When we studied the single-phase circuits, we saw the relation that existed between voltage and current
for circuits containing reactive and purely resistive elements. Now let's take a look at the polyphase circuits.
However, in this site we will restrict our studies to three-phase circuits. At the moment we focus
on the three-phase circuits called BALANCED, that is, when the phase voltages with respect to the
neutral have the same modulus, only varying the angle phase.
Unlike the single-phase power generator, which has only one winding in charge of generating electric power,
in the case of the three-phase generator, we have three independent windings that are mechanically out of
phase with 120°. As a consequence, with each rotation of the generator rotor, there will be three
alternating voltages called phases, which will obviously be electrically offset 120° from
each other. These voltages, as in the case of single phase, obey the sinusoidal or
cosine function.
The three phases produced are identified by the letters of the alphabet.
Nowadays, they are usually designated by the letters A, B, C and in older literatures
the letters R, S, T were used. By default, the direction of rotation of the phases is always
in the counterclockwise direction.
In this way there are two sequences of possible phases.
One is the DIRECT string,
designated as ABC sequence.
It can also be
called BCA or CAB. We just move the initial letter to the end. Then the
sequence is repeated. The other is the sequence INVERSE, known as the sequence
ACB. It can also be CBA or BAC. To get the INVERSE sequence,
we only change the B phase to the C phase and vice versa. The direction of rotation remains counter-clockwise.
Observation
"At any instant of time, the phasor sum of the three phase voltages of a three-phase generator is NULL."
In the Figure 81-01 we see the graphic representation of the three phases.
Notice that at the junction of the phases we have the point N called NEUTRAL.
By default, the phase A is used horizontally, meaning that it has angle ZERO degrees.
Therefore, the other phases will be out of phase by 120° in relation to phase A.
Notice the arrow to the right, indicating the counterclockwise direction of the phases.
We must be very careful in interpreting this "phase rotation".
Many students follow the arrow and end up interpreting, erroneously, as sequence ACB. This is NOT
the correct way to interpret. The RIGHT form is to reference the position of phase A. So we must ask:
If we turn 120° counterclockwise, which phase will go to the phase A position?
It is obvious that it will be the phase B. We turn more 120° and now who is in phase B is the
phase C. Then we get the sequence ABC. And this sequence is known as DIRECT sequence or POSITIVE.
The Figure 81-02 shows the three phases where we use the sine function to represent them. Note that
the maximum voltage (Vmax) occurs at4 volts. From the equation we can read the value of
ω = 2 π f (numerical value that comes before t) and is equal to 314. So the frequency is
50 Hz. Then the period (T = 1 / f) is 20 ms.
In the Figure 81-03 we see the graphic representation of the inverse sequence. Note that the direction
of rotation remains the same. Anticlockwise. What happened was the change of the BN phase through the CN phase.
Then, if we turn from 120°, who will come to the position of
phase A will be the phase C. One more turn of 120° and we will have the phase B in place of the
phase C. And repeat the cycle. Then we get the sequence ACB. This characterizes the INVERSE sequence.
Trigonometrically we can write the phases as:
VAN = 4 sen (314 t + 0°)
VBN = 4 sen (314 t + 120°)
VCN = 4 sen (314 t - 120°)
We should point out that in some problems it may be explicit that the phase AN does not have an angle equal to
zero degree. In this case we must position it in the angle provided by the problem. Obviously, the other two phases
will also suffer equivalent displacement and should be out of phase by
120° in relation to phase AN.
There will always be situations where we should change the reference for line voltage or phase voltage. So let's see how to make these transformations.
3.1 Nomenclature
When calling a voltage, the number or letter of the first index must be that has the highest voltage value.
The second index, in turn, will have the lowest value. In case VAB is reporting that the highest voltage is that of the point A and can also be written as VAB = VA - VB. In addition, the arrow of the phasor points to letter A.
Figure 81-04
In the Figure 81-04 we see examples of some phasors and their correct nomenclature.
In the figure below, we present another way of representing the line voltages in a graph, from the
phase voltages. This system is totally equivalent to that previously seen.
3.2.1 Direct or Positive Sequence
Figura 81-05
Note that to get the voltage VAB, we make the phasor sum of
VAN with - VBN. The most important thing here is that the line voltage resulting
from this sum is advanced by 30° in relation to the corresponding phase voltage, that is, VAN,
and in addition its magnitude is multiplied by
√3. With the other voltages of line the same thing happens.
Thus, in solving problems when there is a need to transform phase voltages to line voltages and vice versa, it is very
important that this be taken into account, otherwise the solution of the problem will be wrong.
A detail: if we are moving from line voltage to phase voltage, then we must DELAY the phase
voltage at 30°, that is, VF ∠φ = VL ∠θ - 30°, where we have
φ = θ - 30° representing the angle of the phase voltage and
θ, the angle of the line voltage. In addition, we must divide the line voltage modulus by
√ 3 to obtain the magnitude of the phase voltage. Next, we will represent the transformation of
phase voltages to line voltages, referring to the graph above which shows the sequence
direct or positive.
VAN = VF ∠0° ⇒ VAB = √3 VF ∠30°
VBN = VF ∠-120° ⇒ VBC = √3 VF ∠-90°
VCN = VF ∠120° ⇒ VCA = √3 VF ∠150°
Of course we could have written VBN = VF ∠240° and
VBC = √3 VF ∠270°. However, it is standard to choose the smallest numeric value of the angle (in modulus).
Now we are going to show the transformation of the line voltages to the phase voltages,
supposing VAB = VL∠0° (the previous graphic should be rotated by 30° clockwise).
In the Figure 81-06 we see the diagram of voltages, but with the sequence inverse or negative. In this case,
there was a change of VBN with VCN and vice-versa.
Figure 81-06
As a consequence, the angles of line voltages have changed markedly.
In the inverse sequence it is easy to see that the line voltages are now delayed by 30° with respect to the corresponding
phase voltages. With reference to this diagram, the voltages are:
VAN = VF ∠0° ⇒ VAB = √3 VF ∠-30°
VBN = VF ∠120° ⇒ VBC = √3 VF ∠90°
VCN = VF ∠-120° ⇒ VCA = √3 VF ∠-150°
Notice that in the inverse sequence there was a change of signal in the angle of all the voltages with respect to the direct sequence,
except in VAN, because this is the reference voltage in both cases.
Now we show the transformation of the line voltages to the phase voltages,
supposing VAB = VL∠0° (the previous graphic should be rotated by 30° counterclockwise).
When we have a balanced circuit connected in Triangle or Delta, we often have to transform phase
currents into line currents and vice versa. To do so, we must pay attention if we are working with a direct or inverse sequence.
For the case of DIRECT sequence, when transforming line current to phase current, the
line current must be DELAY of 30° relative to the phase current.
For the INVERSE sequence case, when transforming line current to phase current, the
line current must be ADVANCED of 30° relative to the phase current.
Let us study how we can represent these phasors mathematically. There are two ways to represent: the
trigonometric form and the phasorial form.
4.1. Trigonometric Form
The trigonometric form is explained by the following function:
VAN = Vmax sin ( ω t + θ)
Let's see an example.
VAN = 100 sin ( 200 t + 30°)
This means that the maximum voltage value is 100 volts. The value of ω is
200 rad / s. And the voltage is out of phase +30°.
We must pay particular attention to the trigonometric form because it requires that the value of the voltage
or current multiplying the sine or cosine function, it must be as voltage or current of peak or
maximum value. So if we are working with effective voltage or current we must multiply the effective value by
root of two to calculate the maximum value.
To represent in the phasor form, we must take into account the modulus of the amplitude and explain whether it is the effective value or the maximum value. In addition we must know the phase angle of the variable. To represent the voltage VAN of the previous example, we should write:
VAN = 100 ∠30°
Note that in the phasor form we do not have the information about ω. This information,
if necessary, should be informed in the problem statement.
The two most commonly used connections for three-phase circuits are the Delta circuit, also known as the
Triangle circuit, and the Star circuit, also called circuit Y. The following analysis considers a balanced circuit.
5.1. Star Circuit
The star circuit is characterized by the presence of NEUTRAL. So we have three phases plus a
fourth element, the neutral. Therefore, all phases are referenced to neutral. See the figure below for an example.
Figure 81-07
Note the presence of the neutral and the three phases. They are called VAN,
VBN and VCN.
Notice that in the denomination the arrow always points to the point of greatest potential.
So, VAN means that the point A
has voltage greater than the point N. Note that the value of the three voltages in modulus are exactly
the same, changing only the phase angle.
And the direction of rotation is always counter-clockwise .These three voltages are called PHASE VOLTAGE,
because each phase is referenced to the neutral.
The Delta circuit is characterized by the absence of neutral. In this way, the voltages are drawn
between the points ABC, and therefore these voltages are called LINE VOLTAGE. There is a mathematical
relation between voltage of phase and voltage of line as we will see later.
Figure 81-08
In the Figure 81-08 we have an example of a three-phase diagram representing the voltages of
LINE.
Note that the ABC points are connected by line segments, giving an equilateral triangle.
The three phase voltages, phase AN , phase BN and CN phase, are represented here by
VF, since all have the same value in modulus.
Like this, |VF| = |VAN| = |VBN| = |VCN|.
With this concept, we can establish the relationship between phase voltage and line voltage.
Notice that the triangles Δ BON and Δ AON are congruent rectangle triangles.
Then, the voltage VFis the hypotenuse of the two triangles. On the other hand, we know that & nbsp; OÂN = 30°.
The same holds for the angle B. Then, using the trigonometric relation, we have:
AO = BO = VF cos (30°) = (√3 / 2) VF
Mas, sabemos que:
AB = AO + BO = (√3 / 2) VF + (√3 / 2) VF
And hence, knowing that AB represents the line voltage and VF the
phase voltage, we can easily conclude that, in modulus, we have:
eq. 81-01
And of course, from this expression we can write that:
Similar process occurs for currents in three-phase systems. In the Star circuit, the line
currents are equal to the phase currents. And in the Delta circuit, the equivalence between
currents follows the same pattern for electric voltages. Therefore, in module, it is worth the following relations:
eq. 81-03
And of course, from this expression we can write that:
