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# Three phase transformer losses

three phase transformer losses, transformer loss FEA finite element analysis, core loss calculation, copper loss calculation

Unbalanced load is connected to the three phase transformer.

Problem Type
Plane-parallel problem of AC magnetics.

Geometry  Given
Core permeability μ = 1000,
Core mass density ρ = 7650 kg/m³,
Core losses Cm = 1.5 W/kg (at f=50 Hz and B=1.5 T),
Copper conductivity g=56 MS/m,
Primary winding Y: 6.25 mm² x 2560 turns,
Secondary winding Δ: 16 mm² x 150 turns,
Frequency f=50 Hz.

Calculate magnetic and electric losses in the three phase transformer.

Solution
Winding (copper) losses volume density:
pe = j² / g [W/m³].
Steinmetz equation to calculate core (steel) losses volume density:
pm = Cm · (f/50)α · (B/1.5)β · ρ [W/m³],
where α = 1, β = 2, B - average flux density in the core (peak value).

Results
Average flux density in the core (peak value): B = √0.746 · √2 = 1.22 T.
Power losses in the core pm = 1.5 · (50/50)1 · (1.22/1.5)² · 7650 = 7.59 kW/m³. Winding (copper) losses volume density
pe = (I/S)² / g:

 Winding name Conductor cross section, S Phase current (RMS), I Joule heat losses, pe A1 6.25 mm² 19 A 165 kW/m³ B1 6.25 mm² 13.4 A 82 kW/m³ C1 6.25 mm² 9.4 A 40 kW/m³ A2 16 mm² 45.7 A 146 kW/m³ B2 16 mm² 35.3 A 87 kW/m³ C2 16 mm² 23.2 A 38 kW/m³ * Reference: http://en.wikipedia.org/wiki/Transformer Product Order Evaluation request Editions Version history Packages Components Programming Consulting Applications Industrial Educational Scientific Sample problems Success stories Customers Support Webinars Virtual Classroom Online courses Customer login Glossary QuickField help FAQ
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