02-Alternator on Load

8/11/2019 02-Alternator on Load

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02-Alternators

Prepared by Dr. M. A. Mannan Page 1 of 12

AlternatorsAlternator on Load [1/37.16/p.1422]

As the load on an alternator is varied, its terminal voltage is also found to vary as in

DC generators. This variation in terminal voltage Vis due to the following reasons:

1.

Voltage drop due to armature resistance,Ra,2.

Voltage drop due to armature leakage reactance,XL, and

3. Voltage drop due to armature reaction

Armature ResistanceThe armature resistance/phaseRacauses a voltage drop/phase ofIRawhich is in phase

with the armature currentI. However, this voltage drop is practically negligible.

Armature Leakage ReactanceWhen current flows through the armature conductors, fluxes are set up which do not

cross the air-gap, but take different paths. Such fluxes are known as leakage fluxes. Various

types of leakage fluxes are shown in Fig. 35.22.The leakage flux is practically independent of saturation, but is depend on Iand its

phase angle with terminal voltage V. This leakage flux sets up an emf of self-induced which

is known as reactance emf and which is ahead of I by 90o. Hence, armature winding is

assumed to posses leakage reactance XL (also known as Potier reactance XP) such that

voltage drop due to this equalIXL. A part of the generated emf is used up in overcoming this

reactance emf

)( La jXR ++= IVE

This fact is illustrated in the vector diagram of Fig. 35.23.

Armature reaction

Armature reaction is the effect of armature flux on the main field flux. In the case ofalternators, the power factor of the load has a considerable effect on the armature reaction.

We will consider three cases: (i) when load of p.f. is unity, (ii) when p.f. is zero lagging, and

(iii) when p.f. is zero leading.

Unity Power FactorIn the case of three phase stator, the three phase currents set up three fluxes. The

armature phase currents have the same magnitude and are 120oapart in time as shown in Fig.


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19.15. In accordance of Fig. 19.15. the expression of fluxes can be written as follows (taking

the direction of flux due to phase aas reference direction):

)120sin()120sin120(cos

);120sin()240sin240(cos

;sin)0sin0(cos

ooo

ooo

oo

++=

+=

+=

tj

tj

tj

mc

mb

ma

Expanding and adding the above equations, we get

)90(2

3)cos(sin

2

3 ottjt mmr =+=

It is seen from the above equation that the resultant field set up the current in the

armature remains constant in magnitude and rotates at synchronous speed.

Moreover it is seen that when the current is in phase with the induced voltage, thearmature-reaction field always lags the main field by 90oas shown in Fig. 35.25(a). This said

to be a cross-magnetizingfield. In the other words, armature-reaction for unity power factor

is distortional.

If we consider the armature-reaction field to act independently, this field induces

another voltage in each phase of the armature which lags the respective phase current by 90o.

Zero power Factor LaggingIn this case, the phase current lags the phase voltage by 90 o. This resultant armature-

reaction flux is moved backward by 90o. Thus, the direction of the armature-reaction flux is

now seen to be 180o behind the main filed flux as shown in Fig. 35.25(b). The armature-

reaction flux directly opposes and weakens the main field flux and this is said to be

demagnetizing.To keep the value of generated emf the same, field excitation will have to be

increased to compensate for this weakening.

Zero power Factor LeadingIn this case, the phase current leads the phase voltage by 90

o. This resultant armature-

reaction flux is moved forward by 90o. Thus, the direction of the armature-reaction flux is

now seen to be in phase with the main filed flux as shown in Fig. 35.25(c). This results in


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added main flux. Hence, armature-reaction is wholly magnetizing, which results in greater

induced emf.

To keep the value of generated emf the same, field excitation will have to be reduced

somewhat.

(a) For unity power factor (b) For zero lagging power factor

(c) For zero leading power factor

Fig. 35. 25

For intermediate power factor between unity power factor and zero lagging powerfactor, the effect is partly cross-magnetizing(or distortional) and partly demagnetizing. If

represents the phase angle between phase current and induced phase voltage, then sin

represents the demagnetizing component, while cos represents the cross-magnetizing

component.

For intermediate power factor between unity power factor and zero leading power

factor, the effect is partly cross-magnetizing (or distortional) and partly magnetizing. If

represents the phase angle between phase current and induced phase voltage, then sin

represents the magnetizing component, while cos represents the cross-magnetizing

component.

Synchronous Reactance [1/37.17/1425]From the above discussion, it is clear that for the same field excitation, terminal

voltage is decreased from its no-load valueE0to V(for lagging power factor). This is because

of

1.

drop due to armature resistance,IRa;

2.

drop due to leakage reactance,IXL; and

3. drop due to armature reaction.


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The drop in voltage due to armature reaction may be accounted for by assuming the

presence of afictitious reactanceXain the armature winding. The value of Xais such thatIXa

represents the voltage drop due to armature reaction.

The leakage reactanceXL(orXP) and the armature reactanceXamay be combined to

give synchronous reactanceXS.

HenceXS=XL+Xa.

Therefore, total voltage drop inan alternator under load is

SSaSa IjXRIjIXIR Z=+=+= )(

Where ZS is known as synchronous

impedance of the armature, the word

synchronous being used merely as an

indication that it refers to the working

conditions.

Hence, we learn that the vector

difference between no-load voltage E0

and terminal voltage Vis equal toIZS,

as shown in Fig. 35.26.

Vector Diagram of a Loaded Alternator [1/37.18/1426]

(a) unity power factor

ajIXEE +=0 ; La jIXIRVE ++=

aLa jIXjIXIRVE +++=0

)()(0 aLa IXIXjIRVE +++=

sa jIXIRVE ++= )(0

22

0 )()( sa IXIRVE ++=

(b) laging power factor

)sin()cos(0 sa IXVjIRVE +++=

22

0 )sin()cos( sa IXVIRVE +++=

(c) leading power factor

)sin()cos(0 sa IXVjIRVE ++=

220 )sin()cos( sa IXVIRVE ++=

Fig. 35.27Before discussing the diagrams, following symbols should be clearly kept in mind.


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E0= No-load emf. This being the voltage induced in the armature in the absence of

three factors discussed in Art. 35.16. Hence it is represents the maximum

value of the induced emf.

E = Load induced emf. It is the induced emf after allowing for armature reaction.Eis

vectorially less thanE0byIXa. Sometimes, it is written asEa.

V = Terminal voltage. It is vectorially less thanE0by IZSor it is vectorially less than

EbyIZ. where, 22 La XRZ += . It may also be written asZa.

I = Armature current/phase and = load power factor angle.

In Fig. 35.27(a) is shown thee case for unity power factor, in Fig. 35.27(b) for lagging power

factor and in Fig. 35.27(c) for leading power factor. All these diagrams apply to one phase of

a 3-phase machine. Diagrams for the other phases can also be drawn similarly.

Example 37.16. A 3-phase, star-connected alternator supplies a load of 10 MW at PF 0.85

lagging and at 11 kV (terminal voltage). Its resistance is 0.1 ohm per phase and

synchronous reactance 0.66 ohm per phase. Calculate the line value of EMF

generated.

Voltage Regulation [1/37.19/1427]It is clear that with change

in load, there is a change in

terminal voltage of an alternator.

The magnitude of this change

depends not only to the load but

also on the load power factor.

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.

100upregulation% 0

=V

VE

Voltage characteristics of an

alternator are shown in Fig. 35.29.

At leadingloads, the armature-reaction is magnetizingand tends to produce additional

generated voltage as load is applied, producing a negative voltage regulation.

At laggingloads, the armature-reaction is demagnetizingand its effect in reducing the

generated voltage, results in a rapid decrease in terminal voltage as load is applied, producing

apositive voltage regulation.

A sufficient leadingpower factor produces a voltage rise. Unity and laggingpowerfactors produce a dropin terminal voltage with application of load.

Determination of Voltage Regulation [1/37.20/1427]In the case of small machines, the regulation may be found by direct loading. The

procedure is as follows:

The alternator is driven at synchronous speed and the terminal voltage is adjusted to

its rated value V.


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The load is varied until the wattmeter and ammeter (connected for the purpose)

indicate the rated values of desired power factor.

Then the entire load is thrown off while the speed and fixed excitation are kept

constant.

The open-circuit or no-load voltageE0is read.

Hence, regulation can be found from: 100regulation%

0

= V

VE

In the case of large machines, the cost of finding the regulation by direct loading

becomes prohibitive. Hence other methods are used where all these methods differ chiefly in

the way the no-load voltageE0is found in each case.

1. Synchronous Impedance or EMF Method: It is due to Behn Eschenberg.

2. The Ampere-Turn or MMF Method: This is due to Rothert.

3. Zero power factor (PF) or Potier Method: It is due to Potier.

4. old A.I.E.E. Method

5. The American Standards Association method

All these methods require:

1. Armature (or stator) resistance,Ra,

2. Open-circuit or no-load characteristics

3.

Short-circuit characteristics (but zero PF lagging characteristic for Potier method).

Value ofRaArmature resistanceRaper phase can be measured directly by voltmeter and ammeter

method or by using Wheatstone bridge. However, under working conditions, the effective

value ofRais increased due to skin effect. The value ofRaso obtained by 60% or so to allow

for this effect. Generally, a value of 1.6 times the DC value is taken.

Open-Circuit (OC) CharacteristicsThis is plotted by running the machine on no-load and by noting the values of induced

voltage and field excitation current. It is just like theB-Hcurve.

Short-Circuit (SC) characteristicsIt is obtained by short-circuiting the armature (i.e. stator) winding through a low-

resistance ammeter. The excitation so adjusted as to give 1.5 to 2 times the value of full-load

current. During this test, the speed which is not necessarily synchronous, is kept constant.

Example 37.17(b). A 60 kVA, 200 V, 50 Hz, 1-alternator has effective resistance of 0.016

ohm and an armature leakage reactance of 0.07 ohm. Compute the voltage induced inthe armature when the alternator is delivering rated current at a load PF of (a) unity,

(b) 0.7 lagging, and (c) 0.7 leading.

Example 37.18(a). In a 50 kVA, star connected, 440 V, 3-phase, 50 Hz alternator, the

effective armature resistance is 0.25 ohm per phase. The synchronous reactance is 3.2

ohm per phase and leakage reactance is 0.5 ohm per phase. Determine at rated load

and unity PF: (a) internal EMF E, (b) no load EMF E0, (c) percentage regulation on

full-load, (d) value of synchronous reactance which replaces armature reaction.


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Example 37.19(a). Find the synchronous impedance and reactance of an alternator in which

a given field current produces an armature current of 200A on short-circuit and a

generated EMF of 50V on open-circuit. The armature resistance is 0.1 ohm. To what

induced voltage must the alternator be excited if it is to deliver a load of 100A at a p.f.

of 0.8 lagging, with a terminal voltage of 200V.

Problem. Find the synchronous impedance and synchronous reactance of an alternator in

which a given field current of 20A produces an armature current of 200A on short-circuit and a generated e.m.f of 50 V on open-circuit. The effective armature

resistance is 0.1 ohm. Find the voltage regulation if it is to deliver a load of 100A at

i) 0.8 p.f. lagging ii) 0.8 p.f. leading and iii) unity p.f. with a terminal voltage of

200V. Comment on the result.

Synchronous Impedance or EMF Method [1/37.21/1431]Following procedural steps are involved in this method:

1.

OCC is plotted from the given data as shown in Fig. 37.33(a).

2.

Similarly, SCC is drawn from the data given by the short-circuit test. It is a

straight line passing through the origin. Both these curves are drawn on a common

field-current base.

Consider a field currentIf1. The OC voltage corresponding to this field current is E1.

When winding is short circuited, the terminal voltage is zero. Hence, it may be assumed that

the whole of this voltageE1is being to circulate the armature short-circuit current I1against

the synchronous impedanceZs.

circuit)(short

circuit)(open

1

111

==

I

EZZIE

ss

3.

SinceRacan be found as discussed earlier,22

ass RZX =

4.

Knowing Ra and Xs, vector diagram Fig. 37.33(b) can be drawn for any load and

any PF.

Here, OD=E0; 220 BDOBE += ;

Or 220 )sin()cos( sa IXVIRVE +++=

100regulation% 0

=V

VE


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Voltage regulation for unity PF or leading PF can also be found in a similar way.

This method is not accurate because the value of Zsso found is always more than its value

under normal voltage conditions and saturation. Hence the value of regulation so obtained is

always more than that found from an actual test. That is why it is calledpessimisticmethod.

The value ofZsis not constant but varies with saturation. At low saturation, its value is larger

because then the effort of a given armature ampere-turns is much more than at highsaturation.

Now under short-circuit conditions, saturation is very low, because armature MMF is directly

demagnetizing. Different values of Zscorresponding to different values of field current are

also plotted in Fig. 37.33(a).

The value of Zsusually taken is that obtained from the full-load current in the short-circuit

test.

Here, armature reactanceXahas not been treated separately but along with leakage reactance

XL.

Example 37.21 (a). A 100 kVA, 300 V, 50 Hz, 3-phase star-connected alternator has

effective armature resistance of 0.2 ohm. The field current of 40 A produces short-

circuit current of 200 A and an open-circuit EMF of 1040 V (line value). Calculate the

full-load voltage regulation at, unity PF, 0.8 PF lagging, and 0.8 PF leading. Draw

phasor diagram.

Problem. Find the power angle () and the no load generated emf (Eo) when a 1500-KVA,

6.6-kV, 3-phase, Y-connected alternator having a resistance of 0.4 ohm, and a

reactance of 6 ohm per phase delivers full- load current at normal rated voltage and

0.8-pf lagging. Also draw the phasor diagram

Example 126 [2/ p.166] A 3-phase Y-connected alternator is of the following rating

164 KVA, 3600rpm, 2200V, 2 poles and 60 cycles.

For a particular value of field current of 31A, the maximum open circuit voltage is3200V, and the same excitation produces a short circuit current of 70A. Effective

resistance of the armature per phase is 0.60 ohm. Determine the voltage regulations

for unity power factor and also for 0.8 lagging power factor. (Use synchronous

impedance method).

MMF or Ampere-Turn Method 1/37.22/1439]This method also utilizes OC and SC data, but is the converse of the EMF method in

the sense thatarmature leakage reactance is treated as an additional armature reaction. In

other wards, it is assumed that the change in potential difference (PD) on load is due entirely

to armature reaction (and due to the ohmic resistance drop, which, in most cases, is

negligible). This fact is shown in Fig. 37.43.Now, ampere-turn (AT) required to produce Von full-load is the vector sum of the

following:


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(i) Field AT required to produce V(or

Ra is to be taken into account, then

V+IRacos) on no-load. This can be found

from OCC. and

(ii) Field AT required toovercome the

demagnetizing effect of armature reaction on

full-load. This value is found from short-circuit test. The field AT required to produce

full-load current on short-circuit balances the

armature reaction and the impedance drop.

The impedance drop can be neglected because Rais usually very small and Xsis also

small under short-circuit conditions. Hence, PF on short-circuit is almost zero lagging and the

field AT wholly demagnetizing.

In other words,the demagnetizing armature AT on full-load are equal and opposite

to the field AT required to produce full-load current on short circuit.

Now, if the alternator, instead of being on short-circuit, is supplying full-load current

at its normal voltage and zero PF lagging, then total field AT required are the vector sum of

(i)

The field AT=OA (as shown in Fig. 37.44a) necessary to produce normal

voltage (as obtained from OCC) and

(ii)

The field AT necessary to neutralize the armature reaction AB1(as shown

in Fig. 37.44a). The total field AT are represented by OB1in Fig 37.44(a)

and equals the vector sum of OAand OB1.

If the PF is zero leading, the armature reaction is wholly magnetizing. Hence, in that

case, the field AT required is OB2which is less than OAby the field AT=AB2 required to

produce full load current on short-circuit [Fig. 37.44(b)].

If the PF is unity, the armature reaction is wholly cross-magnetizing. Hence, in that

case, the field AT required is OB3 i.e.vector sum of OA and AB3which is drawn at right

angles to OA as in Fig. 37.44(c).

General Case [1/37.23/1440]Let us consider the general case when the PF has any value between zero (lagging and

leading) and unity. Field AT OAcorresponding to V(or V+IRacos) is laid of horizontally.

Then AB1 representing full-load (FL) short-circuit (SC) field AT=AB1 is drawn at angle of

(90o+) for a lagging PF. The total field AT are given by OB1 as in Fig. 35.45(a). For a


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02-Alternators


leading PF, SC AT=AB2 is drawn at an angle of (90o+) as shown in Fig. 37.45(b) and for

unity PF,AB3is drawn at right angles as shown in Fig. 37.45(c).

In those cases where the number of turns on the field coil is not known, it is usual to

work in terms of the field current as shown in Fig 37.46.

In Fig. 37.46 is shown the complete diagram along with OC and SC characteristics.

OA represents field current for normal voltage V. OC represents field current required for

producing FL current on SC.VectorAB=OCis drawn at an angle of (90

o+) to OA (if then PF is lagging). The total

field current is OBfor which the corresponding OCvoltage isE0.

100regulation% 0

=V

VE

This method gives results which are less than the actual results, that is why it is

sometimes referred to as optimisticmethod.

Zero PF or Potier Method 1/37.24/1444]This method is based on the separation of armature leakage reactance drop and the

armature reaction effects. Hence, it gives more accurate results. It makes use of the first two

methods to some extent.

The experiment data required is:


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(i) NL curve, and

(ii)

FL zero PF curve also called wattles load characteristic. It is the curve of

terminal volts against excitation when armature is delivered FL curve at zero

PF.

The reduction in voltage due to armature-reaction is found from above and (ii) and

voltage drop due to armature leakage reactance XL (also called Potier reactance) is found

from both. By combining these two,E0can be calculated.If we vectorially add to Vthe drop due to the resistance and leakage reactanceXL, we

getE. if toEis further added the drop due to armature reaction (assuming lagging PF) then

we getE0.

The zero PF lagging curve can be obtained

(a)

if a similar machine is available which may be driven at NL as a

synchronous motor at practically zero PF, or

(b)

by loading the alternator with pure reactors,

(c)

by connecting the alternator to 3-phase line with ammeters and wattmeters

connected for measuring current and power and by so adjusted the field

current that we get FL current with zero wattmeter reading.

PointB (Fig. 37.56) was obtained

in this manner when wattmeter wasreading zero. Point A is obtained from

the SC test with FL armature current.

Hence, OA represents field current

which is equal and opposite to the

demagnetizing armature reaction and

for balancing leakage reactance drop at

FL. Knowing these two points, FL zero

PF curveABcan be drawn as under.

From B, BH is drawn equal to

and parallel to OA. From H, HD is

drawn parallel to initial straight part of

NL curve i.e. parallel to OC, which is

tangential to NL curve. Hence, we get

point D no NL curve, which

corresponds to point B on FL zero PF

curve. The triangle BHD is known as

Potier triangle.

This triangle is constant for a

given armature current and hence can

be transferred to give us other points

like M, L etc. Draw DE perpendicular

to BH. The length DE represents the

drop in voltage due to armature leakagereactance XL i.e. IXL. BE gives field

current necessary to overcome

demagnetizing effect to armature

reaction at FL andEHfor balancing the

armature leakage reactance dropDE.


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P d b D M A M P 12 f 12

Let Vbe the terminal voltage on FL, then if we add to it vectorially the voltage drop

due to armature leakage reactance alone (neglecting Ra), then we get voltageE=DF(and not

E0). Obviously, field excitation corresponding toEis given by OF.NA (=BE) represents the

field current needed to overcome armature reaction. Hence, if we add NAvectorially to OF

(as in AT method) we get excitation forE0whose value can be read from NL curve.

In Fig. 37.56, FG (=NA) is drawn at an angle of (90o+) for a lagging PF (or it is

drawn at an angle of 90o- for leading PF). The voltage corresponding to this excitation isJK=E0.

100regulation% 0

=V

VE

The vector diagram is also shown separately in Fig. 37.57.

Assuming a lagging PF with angle , vector forIis drawn at angle of to V. IRa is

drawn parallel to current vector and IXLis drawn perpendicular to it. FG(=NA=BE in Fig.

37.56) representing field current equivalent to FL armature reaction, is drawn parallel to

current vector OI. The closing side OG gives field excitation for E0. Vector for E0 is 90o

lagging behind OG.DLrepresents voltage drop due to armature reaction.

Procedural Steps for Poteier method [1/37.25/1445]

1.

Suppose we are given V-the terminal voltage/phase.2.

We will be given or else we can calculate armature leakage reactance XLand hence

can calculateIXL.

3.

AddingIXL(andIRaif given) vectorially to V, we get voltageE.

4.

We will next find from NL curve, field excitation for voltageE. Let it beIf1.

5.

Further field currentIf2necessary for balancing armature reaction is found from Potier

triangle.

6. CombineIf1 andIf1vectorially (as in AT method) to getIf.

7.

Read from NL curve, the EMF corresponding to If. This gives us E0. Hence,

regulation can be found.

References[1] B. L. Theraja, A. K. Theraja, A Textbook of ELECTRICAL TECHNOLOGY in SI Units

Volume II, AC & DC Machines, S. Chand & Company Ltd., (Multicolour illustrative

Edition).

[2] A. F. Puchstein, T. C. Lloyd, A.G. Conrad, Alternating Current Machines, 1942,

Asia Publishing House, Third Edition (Fully revised and corrected Edition 2006-07).

[3] Jack Rosenblatt, M. Harold Friedman, Direct and Alternating Current Machinery,

Indian Edition (2nd

Edition), CBS Publishers & Distributors.

[4] A. E. Fitzgerald, Charles Kingsley, Jr. Stephen D. Umans, Electric Machinery, 5th

Editionin SI units, 1992 Metric Edition, McGraw Hill Book Company.

[5] Irving L. Kosow,Electrical Machinery and Transformers, Second Edition, Prentice Hall

India Pvt. Limited.

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02-Alternator on Load