No-load tests are tests that apply rated voltage in the primary side at the no-load state of the secondary side. Current only flows to the primary side in the no-load test, but this current causes excitation and iron loss of the iron core.
If it is an iron core that has a tripod structure, the magnetic circuit of each phase is not completely equivalent, and thus causes unbalanced current. Also, as the iron core uses anisotropic magnetic steel sheets, iron loss will concentrate in the joint of the leg and yoke.
This example focuses on modeling in the analysis of no-load tests in JMAG and obtains the iron loss distribution, current, excitation conductance, and excitation susceptance that can be acquired from there.
Magnetic flux density distribution, Iron Loss Density Distribution
Magnetic flux density distribution of the core is shown in fig. 1, iron loss density distribution in fig. 2, and the iron loss density waveform in the center axis of the central leg in fig. 3. Directional magnetic steel sheets are used, so effects of the hard axis due to flux bending occur in the joint area of the leg and yoke. A concentration of magnetic flux density and iron loss density can be seen especially in the connecting area of the central leg. Furthermore, regions where iron loss density is concentrated are spreading as it gets deeper into the core, and as shown in fig. 3, peak values are 1.6 times larger than the average value. We learn that a design accounting for the effects on heat generation inside the iron core, neiboring winding and insulating material is necessary.
Iron Loss Current, Magnetization Current
Excitation conductance, Excitation susceptance
Excitation conductance and excitation susceptance are shown in table. 2, and the T-shaped equivalent circuit diagram in the transformer is shown in fig. 4. In the no-load test, the circuit constant in the dotted lines are obtained in the equivalent circuit shown in fig. 4. Voltage decreases due to the primary side coil resistance and the primary side leakage reactance are considered to be sufficiently small, so V’=V is assumed.