Question:hard
In laser beam machining, the time (\(t_m\)) required for the material to attain the melting temperature from a room temperature (\(\theta_0\)) of 32°C is expressed by the following expression:
\[ t_m = \frac{\pi}{\alpha} \left( \frac{(\theta_m - \theta_0) k}{2H} \right)^2 \]
where \(\alpha\) is thermal diffusivity, \(\theta_m\) is melting temperature, \(k\) is thermal conductivity, \(H\) is heat flux.
If a uniformly distributed 1 kW power laser beam with a beam diameter of 0.1 mm is used for machining tungsten carbide, and 10% of beam absorption is assumed, the time \(t_m\) is ______ \(\mu s\) (rounded off to one decimal place).
Note: Thermal properties of tungsten carbide: melting temperature = 3400°C; thermal conductivity = 2.15 W/cm-°C; diffusivity = 0.79 cm\(^2\) s\(^{-1}\); assume \(\pi\) = 3.14.
In laser beam machining, the time (\(t_m\)) required for the material to attain the melting temperature from a room temperature (\(\theta_0\)) of 32°C is expressed by the following expression:
\[ t_m = \frac{\pi}{\alpha} \left( \frac{(\theta_m - \theta_0) k}{2H} \right)^2 \]
where \(\alpha\) is thermal diffusivity, \(\theta_m\) is melting temperature, \(k\) is thermal conductivity, \(H\) is heat flux.
If a uniformly distributed 1 kW power laser beam with a beam diameter of 0.1 mm is used for machining tungsten carbide, and 10% of beam absorption is assumed, the time \(t_m\) is ______ \(\mu s\) (rounded off to one decimal place).
Note: Thermal properties of tungsten carbide: melting temperature = 3400°C; thermal conductivity = 2.15 W/cm-°C; diffusivity = 0.79 cm\(^2\) s\(^{-1}\); assume \(\pi\) = 3.14.
\[ t_m = \frac{\pi}{\alpha} \left( \frac{(\theta_m - \theta_0) k}{2H} \right)^2 \]
where \(\alpha\) is thermal diffusivity, \(\theta_m\) is melting temperature, \(k\) is thermal conductivity, \(H\) is heat flux.
If a uniformly distributed 1 kW power laser beam with a beam diameter of 0.1 mm is used for machining tungsten carbide, and 10% of beam absorption is assumed, the time \(t_m\) is ______ \(\mu s\) (rounded off to one decimal place).
Note: Thermal properties of tungsten carbide: melting temperature = 3400°C; thermal conductivity = 2.15 W/cm-°C; diffusivity = 0.79 cm\(^2\) s\(^{-1}\); assume \(\pi\) = 3.14.
Show Hint
Convert all lengths to cm to match the given thermal conductivity and diffusivity units, then substitute directly into the given formula.
Updated On: Aug 5, 2026
Show Solution
Correct Answer: 32.1
Solution and Explanation
Step 1: Working through the beam radius instead of diameter:
Absorbed power: $ P_{abs} = 0.10 \times 1000 \text{ W} = 100 \text{ W} $.
Beam radius: $ r = 0.05 \text{ mm} = 0.005 \text{ cm} $.
Step 2: Heat flux from the radius:
Using area as $ \pi r^2 $ instead of $ \frac{\pi}{4} d^2 $, which gives the same value through a different route:
\[ A = \pi r^2 = 3.14 \times (0.005)^2 = 3.14 \times 2.5 \times 10^{-5} = 7.85 \times 10^{-5} \text{ cm}^2 \]
\[ H = \frac{100}{7.85 \times 10^{-5}} = 1.274 \times 10^{6} \text{ W/cm}^2 \]
This matches the diameter based route exactly, confirming the area calculation.
Step 3: Building the bracket term step by step:
Temperature difference: $ \Delta\theta = 3400 - 32 = 3368 $°C.
\[ \frac{\Delta\theta \cdot k}{2H} = \frac{3368 \times 2.15}{2 \times 1.274 \times 10^{6}} = \frac{7241.2}{2.548 \times 10^{6}} = 2.842 \times 10^{-3} \]
Squaring this ratio:
\[ \left( 2.842 \times 10^{-3} \right)^2 = 8.077 \times 10^{-6} \]
\[ \frac{\pi}{\alpha} = \frac{3.14}{0.79} = 3.975 \]
\[ t_m = 3.975 \times 8.077 \times 10^{-6} = 3.211 \times 10^{-5} \text{ s} = 32.11 \ \mu s \]
Final Answer:
Rounding to one decimal place, $ t_m = 32.1 $ $\mu s$, matching the diameter based calculation.
\[ \boxed{32.1 \ \mu s} \]
Absorbed power: $ P_{abs} = 0.10 \times 1000 \text{ W} = 100 \text{ W} $.
Beam radius: $ r = 0.05 \text{ mm} = 0.005 \text{ cm} $.
Step 2: Heat flux from the radius:
Using area as $ \pi r^2 $ instead of $ \frac{\pi}{4} d^2 $, which gives the same value through a different route:
\[ A = \pi r^2 = 3.14 \times (0.005)^2 = 3.14 \times 2.5 \times 10^{-5} = 7.85 \times 10^{-5} \text{ cm}^2 \]
\[ H = \frac{100}{7.85 \times 10^{-5}} = 1.274 \times 10^{6} \text{ W/cm}^2 \]
This matches the diameter based route exactly, confirming the area calculation.
Step 3: Building the bracket term step by step:
Temperature difference: $ \Delta\theta = 3400 - 32 = 3368 $°C.
\[ \frac{\Delta\theta \cdot k}{2H} = \frac{3368 \times 2.15}{2 \times 1.274 \times 10^{6}} = \frac{7241.2}{2.548 \times 10^{6}} = 2.842 \times 10^{-3} \]
Squaring this ratio:
\[ \left( 2.842 \times 10^{-3} \right)^2 = 8.077 \times 10^{-6} \]
\[ \frac{\pi}{\alpha} = \frac{3.14}{0.79} = 3.975 \]
\[ t_m = 3.975 \times 8.077 \times 10^{-6} = 3.211 \times 10^{-5} \text{ s} = 32.11 \ \mu s \]
Final Answer:
Rounding to one decimal place, $ t_m = 32.1 $ $\mu s$, matching the diameter based calculation.
\[ \boxed{32.1 \ \mu s} \]
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