Line Losses (1-phase 2-wire Mid-point Earthed) Calculator

Formula Used:

\[ \text{Line Losses} = 2 \times (I_{\text{Underground AC}}^2) \times R_{\text{Underground AC}} \]

Line Losses (Ploss):

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1. What is Line Losses (1-phase 2-wire Mid-point Earthed)?

Line Losses in a 1-phase 2-wire mid-point earthed system refer to the power dissipated as heat due to the resistance of the conductors. This loss occurs when current flows through the underground AC supply wires and is an important factor in electrical system efficiency.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ \text{Line Losses} = 2 \times (I_{\text{Underground AC}}^2) \times R_{\text{Underground AC}} \]

Where:

\( I_{\text{Underground AC}} \) — Current flowing through the underground AC supply wire (Ampere)
\( R_{\text{Underground AC}} \) — Resistance of the underground AC wire (Ohm)

Explanation: The formula calculates power loss by considering the squared current multiplied by resistance, with a factor of 2 accounting for both conductors in the system.

3. Importance of Line Losses Calculation

Details: Calculating line losses is crucial for designing efficient electrical systems, determining energy efficiency, and optimizing conductor sizing to minimize power dissipation.

4. Using the Calculator

Tips: Enter the current in Amperes and resistance in Ohms. Both values must be positive numbers greater than zero for accurate calculation.

5. Frequently Asked Questions (FAQ)

Q1: Why is there a factor of 2 in the formula?
A: The factor of 2 accounts for power losses in both conductors of the 1-phase 2-wire system.

Q2: How does current affect line losses?
A: Line losses increase with the square of the current, making current reduction highly effective for minimizing losses.

Q3: What factors influence resistance in underground AC systems?
A: Conductor material, cross-sectional area, length, temperature, and installation conditions affect resistance.

Q4: How can line losses be reduced in practice?
A: Using larger conductors, reducing cable length, improving installation techniques, and optimizing load distribution.

Q5: Are these calculations applicable to all electrical systems?
A: This specific formula is designed for 1-phase 2-wire mid-point earthed underground AC systems. Other system configurations require different formulas.