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Power Transmitted Using Volume Of Conductor Material (2 Phase 3 Wire US) Calculator

Formula Used:

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

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1. What Is Power Transmitted Using Volume Of Conductor Material?

This calculation determines the amount of electrical power that can be transmitted through an underground AC system (2 Phase 3 Wire US configuration) based on the volume of conductor material used, while accounting for various electrical parameters and losses.

2. How Does The Calculator Work?

The calculator uses the formula:

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

Where:

Explanation: This formula calculates the maximum power that can be transmitted while considering conductor volume, electrical losses, and system parameters for a 2-phase 3-wire underground AC system.

3. Importance Of Power Transmission Calculation

Details: Accurate power transmission calculation is crucial for designing efficient electrical distribution systems, optimizing conductor material usage, minimizing energy losses, and ensuring system reliability and safety.

4. Using The Calculator

Tips: Enter all values in appropriate units. Line losses, volume, maximum voltage, resistivity, and length must be positive values. Phase difference should be in radians (0 to 2π).

5. Frequently Asked Questions (FAQ)

Q1: Why is conductor volume important in power transmission?
A: Conductor volume directly affects the current-carrying capacity and resistance of the transmission line, which impacts both power handling capability and energy losses.

Q2: What factors affect line losses in underground systems?
A: Line losses are influenced by conductor material, cross-sectional area, length, current magnitude, operating temperature, and system configuration.

Q3: How does phase difference affect power transmission?
A: Phase difference (power factor) affects the real power component of the transmitted power. Lower power factors result in higher losses for the same apparent power.

Q4: What are typical resistivity values for common conductor materials?
A: Copper: ~1.68×10⁻⁸ Ω·m, Aluminum: ~2.82×10⁻⁸ Ω·m, Silver: ~1.59×10⁻⁸ Ω·m at 20°C.

Q5: Why use the (2+√2) factor in the denominator?
A: This factor accounts for the specific geometry and configuration of the 2-phase 3-wire underground AC system in the calculation.

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