Question 4.10: The plate in Example 4.9 is to be surface hardened by scanni......
The plate in Example 4.9 is to be surface hardened by scanning or rastering the beam over the surface. The temperature at the surface is not to exceed 1300°C after 1.0 m/sec.
Determine the power density required. Also determine the temperature along the beam axis at a depth of 200 µm after 10^−3 sec on the application of the power density calculated above.
The equation for surface temperature can be obtained by putting z = 0 in the expression
for T_{(z,t)} in Example 4.8
\begin{aligned}T_{(0, t)} & =\frac{2 q_o \sqrt{\alpha t}}{\lambda}\left(i e r f c o-i \operatorname{erfc} \frac{r_f}{2 \sqrt{\alpha t}}\right) \\& =\frac{2 q_o \sqrt{\alpha t}}{\lambda}\left(1.1284-i \operatorname{erfc} \frac{r_f}{2 \sqrt{\alpha t}}\right)\end{aligned}Given τ = 10^{-3} sec, α = 0.08 cm²/sec, λ = 0.51 w/cm°C, r_f = 0.015 cm substituting
\begin{aligned} 1300 & =\frac{2 q_o \sqrt{0.08 \times 10^{-3}}}{0.51}\left[1.1284-i e r f c \frac{0.015}{2 \sqrt{0.08 \times 10^{-3}}}\right] \\ & =q_o \frac{17.88 \times 10^{-3} \times 0.9611}{0.51}=3.37 \times 10^{-2} \\ q_o & =\frac{1300 \times 10^2}{3.37} \\ & =3.86 \times 10^4 \ \mathrm{w} / \mathrm{cm}^2 \end{aligned}The temperature at z = 200 µm or 0.02 cm
\begin{aligned}t\left(0.02,10^{-3}\right)= & \frac{2 \times 3.86 \times 10^4 \times 8.94}{0.51} \\& \times\left[\operatorname{ierfc} \frac{0.02}{17.88 \times 10^{-3}}-i e r f c \frac{\sqrt{0.02^2+0.15^2}}{17.88 \times 10^{-3}}\right] \\& \simeq 60^{\circ} \mathrm{C}\end{aligned}The beam power required is
Q = power density × spot area
= 3.86 \times 10^4 \times \pi \times 0.015^2 = 27.3 \ w
If the time t is reduced to an order of 10^{-4} to 10^{-6}sec, the power required will increase.