 3 × 400 × 100 × 0.409

𝑾𝑺𝑵 = 𝟐𝟖𝟑𝟑𝟔𝑾

Choose current scale say 1 CM=5A

1. Draw vector OO’=𝐼0 = 9 𝐴 𝑖. 𝑒 1.8 𝐶𝑚 𝑎𝑠 𝑝𝑒𝑟

2. Draw horizontal line from O’ parallel to X axis.

3. Draw vector OA=𝐼𝑆𝑁 = 1000 𝐴

4.Join O’A this is output Line.

5.draw perpendicular bisector of O’A to meet horizontal line drawn from O’at point C.This is center of

circle.

6.With C as center and CO’ as radius,draw a semicircle to meet horizointal line from O’ at B.

40


7.Draw perpendicular from A on the X-axis meeting it at D.

l(AD)=8.5cm=𝑊𝑆𝑁

Power scale=𝑊𝑆𝑁

𝑙(AD )=

28336

8.5

Power scale=3333.6 W/Cm

8.Now rotors cu loss are half the total copper loass

𝐴𝐸

𝐴𝐹=

1

2𝑖. 𝑒 𝐴𝐸 =

1

2𝐴𝐹

Join O’E this is the torque line.

9.To locate full load point,draw AA’such that AA’ represents full load output to the power scale.

AA’=15000

3333 .6= 4.49 𝑐𝑚

10) draw parallel to the output line from A’ to meet circle at point P.This is full load point.

11) Draw vertical line from P to intersects output line at Q torque at R,base line at S and X-axis at T

So at Full load .

i) Line Current =l(OP)×Ccurrent Scale

= 5 .3 × 5

Line Current = 26.5 A

ii) Power Factor=Cos∅ = cos ∟ 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑂𝑃 𝑤𝑖𝑡𝑕 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑥𝑖𝑠

=cos(32°)

Power Factor= 0.84 (lag)

iii) Slip (S)=𝑄𝑅

𝑃𝑅=

𝑅𝑜𝑡𝑜𝑟 𝑐𝑢 𝑙𝑜𝑠𝑠𝑒𝑠

𝑅𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡=

0.35 𝑐𝑚

4 𝑐𝑚

S=8.75%

iv) 𝜂 =𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡=

𝑃𝑄

𝑃𝑇=

3.6 𝑐𝑚

4.5 𝑐𝑚= 80%

v) Maximum Torque

𝑙(JK)=8 cm=8× 3333.6

Maximum Torque =26668.8 Synchronous watts.

41


42


17. An induction motor has an efficiency of 0.9 when the shaft load is 45 kW. At this load, stator

ohmic loss and rotor ohmic each is equal to the iron loss. The mechanical loss is one-third of the no-

load losses. Neglect ohmic losses at no-load. Calculate the slip. Dec-2014, NOV-DEC 2017

Soloution:

Total losses for an output of 45 KW are

= 1

0.9− 1 × 45000 = 5000 𝑊𝑎𝑡𝑡𝑠

Total losses=stator 𝐼2𝑅 Loss+Stator core Loss+Rotor 𝐼2𝑅 +Mechanical Loss……(i)

At no load,the losses include mechanical loss,stator core loss and a small amount of 𝐼2𝑅loss in stator and

rotor.As 𝐼2𝑅 𝐿𝑜𝑠𝑠𝑒𝑠 𝑎𝑟𝑒 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑒𝑑 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑, 𝑤𝑒 𝑕𝑎𝑣𝑒

No load losses=stator core loss + mechanical loss (ii)

It is given that mechanical loss is one third of no load loss in 3 times the mechanical loss.Therfore,from

(ii)we have

3(mechanical loss)=stator core loss+mechanical loss

Or

Mechanical loss=1

2= (𝑆𝑡𝑎𝑡𝑜𝑟 𝑐𝑜𝑟𝑒 𝑙𝑜𝑠𝑠)

Let stator 𝐼2𝑅 𝑙𝑜𝑠𝑠(= 𝑅𝑜𝑡𝑜𝑟 𝐼2𝑅 Loss=Stator core Loss)be A.Then from (i)

Total losses=A+A+A+𝐴

2 or 5000=

7𝐴

2

∴ 𝑅𝑜𝑡𝑜𝑟 𝑜𝑕𝑚𝑖𝑐 𝐿𝑜𝑠𝑠 = 𝐴 = 1428.57 𝑊

Air gap power ,𝑃𝑔 = 45000 + 1428.57 +1428 .57

2= 47142.86𝑊

∴ 𝑆𝑙𝑖𝑝 =𝑅𝑜𝑡𝑜𝑟𝐼2𝑅 Loss

𝑃𝑔=

1428.57

47142.86= 0.0303

18. A 400 V, 6 pole, 3-phase, 50 Hz star connected induction motor running light at rated voltage

takes 7.5 A with a power input of 700 W. With the rotor locked and 150 V applied to the stator, the

input current is 35 A and power input is 4000 W; the stator resistance/phase being 0.55 ohms

under these conditions. The standstill reactance of the stator and rotor as seen on the stator side

are estimated to be in the ratio of 1:0.5. Determine the parameters of the equivalent circuit.

May-2015

From blocked rotor test,

𝑍𝐵𝑅 =150 3

35= 2.47Ω

𝑅𝐵𝑅 =4000 × 3

35 2= 1.09 Ω

𝑋𝐵𝑅 = 2.47 2 − 1.09 2 =2.22 Ω

𝑋1 + 𝑋2′ = 2.22

𝑋1 + 0.5𝑋1 = 2.22 Ω

43


𝑋1 = 1.48 Ω, 𝑋2′ = 0.74 Ω

𝑅2′ = 𝑅𝐵𝑅 − 𝑅1

𝑋𝑚 + 𝑋2′

𝑋𝑚

2

𝑋𝑚 = 29.03 Ω(Obtained from NL test below)

𝑅2′ = (1.09 − 0.55)

29.03 + 0.74

29.03

2

𝑅2′ =0.568 Ω

From no-load test,

𝑍𝑜 =

400 3

7.5= 30.79 Ω

𝑅0 =700

3

7.5 2= 4.15Ω

𝑋𝑜 = 30.79 2 − 4.15 2 =30.51 Ω

𝑋𝑚 = 𝑋0 − 𝑋1 = 30.51 − 1.48 = 29.03 Ω

𝑍𝑓 = 𝑗𝑋𝑚 ∥ 𝑅2

′

𝑠 + 𝑗𝑋2′

= j29.03 ∥(14.2+j0.74)

=10.98+j5.96=𝑅𝑓 + 𝑗𝑋𝑓

𝑅𝑓 = 10.98 Ω

𝑍𝑖𝑛 = 0.55 + 𝑗1.48 + 10.98 + 𝑗5.98

=11.53+j7.44=13.72< 32.8 ∘ Ω

𝐼1 =231

13.72= 16.84 𝐴

𝑃𝑓 = cos 32.8∘ = 0.84 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

Power input 𝑷𝒊𝒏= 𝟑 × 𝟒𝟎𝟎 × 𝟏𝟖. 𝟖𝟒 × 𝟎. 𝟖𝟒

Power input 𝑷𝒊𝒏= 9.80 KW

44


19) A 40 KW,3phase slip ring induction motor of negligible stator impedance runs at a speed of

0.96 times synchronous speed at rated torque. the slip ring at maximum torque is 4 times the full

load value. If the rotor resistance of the motor is increased by 5 times determine:

i) The speed, power output and rotor copper losses at rated torque;

ii)The Speed Corresponding to maximum Torque. May-June 2016

Solution:

𝑃𝑜𝑢𝑡 = 40 𝐾𝑊; 𝑁 = 0.96𝑁𝑠; 𝑠𝑚𝑇 = 4𝑠𝑓

Full load slip,𝑺𝒇 =𝑵𝒔−𝑵

𝑵𝒔=

𝑵𝒔−𝟎.𝟗𝟔𝑵𝒔

𝑵𝒔= 𝟎. 𝟎𝟒

∴ 𝑠𝑚𝑇 = 4𝑠𝑓 = 4 × 0.04 = 0.16

Now,𝑻𝒇

𝑻𝒎=

𝟐𝒔𝒇𝒔𝒎𝑻

𝒔𝒇𝟐+𝒔𝟐𝒎𝑻=

𝟐×𝟎.𝟎𝟒×𝟎.𝟏𝟔

𝟎.𝟎𝟒 𝟐+ 𝟎.𝟏𝟔 𝟐= 𝟎. 𝟒𝟕𝟎𝟔

When the rotor circuit resistance is increased 5 times,the magnitude of maximum torque will remain

unchanged because it is independent of load but the slip corresponding to the maximum torque will

change. Let the new slip corresponding to maximum torque be 𝑆𝑚𝑇𝑛 .

Since slip corresponding to maximum torque is proportional to rotor resistance provided its standstill

reactance is fixed,So

𝑠𝑚𝑇𝑛 = 𝑠𝑚𝑇 ×𝑅2𝑛

𝑅2= 𝑠𝑚𝑇 × 5 = 0.16 × 5 = 0.8

Now,𝑻𝒇

𝑻𝒎=

𝟐𝒔𝒇𝒔𝒎𝑻

𝒔𝒇𝟐+𝒔𝟐𝒎𝑻

Or,𝟎. 𝟒𝟕𝟎𝟔 =𝟐𝒔𝒇×𝟎.𝟖

𝒔𝒇𝟐+ 𝟎.𝟖 𝟐 𝒐𝒓 𝒔𝟐𝒇𝒏 + 𝟎. 𝟔𝟒 = 𝟑. 𝟒𝑺𝒇𝒏

Or 𝒔𝟐𝒇𝒏 − 𝟑. 𝟒𝒔𝒇𝒏 + 𝟎. 𝟔𝟒 = 𝟎

Or 𝒔𝒇𝒏 =3.4± 3.42−4×0.64

2=

3.4±3

2= 𝑜. 2, 𝑅𝑒𝑗𝑒𝑐𝑡𝑖𝑛𝑔 𝐻𝑖𝑔𝑕𝑒𝑟 𝑉𝑎𝑙𝑢𝑒.

i) The speed, power output and rotor copper losses at rated torque:

New speed at full load,𝑁 , = 1 − 𝒔𝒇𝒏 𝑵𝒔 = 𝟏 − 𝟎. 𝟐 𝑵𝒔 = 𝟎. 𝟖𝑵𝒔

Gross torque at full load=40×1000

(2𝜋𝑁

60)

=40×1000 ×60

2𝜋×0.96𝑁𝑠

∴ 𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 = 𝑔𝑟𝑜𝑠𝑠 𝑡𝑜𝑟𝑞𝑢𝑒 𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 ×2𝜋𝑁

60

=40 × 1000 × 60

2𝜋 × 0.96𝑁𝑠×

2𝜋 × 0.8𝑁𝑠

60= 33.333𝑘𝑤.

Rated copper losses at rated torque=𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡

1−𝑆𝑓𝑛× 𝑆𝑓𝑛

=33.333

1−0.2 × 0.2 = 8.333 𝐾𝑤

ii)The Speed Corresponding to maximum Torque.

Speed corresponding to maximum Torque

= 𝑁𝑠 1 − 𝑠𝑠𝑇𝑛 = 𝑁𝑠 1 − 0.8 = 0.2𝑁𝑠

1

Dr.A.Jeralidine Viji & M. Vijayaragavan / EEE / MEC Unit-04

UNIT – IV

Starting and Speed Control of Three Phase Induction Motor

PART-A

1. Give the functions performed by induction motor starter? (Dec - 2014)

To limit the starting current

To start the motor

To protected from over load condition and low voltage condition.

2. List out four methods of speed control in 3 phase induction motor.Dec-2012, 2015, May-2013

Stator voltage control.

Stator frequency control

Rotor resistance control.

Pole changing method.

Slip power recovery scheme.

3. Why is starter necessary to start a 3 phase induction motor?May-2012,Nov-Dec 2016,APR-

MAY 2018

When a 3-phase induction motor is switched on at normal supply voltage, heavy current will

flow through the motor because at the time of starting, there is no back emf. An induction

motor,when directly switched on, takes five to seven times its full load current and it develops only

1.5 to 2.5 times full load torque.

This initial inrush of excessive current is objectionable because it will produce large line

voltage drop. This will affect the operation of other electrical equipment’s connected to the same

line. Due to this, starters are used for starting the three phase induction motor.

4. Name the different types of starters to be used in 3-induction motor?Dec-2013,Nov-Dec 2016.

DOL starter.

Primary resistance starte3r.

Autotransformer starter.

Star-Delta starter

Rotor resistance starter.

5. What are the advantages and disadvantages of auto transformer starter?

Advantages:

Reduced line current

Smooth starting

High acceleration

Disadvantages

Cost is high

2


It is used for large motors only.

6. What is meant by stator voltage control?

The induction motor speed can be controlled by varying the stator voltage. This method of

speed control is known as stator voltage control. Here, the supply frequency is constant. The stator

voltage can be controlled by two methods.

7. What are the different methods of stator voltage control?

Using auto transformer

Primary resistors connected in series with stator winding.

8. What is voltage /frequency method?

The voltage/frequency control is one method of speed control of three-phase induction motor.

From the emf equation, the airgap flux is given by,

𝜑 =1

2 𝜋 𝑇1𝐾𝑤 𝑉

𝑓

From this expression, by varying the supply frequency the airgap flux changes. This will lead

to saturation of the motor. To avoid this, the air gap flux should be maintained constant.

To maintain airgap flux constant, parameters V and f must be changed so as to maintain (V/ f)

ration constant. This is known as V/f control.

9. What is the purpose of adding external resistance in the rotor circuit?

Starting torque can be improved.

Starting current will be controlled.

Motor speed can be controlled.

10. What are the advantages and disadvantages of rotor resistance control?

Advantages:

Smooth and wide range of speed control.

Absence of in-rush starting current.

Available of full-rated torque at starting.

High line power factor and absence of line current harmonics.


Disadvantages:

Reduced efficiency because the slip energy is wasted in the rotor circuit resistance.

Speed changes very widely with load variation.

Unbalance in voltage and current if rotor circuit resistance are not equal.

11. What are the disadvantages of cascade control?

This method requires two motors.

More expensive.

Wire range of speed control is not possible.

It cannot be operated when P1=P2 or P1>P2.

3


12. What is meant by slip power recovery scheme? Dec-2013

Slip power can be returned to the supply source and can be used to supply an additional motor

which is mechanically coupled to the main motor. This type of drive is known as slip power

recovery system and improves the overall efficiency of the system.

13. Why is speed control by pole changing technique suitable only to squirrel cage induction

motors?

Cage rotor is not wound for any specific number of poles as stator winding has. Therefore, in a

squirrel cage induction motor, an arrangement is required only for changing the number of poles

stator. In slip ring induction motor arrangement for changing the number of poles in rotor is also

required, which complicates the machine. Therefore, this method of speed control is used with

squirrel cage induction motors only.

14. Mention the various methods of starting a 3-phase squirrel cage induction motor.

Auto transformer starter

Star/delta starter

DOL starter

Primary resistance starter.

15. What are the various methods of speed control of 3-phase induction motor from stator side?

stator voltage control

stator frequency control

v/f method

pole changing method

16. What are the speed control of 3-phase induction motor from rotor side?

rotor resistance control

slip power recovery scheme.

cascaded control

17. What are the types of slip power recovery scheme?

Kramer system

Scherbiussystem

18. What is the cheapest method of starting a 3-phase induction motor?

Direct-on-line starter

19. What is the effect of change in supply on starting torque? Dec 2015, May 2016,May 2017.

Large reduction in starting torque because the starting torque varies as the square of voltage

applied to the stator.

20. A three phase induction motor is to be started first using an auto transformer with 70%

tapping and next direct-on-line. What will be the ratio of starting torque?

X=70% or 0.7

4


𝐬𝐭𝐚𝐫𝐭𝐢𝐧𝐠 𝐭𝐨𝐫𝐪𝐮𝐞 𝐰𝐢𝐭𝐡 𝐚𝐮𝐭𝐨 − 𝐭𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦𝐞𝐫

𝐬𝐭𝐚𝐫𝐭𝐢𝐧𝐠 𝐭𝐨𝐫𝐪𝐮𝐞 𝐰𝐢𝐭𝐡 𝐝𝐢𝐫𝐞𝐜𝐭 𝐬𝐰𝐢𝐭𝐜𝐡𝐢𝐧𝐠= 𝐱𝟐 = 𝟎.𝟕𝟐 = 𝟎.𝟒𝟗

21. Is it possible to add an external resistance in the rotor circuit of a 3-phase cage induction

motor? Give reasons for your answer. Or why is rotor rheostat starter unsuited for a

squirrel cage motor ? NOV-DEC 2017

No, because the rotor winding is permanently short –circuited.

22. Rotor resistance starter is preferred to reduce voltage starting of a wound rotor induction

motor why?

Because rotor resistance of a wound round induction motor not only reduces the starting current

at the starting instant but increases the starting torque also improvement in power factor whereas

the reduced voltage starting reduces only the starting current at the starting instant.

23. What are the advantages of star-delta starter?

This method of starting is simple,cheap and effective since no power is lost in additional

components.

24. While controlling induction motor speed, how sub-synchronous speed is achieved?

Sub-synchronous speed can be achieved, while controlling the speed of an induction motor,

by injection a slip frequency emf in phase opposition with emf induced in the rotor circuit.

25. What is the torque developed by an induction motor when applied voltage is reduced to half

supply frequency remaining unchanged? Dec-2012, 2011

Torque developed by an induction motor with half-rate voltage,supply frequency remaining

unchanged, will be reduced to one-fourth because T α V2 .

26. What are the advantage of slip power schemes? APR-MAY 2018

The slip power can be recovered and fed back to the supply. The overall efficiency also improved.

The main advantage of this method is that any speed, within the working range, can be obtained

instead of only two or three, as with other methods of speed control.

If the rotor converter is over excited, it will take a leading current which compensates for the

lagging current drawn by SRIM & hence improves the power factor of the system.

27. Why it is that the v/f ratio kept constant while controlling the speed of a 3-phase induction

motor by varying the supply frequency?

By keeping the ration v/f constant, the flux density is kept constant so that the performance of

the machine is not affected.

28. Define slip power in an induction motor.

The portion of air gap, which is not converted into mechanical power is called slip power.

Slip power is nothing but multiplication of slips(s) and air gap power(Pag).Slip power = S x Pag

5


29. Why most of the 3ɸIM constructed with delta connected stator winding? May - 2012

For the same power rating, the current drawn by the delta connected stator winding, from

the supply is much less than the star connected stator winding, in running condition. Due to this, the

stator copper losses are less and performance of induction motor is better. Hence most of the three

phase induction motors are constructed with delta connected stator winding.

30. What is the effect of increasing the rotor resistance on starting current and torque?Dec-2012

To increase the starting torque,

To limit the high starting stator current

To obtain the speed control

31. Why it is objectionable to start large 3ɸIM by switching it directly on the line?May-2013

In this type of starter connected directly to the supply lines without any reduction voltage, hence

this stator does not reduce the apply voltage that’s why it is objectionable to start large three phase

induction motor by switching it directly on the line.

32. Which is the cheapest method of starting a three phase induction motor? May-2014

Stator resistance starter

33. What is meant by plugging? May-2014

The IM can be stopped immediately by just interchanging any two of the stator leads by

doing this,it reverses the direction of the revolving flux,which produces a torque in the reverse

direction,thus causing a breaking effect on the rotor. This breaking period is called the plugging.

34. While controlling the speed of an induction motor, how is super-synchronous speed

achieved? Dec-2014, May 2017.

Super synchronous speed can be achieved, while controlling the speed of an induction

motor, by injecting a slip frequency emf in phase with emf induced in the rotor circuit.

35. State an important distinguishing factor of induction generator and alternator. May-2015

Induction generator does not require DC excitation.

It does not hunt or drop out of synchronism.

36. Draw the torque-speed characteristics of an induction motor whose rotor resistance is very

large compared to rotor inductance. May-2015

6


37. How can the direction of a capacitor run motor be reversed? (May - 2016) By interchanging the connections of main winding or auxiliary winding.

38. What are the conditions for regenerative braking of an induction motor to be possible?

NOV-DEC 2017

When the speed of the motor is more than the synchronous speed, relative speed between

the motor conductors and air gap rotating field reverses, as a result the phase angle because greater than

90o and the power flow reverse and thus regenerative braking takes place.

PART-B

1) Write short notes on Necessity of starters. Dec-2012

In a three phase induction motor, the magnitude of an induced emf in the rotor circuit depends

on the slip of the induction motor. This induced emf effectively decides the magnitude of the rotor

current. The rotor current in the running condition is given by,

𝐼2𝑟 = 𝑠 𝐸2

𝑅22 + 𝑠 𝑋2 2

But at start, the speed of the motor is zero and slip is at its maximum i.e. unity. So

magnitude of rotor induced e.m.f. is very large at start. As rotor conductors are short circuited, the

large induced e.m.f. circulates very high current through rotor at start.

The condition is exactly similar to a transformer with short circuited secondary. Such a

transformer when excited by a rated voltage, circulates very high current through short circuited

secondary. As secondary current is large, the primary also draws very high current from the supply.

Similarly in a three phase induction motor, when rotor current is high, consequently the

stator draws a very high current from the supply.

Similarly in a three phase induction motor, when rotor current is high, consequently the

stator draws a very high current from the supply. This current can be of the order of 5 to 8 times the

full load current, at start.

Due to such heavy inrush current at start there is possibility of damage of the motor

winding. Similarly such sudden inrush of current causes large line voltage drop. Thus other

appliances connected to the same line may be subjected to voltage spikes which may affect their

working. To avoid such effects, it is necessary to limit the current drawn by the motor at start.

The starter is a device which is basically used to limit high starting current by supplying

reduced voltage to the motor at the limit of starting. Such a reduced voltage is applied only for short

period and once rotor gets accelerated, full normal rated voltage is applied.

Not only the starter limits the starting current but also provides the protection to the

induction motor against overloading loading and low voltage situations.

7


The protection against single phasing is also provided by the starter. The induction motor

having rating below 5 HP can withstand starting currents hence such motors can be started directly

on line. But such motors also need overload, single phasing and low voltage protection which is

provided by a starter. Thus all the three phase induction motors need some or the other type of

starter.

2) Illustrate any two methods used for starting an induction motor. (Dec – 2012),Nov-Dec

2016,APR-MAY 2018

Types of Starters

From the expression of rotor current it can be seen that the current at start can be controlled

by reducing E2 which is possible by supplying reduced voltage at start or by increasing the rotor

resistance R2 at start. The second method is possible only on case of slip ring induction motors. The

various types of starters based on the above two methods of reducing the starting current are,

1. Stator resistance starter

2. Autotransformer starter

3. Star-delta starter

4. Rotor resistance starter

5. Direct on line starter

Auto Transformer Starter

A three phase star connected autotransformer can be used to reduce the voltage applied to

the stator. Such a starter is called an autotransformer starter. The schematic diagram of

autotransformer starter is shown in the Fig 1. It consists of a suitable change over switch.

When the switch is in the start position, the stator winding is supplied with reduced voltage.

This can be controlled by tappings provided with autotransformer.The reduction in applied voltage

by the fractional percentage tapping’s x, used for an autotransformer is shown in the Fig. 2.

8


When motor gathers 80% of the normal speed, the changeover switch is thrown into run position.

Due to this, rated voltage gets applied to stator winding. The motor starts rotating with

normal speed. Changing of switch is done automatically by using relays. The power loss is much

less in this type of starting. It can be used for both star and delta connected motors. But it is

expensive than stator resistance starter.

Relation between Tst and TF.L.

Let x be the fractional percentage tappings used for an autotransformer to apply reduced

voltage to the stator.

So if, Isc= Starting motor current at rated voltage and

Ist = Starting motor current with starter

then Ist= x Isc .....Motor side ............(1)

But there is exists a fixed ratio between starting current drawn from supply Ist(supply) and

starting moor current Ist(motor) due to autotransformer, as shown in the Fig.3.

Autotransformer ratio x = Ist (supply)/ Ist(motor)

Ist(supply) = x Ist(motor) .............(2)

Substituting Ist(motor) from equation (1),

... Ist(supply) = x .x Isc = x

2Isc ............(3)

Now Tst α Ist2 (motor) α x

2Isc

2

and TF.L. α (IF.L.)2/sf

9


Note: Thus starting torque reduces by x

2where x is the transformer ratio.

Direct on line (DOL) starter (Dec – 2013)

DOL starter just connects the motor terminals to the supply directly.

Two push buttons for start and stop and the over load protection.

The contactor M has four poles, three for the motor to be connected to the three phase supply

and one for the contactor coil.

Even if the operator releases the start push button, the coil continues to remain energized

through the fourth pole of the contactor M, and the motor keeps on running.

When the operator wishes to stop the motor, we just pushes the stop button (Normally Red in

colour) in, and the coil circuit is broken, the coil thus de-energized and the contactor M opens and

the motor is isolated from the supply and hence it stops.

Even if the operator releases the start push button, the coil continues to remain energized

through the fourth pole of the contactor M, and the motor keeps on running.

When the operator wishes to stop the motor, we just pushes the stop button (Normally Red in

colour) in, and the coil circuit is broken, the coil thus de-energized and the contactor M opens

and the motor is isolated from the supply and hence it stops.

In case of overload, the overload relay (usually thermal type in the form of a bimetallic strip)

also opens the circuit of the coil and the contactor opens.

10


In case of a supply failure while the motor is running, the coil is de-energized and the motor is

isolated from the supply lines. After the supply is returned, the operator has to operate the

START push button for running the motor.

Over Load Protection:

When the line current exceeds the preset value, OLRC is energized more and causes the

contactor S4 to open. When S4 opens, the UVRC is disconnected from the supply. Therefore, it

will release the main contactors.

The relation between starting torque(Tst) and full load torque(Tfl) is given by

𝑇𝑠𝑡𝑇𝑓𝑙

= 𝐼𝑠𝑐𝐼𝑓𝑙

2

𝑠𝑓

Where Isc=Ist = short circuit current,

Sf = full load slip.

Stator Resistance Stator.

In order to apply the reduced voltage to the stator of the induction motor, three resistances are

added in series with each phase of the stator winding.

Initially the resistances are kept maximum in the circuit. Due to this, large voltage gets dropped

across the resistances.

Hence a reduced voltage gets applied to the stator which reduces the high starting current.

The schematic diagram showing stator resistances is shown in the Fig.

When the motor starts running, the resistances are gradually cut off from the stator circuit.

When the resistances are entirely removed from the stator circuit i.e. rheostats in RUN position

then rated voltage gets applied to the stator.

Motor runs with normal speed.

The starter is simple in construction and cheap. It can be used for both star and delta connected

stator but there are large power losses due to resistances.

Also the starting torque of the motor reduces due to reduced voltage applied to the stator.

If the voltage across the terminal is reduced by 50%, then the starting current is reduced by

50%, but torque is reduced to 25% of the full voltage value.

Let reduced per phase voltage = xV1

Per phase starting current Ist = 𝑥𝑉1

𝑍𝑆𝐶= 𝑥𝐼𝑠𝑐

We know that 𝑇𝑠𝑡

𝑇𝑓𝑙=

𝐼𝑠𝑐

𝐼𝑓𝑙

2

𝑆𝑓

= 𝑥𝐼𝑆𝐶

𝐼𝑓𝑙

2

𝑆𝑓

11


𝑇𝑠𝑡𝑇𝑓𝑙

= 𝑥2 𝐼𝑠𝑐𝐼𝑓𝑙

2

𝑠𝑓

In an induction motor, torque α (voltage)2

𝑆𝑡𝑎𝑟𝑡𝑖𝑛𝑔𝑡𝑜𝑟𝑞𝑢𝑒𝑤𝑖𝑡𝑕𝑟𝑒𝑎𝑐𝑡𝑜𝑟𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔

𝑆𝑡𝑎𝑟𝑡𝑖𝑛𝑔𝑡𝑜𝑟𝑞𝑢𝑒𝑤𝑖𝑡𝑕𝑑𝑖𝑟𝑒𝑐𝑡𝑠𝑤𝑖𝑡𝑐𝑕𝑖𝑛𝑔=

𝑥𝑉1

𝑉1

2

= 𝑥2

ADVANTAGES:

Smooth acceleration

High power factor during start

Less expensive

Closed transition starting

DISADVANTAGES:

Power lost in resistors

Low starting torque

Less efficiency

-----------------------------------------------------------------------------------------------------------------------------

3) With the help of a neat diagram, explain the working of a star-delta starter for a three-phase

induction motor.May-2012,2013,Dec- 2013, 2014, Nov-Dec 2016,2017

This is the cheapest starter of all and hence used very commonly for the induction motors. It

uses tripple pole double throw (TPDT) switch. The switch connects the stator winding in star at

start. Hence per phase voltage gets reduced by the factor 1/√3. Due to this reduced voltage, the

starting current is limited.

When the switch is thrown on other side, the winding gets connected in delta, across the

supply. So it gets normal rated voltage. The windings are connected in delta when motor gathers

sufficient speed.

The arrangement of star-delta starter is shown in the Fig.

12


The operation of the switch can be automatic by using relays which ensures that motor will

not start with the switch in Run position. The cheapest of all and maintenance free operation are the

two important advantages of this starter. While its limitations are, it is suitable for normal delta

connected motors and the factor by which voltage changes is 1/√3 which cannot be changed.

Ratio of Tst to TF.L.

We have seen in case of autotransformer that if x is the factor by which the voltage is

reduced then,

Now the factor x in this type of starter is 1/√3.

Where Isc= Starting phase current when delta connection with rated voltage

IF.L. = Full load phase current when delta connection

----------------------------------------------------------------------------------------------------------------------------- --

4) Explain briefly rotor resistance starter for slip ring induction motors and also explain the

calculation of steps of rotor resistance. (Dec – 2015) May-2012,2014, 2016

This starter is used in slip-ring induction motors.

To limit the rotor current which consequently reduces the current drawn by the

motor from the supply, the resistance can be inserted in the rotor circuit at start.

13


This addition of the resistance in rotor is in the form of 3ɸ star connected rheostat.

The arrangement is shown in the Fig.

The external resistance is inserted in each phase of the rotor winding through slip

ring and brush assembly. Initially maximum resistance is in the circuit.

As motor gathers speed, the resistance is gradually cut-off. The operation may be

manual or automatic.

We have seen that the starting torque is proportional to the rotor resistance.

Hence important advantage of this method is not only the starting current is limited but

starting torque of the motor also gets improved.

Calculation of Steps of Rotor Resistance Starter

The calculation of steps of rotor resistance starter is based on the assumptions that,

1. The motor starts against a constant torque

2. The rotor current fluctuates between two fixed values, a maximum and a minimum,

denoted as I2max and I2min.

The Fig shows a single phase of a three phase of a three phase rheostat to be inserted

in the rotor. The starter has n steps, equally divided into the section AB. The contact point

after each step is called stud. The total resistances upto each stud from the star point of star

connected rotor as denoted as R1, R2, ....Rn-1.

14


It consists of rotor resistance r2 and the external resistances Rx1, Rx2...etc. At the time of

reaching to the next step, current is maximum. Then motor gathers speed, slip reduces and hence

while leaving a stud, the current is I2min.

Let E2 = Standstill rotor e.m.f. per phase

When handle is moved to stud 1, the current is maximum given by,

where s1 = Slip at start = 1

while moving to stud 2, the current reduces to I2min given by,

Just reaching to stud 2, the current again increases to I2min as the part of external resistance Rx1 gets

cut-off.

While leaving stud 2, the slip changes to s3 and current again reduces to,

While just reaching to stud 3, Rx2 gets cut off completely and current again increases to,

Hence at the last n

th stud, the maximum current is,

Wheresn = Slip under normal running condition

At nth stud no external resistance is in series with rotor.

http://3.bp.blogspot.com/-lNqY-ciE2gw/TnprP36oeaI/AAAAAAAABiQ/1IZtYg22akI/s1600/ABB168.jpeg

15


From (1) and (2) we can write,

where K = Constant

From (1), R1 = s1r2/sn but s1 = 1 at start

Once R1 is known, other resistances can be calculated.

R2 = KR1, R3 = K R2 = KKR1 = K2 R1

R4 = K3 R1, .... .... r2 = K

n-1 R1

From last expression of r2,

where n = Number of starter studs

Thus the resistances of various sections can be obtained as,

-----------------------------------------------------------------------------------------------------------------------------

5) Explain the different speed control schemes available for poly phase induction motor from

stator side?May-2013,2016,Dec 2015,APR-MAY 2018, NOV-DEC 2017

Need for speed controlof IM:

A three phase induction motor is practically a constant speed motor like a d.c. shunt motor. But

the speed of d.c. shunt motor can be varied smoothly just by using simple rheostats. This maintains

the speed regulation and efficiency of d.c. shunt motor. But in case of three phase induction motors

it is very difficult to achieve smooth speed control. And if the speed control is achieved by some

16


means, the performance of the induction motor in terms of its power factor, efficiency etc. gets

adversely affected.

For the induction motor we know that,

From this expression it can be seen that the speed of induction motor can be changed either

by changing its synchronous speed or by changing the slip s.

Similarly torque produced in case of three phase induction motor is given by,

N = Ns (1 - s)

So as the parameters like R2, E2 are changed then to keep the torque constant for constant

load condition, motor reacts by change in its slip. Effectively its speed changes.

Thus speed of the induction motor can be controlled by basically two methods:

1. From stator side and

2. From rotor side

From stator side, it includes following methods:

Supply frequency control to control Ns, called V / f control.

Supply voltage control.Refer Page No. 21

Controlling number of stator poles to control Ns. Refer Page No. 18

Adding rheostats in stator circuit.Refer Page No. 22

From rotor side, it includes following methods:

Adding external resistance in the rotor circuit.Refer Page No. 26

Cascade control.Refer Page No. 24

Injecting slip frequency voltage into the rotor circuit.Refer Page No. 27

Supply frequency V/F control:

The synchronous speed is given by,

Ns = 120f / P

Thus by controlling the supply frequency smoothly, the synchronous speed can be

controlled over a wide range. This gives smooth speed control of an induction motor.

But the expression for the air gap flux is given by,

17


This is according to the e.m.f. equation of a transformer where,

K1 = Stator winding constant

Tph1 = Stator turns per phase

V = Supply voltage

f = Supply frequency

It can be seen from this expression that if the supply frequency f is changed, the value of air

gap flux also gets affected. This may result into saturation of stator and rotor cores. Such a

saturation leads to the sharp increase in the (magnetization) no load current of the motor. Hence it is

necessary to maintain air gap flux constant when supply frequency f is changed.

To achieve this, it can be seen from the above expression that along with f, V also must be

changed so as to keep (V/f) ratio constant. This ensures constant air gap flux giving speed control

without affecting the performance of the motor. Hence this method is called V / f control.

Hence in this method, the supply to the induction motor required is variable voltage variable

frequency supply and can be achieved by an electronic scheme using converter and inverter

circuitry. The scheme is shown in the Fig. 1.

The normal supply available is constant voltage constant frequency a.c. supply. The

converter converts this supply into a d.c. supply. This d.c. supply is then given to the inverter. The

inverter is a device which converts d.c. supply, to variable voltage variable frequency a.c. supply

which is required to keep V / f ratio constant. By selecting the proper frequency and maintaining V

/ f constant, smooth speed control of the induction motor is possible.

If f is the normal working frequency then the Fig. 2 shows the torque-slip characteristics for

the frequency f1 >f and f2 <f i.e. for frequencies above and below the normal frequency.

Another disadvantage of this method is that the supply obtained cannot be used to supply other

devices which require constant voltage. Hence an individual scheme for a separate motor is required which

makes it costly.

18


6) Describe the method of speed control of a 3-phase squirrel cage induction motor by changing

the number of stator poles and state the applications of this method. Nov-Dec-2014,2017

The method is called pole changing method of controlling the speed. In this method, it is

possible to have one, two or four speeds in steps, by the changing the number of stator poles. A

continuous smooth speed control is not possible by this method.

The stator poles can be changed by following methods :

1. Consequent poles method

2. Multiple stator winding method

3. Pole amplitude modulation method.

Consequent Poles Method

In this method, connections of the stator winding are changes with the help of simple

switching. Due to this, the number of stator poles get changed in the ratio 2:1. Hence either of the

two synchronous speed can be selected.

Consider the pole formation due to single phase of a three phase winding, as shown in the

Fig.1. There are three tapping points to the stator winding. The supply is given to two of them and

third is kept open.

It can be seen that current in all the parts of stator coil is flowing in one direction only. Due

to this, 8 poles get formed as shown in the Fig. 1. So synchronous speed possible with this

arrangement with 50 Hz frequency is Ns= 750 r.p.m.

19


If now the two terminals to which supply was given either are joined together and supply is

given between this common point and the open third terminal, the poles are formed as shown in the

Fig. 2.

It can be seen that the direction of current through remaining two. Thus upward direction is

forming say S pole and downward say N. it can be observed that in this case only 4 poles are

formed. So the synchronous speed possible is 1500 r.p.m. for 50 Hz frequency.

Thus series/parallel arrangements of coils can produce the poles in the ratio 2:1. But the

speed change is in step and smooth speed control is not possible. Similarly the method can be used

only for the squirrel cage type motors as squirrel rotor adjusts itself to same number of poles as

stator which is not the case in slip ring induction motor.

Multiple Stator Winding Method

In this method instead of one winding, two separate stator winding are placed in the stator

core. The windings are placed in the stator slots only but are electrically isolated from each other.

Each winding is divided into coils to which, pole changing with consequent poles, facility is

provided.

Thus giving supply to one of the two windings and using switching arrangement, two speeds

can be achieved. Same is true for other stator winding. So in all four different speeds can be

obtained.

The various limitations of this method are,

1. Can be applied only to squirrelcage motor.

2. Smooth speed control is not possible. Only step changes in speed are possible.

3. Two different stator windings are required to be wound which increases the cost

of the motor.

4. Complicated from the design point of view.

Typical speed-torque characteristics of pole changing induction motor are shown in the Fig.

20


Pole Amplitude Modulation Method

The basic disadvantage of other methods which is nonavailability of smooth speed control,

is eliminated by this method. The ratio of two speeds in this method, need not be necessarily

2:1.

The basic principle of this method is the modulation of two sinusoidally varying m.m.f.

waves, with different number of poles.

Consider sinusoidally distributed m.m.f. wave one phase of the stator as,

P=No. of.Polesandθ=Mechanical angle

This wave is modulated by another sinusoidal m.m.f. wave having PM number of poles, expressed

as,

The resultant m.m.f. wave after modulation is,

Thus the resultant wave is equivalent to two m.m.f. waves having two separate number of poles as,

P1=P-PMandP2=P+PM

This is called suppressed carrier modulation.

If we succeed in suppressing one of the two poles then there exists rotating magnetic field

with number of poles as P1 or P2. And while suppressing, the method can be used such that the

resultant number of poles retained is as required from the speed point of view.

Now if the three stator windings are placed such that angle between their phase axes is

(2π/3)r radians where r is an integer which is not divisible by 3 then the phase axes angle for

modulated poles is given by,

Now to suppress one of the two poles, the angle between its phases axes must be multiple of 2π.

21


So if r and n are selected so as to satisfy one of the above relations, then either P1 or P2 get

suppressed and field corresponding to other pole exists. So speeds corresponding to P poles without

modulation and corresponding to either P1 or P2 with modulation, can be achieved. The negative

sign in equation (1), gives suppression of P2 and existence of P2 = P + PM while positive sign in

equation (1), gives suppression P2 of and existence of P1 = P - PM poles.

For example, stator has 8 poles while values of n and r are selected as 1 and 4 respectively. r

is not divisible by 3.

Let poles of modulation function PM are 2.

From equation (1) we can see that,

Thus P1 gets suppressed and we get poles P2 = P + PM = 10.So two speeds corresponding to

P and P2 can be obtained.

Similarly if the poles of modulation function PM are 4 and n and r are selected as 1 and 2 then,

In this case gets suppressed and we get poles P1= P-PM=4.

This method is advantages as it reduced the size to a great extent and hence cost of the

machine.The limitation that it can be used only for squirrel cage motors still continues.

Practically the rectangular wave is used for modulation. This is achieved by dividing stator

coil into groups and then by dropping alternate group, other groups are connected in series

opposition.

CHANGE IN STATOR VOLTAGE:

The speed of the induction motor can be controlled by varying the stator voltage.This

method of speed control is known as stator voltage control. Here, the supply frequency is constant.

The stator voltage can be controlled by two methods.

Using auto transformer

Primary resistors connected in series with stator winding.

22


Using auto transformers:

The speed of the induction motor can be controlled by using auto transformer. The input to

the auto transformer is a fixed ac voltage. By varying the auto transformer, we can get variable ac

output voltage without change in supply frequency. The variable voltage is fed to the induction

motor. Then the induction motor speed also changes.

Primary resistors connected in series with stator winding:

The primary resistors are connected in series with stator windings.

By varying the primary resistance, the voltage drop across the motor terminal is

reduced.Thatis,reducedvoltage is fed to the motor. Then the motor speed can be reduced. It is one

method of conventional speed control of induction motor. The control method is very simple. The

main disadvantage is that more power loss occurs in the primary resistors.

23


The torque is proportional to the square of its stator voltage. i.e., T α V

2. By varying the

voltage the torque also changes.

The speed torque characteristics of induction motor under stator voltage control. Here the

slip at the maximum torque remains unchanged since it is not a function of voltage. By varying the

stator voltage, the maximum torque and starting torque also change. For a low sip motor, the speed

range is very narrow. So this method is not used for wide range of speed control and constant

torque load. This applicable where load requiring low starting torque and a narrow speed range at

relatively low slip are required.

Change in stator frequency:

The stator frequency control is one of the methods of speed control for a 3-phase induction

motors. Here, we can vary the input frequency of the motor. Under steady state condition, the

induction motor operates in the small-slip region, where the speed of the induction motor is always

close to the synchronous speed of the rotating flux. The synchronous speed of the induction motor

is given by

𝑵𝑺 =

𝟏𝟐𝟎𝒇

𝑷

Where,f = frequency of the supply voltage, P = Number of poles

In this equation, synchronous speed of the motor is directly proportional to the frequency of

the supply voltage. When the supply frequency changes, the motor speed also changes. It is possible

only, by controlling the speed of the prime movers of the generators. This method is rarely used.

The emf V induced in the stator winding of the induction motor is given by,

V = 2πf T1 Ф Kw

Where, Ф = flux / pole

Kw = winding factor

f = frequency of stator supply,

T1= no. of turns in the stator winding.

Here we consider two cases.

Low frequency operation at constant voltage.

High frequency operation at constant voltage.

24


Low frequency operation at constant voltage:

by decreasing the supply frequency at constant voltage V the value of air gap flux increases

and the induction motor magnetic circuit gets saturated. Consider the emf equation.

V = constant

F = frequency

Ф = increases

Due to this low frequency operation, the following effects take place.

The reactance will be low leading to high motor currents.

More losses

Very low efficiency

High frequency operation at constant voltage:

With the constant input voltage, if the stator frequency is increased, the motor speed slso

increases. Due to increase in frequency, flux and torque are reduced.

V = constant

F = increases

Ф = decreases

By increasing the supply frequency of the motor, the following effects will follow

The no-load speed increases

The maximum torque decreases

Starting torque reduces

This type of frequency control is not normally used because of the above disadvantages.

---------------------------------------------------------------------------------------------------------------------------

7) Explain the different speed control schemes available for poly phase induction motor from

rotor side? May- 2012,2013,2014, Dec-2012, 2013

i) Cascade control:

Another method of speed control of slip ring induction motor is cascade control. It is also

known as tandem control.

It consists of two slip ring induction motors. The 1st motor is called motor M1.this motor is

coupled with second motor. The second motor is called auxiliary motor M2.

A three phase supply is fed to the stator of the main motor M1. This method of connection

is called cascade connection or tandem connection.

In this cascading method, if both motors produce the torque in the same direction means,

cumulative cascading and opposite direction means, differential cascading.

The expression for the speed of the set is derived as follows.

Let P1 = Number of poles of main motor M1.

25


P2 = Number of poles of auxiliary motor M2.

f = supply frequency

f1 = slip frequency of main motor M1.

f2 = supply frequency of auxiliary motor M1

N = speed of both motors.

Synchronous speed of the main motor Ns1 is given by

𝑁𝑆1 = 120𝑓

𝑃1

N = speed of both motors

Slip for main motor M1, 𝑆1 = 𝑁𝑆1−𝑁

𝑁𝑆1

f1 = frequency of rotor induced emf of main motor M1

f1 = s1f

the supply frequency of the auxillary motor M2 is f1 i.e., f2 = f1

Ns2 = 120𝑓2

𝑃2 =

120𝑓1

𝑃2

= 120𝑠1𝑓

𝑃2 =

120𝑓

𝑃2 𝑁𝑆1−𝑁

𝑁𝑆1

Under no-load condition, the speed of the auxiliary motor M2 is N. it is approximately

equal to its synchronous speed Ns2.

Ns2 = N

N = 120𝑓

𝑃2 𝑁𝑆1−𝑁

𝑁𝑆1

N = 120𝑓

𝑃2 1 −

𝑁

𝑁𝑆1

N = 120𝑓

𝑃2 1 −

𝑁

120𝑓

𝑃1

N = 120𝑓

𝑃2 1 −

𝑁𝑃1

120𝑓 =

120𝑓

𝑃2 -

𝑁𝑃1

𝑃2

N 1 +𝑃1

𝑃2 =

120𝑓

𝑃2 (or) N =

120𝑓

𝑃1+𝑃2

From this equation, the number of poles is equal to the sum of the number of poles of two machines.

This method can give four different speeds,

Main motor alone : Ns = 120𝑓

𝑃1

Auxiliary motor alone : Ns = 120𝑓

𝑃2

Cumulative cascade connection : N = 120𝑓

𝑃1+𝑃2

Differential cascade connection : N = 120𝑓

𝑃1−𝑃2 [P2<P1]

The main disadvantages are

This method requires two motors.

More expensive

Wide range of speed control is not possible

It cannot be operated when P1 = P2 (or) P1< P2.

26


ii) Adding external resistance in the rotor circuit:

This method is applicable only for slip ring induction motor. The external resisitance can be

added in the rotor circuit.

A simple and primitive method of speed control of a slip ring induction motor is by

mechanical variation of the rotor circuit resistance

The torque equation of the induction motor is T α 𝑆𝐸2

2𝑅2

𝑅22+ 𝑆𝑋2

2

The slip corresponding to maximum torque is given by sm=𝑅2

𝑋2 , sm α R2

If we add external resistance in the rotor circuit, the slip increase and speed decreases.

The maximum torque equation is Tmaxα𝐸2

2

2𝑋2

This equation is independent of rotor resistance i.e. by varying the rotor resistance the

maximum torque is not affected. It is clearly indicated.

The starting torque of the induction motor is Tst α 2𝐸2

2𝑅2

𝑅22+𝑋2

2

From this equation, it is seen that by increasing the rotor circuit resistance the rotor circuit

resistance, the starting torque also increases. It is clearly indicated in the speed- torque

characteristics.

Advantages:

Smooth and wide range of speed control.

Absence of in-rush starting current.

Availability of full-rated torque at starting

High line power factor

Absence of line current harmonics.


Disadvantages:

Reduced efficiency because the slip energy is wasted in the rotor circuit resistance.

Speed changes vary widely with load variation.

27


Unbalance in voltage and current if rotor circuit resistances are not equal.

iii) Slip power recovery system:

8) With neat diagrams, explain the slip power recovery scheme as applied to wound rotor

induction motors. Dec-2014,2015,May 2016,2017,Nov-Dec 2017

This system is mainly used for speed control of slip ring induction motor. The speed ofslip

ring IM can be controlled either by varying the stator voltage or by controlling the power flow in

the rotor circuit.

Rotor air gap power = mechanical power + rotor copper loss

Pag = Pm + Pcu

Pag= ωsT

Pm = ωT

ω = ωs(1 - s)

Pcu = sωsT

sPag = slip power

Pm =(1 - s)Pag

Where T = electromagnetic torque developed by the motor

ωs= synchronous angular velocity

The air gap flux of the machine is established by the stator supply and it remains practically

constant if the stator impedance drops and supply voltage fluctuations are neglected. The rotor

copper loss is proportional to slip. The speed control of a slip ring induction motor is achieved by

connecting external resistance in the rotor side. The main drawback of the system is that large

amount of slip power is dissipated in the resistance and this reduced the efficiency of the motor at

low speed.

This slip power can be recovered and fed back to the supply of an additional motor which is

mechanically coupled to the main motor. This type of drive is known as slip power recovery system

and improves the overall efficiency of the system.

TYPES OF SLIP POWER RECOVERY SYSTEM:

This slip power recovery system can be classified into two types of principle.

Kramer system

Scherbius system

KRAMER SYSTEM:

The Kramer system is applicable only for sub-synchronous speed operation.

It consists of main induction motor M, the speed of which is to be controlled. The two

additional equipments are, d.c. motor and a rotary converter. The slip rings of the main

motor are connected to the a.c. side of a rotary converter. The d.c. side of rotary converter

feeds a d.c. shunt motor commutator, which is directly connected to the shaft of the main

motor. A separate d.c. supply is required

28


to excite the field winding of d.c. motor and exiting winding of a rotary converter. The

variable resistance is introduced in the field circuit of d.c. motor which is acts as a field regulator.

Slip Rings.ccoverter

The speed of the set is controlled by varying the field of the d.c. motor with the rheostat R.

When the field resistance is changed, the back e.m.f. of motor changes. Thus the d.c. voltage at the

commutator changes. This changes the d.c. voltage on the d.c. side of a rotary converter. Now

rotary converter has a fixed ratio between it a.c. side and d.c. side voltages. Thus voltage on its a.c.

side also changes. This a.c. voltage is given to the slip rings of the main motor. So the voltage

injected in the rotor of main motor changes which produces the required speed control.

Advantages:

The main advantage of this method is that a smooth speed control is possible. Similarly

wide range of speed control is possible.

The main advantage of this method is that any speed, within the working range can be

obtained.

Another advantage of the system is that the design of a rotary converter is practically

independent of the speed control required.

The slip power is converted into dc by a 3-phase diode bridge rectifier. This dc power is fed

to the dc motor. This dc motor is mechanically coupled to SRIM. This slip power is

converted to mechanical power and fed back to the SRIM shaft.

---------------------------------------------------------------------------------------------------

SCHERBIUS SYSTEM:

The scherbius system is similar to Kramer system but only difference is that in the Kramer

system the feedback is mechanical and in the scherbius system the return power is electrical.

The method requires an auxiliary 3 phase or 6 phase a.c. commutator machine which

is called Scherbius machine.

29


The Scherbius machine is excited at slip frequency from the rotor of a main motor

through a regulating transformer. The taps on the regulating transformer can be varied, this changes

the voltage developed in the rotor of Scherbius machine, which is injected into the rotor of main

motor. This controls the speed of the main motor.

The scherbius machine is connected directly to the induction motor supplied from main line

so that its speed deviates from a fixed value only to the extent of the slip of the auxiliary induction

motor.

For any given setting of the regulating transformer, the speed of the main motor remains

substantially constant irrespective of the load variations.

Advantage:

This method is similar to the Kramer system.

Disadvantage:

This method can be used only for slip ring induction motor.

---------------------------------------------------------------------------------------------------------------

9) Discuss about the methods of electric braking in induction motor

Electric braking of an induction motor:

The mechanical brakes or electric brakes can be used to bring an electric motor to test

quickly.But with the mechanical brakes, smooth stop is not possible. Similarly the linings levers

and other Mechanical arrangements are necessary to apply mechanical brakes. Mechanical brakes

also,Depends on the skill of the operator. As against this,an electric braking is easy and reliable

hence, it is used to stop the induction motors very quickly.Though the motor is brought to rest

electrically, to maintain its state to rest a mechanical brake is must.

30


Types of electric braking:

Dynamic or Rheostatic braking

Plugging

DC dynamic braking

Regenerative braking

Dynamic or rheostatic braking: Nov-Dec 2017

In rheostatic braking,one supply line out of R,Y or B is disconnected from the supply.

Depending upon the condition of this disconnected line ,two types of rheostatic braking can be

achieved.

1. Two lead connections:

In this method,the disconnected line is kept open.This is shown in Fig.2.54

(a) and is called two lead connections

2.Three lead connections:

In this method, the disconnected line is connected directly to the other line of the

machine. This is shown in the Fig.2.54 (b).

In both cases, a high resistance is inserted in the rotor circuid, with the help of rheostat.

As one of the motor terminal is not connected to the supply, the motor continues to run as a

single phase motor. In this case the breakdown torque i.e. maximum torque decreases to 40% of its

original value and motor develops no starting torque at all. And due to high rotor resistance, the net

torque produced becomes negative and the braking operation is obtained.

In two lead connections, the braking torque is small while in three lead connections, the

braking torque is high at high speeds. But in three lead connections there is possibility of inequality

between the contact resistance in connections of two paralleled lines.

31


This might reduce the braking torque and even may produce the motoring torque again.

Hence inspite of low braking torque, two lead connections is preferred over three lead connections.

The torque-slip characteristics for motoring and braking operation is shown in the Fig.2.55.

Plugging:

The reversal of direction of rotation of motor is the main principle in plugging of motor. In

case of an induction motor, it can be quickly stopped by inter changing any two stator leads. Due to

this, the direction of rotating magnetic field gets reversed suddenly. This produces a torque in the

reverse direction and the motor tries to rotate in opposite direction. Effectively the brakes are

applied to the motor. Thus during the plugging, the motor acts as a brake.

One important aspect about plugging is produced of very high heat in the rotor. While

plugging, the load keeps on revolving and rotor absorbs kinetic energy from the revolving load,

causing speed to reduce. The corresponding gross mechanical power Pm is entirely dissipated as

heat in the rotor. Similarly as stator is connected to supply, rotor continues to receive power p2 from

stator which also gets dissipated as heat in the rotor. This is shown in the Fig 2.56.

The plugging should not be done frequently as due to high heat produced rotor may attain

high temperature which can melt the rotor bars and even may over the stator as well.

In some industrial applications where quick stop motor and its load is necessary, the

plugging method is used.

32


DC dynamic braking:

A quick stopping of an induction motor and its high inertia load can be achieved by

connecting stator terminals to d.c. supply. Any two stator terminals can be connected to a d.c.

supply and third terminal may be connected directly to other stator terminals. This is called d.c.

dynamic braking. If third terminal is kept open it is called two lead connections while if it is shorted

directly with other stator terminal it is called three lead connections. A diode bridge can be used to

get d.c. supply. The Fig. 2.57 shows two lead connections with diode bridge for a d.c. dynamic

braking of an induction motor.

When d.c. supplied to the stator, stationary poles N, S are produced in stator. The number of

stationary poles is P for which stator winding is wound. As rotor is rotating, rotor cuts the flux

produced by the stationary poles. Thus the a.c. voltage gets induced in the rotor. This voltage

produced an a.c. current in the rotor. The motor works as a generator and the I2

R losses are

dissipated at the expenditure of kinetic energy stored in rotating parts. Thus dynamic braking is

achieved. When all kinetic energy gets dissipated

as heat in the rotor, the induction motor comes rest.

Advantages:

The heat produced less compared to plugging.

The energy dissipated in the rotor is not dependent on the magnitude of the d.c.

current.

The braking torque is proportional to the square of the d.c. current.

The method can be used for wound rotor or squirrel cage rotor induction motors.

Regenerative braking:

The input power to a three phase induction motor is given by,

Where,

Pin = 3 VphIphcos φ

Φ = Angle between stator phase voltage and phase current.

This Φ is less than 90othe motoring action.

33


If the rotor speed is increased greater than the synchronous speed with the help of external

device, it acts as an induction generator. It converts the input mechanical energy to an electrical

energy which is given back to supply. It delivers active power to the 3 phase line. The Φ becomes

greater than 90o. The power low reverses hence rotor induced e.m.f. and rotor current also reverse.

So rotor produces torque in opposite direction to achieve the braking. As the electrical energy is

given back to the lines while braking, it called regenerative braking. The arrangement for

regenerative braking is shown in the Fig 2.58.

The torque-slip characteristic for motoring and generating action is shown in the Fig. 2.59.

The main advantage is that the generator power can be used for useful purposes. While the

dis-advantages is that for fixed frequency supply it can be used only for speeds above synchronous

speed.

--------------------------------------------------------------------------------------------------------------------------

10) Illustrate the phenomenon of cogging and crawling in induction motor. May-2015

A special behavior is shown by squirrel cage induction motor during starting for certain

combinations of number of stator and rotor slots. If number of stator slots S1 are equal to number of

rotor slots S2 or integral multiple o rotor slots S2 then variation of reluctance as a function of space

will have pronounced effect producing strong forces than the accelerating torque. Due to this motor

fails to start. This phenomenon is called cogging. Such combination of stator and rotor slots should

be avoided while designing the motor.

34


Certain combination of S1 and S2 cause accentuation of certain space harmonics of the mm wave,

e.g. fifth and seventh harmonic which correspond to poles five and seven times that of the

fundamental.

Since the space-phase difference between fundamental poles of the winding phase is (0o,

120o,

240o), this (space-phase) difference is (0

o,120

o,240

o) for the fifth harmonic poles and

(0o,120

o,240

o) for the seventh. Hence the fifth harmonic poles rotate backward with synchronous

speed is= ns/5 and the seventh the seventh harmonic poles rotate forward at ns/7. These harmonic

mmfs produce their own asynchronous(induction) torque of the same general torque-slip shape as

that of the fundamental. Figure 9.36 shows the superimposition of the fundamental, fifth and

seventh harmonic torque-slip curve.

A marked saddle effect is observed with stable region of operation(negative torque-slip

slope) around 1/7th normal motor speed(s = 6/7). In Fig. 9.36. the load torque curve intersects the

motor torque curve at the point M resulting in stable operation. This phenomenon is known as

crawling(running stably at low speed)

Certain slot combinations, e.g. S1 = 24 and S2 = 18 cause the stator mmf to possess a

reversed 11th

and a forward 13th harmonic mmf while the rotor has a reversed 13

th and a forward

15th

. The stator 13th harmonic mmf rotates at speed +ns/13 with stator and the rotarmmf of the 13

th

harmonic rotates at –(ns—n)/13 with respect to the rotor when the rotor is running at speed n. These

two mmf’s lock into each other to produce a synchronous torque when

ns /13 = n-(ns-n)/(13)

n=ns/7

or

Thus there is a discontinuity at ns/7 in the torque-slip characteristic produced by not the

seventh but the 13th harmonic as shown in Fig. 9.37.

The stator slot harmonic of order 2S1/P ±1 may interact with rotor slot harmonic of order

2S2/P ±1

To develop the harmonic synchronous torque.

2S1/P ±1 = 2S2/P ±1

35


S1= S2

And

2S1/P -1 = 2S2/P +1

S1- S2 = P

It can be thus seen that if S1= S2 or S1- S2 = P then cogging will be definitely observed in

the induction motor.

The cogging and crawling is not predominant in slip ring induction motor as these motors

are started with higher starting torques with external resistance in rotor circuit.

The crawling effect can be reduced by taking proper care during the design .Still if crawling

is observed then it can be overcome by applying a sudden external torque to the driven load in the

direction of rotor. If there is reduction is supply voltage then torque also decreases (T∞V12). Hence

asynchronous crawling may be observed which is absent under rated voltage conditions. Thus the

asynchronous torque cannot be avoided but can be reduced by proper choice of coil span and by

skewing the stator or rotor slots.

The synchronous harmonic torque can be totally eliminated by proper combination of stator

and rotor slots.

---------------------------------------------------------------------------------------------------------

11) Determine approximately the starting torque of an induction motor in terms of full load

torque when started by Auto –starter with 50% tapings.The short circuit current of the motor

at normal voltage is 5 times the full load current and the full load slip is 4 %. May-2012

Isc=5 IF.L and Sf =4%=0.04

i) Star – delta starter

𝑇𝑠𝑡

𝑇𝐹 .𝐿=

1

3 𝐼𝑠𝑐

𝐼𝐹 .𝐿

2

× 𝑠𝑓 = 1

3 5 2 × 0.04

Tst = 33.33% of TFL

ii) Autotransformer with 50% tapping

K = 0.5, K2 = 0.25

𝑇𝑠𝑡

𝑇𝐹 .𝐿= 𝑘2

𝐼𝑠𝑐

𝐼𝐹 .𝐿

2

× 𝑠𝑓 = 0.25 × 25 × 0.04

Tst = 25% of TFL

-----------------------------------------------------------------------------------------------------------------------------

12) A three phase induction motor takes a starting current which is 5 times full load current at

normal voltage.Its full load slip is 4 percent.What auto-transformer ratio would enable the

motor to be started with not more than twice the full load current drawn from the supply?

What would be the starting torque under this conditions? May-2014

Solution: Starting current at rated voltage=Isc

Isc=5 IF.L and Sf =4%=0.04

Let x=Tapping on auto transformer

TF.L=Tst

36


𝑇𝑠𝑡𝑇𝐹.𝐿

= 𝑋2 𝐼𝑠𝑐𝐼𝐹.𝐿

2

× 𝑠𝑓

1=𝑋2 5

1

2

× 0.04

x=1

Thus 100% tapping is required

Now 𝐼𝑠𝑡 𝑠𝑢𝑝𝑝𝑙𝑦 = 𝑥𝐼𝑠𝑡 𝑚𝑜𝑡𝑜𝑟 = 𝑥 𝑥𝐼𝑠𝑐 =𝑥2𝐼𝑠𝑐

=𝑥2 × 5𝐼𝐹.𝐿 = 5𝐼𝐹.𝐿

Thus supply starting current is 5 times the full load current.

------------------------------------------------------------------------------------------------------------------------

13) A 3-phase 440 V distribution circuit is designed to supply not more than 1200 A. Assuming

that a 3-phase squirrel cage induction motor has full load efficiency of 0.85 and a full load

power factor of 0.8 and that the starting current at rated voltage is 5 times the rated full load

current. What is the maximum permissible kW rating of the motor if it is to be started using

an auto-transformer stepping down the voltage to 80%. Dec-2014,May 17

Solution

Maximum permissible line current that the 3 phase induction motor can take from

the distribution circuit is 1200A at the time of starting. It is given that the starting current at rated

voltage is 5 times the rated current of the induction motor.

Therefore the rated line current of 3 phase induction motor with full voltage starting is

1200/5 = 240A.

Thus the maximum permissible induction motor rating when started at full voltage

= 3 V1 I1 cos Ɵ1× efficiency

= 3 × 440 × 240 × 0.8 × 0.85 = 124.371 𝑘𝑤

Maximum permissible starting current from the supply mains,

Ist = 1200 = 𝑥2Isc= 𝑥2(5IFL)

1200 = (0.8)2 (5IFL)

IFL = 1200

0.82 × 5 = 375 A

Maximum permissible induction motor rating

= 3 × 440 × 375 × 0.8 × 0.85 = 194.33 𝑘𝑤

----------------------------------------------------------------------------------------------------------------------------

14) A small squirrel cage induction motor has a starting current of six times the full load current

and a full load slip of 0.05. Find in p.u of full load values, the current (line) and starting

torque with the following methods of starting.

(a) direct switching

(b) stator resistance starting with motor current limited to 2 pu

(c) auto transformer starting with motor current limited to 2 pu

(d) star delta starting May-2015, 2016

(e) What auto transformer ratio would give 1 pu starting torque?

37


(a) Direct switching

Is = 6 pu

Ts = 62 × 0.05 = 1.8

(b) Stator resistance starting

Is = 2 pu (limited to)

Ts = 22 × 0.05 = 0.2 pu

(c) Autotransformer starting

X = 2

6 =

1

3

Is(motor) = 2 pu

Is (line) = 1

3 × 2 pu = 0.67 pu

Ts = 22 × 0.05 = 0.2 pu

(d) Star - delta starting

Is = 1

3 × 6 pu = 2 pu Is = x Isc

Ts = 1

3 × 62

× 0.05 = 0.6 pu

(e) Autotransformer starting

Ts = 𝑥2 × 62 × 0.05 = 1.0 pu

x = 0.745 ( ≅75% tap)

-------------------------------------------------------------------------------------------------------------------------

15) The impedances at standstill of the inner and outer cages of a double cage rotor are (0.01+j

0.5) and (0.05+j 0.1) respectively. The stator impedance may be assumed to be negligible.

Determine the ratio of the torques due to the two cages (1) at starting and (2) when running

with a slip of 5%. May-2015

Ts = 3

𝑊𝑠

𝑉2

2𝑅22

𝑅22 + 𝑥22

V2 - rotor induced emf

Substituting values,

Tso = 3

𝑊𝑠

𝑉22 (0.05)

(0.05)2+ (0.1)2

Tst = 3

𝑊𝑠

𝑉22 (0.01)

(0.01)2+ (0.5)2

𝑇𝑠0

𝑇𝑠𝑡=

(0.01)2+(0.5)2

(0.05)2+(0.1)2 ×0.05

0.01 = 100

T = 3

𝑊𝑠

𝑉2

2𝑅2/ 𝑆

( 𝑅2/ 𝑆 )2+ 𝑥22

Substituting values,

To = 3

𝑊𝑠

𝑉22(0.05/0.05)

(0.05/0.05)2+ (0.1)2

38


Ti = 3

𝑊𝑠

𝑉22 (0.01/0.05)

(0.01/0.05)2+ (0.3)2

𝑇0

𝑇𝑖=

(0.01 / 0.05)2+(0.5)2

(0.05 / 0.05)2+(0.1)2 ×0.05

0.01 = 1.436

----------------------------------------------------------------------------------------------------------------------------

1

Dr. A.Jeraldine Viji & M. Vijayaragavan / EEE / MEC Unit-05

UNIT - V

Single Phase Induction Motors and Special Machines

PART-A

1. Why capacitor –start induction motors advantageous? May-2012

In capacitor start induction motors capacitor is connected in series with the auxiliary winding.

When speed of the motor approaches to 75 to80%of the synchronous speed the starting winding

gets disconnected due to the operation of the centrifugal switch.The capacitor remains in the circuit

only at start.The starting torque is proportional to phase angle α and hence such motors produce

very high starting torque.

2. List out 4 applications of shaded pole induction motor.Nov-Dec 2016.

Shaded pole motors have very low starting torque, low power factor and low efficiency. The

motors are commonly used for small fans, toy motors, advertising displays, film projectors,

record players, gramophones,hair dryers, photocopying machines etc.

3. What are the drawbacks of the presence of the backward rotating field in a single phase

induction motor?

Net flux will be zero

No starting in the motor

4. Is single phase induction motor self starting? Why? Dec-2006, 2015, May 2016,2017

Due to cutting of flux, emf gets induced in the rotor which circulates rotor current.the rotor

current produces rotor flux. This flux interacts with forward component ɸf to produce a torque in

one particular direction say anticlockwise direction. While rotor flux interacts with backward

component ɸb to produce a torque in the clockwise direction. So if anti clock wise torque is

positive then clockwise torque is negative thus net torque experienced by the rotor is zero at

start. Hence net torque experienced by rotor is zero at start and so single phase induction motors

are not self starting.

5. Why is hysteresis motor free from mechanical and magnetic vibrations?

The stator of hysteresis motor carries main and auxiliary windings to produce rotating

magnetic field or of shaded pole type also. The rotor is smooth cylindrical type made up of hard

magnetic material.The torque in this motor is constant at all speeds it runs at synchronous speed

there is no relative motion between stator and rotor field so the torque due to eddy current

vanishes. Only hysteresis torque is present which keeps rotor running at synchronous speeds. The

high retentivity ensures continuous magnetic locking between stator and rotor. Hence it is

free from magnetic vibrations.

6. In which direction a shaded pole motor runs? May-2013

The rotor starts rotation in the direction from unshaded part to the shaded part.

2


7. What types of motor is used in computer drives and wet grinders? May-2008

For computer drives permanent magnet dc motors are used while in wet grinders universal

motor may be used.

8. what is the principle of operation of a linear induction motor ?and write List some

applications of linear induction motor

.Linear induction motor work at magnetic levitation principle. They are used in machine tool

industry and in robotics.They are used in trains operated on Magnetic levitation, reciprocating

compressors can also be driven by linear motors.

9. What are the specific characteristic features of the repulsion motor?

Repulsion motors give excellent performance characteristics. A very high starting torque of

about 300 to350% of full load can be obtained with starting currents of about 3 to 4 times the full

load current. Thus it has got very good operating characteristic. The speed of the motor changes

with load. With compensated type of repulsion motor the motor runs with improved power factor

as the quadrature drop in the field winding is neutralised. Also the leakage between armature and

field is reduced which gives better regulation.

10. Discuss characteristics of single phase series motor? Dec-2013

a. To reduce the eddy current losses,yoke and pole core construction is laminated.

b. The power factor can be improved by reducing the number of turns. But this reduces the

field flux. But this reduction in flux increases the speed and reducing the torque. To keep

the torque same it is necessary to increase the armature turns proportionately. This

increases the armature inductance.

11. What are the demerits of repulsion motor?

very expensive

speed changes with load

on no load speed is very high causing sparking at brushes

low power factor on no load

12. List four applications of reluctance motors.

This motor is used in signalingdevices, control apparatus,automatic regulators, recording

instruments, clocks and all kinds of timing devices, tele-printers,gramophones.

13. What is a universal motor? Dec-2009

There are small capacity series motors which can be operated on dc supply or single phase

ac supply of same voltage with similar characteristics called universal motors. The construction of

this motor is similar to that of ac series motor.

14. Name the two winding ofsingle phase induction motor?

Running and starting winding.

3


15. What are methods available for making 1ɸ induction motor a self starting? Dec 2015

By splitting the single phase, by providing shading coil in the poles.

16. What is the function of capacitor in single phase induction motor?

To make phase difference between starting and running winding, to improve PF and to getmore

torque.

17. State any 4 use of single phase induction motor.

Fans, wet grinders, vacuum cleaner, small pumps, compressors, drills.

18. What kind of motors is used in ceiling fan and wet grinders? May-2014, 2016

Ceiling fan- Capacitor start and capacitor run single phase induction motor,

Wet grinders-Capacitor start capacitor run single phase induction motor.

19. What is the application of shaded pole induction motor?

Because of its small starting torque, it is generally used for small toys, instruments, hair driers,

ventilatorsetc.

20. Define the term step angle in a stepper motor? Nov-Dec 2017

𝑁𝑟 = 𝑁𝑠 ± 𝑁𝑠𝑞

𝑁𝑟 ,𝑁𝑠 −𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑆𝑡𝑎𝑡𝑜𝑟 𝑝ℎ𝑎𝑠𝑒𝑠,𝑅𝑜𝑡𝑜𝑟 𝑝ℎ𝑎𝑠𝑒𝑠.

𝑞 − 𝑁𝑜. 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒𝑠.

21. Differentiate between “capacitor start” & “Capacitor start capacitor run” single phase

induction motor? May-2010

Capacitor start – capacitor is connected series with starting winding, but it will be disconnected

from supply when motor pick up its speed.

Capacitor start capacitor run-starting winding and capacitor will not be disconnected from

supply even though motor pickup its speed.

22. State the double revolving field theory.Dec-2013,May 2017.

According to double revolving field theory, an alternating stator flux can be resolved into two

rotating components which rotate in opposite directions i.e the forward component rotating in

anticlockwise while is the backward component rotating in clockwise directions. Both components

produce rotor flux. The rotor flux interacts with to produce torque in anti-clockwise direction while

it reacts with to produce torque in clock wise directions. At start these two torques are equal in

magnitude but opposite in direction. Thus net torque experienced by the rotor is zero at start and

hence the single phase induction motors are not self starting.

23. Distinguish the terms rotating and pulsating magnetic fields. May-2015

Rotating magnetic fields:

The field or flux having constant amplitude but whose axis is continuously rotating

in a plane with a certain speed.

4


Pulsating magnetic fields

The m.m.f. wave produced by the single phase a.c. winding is pulsating, whose

amplitude varies sinusoidally with time.

24. Draw the torque-slip characteristics of single phase induction motors.May-2013

25. Name the method of starting single phase inductions motors. Dec-2012 Or What are the

various methods available for making a single phase motor self starting? APR-MAY 2018

Spilt phase induction motors

Capacitor start induction motor

Capacitor start capacitors run induction motor

Shaded pole induction motor

26. What is mean by single phasing? Dec-2012

If the motor runs two phases instead of three phases is called single phasing

27. Why are centrifugal switches provided in many 1-phases induction motors? May-2012

Centrifugal switch is the mechanical switch.If the motor reach 75% of the rated speed this

mechanical switch disconnect the auxiliary winding.For this purpose centrifugal switches provided

in many 1-phase induction motors.

28. How can the direction of a capacitor run motor be reversed? May-2014, 2016, Dec- 2015

By interchanging the connections of main winding or auxiliary winding.

29. How is the direction of rotation of a single phase induction motor reversed? Dec-2014,Nov-

Dec 2017

The direction of rotation of a single phase induction motor is reversed by reversing the

leads to the main or auxillary winding, but not both.

30. What is the principle of reluctance motor?Dec-2014 , APR-MAY 2018

Whenever a piece of ferromagnetic material is located in a magnetic field, a force is

exerted upon the material, tending to ring it into the position of the densest portion of the field. The

force tends to align the specimen of material so that the reluctance of the magnetic path passing

through the material will be at a minimum.

5


31. State the limitations of shaded pole motors. May-2015

Starting torque is poor

Power factor is very low.

Due to copper losses in the shading ring the efficiency is very low.

Speed reversal is very difficult

Size and power rating is very small

PART-B

1) Write short notes on single phase induction motor.

Single-Phase Motors

Introduction

Single phase motors are the most familiar of all electric motors because they are

extensively used in home appliances, shops, offices etc.It is true that single phase motors are less

efficient substitute for 3-phase motors but 3-phase power is normally not available except in large

commercial and industrial establishments.

Since electric power was originally generated and distributed for lighting only, millions of

homes were given single-phase supply. This led to the development of single-phase motors. Even

where 3-phase mains are present, the single-phase supply may be obtained by using one of the

three lines and the neutral.

Types of Single-Phase Motors

Single-phase motors are generally built in the fractional-horsepower range and may be

classified into the following four basic types:

1. Single-phase induction motors

split-phase type

capacitor type

shaded-pole type

2. A.C. series motor or universal motor

3. Repulsion motors

Repulsion-start induction-run motor

Repulsion-induction motor

4. Synchronous motors

Reluctance motor

Hysteresis motor

Single-Phase Induction Motors

6


A single phase induction motor is very similar to a 3-phase squirrel cage induction motor. It has

(i) a squirrel-cage rotor identical to a 3-phase motor and

(ii) a single-phase winding on the stator.

Unlike a 3-phase induction motor, a single-phase induction motor is not self starting but

requires some starting means. The single-phase stator winding produces a magnetic field that

pulsates in strength in a sinusoidal manner.

The field polarity reverses after each half cycle but the field does not rotate. Consequently,

the alternating flux cannot produce rotation in a stationary squirrel-cage rotor. However, if the

rotor of a single-phase motor is rotated in one direction by some mechanical means, it will

continue to run in the direction of rotation.

As a matter of fact, the rotor quickly accelerates until it reaches a speed slightly below the

synchronous speed. Once the motor is running at this speed, it will continue to rotate even though

single-phase current is flowing through the stator winding. This method of starting is generally not

convenient for large motors. Nor can it be employed fur a motor located at some inaccessible spot.

Single-phase Induction Motor

The above figure shows single-phase induction motor having a squirrel cage rotor and a

single phase distributed stator winding.

Such a motor inherently docs not develop any starting torque and, therefore, will not start

to rotate if the stator winding is connected to single-phase A.C. supply.

However, if the rotor is started by auxiliary means, the motor will quickly attain me final

speed. This strange behavior of single-phase induction motor can be explained on the basis of

double-field revolving theory.

--------------------------------------------------------------------------------------------------------- 2) Explain why single phase induction motor is not self starting, with the help of double field

revolving theory. May-2014, 2013, 2012, 2009, Dec-2012, 2015,Nov-Dec 2016,2017

Or illustrate the operation of single phase induction motor with double field revolving

theory. May-2015, 2017

Double-Field Revolving Theory

The double-field revolving theory is proposed to explain this dilemma of no torque at start

and yet torque once rotated. This theory is based on the fact that an alternating sinusoidal flux (ɸ =

ɸm cos ωt) can be represented by two revolving fluxes, each equal to one-half of the maximum

7


value of alternating flux (i.e., ɸm/2) and each rotating at synchronous speed (Ns = 120f/P, w = 2f )

in opposite directions.

The above statement will now be proved. The instantaneous value of flux due to the stator

current of a single-phase induction motor is given by;ɸr = ɸm cos ωt

From the above figure, consider two rotating magnetic fluxes ɸ1 and ɸ2 each of magnitude

ɸm/2 and rotating in opposite directions with angular velocity .Let the two fluxes start rotating from

OX axis at t = 0. After time t seconds, the angle through which the flux vectors have rotated is at.

Resolving the flux vectors along-X-axis and Y-axis,

We have,

Total X-component = ɸ𝑚

2 cos wt +

ɸ𝑚

2 cos wt = ɸm cos wt

Total Y-component = ɸ𝑚

2 cos wt -

ɸ𝑚

2 cos wt = 0

Resultant flux = (ɸ𝑚 cos𝑤𝑡) 2 + 0 = ɸm cos wt

Thus the resultant flux vector is ɸ= ɸm cos wt along X-axis. Therefore, an alternating field

can be replaced by two relating fields of half its amplitude rotating in opposite directions at

synchronous speed.

Note that the resultant vector of two revolving flux vectors is a stationary vector that

oscillates in length with time along X-axis. When the rotating flux vectors are in phase [See below

fig. (i)], the resultant vector is ɸr= ɸm; when out of phase by 180° [See below fig.(ii) ], the

resultant vector ɸr= 0

Let us explain the operation of single-phase induction motor by double-field revolving theory.

(i) Rotor at standstill

Consider the case that the rotor is stationary and the stator winding is connected to a single-

phase supply. The alternating flux produced by the stator winding can be presented as the sum of

8


two rotating fluxes ɸ1 and ɸ2, each equal to one half of the maximum value of alternating flux and

each rotating at synchronous speed (Ns = 120 f/P) in opposite directions as shown in the below fig.

(i).

Let the flux ɸ1 rotate in anti clockwise direction and flux ɸ2 in clockwise direction. The

flux ɸ1 will result in the production of torque T1 in the anti clockwise direction and flux ɸ2 will

result in the production of torque T2 In the clockwise direction.

At standstill, these two torques are equal and opposite and the net torque developed is zero.

Therefore, single-phase induction motor is not self-starting. This fact is illustrated in the below fig.

(ii).

Note that each rotating field tends to drive the rotor in the direction in which the field

rotates. Thus the point of zero slip for one field corresponds to 200% slip for the other as explained

later. The value of 100% slip (standstill condition) is the same for both the fields.

(ii) Rotor running

Now assume that the rotor is started by spinning the rotor or by using auxiliary circuit, in

say clockwise direction. The flux rotating in the clockwise direction is the forward rotating flux

(ɸf) and that in the other direction is the backward rotating flux (ɸb). The slip w.r.t. the forward

flux will be

Sf = 𝑁𝑠−𝑁𝑁𝑠

= S

Where, Ns = synchronous speed

N = speed of rotor in the direction of forward flux

The rotor rotates opposite to the rotation of the backward flux. Therefore, the slip w.r.t. the

backward flux will be

Sb=𝑁𝑠−(−𝑁)

𝑁𝑠 =

𝑁𝑠+𝑁

𝑁𝑠 =

2𝑁𝑠−𝑁𝑠+𝑁

𝑁𝑠

= 2𝑁𝑠

𝑁𝑠 - 𝑁𝑠−𝑁

𝑁𝑠= 2- S

Sb = 2- S

Thus for forward rotating flux, slip is s (less than unity) and for backward rotating flux, the

slip is 2 -s (greater than unity). Since for usual rotor resistance/reactance ratios, the torques at slips

9


of less than unity arc greater than those at slips of more than unity, the resultant torque will be in

the direction of the rotation of the forward flux.

Thus if the motor is once started, it will develop net torque in the direction in which it has

been started and will function as a motor. Fig. (a) shows the rotor circuits for the forward and

backward rotating fluxes.

Note that r2 = R2/2, where R2 is the standstill rotor resistance i.e., r2 is equal to half the

standstill rotor resistance. Similarly, x2 = X2/2 where X2 is the standstill rotor reactance.

At standstill, s = 1 so that impedances of the two circuits are equal. Therefore, rotor

currents are equal i.e., I2f = I2b. However, when the rotor rotates, the impedances of the two rotor

circuits are unequal and the rotor current I2b is higher (and also at a lower power factor) than the

rotor current I2f. Their m.m.f.s, which opposes the stator m.m.f.s, will result in a reduction of the

backward rotating flux.

Consequently, as speed increases, the forward flux increases, increasing the driving torque

while the backward flux decreases, reducing the opposing torque. The motor-quickly accelerates to

the final speed.

Figure (a)

---------------------------------------------------------------------------------------------------------

3) Draw and explain the equivalent circuit of single phase induction motor. May 2009, 2014,

Or Derive the equivalent circuit of a single phase IM with the help of double field revolving

theory. Dec-2014, 2015

Equivalent Circuit of Single-Phase Induction Motor

When the stator of a single-phase induction motor is connected to single-phase supply, the

stator current produces a pulsating flux that is equivalent to two-constant-amplitude fluxes

revolving in opposite directions at the synchronous speed (double-field revolving theory).

Each of these fluxes induces currents in the rotor circuit and produces induction motor

action similar to that in a 3-phase induction motor Therefore, a single-phase induction motor can to

imagined to be consisting of two rotors, having a common stator winding but with their respective

rotors revolving in opposite directions. Each rotor has resistance and reactance half the actual rotor

values.

Let R1 = resistance of stator winding

X1 = leakage reactance of stator winding

10


Xm = total magnetizing reactance

R'2 = resistance of the rotor referred to the stator

X'2 = leakage reactance of the rotor referred to the stator revolving theory.

(i) At standstill. At standstill, the motor is simply a transformer with its secondary short-circuited.

Therefore, the equivalent circuit of single-phase motor at standstill will be as shown in the figure

below. The double-field revolving theory suggests that characteristics associated with each

revolving field will be just one-half of the characteristics associated with the actual total flux.

Therefore, each rotor has resistance and reactance equal to R'2/2 and X'2/2 respectively.

Each rotor is associated with half the total magnetizing reactance. Note that in the equivalent

circuit, the core loss has been neglected. However, core loss can be represented by an equivalent

resistance in parallel with the magnetizing reactance.

Now Ef = 4.44 f N ɸf; Eb = 4.44 f N ɸb

At standstill, ɸf = ɸb. Therefore, Ef = Eb

V1 = Ef + Eb = I1Zf + I1Zb

Where, Zf = impedance of forward parallel branch

Zb = impedance of backward parallel branch

(ii) Rotor running. Now consider that the motor is running at some speed in the direction

of the forward revolving field, the slip being s. The rotor current produced by the forward field will

have a frequency sf where f is the stator frequency. Also, the rotor current produced by the

backward field will have a frequency of (2 - s)f.

Figure shows the equivalent circuit of a single-phase induction motor when the rotor is

rotating at slip s. It is clear, from the equivalent circuit that under running conditions, Ef becomes

much greater than Eb because the term R'2/2s increases very much as tends towards zero.

Conversely, E^ falls because the term R'2/2(2-s) decreases since (2-s) tends toward 2.

Consequently, the forward field increases, increasing the driving torque while the backward field

decreases reducing the opposing torque.

11


Total impedance of the circuit is given by;

--------------------------------------------------------------------------------------------------------- 4) Explain the no-load test and blocked rotor test for obtaining the equivalent circuit

parameters of a single phase IM. Dec-2014, 2015, NOV-DEC 2017

No-Load test:

The test is conducted by rotating the motor without the load.The input current,voltage and

power are measured by connecting the ammeter , voltmeter and wattmeter in the circuit.These

readings are denoted as V0,I0and W0.

W0 = V0 I0 cos ɸ0

Cos ɸ0=𝑊0

𝑉0 𝐼0= No load power factor

.Now,

The motor speed on no load is almost equal to its synchronous speed hence for practical

purposes, the slip can be assumed zero. Hence r2/sbecomes ∞ and acts as open circuit in the

equivalent circuit. Hence for forward rotor circuit, the branch r2/s + jx2 gets eliminated.

While for a backward rotor circuit, the term r2/(2-s) tends to r2/2. Thus xo is much higher than

the impedance r2/2+jx2 Hence it can be assumed that no current flow through xm and that branch

can be eliminated. So circuit reduces to as shown in the Fig . 8.11.1.

12


Blocked rotor test:

In blocked rotor test, the rotor is held fixed so that it will not rotate. A reduced voltage is

applied to limit the short circuit current. This voltage is adjusted with the help of autotransformer

so that the rated current flows through main winding. The input voltage, current and power are

measured by connecting the voltmeter, ammeter and wattmeter respectively.These readings are

denoted as VSC, ISC and WSC.

Now as rotor is blocked, the slip s = 1.Hence the magnetizing reactance X0 is much higher

than the rotor impedance and hence it can be neglected as connected in parallel

13


The stator resistance R1 is measured by voltmeter – ammeter method, by disconnecting the

auxiliary winding and capacitors present if any. Due to skin effect, the a.c. resistance is 1.2 to 1.5

times more than d.c. resistance. Thus with these two tests, all the parameters of single phase

induction motor can be obtained.

----------------------------------------------------------------------------------------------------------------------------- -

5) Explain how to start the single phase induction motor. Dec-2013, 2007

Making Single-Phase Induction Motor Self-Starting

The single-phase induction motor is not self starting and

it is undesirable to resort to mechanical spinning of the shaft or

pulling a belt to start it.

To make a single-phase induction motor self-starting, we

should somehow produce a revolving stator magnetic field. This

may be achieved by converting a single-phase supply into two-

phase supply through the use of an additional winding.

When the motor attains sufficient speed, the starting

means (i.e., additional winding) may be removed depending

upon the type of the motor. As a matter of fact, single-phase

induction motors are classified and named according to the

method employed to make them self-starting.

(i) Split-phase motors-started by two phase motor action through the use of an auxiliary or

starting winding.

(ii) Capacitor motors-started by two-phase motor action through the use of an auxiliary

winding and a capacitor.

(iii) Shaded-pole motors-started by the motion of the magnetic field produced by means of a

shading coil around a portion of the pole structure.

------------------------------------------------------------------------------------------------------

6) Describe the construction and principle of operation of split phase induction motor.

Split-Phase Induction Motor:

The stator of a split-phase induction motor is provided with an auxiliary or starting winding

S in addition to the main or running winding M. The starting winding is located 90° electrical from

the main winding [from the below fig (i)] and operates only during the brief period when the motor

starts up.

14


The two windings are so resigned that the starting winding S has a high resistance and

relatively small reactance while the main winding M has relatively low resistance and large

reactance as shown in the schematic connections [from the below fig (ii)].

Consequently, the currents flowing in the two windings have reasonable phase difference c

(25° to 30°) as shown in the phasor diagram [from the below fig (iii)]

Split-phase induction motor

Operation

(i) When the two stator windings are energized from a single-phase supply, the main winding

carries current Im while the starting winding carries current Is.

(ii) Since main winding is made highly inductive while the starting winding highly resistive, the

currents Im and Is have a reasonable phase angle a (25° to 30°) between them as shown in the

above figure(iii). Consequently, a weak revolving field approximating to that of a 2-phase

machine is produced which starts the motor. The starting torque is given by;

Ts= kIm Is sin θ

wherek is a constant whose magnitude depends upon the design of the motor

(iii) When the motor reaches about 75% of synchronous speed, the centrifugal switch opens the

circuit of the starting winding. The motor then operates as a single-phase induction motor and

continues to accelerate till it reaches the normal speed. The normal speed of the motor is below

the synchronous speed and depends upon the load on the motor.

Characteristics

a) The sinning torque is 15 to 2 times the full-loud torque mid (lie starting current is 6 to 8

timesthe full-load current.

b) Due to their low cost, split-phase IM are most popular single phase motors in themarket.

c) Since the starting winding is made of fine wire, the current density is high and the winding

heats up quickly. If the starting period exceeds 5 seconds, the winding may burn out unless

the motor is protected by built-in-thermal relay. This motor is, therefore, suitable where

starting periods are not frequent.

d) An important characteristic of these motors is that they are essentially constant-speed

motors. The speed variation is 2-5% from no-load to full load.

e) These motors are suitable where a moderate starting torque is required and where starting

periods are infrequent e.g., to drive:

Fans

Washing machines

15


Oil burners

Small machine tools etc.

The power rating of such motors generally lies between 60 W and 250 W.

-------------------------------------------------------------------------------------------------------------------------

7) Describe the construction and principle of operation of capacitor start and run single phase

induction motor. May- 2016, 2013, 2009, Dec-2013,APR-MAY 2108

Capacitor-Start Motor

The capacitor-start motor is identical to a split-phase motor except that the starting winding

has as many turns as the main winding. Moreover, a capacitor C is connected in series with the

starting winding as shown in Fig.(i)

The value of capacitor is so chosen that Is leads Im by about 80° (i.e., ~ 80°) which is

considerably greater than 25° found in split-phase motor [Fig.(ii)]. Consequently, starting

torque(Ts = k Im Is Sin θ) is much more than that of a split-phase motor Again, the starting

winding is opened by the centrifugal switch when the motor attains about 75% of synchronous

speed. The motor then operates as a single-phase induction motor and continues to accelerate till it

reaches the normal speed.

Characteristics

a) Although starting characteristics of a capacitor-start motor are better than those of a split-

phase motor, both machines possess the same running characteristics because the main

windings are identical.

Capacitor-Start Motor

b) The phase angle between the two currents is about 80° compared to about 25° in a split-

phase motor. Consequently, for the same starting torque, the current in the starting winding

is only about half that in a split-phase motor. Therefore, the starting winding of a capacitor

start motor heats up less quickly and is well suited to applications involving either frequent

or prolonged starting periods.

c) Capacitor-start motors are used where high starting torque is required and where the

starting period may be long e.g., to drive:

16


compressors

large fans

pumps

high inertia loads

The power rating of such motors lies between 120 W and 7-5 kW.

Capacitor-Start Capacitor-Run Motor

This motor is identical to a capacitor-start motor except that starting winding is not opened

after starting so that both the windings remain connected to the supply when running as well as at

starting. Two designs are generally used.

(i) In one design, a single capacitor C is used for both starting and running as shown in the below

fig (i). This design eliminates the need of a centrifugal switch and at the same time improves

the power factor and efficiency of the motor.

(ii) In the other design, two capacitors C1 and C2 are used in the starting winding as shown in the

below fig. (ii). the smaller capacitor C1 required for optimum running conditions is

permanently connected in series with the starting winding. The much larger capacitor C2 is

connected in parallel with C1 for optimum starting and remains in the circuit during starting.

The starting capacitor C1 is disconnected when the motor approaches about 75% of

synchronous speed. The motor then runs as a single-phase induction motor.

Characteristics

(i) The starting winding and the capacitor can be designed for perfect 2-phase operation at

any load. The motor then produces a constant torque and not a pulsating torque as in other single-

phase motors.

Capacitor-Start Capacitor-Run Motor

(ii) Because of constant torque, the motor is vibration free and can be used in:

hospitals

studios and

Other places where silence is important.

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8) Explain the operating principle of shaded pole induction motor with neat diagram. Dec -

2012, 2008, May-2012,Nov-Dec 2016, NOV-DEC 2017

17


Shaded-Pole Motor

The shaded-pole motor is very popular for ratings below 0.05 H.P. (~ 40 W) because of its

extremely simple construction. It has salient poles on the stator excited by single-phase supply and

a squirrel cage rotor as shown in the below figure. A portion of each pole is surrounded by a short-

circuited turn of copper strip called shading coil.

Shaded-Pole Motor

Operation

The operation of the motor can be understood by referring to Fig. Which shows one pole of

the motor with a shading coil.

During the portion OA of the alternating-current cycle [fig (i)], the flux begins to increase and an

e.m.f. is induced in the shading coil. The resulting current in the shading coil will be in such a

direction (Lenz’s law) so as to oppose the change in flux. Thus the flux in the shaded portion of the

pole is weakened while that in the unshaded portion is strengthened as shown in the below fig. (ii).

During the portion AB of the alternating-current cycle, the flux has reached almost maximum

value and is not changing. Consequently, the flux distribution across the pole is uniform (fig. (iii))

since no current is flowing in the shading coil.

As the flux decreases (portion BC of the alternating current cycle), current is induced in the

shading coil so as to oppose the decrease in current. Thus the flux in the shaded portion of the pole

is strengthened while that in the unshaded portion is weakened as shown in the below fig. (iv)

The effect of the shading coil is to cause the field flux to shift across the pole face from the

unshaded to the shaded portion. This shifting flux is like a rotating weak field moving in the

direction from unshaded portion to the shaded portion of the pole.

The rotor is of the squirrel-cage type and is under the influence of this moving field. Consequently,

a small starting torque is developed. As soon as this torque starts to revolve the rotor, additional

torque is produced by single-phase induction-motor action. The motor accelerates to a speed

slightly below the synchronous speed and runs as a single-phase induction motor.

18


Characteristics

The salient features of this motor are extremely simple construction and absence of centrifugal

switch.

Since starting torque, efficiency and power factor are very low, these motors are only suitable for

low power applications e.g., to drive.(The power rating of such motors is upto about 30 W.)

Small fans

Toys

Hair driers and Desk fans etc.

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9) With neat diagram explain the construction and working principle of linear induction motor.

Dec – 2015, May - 2016

The linear induction motor works on the same principle as that of normal induction motor

with the difference that instead of rotational movement, the rotor moves linearly. If the stator and

rotor of induction motor are made flat then it forms the linear induction motor. The flux produced

by the flat stator moves linearly with the synchronous speed is given by,

Vs = 2wf

Where, Vs = Linear synchronous speed (m/s)

W = Width of one pole pitch (m)

f = Frequency of supply (Hz)

It can be that the synchronous speed is independent of number of poles but depend on only

width of pole pitch and supply frequency. The schematic if linear induction motor is shown in the

fig.9.6.1.

The flux moves linearly and forces the rotor to move in straight line in the same

directions.In many of the practical applications the rotor plate is a stationry member whereas the

stator moves.The analysis of linear machines is nearly same as that of rotating machines. All the

angular dimensions and displacements are displaced by linear once and torque is replaced by the

force. The expressions for the machine parameters are divided analogously and the results are in

the similar in form.Some of the typical results are as given below,

19


The linear induction motors are widely used in the transportation fields i.e. electric

trains.The stator is mounted on moving vehicle and a conducting stationary rotor forming the rails.

The induced current in the rail not only forced the stator but also provide the magnetic levitation in

which the train floats in the air above the track. This mechanism proves better for high speed

transportation without the difficulties associated with the wheel-rail interactions present in

conventional rail transport. Thus the trains may have speed of about 300km/hr. A powerful

electromagnet fixed the currents in the rail which provides levitation so that the train is pushed up

above the track in the air. The operations of such the system is automatic and system is reliable and

safe.

Linear motors also find the applications in the machine tool industry and in robotics where

linear motion is requires for positioning and for operation of the manipulators. In addition to this,

the reciprocating compressors are also driven by the linear machines.

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10) Describe the constructional and working principle of repulsion motor. Dec - 2012, 2013

Repulsion motors are similar to the series motor except the stator and the rotor windings are

inductively coupled i.e. the rotor current is obtained by transformer action from the stator.

TYPES OF REPULSION MOTOR

1.Two stator winding repulsion motor 2.Compensated repulsion motor.

3.Repulsion start induction motor. 4. Repulsion induction motor.

Repulsion motor is series to d.c. series motor with the rotor energized inductively.

The single phase winding is placed in stator slots similar to the main winding of a single phase

20


induction motor.The rotor consists of anordinary distibuted d.c. winding connected to the

commutator at one end.The brushes are in touch with commutator and are short circuited to

provide a closed current path.Figure 5.34 shows a basic circuit diagram for a repulsion motor.

If the brushes are located in the d-axis, the emf in the the effective armature winding will

be maximum and so would be short circuit current I2. Due to this, no torque will be produced and

stator and rotor fields are aligned.If the brushes are located are located in the q-axis the emf

induced in the armature winding would add up to zero and current I2 is zero and hence no torque

will be produced.

As it is essential that both an armature current should flow and an angular displacement

must exits between the two field the brushes are located in an intermediate position making a large

angle of about of about 70o with the d-axis .This appears like repulsion between the stator and

rotor fields. That iis why it is called as repulsion motor. The direction of rotation of the motor will

be reversed if the brushes are displaced on the other side of the q-axis.

Characteristics

Repulsion motors have characteristics similar to those of the series motor i.e. high starting

torque and high no-load speed. The motor is a reversing type, and the direction may be changed

during rotation.

Disadvantages

The draw-backs of repulsion motors are

a) Speed variations with the variations in load dangerously high at no load.

b) Low power factor, except at high speeds.

c) Requires frequent maintenance.

d) Higher cost.

e) Sparking at the brushes.

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21


11) Explain the operating principle of reluctance motor with neat diagram.

May-2009, 2012, 2014, Dec-2012

Single-Phase Synchronous Motors

Very small single-phase motors have been developed which run at true synchronous speed.

They do not require D.C. excitation for the rotor. Because of these characteristics, they are called

unexcited single-phase synchronous motors.

The most commonly used types are:

Reluctance motors

Hysteresis motors

The efficiency and torque-developing ability of these motors is low; the output of most of

the commercial motors is only a few watts.

Reluctance Motor

It is a single-phase synchronous motor which does not require D.C. excitation to the rotor.

Its operation is based upon the following principle:

“Whenever a piece of ferromagnetic material is located in a magnetic field; a force is

exerted on the material, tending to align the material so that reluctance of the magnetic path that

passes through the material is minimum”.

Construction

A reluctance motor (also called synchronous reluctance motor)

(i) a stator carrying a single-phase winding along with an auxiliary winding to produce a

synchronous-revolving magnetic field.

(ii) a squirrel-cage rotor having unsymmetrical magnetic construction. This is achieved by

symmetrically removing some of the teeth from the squirrel cage rotor to produce salient

poles on the rotor. As shown in the figure shown below 4 salient poles have been produced

on the rotor. The salient poles created on the rotor must be equal to the poles on the stator.

Reluctance Motor

(i) The rotor salient poles offer low reluctance to the stator flux and, therefore, become

strongly magnetized.

Operation

22


a) When single-phase stator having an auxiliary winding is energized, a synchronously-

revolving field is produced. The motor starts as a standard squirrel-cage induction motor

and will accelerate to near its synchronous speed.

b) As the rotor approaches synchronous speed, the rotating stator flux will exert reluctance

torque on the rotor poles tending to align the salient-pole axis with the axis of the rotating

field. The rotor assumes a position where its salient poles lock with the poles of the

revolving field. Consequently, the motor will continue to run at the speed of revolving flux

i.e., at the synchronous speed.

c) When we apply a mechanical load, the rotor poles fall slightly behind the stator poles,

while continuing to turn at synchronous speed. As the load on the motor is increased, the

mechanical angle between the poles increases progressively. Nevertheless, magnetic

attraction keeps the rotor locked to the rotating flux. If the load is increased beyond the

amount under which the reluctance torque can maintain synchronous speed, the rotor drops

out of step with the revolving field. The speed, then, drops to some value at which the slip

is sufficient to develop the necessary torque to drive the load by induction-motor action.

Characteristics

These motors have poor torque, power factor and efficiency.

These motors cannot accelerate high-inertia loads to synchronous speed.

The pull-in and pull-out torques of such motors are weak.

Despite the above drawbacks, the reluctance motor is cheaper than any other type of

synchronous motor. They are widely used for constant-speed applications such as timing devices,

signaling devices etc.

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12) Explain the constructional details, principle of operation and the applications of Hysteresis

motor.Dec-2014, 2015, May 2016 ,Nov-Dec 2016,APR-MAY 2018,NOV-DEC 2017

Hysteresis Motor

It is a single-phase motor whose operation depends upon the hysteresis effect i.e., magnetization

produced in a ferromagnetic material lags behind the magnetizing force.

Construction

a) a stator designed to produce a synchronously-revolving field from a single-phase supply. This is

accomplished by using permanent-split capacitor type construction. Consequently, both the

windings (i.e., starting as well as main winding) remain connected in the circuit during running

operation as well as at starting. The value of capacitance is so adjusted as to result in a flux

revolving at synchronous speed.

b) a rotor consisting of a smooth cylinder of magnetically hard steel, without winding or teeth.

Operation

a) When the stator is energized from a single-phase supply, a synchronously revolving field

(assumed in anti-clockwise direction) is produced due to split-phase operation.

23


b) The revolving stator flux magnetizes the rotor. Due to hysteresis effect, the axis of

magnetization of rotor will lag behind the axis of stator field by hysteresis lag angle a as shown

in fig. given below. Thus the rotor and stator poles are locked. If the rotor is stationary, the

starting torque produced is given by:

Ts =ɸsɸrsinɸ

Where ɸs = stator flux.

ɸr = rotor flux.

From now onwards, the rotor accelerates to synchronous speed with a uniform torque.

After reaching synchronism, the motor continues to run at synchronous speed and adjusts its torque

angle so as to develop the torque required by the load.

Hysteresis Motor

Characteristics

A hysteresis motor can synchronize any load which it can accelerate, no matter how great the

inertia. It is because the torque is uniform from standstill to synchronous speed.

Since the rotor has no teeth or salient poles or winding, a hysteresis motor is inherently quiet and

produces smooth rotation of the load.

The rotor takes on the same number of poles as the stator field. Thus by changing the number of

stator poles through pole-changing connections, we can get a set of synchronous speeds for the

motor.

Applications

Due to their quiet operation and ability to drive high-inertia toads, hysteresis motors are

particularly well suited for driving

electric clocks

timing devices and tape-decks

From-tables and other precision audio-equipment

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13) Explain the operating principle of AC Series motor with neat diagram.

24


May- 2009, 2014, 2013, 2016, Dec-2013,Nov-Dec 2016,May 2017, APR-MAY 2018

A.C. Series Motor or Universal Motor

A D.C. series motor will rotate in the same direction regardless of the polarity of the

supply. One can expect that a D.C. series motor would also operate on a single-phase supply. It is

then called an A.C. series motor. However, some changes must be made in a D.C. motor that is to

operate satisfactorily on A.C. supply. The changes effected are:

The entire magnetic circuit is laminated in order to reduce the eddy current loss. Hence an A.C.

series motor requires a more expensive construction than a D.C. series motor.

The series field winding uses as few turns as possible to reduce the reactance of the field winding

to a minimum. This reduces the voltage drop across the field winding.

A high field flux is obtained by using a low-reluctance magnetic circuit.

There is considerable sparking between the brushes and the commutator when the motor is used on

A.C. supply. It is because the alternating flux establishes high currents in the coils short-circuited

by the brushes. When the short-circuited coils break contact from the commutator, excessive

sparking is produced. This can be eliminated by using high-resistance leads to connect the coils to

the commutator segments.

Construction

The construction of an A.C. series motor is very similar to a D.C. series motor except that

above modifications are incorporated [See Fig.]. Such a motor can be operated either on A.C. or D.C.

supply and the resulting torque-speed curve is about the same in each case. For this reason, it is

sometimes called a universal motor.

A.C. Series Motor or Universal Motor

25


Operation

When the motor is connected to an A.C. supply, the same alternating current flows through

the field and armature windings. The field winding produces an alternating flux ɸthat reacts with

the current flowing in the armature to produce a torque. Since both armature current and flux

reverse simultaneously, the torque always acts in the same direction. It may be noted that no

rotating flux is produced in this type of machines; the principle of operation is the same as that of a

D.C. series motor.

Characteristics

The operating characteristics of an a.c. series motor are similar to those of a D.C. series motor.

The speed increases to a high value with a decrease in load. In very small series motors, the losses

are usually large enough at no load that limits the speed to a definite value (1500 - 15,000 r.p.m.).

The motor torque is high for large armature currents, thus giving a high starting torque.

At full-load, the power factor is about 90%. However, at starting or when carrying an overload, the

power factor is lower.

Applications

The fractional horsepower A.C. series motors have high-speed (and corresponding small

size) and large starting torque. They can, therefore, be used to drive:

high-speed vacuum cleaners

sewing machines

electric shavers

drills

Machine tools etc.

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14) What modifications have to be done on a DC series motor to make it to work with single

phase AC supply? State the applications of AC series motor. Dec-2014

In a normal d.c. motor if direction of both the field and armature current is reversed, the

direction of torque remains unchanged. So when normal d.c. series motor is connected to an a.c.

supply both field and armature current get reversed and unidirectional torque gets produced in the

motor can work on a.c. supply.

But the performance of such motor is not satisfactory due to the following reasons :

There are tremendous eddy current losses in the yoke and field cores,which causes overheating.

Armature and field winding offer high reactance to a.c. due to which operating power is very low.

The sparking at brushes is a major problem because of high voltage and current induced in the

short circuited armature coils during the commutation period.

Some modifications are required to have the satisfactory of d.c. series motor on a.c. supply, when

it is called a.c.series motor .The modifications are:

26


(i) To reduce the eddy current losses,yoke and pole construction is laminated.

(ii) The power factor can be improved by reducing the magnitudes of field and armature

reactances. Fields can be decreased by reducing the number of turns .But this reduction in

flux[Nα1/ ɸ],increases the speed and reducing the torque. To keep the torque same it is

necessary to increase the armature turns proportionately.This increases the armature

inductance.

Now to compensate for increased armature flux which produces severe armature reaction, it is

necessary to use compensating winding. The flux produced by this winding is opposite to that

produced by armature and effectively neutralizes the armature reaction.

If such a compensating winding is connected in series with the armature as shown in the

Fig. (a),the motor is said to be ‘conductively compensated’. For the motors to be operated on a.c.

and d.c. both, the compensation should be conductive. If the compensating winding is

shortcircuited on itself as shown in the Fig. (b),the motor is said to be ‘inductively

compensated’.In this compensating winding acts as a secondary of transformer and armature as its

primary.The ampere turns produced by compensating winding th armature ampere turns.

To reduce the induced emf due to transformer action in the armature coils while commutation

period, the following measures are taken:

(i) The flux per pole is reduced and number of poles are increased.

(ii) The frequency of supply used is reduced.

(iii) Preferably single turn armature coils are used.

The characteristics of such motor are similar to that of d.c. series motor. The torque varies

as square of the armature current and speed varies inversely as the armature current. The speed of

27


such motor can be dangerously high on no load condition and hence it is always started with some

load. Starting torque produced is high which is 3 to 4 times the full load torque.

Applications:

Because of high starting torque it is used in electric traction, hoists, locomotives etc.

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15) Explain the operating principle of servo motors.

They are also called control motors and have high-torque capabilities.Unlike large industrial

motors, they are not used for continuous energy conversion but only for precise speed and precise

position control at high torques. Of course, their basic principle of operation is the same as that of

other electromagnetic motors.

However, their construction design and mode of operation are different. Their power

ratings vary from a fraction of a watt upto a few 100w. Due to their low – inertia, they have high

speed of response. That is why they are smaller in diameter but longer in length. They generally

operate at very low speed or sometimes zero speed. They find wide applications in radar,tracking

and guidance systems , process controllers, computers and machine tools, Both dc and ac (2-phase

and 3-phase) servomotors are used at present.

Servomotors differ in application capabilities from large industrial motors in the following

respects:

1. They produce high torque at all speeds including zero speed.

2. They are capable of holding a static (i.e.no motion) position.

3. They do not overheat at standstill or lower speeds.

4. Due to low inertia they are able to reverse directions quickly.

5. They are able to accelerate and deaccelerate quickly.

DC Servomotors:

These motors are either separately excited dc motors or permanent – magnet dc motor. The

schematic diagram of a separately excited dc motor along with its armature and field MMFs and torque /

speed characteristics is shown in fig. The speed of d.c. servomotors is normally controlled by varying the

armature voltage. Their armature is deliberately designed to have large resistance so that torque-speed

characteristics are linear and have a large negative slope as shown in fig. The negative slope serves the

purpose of providing the viscous damping for the servo drive system.

28


The armature mmf and field mmf are in quadrature. This fact provides a fast torque

response because torque and flux become decoupled. Accordingly, a step change in the armature

voltage or current produces a quick change in the position or speed of the rotor.

AC servo motors:

Most of the AC servomotors are of the two-phase squirrel cage induction type and used for

low power applications. However, recently three-phase induction motorshave been modified for

high power servo systems which had so far been using high power DC servomotors.

(a) Two phase AC servo motors:

Such motors normally run on a frequency of 60 Hz or 400 Hz (for airborne system). The

stator has two distributed windings which are displaced from each other by 900 (electrical). The

main winding (also called reference or fixed phase) is supplied from a constant voltage source,

VmL0. The other winding (also called the control phase) is supplied with a variable voltage of the

same frequency as the reference phase but is phase displaced by 900 (electrical).

The control phase voltage is controlled by an electronic controller. The speed and torque of

the rotor are controlled by the phase difference between the main and control windings. Reversing

the phase difference from leading to lagging (or vice-versa) reverses the motor direction. Since the

rotor bars have high resistance, the torque-speed characteristics for various armature voltage are

almost linear over a wide speed range particularly near the zero speed. The motor operation can be

controlled by varying the voltage of the main phase while keeping tat of the reference phase

constant.

29


(b) Three phase AC servo motors:

A great deal of research has been to modify a three phase squirrel cage induction motor for

use in high power servo systems. Normally, such a motor is a highly non – linear coupledcircuit

device. Recently, this machine has been operated successfully as a linear decoupled machine (like

a d.c. machine) by using a control method called vector control or field oriented control. In this

method, the currents fed to the machine are controlled in such a way that its torque and flux

become decoupled as in d.c.machnine. This results in a high speed and a high torque response.

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16) Describe the working principle of stepper motor. Or Write short notes on stepper motor.

May-2015, NOV-DEC 2017

Stepper motor is known by its important property to convert a train of input pulses i.e. a

square wave pulses into a precisely defined increment in the shaft position.Each pulse moves the

shaft through a fixed angle. So the stepper motor is an electromechanical device which actuates a

train of step movements of shaft. Each pulse moves the shaft through a fixed angle. So the stepper

motor is an electromechanical device which actuated a train of step movements of shaft in

response to train of input pulses. The step movement may be angular or linear. The is one-one

relationship between an input pulse and step movement of the shaft. Eachpulse input actuates one

step movement of the shaft. When a given number of drive pulse are supplies to the motor, the

shaft gets turned through a known angle. The angle through which the motor turns or shaft moves

for each pulse is known as the step angle, expressed in degrees.

30


As such angle is dependent on the number of input pulsed, the motor is suitable for

controlling position by controlling the number of input pulses.Such system, used to control the

position is called position control system. The average motor speed is proportional to the rate at

which the input pulse command is delivered. When the rate is low, the motor in steps but for high

rate of pulses, due to inertia, it rotates smoothly like d.c. motors. Due to this property it is also used

in speed control systems. These motors are available in sub-fractional horse power ratings. As the

input command is in pulse, the stepper motor is compayible with modern digital equipments.

Due to its compatibility with digital equipments, its market is greately increased in recent

times. The stepper motors are widely used in X-Y plotters, floppy disk drives, machine tools,

process control systems, robotics, printers, tape drivers and variety of other industrial applications.

Types of stepper motors:

Variable reluctance stepper motor

Permanent magnet stepper motor

Hybrid stepper motor

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17) Explain the principle of magnetic levitation system.

When a moving permanent magnet sweeps across a conducting ladder, it tends to drag the

ladder along with, because it applies a horizontal tractive force F = B I l. It will now be shown that

this horizontal force is also accompanied by a vertical force, which tends to push the magnet away

from the ladder in the upward direction.

A portion of the conducting ladder of Fig 34.43(a) has been shown in fig 34.43(b). The

voltage induced in the conductor (or bar) A maximum because flux is greatest at the centre of the

N pole.If the magnet speed is very low,the induced current reaches its maximum value in A at

virtually the same time (because delay conductor inductance is negligible). As this current flows

via conductor B and C, it produces induced SSS and NNN poles as shown. Consequently the front

half of the magnet is pushed upwards while the rear half is pull downwards.

Since distribution of NNN and SSS is symmetrical with respect to the centre of the magnet

the vertical force of attraction and repulsion being equal and opposite each other and cancel out

leaving behind only horizontal tractive force.

31


Magnetic levitation is being used in ultra-high speed trains(upto 300 km/h) which float in

which in the air about 100mm to 300mm above the metallic track. They do not have any wheels

and do not require the traditionals steel rail. A powerful electromagnet (whose coils are cooled to

about 4o

K by liquid helium) fixed underneath the train moves across the conducting rail, there by

including current in the rail. This gives rise to vertical force(called force of levitation) which keeps

the train pushed up in the air above the track. Linear motors are used to propel the train.

A similar magnetic leviation system of transit is being considered for connecting Vivek

Viher in East Delhi to Vikaspuri in WestDelhi The system popularly known as Magneto – Bahm

(M-Bahm) completely eliminates the centuries – old ‘ steel – wheel – over steel rail’ traction. The

M-Bahn train floats in the air through the principle of magnetic leviation and propulsion is by

linear induction motors. There is 50% decrease in the train weight and 60% reduction in energy

consumption for propulsion purposes. The system is extraordinary safe (even during an

earthquake) and the operation is fully automatic and computer based.

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18) Explain the theory of Brushless DC machine. May-2015

In conventional DC motors, the armature is the rotor, and the field magnets are placed in

the stator. A brushless DC motor of this structure is very difficult to make. The construction of

modern brushless DC motors is very similar to the AC motor,known as the permanent magnet

synchronous motor(see Fig. 4.1). The armature windings are the part of the stator, and the rotor is

composed of one or more magnets.

The windings in brushless DC motor are similar to those in a polyphase AC motor, and the

most orthodox and efficient motor has a set of three- phase windings and is operated in bipolar

excitation (see Fig.4.2). Brushless DC motors are different from AC synchronous motors in that

the former incorporates some means to detect the rotor position (or magnetic poles) to produce

signals to control the electronic switches. The most common position/pole sensor is the Hall

element, but some motors use optical sensors.

By examining a simple three- phase unipolar- operated motor, one can easily understand

the basic principles of brushless DC motors. Figure illustrates a motor of this type that uses optical

sensord (phototransistors) as position detectors. Three phototransistors PT1,PT2 and PT3 are

placed on the end-plate at 120o intervals, and are exposed to light in sequence through a revolving

shutter coupled to the motor shaft.

As shown in Fig. the south pole of the rotor now faces the salient pole P2 of the stator and

the phototransistor PT1 detects the light and turns transistor Tr1 on. In this state, the south pole

which is created at the salient pole P1 by the electrical current flowing through the winding W1 is

attracting the north pole of the rotor to move it in the direction of the arrow(CW).When the south

pole comes to the position to face the salient pole P1 the shutter which is coupled to the rotor shade

PT1 and the PT2 will exposed on the light and the current will flow through the transistot Tr2.

When the current flows throush the winding W2 and creates a south pole on salient P2 and

the north pole in thwe rotor will revolve in the direction of the arrow and face the salient pole P2,

at this moment the shutter shades PT2,and the phototransistor PT3 is exposed to light. These

32


actions steer the current from winding W2 to W3.Thus the salient pole P2 is de-energised,while the

salient pole P3 is energised and creates the south pole.Hence the north pole on the rotor further

travels from P2 to P3 without stopping. By repeating such a switching action in the sequence

givenin Fig. the permanent magnet rotor revolves continuosly.

Fig. Three phase unipolar driven brushless DC motor

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19) A 220 V, 6 pole, 50 Hz, single-winding single phase induction motor has the following

equivalent circuit parameters as referred to the stator.

R1m = 3.0 Ω X1m = 5.0 Ω

R2 = 1.5 Ω X2 = 2.0 Ω

Neglect the magnetizing current. When the motor runs at 97% of the synchronous speed,

compute the following:

(a) the ratio Emf / Emb (b) the ratio Vf / Vb

(c) the ratio Tf / Tb (d) the gross total torque. May-2015

(e) the ratios Tf / Total torque and Tb / Total torque

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MAILAM ENGINEERING COLLEGE · 16. Name the various methods for predetermining the voltage...

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Transcript of MAILAM ENGINEERING COLLEGE · 16. Name the various methods for predetermining the voltage...