Question:
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 3, 2026
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Correct Answer: 32.1
Solution and Explanation
Step 1: Understanding the Question:
We are given a formula for the time needed to heat a spot on tungsten carbide up to its melting point using laser energy, and we just need to plug in the numbers carefully with consistent units.
Step 2: Setting up the heat flux H:
Absorbed power is 10% of the 1000 W beam:
\[ P_{abs} = 0.10 \times 1000 = 100 \text{ W} \]
The beam diameter is 0.1 mm, which is 0.01 cm, so the beam spot area is:
\[ A = \frac{\pi}{4} d^2 = \frac{3.14}{4} (0.01)^2 = 7.85 \times 10^{-5} \text{ cm}^2 \]
Heat flux is absorbed power divided by area:
\[ H = \frac{P_{abs}}{A} = \frac{100}{7.85 \times 10^{-5}} = 1.2739 \times 10^{6} \text{ W/cm}^2 \]
Step 3: Substituting into the given formula:
The temperature rise needed is \( \theta_m - \theta_0 = 3400 - 32 = 3368 \)°C, so the numerator of the bracket is:
\[ (\theta_m - \theta_0) k = 3368 \times 2.15 = 7241.2 \]
The denominator is:
\[ 2H = 2 \times 1.2739 \times 10^{6} = 2.5478 \times 10^{6} \]
So the bracket term and its square are:
\[ \frac{7241.2}{2.5478 \times 10^{6}} = 2.8422 \times 10^{-3}, \quad (2.8422 \times 10^{-3})^2 = 8.078 \times 10^{-6} \]
Multiplying by \( \pi / \alpha = 3.14 / 0.79 = 3.9747 \):
\[ t_m = 3.9747 \times 8.078 \times 10^{-6} = 3.211 \times 10^{-5} \text{ s} \]
Final Answer:
Converting to microseconds by multiplying by \(10^6\) gives \( t_m = 32.1 \) \(\mu s\).
\[ \boxed{32.1 \ \mu s} \]
We are given a formula for the time needed to heat a spot on tungsten carbide up to its melting point using laser energy, and we just need to plug in the numbers carefully with consistent units.
Step 2: Setting up the heat flux H:
Absorbed power is 10% of the 1000 W beam:
\[ P_{abs} = 0.10 \times 1000 = 100 \text{ W} \]
The beam diameter is 0.1 mm, which is 0.01 cm, so the beam spot area is:
\[ A = \frac{\pi}{4} d^2 = \frac{3.14}{4} (0.01)^2 = 7.85 \times 10^{-5} \text{ cm}^2 \]
Heat flux is absorbed power divided by area:
\[ H = \frac{P_{abs}}{A} = \frac{100}{7.85 \times 10^{-5}} = 1.2739 \times 10^{6} \text{ W/cm}^2 \]
Step 3: Substituting into the given formula:
The temperature rise needed is \( \theta_m - \theta_0 = 3400 - 32 = 3368 \)°C, so the numerator of the bracket is:
\[ (\theta_m - \theta_0) k = 3368 \times 2.15 = 7241.2 \]
The denominator is:
\[ 2H = 2 \times 1.2739 \times 10^{6} = 2.5478 \times 10^{6} \]
So the bracket term and its square are:
\[ \frac{7241.2}{2.5478 \times 10^{6}} = 2.8422 \times 10^{-3}, \quad (2.8422 \times 10^{-3})^2 = 8.078 \times 10^{-6} \]
Multiplying by \( \pi / \alpha = 3.14 / 0.79 = 3.9747 \):
\[ t_m = 3.9747 \times 8.078 \times 10^{-6} = 3.211 \times 10^{-5} \text{ s} \]
Final Answer:
Converting to microseconds by multiplying by \(10^6\) gives \( t_m = 32.1 \) \(\mu s\).
\[ \boxed{32.1 \ \mu s} \]
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