Nichrome wire size for 4.5 kW resistance oven

The wire must have the resistance that gives 4.5 kW at 200 V, and enough surface area to radiate that power at the given temperatures.

Given: $P = 4500$ W, $V = 200$ V, $T_1 = 1000 + 273 = 1273$ 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 unit surface area (Stefan's law)

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right]\ \text{W/m}^2 \\ &= 5.72 \times 0.9 \times 1.0\,[12.73^4 - 7.73^4] \\ &= 5.148 \times (26\,261 - 3\,570) = 1.168 \times 10^5\ \text{W/m}^2 \end{aligned}$$

Resistance condition

$$R = \frac{V^2}{P} = \frac{200^2}{4500} = 8.889\ \Omega, \qquad R = \frac{4\rho l}{\pi d^2}$$ $$\frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 8.889}{4 \times 42.5\times10^{-6}} = 1.6427 \times 10^5 \quad (1)$$

Heat-dissipation condition

$$P = \pi d l H \Rightarrow d\,l = \frac{4500}{\pi \times 1.168 \times 10^5} = 0.012262 \quad (2)$$

Solving

Dividing (2) by (1): $d^3 = \dfrac{0.012262}{1.6427 \times 10^5} = 7.465 \times 10^{-8}$

$$\begin{aligned} d &= 4.21 \times 10^{-3}\ \text{m} = 4.21\ \text{mm} \\ l &= \frac{0.012262}{4.21 \times 10^{-3}} = 2.91\ \text{m} \end{aligned}$$

Answer: Diameter $\approx 4.21$ mm, length $\approx 2.91$ m.

Note: the given resistivity is used as printed; real nichrome is about $1.1\ \mu\Omega$m, which would give a thinner, longer wire.