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