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Question :
A 325-MVA, 26-kV, 60-Hz, three-phase, salient-pole synchronous generator is observed to be operating at a power output of 250-MW and a
lagging power factor of 0.89 at a terminal voltage of 26 kV. The generator synchronous reactances are X = 1.95 and X = 1.18, both in per
unit. The generator achieves rated-open-circuit voltage at a field current AFNL = 342 A. Calculate (a) the angle xcexb4 between the generator
terminal voltage and the generated voltage, (b) the magnitude of the generated voltage E in per unit, and (c) the required field current in
amperes.
Answer (Ignore Step 0) :
Step 1
(a)
Consider 325MVA and 26KV as base values.
Thus,
Write the expression for the three-phase real power.
...... (1)
Here,
P is 3-phase power
Is the line-line terminal voltage of machine,
Is the armature current of machine,
Is the power factor.
Step 2
Substitute
for
,
for
and
for
in equation (1).
✌ ✌ Lucky Carrot ✌ ✌ 
d q
af
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In polar form,
The negative sign is taken before the angle because current is lagging the terminal voltage by
Step 3
Calculate base current.
Calculate the armature current in per unit.
Thus,
In polar form,
Step 4
Refer to Figure 5.27 for phasor diagram of generator with lagging load.
The Quadrature axis can be located by the phasor,
Thus,
From the phasor diagram of Figure 5.27, observe that the Quadrature axis is leading the reference phasor by an angle
.
Since we determined the Quadrature axis location.
Thus,
.
Therefore, the angle between the generated voltage and the generator terminal voltage,
is
.
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Step 5
(b)
From the phasor diagram of figure 1, the phase angle between
and
is,
Step 6
Refer to equations 5.59 and 5.60 from the textbook.
Calculate the magnitudes of direct axis component of armature current,
.
From Figure 5.27,
The negative symbol before angle indicates that
lags reference phasor by
.
Calculate the magnitudes of direct axis and Quadrature axis components of armature current,
.
From the phasor diagram of figure 5.27,
The positive symbol before angle indicates that
leads the reference phasor by
.
Step 7
Refer to equation 5.58 from the textbook.
Calculate
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.
Therefore, the generated emf,
is
.
Observe that angle of
is
which is exactly equal to
.
Thus,
along Quadrature axis.
Step 8
(c)
Refer to equation 3.47 from the textbook.
Calculate the field current,
.
Therefore, the field current,
is
.