alternators ii

ALTERNATORS

PART 2

PREPARED BY: JCM

Alternator On Load

• 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

is due to the following reasons:is due to the following reasons:

– Voltage drop due to armature resistance, Ra

– Voltage drop due to armature leakage reactance,

XL

– Voltage drop due to armature reaction

Alternator On Load

Armature Resistance

– The armature resistance/phase causes a voltagedrop/phase of IRa which is in phase with thearmature current. However, this voltage drop ispractically negligible.practically negligible.

Armature Leakage Reactance

– When current flows through the armatureconductors, fluxes are set up which do not crossthe air-gap but take different paths. Such fluxesare known as leakage fluxes.

Alternator On Load

Armature Leakage Reactance (cont...)

– The leakage flux is practically independent of

saturation but is dependent on I and its phase angle

with terminal voltage V.

– This leakage flux sets up an emf of self inductance

which is known as reactance emf which is ahead of I

by 900.

– Hence, armature winding is assumed to possess

leakage reactance, XL (also known as Potier reactance)

such that the voltage drop due to this equal IXL.

Alternator On Load

Armature Reaction

– As in dc generators, armature reaction is the effect

of armature flux on the main field flux.

– In the case of alternators, the power factor of the– In the case of alternators, the power factor of the

load has a considerable effect on the armature

reaction.

– In a 3-phase machine the combined mmf wave is

sinusoidal which moves synchronously. This mmf

wave is fixed relative to the poles, its amplitude is

proportional to the load current, but its position

depends on the pf of the load.

Alternator On Load

Armature Reaction (cont...)

– Unity Power Factor

• The armature flux is cross magnetizing. It is 900 space

degrees with respect to the poles.

• The result is that the flux at the leading tips of the poles

is reduced while it is increased at the trailing tips.

However, these two effects nearly offset each other

leaving the average field strength constant.

• Armature reaction for unity power factor is distortional.

Alternator On Load

Alternator On Load


– Zero Power factor Lagging

• The armature flux whose wave has moved backward by

900 is in direct opposition to the main flux. Hence, the

main flux is decreased.

• The armature reaction in this case is wholly

demagnetizing, with the result that due to the

weakening of the main flux, less emf is generated.

• To keep the value of the generated emf the same, field

excitation will have to be increased to compensate for

this weakening.

Alternator On Load

Alternator On Load


– Zero Power factor Leading

• Armature flux wave has moved forward by 900 so that it

is in phase with the main flux wave. This results in

added main flux.

• In this case, armature reaction is wholly magnetising,

which results in greater emf.

• To keep the value of generated emf the same, field

excitation will have to be reduced somewhat.

Alternator On Load

Alternator On Load


– Intermediate Power Factor

• For lagging power factor, the effect is partly distortional

and partly demagnetizing.

• For leading power factor, the effect is partly distortional

and partly magnetizing.

Synchronous Reactance

• From the above discussion, it is clear that for

the same field excitation, terminal voltage is

decreased or increased from its no-load value

Eg to V. This is because of:Eg to V. This is because of:

– Drop due to armature resistance, IRa

– Drop due to leakage reactance, IXL

– Drop due armature reaction


• The drop in voltage due to armature reaction

may be accounted for by assuming the

presence of a fictitious reactance Xar in the

armature winding. The value of Xar is such thatarmature winding. The value of Xar is such that

I Xar represents the voltage drop due to

armature reaction.

• The leakage reactance XL and the armature

reactance may be combined to give

synchronous reactance XS.


arLS XXX +=• Therefore, the total voltage drop in an • Therefore, the total voltage drop in an

alternator under load is:

( )

impedance us synchronothe called is Zwhere

IZ

jXRI

jIXIRV

S

S

Sa

Sadrop

:

=

+=

+=

Vector Diagrams of Loaded

Alternator


Alternator

Example: A 3-phase, star-connected alternator

supplies a load of 10 MW at 0.85 pf lagging at

11 kV. Its resistance is 0.1 ohm per phase and

synchronous reactance of 0.66 ohm. Calculatesynchronous reactance of 0.66 ohm. Calculate

the line value of emf generated.


Alternator

Solution:

line)-(line kV

laggingfp

MW

11

..85.0

10


Alternator

( )

MW

SII

kV VL

788.3185.0

10

03509.603

11

0

*

00

−∠

===

∠=∠=φ

( )( )

( )( )

( )( ) kVkVE

kV

kVjAE

A

kVV

SII

LL

LL

G

G

L

AL

4757.116255.63

.7154.26255.6

03509.666.01.0788.31487.617

.788.31487.617

0113

85.0

3

0

00

0

0*

==

∠=

∠+Ω+−∠=

−∠=

∠===

φ


Alternator

Alternative Solution:

( )( )( )( )

( ) ( )

( )( ) kVkVE

kV

kVE

VXI

VRI

LLG

G

SL

AL

4757.116255.63

6255.6

212.58sin5414.407788.31sin7487.61212.58cos5414.407788.31cos7487.613509.6

5414.40766.0487.617

7487.611.0487.617

2200

==

=

+−+++=

==

==

φ

Voltage Regulation

• 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 on the load but also on the load poweronly on 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.”

Voltage Regulation

%100% ×−

=FL

FL'L

V

VVVR

Note:Note:

• VNL – VFL is the arithmetical differenece and not the

vectorial one.

• In the case of leading load pf terminal voltage will fall

on removing the full-load. Hence, regulation is negative

in that case.

• The rise in voltage when full-load is thrown off is not

the same as the fall in voltage when full-load is applied.

Determination of Voltage

Regulation

• 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 alternator is driven at synchronous speed and

the terminal voltage is adjusted to its rated value.

– The load is varied until the wattmeter and

ammeter (connected for the purpose) indicate the

rated values at desired pf.


Regulation

– Then the entire load is thrown off while the speed

and excitation are kept constant.

– The open-circuit or no-load voltage is read.

– The regulation can be found from:– The regulation can be found from:

%100% ×−

=FL

FL'L

V

VVVR


Regulation

• In the case of large machines, the cost of

finding the regulation by direct loading

becomes prohibitive. Hence, other indirect

methods are used.methods are used.

• It will be found that all these methods differ

chiefly in the way the no-load voltage is found

in each case.


Regulation

• Indirect Methods

– Synchronous Impedance or EMF Method. It is due

to Behn Eschenberg.

– The Ampere-Turn or MMF Method. This method is– The Ampere-Turn or MMF Method. This method is

due to Rothert.

– Zero Power Factor or Potier Method

• All these methods require :

– Armature Resistance, RA

– Open Circuit/No-Load Characteristics

– Short-Circuit Characteristics (but zero power factor lagging

characteristic for Potier Method


Regulation

• Value of RA

– Armature resistance per phase can be measured

directly by voltmeter-ammeter method or by

using Wheatstone bridge. However, underusing Wheatstone bridge. However, under

working conditions, the effective value of RA is

increased due to “skin effect”. The value of RA so

obtained is increased by 60% or so to allow for

this effect.

– Generally, a value 1.6 times the dc value is taken.


Regulation

• Open-Circuit (OC) Characteristic

– As in dc machines, this is plotted by running the

machine on no-load and by noting the values of

the induced voltage and field excitation current.the induced voltage and field excitation current.

• Short-Circuit (SC) Characteristic

– It is obtained by short-circuiting the armature

windings through a low resistance ammeter. The

excitation is adjusted as to give 1.5 to 2 times the

value of the full-load current. During this test, the

speed which is not necessarily synchronous is

kept constant.


Regulation

Example: The effective resistance of a 2200 V, 50

Hz, 440 kV, 1-phase alternator is 0.5 ohm. On

short circuit, a field current of 40 A gives the

full-load current of 200 A. The emf on openfull-load current of 200 A. The emf on open

circuits with the same field excitation is 1160

V. Calculate the synchronous impedance and

reactance.


Regulation

Ω===1160

:

V

Solution

Ω=−=−=

Ω===

78.55.08.5

8.5200

1160

2222 RZX

I

VZ

SS

SC

OCS


Regulation

Example: A 100-kVA, 3000-V, 50-Hz, 3-phase,

star-connected alternator has effective

armature resistance of 0.2 ohm. The field-

current of 40 A produces a short-circuitcurrent of 40 A produces a short-circuit

current of 200 A and an open-circuit emf of

1040 V (line value). Calculate the generated

line emf at rated load and 0.8 pf lagging.


Regulation

I

VZ

Solution

SC

OC

S 00.3200

3

1040

:

Ω=== φ

φ

( )

( )( ) VE

V

jE

RZX

LLG

G

SS

068.3066196.17703

415.1196.1770

03

300099.22.087.36245.19

99.22.03

0

00

2222

==

∠=

∠

++−∠=

Ω=−=−=

φ

φφφ

Synchronous Impedance Method

• Following procedural steps are involved in this

method:

– OCC is plotted from the given data.

– SCC is plotted from the given data by the short-– SCC is plotted from the given data by the short-

circuit test. Both these curves are drawn on a

common field current base.

– XS is obtained from

22 RZX SS −=


• Knowing RA and XS, a vector diagram can

drawn for any load and any power factor.


• This method is not accurate because the value

of ZS so found is always more than its value

under normal voltage conditions and

saturation. Hence, the value of regulation sosaturation. Hence, the value of regulation so

obtained is always more than that found from

an actual test. That is why it is called a

pessimistic method.

• The value of ZS usually taken is that obtained

from full-load in the short-circuit test.


• Here, the XAR has not been treated separately

but along with leakage reactance XL.

Example: The following test results are obtained from a

3-phase, 6000-kVA, 6.6kV, star connected, 2-pole, 503-phase, 6000-kVA, 6.6kV, star connected, 2-pole, 50

Hz turbo-alternator: With a field current of 125 A, the

open-circuit voltage is 8kV at the rated speed; with

the same field current and rated speed, the short-

circuit current is 800 A. At rated full-load, the

resistance drop is 3%. Find the regulation of the

alternator on full-load and at a pf of 0.8 lagging.


Ω==

6.6

774.5800

38000

:

φS

kV

Z

Solution

( )

Ω=−=

Ω==

=

=

770.5218.0774.5

218.0864.524

315.114

315.1143

6.603.0

22

φ

φ

φ

S

A

AL

X

R

VkV

RI


( )( ) 03

6.677.5218.0869.36864.524

:...)(

00 ∠++−∠=kV

jE

contSolution

Gφ

%536.78%100102.3464

102.346466.6184%

373.2266.6184

30

=×−

=

∠=

VR

V


Example: A 3-phase 50-Hz star-connected 2000

kVA, 2300 V alternator gives a short-circuit

current of 600 A for a certain field excitation.

With the same excitation, the open circuitWith the same excitation, the open circuit

voltage was 900 V. The resistance between a

pair of terminals was 0.12 Ω. Find full-load

regulation at 0.8 pf leading.


Ω== 866.0600

3900

:

φSZ

Solution

Ω=−=

Ω=

=

Ω==

861.0096.0866.0

096.02

12.06.1

866.0600

22

φ

φ

φ

S

A

S

X

R

Z


( )( ) 03

3.2861.0096.0869.36044.502

:...)(

00 ∠++∠=kV

jE

contSolution

Gφ

%981.11%100906.1327

906.1327811.1168%

7.18811.1168

30

−=×−

=

∠=

VR

V


Example: A 3-phase, star-connected alternator is

rated at 1600 kVA, 13500 kV. The armature

resistance and synchronous reactance are 1.5

Ω and 30 Ω respectively per phase. CalculateΩ and 30 Ω respectively per phase. Calculate

the % regulation for a load of 1280 kW at 0.8

leading power factor.


( )( )

( )( ) 013500

305.187.36427.68

87.36427.6887.36135003

8.01280

:

00

00

∠++∠=

∠=∠=

jE

A

kW

I

Solution

L

( )( )

%99.11%100

3

135003

13500624.6859

%

382.14624.6859

03

13500305.187.36427.68

0

00

−=×−

=

∠=

∠++∠=

VR

V

jEGφ


Example: The effective resistance of a 1200-kVA,

3.3 kV, 50-Hz, 3-phase, Y-connected alternator

is 0.25 Ω per phase. A field current of 35 A

produces a current of 200 A on short-circuitproduces a current of 200 A on short-circuit

and 1.1 kV on open circuit. Calculate the

power angle and p.u change in magnitude of

the terminal voltage when the full load of

1200 kVA at 0.8 pf lagging is thrown off. Draw

the corresponding phasor diagram.


175.3200

31100

:

Z

Solution

S Ω==φ

( )( ).87.36946.20987.36

33003

1200

165.325.0175.3

200

00

22

AkVA

I

X

L

S

−∠=−∠=

Ω=−=φ


( )( ) 00 03

3.3165.325.087.36946.209

:...)(

∠++−∠=φ

kVjE

contSolution

G

0

0

034.12

259.0256.1905

256.1905644.2398.

034.12644.2398

=

=−

=

∠=

δ

up

V

MMF or Ampere-Turns Method

(Rothert’s Method)

• This method also utilizes OC and SC data, but

is the converse of the Synchronous Impedance

(EMF) method in the sense that the armature

leakage reactance is treated as an additionalleakage reactance is treated as an additional

armature reaction.

• Therefore, it is assumed that the change in

terminal voltage on load is due entirely to

armature reaction (and due to the ohmic

resistance drop which, in most cases, is

negligible).



• In the MMF method, use is made of a vector

diagram of magneto motive forces. The theory

upon which this method is developed is basedupon which this method is developed is based

on the assumption that for every voltage

vector of the alternator diagram there is a

corresponding magneto motive force.



• Calculation procedure for MMF method

– Terminal voltage per phase (V) is used as the

reference phasor.

– Lay off the armature IARA drop in phase with the– Lay off the armature IARA drop in phase with the

current, at the pf angle for which the regulation is

desired. Determine the field current If’ required to

produce the voltage E1 using the OC curve.

– It is assumed that on short-circuit all the excitation

is opposed by the mmf of armature reaction and

armature reactance.




(cont...)

– Hence A represents the mmf (or field current)

required to produce rated current on short circuit.required to produce rated current on short circuit.

It is the field current required to overcome the IAXS

drop and it is constructed opposite to the current

IA.

– The excitation If required to produce terminal

voltage at no load is then the vector sum of If’

and A. EO is obtained from the OC curve.




(cont...)


(Rothert’s Method)Example: A 3.5-MVA, Y-connected alternator rated at

4160 V at 50-Hz has the OC characteristic given by

the following data:

Field 50 100 150 200 250 300 350 400 450

A field current of 200 A is found necessary to circulate

full-load on short-circuit of the alternator. Calculate

the full-load regulation at 0.8 pf lagging.

Current, A

EMF, V 1620 3150 4160 4750 5130 5370 5550 5650 5750



Solution:

• Neglect RA

• It is seen from the given data that for normal

voltage of 4160 V, the field current needed isvoltage of 4160 V, the field current needed is

150 A.

• The field current necessary to circulate FL

current is 200 A.



Solution (cont...):



Solution (cont...):

( ) ( ) ( )( ) ( )I

87.3690cos1502002150200

87.368.0cos

22

0

+−+=

== −θ

( ) ( ) ( )( ) ( )A.

If

85.313

87.3690cos1502002150200

=

+−+=

The generated emf EO corresponding to this

excitation as found from OCC if drawn is 5440

V.



Solution (cont...):

%10041605440

% ×−

=VR

%78.30

%1004160

%

=

×=VR


(Rothert’s Method)Example: The open-circuit and short-circuit test

readings for a 3-phase, star-connected, 1000-kVA,

2000 V, 50-Hz, synchronous generator are:

Field 10 20 25 30 40 50

The armature effective resistance is 0.2 ohm per phase.

Estimate the full-load %VR at 0.8 pf leading.

Current, A

OC

terminal V

800 1500 1760 2000 2350 2600

SC current

A

____ 200 250 300 ____ ____



Solution:

The OCC and SCC curves are plotted:



Solution (cont...):

Vvoltage phase load-Full 70.11543

2000==

( )( )( ) ( ) VE

A 3

1000kVAcurrent load-Full

4.12016.02.07.2888.02.07.2887.1154

7.2882000

3

22 =××+××+=

==



Solution (cont...):

• From OCC curve, the field excitation necessary

to produce E is 32 A.

• From SCC curve, the field excitation necessary• From SCC curve, the field excitation necessary

to produce full-load current is 29 A.



Solution (cont...):

( ) ( ) ( )( ) ( )I 78.54cos293222932

78.54

22

0

−+=

=φ

( ) ( ) ( )( ) ( )A.

If

19.28

78.54cos293222932

=

−+=

The generated emf EO corresponding to this

excitation as found from OCC if drawn is 1080

V per phase or 1870.61 V line-to-line.



Solution (cont...):

%100200061.1870

% ×−

=VR

%47.6

%1002000

%

−=

×=VR

Zero Power Factor Method (Potier

Method)

• This method is based on the separation of

armature-leakage reactance drop and the

armature reaction effects. Hence, it gives

more accurate results.more accurate results.

• The experimental data required is:

– No-load curve

– Full-load zero pf curve (not the SCC) also called

wattless load characteristic. It is the curve of

terminal voltage against excitation when armature

is delivering FL current at zero pf.


Method)

• The zero pf curve can be obtained:

– If a similar machine is available which may be

driven at no-load as a synchronous motor at

practically zero pfpractically zero pf

– By loading the alternator with pure reactors

– By connecting the alternator to a 3-phase line

with ammeters and wattmeters and so adjusting

the field current that we get full-load current with

zero wattmeter reading.


Method)


Method)

• Procedural Steps for Potier Method

– Suppose we are given V- the terminal voltage per

phase.

– We will be given or else we calculate armature– We will be given or else we calculate armature

leakage reactance XL and hence can calculate IXL.

– Adding IXL (and IRA if given) vectorially to V, we get

voltage E.

– We will next find from OC curve, field excitation

for voltage E. Let this be If1.


Method)

• Procedural Steps for Potier Method (cont...)

– Further, field current If2 necessary for balancing

armature reaction is found from Potier triangle.

– Combine If1 and If2 vectorially to get If.– Combine If1 and If2 vectorially to get If.

– Read from OC curve the emf corresponding to If.

This gives us EO. Hence, regulation can be found.


Method)Example: A 3-phase , 6000-V alternator has the following OCC at

normal speed:

Field

Current, A

14 18 23 30 43

Terminal 4000 5000 6000 7000 8000

With armature short-circuited and FL current flowing the field

current is 17 A and when the machine is supplying FL of 2000-

kVA at zero pf, the field current is 42.5 A and terminal voltage

is 6000 V. Determine the field current required when the

machine is supplying the full-load at 0.8 pf lagging.

Terminal

volts, V

4000 5000 6000 7000 8000


Method)

Solution:


Method)

Solution (cont...):

In the Potier Triangle BDH, line DE represents the leakage

reactance drop (IXL) and is (by measurement) equal to 450 V.

( )( )E 87.126cos10.3464450245010.3464 022 −+=

From OCC curve, it is found that the field amperes required for

this voltage = 26.5 A.

Field amperes required for balancing armature reaction = BE =

14.5 A (by measurement from Potier triangle BDH)

( )( )V.

E

41.3751

87.126cos10.3464450245010.3464 022

=

−+=


Method)

Solution (cont...):

( ) ( ) ( )( ) ( )A.

If

82.37

38.132cos5.145.2625.145.26

38.132

22

0

=

−+=

=φ


Method)Example: A 600-kVA , 3300-V, 8-pole, 3-phase alternator has the

following characteristics:

Amp-

Turns per

pole

4000 5000 7000 1000

Terminal 2850 3400 3850 4400

There are 200 conductors in series per phase. Find the full-load

voltage regulation at 0.8 pf lagging having given that the

inductive voltage drop at full load is 7% and that the

equivalent armature reaction in amp-turns per pole = 1.06 X

ampere-conductors per phase per pole.

Terminal

volts, V

2850 3400 3850 4400


Method)

Solution:

- OC terminal voltages are first converted into phase voltages

and plotted against field amp-turns.


Method)

phase per pole per turns-Amp ingDemagnetiz

AkVA

current load-Full 97.10433003

600=

×=

( )( )( )

( )( ) V 0.07 drop reactance Leakage

V 3

3300voltage phase 'ormal

AT 1.06

pf zero at load-full for

4.13326.1905

26.1905

71.27818

20097.104

==

==

==


Method)

( )( )V.

E

17.1988

87.126cos26.19054.13324.13326.1905 022

=

−+=

From OCC curve, we find that 1988.17 V corresponds to 5100From OCC curve, we find that 1988.17 V corresponds to 5100

AT.


Method)

( ) ( ) ( )( ) ( )AT. 7208.82

mmfO

=

−+=

=

95.129cos71.27815100271.27815100

95.129

22

0φ

AT. 7208.82 =

From OCC curve, it is found that this corresponds to an OC

voltage of 2242 V per phase.

%67.17

%10026.1905

26.19052242%

=

×−

=VR


Method)Example: The following figures give the open-circuit and full-load

zero pf saturation curves for a 15000 kVA, 11000 V, 3-phase,

50-Hz star-connected turbo-alternator:

Field Amp-

turns in 103

10 18 24 30 40 45 50

Find the armature reaction, the armature reactance and the

synchronous reactance. Deduce the regulation for full-load at

0.8 pf lagging.

OC terminal,

kV

4.9 8.4 10.1 11.5 12.8 13.3 13.65

Zero pf full-

load line kV

____ 0 ____ ____ ____ 10.2 ____


Method)

Solution:

- First OCC is drawn between phase voltages

and field ampere-turns.

- Full-load zero pf curve can be drawn because- Full-load zero pf curve can be drawn because

two points are known, i.e., A(18,0) and C(45,

5890). Other points on this curve by

transferring the Potier triangle.


Method)


Method)

Solution (cont...):

– In the Potier Triangle CDE, line EF = GH represents the

leakage reactance drop (IXL) and is (by measurement)

equal to 640 V.

– Field ampere-turns required for balancing armature– Field ampere-turns required for balancing armature

reaction = CF = 15700 AT (by measurement from Potier

triangle CDE).

– Short-circuit ampere-turns required = OA = 18000 AT.


Method)

( )( )

813.03.787

640

3.787110003

15000

LX

A. kVA

current load-Full

Ω==

==

.

16.63.787

38400

813.03.787

φφ

φ

SSA

S

L

ZX negligible is R As

Z

X

≅

Ω==

Ω==


Method

( )( )V.

E

28.6754

87.126cos85.6350640264085.6350 022

=

−+=

From OCC curve, we find that 6754.28 V corresponds to 30800From OCC curve, we find that 6754.28 V corresponds to 30800

AT.


Method

( ) ( ) ( )( ) ( )AT. 42806.73

mmfO

=

−+=

=

22.131cos308001570021570030800

22.131

22

0φ

From OCC curve, it is found that this corresponds to an OC

voltage of 7540 V per phase.

%72.18

%10085.6350

85.63507540%

=

×−

=VR

alternators ii

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