Absorption of heat by laser light into material, calculation

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Anssi
Anssi 2014-9-17
评论： Effy Shafner 2017-2-1
Hello everyone and thank you beforehand if you can offer some insight to this problem!
I have been given the task to create a simple calculation for melting time of n more or less isotropic plastic material (ABS) when targeted with a constant 50 W laser with wavelength about 800 nm. I already have the laser up and running but the calculations I have made are pretty much crap compared to reality. Of course if I could measure the temperature the laser creates on the surface of the plastic, the task would be easier but unfortunately this is not possible.
I used a simple model of energy transfer E = c*m*(T_m-T_r), where c: specific heat capacity, m: mass of material, T_m: melting temperature of material and T_r: ambient temperature.
First calculations that I made on paper I just substituted E = P*t from which t could be calculated but soon noticed that this does not apply to monochromatic light with a specific wavelength but rather to normal heat conduction through a wall (to which I have used it previously).
After this I tried replacing the energy term E with Beer-Lamberts law: E = E_0*e^(-alpha*z), where E_0: incident energy, alpha: absorption coefficient and z: depth inside material (=0,5 mm).
Now the questions: Is the Beer-Lambert law valid for this type of calculation? - If not: Why? How should I approach this problem? - If yes: Can I substitute E_0 = P*t to formula in order to get time, t? How to estimate the coefficient alpha for the material with this specific wavelength of light? Do I have to consider the change of alpha with temperature in these calculations or can I get a decent estimation with an approximate value?
Parameters: Power, P: 50 W, Wavelength: 800 nm, Mass targeted by laser, m: 0,0001243 kg, Specific heat capacity for ABS plastic, c: 1423.5 J/kg*K (cannot remember where I got this from)

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回答（1 个）

Iain
Iain 2014-9-17
编辑：Iain 2014-9-17
You need to get the physics clearer.
1. Reflection, by the plastic, of the laser light will reduce the energy absorbed.
2. Beer-lambert will only allow you to figure out how the energy is deposited within the material. - I'd suspect that for a plastic, that most of the absorption will have occurred within the top mm or so - but some are transparent...
3. Once the plastic starts melting, the absorption will vary if the plastic is at all transmissive to that laser light.
4. The plastic, as it heats up, will radiate heat back, and it will conduct heat away from the laser spot, the amount of conduction will likely increase if the spot itself melts.
5. The power rating may be for power being input to the laser, not output. 50W is pretty HUGE power for a laser.
6. You may have to accomodate how transmission/absorption vary with temperature - that depends on how it varies with temperature.
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Iain
Iain 2014-9-17
Ok, well, the beer-lambert law is usable. It tells you how much of the energy will have passed through 0.5mm of material.
So, your maximum energy absorbed by the material will be 50(1-e^(-alpha*z)), and reflection, it's probably in the order of a few percent.
In 0.1 seconds you won't notice much effect due to reradiating or heat conduction...
Effy Shafner
Effy Shafner 2017-2-1
One more thing to consider: (Latent) heat of fusion – a significant amount of energy is required to make a material undergo the transition from solid to liquid when it melts – this is likely to be the major cause in discrepancy between your calculations and measurements. (From my colleague, Shimon Elstein, senior physicist at Ophir Photonics.)

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