How to Build a Laser Death Ray
Material Response to Average Intensity
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How to Build a Laser Death RayMaterial Response to Average Intensity |
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Light contains energy. When that light is absorbed by a material, its energy goes toward heating the material. However, the faster the material gains heat, the faster it sheds heat. The temperature to which a beam of light can bring a material depends on the intensity of the light and the rates and mechanisms by which the material responds to the heat.
One easy estimate allows you to get a rough upper limit to the temperature of a surface exposed to your death ray. Find the absorptivity, a, of the illuminated material. This is a number between 0 and 1 that determines the fraction of light that is absorbed. The section on diffraction allows to to find the diameter of the spot, S, into which you can focus your beam. If you know the time average power, P, of the beam, the total intensity (power divided by area) of the beam is I = P / (π S2 / 4).
The intensity absorbed by the surface is
Ia = I × a.
One of the methods by which hot surfaces dispose of their heat is by radiating it as electromagnetic waves. The total radiated power is given by Ir = e σ T4,
where T is the temperature in Kelvins, e is the emissivity (a number between 0 and 1 that determines how well the surface can radiate), and σ is a constant of nature
σ = 5.67×10-8 W m-2 K-4.
Note that σ is a constant - the letters after it just tell you want units to use in the equation. If you use temperature in kelvins, for example, it gives you intensity in watts per square meter.
For this rough estimate, you can set the absorptivity equal to the emssivity (this is strictly always true for absorption and emission at the same wavelength, but both a and e vary with wavelength). If we assume that radiation is the only mechanism through which a surface sheds its heat, then the surface will gain heat until the intensity of radiation equals the intensity of beam energy being absorbed. In other words, Ia = Ir. This allows us to solve for the temperature the surface will reach T = (I/σ)1/4.
This estimate ignores other mechanisms of shedding heat, such as heat conduction from the surface into the bulk material, or transfer of heat to the surrounding air or water which is then carried away by a rising plume of this warmed fluid. When a surface gets very hot, radiation is the dominant method of disposing of heat. However, for cooler temperatures (like "just" the heat of a stove burner on high), the transfer of heat by conduction to cooler matter in contact with the material is much faster than radiation. In the vacuum of space, where a wide beam is irradiating a thin slab or a good thermal insulator, this approximation is good. For an intense beam that heats a material to extreme temperatures, this approximation is good. For relatively weak beams illuminating surfaces of materials with good thermal conductivity or in air, this approximation is not so good and the final temperature will be lower. If the above estimated temperature is higher than the melting temperature, the material can start to melt. It takes heat to melt matter. So long as the material can continue melting, the process of melting will absorb the energy of the beam and keep the material from getting hotter. The rate at which matter will melt rm can be found if you know the melting temperature Tm, the initial temperature, T0, the specific heat, cv, the heat of fusion Hf and the density ρ rm = a × P / (ρ × (Hf + (Tm-T0) × cv)).
Tm, cv, Hf, and ρ can be found on the web for most common materials these days, but you may need to convert from inconvenient units. If Hf is in joules per kilogram, cv is in joules per kilogram per kelvin, Tm and T0 are in kelvin, and ρ is in kilograms per cubic meter, then rm is the rate at which matter is removed in cubic meters per second. Divide by the laser's spot size to find the rate at which the melt front propagates into the material.
If the estimated temperature is above the boiling point of the material, the material may begin to vaporize if it cannot conduct heat from its molten surface to unmelted material fast enough to lower the temperature. If you know the heat of vaporization, Hv, and the boiling temperature, Tb, in addition to the other materials known for melting (and assuming the material starts off solid), the rate at which material is vaporized, rv is given by rm = a × P / (ρ × (Hv + Hf + (Tb-T0) × cv)).
(if the material is already liquid, ignore Hf in the above equation. For materials that don't melt at all but directly sublimate into vapor, you should also ignore Hf). As with melting, vaporization uses energy, and the temperature will remain at the boiling temperature unless heat is being added at an extraordinary rate. This is because the vaporized material does not stick around to insulate the rest of the target, but rather immediately escapes, taking its heat with it. Note that boiling temperature depends on pressure. This will not matter so much for most high boiling temperature materials, but for low boiling points it can make a large difference whether the material is in an atmosphere or in vacuum. Also note that boiling matter can splatter molten material out of the way, resulting in faster drilling.
Experience with laser machining tells us that the absorptivity of materials under intense laser irradiation is typically high. From this reference we find
Im = σ × (1811 K)4 = 6.1×105 W m-2.
The minimum intensity required to vaporize iron under perfect conditions is
Iv = σ × (3134 K)4 = 5.5×106 W m-2.
In practice, iron is a good thermal conductor so heat loss into the bulk significantly increases the above required intensities for small spot sizes.
If we input cv=0.449 kJ/(kg K), Hf=0.247 MJ/kg, Hv=6.088 MJ/kg, and ρ=7874 kg/m, we find that a 1 MW laser can evaporate 17 cm3 of iron per second, and it can melt 140 cm3 of iron per second.
Rough empirical values used for laser machining and welding are described below (Lasers and their Applications, A. Sona, ed, chapter "Machining with lasers" by Dieter Roess)
At a somewhat higher power flux, between 106 to 108 Wcm2 [sic], the thin absorbing surface layer is heated up to its evaporation temperature before the solid-fluid interface has progressed appreciably into the material by heat conduction. Thus, with continuing laser power, a less than μ-thick layer of material will continually evaporate, with the material-air interface progressing into the material. Typically a gas jet develops [...]. The gas jet ejects also part of the molten material, thus that the progress rate of the hole is faster than with evaporation of all the material. [...]
At a still higher flux of 109 to 1010 Wcm-2, after initial evaporation of the surface layer, the gas jet is thermally ionized and absorbs most of the incident radiation, which is such blocked away from the material. The surface layer explodes with an ultrasonic jet, its temperature may rise beyond 10 105 ° K [sic], its pressure beyond 103 at.
Below is a calculator to determine the black body temperature and rate of melting for a heat ray incident on a surface of known material properties (or select from a set of pre-defined materials). It uses the approximations given above, and assumes absorptivity=1.
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