Nichrome wire size for 5.5 kW resistance oven

The wire must give 5.5 kW at 220 V (resistance condition) and must radiate this power at 1100°C (surface condition).

Given: $P = 5500$ W, $V = 220$ V, $T_1 = 1100 + 273 = 1373$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 1.0$, $e = 0.9$.

Heat dissipated per m² of wire surface

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 1.0 \times (13.73^4 - 7.73^4) \\ &= 5.148 \times (35\,537 - 3\,570) = 1.6456 \times 10^5\ \text{W/m}^2 \end{aligned}$$

Resistance condition

$$R = \frac{V^2}{P} = \frac{220^2}{5500} = 8.8\ \Omega$$ $$R = \frac{\rho l}{\pi d^2/4} \Rightarrow \frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 8.8}{4 \times 42.5 \times 10^{-6}} = 1.6262 \times 10^5 \quad (1)$$

Surface (heat) condition

$$P = (\pi d l)H \Rightarrow d\,l = \frac{5500}{\pi \times 1.6456\times10^5} = 0.010638 \quad (2)$$

Solving (1) and (2)

$$\begin{aligned} d^3 &= \frac{0.010638}{1.6262 \times 10^5} = 6.542 \times 10^{-8} \\ d &= 4.03 \times 10^{-3}\ \text{m} = 4.03\ \text{mm} \\ l &= \frac{0.010638}{4.03 \times 10^{-3}} = 2.64\ \text{m} \end{aligned}$$

Answer: Wire diameter $\approx 4.03$ mm and length $\approx 2.64$ m.

Note: the resistivity is used as given; real nichrome is about $1.1\ \mu\Omega$m.