|
V
is found out from the node equation at the load neutral.
V
V
V
−
∠ °
+
−
∠ −
°
+
+
−
∠
°
−
=
230 9 0
10
230 9
120
8
6
230 9 120
8
6
0
.
.
.
j
j
i e
j
j
j
. .,
.
.
V
1
10
1
8
6
1
8
6
230 9 0
10
230 9
120
8
6
23
+
+
+
−
=
∠ °
+
∠ −
°
+
+
00 9 120
8
6
.
∠
°
−
j
Solving for
V
, we get,
V
=
-
74.55 V rms.
Observe the unacceptably large neutral-shift voltage when the source neutral and load neutral are
isolated.
Now, we can find wire currents as
I
I
R
Y
=
∠ ° +
=
∠ °
=
∠ −
° +
=
230 9 0
74 55
10
30 545 0
230 9
120
74 55
10
.
.
.
.
.
A rms
220 414
138 43
230 9 120
74 55
8
6
20 414 138 4
.
.
.
.
.
.
∠ −
°
=
∠
° +
−
=
∠
A rms
I
B
j
33
0
°
=
+
+
=
A rms
A rms.
.
I
I
I
I
N
R
Y
B
The load phase voltages can be determined by multiplying line current phasors by phasor impedance
of each phase.
V
V
RN
YN
=
∠ ° ×
−
=
∠ −
°
=
∠ −
30 545 0
9 9
0 3
302 53
1 74
20 414
1
.
.
. )
.
.
.
(
V rms
j
338 43
7 9
5 7
198 87
102 62
20 414 138 43
.
.
. )
.
.
.
.
° ×
+
=
∠ −
°
=
∠
(
V rms
j
V
BN
°° ×
−
=
∠
°
(
V rms
7 9
6 3
206 27 99 86
.
. )
.
.
j
Observe the over-voltage in R-phase. The line voltages across the load can be determined in terms
of the phase voltage phasors now.
V
= V
V
RY
RN
YN
−
=
∠ −
° −
∠ −
° =
∠
°
302 53
1 74
198 87
102 62
392 16
28 13
.
.
.
.
.
.
V rrms
V
= V
V
YB
YN
BN
−
=
∠ −
° −
∠
° =
∠ −
198 87
102 62
206 27 99 86
397 37
91
.
.
.
.
.
.22
206 27 99 86
302 53
1 74
398 94 147
°
−
=
∠
° −
∠ −
° =
∠
V rms
V
= V
V
BR
BN
RN
.
.
.
.
.
..83
°
V rms
8.22
SinusoidalSteady-StateinThree-PhaseCircuits
The active power and reactive power delivered to the load are determined by adding up the
corresponding power delivered to each phase of the load.
P
P
P
P
Q
Q
Q
Q
=
+
+
=
=
+
+
= −
R
Y
B
R
Y
B
kW
kVAr
15 82
0 53
.
.
Apparent power is 17.11 kVA and average power factor is 0.925.
Note that, in all the cases, the three line voltages add up to zero. The sum of line voltages in any
three-phase system is zero since these three voltages are voltages in a loop formed by R to Y, Y to B
and B to R.
This example demonstrated the need for a strong neutral tie in unbalanced four-wire circuits to
maintain the phase voltages at load terminal at near-nominal values.
8.4.2
circulating current in unbalanced delta-connected sources
A delta-connected source comprises three sinusoidal sources in a closed loop. If the sources are
ideal independent voltage sources they have to add up to zero at all instants. Otherwise, there
will be infinitely large circulating current in the closed loop. The sources will have some non-
zero impedance in practice and this source impedance will limit the circulating current arising
out of residual voltage in the loop. However, if the source impedance is small (and, it should be
small for a good source), the circulating current will consume the current carrying capacity
of the source and leave only a portion of its capacity to deliver useful power to the external
load. Moreover, large circulating current in a delta loop produces large dissipation within the
sources.
Therefore, the phase-source voltage phasors in a delta-connected source should add up to zero
preferably. This happens naturally in the case of a balanced source. But it need not happen always in
the case of unbalanced source though it can for certain combinations of magnitudes and phase angles.
Essentially, the magnitudes and phase angles of the three phase-sources must be such that the phasors
form a closed triangle in complex plane. This is difficult to arrange in a synchronous generator. That
is why generators are always Y-connected.
It is rarely that we encounter a delta-connected source. One situation where it can appear is when
a transformer secondary is delta-connected. We can use a delta-connected voltage source, with each
arm of delta containing a source in series with the transformer series impedance, as a circuit model
for the secondary of the transformer. Exact symmetry in a three-phase equipment is hard to achieve
in practice. For instance, there can be small differences in the number of turns employed in the three
phase-windings of the transformer. This may produce unbalanced voltages in the three secondary
windings. Further, those voltage phasors may not ‘close’ into a triangle. Circulating current in the
secondary winding of the transformer is the result.
The next example illustrates this problem as well as the application of mesh analysis and node
analysis in solving unbalanced three-phase circuits.
example: 8.4-2
A delta-connected unbalanced source delivers power to a balanced Y-connected resistive load as
shown in Fig. 8.4-2. Solve the circuit completely by using mesh analysis.
AnalysisofUnbalancedThree-PhaseCircuits
8.23
400
∠
0° V rms
370
∠
–120°
V rms
370
∠
120°
V rms
–
–
–
+
+
+
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