# Sintering math -- how much power is needed anyway?

Posted by Igor Lobanov
 Sintering math -- how much power is needed anyway? August 01, 2013 05:10PM Registered: 8 years ago Posts: 43
Guys,

I had been doing some modelling to find out how much laser power is required to do selective sintering of nylon powder, and, to be honest, I'm a bit surprised with the results. It turns out that to sinter nylon powder heated to the temperature of 8 degrees below melting point with a beam focused down to 0.2 mm spot and using layer height of 0.2 mm with feed rate of 300 mm/s one needs only about 140 mW, which is easily achievable with commodity laser diodes. In fact, less than one in 16×DVD-R burner which is 250 mW according to Wikipedia.

The model takes into account heat losses due to contact heat transfer within the material (negligible, in fact), but assumes zero laser beam dissipation/reflection (i.e. black powder and visible light laser), and round laser spot (i.e. some corrective optics for the diode).

Interestingly, according to the model, the biggest controllable factor influencing the power requirement for laser is the temperature of the build chamber. If the chamber itself is kept with room temperature, the laser power requirement goes up to 2.6 W. But we know that such a big temperature gap is undesirable as it causes wrapping, so some kind of heated volume is needed anyway.

The model itself in the form of Excel spreadsheet is attached, so you can take a look and play with the parameters. I've included pseudo-formulaic explanations for all the derived values including dimensions for clarity.

Looking for the feedback if:
- the model is actually correct mathematically, i.e. I didn't drop couple of zeroes or something like that
- the assumptions above are sensible in real life
- there are factors which also very important that I haven't included in the model
 Re: Sintering math -- how much power is needed anyway? August 16, 2013 07:25PM Registered: 8 years ago Posts: 43
Since no one has given me a feedback, I'm assuming there's no obvious mistakes So I'm adding the information to the DIY SLS FAQ.
 Re: Sintering math -- how much power is needed anyway? September 18, 2013 10:59AM Registered: 8 years ago Posts: 206
I quickly went through it, but you are saying your need "140mW". But which wavelength? You realize certain material absorb more or less depending on the wavelength?
But yeah, you don't need 1kW laser to burn plastic or metal if the laser is focused.
 Re: Sintering math -- how much power is needed anyway? September 18, 2013 05:02PM Registered: 8 years ago Posts: 43
tleneel,

I did realise that. The model "assumes zero laser beam dissipation/reflection" like with "black powder and visible light laser". If you've got a powder which is transparent or highly reflective at the wavelength of the laser, it wouldn't barely pick any power from the beam.

Pragmatically, it means either CO2 laser as clear plastics are opaque at 10um, or near IR/visible light diode laser with pigmented/specialist thermoplastics. There is a post nearby where VDX shares a link to some material on that - [forums.reprap.org]
 Re: Sintering math -- how much power is needed anyway? October 01, 2013 03:37PM Registered: 7 years ago Posts: 4
Hi Igor

Thanks for the wonderful information you have provided in the FAQ. I have a few questions in mind.
For a powder specifically Nylon 12 (PA 12), would 808 nm be ideal? How does this wavelength parameter play a role in the calculations you have done? Apart from increase in weight of the system , what would be the problem if I go for higher wavelengths? Also I came across lasers of types Continuous wave and quasi-continuous wave. Do you have any information you can share about how these two types would effect the sintering process.

Sorry for the barrage of questions!

Thanks

Azhar
 Re: Sintering math -- how much power is needed anyway? October 04, 2013 07:12PM Registered: 8 years ago Posts: 43
Azhar,

I am by no means an expert in practicalities of the sintering process, so the following is merely a speculation based on my understanding of the principles.

The model does assume full absorption of the beam energy by the powder, which is close to what one could expect with black powder and laser in a visible or near-IR wavelengths range. If you've got white powder it is going to reflect back or dissipate most of the energy and you might need eiter more of power in the beam, or use something like coal dust to darken the powder. Or, alternatively, go further into IR wavelenghts and use something like CO2 laser with 10um. At that wavelength most plastics are opaque and their colour does not really matter, but CO2 laser is much harder to hande (and more dangerous) than a laser diode.

As of laser types, to my understanding, it does not make much difference for sintering process as the only thing you're really interested in is to deliver enough energy to a small spot to raise its temperature. Sintering is effectively about the integration of beam power over time, so mode does not really matter.

Hope this helps.

Edited 1 time(s). Last edit at 10/05/2013 08:56AM by Igor Lobanov.
 Re: Sintering math -- how much power is needed anyway? October 07, 2013 04:19PM Registered: 7 years ago Posts: 4
Igor,

Thanks a lot for the response! Although I understood the concept of how the total energy delivered that matters for the sintering to takeplace, could you tell me which of them is popularly used, i.e, continuous wave (CW) or quasi- continuous (QCW) or pulsed? If I am going to purchase a laser diode system, I am faced with this question when I look at the laser diode specs!

Thanks again!

Azhar
 Re: Sintering math -- how much power is needed anyway? October 07, 2013 05:15PM Registered: 8 years ago Posts: 43
Azhar,

I wasn't sure I knew exactly was quasi continuous wave laser is, so I looked it up here:
Quote
Quasi-continuous-wave Operation
Quasi-continuous-wave (quasi-cw) operation of a laser means that its pump source is switched on only for certain time intervals, which are short enough to reduce thermal effects significantly, but still long enough that the laser process is close to its steady state, i.e. the laser is optically in the state of continuous-wave operation. The duty cycle (percentage of “on” time) may be, e.g., a few percent, thus strongly reducing the heating and all the related thermal effects, such as thermal lensing and damage through overheating

If I'm reading it correctly it means that quasi-CW is not different from CW as far as power delivery is concerned, but the laser cannot be switched on for longer than certain period of time, after which it has to be left in off state to cool down. I think you should have two numbers for a quasi-CW laser: duty cycle precentage and max duration for CW operation. If, for example, max duration is 1 second and max duty cycle is 20%, then after burning for one second the laser has to be turned off for 4 seconds to let it cool down.

I wouldn't be surprised if a quasi-CW laser could be used as CW laser by adding proper heatsink and active cooling. I know that most laser diodes can't really work properly in CW without at least a heatsink.
 Re: Sintering math -- how much power is needed anyway? October 08, 2013 02:52AM Admin Registered: 13 years ago Posts: 12,892
... I'm driving my diodes with pulsing or QCW, but not for cooling reasons (the big coolers are sufficient), but for modulating them in respect to moving speed.

Even with a 0-10V analog output from a servo-CNC calculated for perfect speed-djustment I'm converting the analog signal into PWM for better calibrating capabilities ...

Viktor
--------
Aufruf zum Projekt "Müll-freie Meere" - [reprap.org] -- Deutsche Facebook-Gruppe - [www.facebook.com]

Call for the project "garbage-free seas" - [reprap.org]
 Re: Sintering math -- how much power is needed anyway? October 08, 2013 07:28AM Registered: 8 years ago Posts: 43
Victor,

Is there any particular reason why you're using servos and not steppers?
 Re: Sintering math -- how much power is needed anyway? July 07, 2014 04:21PM Registered: 7 years ago Posts: 4
Hi again Igor, after a long time! Hope you are doing well.

As you mentioned before that white powders reflect most of the laser power, I am currently facing the exact same issue. You suggested I mix the powder with coal dust to make it absorb more power. Is there any other alternative, like may be dark powders or something. Or if you happen to know where I can find coal dust to purchase, could you share it with me?

Thanks a lot again!

Azhar
 Re: Sintering math -- how much power is needed anyway? July 07, 2014 08:57PM Registered: 7 years ago Posts: 1,381
WEST SYSTEM
#423 Graphite Powder - 12 OZ
Model # 323659 | Mfg # 423
\$17.99
[www.westmarine.com]

I've used West System #423 Graphite Powder with epoxy, and found it to be a very fine dust/powder.

Edited 1 time(s). Last edit at 07/07/2014 08:57PM by A2.
 Re: Sintering math -- how much power is needed anyway? July 08, 2014 06:55PM Registered: 7 years ago Posts: 4
Thanks a lot A2! You saved the day!

Azhar
 Re: Sintering math -- how much power is needed anyway? October 28, 2014 05:52AM Registered: 6 years ago Posts: 1
Hey, I checked the excel (thanks for all the formulas btw ) some references in formula explanations and the cells in excel references don't seem match. (For instance B26)
 Re: Sintering math -- how much power is needed anyway? April 29, 2015 04:14AM Registered: 8 years ago Posts: 11
I did not look at your equations, but your power is way off.
Given your stated conditions of a 0.2mm spot with a layer thickness of 0.2mm at a scan rate of 300mm/sec, the fused volume generated in 1 second is a long box with a square cross section 0.2mm on a side and 300mm long. The volume is 0.2mm*0.2mm*300mm = 0.012cc
The heat of fusion of Nylon 12 (sintering powder) is 215 kJ/kg (Page 648 Handbook of Thermoplastics edited by Olagoke Olabisi, Kolapo Adewale)
As the density of fused nylon is very close to 1 gram/cc, it takes 215 J/cc * 0.012 cc = 2.58 Joules to melt the volume fused in one second, or 2.5 Watts of laser power at 100% efficiency.
Lastly 300mm/sec is pretty slow, commercial SLS printers scan at a rate of over 12000 mm/sec which is why they use 100 Watt or 200 Watt CO2 Lasers
Ken
 Re: Sintering math -- how much power is needed anyway? January 08, 2017 06:15PM Registered: 3 years ago Posts: 3
Quote
kburgess
I did not look at your equations, but your power is way off.
Given your stated conditions of a 0.2mm spot with a layer thickness of 0.2mm at a scan rate of 300mm/sec, the fused volume generated in 1 second is a long box with a square cross section 0.2mm on a side and 300mm long. The volume is 0.2mm*0.2mm*300mm = 0.012cc
The heat of fusion of Nylon 12 (sintering powder) is 215 kJ/kg (Page 648 Handbook of Thermoplastics edited by Olagoke Olabisi, Kolapo Adewale)
As the density of fused nylon is very close to 1 gram/cc, it takes 215 J/cc * 0.012 cc = 2.58 Joules to melt the volume fused in one second, or 2.5 Watts of laser power at 100% efficiency.
Lastly 300mm/sec is pretty slow, commercial SLS printers scan at a rate of over 12000 mm/sec which is why they use 100 Watt or 200 Watt CO2 Lasers
Ken

Is this true? This was never confirmed, but it sounds plausible. I know it's been awhile since this thread has had a response, but it is linked in the DIY SLS FAQ so it would be good to get it sorted out.

I was following through with what kburgess was saying and I wanted to be sure it was on point.
 Re: Sintering math -- how much power is needed anyway? January 08, 2017 07:01PM Admin Registered: 13 years ago Posts: 12,892
... you'll need not only the power to melt the volume in the spot, but the solidified cold surface under the powder slice too, so the numbers should be a bit more.

But yes, for melting dark plastic powder with 200 to 400mm/s it should be more in the "low power" range of 2 to 5 Watts than 100 Watts or more ...

Viktor
--------
Aufruf zum Projekt "Müll-freie Meere" - [reprap.org] -- Deutsche Facebook-Gruppe - [www.facebook.com]

Call for the project "garbage-free seas" - [reprap.org]
 Re: Sintering math -- how much power is needed anyway? January 09, 2017 01:38PM Registered: 3 years ago Posts: 3
Ok, now I understand.

I'll try to incorporate that into the calculator Igor originally made and up and upload it for review.

Thanks VDX.
 Re: Sintering math -- how much power is needed anyway? January 13, 2017 11:40PM Registered: 3 years ago Posts: 3
Alright, I think I got most of it. I would greatly appreciate someone looking over it to make sure it's solid. I just had a few concerns that are highlighted in a salmon color.

I should have started from scratch, but I just kept adding to Igor's. I still cleaned it up a bit and made it a bit easier to read and reference.

I would love to draw a diagram and figure out shortening the terms to variables. That might still be to come.

For those of you interested in hearing, this is a senior project for a group I am working with in school. We hope to manufacture an SLS printer for under \$2,000.

The laser will be low powered and will have a slow feed rate as our goal is to sinter PA 12 cost effectively.

Thanks.

- Gurr
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