Nichrome strip width for 40 kW 3-phase oven

For a thin strip, the radiating surface is both faces, $2wl$ (edges neglected), and the cross-section is $wt$.

Given: $P = 40$ kW, 3-phase, 400 V, star connected; $t = 0.0025$ cm $= 2.5 \times 10^{-5}$ m; $T_1 = 1100 + 273 = 1373$ K; $T_2 = 700 + 273 = 973$ K; $k = 0.6$; $e = 0.9$; $\rho = 1.03 \times 10^{-6}\ \Omega$m.

Per-phase values

$$\begin{aligned} P_{ph} &= \frac{40\,000}{3} = 13\,333\ \text{W}, \quad V_{ph} = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ R &= \frac{V_{ph}^2}{P_{ph}} = \frac{230.94^2}{13\,333} = 4.0\ \Omega \end{aligned}$$

Heat dissipated per m² (Stefan's law)

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 0.6 \times (13.73^4 - 9.73^4) \\ &= 3.0888 \times (35\,537 - 8\,963) = 82\,082\ \text{W/m}^2 \end{aligned}$$

Equations for the strip

$$\begin{aligned} R = \frac{\rho l}{w t} &\Rightarrow \frac{l}{w} = \frac{R t}{\rho} = \frac{4 \times 2.5\times10^{-5}}{1.03\times10^{-6}} = 97.09 \quad (1) \\ P_{ph} = 2wlH &\Rightarrow wl = \frac{13\,333}{2 \times 82\,082} = 0.08122\ \text{m}^2 \quad (2) \end{aligned}$$

Solving

$$\begin{aligned} w^2 &= \frac{0.08122}{97.09} = 8.365 \times 10^{-4} \\ w &= 0.02892\ \text{m} = 28.9\ \text{mm} \\ l &= 97.09 \times 0.02892 = 2.81\ \text{m (per phase)} \end{aligned}$$

Answer: Strip width $\approx 28.9$ mm (about 2.9 cm), with length $\approx 2.81$ m per phase.