Three phase transformer losses

QuickField simulation example

We perform 2 classical experiments to determine the transformer parameters:
1. Nominal voltage is applied to the primary winding with the secondary winding open;
2. Nominal current is applied to the primary winding with the secondary winding short-connected.
For each regime we calculate the average magnetic flux density, core loss and copper loss.

Problem Type
Plane-parallel problem of AC magnetics.

Geometry
Model z-length is 40 mm.

Given
Primary winding Δ 400 V, 3 A, 1.43 mm² x 1092 turns,
Secondary winding Y 230 V, 2.54 mm² x 630 turns,
Frequency f=50 Hz, power 500 W.
Copper conductivity σ=56 MS/m.

Core permeability μ = 1000, core loss C_m = 1.5 W/kg at f=50 Hz and B=1.5 T.
Core steel density ρ = 7650 kg/m³.

Task
Calculate magnetic and electric losses in the three phase transformer.

Solution
Only the core and portion of the winding are present in the 2D model. To account for the resistance of the parts of the conductors not included in the model we reduce the conductivity of the material. For the primary winding the corrected conductivity value is 17 MS/m, for the secondary winding it is 9.73 MS/m.
QucikField uses the Bertotti core loss model. It includes three components: hysteresis loss, eddy current loss and excess loss. As no specific information is provided in the input data we classify the losses as hysteresis loss: P_hyst = k_hyst · B² · f [W/m³].
Taking into account that P_hyst[W/m³] = C_m[W/kg] * ρ[kg/m³], we can calculate the value of k_hyst = (1.5*7650) / (1.5)² / 50 = 102.

Results

Reference:
* http://en.wikipedia.org/wiki/Transformer

• Video: Three phase transformer losses. Watch on YouTube.
  
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