Mindyard | In a single phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are 1600 W and 900 W , 900 W and 1600 W , 250 W and 600 W , 600 W and 250 W

In a single phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

Options

1600 W and 900 W

900 W and 1600 W

250 W and 600 W

600 W and 250 W

Correct Answer :

Option B

900 W and 1600 W

Solution :

The correct answer is 900 W and 1600 W.

Note on the attached image: The attached image shows a circuit diagram with a voltage source $V$ operating at $ω = 1 k rad/sec$ in series with a $4 mH$ inductor and $0.5 mF$ capacitor, and a load resistor $R_{L}$ in parallel with a $2 Ω$ resistor. This circuit diagram is unrelated to the single-phase transformer problem described in the text. The solution below is derived strictly from the text of the question.

The total iron loss ( $W_{i}$ ) in a transformer is the sum of the hysteresis loss ( $W_{h}$ ) and the eddy current loss ( $W_{e}$ ):
$W_{i} = W_{h} + W_{e}$

Let us calculate the ratio of voltage ( $V$ ) to frequency ( $f$ ) for both cases:
Case 1 (Nominal): $\frac{V_{1}}{f_{1}} = \frac{440}{50} = 8.8$
Case 2: $\frac{V_{2}}{f_{2}} = \frac{220}{25} = 8.8$

Since the ratio $V / f$ is constant, the maximum flux density ( $B_{m}$ ) in the core remains constant.
Under constant maximum flux density, the relationships for hysteresis and eddy current losses are:
Hysteresis loss: $W_{h} = A \cdot f$
Eddy current loss: $W_{e} = B \cdot f^{2}$
where $A$ and $B$ are constants.

Thus, the total iron loss can be written as:
$W_{i} = A \cdot f + B \cdot f^{2}$
Dividing both sides by the frequency $f$ yields:
$\frac{W_{i}}{f} = A + B \cdot f$

Now, substitute the values for the two cases to form a system of linear equations:
For Case 1: $W_{i 1} = 2500 W$ at $f_{1} = 50 Hz$
$\frac{2500}{50} = A + 50 B \Rightarrow A + 50 B = 50$ --- (Equation 1)

For Case 2: $W_{i 2} = 850 W$ at $f_{2} = 25 Hz$
$\frac{850}{25} = A + 25 B \Rightarrow A + 25 B = 34$ --- (Equation 2)

Subtract Equation 2 from Equation 1:
$(A + 50 B) - (A + 25 B) = 50 - 34$
$25 B = 16$
$B = \frac{16}{25} = 0.64$

Substitute the value of $B$ back into Equation 2:
$A + 25 \cdot 0.64 = 34$
$A + 16 = 34$
$A = 18$

Now, evaluate each loss at the nominal frequency of $f = 50 Hz$ :
Hysteresis loss:
$W_{h} = A \cdot f = 18 \cdot 50 = 900 W$

Eddy current loss:
$W_{e} = B \cdot f^{2} = 0.64 \cdot 50^{2} = 0.64 \cdot 2500 = 1600 W$

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In a single phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

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