AnalysisofUnbalancedThree-PhaseCircuits
8.17
115.5
∠
–30°
V rms
115.5
∠
90°
V rms
115.5
∠
–150° V rms
–
–
–
+
+
+
I
R
I
B
I
Y
R
N
Y
B
0.1 +
j
0.3
Ω
0.1 +
j
0.3
Ω
0.1 +
j
0.3
Ω
5.9 +
j
7.7
Ω
5.9 +
j
7.7
Ω
5.9 +
j
7.7
Ω
Fig. 8.3-8
ThestarequivalentofthecircuitinFig.8.3-6
Hence, line voltage (RY) across the load
=
112
√
3
∠
(
-
30.6
° +
30
°
)
=
194
∠-
0.6
°
V rms.
(ii) The line current (R) is 115.5
∠-
30
°
÷
6
+
j8
=
11.55
∠-
83.13
°
A rms. Refer to Fig. 8.3-6. The
current
marked as
I
1
=
[200
∠
0
°
- line voltage (RY) across
load]
÷
(0.3
+
j0.9)
=
[200
∠
0
°
-
194
∠-
0.6
°
]
÷
0.9487
∠
71.57
°
=
(6
+
j2)
÷
0.9487
∠
71.57
°
=
6.68
∠-
51.13
°
A rms. Then,
I
2
will be 6.68
∠-
171.13
°
A rms and
I
3
will be 6.68
∠
68.87
°
A rms
(iii) The active power delivered to the load
=
√
3
×
194
×
11.55
×
cos(
-
30
°-
(
-
83.13
°
))
=
2.33 kW.
The reactive power delivered to load
=
√
3
×
194
×
11.55
×
sin(
-
30
°-
(
-
83.13
°
))
=
3.11 kVAr.
The internal impedance of the source is 0.3
+
j0.9
W
and it carries an rms current of 6.68 A.
Therefore, the active power consumed by the internal impedance of the delta-connected source
=
3
×
6.68
2
×
0.3
=
40.2 W. The reactive power consumed by this impedance
=
3
×
6.68
2
×
0.9
=
120.6 VAr.
8.4
analysIs of unBalanced three-Phase cIrcuIts
An unbalanced three-phase circuit is one that contains at least one source or load that does not possess
three-phase symmetry. A source with the three source-function magnitudes unequal and/or the phase
displacements different from 120
°
can make a circuit unbalanced. Similarly, a three-phase load with
unequal phase impedances can make a circuit unbalanced.
The single-phase equivalent
circuit technique of analysis does not work for unbalanced three-
phase circuits. General circuit analysis techniques like mesh analysis or nodal analysis will have to be
employed for analysing such circuits.
8.4.1
unbalanced y–y circuit
Y–Y connection is typically used in the
last mile in a power system,
i.e., in the low-tension (LT)
distribution system.
The primary distribution system usually runs at 11 kV line voltage level. Distribution transformers
that are rated for 11 kV/400 V and are
D
-connected in the primary
side and Y-connected in the
secondary side are used to step down the primary distribution voltage to the LT distribution level. Such
transformers are located at the load centre location of the area that a particular transformer is expected
to serve.
The neutral point of secondary side of a distribution transformer is usually earthed solidly.
115.5
∠
–30°
V rms
–
+
I
R
R
0.1 +
j
0.3
Ω
5.9 +
j
7.7
Ω
Fig. 8.3-9
Single-phaseequivalent
ofcircuitinFig.8.3-6
8.18
SinusoidalSteady-StateinThree-PhaseCircuits
4-wire LT lines emanating from the secondary of transformer distribute power to various subareas
of the service area. Individual consumers are provided service by means of service drops tapped from
these lines. Both single-phase loads and three-phase loads are provided power from these LT lines at
various points. Thus, the load on the LT line will be unbalanced in general. Even if the single phase
loads are to get distributed equally among three lines (R, Y and B), only the transformer secondary will
perceive a balanced load and various sections of the LT line will see varying degree of unbalance in
load. This is due to the spatially dispersed nature of loads. Thus, the neutral wire in an LT distribution
line will invariably carry current – that too, different currents in different locations – even if the
aggregate load is balanced.
The system in which the neutral wire is available for connecting single-phase loads and three-phase
loads that require a neutral tie-up is called a
four-wire system. The system in which neutral wire is
not
available for load connection is called a
three-wire system. This could be Y-Y system with neutrals
isolated, Y
-D
,
D-
Y or
D-D
system.
The exposed metal parts of all electrical equipment at the consumer location are tied together and
earthed locally. This earth point is made available as the third pin of all three-pin power points installed in
the consumer electrical system. If the earthing resistances at the transformer neutral and at the consumer
location are negligibly small, then, the earth-pin at all consumer locations will be at the same potential
as the transformer secondary neutral. However, the neutral-pin at the consumer locations will differ in
potential from the transformer neutral potential due to the neutral wire drop. This difference will be more
if there is unbalance in the load and neutral is carrying heavy currents. Moreover, this potential difference
between the earth-pin and neutral-pin will be different for different consumers due to spatially distributed
nature of line load. This problem of neutral – earth voltage difference is called the
neutral- shift problem.
Electronic equipment in general, and, computing equipment in particular, are sensitive to this neutral-
shift voltage and often malfunction when it exceeds certain pre-specified levels. Damage to components
can also take place due
differences in neutral-shift potential that exist among various power points when
different components of an interconnected computing system are powered from different power points.
example: 8.4-1
A balanced three-phase source of 400 V, 50 Hz supplies an unbalanced load through an LT 4-wire line
as in Fig. 8.4-1. Each wire in the line has impedance of 0.1
+
j0.3
W
. (a) Find (i) phase and line voltages
at load location and neutral-shift voltage, (ii) wire currents, (iii) power and reactive power delivered
to the load, (iv) three-phase apparent power and power factor. (b) Repeat (a) assuming that the neutral
wire is a thick conductor with zero impedance. (c) Repeat (a) assuming that the neutral wire is open.
230.9
∠
120°
V rms
230.9
∠
0°
V rms
230.9
∠
–120°
V rms
–
–
–
+
+
+
Do'stlaringiz bilan baham: 
