1
$\begingroup$

Could concentrated sunlight be used to build a metal 3D printer, sintering or melting metal powder?

Related question on physics stack exchange https://physics.stackexchange.com/questions/143049/what-temperature-is-achieved-in-focus-point-by-5000-flat-1x1cm-mirrors-onto-a-sa

$\endgroup$
3
  • 2
    $\begingroup$ Yes that could be possible, I've seen that before, but that printer used a Fresnel lens that melted sand to glass. But I doubt whether it would be practical. Are you coming from a planet builder perspective? $\endgroup$
    – 0scar
    Jun 4 '20 at 21:43
  • $\begingroup$ I was mainly wondering whether oi could be practical to make cheaper printers without an expensive laser $\endgroup$
    – stckmower
    Jun 5 '20 at 12:46
  • 1
    $\begingroup$ That expensive laser will create a highly concentrated focused bundle of energy (light), it is probably undoable to positioning 5000 mirrors to achieve the same accurate focal point. $\endgroup$
    – 0scar
    Jun 5 '20 at 13:52
3
$\begingroup$

Let's start with the obvious: this printer would need to be really big. Not because of a large print volume, but because it needs to collect a lot of sunlight or needs a really big focussing array. The linked question states that the array there, about 0.6 m² large, has roundabout 600 W of power to focus on that one point.

Power draw needs

What powers are we dealing with?

A typical laser cutter uses a laser tube that at least 20 W for thin material and up to 300 W for thicker material. But we need to weld steel, so we need to be roughly equivalent to a cutting laser for the same material. We are not talking mere hundreds of watts, we are talking an industrial 2000 to 20000 W in a laser of less than a millimeter in diameter. Why do I say the later? Well, power need scales with the area, which scales with the square of the radius.

Let's use a ballpark, a nice round 10000 W Laser with a somewhat large 1 mm² crossection. We're talking about the ballpark of 100 Gigawatt per square meter.

$L=P/A=\frac{10000\ \text W}{0,001\times 0.001\ \text m^2}=10^{10}\frac{\text W}{\text m^2}$

Solar power harvester size

Luckily, we don't need to illuminate a whole square meter, so we only need some 10 Kilowatts of sunlight for our application. But we need this number to calculate how much sunlight we need to harvest in our smelting machine.

At the stratosphere, Earth gets about 1400 W/m², and on a sunny day, about 1 kW/m² makes it to the surface, we call this the solar constant $S$. Now, if we compare the ballparks, we get to quickly see the ballpark size of our machine:

$P/S=A=\frac{10 \text{ kW}}{1 \frac {\text{kW}} {\text m^2}} = 10\ \text m^2$

10 square meters of harvesting area gets us the same power. Incidentally, this scales linearly with the item discussed in the linked question, as that one already uses square meters.

Sizing down

But 10 square meters of mirrors into a focal point and then lenses to a focal point is huge, can we make it smaller? And to that I must say somewhat. First of all, we could get our focal point smaller: We need to get 10 Gigawatt per square meter for our $L$, but we can turn two screws here: what if we get from 1 mm² to only 0.1 mm²?

$P=L*A=10^{10}*(0.0001\times 0.0001)\ \text W=100 W$

100 Watts focussed on 0.1 mm² is a sixth of the power that array can deliver, so totally feasible in terms of power, as long as you can achieve such a small focal point.

Conclusion

Yes, with a focussing array good enough or a mirror array large enough you could achieve the powers needed to melt metals on a spot focus. Note though, that you need to have a really good focus setup that creates pretty much a solar-powered laser, which means that such a machine will be extremely expensive due to the high precision machinery needed for that - and that this focusing aperture will be most likely the largest part of your machine. We're talking building-size scale. I'd be cheaper and easier to just put lots of solar panels onto the roof of the building you run a conventional metal 3D printer in.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.