Three-PhaseSourcesandThree-PhasePower
8.9
V
BN
V
YB
V
YN
V
RY
V
RN
V
BR
90º
=
90º
90º
(a)
30º
3
30º
30º
V
P
V
P
V
L
V
BN
V
YN
V
RN
(b)
θ
θ
θ
120º
V
P
I
R
I
Y
I
B
Fig. 8.2-3
(a)PhasordiagramofphaseandlinevoltagesofaY-connectedsource(b)Phasor
diagramoflinecurrentsandphasevoltagesinY-connectedsource
3. Each source contributes a reactive power of
V
P
I
L
sin
q
W. Hence, total three-phase reactive power
delivered
by the three-phase source
=
3
V
P
I
L
sin
q
=
√
3
V
L
I
L
sin
q
VArs, where
q
is the angle by
which the ‘negative of phase current defined as per passive sign convention’ (
i.e.,
-
I
RN
in the
case of R-line) lags behind the phase voltage. Equivalently,
q
is the angle by which the current
delivered by a phase source lags behind the phase voltage.
4. Therefore, the complex power delivered
by the three-phase source
S
=
√
3
V
L
I
L
∠
q
VA.
8.2.2
the
D
-connected source
The generators in a power system are almost invariably Y-connected. However, power systems use Y-
D
transformers and a
D
-connected three-phase source can be used to model the secondary side of such
a transformer. A
D
-connected three-phase source and the phasor diagram of its voltages and currents
are shown in Fig. 8.2-4.
We assume that the load on the source is balanced.
V
RY
is taken as the reference phasor for phasor
diagrams.
Note that
phase voltage and
line voltage are the same for this source and that there is no neutral
point and hence no
line-to-neutral voltage can be defined for this source. Applying KCL at the three
corners
of delta, we get the following equations.
I
I
I
I
I
I
I
I
I
BR
RY
R
YB
RY
Y
YB
BR
B
−
=
−
+
=
−
=
This system of equations has no unique solution since
we can add a constant to
I
RY
,
I
YB
and
I
BR
without affecting these equations. Such a constant added to the three currents will represent a
circulating current within the delta-loop. Such a circulating current is possible only since three ideal
sources are connected in a loop. However, in practice, all the sources in the delta will have some small
impedance or other which will force the condition that (
I
RY
+
I
YB
+
I
BR
)
Z
=
0 where
Z
is the small
source impedance of the sources. Assuming that the circulating current component in delta is zero, let
the current
-
I
RY
be
I
P
∠-
q
. Then, the remaining two phase currents will be
-
I
YB
=
I
P
∠-
q
-
120
°
and
-
I
BR
=
I
P
∠-
q
+
120
°
, where
I
P
is the magnitude of phase currents.
8.10
SinusoidalSteady-StateinThree-PhaseCircuits
Now,
I
R
can be constructed by
I
R
=
-
I
RY
– (
-
I
BR
). This is shown in the phasor diagram in Fig. 8.2-4.
The
resulting
I
R
=
√
3
I
P
∠-
(30
° +
q
) A rms.
V
RY
V
BR
V
YB
V
L
V
RY
I
RY
I
YB
I
R
I
Y
I
B
I
BR
V
BR
V
YB
V
L
∠
120º
0º
–120º
V
L
V
L
+
–
+
+
B
Y
R
–
–
– I
BR
– I
RY
– I
YB
I
R
I
L
30º
θ
Fig. 8.2-4
Adelta-connectedsourceanditsphasordiagrams
Hence,
the line current magnitude is I
L
=
√
3
I
P
in a delta-connected three-phase source.
The line currents delivered by a balanced
D
-connected source form a balanced three-
phase set of currents that are
√
3 times in magnitude and 30
°
behind in phase with
respecttophasecurrentsdeliveredbythephase-sources.
We make the following observations from these phasor diagrams.
1. The delivered phase current lags the phase-source voltage by
q
and line current lags behind the
corresponding line voltage by (30
° +
q
).
2. Each source contributes an active power of
V
L
I
P
cos
q
W. Hence, total three-phase active power
delivered
by the three-phase source
=
3
V
L
I
P
cos
q
=
√
3
V
L
I
L
cos
q
W, where
q
is the angle by which
the ‘negative of phase current defined as per passive sign convention’ (
i.e.,
-
I
RY
in the case of RY-
phase source)
lags behind the phase voltage.
3. Each source contributes a reactive power of
V
L
I
P
sin
q
W. Hence, total three-phase reactive power
delivered
by the three-phase source
=
3
V
L
I
P
sin
q
=
√
3
V
L
I
L
sin
q
VArs where
q
is the angle by
which the ‘negative of phase current defined as per passive sign convention’ (
i.e.,
-
I
RY
in the
case of RY-phase source)
lags behind the phase voltage. Equivalently,
q
is
the angle by which the
current
delivered by a phase-source lags behind the phase voltage.
4. Therefore, the complex power delivered by the three-phase source
S
=
√
3
V
L
I
L
∠
q
VA.
Therefore, we state the following conclusion:
Thethree-phasecomplexpowerdeliveredbyathree-phasesourceisgivenby
S
=
√
3
V
L
I
L
∠
q
=
[
√
3
V
L
I
L
cos
q
+
j
√
3
V
L
I
L
sin
q
]VA
where
q
istheanglebywhichthecurrentphasor
delivered by
aphasesourcelagsbehind
itsvoltagephasor.ThisrelationshipisindependentofwhetherthesourceisY-connected
or
D
-connected.Itshouldbenotedthat
q
is not
theanglebetweenlinevoltagephasor
andlinecurrentphasor.
AnalysisofBalancedThree-PhaseCircuits
8.11
Electrical Power Engineering adopts a convention to specify a three-phase source by specifying its
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