1. **Problem statement:** We need to find the time required for nitrogen diffusion in steel during a nitriding process at 520 °C. The initial nitrogen content is 0.003 wt%, surface content is 0.60 wt%, and at 0.45 mm depth, nitrogen content is 0.20 wt%. 2. **Formula used:** For diffusion in a semi-infinite solid, the concentration profile is given by the error function solution: $$\frac{C_x - C_0}{C_s - C_0} = 1 - \operatorname{erf}\left(\frac{x}{2\sqrt{Dt}}\right)$$ where: - $C_x$ is concentration at depth $x$ - $C_0$ is initial concentration - $C_s$ is surface concentration - $D$ is diffusion coefficient - $t$ is time - $x$ is depth 3. **Known values:** - $C_0 = 0.003$ - $C_s = 0.60$ - $C_x = 0.20$ - $x = 0.45$ mm = 0.045 cm (convert to cm for consistency) 4. **Calculate the normalized concentration:** $$\frac{C_x - C_0}{C_s - C_0} = \frac{0.20 - 0.003}{0.60 - 0.003} = \frac{0.197}{0.597} \approx 0.33$$ 5. **Find the error function value:** $$1 - \operatorname{erf}\left(\frac{x}{2\sqrt{Dt}}\right) = 0.33 \implies \operatorname{erf}\left(\frac{x}{2\sqrt{Dt}}\right) = 0.67$$ 6. **Find the argument of the error function:** From error function tables or calculator, $\operatorname{erf}(z) = 0.67$ corresponds to $z \approx 0.68$. 7. **Solve for time $t$:** $$z = \frac{x}{2\sqrt{Dt}} \Rightarrow \sqrt{Dt} = \frac{x}{2z}$$ $$Dt = \left(\frac{x}{2z}\right)^2$$ $$t = \frac{x^2}{4z^2 D}$$ 8. **Determine diffusion coefficient $D$:** For nitrogen in BCC steel at 520 °C, approximate $D = 1.2 \times 10^{-11}$ cm$^2$/s (typical literature value). 9. **Calculate $t$:** $$t = \frac{(0.045)^2}{4 \times (0.68)^2 \times 1.2 \times 10^{-11}} = \frac{0.002025}{4 \times 0.4624 \times 1.2 \times 10^{-11}} = \frac{0.002025}{2.22 \times 10^{-11}} \approx 9.12 \times 10^{7} \text{ seconds}$$ 10. **Convert time to hours:** $$\frac{9.12 \times 10^{7}}{3600} \approx 25333 \text{ hours}$$ **Final answer:** The time required for the nitriding diffusion treatment is approximately **25333 hours**.