3.4 Analysis of the single-phase transformer parameters
3.4.2 Simulation of single-phase transformer short circuit condition
3.4.2 Simulation of single-phase transformer short circuit condition
According to the theory of electrical machines, the short circuit test is made at transformer, if the secondary terminals are short circuited and a fraction of the rated voltage sufficient to produce rated currents, at rated frequency is applied to the primary terminals. At the simulation of this condition, both windings are fed by their rated currents; it means that magnetomotive forces are equal and current density corresponds to the values of currents and the next equation, at which the magnetizing current is neglected Iμ≈0 A, is valid, N1I1=N2I2, which results in the fact that no magnetizing flux is created in the core. Then only leakage flux occurs, which is closed through the leakage paths, which means by air, insulation, and nonmagnetic materials [1]. The 2D analysis is done in a different way in
Figure 63.
Distribution of the magnetic flux lines and calculation of the surface integral A.J (integral result) of the investigated transformer.
Figure 64.
Comparison of the magnetizing inductances obtained by measurement and FEMM simulation.
comparison with no load condition, when the whole magnetic flux was in the ferromagnetic core.
A procedure described in [11] is applied in the analysis of the short circuit condition. The value of the leakage inductance is calculated from the magnetic field energy, because in short circuit condition no saturation of ferromagnetic circuit occurs. The value of the energy is calculated as follows:
W ¼lav
ð
S
1 2μH2dS
where S is a surface of the whole transformer cross-section area and lavis an average length of the conductor or a half of the average length of the turn. It can be calculated as an average length between both windings of the primary and second- ary coils (Figure 65). Nevertheless, based on experience, this calculated value should be increased about 5 till 10%, because during the manufacturing, the coils are not wound exactly and this dimension is very important for leakage inductance calculation. An increase of about 7.5% is used here. Based onFigures 56and65, this value can be calculated as lav= 1.075�(50 + 2�2 + 2�10 + 2�0.25 + 49.6 + 2�2 + 2�10 + 2�0.25) = 159.7 mm, and the measured value is lav= 161 mm, which is employed during further calculation. The calculation of the leakage inductance can be made by two ways:
1.The first one is setting of the z-coordinate in Problem (Depth) on the value lav. The calculation is then made as follows:
Figure 65.
Illustration figure of the transformer cross-section area to define the average length of the turn.
Lσ1þLσ20¼Lσ ¼2W I1N2
or similarly as in the no load condition:
Lσ1þLσ20¼Lσ¼ ψ
I21N ¼∮A�JdV I21N
2.The second approach is such, that z-coordinate is set in Problem (Depth) on the value 1 mm and the values obtained in the postprocessor must be multiplied by the average length of the conductor or by one half of the average length of the coil turn lav. Then the calculation is as follows:
Lσ1þLσ20¼Lσ¼lav2W I1N2
or
Lσ1þLσ20¼Lσ ¼lav ψ
I21N¼lav∮A�JdV I21N :
The purpose of this simulation is to calculate the total value of the leakage inductance Lσ1þLσ20¼Lσ. In transformers that have only small number of the turns on the secondary side, e.g., there is only one layer, it is needed to draw individual turns and, in each turn, to define current or current density. If that few turns are replaced by only one turn to simplify it, a significant error could appear, because of leakage flux is flowing around the individual turns. In investigated transformer here, the secondary side is created by some layers; therefore the solution is made by one block of the turns.
A procedure of the calculation is as follows: The current densities J,
corresponding to the windings, are calculated. It must be valid N1I1=�N2I2. Then the current density is equal in both windings, but with opposite signs. The rated current in the primary winding is I1N= 2.75 A, and current density for the windings is as follows:
Primary winding, left side:
Jpwl¼N1I1N
Sp ¼354�2:75
0:00065 ¼1:497693 MA=m2 Primary winding, right side:
Jpwr¼ �N1I1N
Sp ¼ �354�2:75
0:00065 ¼ �1:497693 MA=m2
In coincidence with the equation N1I1=�N2I2, the secondary winding, left side, is:
Jswl¼ �N1I1N
Sp ¼ �354�2:75
0:00065 ¼ �1:49769 MA=m2 and secondary winding, right side, is:
Jswr¼N1I1N
Sp ¼354�2:75
0:00065 ¼1:49769 MA=m2
These values are introduced to the preprocessor in the corresponding coils. It is recommended to refine the mesh and to increase the number of finite elements around the windings and in the air because there is a main part of the leakage flux.
Then calculation and analysis of the results can be made. InFigure 66, the distri- bution of the flux lines, which correspond to the short circuit condition, is seen. As it was supposed, the flux lines are in the surrounding of the coils because the main flux is neglected. In this condition, no saturation occurs; therefore the calculation of the leakage inductance can be made by means of the energy of electromagnetic field and for comparison also from the linkage magnetic flux. It is recommended to do it by both ways.
The first way starts at setting of the z-coordinate on the value lav= 161 mm.
Then the calculation is launched to calculate the total leakage inductance. At the calculation based on the magnetic field energy, the whole cross-section area of transformer is marked in postprocessor. Then it is valid that:
Lσ1þLσ20¼Lσ¼2W
I1N2¼2�0:0193107
2:752 ¼5:1 mH
or based on the linkage magnetic flux, but in postprocessor only the areas corresponding primary and secondary windings are marked and calculate A.J.
Then it is valid that:
Lσ1þLσ20¼Lσ¼ ψ
I21N¼∮A�JdV
I21N ¼0:0386215
2:752 ¼5:1 mH
The second way is based on the fact that z-coordinate is set on 1 mm. Therefore, the value of energy must be multiplied by the average length of the conductor lav= 161 mm. For the calculation from the magnetic field energy, it is valid that:
Lσ1þLσ20¼Lσ¼lav2W
I1N2¼1612�0:000119942
2:752 ¼5:1mH
Figure 66.
Magnetic flux distribution in transformer under the short circuit condition.
Lσ1þLσ20¼Lσ¼2W I1N2
or similarly as in the no load condition:
Lσ1þLσ20¼Lσ ¼ ψ
I21N¼∮A�JdV I21N
2.The second approach is such, that z-coordinate is set in Problem (Depth) on the value 1 mm and the values obtained in the postprocessor must be multiplied by the average length of the conductor or by one half of the average length of the coil turn lav. Then the calculation is as follows:
Lσ1þLσ20¼Lσ ¼lav2W I1N2
or
Lσ1þLσ20¼Lσ ¼lav ψ
I21N¼lav∮A�JdV I21N :
The purpose of this simulation is to calculate the total value of the leakage inductance Lσ1þLσ20¼Lσ. In transformers that have only small number of the turns on the secondary side, e.g., there is only one layer, it is needed to draw individual turns and, in each turn, to define current or current density. If that few turns are replaced by only one turn to simplify it, a significant error could appear, because of leakage flux is flowing around the individual turns. In investigated transformer here, the secondary side is created by some layers; therefore the solution is made by one block of the turns.
A procedure of the calculation is as follows: The current densities J,
corresponding to the windings, are calculated. It must be valid N1I1=�N2I2. Then the current density is equal in both windings, but with opposite signs. The rated current in the primary winding is I1N= 2.75 A, and current density for the windings is as follows:
Primary winding, left side:
Jpwl¼N1I1N
Sp ¼354�2:75
0:00065 ¼1:497693 MA=m2 Primary winding, right side:
Jpwr¼ �N1I1N
Sp ¼ �354�2:75
0:00065 ¼ �1:497693 MA=m2
In coincidence with the equation N1I1=�N2I2, the secondary winding, left side, is:
Jswl¼ �N1I1N
Sp ¼ �354�2:75
0:00065 ¼ �1:49769 MA=m2 and secondary winding, right side, is:
Jswr ¼N1I1N
Sp ¼354�2:75
0:00065 ¼1:49769 MA=m2
These values are introduced to the preprocessor in the corresponding coils. It is recommended to refine the mesh and to increase the number of finite elements around the windings and in the air because there is a main part of the leakage flux.
Then calculation and analysis of the results can be made. InFigure 66, the distri- bution of the flux lines, which correspond to the short circuit condition, is seen. As it was supposed, the flux lines are in the surrounding of the coils because the main flux is neglected. In this condition, no saturation occurs; therefore the calculation of the leakage inductance can be made by means of the energy of electromagnetic field and for comparison also from the linkage magnetic flux. It is recommended to do it by both ways.
The first way starts at setting of the z-coordinate on the value lav= 161 mm.
Then the calculation is launched to calculate the total leakage inductance. At the calculation based on the magnetic field energy, the whole cross-section area of transformer is marked in postprocessor. Then it is valid that:
Lσ1þLσ20¼Lσ¼2W
I1N2¼2�0:0193107
2:752 ¼5:1 mH
or based on the linkage magnetic flux, but in postprocessor only the areas corresponding primary and secondary windings are marked and calculate A.J.
Then it is valid that:
Lσ1þLσ20¼Lσ¼ ψ
I21N¼∮A�JdV
I21N ¼0:0386215
2:752 ¼5:1 mH
The second way is based on the fact that z-coordinate is set on 1 mm. Therefore, the value of energy must be multiplied by the average length of the conductor lav= 161 mm. For the calculation from the magnetic field energy, it is valid that:
Lσ1þLσ20¼Lσ¼lav2W
I1N2¼1612�0:000119942
2:752 ¼5:1mH
Figure 66.
Magnetic flux distribution in transformer under the short circuit condition.
or from the linkage magnetic flux by means of integral A.J:
Lσ1þLσ20¼Lσ¼lav ψ
I21N ¼lav∮A�JdV
I21N ¼1610:000239885
2:752 ¼5:1 mH The measured value of total leakage inductance is 5.2 mH (seeTable 7), which means a very good coincidence of the results.
In this FEMM program, it is possible to calculate approximately the resistance of primary and secondary winding. For more precise calculation, the 3D program would be needed.
The resistance depends on the electrical conductivity of copper from which the windings are made. In simulating the value of copper, specific electrical conductiv- ityσ= 58 MS/m is used or can be set based on the material library in the FEMM program. The calculation starts from the short circuit simulation, whereby z-coor- dinate is set on 1 mm. Now the average length of the turn of primary lavpand secondary lavswinding must be calculated. The calculation is made based on the geometrical dimensions inFigures 56and65. Then lavp= 255.2 mm and
lavs= 339.2 mm. After the calculation, in postprocessor, the blocks must be marked, which correspond to the primary winding, and by means of the command Resistive losses, the Joule losses in the primary windingΔPjare calculated. In the same way, the losses in the secondary winding are calculated. Nevertheless, it is the same value because the cross-section area of the winding and current density is the same. The resistance is then calculated at 20°C for both windings as follows:
Rp¼lavpΔPj
I1N2¼255:20:050276
2:752 ¼1:71Ω Rs¼lavsΔPj
I2N2¼339:20:050276
26:32 ¼24:6 mΩ
For comparison the measured values are Rp= 1.91Ωand Rs= 20 mΩ.
In the end the simulated and measured values of equivalent circuit parameters are summarized inTable 8. It can be proclaimed that the values obtained by simulation and measurement are in good coincidence.