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Question

A transformer core is wound with a coil carrying an alternating current at a frequency of 50 Hz. The hysteresis loop has an area of 70000 units, when the axes are drawn in units of $10^{-4} Wb-m^{-2}$ and $10^2 A-m^{-1}$. What is the hysteresis loss by assuming the magnetization to be uniform throughout the core volume of $0.02 m^3$?

The correct answer is
700 W

Transformer Hysteresis Loss Calculation

This solution calculates the hysteresis loss in a transformer core based on its properties.

Transformer Core Parameters

  • Frequency ($f$): 50 Hz
  • Hysteresis loop area (numerical value): 70000 units
  • Area units scaling: B-axis in $10^{-4} \frac{Wb}{m^2}$, H-axis in $10^2 \frac{A}{m}$
  • Core Volume ($V$): $0.02 m^3$

Actual Hysteresis Loop Area Calculation

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 Volume Determination

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} $

Total Hysteresis Loss Calculation

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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