How to Caculate the Nichrome80 Temperature?

To determine the operating temperature of a Ni80 ( Nichrome 80) flat wire with the given specifications, follow these steps:

Given Data:

• Nichrome Wire Dimensions: 3.0 mm × 0.5 mm

• Wire Length: 600 mm (0.6 m)

• Applied Voltage: 220V

• Material: Ni80 (Nichrome 80)

• Resistivity of Ni80: 1.09 × 10⁻⁶ Ω·m (at room temperature)

• Specific Heat Capacity: 460 J/kg·°C

• Density of Ni80: 8,400 kg/m³

• Emissivity (estimated): 0.7 (for oxidized Nichrome)

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Step 1: Calculate Wire Resistance

The resistance $R$ of the wire is given by:

$$R = \rho \frac{L}{A}$$

where:

• $\rho = 1.09 \times 10^{-6}$ Ω·m (resistivity of Ni80)

• $L = 0.6$ m (wire length)

• $A = 3.0 \times 0.5 = 1.5$ mm² = 1.5 × 10⁻⁶ m² (cross-sectional area)

$$R = \frac{(1.09 \times 10^{-6}) \times 0.6}{1.5 \times 10^{-6}}$$$$R \approx 0.436 \, \Omega$$

Step 2: Calculate Power Dissipation

Using Ohm’s Law and Power Formula:

$$P = \frac{V^2}{R}$$$$P = \frac{220^2}{0.436}$$$$P \approx 111,000 \text{ W} = 111 \text{ W}$$

So, the wire dissipates 111 watts as heat.

Step 3: Calculate Steady-State Temperature

At steady state, heat dissipation due to radiation and convection balances power input. Using the Stefan-Boltzmann Law:

$$P = \epsilon \sigma A (T^4 - T_{\text{ambient}}^4)$$

where:

• $\epsilon$ = 0.7 (Nichrome emissivity)

• $\sigma = 5.67 \times 10^{-8}$ W/m²·K⁴ (Stefan-Boltzmann constant)

• $A = \text{Surface Area} = (2w + 2h) \times L = (2 \times 3 + 2 \times 0.5) \times 0.6 = 4.2 \times 0.6 = 0.00252$ m²

• $T_{\text{ambient}}$ = 25°C = 298 K

• $P = 111$ W

Rearranging for $T$:

$$T = \left( \frac{P}{\epsilon \sigma A} + T_{\text{ambient}}^4 \right)^{\frac{1}{4}}$$

Substituting the values:

$$T = \left( \frac{111}{(0.7 \times 5.67 \times 10^{-8} \times 0.00252)} + 298^4 \right)^{\frac{1}{4}}$$

Solving this numerically:

$$T \approx 1130 \text{ K} \approx 857^\circ C$$

Final Answer:

The operating temperature of the Ni80 wire under these conditions is approximately 857°C (1130 K).

This calculation assumes heat loss is mainly through radiation. If convection cooling is significant (e.g., forced airflow), the temperature will be lower.