Question

How do you calculate line losses due to temperature variations?

Answer 1

Calculating line losses due to temperature variations is essential for understanding how heat affects the efficiency of power transmission systems. These losses occur primarily due to **resistive heating**, also known as **I²R losses**. The resistance of conductors, like transmission lines, increases with temperature, and as resistance rises, more energy is lost as heat.

Here's a step-by-step guide on how to calculate line losses with temperature variations:

### 1. **Calculate the Initial Resistance (R₀)**
The initial resistance of a conductor can be calculated using the formula:
\[
R_0 = \frac{\rho_0 \cdot L}{A}
\]
Where:
- \( R_0 \) = Initial resistance at reference temperature (in ohms)
- \( \rho_0 \) = Resistivity of the material at reference temperature (e.g., 20°C) (in ohm-meter)
- \( L \) = Length of the conductor (in meters)
- \( A \) = Cross-sectional area of the conductor (in square meters)

### 2. **Calculate the Resistance at the New Temperature (Rₜ)**
Resistance increases with temperature, and the relationship between temperature and resistance is given by:
\[
R_t = R_0 \cdot \left(1 + \alpha \cdot (T_t - T_0)\right)
\]
Where:
- \( R_t \) = Resistance at the new temperature \(T_t\) (in ohms)
- \( R_0 \) = Initial resistance at the reference temperature \(T_0\) (in ohms)
- \( \alpha \) = Temperature coefficient of resistance (for most metals, it is around 0.0039 per °C for copper)
- \( T_t \) = New temperature (in °C)
- \( T_0 \) = Reference temperature (usually 20°C)

### 3. **Calculate the Power Loss (Pₗ)**
The power loss due to resistance is primarily calculated using the **I²R** formula:
\[
P_l = I^2 \cdot R_t
\]
Where:
- \( P_l \) = Power loss (in watts)
- \( I \) = Current flowing through the conductor (in amperes)
- \( R_t \) = Resistance at the operating temperature \(T_t\) (in ohms)

### 4. **Consider Temperature Rise Due to Line Current**
The actual temperature of the conductor depends not only on ambient temperature but also on the heating effect of the current passing through it. The relationship between the temperature rise and current flow can be derived from thermal properties, but it often involves simulation or empirical data to determine how much the current raises the conductor's temperature.

### 5. **Total Line Losses over a Distance**
For long transmission lines, line losses are calculated over the length of the line. If the line is non-uniform in temperature, an integral over the length must be performed, but for uniform conditions, you can multiply the power loss by the line's length.

### **Example Calculation**
Let's calculate the power loss for a 100-meter copper line (with a cross-sectional area of 50 mm²) carrying a current of 100 A, with a temperature rise from 20°C to 50°C.

1. **Initial resistance calculation** (at 20°C):
   - \( \rho_0 \) for copper ≈ \( 1.72 \times 10^{-8} \, \Omega \cdot m \)
   - Cross-sectional area \( A = 50 \, \text{mm}^2 = 50 \times 10^{-6} \, m^2 \)
   - Length \( L = 100 \, m \)

   \[
   R_0 = \frac{1.72 \times 10^{-8} \cdot 100}{50 \times 10^{-6}} = 0.0344 \, \Omega
   \]

2. **Resistance at 50°C**:
   - \( \alpha \) for copper ≈ 0.0039/°C
   \[
   R_t = 0.0344 \cdot \left(1 + 0.0039 \cdot (50 - 20)\right)
   \]
   \[
   R_t = 0.0344 \cdot (1 + 0.117) = 0.0384 \, \Omega
   \]

3. **Power loss**:
   \[
   P_l = I^2 \cdot R_t = 100^2 \cdot 0.0384 = 384 \, \text{watts}
   \]

### Key Takeaways:
- As temperature increases, the resistance of the conductor rises, leading to higher line losses.
- Proper material selection and understanding of temperature effects are crucial for minimizing losses.
- This calculation assumes uniform temperature and steady-state conditions. In real-world scenarios, variations in current and environmental factors may need to be considered.

Answer 2

Calculating line losses due to temperature variations involves understanding how temperature affects the resistance of conductors. Here’s a general approach:

### 1. **Understand the Temperature Coefficient of Resistance (TCR)**:
   - The resistance of a conductor changes with temperature. The formula for resistance \( R_t \) at a temperature \( T \) is:
     \[
     R_t = R_0 \times [1 + \alpha (T - T_0)]
     \]
     where:
     - \( R_0 \) = resistance at reference temperature \( T_0 \) (usually 20°C).
     - \( \alpha \) = temperature coefficient of resistance (in °C⁻¹) for the material (e.g., copper or aluminum).
     - \( T \) = temperature in °C.

### 2. **Calculate the Resistance**:
   - If the temperature changes, calculate the new resistance using the formula above.

### 3. **Determine Current Flowing Through the Line**:
   - The line losses are typically calculated using the current \( I \) flowing through the conductor.

### 4. **Calculate Power Losses**:
   - The power loss \( P \) due to resistance is given by:
     \[
     P = I^2 \times R_t
     \]
   - This represents the heat loss in the conductor due to its resistance at the new temperature.

### 5. **Adjust for Variations**:
   - If you're analyzing multiple temperatures or need to account for operational ranges, repeat the calculations for each relevant temperature scenario.

### Example Calculation:
1. Assume a copper conductor with:
   - \( R_0 = 10 \, \Omega \) (at 20°C),
   - \( \alpha = 0.00393 \, \text{°C}^{-1} \),
   - \( T = 50 \, \text{°C} \),
   - Current \( I = 10 \, A \).

2. Calculate \( R_t \):
   \[
   R_t = 10 \, \Omega \times [1 + 0.00393 \times (50 - 20)]
   \]
   \[
   R_t = 10 \, \Omega \times [1 + 0.00393 \times 30]
   \]
   \[
   R_t \approx 10 \, \Omega \times 1.1179 \approx 11.18 \, \Omega
   \]

3. Calculate power loss:
   \[
   P = 10^2 \times 11.18 \approx 1118 \, W
   \]

This process will give you a clear picture of how temperature variations affect line losses.