Nichrome strip resistor length and width

The strip must have the resistance that gives 1 kW at 250 V, and enough surface area to radiate 1 kW at the allowed temperature.

Assumption: ambient temperature 25 °C, so the strip temperature is 25 + 300 = 325 °C. Heat is lost by radiation only (Stefan's law), from both faces of the strip; the edges are neglected.

Step 1: resistance required

$$R = \frac{V^2}{P} = \frac{250^2}{1000} = 62.5\ \Omega$$

Step 2: heat dissipated per m² of surface

Stefan's law with emissivity $e$ and radiating efficiency $\eta$:

$$\begin{aligned} q &= 5.7\times10^{-8}\, e\,\eta\,(T_1^4 - T_2^4) \\ T_1 &= 325 + 273 = 598\ \text{K},\quad T_2 = 25 + 273 = 298\ \text{K} \\ q &= 5.7\times10^{-8}\times0.9\times0.75\times(598^4 - 298^4) \\ &= 3.8475\times10^{-8}\times(1.2788\times10^{11} - 0.0789\times10^{11}) \\ &= 4617\ \text{W/m}^2 \end{aligned}$$

Step 3: surface area needed

For length $l$ and width $w$, the radiating surface is $2lw$:

$$\begin{aligned} 2\,l\,w\,q &= 1000 \\ l\,w &= \frac{1000}{2\times4617} = 0.10830\ \text{m}^2 \quad (1) \end{aligned}$$

Step 4: length–width ratio from resistance

$$\begin{aligned} R &= \frac{\rho\, l}{w\,t} \\ \frac{l}{w} &= \frac{R\,t}{\rho} = \frac{62.5\times0.2\times10^{-3}}{1\times10^{-6}} = 12500 \quad (2) \end{aligned}$$

Step 5: solve (1) and (2)

$$\begin{aligned} l^2 &= 12500\times0.10830 = 1353.75 \\ l &= 36.79\ \text{m} \\ w &= \frac{l}{12500} = 2.943\times10^{-3}\ \text{m} = 2.94\ \text{mm} \end{aligned}$$

Check: $R = \dfrac{10^{-6}\times36.79}{2.943\times10^{-3}\times0.2\times10^{-3}} = 62.5\ \Omega$.

Answer: length ≈ 36.8 m, width ≈ 2.94 mm (for 25 °C ambient). With a 20 °C ambient the same method gives about 37.4 m and 2.99 mm.