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There is a lot of misunderstanding about 3D printer filament drives so I thought it might be a good idea to ask here :)

Bernouilli Effect

A FDM 3D printer basically has a drive gear pushing the filament (1.75mm or 3.0mm) down into a heater block where it becomes liquid. There is a nozzle (usually between 0.2 and 0.6mm diameter hole) where the filament comes out.

Cutaway of an extruder

So, this is where I believe that the Bernouilli Equation comes into play: for any given "flow rate" the pressure required to push molten plastic out of, say, a 0.2mm nozzle, will be exactly the same (friction etc. being ignored from consideration) regardless of the "incoming" filament width... but what about the incoming PRESSURE (and filament speed)?

So this is where the confusion lies in the 3D printing world. There is the belief that the pressure that has to be exerted by the drive-gear is either the SAME or LESS for a 3mm filament than it is for a 1.75mm filament drive system. Intuitively it is understood that the drive speed has to be greater, but it is not clear in the community by how much.

Looking at the Bernouilli Equation I believe that it is the case that the 1.75mm filament's pressure is reduced by a factor of (3.0 / 1.75) squared, and the speed at which the filament has to be pushed is also raised by the same factor.

However now that I think about it, I am not so sure that that is actually true, and that instead we may simply consider this to be a "hydraulic system" and to consider the areas instead. Perhaps the pressure is covered by a linear law but the velocity by a square law: I do not know, hence why I am asking :)

So the questions are (assuming friction and other losses to be zero):

(1) what is the relative ratio of the speeds at which 1.75mm and a 3.0mm filament have to be pushed into the hot-end in order to achieve the same rate of flow of molten filament out of the same sized nozzle?

(2) what is the relative ratio of the pressures exerted on each of the two filaments, under the same conditions in (1)

(3) Bringing the actual drive gear into the picture: assuming that the centre of the filament is an equal distance from the centre of the drive gear shaft, what is the relative ratio of the torque that has to be exerted by the drive motor for each of the two filament sizes, again under the same conditions as (1)

This has been a thorny question in the 3D printing community for some time, so you would be doing a lot of people a huge favour by helping to clear this up :)

(p.s. if anyone knows any better tags and can edit this question please feel free)

lkcl
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  • This might be better suited to physics.stackexchange.com. – Chappers Aug 30 '17 at 13:37
  • hmmm, maybe! although this is a "practical" question, do you think people there would know as much about the bernouilli equation and hydraulics as on here? – lkcl Aug 30 '17 at 13:40
  • There's about 6 times as many questions tagged fluid-dynamics over there: https://physics.stackexchange.com/questions/tagged/fluid-dynamics – Chappers Aug 30 '17 at 14:11
  • They also have a hydraulics tag. – Chappers Aug 30 '17 at 14:16

1 Answers1

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You are missing a drive term in Bernoulli's equation. For our purposes it is better to include a energy gain of the left side that represents the drive gear's input. Note that mechanical energy is given by $E = \tau \omega$ where $\tau$ is the torque of the gear and $\omega$ is the rotational speed.

1) $Q = Av$ where $Q$ is the volumetric flow rate, $A$ is the cross sectional area of the pipe, $v$ is flow velocity. Since the outlet and inlet volumetric flow is conserved, we have: $$A_1v_1 = A_2v_2$$ $$\frac{A_1}{A_2} = \frac{v_2}{v_1}$$ $$\frac{r^2_1}{r^2_2} = \frac{v_2}{v_1}$$

2) You can have the pressure difference pretty easily by arranging Bernoulli's equation, but the pressure ratio can only be found if you know one of the pressures. I remember there are ways to do this with extrusion equations, so let me do some research and come back to this.

3) With the change made to the equation, torque should now be easy to find. The problem is that torque is free, and you can set it to anything you like. There may be some missing constraints here. One big issue is that the system is not lossless and the energy drains may be significant. It may be worth computing these.

Kaynex
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  • very very interesting! ok so i initially thought that because the question is related to relative pressure / torque / velocity (i.e. ratios), that, assuming no losses due to friction that there would be no free constraints. yes the torque is free but the relative torque (1.75mm compared to 3mm) is not. one thing that may help in answering Q1 and Q2: if you take two bernouilli diagrams above, one with an input (P1 and A1) associated with 3mm and the other with 1.75, then cut them in half at P2 (nozzle, 0.2mm diameter) then put them back-to-back, we can eliminate the nozzle part. – lkcl Aug 31 '17 at 08:50
  • sigh have to continue on another comment.... so, assuming losses to be zero the actual nozzle may be eliminated. we may simply consider $$P1_'$$ for 1.75mm and $$P1_''$$ for 3.0mm or, to match the equations you wrote, use $$P_1$$ and $$P_2$$. thus using those equations, $$A_1 = 3.0^2"" and $$A_2 = 1.75^2$$, so the extrusion speed will be proportional to the... square of the diameter, yes! that's what i thought! :) – lkcl Aug 31 '17 at 08:56
  • ok so regarding (2), i notice in bernouilli's equation that there are three terms (one of which includes potential energy). i don't know what effect that has, but intuitively i feel that a gas would cause a bit of an issue and cause the equations to be non-linear. might it be reasonable to assume that the "liquid" is incompressible (i.e. non-elastic) and would that result in a linear relationship with no unknowns? – lkcl Aug 31 '17 at 11:21
  • Liquid is considered incompressable in a low pressure situation like this one. If we are working with a compressible fluid, there's a compressible version of Bernoulli's equation which you may find on google (but it's not very fun so I've never actually used it). However note that we do have some unknowns, like $\omega$, $\tau$, either $P$, $z$, ect. We're missing designer variables. There might be other reasons why one might set some, for example a lower $\tau$ will mean a smaller gear, but a beefier circuit. – Kaynex Aug 31 '17 at 14:30
  • thanks kaynex, i thought it might be reasonable to ignore compression in liquids. regarding the unknowns: i believe the actual values aren't important because we're comparing relative pressure, relative torque and so on. however what i don't know is if Bernouilli's Equation actually does depend non-linearly on, say, the pressure. so if set the pressure arbitrarily in the 3mm case at say... 10psi (no i don't like US units, but hey :) ) then from the equations we may say that the 1.75mm case would be 10psi * (some ratio). carrying on in another comment... sigh... – lkcl Aug 31 '17 at 16:12
  • ... but what i don't know is: if you then set it to 20psi (say) in the 3mm case, i would expect the 1.75mm case to also be 20psi * (exactly same ratio).... or is it? that's what i don't know. is there some term in Bernouilli's Equation (assuming zero friction losses) which makes the ratio between the two cases a non-linear relationship? specifically: there's that "Potential Energy" term. is that related to the compressibility of the material? if so can it be ignored such that we get a linear relationship betweeen the two cases? it's really quite fascinating :) – lkcl Aug 31 '17 at 16:17
  • Pressure difference is related to flow charicteristics, but pressure ratio is not. Is there any reason why you may need the ratio? And the potential energy has nothing to do with compressibility, but instead depends on gravitational effect on fluid. In this case, since fluid flow is down, the potential energy is probably significant. – Kaynex Aug 31 '17 at 18:03
  • i need the ratio because that's what provides a meaningful comparison. now, we could specify a fixed pressure to make it easier if that helps, but what i would then need to know is: if you double the pressure, what happens? so it all comes down to ratios as opposed to fixed quantities. regarding the gravitational effect - which i now understand so thank you - it's worth pointing out that the pressures inside 3d filament printers are absolutely ENORMOUS. the filament is not that viscous, and the "upgearing" from say the 3mm to 0.2mm as far as surface area is concerned is .... argh end of sp – lkcl Sep 01 '17 at 14:13
  • the upgearing from the surface area is a whopping 25:1. an 84 oz-in NEMA-17 motor (which is overkill btw) produces 0.5931703539999972 of torque, and a 4:1 gearing would make that around 2Nm of torque, and the distance from the centre of the hobbed gear to the filament is around 0.08 metres. sooo.... we get... 25 Newtons of force i believe, out of those over-spec'd motors (most extruders only use about 1/2 that). so, 25 Newtons vs 1 Newton (gravity), you see why it's not hugely significant, but yes does need to be taken into account in practical terms (retracting the filament, stop oozing). – lkcl Sep 01 '17 at 14:19
  • .. .however now that you mention what the actual effect of potential energy is, that's actually really really helpful to know and so you've given me a much better understanding of what's going on. enough to say that yes, for the purposes of this ratio-based analysis, we may eliminate the potential energy from the equation on both sides. – lkcl Sep 01 '17 at 14:22
  • wait... sorry... 8mm diameter hobbed gear, so that's 0.04 metres. so 2Nm of torque turns into 50 Newtons of pressure. sorry! i do know of extruders where they take a 5mm bolt and put cuts in it, then add a 4:1 gear ratio on it. that would be about 0.025 mm which would be about 80 Newtons. it's not a small amount of pressure, in other words! – lkcl Sep 01 '17 at 14:42