Resistivity using Line Losses (Two-Phase Three-Wire OS) Calculator

Formula Used:

\[ \rho = \frac{2 \times P_{loss} \times A \times (V_m \times \cos(\Phi))^2}{(2 + \sqrt{2}) \times P^2 \times L} \]

Line Losses (P_loss):

Watt

Area of Overhead AC Wire (A):

m²

Maximum Voltage Overhead AC (V_m):

Volt

Phase Difference (Φ):

Radians

Power Transmitted (P):

Watt

Length of Overhead AC Wire (L):

Meter

Resistivity (ρ):

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1. What is Resistivity using Line Losses (Two-Phase Three-Wire OS)?

Resistivity using Line Losses (Two-Phase Three-Wire OS) calculates the electrical resistivity of a material based on power line losses in a two-phase three-wire overhead system. It provides an accurate assessment of material properties that affect current flow resistance.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ \rho = \frac{2 \times P_{loss} \times A \times (V_m \times \cos(\Phi))^2}{(2 + \sqrt{2}) \times P^2 \times L} \]

Where:

\( \rho \) — Resistivity (Ω·m)
\( P_{loss} \) — Line losses (Watt)
\( A \) — Area of overhead AC wire (m²)
\( V_m \) — Maximum voltage overhead AC (Volt)
\( \Phi \) — Phase difference (Radians)
\( P \) — Power transmitted (Watt)
\( L \) — Length of overhead AC wire (Meter)

Explanation: The formula accounts for the relationship between power losses, material properties, and system parameters in a two-phase three-wire overhead system.

3. Importance of Resistivity Calculation

Details: Accurate resistivity calculation is crucial for designing efficient power transmission systems, selecting appropriate conductor materials, and minimizing energy losses in overhead power lines.

4. Using the Calculator

Tips: Enter all values in appropriate units (line losses in Watt, area in m², voltage in Volt, phase difference in Radians, power in Watt, and length in Meter). All values must be positive numbers.

5. Frequently Asked Questions (FAQ)

Q1: Why is resistivity important in power transmission?
A: Resistivity determines how much a material resists electric current flow, directly affecting power losses and efficiency in transmission lines.

Q2: What are typical resistivity values for common conductors?
A: Copper has resistivity of about 1.68×10⁻⁸ Ω·m, aluminum about 2.82×10⁻⁸ Ω·m, and silver about 1.59×10⁻⁸ Ω·m at 20°C.

Q3: How does temperature affect resistivity?
A: Resistivity generally increases with temperature for most conductors due to increased atomic vibrations that impede electron flow.

Q4: What factors influence line losses in overhead systems?
A: Line losses are influenced by conductor material, cross-sectional area, length, current magnitude, operating temperature, and system configuration.

Q5: When is this specific formula applicable?
A: This formula is specifically designed for two-phase three-wire overhead AC systems and may not be accurate for other system configurations.