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
- Number of primary turns: 19 turns
- Rating of the transformer: 4.916 kVA