Question

The following data about a single-phase transformer is known:
Primary voltage =250V
Supply frequency =50Hz
Maximum flux density in core= 1,66T
Net core area=0,035m²
Diameter of primary conductor=2,5mm
Current density of the conductor=4A/mm²
Calculate the following:
1) the number of primary turns
2) the rating of a transformer

Answer 1

To calculate the number of primary turns and the rating of the transformer, we can use the respective formulas.

### 1) Number of Primary Turns

The formula to calculate the number of turns in the primary winding of a transformer is:
\[
N = \frac{V}{4.44 \cdot f \cdot \Phi_{\text{max}}}
\]
Where:
- \(N\) = number of turns
- \(V\) = primary voltage (250 V)
- \(f\) = supply frequency (50 Hz)
- \(\Phi_{\text{max}}\) = maximum flux (can be calculated from flux density and core area)

First, we need to calculate the max flux \(\Phi_{\text{max}}\):
\[
\Phi_{\text{max}} = B \cdot A
\]
Where:
- \(B\) = maximum flux density (1.66 T)
- \(A\) = net core area (0.035 m²)

\[
\Phi_{\text{max}} = 1.66 \, \text{T} \cdot 0.035 \, \text{m}^2 = 0.0581 \, \text{Wb}
\]

Now we can substitute the values into the formula for \(N\):

\[
N = \frac{250}{4.44 \cdot 50 \cdot 0.0581}
\]

Calculating the denominator:

\[
4.44 \cdot 50 \cdot 0.0581 \approx 12.87
\]

Now, calculate \(N\):

\[
N \approx \frac{250}{12.87} \approx 19.42
\]

Since the number of turns must be a whole number, we round it to:

\[
N \approx 19 \text{ turns}
\]

### 2) Rating of the Transformer

The rating (power) of a transformer can be calculated using the formula:
\[
S = V \cdot I
\]
Where:
- \(S\) = apparent power (in VA or kVA)
- \(V\) = primary voltage (250 V)
- \(I\) = primary current (to be calculated from conductor information)

First, we need to find the primary current \(I\). We can calculate the current using the current density formula:

\[
J = \frac{I}{A_c}
\]

Where:
- \(J\) = current density (4 A/mm²)
- \(A_c\) = cross-sectional area of the conductor (in mm²)

Calculation of the conductor cross-sectional area:

\[
A_c = \frac{\pi d^2}{4} = \frac{\pi (2.5 \, \text{mm})^2}{4} = \frac{\pi (6.25 \, \text{mm}^2)}{4} \approx 4.91 \, \text{mm}^2
\]

Now convert mm² to m²:

\[
A_c = 4.91 \times 10^{-6} \, \text{m}^2
\]

Calculate the primary current \(I\):

\[
I = 4 \cdot A_c \quad \text{(since J is in A/mm², we need area in mm²)}
\]
\[
I = 4 \cdot 4.91 \approx 19.64 \, \text{A}
\]

Now we can calculate the apparent power:

\[
S = 250 \cdot 19.64 \approx 4916 \, \text{VA}
\]

Converting VA to kVA:

\[
S \approx 4.916 \, \text{kVA}
\]

### Summary

1. Number of primary turns: **19 turns**
2. Rating of the transformer: **4.916 kVA**