Optimization of Mechanical Efficiency Models for 2D Piston Pumps with a Stacked Taper Roller Set

Abstract

Since the clearance between the guide rail and the roller reduces the efficiency of the 2D pump, this paper proposes a novel 2D piston pump with stacked taper roller sets to eliminate the effect of the clearance. The structure of the 2D pump is introduced, and the mathematical model of the torque and the mechanical efficiency of the bilateral force on the guide rail are established and analyzed. The model takes into account the change in the oil viscosity, the spatial angle and the oil churning loss. A test rig was built to test the mechanical efficiency under different operating conditions. The unilateral and bilateral force models of the guide rail were compared, which proved that the bilateral force model of the guide rail can predict the mechanical efficiency more accurately than the unilateral force model. In the case of high load pressure, there was a clearance between the test results and the model calculation results. It is speculated that the main reason for this is that the greater oil pressure causes the size of the contact area between the two taper rollers and between the taper roller and the guide rail to become larger. The resulting rolling friction coefficient becomes larger, which affects the mechanical efficiency.

1. Introduction

The hydraulic pump is the core component of the hydraulic system, which converts mechanical energy into hydraulic energy. Axial piston pumps are widely used in hydraulic systems because of their compact structure and high pressure [1,2,3]. However, the inherent overturning movement [4,5] of the axial piston pump and the three major sliding friction pairs [6,7,8] cause leakage and other problems during operation [9,10].

Hydraulic pumps are developing towards high speed, miniaturization and energy saving [11,12]. Therefore, Ruan’s team proposed a novel piston pump with a design based on the “two-dimensional” concept [13,14,15], which has a single piston [16,17,18] and double rotor type [19,20,21]. The 2D piston pump with a balanced force uses the rolling friction of the roller–guide-rail mechanism instead of the sliding friction of the swash plate-slippers mechanism. During high-speed movement, the components in the 2D piston pump are balanced. In addition, the two guide rail sets are used to move in opposite directions in the axial direction to balance the inertial force on the cylinder block, in order to reduce its vibration and make it easier to achieve high speed. The two guide rails drive the piston and the piston ring to move in opposite directions, which indirectly increases the oil discharge stroke without increasing the volume of the pump, thereby further increasing the power-density of the pump. The specific structure is shown in Figure 1a.

Figure 1. 2D piston pump with a balanced force. (a) Structure diagram; (b) clearance between the guiding rail and the taper rollers.

In the experiment, it was found that the 2D piston pump with a balanced force generated strong vibration when it rotated at high speed, which affected its efficiency. This was mainly caused by the clearance between the guide rail group and the taper roll caused by machining and assembly errors [22], as shown in Figure 1b.

In order to eliminate the assembly clearance between the guide rail and the taper roller, and to reduce the problem of vibration and small flow, the author proposes a taper roller overlap type 2D piston pump [23]. Figure 2 shows the structure of the 2D piston pump with a stacked taper roller set.

Figure 2. 2D piston pump with a stacked taper roller set. (a) Structure diagram; (b) picture.

The structure of the 2D piston pump with a stacked taper roller set, from left to right, consists of a driver guide set, a taper roller group, a balanced guide set and a distribution cylinder. The driver guide set includes a left driver guide, a right driver guide and a piston. The balanced guide set includes a left balanced guide, a right balanced guide and a piston ring. The guide surfaces are made into a constant acceleration and constant deceleration. When the two guide sets rotate, due to the constraints of the roller-guide mechanism, the two guide sets perform equal acceleration and deceleration in the axial direction. There are ten distribution grooves on the distribution cylinder, and five oil intake windows and five oil discharge windows are staggered on the cylinder block. During the flow distribution process, the distribution groove communicates alternately with the 10 oil-intake and discharge ports on the casing to achieve continuous oil intake and discharge.

When the top of the taper roller is pressurized by oil, the taper roller set closely adheres to the surfaces of the guide rail, eliminating the clearance between them that may be caused by machining or assembly errors, and ensuring that the strokes of the two guide sets always meet the working requirements. The phases of the two guide rail groups are staggered by 36°. When the two guide rail groups move toward the middle position, the distribution grooves communicates with the oil intake and discharge ports, the low-pressure chamber sucks oil and the high-pressure chamber discharges oil, in order to realize the flow-distribution function.

Through the test, it was found that the left and right taper rollers were in close contact with the guide rails relying on the oil pressure, and both left and right guide rails were stressed. There was a large deviation between the unilateral force model of the guide rail and the actual results, so it was necessary to further optimize the mechanical model of the 2D piston pump with a stacked taper roller set. In this paper, the mechanical efficiency model was re-modeled and compared with the unilateral force model of the guide rail. The model took into account the churning loss of the two guides in the oil [24] and the oil resistance during axial movement. Finally, the theoretical calculation and analysis results are compared with the experimental results herein, and the reasons for the differences in the results are explained.

2. Mathematical Model

2.1. Spatial Angle Relationship

The spatial angle formed by the contact force between the taper roller and the guide rail and the axial and circumferential directions of the 2D pump is very important for the calculation of the mechanical equation, so it is necessary to firstly establish the spatial angle relationship of the roller-guide-rail mechanism [25]. Since the taper roller is affected by the guide rail in the actual working process, its posture will change slightly. In order to establish the space angle transformation more conveniently, it is assumed that the taper roller and the guide rail group always rotate around a central axis of rotation without changing. After coordinate homogeneous transformation, the motion coordinate system O₃-X₃Y₃Z₃ is transformed to the global coordinate system O₀-X₀Y₀Z₀. The coordinate system is shown in Figure 3.

Figure 3. Coordinates system.

In the motion coordinate system O₃-X₃Y₃Z₃, point O₃ coincides with the vertex of the taper roller and falls on the Z₀ axis, the Z₀ axis is collinear with the rotation axis of the guide rail, and the Z₃ axis is collinear with the rotation axis of the taper roller. The coordinates O₁-X₁Y₁Z₁ are obtained by translating s along the Z₀ axis from the coordinates O₀-X₀Y₀Z₀, the coordinates O₂-X₂Y₂Z₂ are obtained by rotating the coordinates O₁-X₁Y₁Z₁ around the Z₁ axis by θ_g degrees, and the coordinates O₃-X₃Y₃Z₃ are obtained by rotating the coordinates O₂-X₂Y₂Z₂ around the X₂ axis by π/2-θ_T degrees.

According to the above coordinate homogeneous transformation, the expression of the surface of the taper roller in the O₀-X₀Y₀Z₀ is as follows:

$$\left\{\begin{array}{l}x=L\cos {\theta }_{\mathrm{T}}\sin {\theta }_g+L\cos \theta \tan \beta \cos {\theta }_{\mathrm{sr}}-L\tan \beta \sin {\theta }_{\mathrm{T}}\sin {\theta }_g\sin {\theta }_{\mathrm{sr}}\\ y=L\tan \beta \sin {\theta }_g\cos {\theta }_{\mathrm{sr}}-L\cos {\theta }_{\mathrm{T}}\cos {\theta }_g+L\cos {\theta }_g\tan \beta \sin {\theta }_{\mathrm{T}}\sin {\theta }_{\mathrm{sr}}\\ z=s+L\sin {\theta }_{\mathrm{T}}+L\cos {\theta }_{\mathrm{T}}\tan \beta \sin {\theta }_{\mathrm{sr}}\end{array}\right.$$

(1)

where θ_g is the rotation angle of the guide rail, θ_sr and θ_T are the rotation angle and inclination angle of the taper roller, h is the design stroke of two guide sets, L is the distance from the surface of the taper roller to the X₃O₃Y₃ plane, and s is the axial displacement of two guide set.

Since the guide-rail surface is composed of 5 cycles of equal acceleration and deceleration curved surface circular array, only one cycle of the surface needs to be analyzed. The motion law of the axial displacement s of the guide rail can be expressed by Equation (2)

$$s=\left\{\begin{array}{l}\frac{50h}{{\pi }^2}{\theta }_{\mathrm{g}}{}^2,{\theta }_{\mathrm{g}}\in [0,\frac{\pi }{10})\\ -\frac{50h}{{\pi }^2}{\theta }_{\mathrm{g}}{}^2+\frac{20h}{\pi }{\theta }_{\mathrm{g}}-h,{\theta }_{\mathrm{g}}\in [\frac{\pi }{10},\frac{3\pi }{10})\\ \frac{50h}{{\pi }^2}{\theta }_{\mathrm{g}}{}^2-\frac{40h}{\pi }{\theta }_{\mathrm{g}}+8h,{\theta }_{\mathrm{g}}\in [\frac{3\pi }{10},\frac{2\pi }{5}]\end{array}\right.$$

(2)

Equation (3) represents the relationship between the surface of the taper roller and the guide rail. Equation (1) can be expressed as a function f(θ_sr,θ_g,L) according to the envelope theory of the single parameter surface family. When f(θ_sr,θ_g,L) satisfies the constraint equation of Equation (3), the variable θ_sr can be eliminated to obtain the parametric guide rail surface, namely f(θ_g,L).

$$\frac{\partial f({\theta }_{\mathrm{sr}},{\theta }_{\mathrm{g}},L)}{\partial {\theta }_{\mathrm{sr}}}\times \frac{\partial f({\theta }_{\mathrm{sr}},{\theta }_{\mathrm{g}},L)}{\partial L}·\frac{\partial f({\theta }_{\mathrm{sr}},{\theta }_{\mathrm{g}},L)}{\partial {\theta }_{\mathrm{g}}}=0$$

(3)

f₁(θ_sr,θ_g,L) = ∂f(θ_sr,θ_g,L)/∂θ_sr and f₂(θ_sr,θ_g,L) = ∂f(θ_sr,θ_g,L)/∂L is the tangent vector at the contact point of the taper roller. n_n = (f₁ × f₂)/∣f₁·f₂∣ is the normal vector at the contact point of the taper roller, which is used to indicate the direction of the positive pressure. The vector n_n′ is the projection of the normal vector on the bottom surface.

Suppose the circumferential unit vector of the guide rail at the contact point is e_c, the Y₃ axial unit vector is transformed into the unit vector in the global coordinate system as j₃, and the Z₀ axial unit vector is k₀. Using the above-mentioned angle solution method and combining the constraint condition Equation (3), the following relationship between the spatial angle and the guide rail rotation angle θ_g and the distance L can be obtained.

The angle between the positive pressure on the inner guide rail and the movement direction, namely the pressure angle α_n, and the angle α_ne between the positive pressure on the inner guide rail and the circumferential vector at the contact point are, respectively.

$${\alpha }_{\mathrm{n}}=\arccos (\frac{n_{\mathrm{n}}·k_0}{\left|n_{\mathrm{n}}\right|})$$

(4)

$${\alpha }_{\mathrm{ne}}=\arccos (\frac{n_{\mathrm{n}}·e_{\mathrm{c}}}{\left|n_{\mathrm{n}}\right|})$$

(5)

The angle α_f between the friction force on the inner guide rail and the movement direction, and the angle α_fe between the friction force on the inner guide rail and the circumferential vector at the contact point are, respectively.

$${\alpha }_f=\arccos (\frac{f_1·k_0}{\left|f_1\right|})$$

(6)

$${\alpha }_{f\mathrm{e}}=\arccos (\frac{f_1·e_{\mathrm{c}}}{\left|f_1\right|})$$

(7)

The angle θ_N between the component of positive pressure of the driver guide set and the Y₃ axial vector is.

$${\theta }_{\mathrm{N}}=\arccos (\frac{n_{\mathrm{n}}{}^{\prime }·j_3}{\left|n_{\mathrm{n}}{}^{\prime }\right|})$$

(8)

2.2. Mechanical Efficiency

The driver rail group on the far right and the balanced rail group on the far left should be set to their initial positions. In the rotation process from 0° to 18°, the left driver rail and piston accelerate to the left, and the right balanced rail and piston ring accelerate to the right. Figure 4 shows the force of the 2D pump.

Figure 4. Schematic diagram of the force of two guide sets from 0° to 18°.

The mechanical balance equations of the two guide rail groups are as follows:

$$ma=5F_{\mathrm{x}1}-5F_{\mathrm{x}2}-5F_{f1}-5F_{f2}-F_{\mathrm{p}}-F_{\mathrm{po}}-F_{\mathrm{sh}}$$

(9)

$$ma=5F_{\mathrm{x}3}-5F_{\mathrm{x}4}-5F_{f3}-5F_{f4}-F_{\mathrm{p}}-F_{\mathrm{po}}-F_{\mathrm{sh}}$$

(10)

where m and a are the mass and acceleration of the guide set, F_x1, F_f1, F_x4 and F_f4 are the axial support and friction forces of the left taper roller to the left two guide rail, F_x2, F_f2, F_x3 and F_f3 are the axial support and friction force of the right taper roller to the right two guide rail, F_p is the hydraulic force on the piston and piston ring, F_po is the oil resistance on the axial motion of the guide rail, and F_sh is the oil shear force on the piston and piston ring, which is expressed by the following equation:

$$F_{\mathrm{x}1}=F_{\mathrm{s}1}·\cos {\alpha }_{\mathrm{n}}$$

(11)

$$F_{\mathrm{x}2}=F_{\mathrm{s}2}·\cos {\alpha }_{\mathrm{n}}{}^{\prime }$$

(12)

$$F_{\mathrm{x}3}=F_{\mathrm{s}3}·\cos {\alpha }_{\mathrm{n}}$$

(13)

$$F_{\mathrm{x}4}=F_{\mathrm{s}4}·\cos {\alpha }_{\mathrm{n}}{}^{\prime }$$

(14)

$$F_{f1}=F_{\mathrm{s}1}{\mu }_f·\cos {\alpha }_f$$

(15)

$$F_{f2}=F_{\mathrm{s}2}{\mu }_f·\cos {\alpha }_f{}^{\prime }$$

(16)

$$F_{f3}=F_{\mathrm{s}3}{\mu }_f·\cos {\alpha }_f$$

(17)

$$F_{f4}=F_{\mathrm{s}4}{\mu }_f·\cos {\alpha }_f{}^{\prime }$$

(18)

$$F_{\mathrm{p}}=p\frac{\pi }{4}(D_1{}^2-d^2)$$

(19)

$$F_{\mathrm{sh}}=\mu \pi D_1{L}_1\frac{v}{\delta }$$

(20)

$$F_{\mathrm{po}}=2·\frac{1}{2}\rho ·(S_1+S_2)·v^2+\mu ·2\pi ·v·l_{\mathrm{g}}·(\frac{R_1}{{\delta }_1}+\frac{R_2}{{\delta }_2})$$

(21)

$$a={(\frac{n}{3})}^{2}h$$

(22)

where F_s1 and F_s4 are the positive pressures of the left taper roller on the left driver guide and the left balanced guide, F_s2 and F_s3 are the positive pressures of the right taper roller on the right driver guide and the right balanced guide, μ_f is the friction coefficient between the guide rail and the taper roller, μ is the oil dynamic viscosity, p is the load pressure, D₁ and L₁ are the diameter and the width of the piston and the piston ring, δ is the clearance between the piston, the piston ring and the distribution cylinder, v is the axial movement speed of the two guide sets, ρ is the oil density, S₁ and S₂ are the oil-blocking area during axial movement of the driver guide rails on both sides, R₁ and R₂ are the radius of the driver guide rails on both sides, d is the diameter of the piston rod, l_g is the length of the guide rail, and n is the rotation speed of the two guide sets. δ₁ represents the distances from the edge of the driver guide to the balanced guide. δ₂ represents the distances from the edge of the driver guide to the cylinder block. α_n′ and α_f′ represent the pressure angle of the left and right outer guide rails and the angle between the friction force and the movement direction. Their radius is different from the inner guide rail, so it is different from α_n and α_f.

Equation (23) expresses the axial movement speed v of the two guide rail sets.

$$v=\left\{\begin{array}{l}\frac{100h}{{\pi }^2}\frac{\pi n}{30}{\theta }_{\mathrm{g}},{\theta }_{\mathrm{g}}\in (0,\frac{\pi }{10})\\ \frac{\pi n}{30}(-\frac{100h}{{\pi }^2}{\theta }_{\mathrm{g}}+\frac{20}{\pi }h),{\theta }_{\mathrm{g}}\in (\frac{\pi }{10},\frac{2\pi }{10})\end{array}\right.$$

(23)

As shown in Figure 5, F_N1 and F_N2 are the oil pressure on the top of the left and right taper rollers, F_a and F_b are the forces between the taper rollers, and F_a = F_b. The force balance equations of the left and right taper rollers on the y-axis and z-axis was established as shown in the following equation:

$$(F_a\cos \frac{b}{2}+F_b\cos \frac{b}{2}-F_{\mathrm{s}2}\cos {\theta }_N-F_{\mathrm{s}3}\cos {\theta }_N{}^{\prime })\cos \beta +F_{f3}\sin {\theta }_N{}^{\prime }-F_{f2}\sin {\theta }_N=0$$

(24)

$$(F_a\cos \frac{b}{2}+F_b\cos \frac{b}{2}-F_{\mathrm{s}1}\cos {\theta }_N-F_{\mathrm{s}4}\cos {\theta }_N{}^{\prime })\cos \beta +F_{f1}\sin {\theta }_N-F_{f4}\sin {\theta }_N{}^{\prime }=0$$

(25)

$$(F_{\mathrm{s}1}+F_{\mathrm{s}4}+F_a+F_b)\sin \beta =F_{\mathrm{N}1}=\frac{\pi }{4}{D}_2{}^2·p$$

(26)

$$(F_{\mathrm{s}2}+F_{\mathrm{s}3}+F_a+F_b)\sin \beta =F_{\mathrm{N}2}=\frac{\pi }{4}{D}_3{}^2·p$$

(27)

where b is the angle of the projection of the contact line onto the bottom surface of the tapered roller, β is the half taper angle, D₂ and D₃ are the effective pressing surface diameters of the left and right taper rollers, and θ_N′ is the angle between the component of positive pressure of the balanced guide set and the Y₃ axial unit vector. Because of the π/2 phase difference between the two guides group, it is different from θ_N.

Figure 5. The force diagram of the taper rollers on the left and right sides.

Combining Equations (9), (10) and (24)–(27), the expressions of F_s1, F_s2, F_s3, and F_s4 can be obtained.

In the course of 18° to 36°, the two guide sets keep the original direction of movement and perform the deceleration movement. Because the bilateral force model does not have problems related to the direction of the pressure, the rotation speed and the load in the unilateral force model, the force of the two guide rails is similar to that of the equal acceleration phase, as shown in Figure 6.

Figure 6. Schematic diagram of the force of two guide sets from 18° to 36°.

The mechanical balance equations of the two guide sets are shown in Equations (28) and (29).

$$-ma=5F_{\mathrm{x}1}-5F_{\mathrm{x}2}-5F_{f1}-5F_{f2}-F_{\mathrm{p}}-F_{\mathrm{po}}-F_{\mathrm{sh}}$$

(28)

$$-ma=5F_{\mathrm{x}3}-5F_{\mathrm{x}4}-5F_{f3}-5F_{f4}-F_{\mathrm{p}}-F_{\mathrm{po}}-F_{\mathrm{sh}}$$

(29)

The force of the taper roller remains unchanged. Therefore, the torque T_i received by the two guide sets during the rotation from 0° to 36° is as shown in Equation (30).

$$\begin{array}{l}{T}_{\mathrm{i}}=R_1·(F_{s1}\cos {\alpha }_{\mathrm{ne}}+F_{s3}\cos {\alpha }_{\mathrm{ne}}{}^{\prime }+F_{f1}\cos {\alpha }_{f\mathrm{e}}+F_{f2}\cos {\alpha }_{f\mathrm{e}}{}^{\prime })\\ +R_2·(-F_{s2}\cos {\alpha }_{\mathrm{ne}}-F_{s4}\cos {\alpha }_{\mathrm{ne}}{}^{\prime }+F_{f3}\cos {\alpha }_{f\mathrm{e}}+F_{f4}\cos {\alpha }_{f\mathrm{e}}{}^{\prime })\end{array}$$

(30)

The 2D piston pump is subject to the oil churning loss torque T_c of the guide rails in the oil and the oil shear torque T_s in the clearance between the cylinder block and distribution cylinder when functioning. T_c fits the curve formula through simulation and experiment [21], which is expressed as Equations (31) and (32).

$$T_{\mathrm{c}}=5.65\times {10}^{-10}·n^2+1.65\times {10}^{-6}·n-4.62\times {10}^{-4}$$

(31)

$$T_{\mathrm{s}}=\mu ·\pi D_4{L}_2·\frac{n\pi D_4}{60{\delta }_3}$$

(32)

where D₄ is the outer diameter of the distribution cylinder, L₂ and δ₃ is the contact width and the clearance between the distribution cylinder and the cylinder block.

Combining the above calculations, Equation (33) expresses the torque T required by the 2D pump when the two guide sets turn from 0° to 36°.

$$T=T_{\mathrm{i}}+T_{\mathrm{c}}+T_{\mathrm{s}}$$

(33)

The mechanical efficiency η_m can be calculated by Equation (34):

$${\eta }_{\mathrm{m}}=\frac{p·V_{\mathrm{D}}}{2\pi ({\overline{T}}_{\mathrm{i}}+T_{\mathrm{s}}+T_{\mathrm{c}})}$$

(34)

where V_D is the 2D pump displacement, and $\overline{T_{\mathrm{i}}}$ is the average value of the torque received by the two guide sets.

3. Simulation Results

According to the mathematical models listed above, the influence of rotation speed and load on torque were analyzed and compared. The relevant setting parameters are shown in Table 1.

Table 1. Simulation parameter.

Description	Value	Description	Value
Oil dynamic viscosity μ (Pa·s)	3.91 × 10−2	The angle between the two contact lines of the taper roller α (°)	36
Diameter of piston D1 (m)	26.8 × 10−3	The inclination angle θT (°)	13.15
Diameter of piston rod d (m)	19.5 × 10−3	The half taper angle β (°)	22.16
Pressing surface diameter of left taper roller D2 (m)	11 × 10−3	The angle of the projection of the contact line onto the bottom surface of the tapered roller b (°)	110
Pressing surface diameter of right taper roller D3 (m)	11 × 10−3	Friction coefficient μf	0.001
Outer diameter of distribution cylinder D4 (m)	33 × 10−3	The oil-blocking area during axial movement of the left driver guide rails S1 (m2)	283.7 × 10−6
Width of piston and piston ring L1 (m)	4 × 10−3	The oil-blocking area during axial movement of the right driver guide rails S2 (m2)	314 × 10−6
The contact length between the distribution cylinder and the cylinder block L2 (m)	18.5 × 10−3	Radius of left driver rail R1 (m)	12.76 × 10−3
Stroke of driver and balanced guide set h (m)	1 × 10−3	Radius of right driver rail R2 (m)	17 × 10−3
Clearance width δ, δ3 (m)	0.05 × 10−3	Length of guiding rail lg (m)	2.6 × 10−3
Distance from the edge of the guide rail to the cylinder block δ1 (m)	2 × 10−3	Pump displacement VD (L/r)	5.3 × 10−3
Distance from the edge of the driver guide to the balanced guide δ2 (m)	1 × 10−3	The mass of driver or balanced guide set m (kg)	90 × 10−3
Intake port pressure (MPa)	0.4

The variation of torque with rotation angle is shown in Figure 7. It can be seen from the curve in the figure that the peak torque increases with the increase in the rotational speed and load. When the guide rail set is rotated to 18°, the torque will suddenly change. The reason for the sudden change in torque is that the acceleration direction switches when the guide sets move to the middle position, and the support force of the taper roller to the guide rail on both sides of the two guide sets changes suddenly. It can be seen from Figure 7a that the torque increases with the increase in the load pressure, mainly due to the increased oil pressure on the piston and the piston ring. Figure 7b shows that the torque increases with the rotation speed, which is mainly due to the increase in the axial acceleration of the transmission part and the increase in the oil shear force.

Figure 7. Torque curve. (a) The speed is 6000 rpm; (b) The load pressure is 6 MPa.

The calculation results of the bilateral force model of the guide rail with the unilateral force model of the guide rail were compared, as shown in Figure 8. It can be seen from the figure that the average and peak values of the torque of the bilateral force model of the guide rail are larger than the unilateral force model. The specific reason is that the two guide sets of the bilateral force model have a supporting force opposite to the rotation speed, so the required rotation torque is increased. When the 0° and 36° guide rails are at the highest and lowest points of the stroke, the torque of the bilateral force model is smaller than that of the unilateral force model. This is because the calculation angle in the bilateral force model of the guide rail uses a space angle to calculate, which is more accurate than the angle conversion of the unilateral force model.

Figure 8. Comparison of the calculation results of the bilateral and unilateral force model of the guide rail. (a) The speed is 6000 rpm; (b) Load pressure is 6 MPa.

Figure 9 shows the mechanical efficiency under different operating conditions. The torque rise speed is less than the load pressure rise speed during the load pressure increases, so that the mechanical efficiency increases. The rotational torque increases during the load pressure increases, thereby reducing the mechanical efficiency. It can be seen from the mechanical efficiency curves of the two models that the torque calculated by the bilateral force model of the guide rail is greater than that of the unilateral force model, and the final mechanical efficiency is significantly lower than that of the unilateral force model.

Figure 9. Mechanical efficiency change curve. (a) The speed is 6000 rpm; (b) Load pressure is 6 MPa.

4. Experimental Research

A test rig was built to conduct experimental research on the prototype. The main parameters of the prototype are shown in Table 2:

Table 2. 2D pump prototype parameters.

Description	Value
Diameter of piston and piston ring/m	26.8 × 10−3
Diameter of piston rod/m	19.5 × 10−3
Outer diameter of distribution cylinder/m	33 × 10−3
Stroke of piston and piston ring/m	1 × 10−3
Pressing surface diameter of taper roller/m	11 × 10−3
Pump’s displacement/(cc/rev)	5.3
Pump’s volume/mm3	104 × 75 × 75
Pump’s weight/kg	1.42

Figure 10 is the principal diagram of the test rig, which includes a supply pump station, a spindle three-phase motor, a torque sensor, a pressure sensor, a flowmeter, and an adjustable pressure valve. The adjustable pressure valve simulates the load pressure, and the torque and the pressure sensor measured data are collected by the NI data acquisition card and display on the compute. The flowmeter is displayed by the matching digital display meter. The flowmeter used in the experiment has a range of 0.05–80 L/min and an accuracy of 0.3%; the torque range of the torque sensor is 0–20 Nm, the speed range is 0–18,000 rpm, and the accuracy is ± 0.1%; the range of the pressure sensor is 0 to 10 MPa with 0.3% accuracy.

Figure 10. Schematic diagram of test rig.

Before the test, adjust the relief valve to an appropriate pressure to ensure that the oil circuit system can quickly build up pressure when the 2D piston pump is started, so that the taper roller set can be compressed, and then adjust the motor frequency converter to change the motor speed, and then read it from the computer Torque data. Take 1000 rpm as a step and increase the speed. The experimental mechanical efficiency curve under different working conditions is shown in Figure 11.

Figure 11. Test results of mechanical efficiency. (a) Speed is 6000 rpm; (b) Load pressure is 6 MPa.

It can be seen from Figure 11 that the mechanical efficiency calculation result of the bilateral force model is lower than that of the unilateral force model and is closer to the test curve. When the speed and load pressure are both low, the experimental mechanical efficiency is basically consistent with the simulation results. However, the difference between the experiment and simulation results gradually increases as the load increases gradually. This difference also appears when comparing the unilateral force model with the experimental results. The possible reason is that the stacked taper roller sets are pressed against each other by oil pressure. When the load gradually increase, the supporting force between the two taper rollers and between the taper rollers and the guide rail also begins to increase, resulting in an increase in the rolling friction coefficient [26], so affecting the mechanical efficiency.

Figure 12 is a graph of mechanical efficiency and rolling friction coefficient at 6000 rpm. When the load pressure is 5 MPa and 8 MPa, the rolling friction coefficient is about 0.001 and 0.004, respectively. In the actual working conditions of the roller–guide-rail mechanism, the linear contact may become point contact and the coexistence of rolling and sliding friction due to the deflection of the taper roller. Therefore, the determination of the relationship between the actual rolling friction coefficient and the load pressure needs to be further studied.

Figure 12. Curve graph of mechanical efficiency and rolling friction coefficient at 6000 rpm.

5. Conclusions

Based on the reassessment of the transmission part of the stacked taper roller set, this paper optimizes the mechanical efficiency mathematical model, and compares it with the unilateral force model of the guide rail. Finally, the experimental results were compared with the simulation results of the bilateral force model, and the differences in the results were explained. The conclusions can be summarized as follows:

(1)

The mechanical efficiency of the bilateral force model of the 2D piston pump with a stacked taper roller set is lower than that of the unilateral force model. The specific reason is that the guide sets are subjected to the opposite support force to the direction of rotation when rotating, which causes the torque of the bilateral force model of the guide rail to be greater than that of the unilateral force model. Therefore, the mechanical efficiency model established based on the bilateral force of the guide rail is closer to the test results than the unilateral force model.

(2)

In the case of the 2D piston pump with a stacked taper roller set at high speed, the transmission loss of the roller–guide-rail mechanism and the oil shear loss are large, so the mechanical efficiency gradually decreases as the speed increases. When the load gradually increases, since the speed of the increasing torque is less than the speed of increasing load pressure, so the mechanical efficiency increases with the increase in the load pressure. When the load pressure is high, the test results of the mechanical efficiency are quite different from the numerical simulation results. The possible reasons are that the higher oil pressure increases the supporting force between two taper rollers and between the taper rollers and the guide rail, which increases the size of the contact area, resulting in an increase in the rolling friction coefficient, which affects the mechanical efficiency.

In the future, the bilateral force model of guide rails with variable pressing forces on both sides of the taper rollers established in this paper can provide a theoretical basis for the design of the 2D piston pump under different pressing forces, with further analysis of the influence of the different sizes of the pressing force of the taper rollers on both sides of its mechanical efficiency.