Question

The eddy current loss and hysteresis loss of 1-\(\phi\), 100 kVA, 50 Hz transformer are 4 kW and 6 kW respectively. If frequency is increased by 10%, then total loss is

• A. \( 11.44\text{ kW} \)
• B. \( 11.66\text{ kW} \)
• C. \( 11.00\text{ kW} \)
• D. \( 12.10\text{ kW} \)

Hint:

When $B_m$ is constant: - Hysteresis loss scales linearly with frequency ($f$). - Eddy current loss scales quadratically with frequency ($f^2$). Always compute the proportional multipliers ($1.1$ and $1.21$) and apply them directly to the separate loss values. {|c|c|c|} Loss Type & Proportionality & New Value
Hysteresis & $\propto f$ & $6 \times 1.1 = 6.6$
Eddy Current & $\propto f^2$ & $4 \times 1.21 = 4.84$

Answer 1

The Correct Option is A

Concept: Core loss in a transformer consists of Hysteresis loss ($P_h$) and Eddy current loss ($P_e$). When calculating changes due to frequency under standard test conditions where the voltage is not specified to change independently, the supply voltage $V$ is assumed constant. Let's look at the dependency formulas on frequency ($f$) and maximum flux density ($B_m$) where $V \propto B_m \cdot f \Rightarrow B_m \propto \frac{V}{f}$:
• Hysteresis Loss: $P_h = k_h \cdot f \cdot B_m^{1.6} = k_h \cdot f \cdot \left(\frac{V}{f}\right)^{1.6} \propto \frac{V^{1.6}}{f^{0.6}}$
• Eddy Current Loss: $P_e = k_e \cdot f^2 \cdot B_m^2 = k_e \cdot f^2 \cdot \left(\frac{V}{f}\right)^2 \propto V^2$ However, in standard practice questions of this specific format, if the voltage is implicitly tracking standard operation or if the question intends to examine basic dependencies directly with constant $B_m$ (where $V/f = \text{constant}$), the relations are $P_h \propto f$ and $P_e \propto f^2$. Let us carefully evaluate the standard interpretation here. If $V$ is constant, $P_e$ remains constant ($4\text{ kW}$) and $P_h$ decreases, which is not supported by the options increasing past $10\text{ kW}$. Therefore, this question employs the standard assumption that maximum core flux density (\(B_m\)) is maintained constant, meaning the voltage scales proportionally with frequency ($V/f = \text{constant}$). Under constant $B_m$: $$P_h \propto f \quad \Rightarrow \quad P_h = A \cdot f$$ $$P_e \propto f^2 \quad \Rightarrow \quad P_e = B \cdot f^2$$

Step 1: Write down the initial loss values at the baseline frequency.
Let the initial baseline frequency be $f_1 = 50\text{ Hz}$. At this frequency, we have: $$P_{h1} = 6\text{ kW}$$ $$P_{e1} = 4\text{ kW}$$

Step 2: Determine the new frequency after a 10% increase.
The frequency is increased by $10%$. Therefore, the new frequency $f_2$ is: $$f_2 = f_1 + 0.10 \cdot f_1 = 1.1 \cdot f_1$$

Step 3: Calculate the new hysteresis loss \( P_{h2} \).
Since $P_h \propto f$, when the frequency increases by a factor of $1.1$, the hysteresis loss increases directly by the same factor: $$P_{h2} = P_{h1} \cdot \left(\frac{f_2}{f_1}\right) = 6\text{ kW} \cdot 1.1 = 6.6\text{ kW}$$

Step 4: Calculate the new eddy current loss \( P_{e2} \).
Since $P_e \propto f^2$, when the frequency increases by a factor of $1.1$, the eddy current loss increases by the square of that factor: $$P_{e2} = P_{e1} \cdot \left(\frac{f_2}{f_1}\right)^2 = 4\text{ kW} \cdot (1.1)^2$$ $$P_{e2} = 4\text{ kW} \cdot 1.21 = 4.84\text{ kW}$$

Step 5: Compute the total core loss at the new frequency.
The new total loss ($P_{\text{total, new}}$) is the sum of the newly calculated individual losses: $$P_{\text{total, new}} = P_{h2} + P_{e2}$$ $$P_{\text{total, new}} = 6.6\text{ kW} + 4.84\text{ kW} = 11.44\text{ kW}$$ This perfectly matches option (1).