A real transformer uses an iron core to increase the coupling coefficient between the primary and secondary windings. This is due to the increased mutual flow that iron provides.
Therefore, for the case of the real transformer, let's consider that the coupling coefficient is
unitary.
The presence of the iron core causes a loss because energy is required to magnetize the core. To represent this loss in the electrical model of the transformer, we add in parallel with the primary winding an inductance symbolized in the model by a reactance designated as Xm.
Moreover, we must provide a resistance in parallel with this reactance to represent the losses that occur due to the presence of the hysteresis and parasitic currents that exist in the iron core. This resistance is symbolized by Rh in the electric model of the transformer, as can be seen in the Figure 92-01.
Figure 92-01
Notice the model above, E1 represents the call Counter Electromotive Force, voltage which opposes the
voltage applied to the primary, V1. If V1 is constant, then E1 so will be,
as will the magnetic flux in the transformer core. This value constancy must be maintained for any electrical current that the
transformer secondary supplies to the load. Thus, the primary winding must absorb from the feeder line, in addition to the
magnetizing current, an electric current I'1. This current is called the
primary reaction current.
The total magnetic flux concatenated with the primary winding can be divided into two components: the resulting mutual
flux which is confined to the iron core and is the result of the combined effect of the primary and secondary electric
currents; and the leakage flux of the primary and secondary, which concatenates with itself. On the other hand,
most of the leakage flux is in the air.
Since air is not saturated, then the leakage flux and the voltage induced by it vary linearly with the primary
current I1. All this effect can be simulated, in the electric model, by a
leakage reactance, represented by Xm.
We must not forget that for making the transformer windings copper or aluminum wires are used, and therefore have
electrical resistance. These resistances cause losses by Joule effect and are represented in the electric model
by R1 (primary) and R2 (secundary).
Therefore, it should be clear that the voltage V1 opposes three phasor voltages: the voltage drop
in the resistance R1 of the primary, the voltage drop in the primary leakage reactance
X1 and the counter electromotive force E1 induced in the primary by the resulting mutual flux.
Resume
In short, in order to study the model of a real transformer we must consider four essential points.
1 -Copper Losses: (R I2) due to the heating of the wires by the passage of the electric current. They are proportional to the square of the electric current.
2 -Losses by Eddy Currents: due to the heating of the transformer core caused by the magnetization of it and are proportional to the square of the voltage applied to the transformer.
3 -Hysteresis Losses: due to the change in the configuration of the transformer core magnetic domains and is a nonlinear function of the voltage applied to the transformer.
4 -Leakage Flux: is the flux that does not include the primary and secondary transformer windings and their losses are represented by the leakage inductance of each winding.
Attention
"It is noteworthy that in a good transformer design, the power dissipated in
the primary winding resistance must be equal to the power dissipated in the
secondary winding resistance."
Thus, by performing two tests (studied below) on the transformer, we were able to determine the value of
resistances of the primary and secondary.
As for reactances , it follows the same principle and with the two tests mentioned above we can determine their values as well as the values of Rh and Xm.
From considerations regarding core losses and magnetization current in power transformers, it is possible to disregard these values. In general, the current IP does not exceed
3% the value of the rated transformer current.
Therefore, we will not take this current into account.
By the point convention, an electric current that enters a winding through the dotted terminal produces a positive magnetomotive force, whereas an electrical current that enters the terminal no point (or exit through the dotted terminal) of a winding produces a negative magnetomotor force.
Thus, when we connect a load to the secondary of the transformer and obeying the polarities of the induced voltage in the secondary, according to the model above, the current that will circulate through the load goes out through the dotted terminal causing a negative magnetomotive force. On the other hand, the primary current enters the dotted terminal generating a positive magnetomotor force. Thus, the total magnetomotive force at the transformer core is equal to:
eq. 92-01
Where the variables are:
FT - Total magnetomotive force
I1 - Electrical current in transformer primary
I2 - Electrical current in transformer secondary
N1 - Number of turns of transformer primary
N2 - Number of turns of transformer secondary
R - Magnetic reluctance of transformer core
Φ - Magnetic flux in the transformer core
But we know that good transformer designs using good quality silicon iron in the core makes the
magnetic reluctance of the core very close to zero . So we can say that the total magnetomotive force is null. As a consequence we obtain the relation between the transformer primary and secondary electric currents given by the equation below:
eq. 92-02
Recalling the equations 91-01 and 91-03 from the previous chapter, we can write the following relation:
Based on the considerations of the previous item, not considering the current IP,
let's turn the previous circuit making a reflection to the primary of the impedance present in the
secondary. In the Figure 92-02 we can see how this transformation went.
Figure 92-02
Notice that the reactance X2 of secondary was multiplied by a2,
as well as resistance R2. As mentioned earlier, we will not take the current
(IP) into account.
Taking this into consideration we can write that:
Another way to analyze the circuit model of a transformer is to reflect on the secondary. In the
Figure 92-03 we can see how this transformation went.
Figure 92-03
Note that in this case the resistances and reactances of the primary are transferred to the secondary by dividing their values by a2. The resistance R2 and the reactance
X2 don't have your values changed.
As for the currents, when we despise Rh and Xm, we have that:
In this test a transformer winding is open circuit, ie there is no connection to it.
The other winding is connected to a voltage source that has the rated line voltage of the transformer.
Under these conditions, all input current must circulate in the transformer excitation circuit. And because the winding resistance and leakage reactance are very small compared to Rh and
Xm, they can be disregarded in the model. Thus, the excitation circuit is subjected to all input voltage.
On the other hand, to be able to perform the measurements correctly, we must connect a power meter, a voltmeter and an ammeter on the side that is connected the voltage source. It should be noted that all measurements should preferably be performed on the low voltage side of the transformer as a safety precaution. In the Figure 92-04 we can appreciate how are the connections in the transformer.
Figure 92-04
Using the wattmeter allows you to measure the power consumed by the transformer. Also, we know that a watmeter only measures real power. Therefore, this power is the power consumed by
Rh, representing the hysteresis losses in the transformer iron. Then we can easily determine the value of this resistance by knowing the values measured by the watmeter and voltmeter. Thus:
eq. 92-04
Also, knowing the nominal voltage (Vnom) and the current Ip we can calculate the
equivalent impedance of the transformer excitation circuit. Soon:
eq. 92-05
Of the three unknowns, we already know two. Remembering that the excitation circuit is formed by a
circuit R-L,in the parallel configuration, we must emphasize that the current that goes through Rh is in phase with Vnom, however the current that goes through Xm is 90° delayed compared to Vnom. So to find the value of
Zeq we must apply the Pythagorean theorem . So, working with the admittances we have the following relation:
(1 / Zeq)2 = (1 / Rh)2 + (1 / Xm)2
But we are interested is in the value of Xm. Then, working algebraically the previous equation we find the following relation:
eq. 92-06
And so, we have all the equations needed to calculate the variables of interest referring
to the excitation or magnetization circuit of the transformer.
In the statement of many problems we find the need to calculate the power factor
from transformer to open circuit. The wattmeter reads the actual power being consumed by the transformer. Since losses in copper are very small as a function of no-current, so they can be disregarded. So we can write that the power read by the wattmeter is given by Wo = V1 Ip cos φo. Hence, we conclude that:
eq. 92-07
Thus, we can calculate the power factor of the transformer under open circuit, represented here by
cos φoand in practice may vary from 0.1 and 0.3.
To determine the components R1, R2,
X1 and X2 we then use the short circuit test.
This test consists of shorting the transformer secondary. In the primary of the transformer, we connect equipment called variac , or another equivalent, which allows the voltage of the primary to be varied until reaching the nominal current specified by the transformer manufacturer on the secondary. This current will be considered the short-circuit current which we will call
Icc. At this point, we note the voltage read by the voltmeter (Vcc ), the current read by the ammeter on the secondary (Icc ) and the power (Wcc ) consumed by the transformer. Note that the primary voltage will be a small portion of the nominal voltage, as a rule, ranging from 2% to 12%.
In the Figure 92-05 we see the equivalent circuit for the short-circuit test with the primary being referred to the secondary. Note that we can disregard the transformer excitation circuit because the current that will pass through this circuit is very small when compared to the nominal current of the transformer.
Figure 92-05
From the above circuit we can express the equivalent impedance of the transformer as:
eq. 92-08
On the other hand, we can calculate this impedance using the information from the ammeter reading
(Icc ) and reading the voltmeter (Vcc ).
eq. 92-09
We know that a wattmeter measures real power, that is, the power dissipated into resistive components.
Thus, this data allows to calculate the value of the equivalent resistance of the circuit together with
the reading made by the wattmeter and ammeter. So we have:
eq. 92-10
We already know two of the three variables involved in the problem. Using this we can calculate the third variable, or:
eq. 92-11
Please note that we are using the statement made previously (item 2. "attention" or
Here!), that in a real
transformer the resistances and reactances of primary and secondary
are designed to be equal or very close, naturally, respecting the relation of transformation.
At this point, we will distinguish the variables when referring to the primary or the
secondary using single quotes or double quotes, respectively . So when we write
R'eq, X'eq or Z'eq we will be referring to values referred to the PRIMARY of resistance, reactance and impedance, respectively. And when we write
R"eq, X"eq or Z"eq we will be referring to
values referred to the SECONDARY of resistance, reactance and impedance, respectively.
Thus, when the impedances are referred to the secondary, we can make the following approximations:
R"eq = ( R1/ a2) + R2
X"eq = ( X1/ a2) + X2
If there is interest in referring to the primary, we can write:
R'eq = R1 + a2 R2
X'eq = X1 + a2 X2
In addition, based on the actual model of the transformer we can write that:
R2 = R1/ a2
X2 = X1/ a2
Therefore, we conclude that the values of R2 e X2 are calculated by:
R2 = R"eq/ 2
eq. 92-12
X2 = X"eq/ 2
eq. 92-13
And obviously, the values of R1 and X1, already referred to the primary, are calculated by:
R1 = a2 R2 = R'eq/ 2
eq. 92-14
X1 = a2 X2 = X'eq/ 2
eq. 92-15
Following the same line of reasoning, we can easily prove the validity of the relations described below:
It is also often necessary to calculate the power factor of the transformer in short circuit. As the wattmeter reads the actual power being consumed by the transformer and, in this case, the losses in the iron are very small due to the short-circuit current, so they can be disregarded. So we can write that the power read by the wattmeter is given by Wcc = Vcc Icc cos φcc, remembering that Icc = I2, where I2 is the nominal current of the secondary of the transformer.
So, we conclude that:
eq. 92-16
Thus, we were able to calculate the power factor of the transformer in short circuit,
represented here by cos φcc.
As seen in the previous item, the parameters of the transformer related to the windings are
constant and independent of the current. However, the output voltage at the secondary of the transformer varies depending on the load.
In this way, we will define Regulation of a transformer for a given load
as:
Definition - "The relationship between the voltage variation that occurs in the secondary with and without load
and the voltage of the secondary with load,
keeping the primary voltage constant V1."
Regulation is usually expressed in percentage.
In the eq. 92-17 we see the relations between these variables.
eq. 92-17
In this equation we have to E2o represents the voltage of the unloaded secondary,
and V2
represents the secondary voltage at full load, that is, it is the nominal voltage value of the transformer.
Thus, regulation
depends on the equivalent impedance of the transformer. The smaller the equivalent impedance of the
transformer, the better the regulation will be.
Another factor that interferes with regulation is the load power factor. We know that the nominal voltage
of the transformer provided by the manufacturer is the voltage that appears at the transformer output terminals
when it operates at full power. In this case, the current I2 supplied to the load by the secondary
is constant.
7.1 Influence of Load Power Factorat the Value of E2
Taking into consideration eq. 92-17, we see that the subtraction of the two tensions that appear in the numerator can be
treated with a ΔV, that is, it is the voltage drop across the transformer's internal impedance. This ΔV
is variable and depends on the value of the load impedance angle (φ2), as we are
considering the constant secondary current.
Thus, we have three possibilities: inductive, resistive or capacitive load.
To study the value of ΔV under different load conditions, we will use the constructed transformer diagram
in an alternative form, known as the Kapp diagram.
To construct this diagram, it is necessary to previously know the values of resistance and
reactance of the transformer, referred to the secondary. Thus, maintaining the adopted convention, we have,
R''eq = R1 / a2 + R2 and
X''eq = X1 / a2 + X2. We will use it as a base
the circuit shown in Figure 92-06. Note that this circuit is the representation of the transformer equivalent circuit
referred to the secondary, as shown in Figure 92-03.
Figure 92-06
To create this diagram, we will use the secondary current (I2) at full load. Thus, the triangle is constructed
OAB, which has the side OA = R''eq I2. Note that this leg is in phase with I2.
The other side, AB, represents the voltage drop across the reactance or, AB = X''eq I2. This cateto is in
quadrature (90°) in relation to I2. It is clear that the hypotenuse (OB) of this triangle represents
the voltage drop across the transformer impedance,
OB = Z''eq I2. This triangle is known as the fundamental triangle of
transformer.
With center at O, we draw a circle (in blue) with a radius equivalent to the no-load tension E2o . Each phasor,
for example, BC, represents the voltage V2 that appears at the transformer terminals
when it supplies electrical current
I2 offset by an angle φ2 in relation to the BC.
Note that the triangle thus constructed
in fact translates the phasor relationship E2o = V2 + R''eq
I2 + X''eq I2,
which is the well-known equation relating to the equivalent circuit referred to the secondary of the transformer and
built from the circuit of Figure 92-06.
Figure 92-07
Now, with center at B, let's draw a second circle (in red) that also has the radius of the no-load tension
E2o. From Figure 92-07, we see that the segment CD represents the arithmetic difference
between E2o and V2, that is, it is the voltage drop across the internal
impedance of the transformer
when operating at full load and the load having a lag angle φ2. Varying the factor
load power,
the voltage drop over the transformer's internal impedance varies according to what is shown in the graph in Figure 92-07.
Note that when
φ2 = 0 the load is resistive and the voltage V2 is represented by the segment BE
being in phase with I2. In this case, the voltage drop is represented by ΔV'.
It is worth noting that when point C passes through point M, we have ΔV = 0, that is, the output voltage at the terminals
of the transformer secondary will have the same value both in no-load and loaded operation. It is also possible to conclude that,
from point E upwards, represents an inductive load, and downwards, represents a capacitive load.
The efficiency of a transformer, represented by the greek letter η, is defined as the relation between electrical power
P2 supplied by the secondary to the load and the electrical power
P1 corresponding, absorbed from the power source by the primary of the transformer.
eq. 92-18
We must remember that a transformer has power losses by Joule effect due to the ohmic resistance of the windings and are generally referred to as losses in copper. For this loss we will use the symbol Pcu. In addition to this, we have losses in the iron used in making the core of the transformers. These losses are due to hysteresis and eddy currents. For this loss we will use the symbol Pfe.
For copper losses in a single-phase transformer, we can clearly determine them by the equation below.
Pcu = R1 I12 + R2 I22
We know that a = I2 / I1, or I2 = a I1. So, replacing in the previous equation, we will find:
Pcu = I12 ( R1 + R2 a2).
But, note that R1 + R2 a2 is the equivalent resistance
(R'eq) of the winding when referring to the primary. In this way, we obtain the equation:
Pcu = R'eq I12
Using the same line of reasoning as the previous paragraph, we can obviously write that:
Pcu = R"eq I22
We conclude, then, that copper losses are directly proportional to the square of the current absorbed by the primary or supplied to the load by the secondary of the transformer. In other words, if the load varies, losses on copper also vary.
Contrary to losses in copper, losses in iron do not depend on the load current. In this way, electrically,
are a constant and represented by Pfe.
In the case of transformers, to reinforce the magnetic field, the coils are wound on low reluctance iron cores.
Naturally, these materials are conductors and are under the action of varying magnetic fields. Due to this fact, in the core
eddy currents are induced, called Foucault currents and generate three effects, as described below.
They heat the material byJoule effect
They generate magnetic fields that oppose the external field, weakening it.
They generate electromagnetic forces.
Eddy currents cause losses that are sought to be reduced by using cores that have high electrical resistance.
The sum of losses due to hysteresis and those due to eddy currents give rise to what is called iron losses and is
are present in all electrical machines.
The real power that the secondary of the transformer delivers to the load is given by:
P2 = V2 I2 cos (φ2)
Where V2 is the voltage on the load, I2 is the current of
the transformer secondary that flows through the load and cos (φ2) is the
power factor of the load. It should be noted that V2 is not the nominal voltage of the
transformer secondary, but the voltage over the load, where we take into account the voltage drop over
R2 and X2.
In this item we are going to rewrite the efficiency equation in a more comprehensive way taking into account what was studied in item 8. We must emphasize that the power that the primary absorbs from the source can be expressed as the power supplied by the secondary plus the
losses in copper and the iron losses. That is:
P1 = P2 + Pfe + Pcu
Substituting the values of these last two equations in eq. 92-14 we can express the efficiency equation as:
eq. 92-19
In the eq. 92-19, write P2 as V2 I2 cos φ2
and copper losses, PCu as R"eq I22 where R"eq
is the equivalent resistance of the transformer referred to the
secondary.
Analyzing the eq. 92-15 we can see that the efficiency of a transformer is dependent on the current of the secondary, I2, current flowing through the load. In order to find the necessary conditions for maximizing efficiency to occur, we must derive eq. 92-15 in relation to I2 and match the result to zero. We leave it to the student to make this derivation. The result to be found will be:
eq. 92-20
Therefore, to maximize the efficiency of a transformer, it is essential to balance copper losses with iron losses.
This is known as the equal loss condition and is a fundamental aspect of transformer design. Losses in copper are caused
by the electrical resistance of the windings, while losses in iron are mainly due to hysteresis and eddy currents in the iron core.
Adjusting the design so that these two losses are equal allows the transformer to operate at its optimum efficiency point,
which not only saves energy but also increases the useful life of the equipment. This balance is achieved through careful
selection of materials, sizing of components and configuration of the magnetic circuit.
The transformer does not always work with the maximum load it can support. As a rule, this does not happen. To evidence this situation, we have the so-called load factor, which is defined as the relation between the power supplied by the transformer to the load and its nominal power. So, for example, for a transformer that has a rated power of 100 kVA, but it supplies a load of 80 kVA we say that the load factor of the transformer is 80%. This implies that:
eq. 92-21
Where the variables are:
Fc - load factor, also known as loading.
SL - power supplied to the load by the transformer.
Snom - rated power of the transformer, that is,
is the maximum power that the transformer can deliver to the load.
Note that the above two equations express the same variable, that is, load factor. The one on the left is a number that varies between 0 and 1 and the one on the right is a percentage value, which varies between 0% and 100%.
Many problem statements present the load factor as a value in percentage.
From all the arguments presented in this item, it is clear that the efficiency of a transformer depends on the load power factor and the load factor of the transformer. So, if we change these two variables we will also be changing the efficiency of the transformer.
Let's make a change to the eq. 92-19 including the load factor. Thus, the equation is:
eq. 92-22
Note that the load factor of the transformer is multiplying all the variables that depend on the current in the secondary of the transformer, except for the losses in the iron, as this, as we have already stated, independent of it.
In this equation the variables are:
Fc - load factor, also known as loading.
cos φ2 - load power factor.
Pfe - losses in the transformer iron.
Pcu - losses in the copper of the transformer windings.
Snom - rated power of the transformer, that is,
is the maximum power that the transformer can deliver to the load.
An alternative formula for calculating efficiency as a value in
percentage is shown below. The variables are the same as in
eq. 92-22.
