This solution calculates the hysteresis loss in a transformer core based on its properties.
The actual area of the hysteresis loop in SI units ($Wb \cdot A / m^3$) is found by multiplying the numerical area by the scaling factors for both axes.
Area scaling factor = (B-axis scale) $\times$ (H-axis scale)
Area scaling factor = $ \left( 10^{-4} \frac{Wb}{m^2} \right) \times \left( 10^2 \frac{A}{m} \right) = 10^{-2} \frac{Wb \cdot A}{m^3} $
Actual Area ($A_{actual}$) = Numerical Area $\times$ Area scaling factor
$ A_{actual} = 70000 \times 10^{-2} \frac{Wb \cdot A}{m^3} = 700 \frac{Wb \cdot A}{m^3} $
Hysteresis loss per unit volume per cycle is equal to the actual area of the hysteresis loop.
Hysteresis loss per unit volume ($P_{vh}$) = $ A_{actual} \times f $
$ P_{vh} = 700 \frac{Wb \cdot A}{m^3} \times 50 Hz = 35000 \frac{W}{m^3} $
The total hysteresis loss ($P_h$) is the hysteresis loss per unit volume multiplied by the total volume of the core.
Total Hysteresis Loss ($P_h$) = $ P_{vh} \times V $
$ P_h = 35000 \frac{W}{m^3} \times 0.02 m^3 $
$ P_h = 35000 \times \frac{2}{100} W = 350 \times 2 W = 700 W $
The total hysteresis loss in the transformer core is 700 W.
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