two reaction theory

03-Alternators

Prepared by Dr. M. A. Mannan of 11

Alternators

Two Reaction Analyses [2, Ch. 16] A multipolar machine with cylindrical rotor has a uniform air-gap, because of which

its reactance remains the same, irrespective of the spatial position of the rotor. However, a synchronous machine with salient pole or projecting poles has non-uniform air-gap due to which its reactance varies with the rotor position.

Blondel Two-Reaction Method [2/150/p.198]

To analyze mathematically and vector diagram of an alternator, it is inherently assumed that the field and armature fluxes are sinusoidally distributed in the air gap. There is a tendency for this method to give results that are somewhat higher than those of actual test.

Sinusoidally flux distribution is nearly true for fields with distributed iron and copper (non-salient pole m/c or smooth cylindrical m/c), but it is not true for salient-pole machine.

In order to remove the criticism of results for salient-pole machines, Professor Andre Blondel published in 1904 the two reaction method. Division of Armature MMF into Sine and Cosine Components [2,150/p.198]

At zero PF lagging load, the armature-current distribution is such that the armature MMF directly opposes the MMF of the main field.

At unity PF load, the armature MMF crowds the flux backward against the rotation. This causes the trailing pole tips to be strengthened and the leading tips to be weakened.

When alternator is operating at normal PFs of 60 to 90 percent, lagging, a combination of both these effects takes place.

Fig. 144 shows this in greater detail. The part of armature ATs which weakens the main flux is shown in diagonal sectioning, and that part which strengthens it is cross-hatched.

The resulting demagnetizing ATs can be found by subtracting the latter (strengthens) area (A2) from the former (weekens) area (A1) and by dividing by the circumferential width ψτ of the pole shoe.

Fig. 144 shows the position of the MMF waves in space w.r.t. the field pole of an alternator when the armature current is lagging the voltage E0, generated by the flux, by an angle θ’ of electrical degree.

This angle θ’ of the time diagram is equal to the angle θ’ on the space diagram of Fig. 144. That is, if the armature current is in phase with E0, the maximum value of the wave A will be shifted to a point 90o to the right (or left, depending on direction of motion of the pole) of the centre line of the pole.

With an armature current that lags E0 by 90o, the maximum value of A will be coincide with the centre line of the poles.

Derivation of the demagnetizing action of the armature reaction for the general case

will be based on the notations and symbols shown in Fig. 144. Expressing τ, θ’, and x in radians, ψ in percent, then

∫= +2

'01 sin

ψπθ xdxAA ;

'2

'

01 ]cos[ψπ

θ +−= xAA ; )]

2'cos(1[1ψπθ +−= AA

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The area of the small part

∫∫−−−

== 2'

0

)'2

(

02 sinsinψπθθψπ

xdxAxdxAA ; 2'

02 ]cos[ψπθ −

−= xAA ;

)]2

'cos(1[2ψπθ −−= AA

Expanding

]2

sin'sin2

cos'[cos1ψπθψπθ −−= AAA

]2

sin'sin2

cos'[cos2ψπθψπθ +−= AAA

]}2

sin'sin2

cos'[cos{]2

sin'sin2

cos'[cos21ψπθψπθψπθψπθ +−−−−=− AAAAAA

]2

sin'sin2

cos'[cos]2

sin'sin2

cos'[cos21ψπθψπθψπθψπθ ++−−−=− AAAAAA

2sin'sin

2cos'cos

2sin'sin

2cos'cos21

ψπθψπθψπθψπθ AAAAAA +++−=−

2sin'sin221

ψπθAAA =−

Dividing this total area by the base ψπ to obtain the average demagnetizing ATs:

ψπ

ψπ

θψπ

2sin2

'sin21 AAA=

− Demagnetizing AT.

This could be written 'sin21 θψπ

AKAA

AD =−

= where ψπ

ψπ2

sin2=K [188]

03-Alternators


This shows that the demagnetizing ATs for a given m/c are proportional to sinθ’. Accordingly, we may conclude that the armature reaction ATs consists of two parts.

One part is a function of Asinθ’, which is demagnetizing. The second part, which is a function of Acosθ’, is cross-magnetizing.

A is the crest (or peak) value of the armature reaction ATs per pole. To calculate the alternator regulation, using Fourier series analysis the value of A is obtained as follows:

pda kk

PZI

NIA2

9.0poleperreactionarmatureof ==

Hence the average demagnetizing ATs per pole are

'sin)2sin2

29.0('sin2

sin2

29.02

sin2'sin θ

ψπ

ψπ

θψπ

ψπ

ψπ

ψπ

θ pda

pda

D kkP

ZIkk

PZI

AA ===

'sinθCkkP

ZIA pd

aD = (188)

where, ψπ

ψπ2

sin2

29.0

=C

The Cross-Component [2/152/p.201]

It is obvious that any sine wave can be separated into two components in quadrature, yielding a resultant equal to the original. In fact, the analysis just followed showed a wave of armature MMF separated into components proportional to sinθ’ and cosθ’. Such components are shown in Fig. 145.

The cross-magnetizing component neither increases nor decreases the field excitation as a whole, but increases the MMF acting over one half of each pole face and decrease by an equal amount the MMF acting over the other half of each pole face. This is shown in Fig. 146(a).

The shaded portion shows approximately that part of the cross MMF that is really effective.

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In Fig. 146(b) the MMF of the cross component is shown and a curve of the flux that it would produce. The greatly increased reluctance of the air path between the salient poles is taken into consideration.

It is assumed that the cross field is of sinusoidal (neglecting harmonics) distribution in space. It rotates at synchronous speed and is in a fixed position w.r.t. the poles.

It will therefore generate a voltage Ec in the armature of fundamental frequency. It would be possible to determined component of the cross-field flux from a Fourier analysis. This approach will not be used here. Instead, a determination will be made of the average value of a fundamental MMF space wave which has the same area as the shaded position between 0 and π of Fig. 146(a).

Evaluation Ac [2/153/p. 203]

xAa sin'cosθ=

∫=∫= 2020 sin'cos22d)Area(shadeψπψπ

θ xdxAadx

[ ] radiansMMFin2

cos1'cos2cos'cos2d)Area(shade 20 ⎥⎦

⎤⎢⎣⎡ −=−=

ψπθθψπ

AxA

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⎥⎦⎤

⎢⎣⎡ −==

2cos1'cos2AreaMMF Average ψπ

πθ

πA

The average value of the cross MMF per pole is

'cos''cos2

cos12 θθψππ

AKAAc =⎥⎦⎤

⎢⎣⎡ −= where, ⎥⎦

⎤⎢⎣⎡ −=

2cos12' ψπ

πK

By substituting the value of pda kk

PZI

A2

9.0=

'cos)2

cos1(9.0 θψππ ⎥⎦

⎤⎢⎣⎡ −= pd

ac kk

PZI

A

'cosθCkkP

ZIA pd

ac = (193)

where )2

cos1(9.0 ψππ

−=C

It should be noted that the value of Ac is based on an MMF wave having the same

area as the two shaded sections of Fig. 146(a). The Vector Diagram [2/154/p.204]

The diagram of Fig. 147 can be built up by noting the following relationship. When the generator delivers a current I, at a terminal voltage V and at a PF angle θ,

there will be set up an armature resistance drop IRa and a leakage reactance drop IXL in phase and in quadrature, respectively, with the current.

These latter values add vectorially to the terminal voltage to give the EMF E, which is generated in armature by the air-gap flux. The generated EMF can be split into two components, in quadrature with each other, of which Ef is due to the main-pole flux and Ec is due to the cross-flux set up in the interpole space and pole edges by the cross component of armature reaction.

To generate Ef requires a resultant MMF along the axis of the field poles which is designated Mf. The initial MMF of the field winding alone is M0, but the direct component of armature reaction AD is subtracted therefrom to yield Mf. This subtraction, of course, presumes a lagging current; for leading current the so-called demagnetizing component would add to that of the field winding.

Fig. 147. Complete vector diagram of alternators showing voltages and MMFs.

03-Alternators


The excitation M0 is presumed to result in a NL voltage of E0 as read from the saturation curve, and hence the regulation would be

%100upregulation% 0 ×−

=∴V

VE

The cross component of armature reaction sets up a flux cut by the armature conductors and so generates the voltage Ec, which may be treated as a voltage drop.

The demagnetizing component of armature AD is an MMF subtracting arithmetically from the field-winding M0.

Angle (φ) calculation between internal induced voltage (E) and NL voltage (E0) The terminal voltage per phase of a generator is designated V, and the regulation is

desired at a load of I, lagging θ degrees behind V. The effective resistance of the armature and leakage reactance are known. V, I, IRa

and IXL can be laid off on the vector diagram, and E can be calculated αθθ +='

v= volts/AT/pole generated on the lower part of the saturation curve. Then it is assumed that

)cos( αθ +== CkkPZIvvAE pdcc here let II a =

But φsinEEc = ; hence, )cos(sin αθφ +== CkkPZIvEE pdc

Or, E

CkkPZIv

EE pd

c)cos(

sinαθ

φ+

==

But λφαθ +=+ ; E

CkkPZIv pd )cos(

sinλφ

φ+

=

φλφλφ

φλφ

sinsinsincoscos

sin)cos( −=

+= Ckk

PZIvCkk

PZIvE pdpd

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= λ

φλ

φλφ

φλφ sin

tancos

sinsinsin

sincoscos Ckk

PZIvCkk

PZIvE pdpd

λφλ sin

tancos Ckk

PZIvCkk

PZIvE pdpd −=

φλλ

tancossin Ckk

PZIvCkk

PZIvE pdpd =+

λφλ costan]sin[ CkkPZIvCkk

PZIvE pdpd =+

λ

λφ

sin

costan

PCkvZIk

E

PCkZIk

v

pd

pd

+=

Since, )sin)(cos( θθ jjXRIVE La −++= The angle (λ-θ) can be determined and likewise the angleλ. Substituting in Eq. (199),

the value of φ is determined and

03-Alternators


λφθ +=' The determination of the angle θ′ locates the direction of the voltage Ef w.r.t. I. This

provides the necessary step in the construction of the diagram of Fig. 147. Modified Vector Diagram [2/156/p.209]

The Blondel two-reaction theory, as just presented, is closely allied to the original form of this development.

An extension of this theory by Doherty and Nickle provides a more complete and accurate analysis.

In deriving the separate components of armature reaction for the Blondel theory, it was found that the directly demagnetizing portion was a function of Asinθ’ and the cross-magnetizing component depended upon Acosθ’.

Since armature reaction ATs A naturally involve armature current, the first step in this process is the association of the current with the terms Asinθ’ and Acosθ’. The vector diagram of Fig. 147 is then redrawn as shown in Fig. 148.

The directly demagnetizing effect resulted in a change in voltage represented by a

difference in the length of Ef and E0. Since it is a function of Asinθ’ and is in quadrature with the component of armature current represented by Isinθ’ it follows that the change in voltage form E0 to Ef might be represented by a fictitious reactance drop.

'0 df IXEE =−

Subscript “d” designated for representing direct components. Suppose Id indicated the direct component of current I, which is in quadrature with the position of the NL voltage.

03-Alternators


In similar manner we find that the cross-magnetizing component Ec will be considered a fictitious reactance drop since it is proportional to cosθ′ and in quadrature with the current component Iq=Icosθ′.

'qqc XIE =

Xq’ is cross reactance.

The idea of associating components of a voltage drop with suitable components of current will be carried a step further as shown in Fig. 149.

Here the leakage reactance drop will be separated into components associated with suitable current components so that vectorially

LqLdL XIXIIX += As shown in Fig. 149, the effect of the reactance Xq

’ merely replaces the effect of the voltage Ec, but by grouping the reactances associated with the current component Iq we obtain a new reactance

Lqq XXX += ' Similarly, associating the reactive drops, using the direct component of current, we

obtain a new reactance Ldd XXX += '

By making of these combined terms we have a comparatively simple diagram as shown in Fig. 150.

Fig. 150. Final step in the transition, resulting in the two-reaction diagram built up with direct axis synchronous reactance ( Ldd XXX += ' ) and quadrature axis synchronous reactance ( Lqq XXX += ' ). The resistance drop can also be divided into direct and quadrature components, or in many cases this can be neglected entirely. Its relative size is exaggerated here.

Now, if Xd and Xq are known the vector diagram can be constructed for the determination of regulation.

An expression for the OC voltage E0 can be obtained from the following relationship. By trigonometric considerations

(motoring)cossin

'tan

g)(generatincossin

'tan

a

q

a

q

IRVIXVIRVIXV

−

−=

+

+=

θθ

θ

θθ

θ

Angle (α) calculation between terminal voltage (V) and NL voltage (E0)

03-Alternators


Also 'sin;'cos θθ IIII dq == then )sin();cos( θαθα +=+= IIII dq From the Fig. 150,

VIRIX

VRIXI aqadqq 'sin'cos

sinθθ

α−

=−

=

VRXI aq )'sin'cos(

sinθθ

α−

=

Substituting (motoring)'

g)(generatin'αθθαθθ

−=+=

in this equation for generating and expanding to sine and cosine functions,

)]sin()cos([sin αθαθα +−+= aq RXVI

}]cossincos{sin}sinsincos{cos[sin θααθαθαθα +−−= aq RXIV ]cossincossinsinsincoscos[sin θααθαθαθα aaqq RRXXIV −−−=

αθθαθθα sin]cossin[cos]sincos[sin aqaq RXIRXIV +−−= αθθαθθα cos]sincos[sin]cossin[sin aqaq RXIRXIV −=++

αθθαθθ cos]sincos[sin}]cossin{[ aqaq RXIRXIV −=++

(motoring)]cossin[

]sincos[cossintan

g)(generatin]cossin[

]sincos[cossintan

θθθθ

ααα

θθθθ

ααα

aq

aq

aq

aq

RXIVRXI

RXIVRXI

+−

−==

++

−==

This is fundamental equation and it will be used when constructing the two-reaction diagram of the alternator.

Also (motoring)cos

g)(generatincos

0

0

ddaq

ddaq

XIRIVE

XIRIVE

−−=

++=

α

α

Direct- and Quadrature-axis Synchronous reactance [2/157/p.212]

A direct-axis quantity is one whose magnetic effect is centered on the axis of the main poles. Direct-axis mmf’s act on the main magnetic circuit.

A quadrature-axis quantity is one whose magnetic effect is centered on the interpoler space.

The replacing of the MMF effects of the direct component of the armature reaction by a fictitious reactance Xd, which contains a leakage reactance component, is, of course, consistent with old concept of synchronous reactance.

Because of this similarity of the two reactances with which we will deal, they are called direct-axis synchronous reactance and quadrature-axis synchronous reactance, respectively. As shown in vector diagram, they must always be used with their respective components of current.

Neglecting the leakage fluxes, the direct component of MMF acts on the main magnetic circuit o the machine. The quadrature component has a magnetic circuit largely through the air gaps and interpolar space. For this reason the quadrature-axis synchronous reactance is smaller than the direct-axis reactance and is less affected by saturation.

In non-salient pole machine, Xd is nearly equal to Xq and would be exactly so were it not for the slight differences in the two magnetic circuits on which they operated.

03-Alternators


Power Developed by a Synchronous Generator [1/37.29/p.1454] If we neglect Ra and hence Cu loss, then the power developed (Pd) by an alternator is

equal to the power output (Pout). Hence, the per phase power output of an alternator is dout PVIP == θcos (150.0)

From Fig. 150.1, qq XIV =αsin (150.1a) αcos0 VEXI dd −= (150.1b)

)cos( θα += II q (150.2a) )sin( θα += II d (150.2b)

Substituting Eq. (150.2a) in Eq. (150.1a), then

θαθαθαα sinsincoscos)cos(sin IXIXIXV qqq −=+= αθαθα sinsinsincoscos VIXIX qq =− (150.3)

Substituting Eq. (150.2b) in Eq. (150.1b), then θαθαθαα sinsincoscos)cos(sin IXIXIXV qqq −=+=

θαθαθαα sincoscossin)sin(cos0 IXIXXIVE ddd +=+=− αθαθα cossincoscossin 0 VEIXIX dd −=+ (150.4)

From Eqs. (150.3) and (150.4)

αθαθα

αθαθα

cossincoscossin

sinsinsincoscos

0 VEIXIX

VIXIX

dd

qq

−=+

=−

αααθααθαα

ααθααθαα

cossinsinsincossincossinsin

sincossinsincoscoscoscos

0 VXXEIXXIXX

VXIXXIXX

qqdqdq

dqdqd

−=+

=−

αααθααθαα

ααθααθαα

cossinsinsincossincossinsin

cossinsinsincoscoscoscos

0 VXEXIXXIXX

VXIXXIXX

qqqdqd

dqdqd

−=+

=−

ααααα

θααθαα

cossinsincossin

cossinsincoscoscos

0 VXEXVX

IXXIXX

qqd

qdqd

−+=

+

αααθ sincossin)(cos 0EXVXXIXX qqdqd +−=

αααθ 2sin2

2sin2

sincos 0 qdqqd XVXVEXIXX −+=

αααθ 2sin2

2sin2

sincos 0

dqd XV

XV

XE

I −+= (150.5)

Substitute (150.5) in (150.0) then

ααααα 2sin2

)(sin2sin

22sin

2sin

20

220

qd

qd

ddqddout XX

XXVX

VEX

VX

VX

VEPP

−+=−+==

03-Alternators


The total power developed would be three times the above power. The power developed consists of two components, the first term represents power due

to field excitation and the second term gives the reluctance power i.e. power due to saliency. If Xd=Xq i.e. the machine has a cylindrical rotor, then the second term becomes zero

and the power is given by the first term only. If, on the other hand, there is no field excitation i.e. E0=0, then the first term in the

above expression becomes zero and the power developed is given by the second term. The value of α is positive (+ve) for a generator and negative (-ve) for a motor.

Example 37.41: A 3-phase Y-connected synchronous generator supplies current of 10A having phase angle of 20o lagging at 400V. Find the load angle (α),the component of armature current Id and Iq, and voltage regulation. If Xd=10 ohm and Xq=6.5 ohm. Assume armature resistance to be negligible. Solution: 94.020coscos =°=θ ; 342.020sinsin =°=θ ; AI 10= ; VV 400=

1447.0342.05.610400

94.05.610sin

costan =

××+××

=+

=θ

θα

q

q

XIVXI

°== − 23.8)1447.0(tan 1α (Ans.) AII d 81.8)2023.8sin(10)sin( =°+°×=+= θα (Ans.) AII q 73.4)2023.8cos(10)cos( =°+°×=+= θα (Ans.)

VXIVE dd 4431073.423.8cos400cos0 =×+°×=+= α

%75.10100400

400443100regulation% 0 =×−

=×−

=∴V

VE (Ans.)

Problem 3: 20KVA, 200V, 50Hz, star connected 3-phase salient pole synchronous generator supplies load at lagging power factor of 45o. The phase constant of the generator are Xd=4 ohm, Xq=2 ohm and Ra=0.5 ohm. Calculate (i) power angle and (ii) voltage regulation under the given conditions.

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 Edition in SI units, ©1992 Metric Edition, McGraw Hill Book Company. [5] Irving L. Kosow, Electrical Machinery and Transformers, Second Edition, Prentice –Hall India Pvt. Limited.

two reaction theory

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