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flow rate estimation

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Dear all,

i found that it is possible to calculate the flow rate (FR) by the following equation:

FR = Nozzle size (mm) x layer height (mm) x print speed (mm/s) = mm3/s.

Nevertheless, it is not clear because of in this equation the surface of the filament it is not taken into account.

In general if I work with a nozzle of 0.5 mm and a layer eight of 0.4 the minimum volume (MV) of filament needed is:

MV = pi*radius^2*height = 3.14*0.25^2*0.4 = 0.0785 mm3.

Let me assume that i'm using a print speed of 30 mm/s and that i'm printing on air obtaining a filament (cylinder) of height = 30 mm for each second of printing.

So, in this way i'm printing a volume of V=pi*radius^2*height = 3.14*0.25^2*30 = 5.88 mm3 that is a flow rate (FR) of 5.88 mm3/s.

May you help me to correct above consideration ?


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Your calculations are both right. The slicer uses the first method because that approaches what is happening when printing. The nozzle prints a line with a more or less rectangular cross section (line width x layer height).

The second method is only valid when you would calculate for printing in air.

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Thank Tomnagel,

but if I'm using a clay extruder with a nozzle having a round shape, i cannot consider a rectangular shape.

In this case the second equation should be more precise because I have ha cylinder. Also, In my opinion the equation should works also if I'm printing not in air.

What is the difference if i'm printing a more o less perfect cylinder on air or if this cylinder is used to build a 3D object ? Briefly, even if after printing the cylider changes is shape, the amount (volume) of printed material should be the same.

Of course the 3D object will have a different volume than the printed material.



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The FDM printers also have a round nozzle, but the extruded plastic is pushed onto the previous layer. The shape of the deposited line has a rectangular cross section.

So the slicer calculates how much plastic is needed to print such lines, and puts this number in the gcode.

You reason the other way around. You calculate with the cross section of your nozzle, and multiply this with the speed. If the slicer would use this calculation, the printer will still flatten the extruded amount of plastic to the chosen layerheight. With normal layerheights (which are normally lower than your nozzle diameter) this will lead to a printed line that is (much) wider than your nozzle, with ugly results.

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Dear Tomnagel,

you consideration are correct and very useful for me. But i'm printing food (vegetable cream) and not plastic filament.

My interest is just to calculate the amount (volume) of material (food) deposited during printing while i have not interest in the total volume of the 3D printed object. Also, let me assume that is not our interest to have a good printing (a good 3D printed object).

I known that by printing on air i have a volume of material greater than the 'real' volume of the printed object. This is because, as you well reported, extruded plastic (or extruded food ) is pushed onto the previous layer producing a modification of its shape (from cylinder to rectangular shape).

But back on the theory, my question is to precisely calculate the amount (volume) of material extruded by considering the nozzle size and print speed as follow. Since the filament (when extruding) have a shape of cylinder the correct total volume of material should be:

V=pi*radius^2*print speed = mm3/s.

Again, obviously this isn't the volume of the printed object because each layer will be squashed between other two layers but, in my opinion, is the correct amount of material estruded through the nozzle thata was necessary to built our 3D object.

What you think on my consideration ?


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I have additional doubt:

considering the above reported equation: V=pi*radius^2*print speed = mm3/s.

But i believe that It is correct only if we are using a flow rate of 100% (by using CURA). It is because if we use a flow of 120%, the volume of deposited material (not of the printed object) should be greater than the 20% ?

Vcorrect=V+V*0.20 ? (Eq.2)

Similarly if we use a flow of 70% the V=V-V*0.30 ? (Eq.3)

What you think ?


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