Transformer Core Loss Components Explained

In any transformer, when alternating current flows through the primary winding, it creates a changing magnetic flux in the core. This alternating flux causes energy losses within the iron core, collectively known as core losses or iron losses. These losses are crucial as they affect the transformer's efficiency. Core losses consist of two main components:

• Hysteresis Loss ($P_h$): This loss occurs due to the molecular friction within the magnetic material of the core as it is repeatedly magnetized and demagnetized by the alternating flux. The energy is lost as heat due to the rotation of magnetic domains. Hysteresis loss is influenced by the core material's properties, the frequency of the applied voltage, and the maximum magnetic flux density ($B_m$) reached in the core.
• Eddy Current Loss ($P_e$): As the magnetic flux changes in the core, it induces voltages within the core material itself. These induced voltages cause small circulating currents, known as eddy currents, to flow within the core. Because the core material has electrical resistance, these currents dissipate energy as heat ($I^2R$ loss). Eddy current loss depends significantly on the frequency, flux density, and the core's construction (specifically, the thickness of the laminations used to increase resistance and reduce these currents).

The total core loss ($P_{core}$) is the sum of these two components: $P_{core} = P_h + P_e$.

Data Analysis for Constant Flux Density

We are provided with core loss measurements under two different operating conditions:

The maximum flux density ($B_m$) in the transformer core is related to the applied voltage ($V$) and frequency ($f$) by the approximate formula $V \approx 4.44 f N K \Phi_m$, where $\Phi_m$ is related to $B_m$. Therefore, we can infer that $B_m$ is approximately proportional to the ratio $\frac{V}{f}$. Let's check this ratio for the given conditions:

• For Condition 1: $\frac{V_1}{f_1} = \frac{230 \text{ V}}{50 \text{ Hz}} = 4.6$
• For Condition 2: $\frac{V_2}{f_2} = \frac{138 \text{ V}}{30 \text{ Hz}} = 4.6$

Since the ratio $\frac{V}{f}$ is identical (4.6) for both test conditions, it indicates that the maximum flux density ($B_m$) in the transformer core remains constant during these two measurements.

Formulating Core Loss Equations

The standard empirical relationships for core losses are:

• Hysteresis Loss ($P_h$): $P_h = K_h f B_m^x$, where $x$ (Steinmetz exponent) is typically between 1.6 and 2.3. A common approximation used is $x=2$. So, $P_h \propto f B_m^2$.
• Eddy Current Loss ($P_e$): $P_e = K_e f^2 B_m^2$. So, $P_e \propto f^2 B_m^2$.

Because we have established that $B_m$ is constant in our given scenarios, we can simplify these relationships by combining the constants:

• Hysteresis Loss: Since $P_h \propto f B_m^2$ and $B_m$ is constant, $P_h$ is directly proportional to frequency $f$. We can write this as $P_h = A f$, where $A = K_h B_m^2$.
• Eddy Current Loss: Since $P_e \propto f^2 B_m^2$ and $B_m$ is constant, $P_e$ is directly proportional to the square of the frequency $f^2$. We can write this as $P_e = B f^2$, where $B = K_e B_m^2$.

The total core loss is the sum of these two components: $P_{core} = P_h + P_e = A f + B f^2$.

Using the data provided, we can establish a system of two linear equations:

1. For the first condition (230 V, 50 Hz, 1050 W): $1050 = A(50) + B(50^2)$ $1050 = 50A + 2500B$
2. For the second condition (138 V, 30 Hz, 500 W): $500 = A(30) + B(30^2)$ $500 = 30A + 900B$

Step-by-Step Calculation of Losses

We now solve the system of equations for the constants $A$ and $B$. The equations are:

1) $1050 = 50A + 2500B$

2) $500 = 30A + 900B$

To simplify, we can divide Equation 1 by 50 and Equation 2 by 10:

1') $21 = A + 50B$

2') $50 = 3A + 90B$

From Equation 1', let's express $A$ in terms of $B$: $A = 21 - 50B$

Now, substitute this expression for $A$ into Equation 2': $50 = 3(21 - 50B) + 90B$ $50 = 63 - 150B + 90B$ $50 = 63 - 60B$

Rearrange to solve for $B$: $60B = 63 - 50$ $60B = 13$ $B = \frac{13}{60}$

Substitute the value of $B$ back into the expression for $A$: $A = 21 - 50B = 21 - 50 \left(\frac{13}{60}\right)$ $A = 21 - \frac{5 \times 13}{6}$ $A = 21 - \frac{65}{6}$ $A = \frac{126}{6} - \frac{65}{6}$ $A = \frac{61}{6}$

With the constants $A = \frac{61}{6}$ and $B = \frac{13}{60}$, we can now calculate the hysteresis and eddy current losses for the specified input condition (230 V, 50 Hz).

Hysteresis Loss ($P_h$): $P_h = A f = \frac{61}{6} \times 50$ $P_h = \frac{3050}{6} = \frac{1525}{3}$ $P_h \approx 508.33 \text{ W}$

Eddy Current Loss ($P_e$): $P_e = B f^2 = \frac{13}{60} \times (50)^2$ $P_e = \frac{13}{60} \times 2500$ $P_e = \frac{13 \times 250}{6} = \frac{3250}{6} = \frac{1625}{3}$ $P_e \approx 541.67 \text{ W}$

The calculated hysteresis loss is approximately 508.33 W, and the eddy current loss is approximately 541.67 W. These values align closely with the option stating 508 W and 542 W.