2.2 - Secondary in "Y" and Load in "Delta"click here!
2.3 - Secondary in "Delta" and Carga in "Y"click here!
2.4 - Secondary in "Delta" and Load in "Delta"click here!
3 - Three-phase Circuits with more than one Loadclick here!
In three-phase circuits there is a power supply, usually a three-phase transformer, which feeds a three-phase load.
This load can be of the balanced load type, which is when we have the
same load impedance in the three phases, thus having the same power factor. Another possibility is to connect an unbalanced load, where the impedances connected to the secondary have different
values in at least one of them.
The unbalanced circuit will be studied in another chapter. For now, let's focus on the study
of the balanced circuit.
Let us analyze the four possible ways of connecting the transformer secondary together with the load, assuming that the sequence is positive or direct, also known as the sequence ABC. For the inverse sequence we must exchange B for C and vice versa. As a consequence, all changes to the negative sequence should be opposite to those in the positive sequence.
In the Figure 83-01 we show the circuit where we use a Y connection on the transformer secondary.
In the case of balanced load, the electric current flowing through the neutral is
equal to zero. Due to this, there is no need to use wire interconnecting the neutral of the load and the feeder transformer (in the figure, indicated by the dashed line between the points N-N ').
Since the circuit is balanced, then ZA = ZB = ZC.
Figure 83-01
Note that Iline = IF. As stated earlier, the current
of the neutral must be zero. Then we can treat the circuit as if it were a
single-phase circuit. See in the Figure 83-02 the simplification of the circuit
that will allow to calculate the current and the power in the load.
Figure 83-02
Note that we use the A phase as a reference. Therefore, the current of
phase A will have the phase angle equal to the difference between the
angle of VAN ∠ θA
and ZA∠ φZ, that is:
IA∠ θIA = (VAN / ZA)
∠ (θA - φZ)
Where, of course,
θIA = ∠ (θA - φZ).
As φZ is the same for all three phases, so θI will depend only on the voltage angle of the phase in which we are calculating the current. Note that since the voltage in the modulus is the same for the three phases as well as the impedances, then the current modulus will be the same for the three phases. It only changes the angle of lag, θI.
Thus, the current in the B phase will be the current of the A phase, in magnitude, plus
120° in the angle θIA.
Same for phase C, by adding 240° (or subtracting 120°, whichever is more convenient) to the angle θIA.
In the Figure 83-03 we present the circuit where we use a Y connection on the transformer secondary and the load is connected as a delta circuit.
Since the circuit is balanced, then ZAB = ZBC = ZCA.
Figure 83-03
Note that in this case, the voltage applied to the load is line voltage and not the phase voltage as in the previous item.
Therefore, on the load ZAB we have the voltage VAB. If, in the problem, the phase voltage is supplied, not the line voltage, then we must make the appropriate transformation. Let us recall the relation between line voltage and phase voltage in a star circuit.
eq. 83-01
We must pay close attention to the fact that we must add 30° to the angle of the phase voltage to get the correct angle of the line voltage, in addition to multiplying its magnitude by √ 3.
Knowing the line voltage you can calculate the phase current on the load by dividing the voltage by the impedance.
On the other hand, you can calculate the line current, if necessary, using the equation below.
eq. 83-02
We must not forget to multiply the phase current modulus by √3,
in addition to
subtract 30° from the angle θF of the phase current
to achieve the correct line current angle.
We will study this configuration from the didactic point of view, since it is not widely used in practice
due to the fact that we do not have neutral reference in the secondary.
See the circuit in the Figure 83-04. Then, using the transformer secondary in the delta configuration, the transformer provides
line voltage . Since the load is connected in Y we must find the phase voltage by dividing the line voltage by √3. So, to find the line current (IL) flowing through the load (in this particular case, the phase current is equal to the line current), we must divide phase voltage by the phase impedance. In a balanced circuit the three currents are equal in magnitude, varying only the angle.
Figure 83-04
Note that this situation is contrary to that presented in item 2.2. Then we must pay attention to the angle that the phase voltage will take. When we pass from the star circuit to delta, we advance the line voltage angle by 30° with respect to the phase voltage angle. Now, as we move from the delta circuit to the star circuit we must delay the phase voltage angle
(VF) in 30° em relação ao ângulo da tensão de linha (VL). Therefore, we must use the equation below.
This configuration also has no reference to neutral. However, it is a configuration widely used in transmission lines in charge of transporting large amounts of energy between two points distant from each other. As a rule, the starting point of the line has a transformer called
voltage elevator transformer. The secondary of this transformer operates with tens, or even hundreds of thousands of volts. On the other side of the transmission line, we have another transformer, called the
voltage drop transformer, with the purpose of reducing the voltage in the network to values used in the electric power distribution system. Then, the load on the secondary of the first transformer is the primary of the second transformer.
See the figure below for a schematic of this configuration.
Figure 83-05
Notice that in this configuration there is no need to change the voltage phase. We should only correct the angle of the currents of phase and line. For this, after calculating the phase current, we can find the line current by multiplying the magnitude of the phase current by √3 and subtract
30° at its angle, that is, we must use the eq. 83-02, repeated here.
eq. 83-02
In this equation, θF represents the angle of the phase current, and subtracting
30° from it, we find the angle of the line current.
As a rule, three-phase circuits feed more than one load. For example, in an industry there are several equipment operating simultaneously, such as motors, lathes, electric furnaces, elevators, etc ... In this way, we must be aware of the type of connection used in the various equipment. Let's look at an example.
Example 3.1
Be an industry powered by a three-phase network at 380 volts. Suppose an electric oven with a total power of 30 kW, connected in a delta configuration and two motors of 20 HP and
cos φ = 0.8 each, connected in the star configuration. Calculate the mains line current and the phase current in the electric oven. Assume that the engines have an efficiency of 90% and that 1 HP = 746 watts.
Solution
When we speak of a three-phase network, it is understood that the voltage supplied is the voltage of
line. Soon, Vline = 380 volts. We have an electric oven with 10 kW of power per phase. Since it is connected in the delta configuration, we know that Vline = VF.
Then we can find the phase current by calculating the quotient between the power and the voltage. Like this:
IF = PF / VF = 10 000 / 380 = 26.32 A
In a balanced circuit we know that Iline = √3 IF. Then:
Iline = √3 x 26.32 = 45.60 A
This is the current in the line only because of the electric oven. Now we have to calculate the current due to two motors. For this problem we will present the equation that allows to calculate the electric current of line of a three-phase electric motor.
eq. 83-04
Meaning of the variables in the equation: We are calculating the line current,
I of the motor. In the numerator of this equation, P represents the three-phase motor power given in HP. We multiplied by 746 because 1 HP = 746 watts. In the denominator we have
VL as the line voltage of the three-phase system. The greek letter η represents the efficiency of the engine, and is a number between zero and one. Finally, we have the power factor represented by cos φ.
Replacing the given numeric values, we find:
I = 40 x 746 / (√3 x 380 x 0.9 x 0.8) = 62.96 A
This current is due to the two 20 HP motors each.
To determine the line current of the network, we must add the value found for
the electric oven. Like this:
Inet = 62.96 + 45.60 = 108.56 A
Observation : If the power of the motor is given in CV, a unit still widely used, we must replace the value 746 with the value735 because
1 CV = 735 watts.
