Local pressure boosting system is a complex and switched system, which is widely used in modern pneumatic systems, to optimize local pressure boosting system; firstly, the basic and the dimensionless mathematical models of the local pressure system were setup. Furthermore, the mathematical models were verified through the experimental study on the local pressure boosting system. Moreover, the influences of the tank’s three main parameters on the performance of local pressure boosting system were studied. It can be seen that the pressure wave amplitude is mainly affected by the dimensionless volume of the tank; its influence degree is 95.1%, and it increases when the later one decreases. The pressure loss of the tank is mainly affected by the dimensionless output pressure, and its influence degree is 68.7%, and it decreases rapidly with the increase of the dimensionless output pressure of the tank. Last, the optimization method of the local pressure boosting system was obtained.
Pressure-boosting technologies have been used in many fields [
Recently, main research on pneumatic booster was focused on its characteristics and factors which influence the characteristics of pneumatic booster [
Therefore, one novel type of booster was proposed, which is called expansion energy used pneumatic booster, short for EEU booster. The wave amplitude of output flow and pressure increases with the increase of the EEU booster’s efficiency and power, and that not only affects the working performance of pneumatic actuator, but also seriously shortens the working span of pneumatic components (such as actuator, regulator, and valve).
To reduce the pulse and the loss of output pressure of the EEU booster, in this paper, the local pressure boosting system (consists of the EEU booster, tank, and pipes, etc.) was studied. Firstly, the working principles of the local pressure boosting system and the EEU booster was introduced, the basic and the dimensionless mathematical models of the local pressure system were set up. Furthermore, pressure boosting pneumatic system was studied experimentally to verify the mathematical models. Last, the influences of the tank’s main parameters on the performance of pressure boosting system were studied. This research lays a foundation for the optimization and energy saving of the local pressure boosting system.
There are several definitions that should be introduced.
The structure of local pressure boosting system can be shown in Figure
Configuration of local pressure boosting system. 1: EEU booster, 2: air tank, 3: pipeline, 4: silencer, 5: solenoid valve, and 6: throttle valve.
As is shown in Figure
A typical EEU booster, as shown in Figure
Structure of EEU Booster. 1: Controller, 2: solenoid valve A, 3: solenoid valve B, 4: magnetic switch A, 5: magnetic switch B, 6: magnetic switch C, 7: magnetic switch D, 8: driving chamber A, 9: piston, 10: magnetic ring, 11: driving chamber B, 12: boosting chamber A, 13: piston rod, 14: boosting chamber B, and 15: check valve.
When piston reaches the left travel destination, Solenoid Valve A and Solenoid Valve B change their position, driving chamber B is connected with the atmosphere, compressed air charged from the primary side flows into boosting chamber A and driving chamber A. Compressed air in boosting chamber A and driving chamber A drives the piston to move toward the right; then the pressure of air in boosting chamber B increases, and higher-pressure compressed air in boosting chamber B is not discharged from the second side until the pressure is higher than the pressure in the second side.
When the piston moves near the magnetic switch C, the solenoid valve A changes its position and compressed air charged from the primary side stops to flow into driving chamber A, depending on its expansion, compressed air in driving chamber A keeps driving piston to move, and finally, arrives the right travel destination (viz., near the magnetic switch D); then part of expansion energy of the air is used. The solenoid valve A and the solenoid valve B are controlled by the controller to change their position, air in driving chamber A flows to atmosphere, and air, charged from the primary side, flows into boosting chamber B and driving chamber B; then air in the boosting chamber B and driving chamber B drives the piston to move towards the left, the pressure of air in boosting chamber A ascends, and finally higher-pressure compressed air in boosting chamber A is discharged from the second side.
When the piston moves near the magnetic switch B, the solenoid valve A changes its position and compressed air charged from the primary side stops to flow into driving chamber B; then part of expansion energy of the air is used to drive the piston continue to move, and arrives the left travel destination (viz., near the magnetic switch A). The Solenoid Valve A and the Solenoid Valve B change their position, air in driving chamber B is exhausted to atmosphere. The booster goes around and repeats the process discussed above, higher-pressure compressed air is discharged continuously.
The piston stroke, when the driving chambers stopped to charge air, is defined to be Piston Stroke-set
Piston stroke-set of EEU booster.
To facilitate this research, the following assumptions were made: The working fluid (air) of the system follows all ideal gas laws. There is no leakage between the chambers, the area of the piston rod end is too small to be considered, and the effective areas of all intake and exhaust ports are the same. Supply temperature is equal to atmosphere temperature. The flow of air moving into and out of the chambers is a stable one-dimensional flow that is equivalent to the flow of air through the nozzle contraction.
Because there is no leakage in either chamber, the chambers do not charge and exhaust air simultaneously. Consequently, the energy equation for the discharge and charge side of each chamber can be illustrated by the following equations:
The value of
From the law of mass conservation, air mass can be given as
Air mass flow is calculated from the flow equation, which is described later on.
The input and output pressures of the booster are below the critical pressure; the temperature of the compressed air is above the critical temperature, according to the ratio
The value of
The velocity of the piston is calculated from Newton’s second law of motion. In this paper, the friction force model is considered to be the sum of the Coulomb friction and viscous friction. The viscous friction force is considered to be a linear function of piston velocity. The forces on the piston of the booster are shown in Figure
The forces on the piston of the booster.
The right side was considered to be the positive direction of the vector. The motion equation of the piston can be given by the following equation:
Pressure changes in the air in each chamber can be obtained by deriving the state equation of ideal gases:
The reference values and the dimensionless variables are shown in Table
Reference values and dimensionless variables.
Variable | Reference value | Dimensionless variable | |
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Time |
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Time to totally exhaust |
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Pressure |
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Supply pressure |
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|
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Temperature |
|
Atmosphere temperature |
|
|
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Air mass flow |
|
Maximum air mass flow at charge side of boosting chamber |
|
|
|||
Air mass |
|
Maximum air mass in boosting chamber |
|
|
|||
Displacement |
|
Maximum displacement |
|
|
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Volume |
|
Maximum volume of boosting chamber B |
|
|
|||
Area of piston |
|
Area of piston in boosting chamber |
|
|
|||
Diameter |
|
Diameter of piston in boosting chamber |
|
The dimensionless energy equation for the discharge side and the charge side becomes
The dimensionless maximum heat transfer area can be calculated by the following equation:
For the charge side:
The dimensionless equation of continuity can be given as the following equation:
The dimensionless flow equation for both sides of the chambers becomes
The average output flow of the booster can be given as
The dimensionless equation of motion can be written as follows:
Here,
Here,
Dimensionless parameter,
From
Because the
The dimensionless Piston Stroke-set
The dimensionless state equation for the discharge side and charge side becomes
The dimensionless output flow and dimensionless cycle time can be attained easily and expediently. These parameters were studied experimentally to verify the dimensionless mathematical model that was set up above.
The experimental apparatus shown in Figure
Configuration of experimental apparatus. 1: Regulator, 2: booster, 3: air power meter, 4: tank, 5: throttle valve, 6: data acquisition card, and 7: computer.
In this experiment, we first opened the compressed air source, adjusted the regulator, and set the pressure to the fixed value (0.6 MPa). Next, we adjusted the throttle valve, making sure that air was exhausted from the tank steadily, and the pressure of the air was approximately the fixed value (0.8 MPa). The last stage was data acquisition and preservation.
The values of the nine dimensionless parameters are shown in Table
The initial values of the parameters.
Parameter |
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Value | 0.67 | 1.33 | 1 | 0.008 | 1 | 0.0008 | 0.0167 | 4.47 | 2.98 |
Curves of output flow of boosters.
Curves of pressure of air in tank.
From Figures
The main reasons for the difference between the simulation results and the experimental results are the fluctuation of the supply pressure, leakage between chambers, temperature of the atmosphere, and the fluctuation of pressure of air in tank, which can be seen in Figure
Assume that the dimensionless output pressure of the tank is 1.33, the dimensionless piston stroke-set of the tank is 0.83, the dimensionless diameter of the outlet of the tank is 1, and the dimensionless volume of the tank is set 1.4, 9.9, 14.2, 21.2, 28.3, 35.4, and 42.4, respectively. The research results are shown in Figure
The influence of the dimensionless volume of the tank.
The dimensionless dynamic characteristics of air pressure in the tank
The dimensionless pressure wave amplitude ratio of the tank
The dimensionless pressure loss in the tank
The influence of the dimensionless output pressure of the tank.
The dimensionless dynamic characteristics of pressure in the tank
The dimensionless pressure wave amplitude in the tank
The dimensionless pressure loss in the tank
The influence of the dimensionless diameter of the outlet of the tank.
The dimensionless dynamic characteristics of pressure in the tank
The dimensionless pressure wave amplitude in the tank
The dimensionless pressure loss in the tank
As shown in Figure
Assume that the dimensionless volume of the tank is 11.33, the dimensionless piston stroke-set of the tank is 0.83, the dimensionless diameter of the outlet of the tank is 1, and the dimensionless output pressure of the tank is set 1.3, 1.29, 1.33, 1.38, 1.42, 1.46, and 1.5, respectively. The research results are shown in Figure
As shown in Figure
Assume that the dimensionless volume of the tank is 11.33, the dimensionless piston stroke-set of the tank is 0.83, the dimensionless output pressure of the tank is 1.33, and the dimensionless diameter of the outlet of the tank is set 0.8, 0.9, 1, 1.1, 1.2, and 1.3, respectively, the research results are shown in Figure
As shown in Figure
Each parameter can be changed within a certain range for comparison of the influence degree while other parameters are kept constant. And the influence degree of the main parameters of the tank on the pressure wave amplitude and the pressure loss influence ratio can be given by the following equations:
Based on the numerical study, the influence degrees of the dimensionless volume, the dimensionless output pressure, and the dimensionless diameter of the outlet of the tank on the pressure wave amplitude and the pressure loss are shown in Figures
The influence degree of the main parameters of the tank on the pressure wave amplitude.
The influence degree of the main parameters of the tank on the pressure loss.
As shown in Figures The pressure wave amplitude of the tank is mainly affected by the dimensionless volume of the tank, and its influence degree is 95.1%. And the influences of the dimensionless output pressure and the diameter of the outlet of the tank can be ignored. The pressure loss of the tank is mainly affected by the dimensionless output pressure and the diameter of the outlet of the tank, and the three parameters’ influence degree is 68.7%, 21.8%, and 9.5%.
In this paper, a new kind of booster, EEU booster, was proposed, the local pressure boosting system was studied, and the basic mathematical model of the system was developed. Appropriate reference values were selected, the basic mathematical model was transferred to a dimensionless expression, and the influence of the tank on the working performance of the local pressure boosting system was analyzed through simulation and experimentation. The conclusions are summarized as follows.
So, aiming at the optimization of the local pressure boosting pneumatic system, we can draw conclusions as follows.
If those measurements were taken, the pressure wave amplitude in the tank and the pressure loss of the compressed air could both be reduced.
This research lays a foundation for the optimization design and energy saving of the local pressure boosting system.
Area of piston [m2]
Effective area of intake and exhaust port [m2]
Viscous friction coefficient [Ns/m]
Specific heat at constant volume
Diameter [m]
Degree
Friction force [N]
Coulomb friction force [N]
Maximum static friction force [N]
Air mass flow [kg/s]
Heat transfer coefficient [W/(m2K)]
Stroke [m]
Piston stroke-set [m]
Mass of piston [kg]
Pressure [Pa]
Gas constant
Heat transfer area [m2]
Time [s]
Time period [s]
Velocity [m/s]
Volume [m3]
Air mass [kg]
Piston displacement [m]
Specific heat ratio
Temperature [K].
Atmosphere
Driving chamber
Boosting chamber
Charge side
Discharge side
Chamber A
Chamber B
Upstream side
Downstream side
Output of booster
Supply of booster
Tank
Pressure loss changes
Pressure wave amplitude changes
Equilibrium conditions.
Dimensionless.
The authors declare that there is no conflict of interests regarding the publication of this paper.
This project is supported by National Natural Science Foundation of China Grant no. (51205008).