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Actuators: hydraulic, pneumatic and electric

Mechatronics (B.E./B.Tech) · Sensors and actuators · free study notes for exams, placements and interviews

In short

An actuator is the output element of a mechatronic system: it converts hydraulic, pneumatic or electrical energy into controlled mechanical motion — linear, rotary or oscillating. The three families differ in power density, stiffness, speed, controllability and cost, so selecting between them is a standard design decision. This topic carries marks in Sensors and Actuators, Hydraulics and Pneumatics, Mechatronics and Industrial Automation papers, and appears in GATE (ME/IN) through force, flow, pressure and air-consumption calculations. In placement interviews at Festo, SMC, Bosch Rexroth, Parker, Wipro PARI, Tata Motors, L&T, Siemens and Rockwell integrators, circuit reading and actuator sizing are asked directly.

Explained

What an actuator is and how it is classified

Hydraulic actuators

Pneumatic actuators

Electric actuators

Choosing between the three

Characteristics and performance terms

Key points

Formulas and examples

F_extend = p × A = p × (π/4)D² — p = gauge pressure (Pa = N/m²), D = bore (m), F in N
F_retract = p × (π/4)(D² − d²) — d = rod diameter (m)
Actual force F_out = η × p × A — η = 0.85–0.95 (hydraulic), 0.80–0.90 (pneumatic) for seal friction
v = Q / A — v = piston speed (m/s), Q = flow (m³/s), A = effective area (m²); 1 L/min = 1.667 × 10⁻⁵ m³/s
Hydraulic power P = p × Q — Pa × m³/s = W; shortcut P(kW) = p(bar) × Q(L/min) / 600
Pump: Q(L/min) = V_d(cm³/rev) × N(rpm) × η_v / 1000; Motor torque T = (V_d × Δp)/(2π) × η_m
Air consumption per double stroke (free air) = [A_e + A_r] × L × (p_g + 1.013)/1.013 — compression ratio form
Intensifier: p_oil = p_air × (A_air / A_oil)
Solenoid force F ≈ (B² A_gap)/(2μ₀), μ₀ = 4π × 10⁻⁷ H/m
Electric linear actuator: T_motor = (F × l)/(2π η) + T_preload — l = screw lead (m), η = screw efficiency (0.9 ball, 0.3–0.5 lead)
Motor: P = 2πNT/60 (N in rpm, T in N·m, P in W); linear speed v = N × l / 60

Example 1 — Pneumatic cylinder sizing. A 63 mm bore, 20 mm rod, 300 mm stroke double-acting cylinder runs at 6 bar gauge.

Example 2 — Hydraulic press force and power. A press cylinder must give 250 kN with oil at 180 bar, overall cylinder efficiency 0.95.

Example 3 — Air-over-oil intensifier. Air piston 200 mm diameter at 6 bar drives an oil piston of 40 mm diameter.

Example 4 — Electric linear actuator. Push 4 kN at 100 mm/s using a ball screw of 10 mm lead, η = 0.90.

Example 5 — Hydraulic motor. A motor of 40 cm³/rev displacement runs on 30 L/min at 150 bar, η_v = 0.95, η_m = 0.90.

CriterionHydraulicPneumaticElectric
Working pressure70–350 bar4–7 bar—
Force densityVery highLow–moderateModerate
Stiffness / load holdingExcellentPoor (springy)Good with brake
Position accuracy0.01–0.1 mm (servo)End-stops mainly0.01 mm or better
SpeedModerate, controllableHighFully programmable
Overall energy efficiencyModerateLowestHighest
CleanlinessOil leaksClean exhaustCleanest
Hazardous areaSuitableBest (no spark)Needs Ex rating
Typical costHighLowMedium–high
Typical usePresses, excavators, mouldingAssembly, packaging, clampingRobots, CNC, pick-and-place

Where it is used

Interview questions

Common mistakes

Related topics

Transducers and sensors basics · Special machines: stepper and BLDC motors · Hydraulics and pneumatics circuits and ISO 1219 symbols · PLC programming and industrial automation · Control systems: closed-loop feedback and PID · Industrial robotics and end effectors.

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