Question 4.9: A continuous laser beam having 600 W power is focused on the......
A continuous laser beam having 600 W power is focused on the surface of a semiinfinite steel plate. Radius of the focal spot is 150 µm.
Diffusivity of steel (α) = 0.08 cm²/sec
Thermal conductivity (λ) = 0.51 W/m°C
Initial temperature throughout the slab = 30°C
Assume that all the incident power is absorbed by the target.
Determine the temperature along the beam axis at a spot located 1.0 mm from the surface after 0.5, 1.0, 1.5, and 2.0 sec of exposure.
The equation to be used is
T_{(z, \tau)}=\frac{2 q_o \sqrt{\alpha \tau}}{\lambda}\left[\operatorname{ierfc} \frac{z}{2 \sqrt{\alpha \tau}}-i e r f c \frac{\sqrt{z^2+r_f^2}}{2 \sqrt{\alpha t}}\right] \\ q_0 = \text{power density} = \text{power/spot area} \\ =\frac{600}{\pi \times 0.015^2}=8.5 \times 10^5 \mathrm{w} / \mathrm{cm}^2 \\ \alpha=0.08 \ \mathrm{~cm}^2 / \mathrm{sec} \ \lambda=0.51 \ \mathrm{w} / \mathrm{m}^{\circ} \mathrm{C}, \quad z=0.1 \ \mathrm{~cm}For \tau = 0.5 sec substituting in the above equation
\begin{aligned}T(0.1,0.5)= & \frac{2 \times 8.5 \times 10^5}{0.51} \sqrt{0.08 \times 0.5} \\& \times\left[\text { ierfc } \frac{0.1}{2 \sqrt{0.08 \times 0.5}}-\text { ierfc }\frac{\sqrt{0.015^2+0.1^2}}{2 \sqrt{0.08 \times 0.5}}\right] \\= & \left.6.67 \times 10^5 \text { (ierfc } 0.25-\text { ierfc } 0.253\right) \\= & 6.67 \times 10^5(0.6982-0.6565) \\= & 6.67 \times 10^5 \times 0.007 \\= & 667^{\circ} \mathrm{C}\end{aligned}Note :The term in brackets gives a small, dimensionless multiplier.
Similarly
\begin{aligned}t(0.1,1.0) & =954^{\circ} \mathrm{C}(t=1.05 \ \mathrm{sec}) \\t(0.1,1.5) & =1150^{\circ} \mathrm{C}(t=1.5 \ \mathrm{sec}) \\t(0.1,2) & =1300^{\circ} \mathrm{C}(t=2.0 \ \mathrm{sec})\end{aligned}If we change the point location, i.e., the z coordinate, the temperature history at that point can be calculated in a similar way
Note :The surface (z = 0) assumes a very high temperature in a very short time and there will be some evaporation loss.