How much power is needed to melt PLA? 5 W

As I started to think about building a 3D-printer I was looking for information about how much power is needed for the hotend. I did not find any usefull information! There are many discriptions for building hotends using resisters, NiCr-wire, heater cartidges, ..... having 20 W, 40W or even more.
I then looked for a PLA datasheet and found this. PLA_SpecificHeat
Based on this spreadsheet I added some calculations.
These are the results.


Please check my calculations.

Edited 1 time(s). Last edit at 10/06/2016 08:58AM by vlorijer.

Hi,

I think you're missing a step in the conversion from nozzle size and layer height to cross sectional area and then volume of extrusion. The cross sectional area is usually treated as a rectangle with two semi circles on the ends, see [manual.slic3r.org] which gives the area as



The equation I think you are using is

=layer height * nozzle diameter * speed * density * energy

would be something of a lower limit as the extrusion width is typically larger than the nozzle diameter.

The size and shape of the extrusion surely isn't relevant? What you want to know is how much PLA goes through the nozzle how fast. The easiest way to measure that is the (maximum) feed rate * filament cross-sectional area

You also need to allow for heat loss from the hot-end... radiation & conduction to the surrounding room (or maybe heated bed?) temperature air which could perhaps be as much as the heat you actually use to melt filament. FWIW, having a 50x15 blower fan blowing onto the hot-end is enough to knock the maximum temperature down to about 185C.

An easy way to measure (as opposed to calculating) the heating requirement would be to wire a 12V clock in parallel with the heater on an existing, working printer. When the heater is on, the clock will run. Print something. Divide your heater-on time by the elapsed time, multiply by the heater wattage, and you'll have the average power usage of the heater.

However, you'll want considerably more than the average available to you, otherwise it'll take an infinite amount of time to get to temperature.

In fact, I think you're over-thinking it. Is there any reason to actually worry about this? Why not go for a standard 40W cartridge? There's no money to be saved by using a lower power cartridge.

Frank

Hi JamesK + Frank.

thanks for checking.

@JamesK: You are rigth. I know the nozle size is not equal to the extrusion width. I used to wrong word. So please replace "nozle diameter" with "extrusion width". The error by supposing a rectangle will be neglectible ( or think of it as being an equivalent rectangle) and will not make a substantial difference to the results.
equation: power = layer height * extrusion width * speed * density * specific heat * temperature difference

@Frank: I am not sure if using a clock is the richt way. Clock speed will be dependant of the mean voltage. So if I would design a clock I would use some kind of voltage stabilizer. Please check if your clock has one. Hotend power is a function of the RMS value. As PWM is used for the heater, mean voltage is proportional to D ( D = puls width in % ) where as RMS is proportional to sqrt(D). My idea for measuring the power is logging the PWM and then calculate power using the input voltage and resistance (wich will change with temperature). Unfortunatly I can not do this myself as I don 't have the required HW and/or SW. Maybe someone else can?

My main point was: how could anyone design a hotend without knowing these numbers?

Ah, that makes perfect sense. I'd love to see comparison figures for abs and nylon.

Quote

how could anyone design a hotend without knowing these numbers?

While I lean towards the logic of design based on an understanding of fundamental principals, it would be fair to say that this is a case where trial and error works very well. The actual energy required to melt a given rate of plastic isn't actually that important as long as you have enough, since the hotend heater is controlled by a feedback loop. Experiment has shown that the common 40W heater cartridges are adequate for typical extrusion rates, but no so much that you can't hit the heater limit if you try hard with a big nozzle. (In my case 33mm³/s of abs was too much!)

I make my own hotends, but stick to a fairly conventional design (E3d v6-like) and use the typical heater cartridges. I've seen several more adventurous designs with co-axial heating elements, but I'm not sure what the pros and cons of those designs are. What do you have in mind, and what problems are you trying to improve on?

Edited 1 time(s). Last edit at 10/12/2016 06:31AM by JamesK.

Hi JamesK,

I am just trying to understand. And did some more calculations.
Suppose you have a 40 W heater and need 5 W to extrude . Marlin and Teacup use a max PWM of 255. (uint_8_t) Please check.
A change of one digit will result in a change of power of 40 W / 255 = 0,156 W. This is 0,156 / 5 = 3,1 %.
Question: Will the PID be able to regulate the temperature?
I know I am neglecting power losses.
Can you have a look at the PWM output? Maybe use a scope?

Edited 1 time(s). Last edit at 10/13/2016 04:34AM by vlorijer.

Ah, I see your point. Too much power and you could overwhelm the resolution of the PWM output. I don't use Marlin or Teacup, I have Repetier firmware configured to use dead time control for the extruders. I confess that I haven't read up on how DTC works, but it seems to give a more stable hotend temperature than I had previously with Marlin, which is one of the reasons I opted to stay with Repetier. Repetier host has a graph of extruder heating as the print progresses, and it seems to be a mixture of modulating the power as well as periods of bang-bang like behaviour. I don't know how accurate the graph is, but I'll try and remember to get a screen grab next time I'm printing.

Out of interest, is the power relationship to the square of the pulse width? I would have thought it was a linear relationship to the duty cycle.

Edit: Found the link to Repetier' dead time control, not much info but it does have a screen grab of the graph I was thinking of:

[www.repetier.com]



From that it looks like the power ramps up to a specified max, with varying durations of max and zero power. It certainly seems to work well.

Edited 2 time(s). Last edit at 10/12/2016 11:59AM by JamesK.

Thanks for the graphs. Is that you blowing or is it a fan?
Vertical resolution is 10 °C / line? Horizontal is minutes:seconds?

Quote

Out of interest, is the power relationship to the square of the pulse width? I would have thought it was a linear relationship to the duty cycle.

Yes. It is a linear relationship to the duty cycle. The RMS value is related to the square root of the pulse width. wikipedia Root_mean_square
Look for "Pulse train".

Edited 1 time(s). Last edit at 10/13/2016 04:50AM by vlorijer.

Not my graphs, those are just links to the image on the Repetier page. It looks very much like the effect you get when a part fan blows on the hotend.

Hmm, I may be misunderstanding, but the reference to A1 sqrt(D) is to give an RMS of the varying signal, in our case voltage. To convert that to power, you'd square the RMS value, so you'd be back to a linear relationship again. Or more intuitively, if the heater is 40W and you run it at 50% duty, it's going to average 20W.

You are absolutely right! I should think a little longer before posting! Please don't tell anyone.
I 've changed the text and removed the wrong files.