2 markshystersis loss is reduced by using cold-rolled-grain-oriented electrical steel for the...

Electrical Engineering Paper - 1ZONE TECH 1

Q.1Sol. When the output voltage is lower than the input voltage, it is called a step-down transformer. when

the output voltage is higher than the input voltage, it is known as a step-up transformer.

Q.2Sol. Unidirectional element:

When element properties and characteristics depend on direction of current then the element iscalled as unidirectional element.Bidirectional Element:When element characteristics and properties are independent of direction of the current then theelement is called as bidirectional element.

Q.3Sol. The material which are known to undergo neutron fission are U235, U233 and PU239. These are fissile

material .U238 and Th232 are not fissionable. These two material are known as fertile material.

Q.4Sol. The iron loss in a transformer is made up of hystersis loss and edy-current loss. Hystersis loss is

reduced by using cold-rolled-grain-oriented electrical steel for the transformer core. Eddy currentloss is minimised by using thin sheets (0.35 mm) of laminations for preparing the transformer core.

2 Marks

Electrical Engineering Paper - 1 2ZONE TECH

Q.5Sol. Electrical conductivity Thermal conductivity

1. The co-efficient of electrical conductivity The Co-efficientg of thermal conductivity isis defined as the quantity of electricity defined as the quantity of heat conductedflowing per unit area per unit time per unit area unit time maintained at unitmaintained at unit potential gradient. termperature gradient.

2. Electrical conductivity is purely due to Thermal conductivity is due to both freenumber of free electrons electrons and phonons.

3. Conduction of electricity takes plane from Conduction of heat takes place from hot end tohigher potential end to the lower potential cold end.end.

4. Unit ohm–1 m–1 Unit Wm–1 K–1

Q.6Sol. A long transmission line draws a substantial quantity of charging current. If such a line is open

circuited or very lightly loaded at the receiving end, the voltage at the receiving end may becomehigher than the voltage at the sending end. This is known as ferranti effect.

Q.7Sol. For maximum transformer efficiency, variable ohmic loss = constant iron loss i.e. n2 P

0 P

i

Fractional load at which maximum efficiency occurs, n i

0

P

P

Q.8

Sol. D.F. Maximum Demand

Connected Load

Q.9Sol. At constant temperature current density is directly proportional to electric field intensity.

aI

l

J E

J Ewhere J............Current density (A/m2)

..........Conductivity 1– m E...........Electric field intensity (V/m)

Q.10Sol. Synchronous speed is the speed of rotationg magnetic field produced when polyphase currents are

applied to a polyphase winding. Synchronous speed Ns is given by.

Ns

120 frpm

P

This shows that synchronous speed depends on the frequency of supply given to polyphase windingsand the number of poles for which polyphase winding is designed.

Q.11Sol. Kirchoff voltage law states that the algebraic sum of all branch voltage around any closed loop of a

Electrical Engineering Paper - 1ZONE TECH 3network is zero.

Q.12Sol. Retentivity (or) remanence:

When the external magnetic field applied to a magnetic material is removed, the magnetic materialwill not loss magnetic property immediately. There exits some residual intensity of magnetisation inthe specimen even when the magnetic field is cut off. This is called residual magnetism or retentivity.Coercivity:The residual magnetism can be completely remoed from the material, by applying a reverse magneticfield. Hence corecivity of the magnetic material is the strength of reverse magnetic field (–H

c) which

is used to completely demagnetise the material.

Q.13Sol. Delta to star (or to T)

R1

12 31

12 23 31

R R

R R R

R2

12 23

12 23 31

R R

R R R

R3

23 31

12 23 31

R R

R R R

Q.14Sol. • Single phase motor have low power factor

• The starting torque is low in single phase motors• Single phase motor have lower efficiency• Single phase motor are castilier than 3-phase motor of the same rating

Q.15Sol. The ACSR (Alluminium Conductor Steel Reinforced) consist of a core of galvanised steel strand

surrounded by a number of aluminium strands.

Q.16Sol. First law: Like charges of electricity repel each other, whereas unlike chrages attract each other.

Second law: According to this law, the force exerted between two point charges (i) is directlyproportional to the product of their strength (ii) is inversely proportional to the square of the distancebetween them.

F 2 1 2

2 2

Q.Q kQ QF

d d

Q.17Sol. The open–circuit voltage across terminals A and B in figure is

VOC drop across R

5 × 2 10 VHence, voltage source has a voltage of 10 V and the resistance of 2 throught connected in series .


A

B

2

10 V

+

–

Q.18Sol. • Single-phase to ground (L-G) fault

• Two phase to ground (2L-G) fault• Phase –to–phase (L – L) fault.

Q.19Sol. By armature reaction of a synchronous machine, we mean how armature flux produced by 3-phase

armature winding affects the distribution of main-field flux created by dc in the field winding.

Q.20

Sol. (i) L 2 2

1 2

1 2

L L – M 4 6 – 3

L L 2M 4 6 – 2 3

15

43.75 H

(ii) L 2

1 2

1 2

L L – M 24 – 9

L L 2M 16

15

160.94 H (approx)

Q.21Sol. Air gap power (P

g): The power that is transfered from the stator to the rotor via the air-gap magnetic

field, is known as the air-gap power (Pg)

Pg

'2 '2 23I .R

s

Subtracting the rotor copper loss from Air-gap power (Pg) gives the internal mechanical power

developed (Pm

)

Pm P

g–

'2 ''2 '22 22 2 2 2

3I R3I .R – 3I .R

s

'22 2 g

13I .R 1 1 s P

s

The internal mechanical power developed (Pm

) is three times the electrical power absorbed in resistance

'2

1R 1

s

.

Shaft power (Psh

) is the power available at shaft after subtracting mechanical losses from Pm

.P

sh Pm

– mechanical losses


Q.22Sol. Type–I (Soft Superconductor) Type–II (Hard) Superconductor

1. The type–I super conductor becomes a Type–II super conductor loss its superbecomes a normal conductor abruptly at conducting property gradually, due to increasecirtical magnetic field in magnetic field.

2. Here we have only one critical field (HC). Here we have two critical fields (i.e.) Lower

critical fields (Hc1

) and Upper critical field (Hc2

).3. No mixed state exists. Mixed (or) vortex state is present.4. Highest known crtitical field is 0.1 Tesla The critical field is greater than type 1 (i.e.) upto

30 Tesls.5. Magnetic phase diagram Magnetic phase diagram

Super Conductor

H

Hc

Normal

TcT

H

TcT

SC

Mixed

Hc

Hc1

Normal

6. Examples: Pb, Zn, Al, Zn, Ga, Hg Examples Niobium, Vanadium

Q.23Sol. The mechanical loss and open circuit core loss combined together are caused by rotor rotation. their

sum is called as no-load rotational losses.

O.C. core loss

F and W loss

Field current

Ro

tati

on

al l

oss

Mechanical Loss: This loss consists of bearing friction, brush friction and windage losses. The windageor wind friction loss includes the power required to circulate air through the machine and ventilatingducts and is approximately proportional to square of the speed.Open Circuit Core Loss: Consists of hystersis and eddy current losses. These occurs in case of statorand rorot carrying an alternating flux.

Hystersis loss Pn K

hxmfB

Eddy current loss 'eP 2 2

e mK f B

No load rotational losses can be determined by running the electrical machine as an unloaded motorat the rotor speed and with armature voltage equal to the normal generator emf. The total power

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input to the unloaded motor minus the no load armature ohmic loss gives the magnitude of no-loadrotational loss.

Q.24Sol. 1. Manufacturers usually specify the impedance values of equipments in per unit of the equipment's

rating. If any data is not available, It is easier to assume its per unit value than its numericalvalue.

2. When expressed in per unit, system parameters tend to fall in relatively narrow numerical ranges.Therefore , any erroneous data can be easily identified.

3. Per unit data representation yields important information about relative magnitudes.4. Power systems contain a large number of transformers. The ohmic value of impedance as referred

to secondary is different from the value as referred to primary. However, if base values areselected properly, the per unit impedance is the same on the two sides of the transformer.

5. The transformer connections in 3-phase circuits do not affect the per unit value of impedancealthough the base voltages on the two sides do depend on the connections.

Q.25Sol. Magnetic circuit Electric circuit

MMFI I

Flux

EMF

Cu

rrent

1. Flux m.m.f.

reluctanceCurrent

e.m.f.

reluctance

2. M.M.F. (Ampere-truns) E.M.F. (volts)3. Flux (webers) Current I (ampers)

4. Flux density B (Wb/m2) Current density (A/m2)

5. Reluctance S 0 rµA µ µ A

l lResistance R

A A

l l

6. Permeance( 1/reluctance) Conductance( 1/resistance)7. Reluctivity Resistivity8. Permeability ( 1/reluctivity) Conductivity ( 1/resistivity)

9. Total m.m.f. 1 2 3S S S .... Total e.m.f. 1 2 3IR IR IR .......

Q.26Sol. Core Type Transformer :

• In core type transformer winding surround the steel core.

• This type of transformer requires less iron but more conductor material• The vertical portion of the core are usually called limbs or leg and the top and bottom portion are

called the yoke. This means that for single-phase transformer, core type has two legged core.• The L.V. winding is placed adjacent to the steel core and H.V. winding out side, in order to

minimise the amount of insulation required.Shell Type Transformer :• In shell type transformer steel core surround a major part of the winding.

Electrical Engineering Paper - 1ZONE TECH 7• Shell type transformer has three legged core.• In shell type transformer, the L.V. and H.V. winding are wound over the center limb and are

interleaved or sandwitch.• Shell type transformer are preferred for low voltage low-power levels.

Q.27

Sol. 24I 20 180 W; I

4 3 A

Since 15 and 20 are in parallel,

I3 × 15 3 × 20

I3 4 A

I2 I

3 +I

4 4 + 3 7A

Now, resistance of the circuit to the right of point A is

10 + 15 × 20/35 130

7

I1 × 25 7 ×

30

7

I1

26

5 A 5.2 A

I I1 + I

2 5.2 + 7 12.2 A

Total circuit resistance

RAE

5 + 25 ||130

7

955

61

V I. RAE 12.2 × 955/61 191 V

Q.28Sol. When a three-phase a.c. voltage is applied to the stator winding, a rotating magnetic field is produced

in the air gap. The stator field rotators at synchronous speed.When a pair of rotating stator poles sweeps across the stationary rotor poles at synchronous speed,the stator poles will tend to rotate the rotor in one direction and then in the other direction. However,because of the rotor inertia, the stator field slides by so fast that the rotor cannot follow it.Consequently, the rotor does not move and we say that the starting torque is zero. In other words,a synchronous motor is not self-starting.

Q.29Sol. Soft Hard

1. They can be easily magnetised and They cannot be easily magnetised (or) demagnetised.demagnetised.

2. Loop area is less and hence the The loop area is large and hence the by stersis loss ishystersis loss is minimum. maximum.

3. Susceptibility and permability is high. Susceptibility and permeability is low.4. Retentivity and coercivity are small Retentivity and coercivity are large.5. They have low eddy current loss. They have high eddy current loss.6. These materials are free from These materials have large amount of impurities and

irregularities like strain or impurities. lattice defects.

Q.30Sol. Voltage Regulation:

The voltage regulation of an alternator is defined as "the rise in voltage when full-load is removed(Field excitation and speed remaining the same) Divided by the rated terminal voltage".


% V.R. 0E – V100

V

Method of Determination of voltage regulation.1. Synchronous Impedance or E.M.F.2. The ampere-turn or m.m.f. method.3. Zero power factor or potier method4. American standarad association method.

Q.31Sol. Applying Thevenin's theorem, after detaching the 6-ohm resistor from terminals A – B.

VTH V

C 15 – 1 × 3 12 VoltsR

TH 4 + (3116) 6 ohmsI

L 12/(6 + 6) 1 A

Q.32Sol. The difference between the synchronous speed N

s and the actual speed N of the rotor is known as

slip. Though it may be expressed in so many revolutions/second, yet it is usual to express it as apercentage of the synchronous speed. Actually, the term 'slip' is descriptive of the way in which therotor 'slips back' from synchronism.

%slip s s

s

N – N100

N

Sometimes, Ns– N is called the slip speed.

Obviously, rotor (or motor) speed is N Ns (1 – s).

It may be kept in mind that revolving flux is rotating synchronous, relative to the stator (i.e. stationaryspace) but at slip-speed relative to the rotor.

Q.33Sol. Figure gives the schematic diagram of a single-phase induction motor with one stator winding and a

squirrel-cage rotor. The winding is distributed in space so that the space fundamental of mmf is themost dominant component of the actual mmf distribution. The sapce harmonics of mmf, as in thecase of a 3-phase induction motor, would then be ignored. When the winding carries a sinusoidalcurrent, it produces a sinusoidally space-distributed whose peak values pulstates with time.

i = i cos tmax

Rotor (squirrel cage)

Axis of phase winding

As seen from the axis of the winding, the mmf at any time t is

F Fpeak

cos ...(i)

where is the angle measured from the winding axis

Now Fpeak F

max cos t ...(ii)

So that the mmf has both space and time distribution expressed as

F Fmax

cos cos t ...(iii)

Electrical Engineering Paper - 1ZONE TECH 9This equation can be trignometrically manipulated into the form

F (1/2)Fmax

maxcos – t 1/2 F cos t ...(iv)

Equation (iv) shows, a pulsating single-phase field be considered as superposition of two rotationgfields rotating at synchronous speed is opposite direction.

(1/2)Fmax

cos – t ; rotating forward rotating field, Ff,

(1/2)Fmax

cos t ; the backward rotating field, Fb.

The forward and backward rotating fields along with the rotor which is rotating at speed n in thedirection of the forward field have slip s and (2 – s) respectively.Under stationary rotor condition, the two rotating field slip past the rotor at the same slip s = 1, andinducing equal currents in the squirrel-cage rotor. The two rotating fields have the same strengthand produce equal and opposite torques resulting is net starting torque of zero value.If however, the rotor is made to run at speed n in the direction of the forward field, the two slips arenow s and (2 - s) > > s and as a consequence the backward field induced rotor currents are muchlarger than at stand still and have lower power factor. The corresponding opposing rotor mmf, inpresence of the stator impedance, causes the backward field to be greatly reduced in strength. Onthe other hand, the low-slip forward rotating field induces smaller currents of a higher power factorin the rotor than at stand still. This lead to great enhancement in the forward flux wave. This reductionin the backward field and strengthening of the forward field is slip-dependent and the differenceincreases as slip reduces or the rotor speed in the forward direction becomes close to the synchronousspeed.

Q.34Sol. 1. Selectivity or discrimination:-

Selectivity, is the quality of protective relay by which it is able to discriminate between a fault inthe protected section and the normal condition. Also, it should be able to distinguish whether afault lies within its zone of protection or outside the zone. Sometimes, this quality of the relay isalso called discrimination.The relay should also be able to discriminiate between a fault and transient conditions like powersurges or inrush of a transformer's magnetising current. The magnetising current of a largetransformer is comparable to a fault current, which may be 5 to 7 times the full load current.

2. Reliability:A protective system must operate reliably when a fault occurs in its zone of protection. Thefailure of a protective system may be due to the failure of any one or more elements of theprotective system. Its important elements are the protective relay, circuit breaker, VT, CT, wiringbattery, etc. To achieve a high degree of reliability, greater attention should be given to thedesign, installation, maintenance and testing of the various elemets of the protective system.Robustness and simplicity of the relaying equipment also contribute to reliability. The contactpressure, the contact material of the relay, and the prevention of contact contamination are alsovery important from the reliability point of view. A typical value of reliability of protectivescheme is 95%.

3. Sensitivity:A protective relay should operate when the magnitude of the current exceeds the preset value.This value is called the pick-up current. The relay should not operate when the current is belowits pick-up value. A relay should be sufficiently sensitive to operate when the operating currentjust exceeds its pick-up value.

4. Stability:A protective system should remain stable even when a large current is flowing through itsprotective zone due to an external fault, which does not lie in its zone. The concerned circuitbreaker is supposed to clear the fault. But the protective system will not wait indefinitely if theprotective scheme of the zone in which fault has occurred fails to operate. After a preset delaythe relay will operate to trip the circuit breaker.

5. Fast Operation:


A protective system should be fast enough to isolate the faulty element of the system as quicklyas possible to minimise damage to the equipment and to maintain the system stability. For amodern power system, the stability criterion is very important and hence, the operating time ofthe protective system should not exceed the critical clearing time to avoid the loss of synchronism.Other points under consideraiton for quick operation are protection of the equipment from burningdue to heavy fault currents, interruption of supply to consumers and the fall in system voltagewhich may result in the loss of industrial loads. The operating time of a protective relay is usuallyone cycle. Half-cycle relays are also available. For distribution system the operating time may bemore than one cycle.

Q.35Sol.

V2N1

N2

I1

e2e1V1

+

–— —

+ I2

LO

AD

+

Supply voltage V1

Emf induced in primary winding e1

Current in primary winding I1

Magnetic flux Turns in primary winding N

1

Turns in secondary winding N2

Emf induced in secondary winding e2

Current in secondary winding I2

Voltage across load V2

When sinusoidal voltage applied to primary winding, a sinusoidal flux is set up in the iron corewhich likns both windings.

Sinusoidal flux variation, max sin t

where max is the maximum value of the magnetic flux

The emf e1 induced in the primary winding by the alternating flux.

e1 1

dN magnitude

dt

The direction of induced emf can be determine by Lenz's law,

e1 1 max

dN sin t

dt 1 maxN cos t

e1 1 max

E cos t 1 maxE sin t 90

where, E1(max)

is the maximum value of induced emfThe emf induced in primary winding is 90° ahead by the core flux.Rms value of emf e

1 induced in primary winding ,

E1

1 max 1 maxE N

2 2


E1

1 maxN 2 f

2

1 max2 fN

4.44 fN1 max ...(i)

Similary emf induced in secondary winding ,

e2 2 2 max

dN N cos t

dt

2 maxN sin t 90

e2 2 max

E sin t 90

Rms value of emf e2 induced in secondary winding

E2 2 max2 max 2 max

N 2 fE N

2 2 2

E2 2 max2 fN 2 max4.44 f N ...(ii)

From eqaution (i) and (ii) we get

1

2

E

E 1

2

N

N or 1 2

1 2

E E

N N emf per turn

i.e. emf per turn in primary winding is equal to emf per turn in secondary winding.

Q.36Sol. 1. General Tariff Form:

Quite a large number of tariff have been proposed from time to time and are in use. They are allderived from the following equation:

A cx + dy + f ....(i) where A Total amount of bill for a certain period (Say one month)

x Maximum demand during the period (kW or kVA)y Total energy consumed during the period (kWh)c Unit charge for maximum demand, Rs per kW (or Rs. per kVA)d Unit cost of energy, Rs, per kWhf Constant charge, Rs.

Thus the total bill consists of three parts, one depending on maximum power demand, the seconddepending on total energy consumed and the third being a constant figure.

2. Flat Demand Rate:The flat demand rate can be expressed in the from, A = cx i.e. the bill depends only on themaximum demand irrespective of the amount of energy consumed.This is the earliest form of tariff and the bill in those days was based on the total number oflamps installed in the permises. Now-a-days the use of this tariff is restricted to sign lightingsignal system, street lighting etc. where the number of hours are fixed and energy consumptioncan be easily predicted. Its use is very common for supplied to irrigaion tube wells since thenumber of hours for which the tube well feeders are switched on are fixed, The charge is madeaccording to the horse power of the motor installed. The cost of metering equipment and meterreading is eliminated by the use of this form of tariff.

3. Straight Meter Rate:This can be represented by the equation

A dy ...(ii)The charges depend on the energy used. This tariff is sometimes used for residential and commercialcustomers. It has the advantage of simplicity. However the main disadvantage of this tariff isthat a customer who does not use energy has zero bill though he has caused the utility to incur adefinite expenditure due to its readiness to serve him. Another disadvantage is that this methoddoes not encourage the use of electricity.

20 Marks


4. Block Meter Rate:To remove the inconsistency of straight meter rate, the block meter rate charges the customerson a sliding scale. A certain unit rate is for a certain block of energy and for each succeedingblock of energy, the corresponding unit charge decreases.

A d1y

1 + d

2y

2 + .........+ d

n[y – (y

1 + y

2 + .......+y

n + ......+y

n - 1)] ...(iii)

where d1, d

2, d

3 are unit charges for energy blocks of magnitude y

1, y

2, y

3 etc. Generally the

charge and energy consumption are divided into three blocks, a high rate for the initial certainnumber of energy units, a lower rate for the next certain number of energy units and a still lowerrate for the remaining energy units. This tariff is very commonly used for residential andcommerical customers.In many states in India, a reverse form of this tariff is being used to restrict the energyconsumption. In this reverse form, the unit energy charges increase with increase in energyconsumption.

5. Hopkinson Demand Rate:This tariff, also known as two part tariff, can be expressed in the form

A cx + dy ...(iv)Thus the total bill includes a demand charge based on the maximum demand plus a charge basedon energy consumed. The factors c and d may be constant or may vary as per sliding scale. Thistariff is used for industrial customers. This tariff introduces the problem of measuring the maximumpower demand of the customers. This maximum demand can either be taken as a certain fractionof the connected load or measured by a maximum demand meter. It is usual to specify a minimumdemand that must be paid for. Sometimes, to discourage the customers from using low powerfactor devices, the demand charges are based on kVA of maximum demand.

6. Doherty Rate:This rate, also kown as three part tariff extends the two part tariff adding a constant term, theform being idential with equation (i). This tariff is suitable for and applied to industrial customers.

7. Wright Demand Rate:Hopkinson demand rate offers an inducement to a customer to keep his maximum demand at alow value. Wright demand rate intensifies this inducement by lowering both the demand chargeand energy charge for a reduction in maximum demand i.e. an improvement in load factor. Thistariff is generally specified for those industrial customers who have a control over their maximumdemands.

Q.37Sol. Given that :

Full load speed of motor,N

r 1440 rpmSynchronous speed of motor,

Ns

120 f

P

Ns

> Nr

Let, Ns 1500 rpm (Nearest standrad synchronous speed for even number of poles).

(i) Number poles 1500 120 50

P

P 4

(ii) Full load slip s s r

s

N – N 1500 – 14400.04

N 1500

(iii) Rotor frequency, f' s.f. (0.04 × 50 2 Hz)(iv) Speed of statory field w.r.t. rotor filed

Ns– N

s 1500 – 1500 0(As two MMFs are stationary w.r.t. each other)

(v) Speed of stator field w.r.t. rotor structure

Electrical Engineering Paper - 1ZONE TECH 13N

s – N

r sN

s

1500 – 1440 60 rpm

Q.38Sol. Power Factor Improvement:

1. Causes of Low Power Factor:• Induction motor is the most commonly used A.C. motor. It is used for a variety of purposes.

At full load a 3-phase induction motor operates at a power factor of around 0.8 lagging. Atpart loads the power factor is still poorter. The power factor of single phase induction motorsis about 0.6 lagging.

• The transformer draw a magnetising current from the system. This current is at a powerfactor of zero lagging.

• Miscellaneous equipment like are lamps. welding equipment etc. also contribute to low powerfactor in the system.

2. Effects of Low power Factor:• To meet the load requirement at a low power factor the capacity of power plant, transmission

and distribution equipment has to be more than that which would be necessary if the loadwere demanded at unity power factor.

• For the same active power operation of an existing power system at a low power factormeans overloading the equipment at times of full load.

• For the same active power, a low power factor means a greater current and hence higherenergy losses.

• Low power factor causes the voltage regulation to be poor.3. Advantages of Power Factor Improvement:

Installation of power factor improvement device, to raise the power factor, results in one ormore of the following effects and advantage:• Reduction in circuit current.• Increase in voltage level at load.• Reduction in copper losses in the system due to reduction in current.• Reduction in investment in the system facilities per kW of the load supplied.• Improvement in power factor of the generators.• Reduction in kVA loading of the generation and circuits. This reduction in kVA loading may

relieve an overloaded conditions or release capacity for additional growth of load.• Reduction in kVA demand charges for large consumers.

To encourage large consumers to install power factor improvement devices at their premises, supplyauthorities charges such customers as per two part tariff, the first part being proportional to themaximum kVA demand. To reduce this charge large industrial consumers install power factorsimprovement devices. The power factor can be improved if the lagging kVAR of the equipment isbalanced by a leading kVAR. This can be done either by the use of static capacitors or synchronouscondensers.

Q.39Sol. Stator Construction: The various parts of the stator include the frame, stator core, stator windings

and cooling arrangement. The frame may be of cast iron for small-size machines and of welded steeltype for large size machines. In order to reduce hysteresis and eddy-current losses, the stator core isassembled with high grade silicon content steel laminations. A 3-phase winding is put in the slots cuton the inner periphery of the stator as shown in Figure. The winding is star connected. The windingof each phase is distributed over several slots. When current flows in a distributed winding it producesan essentially sinusoidal space distribution of emf.


Statorcore

Armatureslot

Statorframe

Alternator Stator

Rotor Construction: There are two types of rotor constructions namely, the salient-pole type and thecylindrical rotor type.Salient-Pole Rotor: The term salient means 'protruding' or 'projecting'. Thus, a salient pole rotorconsists of poles projecting out from the surface of the rotor core. Figure shows the end view of atypical 6-pole salient-pole rotor. Salient-pole rotors are normally used for rotors with four or morepoles.

S N

SN

N S

Fieldwinding

Polebody

Poleface

Rivet

Shaft

Steel spider

Damper bars

Six-pole salient-pole rotor

Electrical Engineering Paper - 1ZONE TECH 15Since the rotor is subjected to changing magnetic fields, it is made of this steel laminations to reduceeddy current losses. Poles of identical dimensions are assembled by stacking laminations to therequired length and then riveted together. After placing the field coil around each pole body, thesepoles are fitted by a dove-tail joint to a steel spider keyed to the shaft. Salient-pole rotors haveconcentrated winding on the poles. Damper bars are usually inserted in the pole faces to damp outthe rotor oscillations during sudden change in load conditions. A salient-pole synchronous machinehas a non-uniform air gap. The air gap is minimum under the pole centres and it is maximum inbetween the poles. The pole face are so shaped that the radial air gap length increases from the polecentre to the pole tips so that the flux distribution in the air gap is sinusoidal. This will help themachine to generate sinusoidal emf.The individual field-pole windings are connected in series to give alternate north and south polarities.The ends of the field windings are connected to a d.c. source (a d.c. generator or a rectifier) throughthe brushes on the slip rings. The slip rings are metal rings mounted on the shaft and insulated fromit. They are used to carry current to or from the rotating part of the machine (usually a.c. machine)via carbon brushes.Salient-pole generators have a large number of poles, and operate at lower speeds. A salient-polegenerator has comparatively a large diameter and a short axial length. The large diameteraccommodates a large number of poles.Salient-pole alternators driven by water turbines are called hydro-alternators or hydrogenerators.Hydrogenerators with relatively higher speeds are used with impulse turbines and have horizontalconfiguration. Hydrogenerators with lower speeds are used with reaction and Kaplan turbines andhave vertical configuration.Cylindrical Rotor: A cylindrical-rotor machine is also called a non-salient pole rotor machine. It hasits rotor so constructed that it forms a smooth cylinder. The construction is such that there are nophysical poles to be seen as in the salient-pole construction. Cylindrical rotors are made from solidforgings of high grade nickel-chrome-molybdenum steel. In about two-third of the rotor periphery,slots are cut at regular intervals and parallet to the shaft. The d.c. field windings are accommodatedin these slots. The winding is of distributed type. The unslotted portion of the rolor forms two (orfour) pole faces. A cylindrical-rotor machine has a comparatively small diameter and long axial length.Such a construction limits the centrifugal forces. Thus, cylindrical rotors are particularly useful inhigh-speed machines. The cylindrical rotor type alternator has two or four poles on the rotor. Such aconstruction provides a greater mechanical strength and permits more accurate dynamic balancing.The smooth rotor of the machine makes less windage losses and the operation is less noisy becauseof uniform air gap.Figure 3.3 shows end views of 2-pole and 4-pole cylindrical rotors. Cylindrical-rotor machines aredriven by steam or gas turbines. Cylindrical-rotor synchronous generators are called turboalternatorsor turbogenerators. Such machines have always horizontal configuraton installation. The machinesare built in a number of ratings from 10 MVA to over 1500 MVA. The biggest size used in India hasa rating of 500 MVA installed in super thermal power plants.

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End view of two-pole and four pole cylindrical rotors

2 markshystersis loss is reduced by using cold-rolled-grain-oriented electrical steel for the...

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