A 40 kVA, 6600/250 V single-phase transformer has the following parameters:
- Primary winding resistance = 10 O
- Secondary winding resistance = 0.02 O
- Total leakage reactance (referred to the primary) = 35 O
Required:
Calculate the full-load voltage regulation at a power factor of 0.08 lagging.
Draw the phasor diagram representing the given transformer conditions.
The correct answer and explanation is :
Solution for Full-Load Voltage Regulation:
To calculate the full-load voltage regulation at a given power factor (pf), we can use the following steps:
Step 1: Determine the rated current of the transformer:
The transformer has a rating of 40 kVA, and it is single-phase. The full-load current on both the primary and secondary side can be calculated using the formula:
[
I = \frac{S}{V}
]
Where:
- ( S = 40 \, kVA = 40,000 \, VA )
- ( V = 250 \, V ) (secondary voltage)
Thus,
[
I_{\text{secondary}} = \frac{40,000}{250} = 160 \, A
]
Step 2: Impedance calculation (refer to the primary side):
The total leakage reactance is given as ( X = 35 \, \Omega ), and the resistance is given for both windings.
- Primary winding resistance (( R_1 )) = 10 Ω
- Secondary winding resistance (( R_2 )) = 0.02 Ω
We need to refer the secondary winding resistance to the primary side using the following relation:
[
R_{\text{primary equivalent}} = R_2 \times \left( \frac{V_1}{V_2} \right)^2
]
[
R_{\text{primary equivalent}} = 0.02 \times \left( \frac{6600}{250} \right)^2 = 0.02 \times (26.4)^2 = 0.02 \times 696.96 = 13.94 \, \Omega
]
Thus, the total primary resistance is:
[
R_{\text{total}} = R_1 + R_{\text{primary equivalent}} = 10 + 13.94 = 23.94 \, \Omega
]
Step 3: Calculate voltage regulation:
Voltage regulation is defined as the difference in secondary voltage at no-load and full-load, divided by the full-load voltage.
At full load, the impedance of the transformer produces voltage drops that reduce the output voltage. The impedance ( Z = R_{\text{total}} + jX ) is the total impedance referred to the primary.
First, calculate the voltage drop across the total impedance:
[
V_{\text{drop}} = I \times Z
]
Where ( Z = \sqrt{R_{\text{total}}^2 + X^2} = \sqrt{23.94^2 + 35^2} = \sqrt{573.27 + 1225} = \sqrt{1798.27} = 42.44 \, \Omega ).
Thus, the voltage drop is:
[
V_{\text{drop}} = 160 \times 42.44 = 6784 \, V
]
At a lagging power factor of 0.08, the voltage regulation can be calculated by the formula:
[
\text{Voltage Regulation} = \frac{V_{\text{no-load}} – V_{\text{full-load}}}{V_{\text{full-load}}}
]
Where ( V_{\text{no-load}} ) is the primary voltage at no load, and ( V_{\text{full-load}} ) is the secondary voltage.
Phasor Diagram:
The phasor diagram would show the voltage drop across the transformer impedance (both resistive and reactive), and the output voltage lagging behind the input voltage due to the inductive reactance. Since we have a low power factor (0.08 lagging), the voltage vector will lag behind the current and the voltage drop will be predominantly reactive.
This process illustrates how the voltage regulation can be calculated for a transformer operating under a specific power factor and load. The full-load voltage regulation is determined by the impedance of the transformer and the phase angle of the current relative to the voltage.