Area Of X-Section(Two-Wire Mid-Point Earthed) Calculator

Formula Used:

\[ A = \frac{P^2 \cdot \rho \cdot L}{2 \cdot P_{loss} \cdot V_m^2} \]

Power Transmitted (P):

Resistivity (ρ):

Ω·m

Length of Wire DC (L):

Line Losses (P_loss):

Maximum Voltage Overhead DC (V_m):

Area of X-Section (A):

Unit Converter ▲

Unit Converter ▼

From:	To:

1. What is the Area of X-Section Formula?

The Area of X-Section formula calculates the cross-sectional area of overhead DC wires in a two-wire mid-point earthed system. This calculation is essential for determining the appropriate wire size to minimize power losses while maintaining efficient power transmission.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ A = \frac{P^2 \cdot \rho \cdot L}{2 \cdot P_{loss} \cdot V_m^2} \]

Where:

\( A \) — Area of X-Section (m²)
\( P \) — Power Transmitted (W)
\( \rho \) — Resistivity (Ω·m)
\( L \) — Length of Wire DC (m)
\( P_{loss} \) — Line Losses (W)
\( V_m \) — Maximum Voltage Overhead DC (V)

Explanation: The formula calculates the optimal cross-sectional area needed to achieve specified power transmission efficiency while considering material properties and system constraints.

3. Importance of X-Section Calculation

Details: Proper wire sizing is crucial for efficient power transmission, minimizing energy losses, preventing overheating, ensuring system reliability, and optimizing material costs in overhead DC systems.

4. Using the Calculator

Tips: Enter all values in appropriate units (power in watts, resistivity in ohm-meters, length in meters, losses in watts, voltage in volts). All values must be positive numbers greater than zero.

5. Frequently Asked Questions (FAQ)

Q1: Why is the formula squared for power?
A: The power term is squared because power losses are proportional to the square of the current, and current is directly related to power in DC systems.

Q2: What is typical resistivity 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.

Q3: How does wire length affect the required cross-sectional area?
A: Longer wires require larger cross-sectional areas to maintain the same level of power losses, as resistance increases with length.

Q4: What are acceptable line loss percentages?
A: Typically 2-5% of transmitted power, though this varies based on system requirements and economic considerations.

Q5: Why is voltage squared in the denominator?
A: Higher voltages reduce current for the same power, which significantly reduces I²R losses, allowing for smaller conductor sizes.