Two Reaction Theory
Two Reaction Theory
Two Reaction Theory
Since the interpolar axis is 900 away from the d-axis, the interpolar axis is called
the quadrature axis or q-axis.
The d-axis involves two small airgaps with a path
of minimum reluctance
E E f Ead Eaq
'
E E f jX ad I d jX aq I q
'
Also the voltage E is equal to the terminal voltage V plus the voltage drops in the
resistance and leakage reactance of the armature.
E V RI a jX l I a
'
The armature current Ia is split into two components, one in phase with excitation
voltage Ef and the other in phase quadrature to it.
=+
Id=Iasin
Iq=Iacos
Ef V (Id Iq )R j(Xad Xl )Id j(Xaq X l )Iq
Ef V (Id Iq )R jXdId jXqIq
Xd = the direct axis synchronous reactance
Xq = the quadrature axis synchronous reactance
=+
Id=Iasin
Iq=Iacos
E f V IaR jX dId jX qIq
To find angle for given values of V, I
a and
Ia Id Iq Iq Ia Id
Substituting Iq in Ef gives
In BCD, BCD =
In BCD,
CD X qIq X qIq
cos BC
BC BC cos
Substituting for Iq,
Xq (Ia cos )
BC X q Ia
cos
In BMQ,
BQ X dId X dId X dIa sin
sin BM X dIa
BM BM sin sin
CM BM BC X d I a X q I a ( X d X q ) I a
In CMN,
CN
sin CN (X d - X q )Iasin
CM
BQ X dId
CN (X d - X q )Ia (X d - X q )Ia (X d - X q )Id
BM X dIa
The point N is the end point of Ef.
Triangle OKC is a right angle triangle.
CK BK BC AL BC V sin X q I a
tan
OK OL LK OL AB V cos RI a
V sin X q I a
tan 1
V cos RI a
=+=-
From phasor diagram, |Ef| = Vcos + RIa +XdId
Determination of
From phasor diagram
E= OC = OA + AB + BC = V + RIa +jXqIa = V + (R+ jXq)Ia
Ia Ia Iacos jIasin
E V (R jXq )(Iacos jIasin )
"
S1~ VI *
Taking Ef as reference
V V Vcos jVsin
Ia Iq - jId
I Iq jId
*
a
X qIq CD AM Vsin
Vsin
Iq
Xq
XdId AC MD OA OM
XdId OA OM Ef Vcos
E f Vcos
Id
Xd
Substituting Id and Iq in S1~
Vsin E f Vcos
S1~ Vcos jVsin j
X X
q d
VE 2 VE
Xd Xq Xd Xq cos2
2
V 1 1 V
S1~ f
sin sin2 j f
cos
X d 2 Xq Xd X d 2X dX q
Also S1~ P1~ jQ1~
VE f V 2 1 1
P1~ sin sin2 Watts
Xd 2 X q X d
Total real power in Watts
3VE f 3V 2 1 1
P3~ sin sin2 Watts
Xd 2 X q X d
Q 1~
VE f
cos
V2
Xd Xq Xd X q cos2 VAr
Xd 2X dX q
Total reactive power in VArs
Q 3~
3VE f
cos
3V 2
X d Xq Xd Xq cos2 VAr
Xd 2X dX q
3VE f 3V 2 1 1
P3~ sin sin2 Watts
Xd
2 Xq Xd
If |E| = 0
3V 2 1 1
P3~ sin2 Watts
then
2 Xq Xd
3V 2 1 1
Pdr3~ Watts
2 Xq Xd
The electromagnetic torque or torque developed for a 3 phase synchronous machine
3P1~
Electromag netic Torque, em
m
3 VEf V2 1 1
em sin sin2
2ns X d 2 X q X d
The resulting torque has two components.
The first term represents the torque due to field excitation, exc
3 VEf
exc sin
2ns X d
The second term is known as reluctance torque, rel
3 V 2 1 1
rel sin2
2ns 2 X q X d
The reluctance torque is independent of excitation and exists only if the machine is
connected to a system receiving reactive power from other synchronous machines
operating in parallel with the terminal voltage V.
The reluctance torque is due to the saliency of the field poles which tend to align the
direct axis with that of the armature mmf.
Q 1~
VE f
cos
V2
Xd Xq Xd X q cos2 VAr
Xd 2X dX q
dQ 1~
For maximum reactive power, 0
d
Differentiating Q1~ w.r.t.
Ef X q
cos
2VX d X q
Substituting cos in Q1~
2 2
V EX
f q
Xd 4Xd X d Xq
Q 1~max
For a cylindrical rotor synchronous generator, Xd = Xq = Xs.
Substituting Xs in Q1~,
VEf V2 VEf V2
Q 1~ cos 2 2X s cos VAr
Xs 2X s Xs Xs
OR
V
Q 1~ E f cos V VAr
Xs
When Efcos = V, (normal excitation) Q = 0; the generator operates at unity power factor
(only P is available).
When Efcos > V, (over excitation) Q is positive; the generator supplies reactive power at
the busbars.
When Efcos < V, (under excitation) Q is negative; the generator absorbs reactive power.
SLIP TEST
Determination of Xd and Xq of a salient pole synchronous machine from a simple no-load
test is known as slip test.
Three phase supply is applied to the armature (stator), having voltage less than the rated
voltage, while the field winding circuit is kept open.
The three phase currents drawn by the armature from a three phase supply produce a
rotating flux.
There will be a small slip between the rotating magnetic field produced by the armature
and the actual salient field poles.
The relative speed between the armature mmf and the field poles is equal to the slip
speed (Ns N).
When the stator mmf moves slowly past the actual field poles, there will be an instant
when the peak of the armature mmf wave is in line with the axis of the actual salient field
poles.
When the stator mmf is aligned with the d-axis of field poles, the reluctance offered by the
small air gap is minimum. This results in minimum magnetizing current Imin as indicated by
the ammeter.
In this position, the armature flux linkage with the field winding is maximum, and the rate
of change of flux is zero.
The d-axis can be located on the oscillogram. From the oscillogram, Xd = (ab/cd). Also, the
ratio of armature terminal voltage per phase to the corresponding armature current per
phase gives Xd.
After one quarter of cycle, the peak armature current is in line with the q-axis.
In this position, the reluctance offered by the long air gap is maximum. The maximum
current is measured from the line ammeter A.
The armature flux linkage with the field winding is zero, and the rate of change of flux is
maximum. The induced voltage across the field winding is maximum.
The q-axis can be located on the oscillogram. From the oscillogram, Xq = (ab/cd). Also,
the ratio of armature terminal voltage per phase to the corresponding armature current
per phase gives Xq.
When the rotor field is aligned with the armature mmf, its flux linkages are maximum, and
hence the voltage induced in the field goes through zero at this instant. This is the
position where alternator offers direct axis synchronous reactance, Xd.
While when rate of change of flux is associated with the rotor is maximum, voltage
induced in the field goes through its maximum. This is the position where alternator offers
quadrature axis synchronous reactance, Xq.