FDM is the most renowned 3D printing method in the market (Kristiawan et al., 2021) due to the variety of possible printable materials from mainstream PLA (Arockiam et al., 2022) to more professional materials such as nylons, reinforced polymers (Wickramasinghe et al., 2020) such as carbon fiber and ABS, and up to engineering materials such as Polyether ether ketone (PEEK) (Wang et al., 2020) and Polypropylene (PP) (Bachhar et al., 2020). 3D printing has been known for its ability in creating complex geometries at low costs, especially for rapid prototyping applications (Kafle et al., 2021). Nevertheless, FDM methodologies are still challenging as their exterior surface quality and internal structure (Kumar et al., 2022) makes these products difficult to be standardized for mechanical applications (Wickramasinghe et al., 2020). Moreover, because of the technology, they result in low productivity (Raza et al., 2022) while they prove to be solid for low-cost end-use parts and small-lot production (Rouf et al., 2022). FDM printing processes have many fields of applications: biomedical (Frizziero et al., 2022a), tissue engineering, human bone repair (Alessandri et al., 2022; Alessandri et al., 2023), antibacterial, bioprinting (Chaturvedi et al., 2022), electrical conductivity (Mora et al., 2020), electromagnetic, sensor, battery, automotive (Romero et al., 2021; Frizziero et al., 2022b; Frizziero et al., 2022c; Frizziero et al., 2023), aviation, smart textile (Chakraborty and Biswas, 2020), environmental, and luminescence applications among other areas (Jandyal et al., 2022).

In summary, the objectives of the work are.

• find a geometry that allows a constant flow rate to be obtained;

A parameterization using geometric data that can be shared among various models of peristaltic pumps is presented and formulas and schematizations for calculating displacement and flow rate are then illustrated.

The peristaltic pump consists of a rotor to which two or more rollers are attached that rotate, squeezing the tube against the case, isolating a volume of fluid, and moving it from the suction to the discharge (Figure 1). Two zones of the machine are defined, the first is the “transfer zone” where the fluid is isolated and moved from the suction to the discharge and is 180° wide; the second is the “disengagement zone,” consecutive to the first and defines the amplitude in which the roller moves away from the tube until it makes no contact.

Beginning with the calculation of the displacement, considering the internal volume of the tube running through the entire transfer zone, which will have to be multiplied by two to consider a full rotation of the rotor. C is the overall length of the circle defined by the middle diameter of the tube called: D a x i s . A int is the internal cross-section area of the tube, defined by the inner diameter d int as reported in Figure 2.

Using the two relationships, the internal volume of the tube ( V t u b e   int . ) is obtained: V t u b e   int . = C A int = π   D a x i s   π d int 2 4 = π 2   D a x i s d int 2 4 (3)

The volume occupied by the z rollers crushing the tube should be subtracted ( V e l e m e n t ; following a full rotation, all the rollers engage once for one V a c t u a l = V t u b e   int − z   V e l e m e n t (4)

The volume occupied by the roller crushing the pipe, considering the pipe locally straight, is obtained from the intersection between two cylinders with different diameters at skewed axes. A schematization is presented in Figure 3.

To derive the volume of the fluid element, begin by calculating the area of the circular segment when the roller crushes the tube; this is done theoretically by considering a tube that adheres perfectly to the roller as depicted in the figure below.

Knowing that: a = d ′ 2 − d int (5) d ′ = d r o l l e r + d e x t − d int (6) r = d ′ 2 (7)

Considering Figure 4, the Pythagorean theorem is applied to derive the segment AH: A H = O A 2 − O H 2 = r 2 − a 2 (8)

From which it is possible to derive the segment AB: A B = 2 r 2 − a 2 (9)

Carnot’s theorem on the AOH triangle: A H 2 = a 2 + r 2 − 2 a r   cos ⁡ β ′ (10)

From which the angle β′ is derived: β ′ = arccos a 2 + r 2 − A H 2 2 a r (11)

Turning then to the ß angle: β = 2 arccos a 2 + r 2 − A H 2 2 a r (12)

With a simple proportion, the length of the arc of the circumference of the circular sector is calculated: L 2 π r = β 360   →   L = β   2 π r 360 (13)

Knowing that the area of the circular sector (yellow area + blue area in Figure 4) and the area of the triangle (yellow area in Figure 4) are: A c i r c u l a r   s e c t o r = L   r 2 (14) A t r i a n g l e = A B   a 2 (15)

The area of the circular segment is easily derived: A c i r c u l a r   s e g m e n t = A c i r c u l a r   s e c t o r − A t r i a n g l e (16)

To obtain the volume of the circular segment, one must multiply the area found by its depth, in this case, it is the inner diameter of the pipe. Mathematically, a solid is generated that has a rectangular cross-section along the longitudinal axis represented in Figure 5 on the left. The theoretical volume to be calculated is represented in Figure 5 on the right.

To arrive at an approximate solution of the volume sought, a corrective term is inserted into the element volume formula, which is the ratio of the area of the circle to the area of the square circumscribed by the circle. A c i r c l e A s q u a r e = π 4 (17)

Through this procedure, the calculated volume differs by approximately 2% from the theoretical volume calculated through the CAD program.

For the exact calculation, one would have to perform a triple integral on the intersection volume of two cylinders with skewed axes and different diameters (similar to Steinmetz volume), it presents too much complexity for the intended purpose.

The volume of the circular segment will be: V e l e m e n t ≅ A c i r c u l a r   s e g m e n t   d int   π 4 (18)

In conclusion, the displacement of the machine will be: V a c t u a l = V t u b e   int − 5 V e l e m e n t (19)

Knowing the rated capacity and displacement of the machine, it is possible to derive the rotational speed of the motor to properly move the pump. n max = Q max V a c t u a l   60   r p m   →   ω max = n max   2 π 60   r a d s (20)

The flow rate has been made independent of the pump rotation speed, so all considerations and comparisons will be comparable without having to define a rotation speed.

To define the flow rate in this machine geometry, the volume of fluid moved with a rotation of one degree by the last fully engaged roller in the conveying zone is estimated, from which the change in volume due to the roller disengaging on the pipe must be subtracted.

As can be seen in Figure 6, the angle ε defines the angular position of the rotor; deriving the angle γ knowing that the sum of the interior angles of a triangle (AOB) always makes 180: γ = 180 − ε + 90 (21)

Knowing the AO side and the two angles ε and γ, the OB side called τ can be derived: τ = r c a s e sin ⁡ 90 sin ⁡ γ (22)

Knowing the radius of the rotor, the cosine theorem is applied: r r o t o r 2 = τ 2 + y 2 − 2 τ y ⁡ cos ⁡ λ (23)

To derive the parameter y that varies as the angle ε changes: y = τ ⁡ cos ⁡ λ   ±   τ 2 cos ⁡ λ 2 − τ 2 + r r o t o r 2 (24)

One must choose the negative sign otherwise it makes no physical sense.

Considering Figure 7, the distance y is used to obtain the parameter f, which is of interest until the roller completely disengages from the tube, and is obtained through the closure equation: y = r r o l l e r + f + d e x t − d int 2   →   f = y − r r o l l e r − d e x t − d int 2 (25)

Needed to derive the parameter a’, which at the point of last detachment will be equal to the radius of the roller: b ′ = d int − f (26) a ′ = y − b ′ − f (27)

Once the parameter aʹ is obtained, one can proceed with the calculation of the volume occupied by the roller on the tube, as carried out for the calculation of the displacement. Starting from Eq. (8), AH (Figure 7) is obtained, to obtain ß and arrive at the surface area value of the circular segment. The calculation of the volume of the fluid element is carried out as in Eq. (18). Plotting the trend on a graph, the curve shown in Figure 8 is obtained.

Knowing the volume occupied by the roller as the angle ε changes, the change in volume between one position (ε) and the next (ε + 1) is calculated: ∆ V e l e m e n t = V e l e m e n t   ε − V e l e m e n t   ε + 1 (28)

Finally, the change in volume of the liquid with one degree rotation is calculated: ∆ V l i q u i d   1 ° = l a r c   1 ° ∙ A t u b e = ∆ ε ∙ 2 π r c a s e 360   π d t u b o   int 2 4 (29)

Considering: l a r c   1 ° = π D a x i s 360 (30)

The flow rate of this pump has two phases.

- The first phase begins when the roller has finished the transfer arc and no longer completely occludes the pipe, generating a passage light for the fluid. From this moment, the roller disengagement phase begins and ends when there is no more contact between the roller and the pipe. The duration of this phase is approximately 22.

The entire cycle repeats every 360/z, where z = 5, so every 72.

Putting this concept back into formulas, getting: Q 1 = Δ V l i q u i d   1 ° − Δ V e l e m e n t (31) Q 2 = Δ V l i q u i d   1 ° (32)

The results of the described relationships are represented on a graph in Figure 9.

Some performance indicators are defined so that results can be evaluated and compared with each other:

Irregularity: I r r e g u l a r i t y = Q max − Q min (33)

Average flow rate: Q a v e r a g e = ∑ i Q i n (34)

Average percentual flow rate: Q a v e r a g e   % = Q a v e r a g e Q max ∙ 100 (35)

From Figure 9, it can be seen that the flow irregularity is due to the disengagement of the roller from the pipe; in particular, there are some instances when the amount of fluid moved toward the discharge is less than the volume change due to the disengagement, resulting in a negative flow rate, i.e., fluid return.

This trend is caused by the disengagement of the roller in an arc of a very short length (≈22°). Such behavior is not permissible.

By varying construction parameters such as roller diameter, tube diameter, number of rollers, and rotor diameter, it is possible to limit the amplitude of negative peaks by obtaining a reasonable irregularity and a good average flow rate.

Unfortunately, none of the parameters allow for a constant flow rate or one that can be considered as such, the problem being due to the abrupt disengagement of the roller that generates a major oscillation in the flow rate.

The concept behind the new geometry is to increase the disengagement zone to reduce the flow irregularity (Figure 10).

The new geometry consists of a first zone, called the “transfer zone,” with a constant radius of amplitude equal to the angle between two successive rollers (e.g., 5-roller rotor, so α = 360/z = 72), which is necessary to isolate the fluid element and prevent the connection between discharge and suction.

The second zone, called the “disengagement zone,” has an eccentric arc to disengage the roller gradually, reducing the irregularity in the flow rate.

A working arc of 180 was also predetermined for this pump geometry so as to have comparable results. From the analyses below it will be clear whether this geometry really brings advantages over the classical pump.

As can be seen in Figure 10, angle α defines the arc of the case with a constant radius centered in O, and angle θ defines the arc of the case with eccentric geometry with the center in O': θ = 0 ÷ 180 − α (36)

Defining a generic point of contact P yields the triangle OPO’ and applying Carnot’s theorem gives: ρ 2 = r c a s e 2 + e 2 − 2 r c a s e e ⁡ cos ⁡ θ   →   ρ = ± r c a s e 2 + e 2 − 2 r c a s e e ⁡ cos ⁡ θ (37)

The positive sign must be chosen otherwise it makes no physical sense.

The parameter ρ defines the diameter of the case according to the value of θ and the other geometric parameters. The graphical representation is shown in Figure 11.

In the triangle OPO,’ while knowing the angle θ which is referred to O′, one can derive the angle φ because it is refers to O and then derive ψ: S e   θ < 90 °   ψ = π − arcsin r c a s e ρ ∙ sin ⁡ θ (38) S e   θ > 90 °   ψ = arcsin r c a s e ρ ∙ sin ⁡ θ (39)

And finally, getting φ: φ = 180 − ψ (40)

As a verification, Carnot’s theorem can be reapplied by referring to the angle φ instead of θ: r c a s e 2 = ρ 2 + e 2 − 2 ρ e ⁡ cos ⁡ ψ   →   ρ = − e cos ⁡ ϕ   ±   e 2 cos ⁡ ϕ 2 − e 2 − r c a s e 2 (41)

The positive sign must be chosen otherwise it makes no physical sense.

To derive the volume of fluid occupied by the roller in all stages of disengagement, as can be seen in Figure 12, it is necessary to know the parameter x: ρ = r r o t o r + r r o l l e r + d e x t − d int 2 + x (42) x = ρ − r r o t o r − r r o l l e r − d e x t − d int 2 (43)

Deriving the b term and finally the a term: b = d int − x (44) a = d ′ 2 − b (45)

At this point, we will refer to the calculations made earlier for the classical peristaltic pump in Eq. (8).

Knowing the volume occupied by the rollers in both the constant-radius phase and the variable-radius phase, it is possible to derive the displacement, which is approximately 11% greater than in the classical pump.

With the new case geometry, the flow rate will be calculated as before, except that the number of rollers that disengage, in some phases, will be more than one at a time (Figure 13).

For clarity, the formulas previously described are given: ∆ V l i q u i d   1 ° = l a r c   1 ° ∙ A t u b e = ∆ ε ∙ 2 π r c a s e 360   d t u b o   int 2 π 4 (46) ∆ V e l e m e n t = V e l e m e n t   ε − V e l e m e n t   ε + 1 (47) Q = Δ V l i q u i d   1 ° − Δ V t o t .   e l e m e n t (48)

The change in volume due to the gripping rollers varies depending on the angle of the rotor considered. ε = 0 ÷ 36 ° → ∆ V t o t .   e l e m . = ∆ V B + ∆ V C (49) ε = 37 ÷ 72 ° → ∆ V t o t .   e l e m . = ∆ V B (50) ε = 72 ÷ 108 ° → ∆ V t o t .   e l e m . = ∆ V B + ∆ V A (51) ε = 108 ÷ 144 ° → ∆ V t o t .   e l e m . = ∆ V A (52) ε = 144 ÷ 180 °   → ∆ V t o t .   e l e m . = ∆ V A + ∆ V E (53)

In this machine geometry, the transfer zone is reduced to the essentials to ensure no communication between suction and delivery and to maximize the roller disengagement zone from the pipe. Representing the instantaneous flow rate values on a graph yields the following (Figure 14).

The flow rate trend is periodic, with a period defined by the angle between two rollers, which in this case is 72.

The peristaltic pump is one of the few pumps that can be made with 3D printing because there are no creeping elements that have to make a seal and need tight tolerances, it has few components and has no parts with complex geometries. Moreover, having the liquid inside the tube allows it to transport a fluid that is not akin to the material the pump is made of. Using CAD software, the eccentric pump was made (Figure 21) where all possible components were 3D printed, so the actual operation described and analyzed above could be verified. Given the small size of the prototype made, steel pins were used for the bearings; for larger machines, they could be printed in one piece with one of the two rotor flanges. Fixed wedges could be used as an alternative to rolling rollers; this configuration must provide for excellent lubrication to avoid excessive tube wear and reduced torque from the motor moving the pump.

Polyethylene terephthalate glycol (PETG), a material with fair mechanical properties in the range of 3D printing materials, was chosen, which has the following characteristics (Table 5).

The design of the machine was structured to minimize supports and facilitate printing while minimizing the components to be assembled. The case is made of one piece and attached to the motor by two diametrically opposed screws, the rotor is composed of two flanges that support the pins and rollers that will be housed on the motor shaft via a form fit. Finally, there is the case closure flange, which prevents the tube from slipping out during operation, and is attached by two screws directly to the motor shell. The printing parameters used are as follows (Table 6).

The final 3D printed result is shown in Figure 22.