Pressure booster

Technical data

Operation notes If the outlet capacity is undersized, pulsation may occur Make sure to install a mist separator at the inlet side of the booster regulator, supply air must be dry - non lubricated The booster regulator has a sliding part inside, and it generates dust. Also, install a cleaning device such as an air filter or a mist separator on the outlet side as necessary Depending on the necessity, install a silencer on the exhaust port of the booster regulator to reduce the exhaustion sound Provide a dedicated pipe to release the exhaust air from each booster regulator. If exhaust air is converged into a pipe, the back pressure that is created could cause improper operation

Working Principle

The IN(Inlet) air passes through the check valve to booster chamber A and B. Meanwhile, air is supplied to drive chamber B via the governor and the switching valve. Then, the air pressure from drive chamber B and booster chamber A are applied to the piston, boosting the air in booster chamber B. As the piston travels, the boosted air is pushed via the check valve to the OUT(Outlet) side. When the piston reaches the switching valve to touch, so that drive chamber B is in the exhaust state and drive chamber A is in the supply state respectively. Then, the piston reverses its movement, this time, the pressures from booster chamber B and drive chamber A boosts the air in booster chamber A and sends it to the OUT side
The process described above is repeated to continuously supply highly pressurized air from the IN to the OUT side. The governor establishes the outlet pressure by handle operation and pressure adjustment in the drive chamber by feeding back the outlet pressure.

Example of circuit usage

In factories where only specific equipment requires high pressure, a booster cylinder can be installed in the corresponding local air circuit. This allows the overall air circuit to maintain low pressure while enabling the use of highpressure equipment in localized areas. This design not only reduces the pressure demand on the overall air circuit but also effectively saves energy and decreases the load on the equipment.

When using two booster cylinders for two-stage boosting, ensure that each booster cylinder is supplied with sufficient airflow to maintain the stability of the inlet pressure at the booster valve. This is critical for ensuring the stability and efficiency of the boosting process.

During the process of charging the air tank, a circuit design incorporating a booster cylinder in parallel with a check valve is used. When the pressure in the air tank is lower than the inlet air source pressure, the air source directly charges the tank through the check valve. This method efficiently utilizes the air source pressure, reducing the charging time and improving overall efficiency.

The inlet pressure (P₁) first passes through a check valve to charge P₂, until P₁ = P₂

When the actuator's output force is insufficient and spatial constraints prevent the use of a larger cylinder, a booster cylinder can be employed to increase output force without replacing the existing actuator.
In cases where miniaturization of the drive component is required, and a small cylinder size is needed while maintaining a high output force, a booster cylinder offers an effective solution.

For single-acting cylinder operations, installing a booster cylinder in the corresponding air supply circuit can reduce compressed air consumption, achieving more efficient operation.

Calculation of the filling time of a tank

Example of calculating the filling time of a tank with a volume of 50 liters when increasing the pressure from a starting pressure of 0.7 MPa to a final pressure of 0.9 MPa using the MVBA4100 pressure booster with an input pressure of 5 bar.
Input pressure: P1 = 0.5 MPa
Initial pressure in the tank: STP = 0.7 MPa
Final pressure in the tank: FTP = 0.9 MPa
Tank volume: V = 50 l
We calculate the ratio STP/P1 and FTP/P1, i.e. 0.7/0.5=1.4 and 0.9/0.5=1.8. From the graph, we determine the time values ​​corresponding to the specified pressure ratios, i.e. for P2/P1=1.4 it is approximately 1 s, for a ratio of 1.8 it is approximately 3 s. We subtract the values, i.e. 3-1=2 s. The time for pressurizing 10 l to a pressure of 0.9 MPa under the specified conditions is 2 s. Pressurizing 50 l will therefore be a multiple of 2 x 5=10 s. The total time for pressurizing 50 l is therefore 10 s.