two reaction theory

1. 03-Alternators
Prepared by Dr. M. A. Mannan Page 1 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
ψπ
θ
xdxAA ;
'
2
'
01 ]cos[
ψπ
θ +
−= xAA ; )]
2
'cos(1[1
ψπ
θ +−= AA

2. 03-Alternators
Prepared by Dr. M. A. Mannan Page 2 of 11
The area of the small part
∫∫
−−−
== 2
'
0
)'
2
(
0
2 sinsin
ψπ
θθ
ψπ
xdxAxdxAA ; 2
'
02 ]cos[
ψπ
θ −
−= xAA ;
)]
2
'cos(1[2
ψπ
θ −−= AA
Expanding
]
2
sin'sin
2
cos'[cos1
ψπ
θ
ψπ
θ −−= AAA
]
2
sin'sin
2
cos'[cos2
ψπ
θ
ψπ
θ +−= AAA
]}
2
sin'sin
2
cos'[cos{]
2
sin'sin
2
cos'[cos21
ψπ
θ
ψπ
θ
ψπ
θ
ψπ
θ +−−−−=− AAAAAA
]
2
sin'sin
2
cos'[cos]
2
sin'sin
2
cos'[cos21
ψπ
θ
ψπ
θ
ψπ
θ
ψπ
θ ++−−−=− AAAAAA
2
sin'sin
2
cos'cos
2
sin'sin
2
cos'cos21
ψπ
θ
ψπ
θ
ψπ
θ
ψπ
θ AAAAAA +++−=−
2
sin'sin221
ψπ
θAAA =−
Dividing this total area by the base ψπ to obtain the average demagnetizing ATs:
ψπ
ψπ
θ
ψπ
2
sin2
'sin21
A
AA
=
−
Demagnetizing AT.
This could be written 'sin21
θ
ψπ
AK
AA
AD =
−
= where
ψπ
ψπ
2
sin2
=K [188]

3. 03-Alternators
Prepared by Dr. M. A. Mannan Page 3 of 11
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:
pd
a
kk
P
ZI
NIA
2
9.0poleperreactionarmatureof ==
Hence the average demagnetizing ATs per pole are
'sin)2
sin2
2
9.0
('sin2
sin2
2
9.02
sin2
'sin θ
ψπ
ψπ
θ
ψπ
ψπ
ψπ
ψπ
θ pd
a
pd
a
D kk
P
ZI
kk
P
ZI
AA ===
'sinθCkk
P
ZI
A pd
a
D = (188)
where,
ψπ
ψπ
2
sin2
2
9.0
=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.

4. 03-Alternators
Prepared by Dr. M. A. Mannan Page 4 of 11
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=
∫=∫= 2
0
2
0
sin'cos22d)Area(shade
ψπψπ
θ xdxAadx
[ ] radiansMMFin
2
cos1'cos2cos'cos2d)Area(shade 2
0 ⎥⎦
⎤
⎢⎣
⎡
−=−=
ψπ
θθ
ψπ
AxA

5. 03-Alternators
Prepared by Dr. M. A. Mannan Page 5 of 11
⎥⎦
⎤
⎢⎣
⎡
−==
2
cos1
'cos2Area
MMFAverage
ψπ
π
θ
π
A
The average value of the cross MMF per pole is
'cos''cos
2
cos1
2
θθ
ψπ
π
AKAAc =⎥⎦
⎤
⎢⎣
⎡
−= where, ⎥⎦
⎤
⎢⎣
⎡
−=
2
cos1
2
'
ψπ
π
K
By substituting the value of pd
a
kk
P
ZI
A
2
9.0=
'cos)
2
cos1(
9.0
θ
ψπ
π ⎥⎦
⎤
⎢⎣
⎡
−= pd
a
c kk
P
ZI
A
'cosθCkk
P
ZI
A pd
a
c = (193)
where )
2
cos1(
9.0 ψπ
π
−=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.

6. 03-Alternators
Prepared by Dr. M. A. Mannan Page 6 of 11
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 +== Ckk
P
ZI
vvAE pdcc here let IIa =
But φsinEEc = ; hence, )cos(sin αθφ +== Ckk
P
ZI
vEE pdc
Or,
E
Ckk
P
ZI
v
E
E pd
c
)cos(
sin
αθ
φ
+
==
But λφαθ +=+ ;
E
Ckk
P
ZI
v pd )cos(
sin
λφ
φ
+
=
φ
λφλφ
φ
λφ
sin
sinsincoscos
sin
)cos( −
=
+
= Ckk
P
ZI
vCkk
P
ZI
vE pdpd
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−= λ
φ
λ
φ
λφ
φ
λφ
sin
tan
cos
sin
sinsin
sin
coscos
Ckk
P
ZI
vCkk
P
ZI
vE pdpd
λ
φ
λ
sin
tan
cos
Ckk
P
ZI
vCkk
P
ZI
vE pdpd −=
φ
λ
λ
tan
cos
sin Ckk
P
ZI
vCkk
P
ZI
vE pdpd =+
λφλ costan]sin[ Ckk
P
ZI
vCkk
P
ZI
vE pdpd =+
λ
λ
φ
sin
cos
tan
P
CkvZIk
E
P
CkZIk
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

7. 03-Alternators
Prepared by Dr. M. A. Mannan Page 7 of 11
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.

8. 03-Alternators
Prepared by Dr. M. A. Mannan Page 8 of 11
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)
cos
sin
'tan
g)(generatin
cos
sin
'tan
a
q
a
q
IRV
IXV
IRV
IXV
−
−
=
+
+
=
θ
θ
θ
θ
θ
θ
Angle (α) calculation between terminal voltage (V) and NL voltage (E0)

9. 03-Alternators
Prepared by Dr. M. A. Mannan Page 9 of 11
Also 'sin;'cos θθ IIII dq == then )sin();cos( θαθα +=+= IIII dq
From the Fig. 150,
V
IRIX
V
RIXI aqadqq 'sin'cos
sin
θθ
α
−
=
−
=
V
RXI aq )'sin'cos(
sin
θθ
α
−
=
Substituting
(motoring)'
g)(generatin'
αθθ
αθθ
−=
+=
in this equation for generating and expanding to sine and cosine functions,
)]sin()cos([sin αθαθα +−+= aq RX
V
I
}]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[
cos
sin
tan
g)(generatin
]cossin[
]sincos[
cos
sin
tan
θθ
θθ
α
α
α
θθ
θθ
α
α
α
aq
aq
aq
aq
RXIV
RXI
RXIV
RXI
+−
−
==
++
−
==
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.

10. 03-Alternators
Prepared by Dr. M. A. Mannan Page 10 of 11
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 += IIq (150.2a)
)sin( θα += IId (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 2sin
2
2sin
2
sincos 0 qdqqd X
V
X
V
EXIXX −+=
αααθ 2sin
2
2sin
2
sincos 0
dqd X
V
X
V
X
E
I −+= (150.5)
Substitute (150.5) in (150.0) then
ααααα 2sin
2
)(
sin2sin
2
2sin
2
sin
2
0
22
0
qd
qd
ddqd
dout
XX
XXV
X
VE
X
V
X
V
X
VE
PP
−
+=−+==

11. 03-Alternators
Prepared by Dr. M. A. Mannan Page 11 of 11
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=
1447.0
342.05.610400
94.05.610
sin
cos
tan =
××+
××
=
+
=
θ
θ
α
q
q
XIV
XI
°== −
23.8)1447.0(tan 1
α (Ans.)
AIId 81.8)2023.8sin(10)sin( =°+°×=+= θα (Ans.)
AIIq 73.4)2023.8cos(10)cos( =°+°×=+= θα (Ans.)
VXIVE dd 4431073.423.8cos400cos0 =×+°×=+= α
%75.10100
400
400443
100regulation% 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.