Topic2/Design1
Designing three phase autotransformer, for 3x400V/3x380V, 10kVA output power and 25% one phase unsymmetrical load
Input parameters
| Input | Voltage | 3 x 400V, +-10%, 50Hz, sine wave |
| Wire | Cu, round, single insulated | |
| Layer insulation | No | |
| Final insulation | No | |
| Output | Nominal output voltage | 3 x 380V, star connection with 3 x 220V between phases and neutral line |
| Nominal output current | 3 x 15.15A and 1 x 3.78A UNSYMMETRICAL load | |
| Wire | Cu, round, single insulated | |
| Layer insulation | No | |
| Core | Steel | M45, alternate stacking, not annealed |
| Assembly | Gaped with 10 mil between E and I for limiting the magnetic flux of the unsymmetrical current. | |
| Bobbin | Type | Single section |
| Design | Insulation class | B, max. operating temperature 120C, |
| Ambient temperature | 40C |
Modifying the wire size from thermal point of view
Due to the fact that the program supports only the three phase symmetrical operation mode we need to compare the currents in both symmetrical and unsymmetrical operation mode, and modify the wire size by hand using the following pictures:
In the unsymmetrical operation mode the current between the neutral and Uo tap is Iun/3 = 3.78/3 = 1.26A. Between Ui and Uo taps the max current is 2Iun/3 = 2.52A
In the symmetrical operation mode the current between the neutral and Uo tap is (1-Uo/Ui)In= 0.635A (view the segment currents on the last page). Between Ui and Uo taps the max current is Ii = 14.6A.
If we compare the currents between the neutral and the Uo tap then we have to follow this procedure in order to design our autotransformer from thermal point of view properly:
1. Run the program in the symmetrical operation mode and select a bigger core in order to get the build approx. 40%
2. Increase the wire cross section of the winding between the neutral and the Uo tap approx. by the factor (1.26/0.635)^2 = 3.9
Calculating the induction during the unsymmetrical operation mode
Finally we need to check the size of the inductions Bo and B of the magnetic fluxes Fo and F.
1. The magnetic fluxes Fo are equal by size and phase. They flow through the legs and close their loops outside the core. They are normally very small and can be neglected.
2. Note that the ampere-turns of the no-load current Wi x Io = 845 x 0.268 = 226 excites over one legs the main induction of 1.3T (set on the input screen).
3. The magnetic flux F is driven by ampere-turns (on the leg with the unsymmetrical load) of the unsymmetrical current are:
(2/3) x (Wi-Wo) x Iun = 0.666 x (845-812)x3.78 = 83
Due to the fact that the phase delay between the main flux (driven by input voltage) and the flux F (driven by unsymmetrical current on the leg with the unsymmetrical load) is 90 degrees the max. induction in the leg during the unsymmetrical operation mode will be:
Bmax = 1.3 x (1+(83/226)^2)^0.5 = 1.3 x 1.06 = 1.385T
4. In order to get a very low influence of the unsymmetrical current the core has to be gaped by 10mil between the E and I part.
