Pneumatics: compressed-air actuation

Pneumatic circuits move loads with compressed air. Because air is compressible, force, speed and timing must be reasoned about separately: pressure sets force, flow sets speed.

Purpose and meaning

In Pneumatic mode the schematic canvas solves a compressed-air network: a pressure source or service unit feeds directional valves, flow controls and cylinders, and every pipe is coloured by its live pressure. The solver reports cylinder position, port pressures and valve states while you operate the push buttons, so you can verify a sequence such as extend, dwell and retract against the circuit you drew.

Mathematical model

Gas state and supply pressure

Air behaves as an ideal gas, $PV = mR_sT$ , with specific gas constant $R_s = 287\text{ J/(kg}\cdot\text{K)}$ for dry air. Here $P$ is absolute pressure, $V$ is volume, $m$ is mass and $T$ is absolute temperature. Industrial networks are normally run near $P_{sys} \approx 6\text{ bar} = 600\text{ kPa}$ gauge, which is the default operating pressure of the supply components. Gauge pressure $P_{gauge}$ is measured above atmosphere, so $P = P_{gauge} + P_{atm}$ with $P_{atm} \approx 101.3\text{ kPa}$ .

Cylinder force

A single-acting cylinder has a return spring, so the available extension force is reduced by the spring: $F_{ext} = PA_{piston} - F_{spring}$ , and the pressure must first exceed $F_{spring}/A_{piston}$ before the rod moves. A double-acting cylinder with bore $D$ and rod diameter $d$ produces $F_{ext} = P\pi D^2/4$ on the full piston face and $F_{ret} = P\pi(D^2 - d^2)/4$ on the annular rod side, so the return force is always smaller. Friction and back pressure are subtracted in practice, so the measured force is always below the ideal figure.

Air consumption and free air delivery

Compressors are sized in free air, the volume the air would occupy at atmospheric pressure. For a swept volume $V_{cyl}$ per cycle, $Q_{FAD} = V_{cyl}\dfrac{P_{gauge} + P_{atm}}{P_{atm}}n_{cycles}$ , where $n_{cycles}$ is the cycle rate. The compression ratio multiplies the swept volume, which is why a modest cylinder run at $6\text{ bar}$ consumes about seven times its swept volume in free air.

Directional valves

Valves are named by ports and positions. A $3/2$ valve (three ports, two positions) drives a single-acting cylinder, a $5/2$ valve drives a double-acting cylinder, and a $5/3$ valve adds a centre position that closes, exhausts or pressurizes both work ports. A valve returns by spring or is held by a pilot or detent, and the symbol shows the current position as the active cell.

Speed control with throttle check valves

Cylinder speed follows $v = Q/A$ . A throttle check valve restricts flow in one direction only, through an adjustable orifice, and lets air pass freely the other way. Meter-in throttles the supply to the cylinder, and the load can then run ahead of the flow when it pulls. Meter-out throttles the exhaust, which keeps back pressure $P_2$ on the rod side so that $P_1A_1 - P_2A_2 = F_{load}$ and motion is stable. Orifice flow becomes choked when $P_2/P_1$ falls below about $0.528$ , after which a lower downstream pressure no longer increases the flow.

Governing equations

PV = m R_s T,\qquad R_s = 287\text{ J/(kg}\cdot\text{K)}

Ideal gas law with the specific gas constant of dry air.

F_{ext} = P\,A_{piston} - F_{spring}

Single-acting extension force reduced by the return spring.

F_{ext} = P\,\frac{\pi D^2}{4}

Double-acting extension force on the full piston face.

F_{ret} = P\,\frac{\pi\left(D^2 - d^2\right)}{4}

Double-acting retraction force on the annular rod side.

Q_{FAD} = V_{cyl}\,\frac{P_{gauge} + P_{atm}}{P_{atm}}\,n_{cycles}

Free air delivery needed for a given swept volume and cycle rate.

v = \frac{Q}{A}

Piston speed from flow and piston area.

P_1 A_1 - P_2 A_2 = F_{load}

Force balance of a meter-out circuit with back pressure.

\frac{P_2}{P_1} \le \left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma}{\gamma - 1}} \approx 0.528

Critical pressure ratio at which orifice flow chokes.

Canvas parameter and control mapping

Medium switch (Pneumatic): Selects air as the working fluid and loads the pneumatic part library. Switching clears the schematic, because the symbols and ratings differ.
Operating Pressure: The supply setting of the pressure source, default $6\text{ bar}$ . It is the gauge pressure $P$ in every force equation.
Bore and Stroke: The advanced cylinder geometry, defaults $D = 25\text{ mm}$ and $100\text{ mm}$ stroke. Bore fixes the areas $A_{piston}$ and $\pi D^2/4$ , and stroke fixes the swept volume $V_{cyl}$ .
Throttle Opening: A slider from $1$ to $100$ percent on flow control valves. It scales the effective orifice area and therefore the flow $Q$ and speed $v$ .
Set Pressure and Switching Pressure: The setting of regulators, relief valves and sequence valves. The regulator holds its outlet at this value whatever the inlet pressure above it.
Reservoir Volume: Vessel volume of an air reservoir or time-delay chamber, from $10$ to $100000\text{ cm}^3$ . A larger $V$ slows the pressure change in the ideal gas relation and lengthens the delay.
Push buttons and Latch: Operators that shift a valve to its active cell. Latch holds the input on, which models a detented lever.
Simulation speed: A slider from $0.1$ to $4$ times real time that scales how fast the solver advances the animation.

Worked numerical example

Sizing a double-acting cylinder and its air consumption

A double-acting cylinder has bore $D = 50\text{ mm}$ , rod $d = 20\text{ mm}$ and stroke $100\text{ mm}$ at $P = 6\text{ bar}$ gauge. Find both forces and the free air needed at $n_{cycles} = 30\text{ min}^{-1}$ .

Piston area and extension force with $P = 600\text{ kPa}$ .
$A = \frac{\pi (0.05)^2}{4} = 1.9635\times10^{-3}\text{ m}^2,\quad F_{ext} = 600\,000 \times 1.9635\times10^{-3} = 1178\text{ N}$
Annular area on the rod side and retraction force.
$A_r = \frac{\pi (0.05^2 - 0.02^2)}{4} = 1.6493\times10^{-3}\text{ m}^2,\quad F_{ret} = 600\,000 \times 1.6493\times10^{-3} = 990\text{ N}$
Swept volume of one full cycle, extend plus retract, with a stroke of $0.1\text{ m}$ .
$V_{cyl} = (1.9635 + 1.6493)\times10^{-3} \times 0.1 = 3.613\times10^{-4}\text{ m}^3$
Compression ratio with $P_{atm} = 1.01325\text{ bar}$ , then free air delivery.
$\frac{6 + 1.01325}{1.01325} = 6.92,\quad Q_{FAD} = 3.613\times10^{-4} \times 6.92 \times 30 = 0.0750\text{ m}^3/\text{min}$

The cylinder pushes with $F_{ext} \approx 1178\text{ N}$ , pulls with $F_{ret} \approx 990\text{ N}$ , and needs $Q_{FAD} \approx 75\text{ L/min}$ of free air.

Industrial applications

Assembly automation: Pick-and-place units, clamps and part feeders use $5/2$ valves and double-acting cylinders because air is clean, fast and cheap to distribute.
Packaging and food lines: Washdown-safe cylinders push, divert and seal products; meter-out throttles set gentle, repeatable stroke speeds.
Robotic end effectors: Pneumatic grippers and vacuum cups give a compliant grip, and the spring-return $3/2$ valve fails safe by releasing or holding on air loss.
Process and mining plant: Pneumatic actuators operate valves and dampers in explosive atmospheres, where air avoids the spark risk of electric drives.

Pneumatics and Hydraulic Drive