Question
A 20kVa, 50Hz, 1100v/240v, single-phase transformer has a gross cross-sectional area of 0,008m². Determine the number of primary and secondary turns if the maximum flux density is 1,5Wb/m². Take the stacking factor to be 0,93.
Answers
Answered by
GPT-4o mini
To determine the number of primary and secondary turns of the transformer, we will use the transformer equations and the given parameters.
### Step 1: Calculate the Maximum Magnetic Flux (Φ)
The formula for the maximum magnetic flux (Φ) is:
\[
Φ = B_{max} \cdot A
\]
Where:
- \( B_{max} \) is the maximum flux density (1.5 Wb/m²)
- \( A \) is the cross-sectional area of the core (0.008 m²)
Substituting the values:
\[
Φ = 1.5 \, \text{Wb/m²} \cdot 0.008 \, \text{m²} = 0.012 \, \text{Wb}
\]
### Step 2: Calculate the Effective (True) Flux (Φ_eff)
The effective flux can be calculated as:
\[
Φ_{eff} = Φ \cdot \text{stacking factor}
\]
Substituting the stacking factor of 0.93:
\[
Φ_{eff} = 0.012 \, \text{Wb} \cdot 0.93 \approx 0.01116 \, \text{Wb}
\]
### Step 3: Calculate the Number of Turns
The number of turns (N) is related to the voltage, frequency, and effective flux as follows:
\[
N = \frac{V}{4.44 \cdot f \cdot Φ_{eff}}
\]
Where:
- \( V \) is the voltage (in this case, we will calculate both primary and secondary voltages)
- \( f \) is the frequency (50 Hz)
- \( Φ_{eff} \) is the effective flux (0.01116 Wb)
#### For Primary Turns:
**Given:**
- Primary Voltage \( V_p = 1100 \, \text{V} \)
Calculating the number of primary turns:
\[
N_p = \frac{1100}{4.44 \cdot 50 \cdot 0.01116}
\]
Calculating the denominator:
\[
4.44 \cdot 50 \approx 222
\]
\[
222 \cdot 0.01116 \approx 2.48
\]
Now substituting back:
\[
N_p \approx \frac{1100}{2.48} \approx 442.74 \approx 443 \text{ turns}
\]
#### For Secondary Turns:
**Given:**
- Secondary Voltage \( V_s = 240 \, \text{V} \)
Calculating the number of secondary turns:
\[
N_s = \frac{240}{4.44 \cdot 50 \cdot 0.01116}
\]
Using the previous calculation for the denominator:
\[
N_s \approx \frac{240}{2.48} \approx 96.77 \approx 97 \text{ turns}
\]
### Conclusion:
The number of turns for the transformer is approximately:
- **Primary Turns (N_p)**: 443 turns
- **Secondary Turns (N_s)**: 97 turns
### Step 1: Calculate the Maximum Magnetic Flux (Φ)
The formula for the maximum magnetic flux (Φ) is:
\[
Φ = B_{max} \cdot A
\]
Where:
- \( B_{max} \) is the maximum flux density (1.5 Wb/m²)
- \( A \) is the cross-sectional area of the core (0.008 m²)
Substituting the values:
\[
Φ = 1.5 \, \text{Wb/m²} \cdot 0.008 \, \text{m²} = 0.012 \, \text{Wb}
\]
### Step 2: Calculate the Effective (True) Flux (Φ_eff)
The effective flux can be calculated as:
\[
Φ_{eff} = Φ \cdot \text{stacking factor}
\]
Substituting the stacking factor of 0.93:
\[
Φ_{eff} = 0.012 \, \text{Wb} \cdot 0.93 \approx 0.01116 \, \text{Wb}
\]
### Step 3: Calculate the Number of Turns
The number of turns (N) is related to the voltage, frequency, and effective flux as follows:
\[
N = \frac{V}{4.44 \cdot f \cdot Φ_{eff}}
\]
Where:
- \( V \) is the voltage (in this case, we will calculate both primary and secondary voltages)
- \( f \) is the frequency (50 Hz)
- \( Φ_{eff} \) is the effective flux (0.01116 Wb)
#### For Primary Turns:
**Given:**
- Primary Voltage \( V_p = 1100 \, \text{V} \)
Calculating the number of primary turns:
\[
N_p = \frac{1100}{4.44 \cdot 50 \cdot 0.01116}
\]
Calculating the denominator:
\[
4.44 \cdot 50 \approx 222
\]
\[
222 \cdot 0.01116 \approx 2.48
\]
Now substituting back:
\[
N_p \approx \frac{1100}{2.48} \approx 442.74 \approx 443 \text{ turns}
\]
#### For Secondary Turns:
**Given:**
- Secondary Voltage \( V_s = 240 \, \text{V} \)
Calculating the number of secondary turns:
\[
N_s = \frac{240}{4.44 \cdot 50 \cdot 0.01116}
\]
Using the previous calculation for the denominator:
\[
N_s \approx \frac{240}{2.48} \approx 96.77 \approx 97 \text{ turns}
\]
### Conclusion:
The number of turns for the transformer is approximately:
- **Primary Turns (N_p)**: 443 turns
- **Secondary Turns (N_s)**: 97 turns