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THREE PHASE NETWORKS
The alternating current networks
that we have studied so far are widely used to model AC mains electric power networks in homes.
However, for industrial use and also for electric power generation, a network of AC generators is more effective. This is realized by
polyphase networks consisting of a number of identical sinusoidal generators
with a phase angle difference. The most common polyphase networks are two- or
three-phase networks. We will limit our discussion here to three-phase networks. Note
that TINA provides special tools for drawing three-phase networks in the Special
component toolbar, under the Stars and Y buttons. A three-phase network can be seen as
a special connection of three single phase or simple AC circuits. Three-phase
networks consist of three simple networks, each having the same amplitude and
frequency, and a 120°
phase difference between adjacent networks. The time diagram of the voltages in
a 120Veff system is shown in the diagram below.
Compared to single-phase systems, three phase networks are superior because both the power stations and the transmission lines require thinner conductors for transmitting the same power. Due to the fact that one of the three voltages is always non-zero, three-phase equipment has better characteristics, and three-phase motors are self-starting without any additional circuitry. It is also much easier to convert three-phase voltages into DC (rectification), due to the reduced fluctuation in the rectified voltage. The frequency of
three-phase electric power networks is 60 Hz in United States and 50 Hz
in Europe. The single phase home network is simply one of the voltages
from a three-phase network.
In
practice, the three phases are connected in one of two ways. 1) The Wye or Y-connection, where
the negative terminals of each generator or load are connected to form the
neutral terminal. This results in a three-wire system, or if a neutral wire is
provided, a four-wire system. The
Vp1,Vp2,Vp3 voltages
of the generators are called phase
voltages, while the voltages VL1,VL2,VL3 between
any two connecting lines (but excluding the neutral wire) are called line
voltages. Similarly, the Ip1,Ip2,Ip3
currents of the generators
are called phase currents while the
currents IL1,IL2,IL3 in
the connecting lines (excluding the neutral wire) are called line
currents. In
Y-connection, the phase and line currents are obviously the same, but the line
voltages are greater than the phase voltages. In the
balanced case:
Let ‘s demonstrate this by a phasor diagram: Let’s
calculate VL for the phasor diagram above using the cosine rule of
trigonometry:
Now let’s calculate the same
quantity using complex peak values: Vp1
= 169.7 ej 0 °
= 169.7 Vp2
= 169.7 ej 120 °
= -84.85+j146.96 VL
= Vp2 - Vp1
= -254.55+j146.96
= 293.9 e j150 °
The
same result with the TINA Interpreter:
Similarly
the complex peak values of the line voltages VL21
= 293.9 ej 150°
V, VL23
= 293.9 ej 270°
V, VL13
= 293.9 ej 30°
V. The
complex effective values: VL21eff
= 207.85 ej 150°
V, VL23eff
= 207.85 ej 270°
V, VL13eff
= 207.85 ej 30°
V. Finally
let’s check the same results using TINA for a circuit with 120
Veff ; VP1 = VP2 = VP3 =169.7 V and
Z1= Z2 =Z3 = 1 ohms 2)
The delta or D-connection
of three phases is achieved by connecting the three loads in series forming a
closed loop. This is only used for three-wire systems.
As
opposed to a Y-connection, in D-connection
the phase and line voltages are obviously the same, but the line currents are
greater than the phase currents. In the balanced case:
Let’s
demonstrate this with TINA for a network with 120 Veff Z=10
ohms.
Result:
Since
either the generator or the load can be connected in D
or in Y, there are four possible interconnections: Y-Y, Y-D,
D-Y
and D-D.
If the load impedances of the different phases are equal, the three-phase
network is balanced. Some
further important definitions and facts: The
phase difference between the phase
voltage or current and the nearest line
voltage and current (if they are not the same) is 30°. If
the load is balanced (i.e. all the
loads have the same impedance), each phase’s voltages and currents are equal.
Furthermore, in the Y-connection,
there is no neutral current even if there is a neutral wire. If
the load is unbalanced, the phase
voltages and currents are different Also, in the Y–Y-connection with no
neutral wire, the common nodes (star points) are not at the same potential. In
this case we can solve for node potential V0 (the common node of the
loads) using a node equation. Calculating V0 allows you to solve for
the phase voltages of the load, current in the neutral wire, etc. Y-connected
generators always incorporate a neutral wire. The
power in a balanced three phase system is PT = 3 VpIp
cos J
=
where
J
is the phase angle between the voltage and the current of the load. The
total apparent power in a balanced three phase system: ST =
The
total reactive power in a balanced three phase system:
QT =
Example
1 The
rms value of the phase voltages of a three-phase balanced Y-connected generator
is 220 V; its frequency is 50 Hz. a/ Find the time function of the
phase currents of the load! b/ Calculate all the average and
reactive powers of the load!
Both
the generator and the load are balanced, so we need to calculate only one phase
and can get the other voltages or currents by changing the phase angles.
In the schematic above we did not draw the neutral wire, but instead
assigned ‘earth’ at both sides. This can serve as a neutral wire; however,
because the circuit is balanced the neutral wire is not needed. The
load is connected in Y, so the phase currents are equal to the line currents:
the peak values: IP1
= VP/(R+jwL) = 311/(100+j314*0.3) = 311/(100+j94.2) = 1.65-j1.55
= 2.26 e-j43.3° A VP1
= 311 V IP2
= IP1 e j 120° =2.26 ej76.7° A
IP3 = IP2 e j 120° =2.26 e-j163.3° A
iP1 = 2.26 cos (w×t – 44.3°) A
iP2 = 2.26 cos (w×t + 76.7°) A The
powers are also equal: P1 = P2 = P3 =
This
is the same as calculated results by hand and TINA’s Interpreter.
Example
2 A
three-phase balanced Y-connected generator is loaded by a delta-connected
three-pole load with equal impedances. f=50 Hz. Find
the time functions of a/ the phase
voltages of the load,
b/ the phase currents of the load,
c/ the line currents!
The
phase voltage of the load equals the line voltage of the generator: VL
=
The phase currents of the
load: I1 = VL/R1+VLjwC = 1.228 + j1.337 = 1.815 ej 47.46° A
I2 = I1 * e-j120° = 1.815 e-j72.54° A = 0.543 – j1.73 A
I3 = I1 * ej120° = 1.815 ej167.46° = -1.772 + j0.394 Seeing the directions:
Ia = I1 – I3 = 3+j0.933 A = 3.14 ej17.26° A. According
to the results calculated by hand and TINA’s Interpreter.
Finally
an example with an unbalanced load: Example
3 The
rms value of the phase voltages of a three-phase balanced Y-connected
generator is 220 V; its frequency is 50 Hz. a/
Find the phasor of the voltage V0 ! b/
Find the amplitudes and initial phase angles of the phase currents
!
Now
the load is an asymmetrical one and we have no neutral wire, so we can expect a
potential difference between the neutral points. Use an equation for the node
potential V0:
hence
V0 = 192.71+ j39.54 V = 196.7 ej11.6° V and: I1 = (V1-V0)*j
w C = 0.125 ej71.5° A; I2
= (V2-V0)*j w C = 0.465 e-j48.43° and I3 =(V3-V0)/R
= 0.417 ej 146.6° A
v0(t)
= 196.7 cos (w×t + 11.6°) V;
i1(t) = 0.125 cos (w×t + 71.5°) A; i2(t) = 0.465 cos (w×t - 48.4°) A;
And, finally, the results calculated
by TINA agree with the results calculated by the other techniques.
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