On the development of heads of 3D FDM-printers. Part 2
Continuation of the first part. In it we will calculate, and I have already built and tested the calculated head.
On the results, conclusions and further possible improvements will be in the third part.
Here we look at: - 1.6. Calculation of friction on linear sections with melt. Calculation of extrusion speed for nozzles of various diameters. The ratio of friction values in the areas of the nozzle and the melting zone. - 1.7 Comparison of calculations of the melting rate (see p. 1.2) and data from experiments. Findings. - 1.8. Deformation zone. Her internal profile. The assumption considered in this paragraph was not confirmed by experience. - 1.9. Calculation smoothing pen, its diameter and height. - Chapter 2. 2.1-2.5 Calculation of the speed head. - 2.6 Calculation of the heater. - 2.7.How to measure the temperature and how to regulate it.
Head type 4.2
')
1.6. Calculation of friction on linear sections with melt. Calculation of extrusion speed for nozzles of various diameters. The ratio of friction values in the areas of the nozzle and the melting zone.
1.6.1. The calculation of the ratio of the resistance of the nozzle and the working area. Effect of spout resistance.
Zone D of Figure 3. Molten plastic rubs against the walls and between its layers. Ultimately, all friction losses occur on the walls. Let's try to count. As a source of wisdom, I used the book by Kasatkin, Basic Processes and Apparatus of Chemical Technology, there in the chapter on Hydraulic Resistance in Pipelines, quite simply, the formula for resistance on straight pipe sections is derived from the Poiseuille equation and the Bernoulli equation. I brought her to mind
ΔP = 32 * μ * w * l / d²
I replaced viscosity and number 32 with K and it turned out to be even simpler: ΔP = K * w * l / d², where w is the speed, l is the length of the resistance section, d is the diameter of the pipe. Fortunately, we can use this equation ... because of the high melt viscosity, the Reynolds criterion favors us. And we can calculate at least at the estimated level (qualitatively, as the chemists say) the degree of friction in the areas of zones D and F too.
Calculate for a particular head type 4.1 the ratio of the contributions of the resistance of the nozzle and the rest of the head. This is useful to us to estimate the minimum acceptable diameter of the nozzle. For a head of this type, after eliminating defects and making changes, the results are quite repetitive and show good accuracy. Up to 1%. Since I have two nozzles of the same diameter, with different spout lengths of 0.4 and 0.8 mm, the difference in the results can be used to calculate the contribution of the nozzle resistance for different cases. Pay attention - it was very difficult to precisely manufacture the nozzle and the head, maintaining all specified dimensions! The results of the experiment for a nozzle of 0.3 mm, with a spout length of 0.4 mm, at a temperature of 260 º - Speed at the nozzle exit 206 mm / s. The slippage coefficient is 80%, the speed at the nozzle exit is 260 mm / s. Filament diameter 3.0 mm, plastic - ABS. The feed rate, taking into account slippage and feed inaccuracy = 2.06 mm / s.
D = 3mm, d = 0.3mm, W = 2.06mm / s, w = 206mm / s, L = 50mm, l = 0.8mm, W2 = 2.60 mm / s, w2 = 260mm / s (extrapol .), l2 = 0,4mm. We consider the resistance:
ΔP1 = K * w * l / d² = K * 206 * 0.8 / 0.09 = 1831.1 * K
ΔP2 = K * w2 * l2 / d² = K * 260 * 0.4 / 0.09 = 1155.6 * K
Performance (and speed) in the second case
Δw = 260 * 100/206 = 1.26 * 100 = 126% of the performance of the first.
Total resistances are:
ΔP1 + X = ΔP2 + X2, where X is the resistance of the head without a nozzle.
Since the speed at the output of the head and at the input increases equally (as many as entered, so much happened)
X2 = X * 1.26
We write the equation for ΔP:
ΔP1 + X = ΔP2 + 1.26X or from the formula above
1831.1 * K + X = 1155.6 * K + 1.26 * X => 1831.1 * K - 1155.6 * K = 1.26 * X - X
0.26X = 675.5K => This sign "follows from this" X = 675.5K / 0.26 = 2598K
Then, in the case of a spout of 0.4 mm, the ratio of resistances will be:
ΔP1 + X = 1831.1 * K + 2598K => 1831.1 / 1831.1: 2598 / 1831.1 => 1: 1.42 => 41%: 59%
Denote them as ΔPn - nozzles and ΔPh - the rest of the head. In this case, the resistance of the nozzle (ΔPn) and the rest of the head (ΔPh) are related as ΔPn = 41% and ΔPh = 59%
1.6.2. Now it is already possible to calculate the effect of a change in the diameter of the nozzle on the ratio of resistances during extrusion , now at a temperature of 280 º. The slip coefficient is about 80%.
D = 3mm, d = 0.2mm, W = 1.75mm / s, w = 393mm / s, L = 50mm, l = l2 = 0.4mm, W2 = 3.02 mm / s, w2 = 302mm / s d2 = 0,3mm
Resistance:
ΔP1 = K * w * l / d² = K * 393 * 0.4 / 0.04 = 3930 * K
ΔP2 = K * w2 * l2 / d² = K * 302 * 0.4 / 0.09 = 1342.2 * K
Performance (feed rate) in the second case
W2 / W = 3.02 / 1.75 = 1.72 = 172% of the performance of the first.
Since the friction resistance is linearly dependent on speed:
X2 = 1.72X
We write the equation for ΔP:
ΔP1 + X = ΔP2 + X2
ΔP1 + X = ΔP2 + 1.26X or from the formula above
3930K + X = 1342.2K + 1.72X => 3930K - 1342.2K = 1.72X - X => X = (3930K-1342.2K) / 0.72 = 3594K
The ratio of resistances will be:
ΔP1 + X = 3930K + 3594K => 3930: 3594 => = 52.2: = 47.8% for a nozzle with a diameter of 0.2 mm,
ΔP2 + 1.72X = 1342.2 * K + 1.72 * 3594K => 1342.2: 6182
Pc = 17.8: Pp = 82.2% for a nozzle with a diameter of 0.3 mm. Thus, the smaller the diameter of the nozzle, the greater the proportion in the total resistance is the resistance of the nozzle.
1.6.3. Consider a nozzle with a diameter of 0.1 mm and an extrusion rate for it. Temperature 280 º.
D = 3mm, d = 0.1mm, W =? mm / s, w =? mm / s, L = 50mm, l = l2 = 0.4mm, W2 = 1.75mm / s, w2 = 393mm / s, d2 = 0.2mm
Resistance:
ΔP1 = K * w * l / d² = K * w * 0.4 / 0.01 = 40 * w * K
ΔP2 = K * w2 * l2 / d² = K * 393 * 0.4 / 0.04 = 3930 * K
The total resistance in both cases is equal - the extruder gives the same force
ΔP1 + X = ΔP2 + X2
But from the previous calculation, we know that in 3930 * K + X, 3930 * K and X relate as 52.2 to 47.8 (for 0.2 mm) and X = 3594K Only now it is necessary to rename X and X2, the sides of the equations have changed.
ΔP1 + X = ΔP2 + X2 => 40 * w * K + X = 3930K + 3594K
but for the remaining part of the head, with a certain degree of confidence, we can apply the same equation as for the nozzle. There and there flow of a viscous fluid in the pipe. Probably some deviation from reality, but since the rates of motion for the melting site are very small, we should expect a good approximation to the theory. Calculations for other diameters showed good agreement with the experiment.
X = K * W * L / D² = K * W * 50/9
since the flow rate of plastic in the nozzle and filament feed are referred to as squares of their diameters:
w = 900W
ΔP1 + X = ΔP2 + X2 => 40 * w * K + X = 3930K + 3594K => 40 * K * w + K * W * 50/9 = 7524K => 40 * K * 900W + K * W * 5.55 = 7524K
36005.55W = 7524K W = 7524 / 36005.55 = 0.209mm / s
w = 0.209 * 900 = 188mm / s. Extrusion speed is not too low, but if you look at the extruded mass ...
Now let's see how the resistance for the case of a 0.1mm nozzle is divided:
ΔP1 + X = K * w * 0.4 / 0.01 + K * W * 50/9 Substitute
ΔP1 + X = K * 188 * 40 / 0.01 + K * 0.209 * 5.55 = 7520K + 1.16K
= 0.999846 = 99.9846% = 0.000154 = 0.0154% w = 188 mm / s
A huge share of all resistance falls on the nozzle.
1.6.4. Let's see what happens if we drastically reduce the length of the nozzle outlet channel - the spout. Up to 0.1 mm. Technically, this is possible - just a razor blade thickness. Holes of such diameter in a thin plate can also be made by a laser, and if not, by the EDM method.
D = 3mm, d = d2 = 0.1mm, W = 0.209 mm / s, w = 188 mm / s, L = 50mm, l = 0.4mm, l2 = 0.1mm, W2 =? mm / s, w2 =? mm / s
Resistance:
ΔP1 = K * w * l / d² = K * 188 * 0.4 / 0.01 = 7520 * K
ΔP2 = K * w2 * l2 / d² = K * w2 * 0.1 / 0.01 = 10 * w2 * K
The total resistance in both cases is equal - the extruder gives the same force
ΔP1 + X = ΔP2 + X2
But from the previous calculation we know that X = 1.16K,
Substitute
ΔP1 + X = ΔP2 + X2 => 7520K + 1.16K = 10 * w2 * K + X2 but
X2 = K * W2 * L / D² = K * W2 * 50/9
since the flow rate of plastic in the nozzle and filament feed are referred to as squares of their diameters:
w2 = 900W2
ΔP1 + X = ΔP2 + X2 => 7520K + 1.16K = 10 * w2 * K + 5.55 * K * W2 =>
7521.16 K = 10 * 900W2 * K + 5.55 * K * W2 => 9005.55 * K * W2 = 7521.16 K
W2 = 7521.16 / 9005.55 = 0.835mm / s
w2 = 0.835 * 900 = 751.5mm / s The speed has increased almost 4 times, due to the fact that such a thin fishing line almost does not increase the speed of the main stream, due to its very small volume, which means that the increased resistance does not reduce the share of force attributable on pushing through the nozzle.
ΔP2 + X2 = 10 * w2 * K + 5.55 * K * W2 = 10 * 751.5 * K + 5.55 * K * 0.835 = 7515K + 4.63K
The ratio of resistance as 7515 to 4.63, which corresponds to 0.999384 to 0.0006157.
Checking the calculations in practice with other diameters - 0.9 and 0.24 showed that the calculations are quite accurate, the error is 10-20%.
Conclusion: The calculations and data obtained allow us to calculate the maximum operating speeds for this particular type of heads, depending on the nozzle diameter, nose length and melt temperature. A calculation is shown for a theoretical nozzle section of 0.1 mm and the conditions under which it will work are shown.
1.7 Comparison of calculations of the melting rate (see section 1.2) and data from experiments. Findings.
Using the above table, the calculated data for the filament of ABS and soft polyamide with a diameter of 3.0 and 1.7 mm were obtained. To check these data were made nozzles with a diameter of 0.9 mm. Using the calculation method from item 1.6, you can calculate the ratio of the nozzle resistance to the head resistance for the nozzle diameter of 0.9 mm. It will be 1.5 to 98.5. Increasing more does not make sense. Let's compare the data of the experiments:
ABS 3.0 estimated 8.4s, from experience 7.65s
Polyamide PA6 3.0 rated 8.64 s, from the experience of 11.04 s
Polyamide PA6 1.7 estimated 2.88 s, from the experience 4.19s
For ABS 1.7 while I do not have data.
And if the ABS results are close to the calculated ones, although not in the wrong direction - faster, but this can be understood, the data on the heat capacity of ABS are very vague, then a sharp increase in the melting time for polyamide would be nice to explain.
Look at the photo
Fig. 12
These are pieces squeezed through a nozzle of 0.9mm with maximum speed and with maximum heating of soft nylon - such trimming line has appeared now. Normal, solid nylon, by all indications close to polyamide 6, PA6, its surface is matte and slightly rough. Its melting point is 220-230º. Soft nylon, different gloss, hand as if sticks, as to the lacquered surface, the melting point below 200º. During the work with heating 280º it actively smokes. It is difficult to judge whether it is due to impurities of low molecular weight polyamide, or plasticizers. Smoke caustic and unpleasant can not be called, and in theory nothing more harmful than cyanide should not work there. Joke. Cyanides there can turn out, but in meager amounts and with very large overheating, even significantly more than I use. There was such research on the web. Serious enough.
To find out the water content, I dried the polyamide trimmer line in the oven. Shrinkage was 5% for a tight trim line with a diameter of 1.6 mm. For soft fishing line with a diameter of 3mm, the shrinkage was 3.7%, which is logical - more diameter, less specific surface, less hydration. Not bad, but for the case of nylon there is an obvious assumption that bubbles formed by heating from the walls make it difficult to further warm up.
1.8. Deformation zone. Her internal profile. The assumption considered in this paragraph was not confirmed by experience. About this site I heard from respected print specialists from the Roboforum suspicions that since a polymer melt is not completely liquid, it is necessary to reckon with its other possible behavior. How to check it? Best experiment.
So, in this area, the flow is reordered from a large diameter to a small one. With decreasing diameter, both the axial velocity and the radial velocity, the rate of flow compression, increases. Assumed that, the resistance increases or is proportional to the speed of deformation, or its square. In heads used in practice, the degree of elongation of the extrusion zone is determined by the angle of sharpening of the drill, which makes the working chamber. For steel drillings, the standard grinding angle is 118º, from which it follows geometrically that the resulting hole has a degree of elongation (the ratio of the depth of the resulting cone to the radius), about 0.7. It turns out that the entire flow rests on a small hole. If we look at the drawings of industrial extruders for the manufacture of woods and cords, we see that the degree of elongation can be 10-15 working diameters. True, while there still placed auger. Alas, it is technically difficult. A nozzle with an elongation of 1 to 12 was tested. These are heads type 3.1, and type 3.2.
Great success is not obtained, but the swelling of the line at the exit there. In recent experiments on the head with interchangeable nozzles type 4.1, it was possible to check the nozzle with an elongation of 1 to 12, and a nozzle with an elongation of 1 to 6.7, and even a “parabolic nozzle”. Based on the fact that the resistance of a visco-elastic fluid is proportional to the velocity along the radius of the hole and along the axis - too. Maybe linear, maybe according to a quadratic law, it doesn't matter. Thus, by doubling the diameter, it is enough to double the length of the deformation section, for the same effect. The radial velocity of the plastic particles will remain the same, which means the resistance and the effects of blowing up - should have remained the same. The blow-up effect is observed at longer extrusion rates due to the fact that viscous fluid layers lying closer to the wall, due to stronger friction, have a lower speed. Friction against the wall is more than any friction against the same fluid. When leaving the nozzle, freezing starts from the surface, and the inner layers, which move more quickly, inflate the walls of the fishing line. In my experiments, I observed, at particularly high extrusion speeds and high heating, thickening up to 3 diameters of the nozzle. So - using these nozzles, a similar effect was almost not observed. True, I see no need to investigate it. In the process of printing this blowing up, it seems almost does not affect the quality. Although, if necessary, it will always be possible to return to it.
Fig.13 Cut of the brass part of the head Type 3.1
In fact, if it is good to count, we will see that the optimal profile of the extrusion section will not be linear at all. I considered four options for changing the cross-sectional area along the axis, that is, what shape an internal opening has. The first is that the hole is made by an ordinary drill with a 118º sharpening angle. The second one - with an angle of 15º, which corresponds to a length of 6.45 mm for every millimeter of radius change - it is more convenient to measure this way. The third is 13 mm cone length per millimeter radius change. The fourth option is parabolic, the working section area decreases with each step by an equal amount. The length remains the same - 10 mm.
Fig.14 Profiles counted nozzles
The first 2 options are shorter - too big a corner. I considered the radial velocity, the radial velocity squared and multiplied by the length of the section - as a quadratic resistance factor, and the radial velocity multiplied by the length of each section - as a linear resistance factor for each section, and then the average.
Conclusion: Alas, experience has shown a significant superiority in the speed of extrusion of the simplest nozzle with an angle of 118º. Probably because the melt properties of the polymer are still closer to a viscous, but Newtonian fluid. At least I could not get confirmation of the opposite. It can be assumed that heads with such a profile (especially parabolic) are good because the fishing line squeezed out of them is subject to bulging to a much lesser extent.
1.9. Calculation smoothing pen, its diameter and height.
Let's look at the classic J-Head Mk 5-V nozzle in the picture.
Fig.15
Around the nozzle opening we see a flat area of much larger diameter. Its purpose is to smooth out the melt that is evolved, it is sometimes called the smoothing pen. Sometimes this area is located on a ledge, sometimes not. We will try to understand what diameter it should be made, whether it is necessary or not, to place it on an elevation, and if so, on which. Let's see how the plastic will be squeezed out of the nozzle and smeared over the substrate.
Figure 16
The upper part of the drawing is an ideal case, when the supplied plastic fills all the space under the penny, the ears on the sides are the result of surface tension, the fluid is viscous, so there are spheres. With the height of the head above the substrate corresponding to this situation, the print quality is very good. The upper surface is completely flat, without zastrugov. If we lower the nozzle below the estimated position for the nickle, small shafts will form along the edges of the drawn fishing line, which is not very good.
Calculate the width of the penny for the nozzle 0.3 mm and the height of the layer of 0.12 mm. 40% of the nozzle diameter.
The cross-sectional area of the line for the nozzle 0.3 mm will be S = πd² / 4 = 3.14 * 0.3² / 4 = 0.07065
In the case of a layer thickness of 0.12 mm, the area in motion must remain the same. In the picture below left. Two semicircles from the sides will have a diameter of 0.12 mm. The remaining area forms a rectangle with a height of 0.12 mm and width X.
0.07065 = 3.14 * 0.0144 / 4 + 0.12 * X => X = 0.059346 / 0.12 = 0.49 mm. Take a close 0.6 mm. These are two nozzle diameters. If it is necessary to make the line already, it will just be necessary to reduce the feed accordingly, although the quality of laying may slightly decrease.
Now consider the case when the head traveled on the substrate with a minimum gap. The picture on the right below. In this case, the entire extruded plastic will be assembled by a ring around the head, the case is more likely that the cross section of this roller will be the same, which means its height will be equal to the diameter of the nozzle. So that the plastic stuck to the sides of the head does not reduce the print quality, a slight elevation is made for the leveling pen. I assume that it should be equal to the diameter of the nozzle. See at the bottom left of the picture, smearing and sticking should be significantly less.
Conclusion: The method of calculating the leveling leveling pen size depending on the diameter of the nozzle and the desired layer thickness is given. The height of the leveling nib, if possible, be equal to the diameter of the nozzle.
Chapter 2. Calculation of the speed head.
2.1 We set the main goals and boundary conditions.
- Select the diameter of the filament. For use with nozzles of sufficiently large sizes (from 0.2 and more) there is reason to choose a filament with a diameter of 3 mm. Based on the results of recent experiments and calculations for the melting rate and viscosity, the volumetric capacity of the heads of equal length will be almost the same, the component of the resistance of the head will be noticeably less, due to the larger diameter. This is of greater importance, the larger the diameter of the nozzle we are going to put. You should also consider a small segment from the hobbol to the entrance to the guide head. In my design it is about 5mm. At high loads, a filament of 1.7 is simply broken, forming a loop. . .
.17 .
— , . 50 , — . — , , 0,3 , 80% , , . 0,3 — 27 22 ³/ ABS , 32 46 ³/. , , , 32 ³/.
— , . 280-320º , .
, , , .
— 3.3. , 5 .
- . 1.6 « », . , . .
— — 15 ( )
— , , . — , — - . , , . .
- , . ! .
— , , . , . Conveniently. — , .
2.2 .
2 : — 3 1,75 . , — -: , . , 1,2 . 1,75, 1,6 — , 1,75 +-0,05 . 0,2 0,3
3,0 .
, . . + . 1,75 0,05, 1,8 . 2, , 1,75 4 — 3 .
3,0 + 0,5 , 3,6 , 4 - .
: 3 — = 4
2.3 . , . , , 3,5 . , ?
: ABS 5-9 10 , 10 , 220º. ? — . ASTM D1238, .
.1.2 . ( ), , , , . .
(. 1.2).
Vv = 32 ³/.
3 , w= 32/(3,1415*9/4) C 4,5 /. , 1.5.5 , 1,5/, 5, 25% . — . , . , 15, 3,33.
Δ = 110-30=80º, 3,33 .
. , .1.2, . Q = M*C*ΔT, Q — , M — , C — , , . 0,895, 1 , , 4,5 / 4,02. , , . 15/4,5=3,33. Δ80º. 1,105, 4,5 / 4,97. Q =3,45 / 15,5. 1, , 15 4 . 40 , .
P= λ*S* ΔT/l, λ =0,25/* — , S= π*D*L=3,14*4*15=188,5 ²=0,0001885² — , l — , 1. ΔT — . 110°, 30° ΔT=110-30=80°.
P= λ*S* ΔT/l= 0,25*0,0001885*80/0,001=3,77, , 4,02 . Δ80º 4,97, 3,7/4,97=76%. Q =3,45 /, 4,5/, 15,5, 3,77*100/15,5 =24% . Very good. — , — . 2 ! , Δ80º, 80*0,76=61º — 76% .
, , 61+30 = 91º.
200º.
: 7,64. , , = 7,64*4,5=34,4. 35 3,5 , .
, L= 35 : Win= 4,5 / = 270 /. Vv= 32 ³/. 0,3 Wout= 450 /. , 20% , .
2.4 , .
, . , . . , , . 3,5-3,6 . , , .
.18
7 — 70,75, 8. 11, .
, , 12 , , 10, 70,75.
— 4 , . , 35-4 = 31 , 4 . 27 10 . , , , , . , . - 0,1 . More on this below.
Select the angle of the sharpening of the head / nozzle - the outer part. In most heads, it is made large - as the angle of sharpening the drill. We need to save space under the edges for a wrench - to change the nozzle. Therefore, the taper of the head from the outside will make 8 to 1.5 diameter to the length. However, this is not particularly important.
The choice of smoothing penny. . , . , . .1.8. 40% — 0,6, . , . 0,2, , . . — .
/ — liner.
3,0 : 8 6 . OD 6 / ID 4. 3,1-3.3 4.1 . . , . 3,5.
: — 8, 70,75, 11, 12, 31 +3,5 . AISI 304, OD/ID 8/6 .
PTFE OD/ID 6/4 .
2.5 , .
. .
. Calc Libre Office. - ( 304/316) K= 9.4 /°,
L=3 D=8, — - - d=4. 280°.
— T = 110°C.
, , 0,25 , 7,5. -!
S= 3.14*D²/4 — 3.14*d²/4 = 3.14*7,5²/4 — 3.14*4²/4 = 31.6 ²
.
ΔT = 280 -110 = 170°
W= K*S*ΔT/L = 9,4*31,6*170/3*1000=16,84
— ( ²) , .
W= 16,84 . .
— L=8
— S= 3.14*D²/4 — 3.14*d²/4 = 3.14*8*8/4 — 3.14*6*6/4 = 21,99 ²
-, , — . 40 , — .
— . // 30°, ΔT = 110-30=80°
W= 9.4*21,99*80/8000 = 2,07
— 16,84 , 2,07, Δ W = 14,77, .
.
— Q=50/√S Q — , , S — . Q, 3. .
, , 14,77 110°.
: 12, 9 16 6.
- , S= 942² = 9,42 ²
— Q=50/√9,42 = 16.29 /.
3 16,29/3=5,43
— 14,77 ,
ΔT = 5,43*14,77 = 80,20° . 30°, 110,2°.
.19
Conclusion:
280° G=(280 — 110)/3=56,7 °/.
= 16 , 6 , 8 , = 12 , 9
= 6 — , 2 , . , — , , 8
2.6 .
, : Q1= 16,84
.2.3 , 15,5. 24% . , 15,5*(100-24)/100=11,78.
:
16,84+ 11,78 = 28,62 . . , , . . 20 , 20 120-140°.
: . , . . 50 .
.
, , — . , , 0,25 , 22,2 .
, , , .
.
:
.20
27 . — 10. 10*3,14 +2 = 33.
. 2,5-3 . 11 22 26 ( 1 ) = 26*22=572 = 0.572. 0,25 , 22,2 , 0,572*22,2 = 12,7 .
.
50 , 12 ,
I=P/U=50/12=4,17. R=U/I=12/4,17=2,88 .
12,7 . 50 25. .
12 24SWG 0,559, , , , . .
2.7. .
… , -. , , . 200, . — 3.1 , , , . , -, .
.21
- ( 3.3,4.1), — , TS922 IRFZ44.
. , - — , .
— 17 , , . , — .
, ABS = 280°, , 280°, 230-240° , , .
2.8 .
, , , , , , , , , , — . . « », , .
.22 « »
, . . , . — , … , , — . , . , . , . . . , .
.23
— . it
, . , . Everything. .
On the results, conclusions and further possible improvements will be in the third part.
Here we look at: - 1.6. Calculation of friction on linear sections with melt. Calculation of extrusion speed for nozzles of various diameters. The ratio of friction values in the areas of the nozzle and the melting zone. - 1.7 Comparison of calculations of the melting rate (see p. 1.2) and data from experiments. Findings. - 1.8. Deformation zone. Her internal profile. The assumption considered in this paragraph was not confirmed by experience. - 1.9. Calculation smoothing pen, its diameter and height. - Chapter 2. 2.1-2.5 Calculation of the speed head. - 2.6 Calculation of the heater. - 2.7.How to measure the temperature and how to regulate it.
Head type 4.2
')
1.6. Calculation of friction on linear sections with melt. Calculation of extrusion speed for nozzles of various diameters. The ratio of friction values in the areas of the nozzle and the melting zone.
1.6.1. The calculation of the ratio of the resistance of the nozzle and the working area. Effect of spout resistance.
Zone D of Figure 3. Molten plastic rubs against the walls and between its layers. Ultimately, all friction losses occur on the walls. Let's try to count. As a source of wisdom, I used the book by Kasatkin, Basic Processes and Apparatus of Chemical Technology, there in the chapter on Hydraulic Resistance in Pipelines, quite simply, the formula for resistance on straight pipe sections is derived from the Poiseuille equation and the Bernoulli equation. I brought her to mind
ΔP = 32 * μ * w * l / d²
I replaced viscosity and number 32 with K and it turned out to be even simpler: ΔP = K * w * l / d², where w is the speed, l is the length of the resistance section, d is the diameter of the pipe. Fortunately, we can use this equation ... because of the high melt viscosity, the Reynolds criterion favors us. And we can calculate at least at the estimated level (qualitatively, as the chemists say) the degree of friction in the areas of zones D and F too.
Calculate for a particular head type 4.1 the ratio of the contributions of the resistance of the nozzle and the rest of the head. This is useful to us to estimate the minimum acceptable diameter of the nozzle. For a head of this type, after eliminating defects and making changes, the results are quite repetitive and show good accuracy. Up to 1%. Since I have two nozzles of the same diameter, with different spout lengths of 0.4 and 0.8 mm, the difference in the results can be used to calculate the contribution of the nozzle resistance for different cases. Pay attention - it was very difficult to precisely manufacture the nozzle and the head, maintaining all specified dimensions! The results of the experiment for a nozzle of 0.3 mm, with a spout length of 0.4 mm, at a temperature of 260 º - Speed at the nozzle exit 206 mm / s. The slippage coefficient is 80%, the speed at the nozzle exit is 260 mm / s. Filament diameter 3.0 mm, plastic - ABS. The feed rate, taking into account slippage and feed inaccuracy = 2.06 mm / s.
D = 3mm, d = 0.3mm, W = 2.06mm / s, w = 206mm / s, L = 50mm, l = 0.8mm, W2 = 2.60 mm / s, w2 = 260mm / s (extrapol .), l2 = 0,4mm. We consider the resistance:
ΔP1 = K * w * l / d² = K * 206 * 0.8 / 0.09 = 1831.1 * K
ΔP2 = K * w2 * l2 / d² = K * 260 * 0.4 / 0.09 = 1155.6 * K
Performance (and speed) in the second case
Δw = 260 * 100/206 = 1.26 * 100 = 126% of the performance of the first.
Total resistances are:
ΔP1 + X = ΔP2 + X2, where X is the resistance of the head without a nozzle.
Since the speed at the output of the head and at the input increases equally (as many as entered, so much happened)
X2 = X * 1.26
We write the equation for ΔP:
ΔP1 + X = ΔP2 + 1.26X or from the formula above
1831.1 * K + X = 1155.6 * K + 1.26 * X => 1831.1 * K - 1155.6 * K = 1.26 * X - X
0.26X = 675.5K => This sign "follows from this" X = 675.5K / 0.26 = 2598K
Then, in the case of a spout of 0.4 mm, the ratio of resistances will be:
ΔP1 + X = 1831.1 * K + 2598K => 1831.1 / 1831.1: 2598 / 1831.1 => 1: 1.42 => 41%: 59%
Denote them as ΔPn - nozzles and ΔPh - the rest of the head. In this case, the resistance of the nozzle (ΔPn) and the rest of the head (ΔPh) are related as ΔPn = 41% and ΔPh = 59%
1.6.2. Now it is already possible to calculate the effect of a change in the diameter of the nozzle on the ratio of resistances during extrusion , now at a temperature of 280 º. The slip coefficient is about 80%.
D = 3mm, d = 0.2mm, W = 1.75mm / s, w = 393mm / s, L = 50mm, l = l2 = 0.4mm, W2 = 3.02 mm / s, w2 = 302mm / s d2 = 0,3mm
Resistance:
ΔP1 = K * w * l / d² = K * 393 * 0.4 / 0.04 = 3930 * K
ΔP2 = K * w2 * l2 / d² = K * 302 * 0.4 / 0.09 = 1342.2 * K
Performance (feed rate) in the second case
W2 / W = 3.02 / 1.75 = 1.72 = 172% of the performance of the first.
Since the friction resistance is linearly dependent on speed:
X2 = 1.72X
We write the equation for ΔP:
ΔP1 + X = ΔP2 + X2
ΔP1 + X = ΔP2 + 1.26X or from the formula above
3930K + X = 1342.2K + 1.72X => 3930K - 1342.2K = 1.72X - X => X = (3930K-1342.2K) / 0.72 = 3594K
The ratio of resistances will be:
ΔP1 + X = 3930K + 3594K => 3930: 3594 => = 52.2: = 47.8% for a nozzle with a diameter of 0.2 mm,
ΔP2 + 1.72X = 1342.2 * K + 1.72 * 3594K => 1342.2: 6182
Pc = 17.8: Pp = 82.2% for a nozzle with a diameter of 0.3 mm. Thus, the smaller the diameter of the nozzle, the greater the proportion in the total resistance is the resistance of the nozzle.
1.6.3. Consider a nozzle with a diameter of 0.1 mm and an extrusion rate for it. Temperature 280 º.
D = 3mm, d = 0.1mm, W =? mm / s, w =? mm / s, L = 50mm, l = l2 = 0.4mm, W2 = 1.75mm / s, w2 = 393mm / s, d2 = 0.2mm
Resistance:
ΔP1 = K * w * l / d² = K * w * 0.4 / 0.01 = 40 * w * K
ΔP2 = K * w2 * l2 / d² = K * 393 * 0.4 / 0.04 = 3930 * K
The total resistance in both cases is equal - the extruder gives the same force
ΔP1 + X = ΔP2 + X2
But from the previous calculation, we know that in 3930 * K + X, 3930 * K and X relate as 52.2 to 47.8 (for 0.2 mm) and X = 3594K Only now it is necessary to rename X and X2, the sides of the equations have changed.
ΔP1 + X = ΔP2 + X2 => 40 * w * K + X = 3930K + 3594K
but for the remaining part of the head, with a certain degree of confidence, we can apply the same equation as for the nozzle. There and there flow of a viscous fluid in the pipe. Probably some deviation from reality, but since the rates of motion for the melting site are very small, we should expect a good approximation to the theory. Calculations for other diameters showed good agreement with the experiment.
X = K * W * L / D² = K * W * 50/9
since the flow rate of plastic in the nozzle and filament feed are referred to as squares of their diameters:
w = 900W
ΔP1 + X = ΔP2 + X2 => 40 * w * K + X = 3930K + 3594K => 40 * K * w + K * W * 50/9 = 7524K => 40 * K * 900W + K * W * 5.55 = 7524K
36005.55W = 7524K W = 7524 / 36005.55 = 0.209mm / s
w = 0.209 * 900 = 188mm / s. Extrusion speed is not too low, but if you look at the extruded mass ...
Now let's see how the resistance for the case of a 0.1mm nozzle is divided:
ΔP1 + X = K * w * 0.4 / 0.01 + K * W * 50/9 Substitute
ΔP1 + X = K * 188 * 40 / 0.01 + K * 0.209 * 5.55 = 7520K + 1.16K
= 0.999846 = 99.9846% = 0.000154 = 0.0154% w = 188 mm / s
A huge share of all resistance falls on the nozzle.
1.6.4. Let's see what happens if we drastically reduce the length of the nozzle outlet channel - the spout. Up to 0.1 mm. Technically, this is possible - just a razor blade thickness. Holes of such diameter in a thin plate can also be made by a laser, and if not, by the EDM method.
D = 3mm, d = d2 = 0.1mm, W = 0.209 mm / s, w = 188 mm / s, L = 50mm, l = 0.4mm, l2 = 0.1mm, W2 =? mm / s, w2 =? mm / s
Resistance:
ΔP1 = K * w * l / d² = K * 188 * 0.4 / 0.01 = 7520 * K
ΔP2 = K * w2 * l2 / d² = K * w2 * 0.1 / 0.01 = 10 * w2 * K
The total resistance in both cases is equal - the extruder gives the same force
ΔP1 + X = ΔP2 + X2
But from the previous calculation we know that X = 1.16K,
Substitute
ΔP1 + X = ΔP2 + X2 => 7520K + 1.16K = 10 * w2 * K + X2 but
X2 = K * W2 * L / D² = K * W2 * 50/9
since the flow rate of plastic in the nozzle and filament feed are referred to as squares of their diameters:
w2 = 900W2
ΔP1 + X = ΔP2 + X2 => 7520K + 1.16K = 10 * w2 * K + 5.55 * K * W2 =>
7521.16 K = 10 * 900W2 * K + 5.55 * K * W2 => 9005.55 * K * W2 = 7521.16 K
W2 = 7521.16 / 9005.55 = 0.835mm / s
w2 = 0.835 * 900 = 751.5mm / s The speed has increased almost 4 times, due to the fact that such a thin fishing line almost does not increase the speed of the main stream, due to its very small volume, which means that the increased resistance does not reduce the share of force attributable on pushing through the nozzle.
ΔP2 + X2 = 10 * w2 * K + 5.55 * K * W2 = 10 * 751.5 * K + 5.55 * K * 0.835 = 7515K + 4.63K
The ratio of resistance as 7515 to 4.63, which corresponds to 0.999384 to 0.0006157.
Checking the calculations in practice with other diameters - 0.9 and 0.24 showed that the calculations are quite accurate, the error is 10-20%.
Conclusion: The calculations and data obtained allow us to calculate the maximum operating speeds for this particular type of heads, depending on the nozzle diameter, nose length and melt temperature. A calculation is shown for a theoretical nozzle section of 0.1 mm and the conditions under which it will work are shown.
1.7 Comparison of calculations of the melting rate (see section 1.2) and data from experiments. Findings.
Using the above table, the calculated data for the filament of ABS and soft polyamide with a diameter of 3.0 and 1.7 mm were obtained. To check these data were made nozzles with a diameter of 0.9 mm. Using the calculation method from item 1.6, you can calculate the ratio of the nozzle resistance to the head resistance for the nozzle diameter of 0.9 mm. It will be 1.5 to 98.5. Increasing more does not make sense. Let's compare the data of the experiments:
ABS 3.0 estimated 8.4s, from experience 7.65s
Polyamide PA6 3.0 rated 8.64 s, from the experience of 11.04 s
Polyamide PA6 1.7 estimated 2.88 s, from the experience 4.19s
For ABS 1.7 while I do not have data.
And if the ABS results are close to the calculated ones, although not in the wrong direction - faster, but this can be understood, the data on the heat capacity of ABS are very vague, then a sharp increase in the melting time for polyamide would be nice to explain.
Look at the photo
Fig. 12
These are pieces squeezed through a nozzle of 0.9mm with maximum speed and with maximum heating of soft nylon - such trimming line has appeared now. Normal, solid nylon, by all indications close to polyamide 6, PA6, its surface is matte and slightly rough. Its melting point is 220-230º. Soft nylon, different gloss, hand as if sticks, as to the lacquered surface, the melting point below 200º. During the work with heating 280º it actively smokes. It is difficult to judge whether it is due to impurities of low molecular weight polyamide, or plasticizers. Smoke caustic and unpleasant can not be called, and in theory nothing more harmful than cyanide should not work there. Joke. Cyanides there can turn out, but in meager amounts and with very large overheating, even significantly more than I use. There was such research on the web. Serious enough.
To find out the water content, I dried the polyamide trimmer line in the oven. Shrinkage was 5% for a tight trim line with a diameter of 1.6 mm. For soft fishing line with a diameter of 3mm, the shrinkage was 3.7%, which is logical - more diameter, less specific surface, less hydration. Not bad, but for the case of nylon there is an obvious assumption that bubbles formed by heating from the walls make it difficult to further warm up.
1.8. Deformation zone. Her internal profile. The assumption considered in this paragraph was not confirmed by experience. About this site I heard from respected print specialists from the Roboforum suspicions that since a polymer melt is not completely liquid, it is necessary to reckon with its other possible behavior. How to check it? Best experiment.
So, in this area, the flow is reordered from a large diameter to a small one. With decreasing diameter, both the axial velocity and the radial velocity, the rate of flow compression, increases. Assumed that, the resistance increases or is proportional to the speed of deformation, or its square. In heads used in practice, the degree of elongation of the extrusion zone is determined by the angle of sharpening of the drill, which makes the working chamber. For steel drillings, the standard grinding angle is 118º, from which it follows geometrically that the resulting hole has a degree of elongation (the ratio of the depth of the resulting cone to the radius), about 0.7. It turns out that the entire flow rests on a small hole. If we look at the drawings of industrial extruders for the manufacture of woods and cords, we see that the degree of elongation can be 10-15 working diameters. True, while there still placed auger. Alas, it is technically difficult. A nozzle with an elongation of 1 to 12 was tested. These are heads type 3.1, and type 3.2.
Great success is not obtained, but the swelling of the line at the exit there. In recent experiments on the head with interchangeable nozzles type 4.1, it was possible to check the nozzle with an elongation of 1 to 12, and a nozzle with an elongation of 1 to 6.7, and even a “parabolic nozzle”. Based on the fact that the resistance of a visco-elastic fluid is proportional to the velocity along the radius of the hole and along the axis - too. Maybe linear, maybe according to a quadratic law, it doesn't matter. Thus, by doubling the diameter, it is enough to double the length of the deformation section, for the same effect. The radial velocity of the plastic particles will remain the same, which means the resistance and the effects of blowing up - should have remained the same. The blow-up effect is observed at longer extrusion rates due to the fact that viscous fluid layers lying closer to the wall, due to stronger friction, have a lower speed. Friction against the wall is more than any friction against the same fluid. When leaving the nozzle, freezing starts from the surface, and the inner layers, which move more quickly, inflate the walls of the fishing line. In my experiments, I observed, at particularly high extrusion speeds and high heating, thickening up to 3 diameters of the nozzle. So - using these nozzles, a similar effect was almost not observed. True, I see no need to investigate it. In the process of printing this blowing up, it seems almost does not affect the quality. Although, if necessary, it will always be possible to return to it.
Fig.13 Cut of the brass part of the head Type 3.1
In fact, if it is good to count, we will see that the optimal profile of the extrusion section will not be linear at all. I considered four options for changing the cross-sectional area along the axis, that is, what shape an internal opening has. The first is that the hole is made by an ordinary drill with a 118º sharpening angle. The second one - with an angle of 15º, which corresponds to a length of 6.45 mm for every millimeter of radius change - it is more convenient to measure this way. The third is 13 mm cone length per millimeter radius change. The fourth option is parabolic, the working section area decreases with each step by an equal amount. The length remains the same - 10 mm.
Fig.14 Profiles counted nozzles
The first 2 options are shorter - too big a corner. I considered the radial velocity, the radial velocity squared and multiplied by the length of the section - as a quadratic resistance factor, and the radial velocity multiplied by the length of each section - as a linear resistance factor for each section, and then the average.
Conclusion: Alas, experience has shown a significant superiority in the speed of extrusion of the simplest nozzle with an angle of 118º. Probably because the melt properties of the polymer are still closer to a viscous, but Newtonian fluid. At least I could not get confirmation of the opposite. It can be assumed that heads with such a profile (especially parabolic) are good because the fishing line squeezed out of them is subject to bulging to a much lesser extent.
1.9. Calculation smoothing pen, its diameter and height.
Let's look at the classic J-Head Mk 5-V nozzle in the picture.
Fig.15
Around the nozzle opening we see a flat area of much larger diameter. Its purpose is to smooth out the melt that is evolved, it is sometimes called the smoothing pen. Sometimes this area is located on a ledge, sometimes not. We will try to understand what diameter it should be made, whether it is necessary or not, to place it on an elevation, and if so, on which. Let's see how the plastic will be squeezed out of the nozzle and smeared over the substrate.
Figure 16
The upper part of the drawing is an ideal case, when the supplied plastic fills all the space under the penny, the ears on the sides are the result of surface tension, the fluid is viscous, so there are spheres. With the height of the head above the substrate corresponding to this situation, the print quality is very good. The upper surface is completely flat, without zastrugov. If we lower the nozzle below the estimated position for the nickle, small shafts will form along the edges of the drawn fishing line, which is not very good.
Calculate the width of the penny for the nozzle 0.3 mm and the height of the layer of 0.12 mm. 40% of the nozzle diameter.
The cross-sectional area of the line for the nozzle 0.3 mm will be S = πd² / 4 = 3.14 * 0.3² / 4 = 0.07065
In the case of a layer thickness of 0.12 mm, the area in motion must remain the same. In the picture below left. Two semicircles from the sides will have a diameter of 0.12 mm. The remaining area forms a rectangle with a height of 0.12 mm and width X.
0.07065 = 3.14 * 0.0144 / 4 + 0.12 * X => X = 0.059346 / 0.12 = 0.49 mm. Take a close 0.6 mm. These are two nozzle diameters. If it is necessary to make the line already, it will just be necessary to reduce the feed accordingly, although the quality of laying may slightly decrease.
Now consider the case when the head traveled on the substrate with a minimum gap. The picture on the right below. In this case, the entire extruded plastic will be assembled by a ring around the head, the case is more likely that the cross section of this roller will be the same, which means its height will be equal to the diameter of the nozzle. So that the plastic stuck to the sides of the head does not reduce the print quality, a slight elevation is made for the leveling pen. I assume that it should be equal to the diameter of the nozzle. See at the bottom left of the picture, smearing and sticking should be significantly less.
Conclusion: The method of calculating the leveling leveling pen size depending on the diameter of the nozzle and the desired layer thickness is given. The height of the leveling nib, if possible, be equal to the diameter of the nozzle.
Chapter 2. Calculation of the speed head.
2.1 We set the main goals and boundary conditions.
- Select the diameter of the filament. For use with nozzles of sufficiently large sizes (from 0.2 and more) there is reason to choose a filament with a diameter of 3 mm. Based on the results of recent experiments and calculations for the melting rate and viscosity, the volumetric capacity of the heads of equal length will be almost the same, the component of the resistance of the head will be noticeably less, due to the larger diameter. This is of greater importance, the larger the diameter of the nozzle we are going to put. You should also consider a small segment from the hobbol to the entrance to the guide head. In my design it is about 5mm. At high loads, a filament of 1.7 is simply broken, forming a loop. . .
.17 .
— , . 50 , — . — , , 0,3 , 80% , , . 0,3 — 27 22 ³/ ABS , 32 46 ³/. , , , 32 ³/.
— , . 280-320º , .
, , , .
— 3.3. , 5 .
- . 1.6 « », . , . .
— — 15 ( )
— , , . — , — - . , , . .
- , . ! .
— , , . , . Conveniently. — , .
2.2 .
2 : — 3 1,75 . , — -: , . , 1,2 . 1,75, 1,6 — , 1,75 +-0,05 . 0,2 0,3
3,0 .
, . . + . 1,75 0,05, 1,8 . 2, , 1,75 4 — 3 .
3,0 + 0,5 , 3,6 , 4 - .
: 3 — = 4
2.3 . , . , , 3,5 . , ?
: ABS 5-9 10 , 10 , 220º. ? — . ASTM D1238, .
.1.2 . ( ), , , , . .
(. 1.2).
Vv = 32 ³/.
3 , w= 32/(3,1415*9/4) C 4,5 /. , 1.5.5 , 1,5/, 5, 25% . — . , . , 15, 3,33.
Δ = 110-30=80º, 3,33 .
. , .1.2, . Q = M*C*ΔT, Q — , M — , C — , , . 0,895, 1 , , 4,5 / 4,02. , , . 15/4,5=3,33. Δ80º. 1,105, 4,5 / 4,97. Q =3,45 / 15,5. 1, , 15 4 . 40 , .
P= λ*S* ΔT/l, λ =0,25/* — , S= π*D*L=3,14*4*15=188,5 ²=0,0001885² — , l — , 1. ΔT — . 110°, 30° ΔT=110-30=80°.
P= λ*S* ΔT/l= 0,25*0,0001885*80/0,001=3,77, , 4,02 . Δ80º 4,97, 3,7/4,97=76%. Q =3,45 /, 4,5/, 15,5, 3,77*100/15,5 =24% . Very good. — , — . 2 ! , Δ80º, 80*0,76=61º — 76% .
, , 61+30 = 91º.
200º.
: 7,64. , , = 7,64*4,5=34,4. 35 3,5 , .
, L= 35 : Win= 4,5 / = 270 /. Vv= 32 ³/. 0,3 Wout= 450 /. , 20% , .
2.4 , .
, . , . . , , . 3,5-3,6 . , , .
.18
7 — 70,75, 8. 11, .
, , 12 , , 10, 70,75.
— 4 , . , 35-4 = 31 , 4 . 27 10 . , , , , . , . - 0,1 . More on this below.
Select the angle of the sharpening of the head / nozzle - the outer part. In most heads, it is made large - as the angle of sharpening the drill. We need to save space under the edges for a wrench - to change the nozzle. Therefore, the taper of the head from the outside will make 8 to 1.5 diameter to the length. However, this is not particularly important.
The choice of smoothing penny. . , . , . .1.8. 40% — 0,6, . , . 0,2, , . . — .
/ — liner.
3,0 : 8 6 . OD 6 / ID 4. 3,1-3.3 4.1 . . , . 3,5.
: — 8, 70,75, 11, 12, 31 +3,5 . AISI 304, OD/ID 8/6 .
PTFE OD/ID 6/4 .
2.5 , .
. .
. Calc Libre Office. - ( 304/316) K= 9.4 /°,
L=3 D=8, — - - d=4. 280°.
— T = 110°C.
, , 0,25 , 7,5. -!
S= 3.14*D²/4 — 3.14*d²/4 = 3.14*7,5²/4 — 3.14*4²/4 = 31.6 ²
.
ΔT = 280 -110 = 170°
W= K*S*ΔT/L = 9,4*31,6*170/3*1000=16,84
— ( ²) , .
W= 16,84 . .
— L=8
— S= 3.14*D²/4 — 3.14*d²/4 = 3.14*8*8/4 — 3.14*6*6/4 = 21,99 ²
-, , — . 40 , — .
— . // 30°, ΔT = 110-30=80°
W= 9.4*21,99*80/8000 = 2,07
— 16,84 , 2,07, Δ W = 14,77, .
.
— Q=50/√S Q — , , S — . Q, 3. .
, , 14,77 110°.
: 12, 9 16 6.
- , S= 942² = 9,42 ²
— Q=50/√9,42 = 16.29 /.
3 16,29/3=5,43
— 14,77 ,
ΔT = 5,43*14,77 = 80,20° . 30°, 110,2°.
.19
Conclusion:
280° G=(280 — 110)/3=56,7 °/.
= 16 , 6 , 8 , = 12 , 9
= 6 — , 2 , . , — , , 8
2.6 .
, : Q1= 16,84
.2.3 , 15,5. 24% . , 15,5*(100-24)/100=11,78.
:
16,84+ 11,78 = 28,62 . . , , . . 20 , 20 120-140°.
: . , . . 50 .
.
, , — . , , 0,25 , 22,2 .
, , , .
.
:
.20
27 . — 10. 10*3,14 +2 = 33.
. 2,5-3 . 11 22 26 ( 1 ) = 26*22=572 = 0.572. 0,25 , 22,2 , 0,572*22,2 = 12,7 .
.
50 , 12 ,
I=P/U=50/12=4,17. R=U/I=12/4,17=2,88 .
12,7 . 50 25. .
12 24SWG 0,559, , , , . .
2.7. .
… , -. , , . 200, . — 3.1 , , , . , -, .
.21
- ( 3.3,4.1), — , TS922 IRFZ44.
. , - — , .
— 17 , , . , — .
, ABS = 280°, , 280°, 230-240° , , .
2.8 .
, , , , , , , , , , — . . « », , .
.22 « »
, . . , . — , … , , — . , . , . , . . . , .
.23
— . it
, . , . Everything. .
Source: https://habr.com/ru/post/366015/
All Articles