Nichrome strip width for 30 kW 3-phase oven

Assumptions: the strip radiates from both faces (edges neglected, surface $= 2wl$); the elements are star connected so each phase gets $V_L/\sqrt3$; the charge absorbs heat as a black body at 700°C; values are used as given ($t = 0.025$ cm, $\rho = 10.3\times10^{-6}\ \Omega$m).

Given: $P = 30$ kW, 400 V, 3-phase; $T_1 = 1373$ K, $T_2 = 973$ K; $k = 0.6$, $e = 0.9$; $t = 2.5\times10^{-4}$ m.

Per-phase values

$$\begin{aligned} P_{ph} &= \frac{30\,000}{3} = 10\,000\ \text{W}, \quad V_{ph} = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ R &= \frac{230.94^2}{10\,000} = 5.333\ \Omega \end{aligned}$$

Heat dissipated per m²

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 0.6 \times (13.73^4 - 9.73^4) \\ &= 3.0888 \times 26\,574 = 82\,082\ \text{W/m}^2 \end{aligned}$$

Strip equations

$$\begin{aligned} R = \frac{\rho l}{w t} &\Rightarrow \frac{l}{w} = \frac{R t}{\rho} = \frac{5.333 \times 2.5\times10^{-4}}{10.3\times10^{-6}} = 129.45 \quad (1)\\ P_{ph} = 2 w l H &\Rightarrow w l = \frac{10\,000}{2 \times 82\,082} = 0.060915\ \text{m}^2 \quad (2) \end{aligned}$$

Solving

$$\begin{aligned} w^2 &= \frac{0.060915}{129.45} = 4.706\times10^{-4} \\ w &= 0.02169\ \text{m} = 21.7\ \text{mm} \\ l &= 129.45 \times 0.02169 = 2.81\ \text{m per phase} \end{aligned}$$

Answer: Suitable strip width $\approx 21.7$ mm (about 2.2 cm), length $\approx 2.81$ m per phase.