Mechatronics · Module 5 of 10
Hydraulic and Pneumatic Actuation
Fluid power turns pressure into large, controllable force. A cylinder's force is pressure times piston area, and its speed is flow divided by that area, whether the fluid is stiff oil or springy air.
Readiness check
This module opens the actuation half of the course. Tick only what you can do closed-notes.
- Compute a circle area from its diameter.
- Recall that pressure is force per unit area.
- Convert litres per minute to cubic metres per second.
- Recall that liquids are nearly incompressible and gases are not.
- Rearrange force equals pressure times area.
The core idea
A fluid-power cylinder makes force by applying pressure over the piston area, so F = P × A with A = πd2/4. It extends at a speed set by the supplied flow divided by that area, v = Q/A. Hydraulics use stiff oil for high force; pneumatics use compressible air for speed and compliance.
A = πd2/4F = P × Av = Q / AFluid power actuation converts the pressure of a fluid into mechanical force and motion. The heart of it is the cylinder: pressure P acting on a piston of area A produces a force F = P × A, and because the piston is circular its area is πd2/4, so the bore diameter sets how much force a given pressure delivers. The same cylinder's extension speed comes from continuity: a volumetric flow Q into the cylinder fills it at v = Q/A, so a bigger bore is stronger but slower for the same pump. The choice of fluid divides the field. Hydraulic systems use nearly incompressible oil at high pressure, giving large, stiff, precisely held forces, ideal for presses and heavy machinery. Pneumatic systems use compressible air at lower pressure, giving fast, springy, and inherently compliant motion that is clean and cheap, ideal for light, repetitive tasks. Directional control valves route the fluid to extend, retract, or hold, and a solenoid on the valve is where the electronic controller meets the fluid power.
The skills, taught in order
Five skills size and control a fluid-power actuator.
5.1 Fluid power and the two families
Hydraulic systems move nearly incompressible oil at high pressure for large, stiff forces; pneumatic systems move compressible air at lower pressure for fast, compliant motion. Compressibility is the key difference: air stores energy like a spring, oil does not.
5.2 Cylinder force
A cylinder converts pressure to force through F = P × A, with the piston area A = πd2/4. A double-acting cylinder is driven both ways; on the rod side the area is reduced by the rod, so the retract force is a little smaller than the extend force.
5.3 Cylinder speed
By continuity the extension speed is v = Q/A: the pump's flow fills the bore. For a fixed pump, a larger bore gives more force but less speed, the fundamental force-speed trade of a cylinder.
| Quantity | Relation | Set by |
|---|---|---|
| Piston area | A = πd2/4 | bore diameter |
| Force | F = P × A | pressure and area |
| Speed | v = Q/A | flow and area |
The three cylinder relations. The area appears in all of them, which is why bore choice is central.
5.4 Directional control valves
A valve is described by its ports and positions, such as a 4/3 valve with four ports and three positions. A solenoid shifts the spool, so a digital output can extend, retract, or centre a cylinder, connecting the controller to the fluid.
5.5 The rest of the circuit
A pump and prime mover supply flow; a relief valve caps pressure for safety; an accumulator stores energy; and in pneumatics a filter, regulator, and lubricator condition the air. These support the actuator and keep it safe.
Engineering connection: an injection-moulding clamp uses a large-bore hydraulic cylinder for stiff tonnage, while the parts are ejected by fast, compliant pneumatic cylinders.
Worked example 1: cylinder force
A hydraulic cylinder has a 40 mm bore and runs at 12 MPa. Find the piston area and the extend force.
- ProblemFind the piston area and extend force for the cylinder in Figure 1.
- Given / findd = 40 mm = 0.040 m, P = 12 MPa. Find A and F.
- AssumptionsForce on the full bore (extend stroke), pressure acts uniformly, friction neglected.
- ModelA = πd2/4; F = P × A.
- EquationsA = π(0.040)2/4F = 12×106 × A
- SolveA = 1.257 × 10−3 m2; F = 12×106 × 1.257×10−3 = 15080 N ≈ 15.1 kN.
- Check12 MPa is 12 N per mm2; the bore is π(20)2 = 1257 mm2; 12 × 1257 = 15080 N, matching.
- ConclusionA modest 40 mm cylinder at 12 MPa lifts about 1.5 tonnes, showing why hydraulics dominate heavy force.
Worked example 2: cylinder speed
The same 40 mm cylinder is fed 15 litres per minute. Find its extension speed.
- ProblemFind the extension speed for the cylinder in Figure 2.
- Given / findQ = 15 L/min, A = 1.257 × 10−3 m2. Find v.
- AssumptionsIncompressible flow, no leakage past the piston.
- ModelContinuity: v = Q/A, with Q in m3/s.
- EquationsQ = 15/60000 = 2.5×10−4 m3/sv = Q/A
- Solvev = 2.5×10−4 / 1.257×10−3 = 0.20 m/s.
- CheckA 0.2 m/s speed over the 1.257×10−3 m2 bore returns 2.5×10−4 m3/s, the given flow.
- ConclusionThe cylinder extends at about 0.2 m/s; doubling the bore for more force would quarter the area's speed for the same flow.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Using diameter as area | Force off by a large factor | "Did I compute πd2/4?" | Force uses the area, not the diameter. |
| Radius and diameter mixed | Area four times wrong | "Is d the bore or the radius?" | Use d in πd2/4, or πr2 with the radius. |
| Flow units left as L/min | Speed 60000 times off | "Is Q in m3/s?" | Convert L/min to m3/s before v = Q/A. |
| Treating air as stiff | Pneumatic position drifts under load | "Is the fluid compressible?" | Air springs; hold precise force with hydraulics. |
Practice ladder
A cylinder has a 50 mm bore at 10 MPa. Find the force.
Show answer
A = π(0.05)2/4 = 1.963 × 10−3 m2; F = 10×106 × A = 19.6 kN.
A 40 mm bore cylinder is fed 20 L/min. Find its speed.
Show answer
Q = 20/60000 = 3.33 × 10−4 m3/s; v = Q/(1.257×10−3) = 0.265 m/s.
You need 20 kN at 16 MPa. What bore diameter is required?
Show answer
A = F/P = 20000/16×106 = 1.25 × 10−3 m2; d = √(4A/π) = 39.9 mm, so a 40 mm bore.
Choose hydraulic or pneumatic actuation for (a) a 60 kN press that must hold position and (b) a fast, light pick-and-place gripper. Justify each.
What good work looks like
(a) Hydraulic: high, stiff force held precisely, since oil is incompressible; (b) pneumatic: fast, clean, compliant, and cheap for light repetitive motion where exact position under load is not critical.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Specify a cylinder for one task: choose a bore and pressure for the force, then check the speed at your pump's flow.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the cylinder force.
F = P × A, with A = πd2/4.
2. Write the cylinder speed.
v = Q/A, with Q in m3/s.
3. Why is hydraulic force stiff?
Oil is nearly incompressible, so it holds position and force precisely.
4. What describes a directional valve?
Its ports and positions, such as a 4/3 valve, shifted by a solenoid.
5. Why choose pneumatics?
Fast, clean, compliant, and cheap motion for light repetitive tasks.
Textbook mapping
This module follows William Bolton, Mechatronics, 6th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Hydraulic and pneumatic systems | Bolton, Chapter 7, Actuation systems |
| Cylinders and force | Bolton, Chapter 7, Cylinders |
| Directional control valves | Bolton, Chapter 7, Directional control valves |
Chapter numbers refer to Bolton's Mechatronics, 6th edition. Any edition with the same chapter titles is equivalent for study.