Sintering math -- how much power is needed anyway? August 01, 2013 05:10PM |
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Re: Sintering math -- how much power is needed anyway? October 07, 2013 05:15PM |
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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
Re: Sintering math -- how much power is needed anyway? October 08, 2013 02:52AM |
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Re: Sintering math -- how much power is needed anyway? October 08, 2013 07:28AM |
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Re: Sintering math -- how much power is needed anyway? January 08, 2017 06:15PM |
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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
Re: Sintering math -- how much power is needed anyway? January 08, 2017 07:01PM |
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Re: Sintering math -- how much power is needed anyway? January 13, 2017 11:40PM |
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