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 :
To calculate the full-load voltage regulation of the given transformer at a power factor of 0.08 lagging, let’s go step by step:
Given:
- Transformer rating = 40 kVA
- Primary voltage ( V_1 = 6600 \, \text{V} )
- Secondary voltage ( V_2 = 250 \, \text{V} )
- Primary resistance ( R_1 = 10 \, \Omega )
- Secondary resistance ( R_2 = 0.02 \, \Omega )
- Leakage reactance referred to primary ( X = 35 \, \Omega )
- Power factor (lagging) ( \text{pf} = 0.08 )
Step 1: Full-load Current
First, calculate the full-load current on the secondary side:
[
I_2 = \frac{\text{Transformer rating}}{\text{Secondary voltage}} = \frac{40,000 \, \text{VA}}{250 \, \text{V}} = 160 \, \text{A}
]
Since the current on the primary side is transformed by the turns ratio, we calculate the primary current as:
[
I_1 = \frac{I_2 \cdot V_2}{V_1} = \frac{160 \cdot 250}{6600} = 6.06 \, \text{A}
]
Step 2: Impedance of the Transformer
The total impedance ( Z_{\text{total}} ) of the transformer is the sum of the primary winding resistance, secondary winding resistance (referred to the primary side), and the total leakage reactance.
First, calculate the referred secondary resistance:
[
R_2′ = \frac{R_2 \cdot V_1^2}{V_2^2} = \frac{0.02 \cdot 6600^2}{250^2} = 0.792 \, \Omega
]
Now, the total impedance on the primary side is:
[
Z_{\text{total}} = R_1 + R_2′ + jX = 10 + 0.792 + j35 \, \Omega
]
Thus, ( Z_{\text{total}} = 10.792 + j35 \, \Omega ).
Step 3: Voltage Regulation
Voltage regulation is calculated as the difference in the primary voltage with full-load and no-load, divided by the no-load voltage:
[
V_{\text{reg}} = \frac{|V_{\text{FL}}| – |V_{\text{NL}}|}{|V_{\text{NL}}|}
]
Where ( V_{\text{FL}} ) is the full-load voltage and ( V_{\text{NL}} ) is the no-load voltage.
At full load, the voltage drop across the impedance of the transformer is:
[
\text{Voltage drop} = I_1 Z_{\text{total}} = 6.06 \cdot (10.792 + j35)
]
For a lagging power factor of 0.08, we calculate the angle of the voltage drop (using the power factor angle):
[
\text{Angle} = \cos^{-1}(0.08) = 85.44^\circ
]
Using this angle and the impedance, we calculate the voltage drop and use that to determine the voltage regulation.
Phasor Diagram Explanation:
- Primary Voltage (( V_1 )): This is the voltage on the primary side of the transformer, represented as a horizontal vector along the real axis.
- Secondary Voltage (( V_2 )): This is the voltage on the secondary side. In the phasor diagram, it would also be represented as a vector at an angle depending on the phase relationship with the primary voltage.
- Current Vectors: The current vectors on both sides (primary and secondary) will be drawn at an angle based on the power factor of 0.08 lagging. For a lagging power factor, the current vector will lag behind the voltage vector by the phase angle corresponding to the power factor of 0.08.
- Voltage Drop Across Impedance: The voltage drop due to the impedance of the transformer will be represented by a vector that combines both the resistive and reactive components of the impedance. The resistive drop will be along the real axis, while the reactive drop will be vertical, representing the inductive reactance.
Steps to Draw the Phasor Diagram:
- Step 1: Draw the primary voltage (( V_1 )) vector along the horizontal axis.
- Step 2: Draw the current vector (( I_1 )) at an angle corresponding to the lagging power factor.
- Step 3: Add the voltage drop due to the total impedance of the transformer.
- Step 4: Use the secondary voltage and current vectors in a similar manner, keeping the same relationship.