Hydraulic pumps, actuators and accumulators
Positive-displacement hydraulic pumps and their efficiencies, cylinders and regenerative circuits, hydraulic motors, and gas-charged accumulator sizing, with pump–cylinder and accumulator–motor examples.
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Why it matters
Hydraulics gives the highest power density of any actuation method: a palm-sized cylinder at 200 bar can push several tonnes. Presses, injection-moulding machines, excavators, aircraft landing gear and machine-tool clamps all depend on choosing the right pump, sizing the cylinder or motor for force and speed, and using accumulators to store energy and smooth pressure. These are routine calculations for a mechatronics engineer.
Key ideas
The basic circuit. A prime mover drives a pump that draws oil from a reservoir through a suction strainer; valves direct the flow to an actuator (cylinder or motor); oil returns to the tank through a filter and cooler. A relief valve limits maximum pressure. Two rules govern everything:
- The pump creates flow; pressure builds only as much as needed to overcome the load (and losses). Pressure is set by load, speed by flow.
- Actuator force or torque depends on pressure × area or displacement; actuator speed depends on flow ÷ area or displacement.
Positive-displacement pumps. Hydraulic pumps trap fixed volumes of oil and push them out, so their output is nearly independent of pressure (unlike centrifugal pumps). Displacement V_d is the volume per revolution. Leakage across internal clearances reduces delivered flow as pressure rises; this is the volumetric efficiency.
- External gear pump: two meshing gears carry oil round the casing. Simple, cheap, tolerant of dirt; fixed displacement; noisy; typically up to about 250 bar.
- Internal gear / gerotor: quieter, compact.
- Vane pump: vanes slide in a slotted rotor inside a cam ring. Balanced designs (elliptical ring) cancel side loads; unbalanced designs with a movable ring give variable displacement and pressure compensation. Quiet, medium pressure.
- Piston pumps (axial swash-plate, bent-axis, radial): highest pressures (about 350–450 bar) and efficiencies; swash-plate angle gives variable displacement, used for load-sensing and pressure-compensated systems. Typical overall efficiencies: gear 0.75–0.85, vane 0.75–0.85, piston 0.85–0.92 (take actual values from the maker's data).
Pump efficiencies. Volumetric η_v = actual flow ÷ theoretical flow; mechanical (torque) η_m = theoretical torque ÷ actual torque; overall η_o = η_v × η_m = hydraulic power out ÷ shaft power in.
Cylinders (linear actuators).
- Single-acting: oil extends the piston; a spring or the load returns it.
- Double-acting: oil drives both strokes. The cap end has full area A; the rod end has annulus area A − a. So for the same pressure and flow, extension is slower with more force; retraction is faster with less force.
- Regenerative circuit: rod-end oil is routed back into the cap end during extension, so the effective area is just the rod area a. Extension becomes fast but force drops to p·a.
- Cushioning throttles the last part of the stroke to prevent impact. Long, thin rods under compression must be checked for buckling (Euler), using a factor of safety from your design code.
Hydraulic motors (rotary actuators). Gear, vane and piston designs work as pumps in reverse. Torque is proportional to pressure drop × displacement; speed is flow ÷ displacement. High-torque, low-speed radial-piston motors drive winches and wheels directly. Limited-rotation (rack-and-pinion or vane) actuators give partial turns.
Accumulators. Store oil under pressure for later use:
- Weight-loaded: constant pressure, bulky; rare now.
- Spring-loaded: pressure rises as it fills; small volumes.
- Gas-charged (bladder, piston or diaphragm, always with nitrogen, never air or oxygen): the most common. Pre-charged to p₀ (usually about 0.9 of minimum working pressure). Uses: supplementing pump flow during short peak demands (allowing a smaller pump), emergency power, maintaining pressure while the pump unloads, compensating leakage and thermal expansion, and absorbing shocks and pump pulsations. Gas behaviour follows p·V^n = constant with n = 1 for slow (isothermal) and about 1.4 for fast (adiabatic) cycles. Accumulators hold stored energy even when the pump is off; they must be discharged before maintenance.
Fluid. Mineral hydraulic oil (grade by viscosity at 40 °C), with water-glycol or synthetic esters for fire resistance. Too low viscosity (hot oil) increases leakage; too high (cold oil) causes sluggish response and pump-inlet cavitation. Cleanliness is critical: most failures come from contamination.
Formulas
Q_th = V_d · N / 60, Q_act = η_v · Q_th
- Q = flow (m³/s), V_d = displacement (m³/rev), N = speed (rpm).
P_hyd = p · Q, P_shaft = p · Q_act / η_o
- P = power (W), p = pump pressure rise (Pa).
T_th = p · V_d / (2π), T_act = T_th / η_m (pump); T_out = η_m · Δp · V_d / (2π) (motor)
- T = torque (N·m), Δp = pressure drop across a motor (Pa).
N_motor = 60 · η_v · Q / V_d
- Motor speed (rpm).
F_ext = p · A, v_ext = Q / A; F_ret = p · (A − a), v_ret = Q / (A − a)
- A = piston area, a = rod area (m²), F = force (N), v = speed (m/s). Back pressure and friction neglected.
F_regen = p · a, v_regen = Q / a (regenerative extension)
p₀·V₀^n = p₁·V₁^n = p₂·V₂^n, ΔV = V₂ − V₁
- Gas-charged accumulator, absolute pressures (Pa); V₀ = accumulator size (gas volume at pre-charge), V₁ and V₂ = gas volumes at maximum and minimum working pressure; n = 1 (isothermal) to 1.4 (adiabatic).
V₀ = ΔV / [ (p₀/p₂)^(1/n) − (p₀/p₁)^(1/n) ]
Worked examples
Example 1 — pump and cylinder sizing (standard). A pump of 40 cm³/rev runs at 1450 rpm with η_v = 0.92 and η_o = 0.85 at 15 MPa. It drives a double-acting cylinder with an 80 mm bore and 45 mm rod. Find the delivered flow, motor power, and the force and speed on extension and retraction.
Q_act = η_v·V_d·N/60= 0.92 × 40 × 10⁻⁶ × 1450/60 = 8.893 × 10⁻⁴ m³/s (53.4 L/min).- Hydraulic power p·Q = 15 × 10⁶ × 8.893 × 10⁻⁴ = 13.34 kW; shaft power = 13.34/0.85 = 15.69 kW.
- Areas: A = π × 0.08²/4 = 5.027 × 10⁻³ m²; a = π × 0.045²/4 = 1.590 × 10⁻³ m²; A − a = 3.436 × 10⁻³ m².
- Extension: F = 15 × 10⁶ × 5.027 × 10⁻³ = 75.4 kN; v = 8.893 × 10⁻⁴/5.027 × 10⁻³ = 0.177 m/s.
- Retraction: F = 15 × 10⁶ × 3.436 × 10⁻³ = 51.5 kN; v = 8.893 × 10⁻⁴/3.436 × 10⁻³ = 0.259 m/s. Answer: 53.4 L/min, 15.7 kW motor; extend 75.4 kN at 0.177 m/s; retract 51.5 kN at 0.259 m/s. In a regenerative circuit the extension would be 0.559 m/s at only 23.9 kN.
Example 2 — accumulator and motor (GATE level). (a) A bladder accumulator must deliver 2 L of oil as the system pressure falls from 21 MPa to 14 MPa; the pre-charge is 9 MPa (treat all as absolute). Find the required size for slow (isothermal) and rapid (adiabatic, n = 1.4) discharge. (b) A motor of 100 cm³/rev receives 60 L/min at a pressure drop of 20 MPa, with η_v = 0.95 and η_m = 0.90. Find its speed, torque and output power.
- (a) Isothermal:
V₀ = ΔV/(p₀/p₂ − p₀/p₁)= 2/(9/14 − 9/21) = 2/(0.6429 − 0.4286) = 9.33 L. - Adiabatic: V₀ = 2/((9/14)^(1/1.4) − (9/21)^(1/1.4)) = 2/(0.7295 − 0.5461) = 10.9 L.
- (b) Speed:
N = η_v·Q/V_d= 0.95 × 60 000/100 = 570 rpm. - Torque:
T = η_m·Δp·V_d/(2π)= 0.9 × 20 × 10⁶ × 100 × 10⁻⁶/(2π) = 286.5 N·m. - Power out = T·ω = 286.5 × 2π × 570/60 = 17.1 kW (input Δp·Q = 20 kW, so η_o = 0.855). Answer: (a) about 9.3 L isothermal, 10.9 L adiabatic; (b) 570 rpm, 286 N·m, 17.1 kW.
Common mistakes
- Using gauge pressures in the accumulator gas law; Boyle's law needs absolute pressures.
- Using the full piston area for the retraction stroke instead of the annulus area.
- Applying volumetric efficiency the wrong way: a pump delivers less than its theoretical flow, a motor needs more than its theoretical flow.
- Saying a pump "creates pressure"; pressure is set by the load or the relief valve.
- Forgetting unit conversions: cm³/rev to m³/rev (× 10⁻⁶), L/min to m³/s (÷ 60 000), bar to Pa (× 10⁵).
- Ignoring that an accumulator stays charged after shut-down.
For GATE ME
Fluid-power hardware is not a regular part of the GATE ME syllabus, but the calculations here — pressure, area, force, flow continuity, power and efficiency chains — use exactly the fluid mechanics GATE tests, and the same positive-displacement versus rotodynamic distinction appears in pump questions. Practise unit conversions and power chains until they are automatic; university and interview questions on cylinder sizing and accumulators follow these patterns.
Quick check
- A pump of 20 cm³/rev at 1500 rpm has η_v = 0.9. What flow does it deliver in L/min?
- Why does a double-acting cylinder retract faster than it extends?
- What gas is used to pre-charge an accumulator, and why?
- What sets the system pressure in a hydraulic circuit?
- A 100 mm bore cylinder works at 5 MPa. What is its extension force?
Answers: 1. 20 × 1500 × 0.9 = 27 000 cm³/min = 27 L/min; 2. the rod-end annulus area is smaller, so the same flow moves the piston faster; 3. dry nitrogen, because it is inert (air or oxygen with oil can ignite under compression); 4. the load, limited by the relief valve setting; 5. 5 × 10⁶ × π × 0.1²/4 ≈ 39.3 kN.
See it move
All Mechatronics animationsAdjust the pump speed and displacement to see how the flow rate changes. Observe how the actuator force responds to pressure and area changes.
Equations used
- Q = V * n — Q flow rate (m³/s), V displacement per revolution (m³/rev), n rotational speed (rev/s)
- F = P * A — F force (N), P pressure (Pa), A area (m²)
Interview questions
All Fluid Mechanics and Fluid Power interview questionsTry answering each one aloud before you open it.
1.What is a hydraulic pump and how does it function in a hydraulic system?Concept
A hydraulic pump is a mechanical device that converts mechanical energy into hydraulic energy by moving fluid from a reservoir into the hydraulic system. It creates a flow of fluid that can be used to perform work, such as moving a piston or rotating a motor. The pump does not create pressure; it only generates flow. Pressure is created by the resistance to fluid flow in the system.
2.Explain the role of hydraulic actuators in a hydraulic system.Concept
Hydraulic actuators are devices that convert hydraulic energy into mechanical energy. They are used to perform work by moving or controlling a mechanism or system. Common types of hydraulic actuators include hydraulic cylinders, which provide linear motion, and hydraulic motors, which provide rotational motion. They are essential for applications requiring high force and precise control.
3.What is a hydraulic accumulator and why is it used in hydraulic systems?Concept
A hydraulic accumulator is a pressure storage reservoir in which a non-compressible hydraulic fluid is held under pressure by an external source, such as a spring, a raised weight, or compressed gas. It is used to store energy, absorb shock, and maintain pressure in the hydraulic system. Accumulators help in smoothing out pulsations in the system and can provide additional fluid flow during peak demand.
4.Why are vane pumps commonly used in hydraulic systems?Application
Vane pumps are commonly used in hydraulic systems because they provide a smooth and consistent flow of fluid, which is essential for precise control. They are also relatively quiet and have a good efficiency range. Vane pumps are suitable for low to medium pressure applications and are often used in industrial machinery and mobile equipment.
5.What happens if a hydraulic system operates with a clogged filter?Application
It depends where the filter is. A clogged suction strainer starves the pump: inlet pressure falls, oil releases air and vapour, and the pump cavitates, becoming noisy and wearing quickly. A clogged pressure or return filter causes a large pressure drop, wasting power as heat, and once the drop reaches the setting of the filter's built-in bypass valve, unfiltered oil flows past the element, so contaminants circulate and wear pumps, valves and seals. Clogging indicators on the filter housing are used to change elements before bypass occurs.
6.How does temperature affect the performance of hydraulic fluid?Application
Temperature affects the viscosity of hydraulic fluid, which in turn affects the performance of the hydraulic system. At high temperatures, the fluid becomes less viscous, which can lead to increased leakage and reduced efficiency. At low temperatures, the fluid becomes more viscous, which can cause sluggish operation and increased energy consumption. Maintaining the correct temperature range is crucial for optimal system performance.
7.Calculate the force exerted by a hydraulic cylinder with a piston diameter of 0.1 m and a system pressure of 5 MPa.Numerical
Piston area A = π × 0.1²/4 = 7.854 × 10⁻³ m². On extension, F = p × A = 5 × 10⁶ × 7.854 × 10⁻³ ≈ 39 270 N, about 39.3 kN, neglecting seal friction and back pressure on the rod side. On retraction the force is lower because only the annulus area (piston minus rod) is pressurised.
8.What are the advantages of using hydraulic systems over pneumatic systems?Application
Hydraulic systems offer several advantages over pneumatic systems, including the ability to generate higher forces due to the incompressibility of hydraulic fluid. They provide precise control and smooth operation, which is essential for applications requiring fine adjustments. Hydraulic systems are also more efficient in terms of energy usage and can operate at higher pressures, making them suitable for heavy-duty applications.
9.What causes cavitation and aeration in hydraulic pumps, and how are they prevented?Concept
Cavitation occurs when inlet pressure drops so low that oil vapour and dissolved air come out of solution; the bubbles then collapse violently at the pressure side, eroding port plates and gears and causing a high-pitched whine. Causes include undersized or long suction lines, a clogged strainer, cold viscous oil, excessive pump speed or a pump mounted too high above the oil level. Aeration is air drawn in through leaking suction joints or a vortex in a low reservoir, giving foamy oil and erratic actuators. Prevention: short, large-bore suction lines with low velocity (about 1 m/s or less), flooded inlets, correct oil viscosity and temperature, sound fittings and adequate reservoir level and baffling.
10.A hydraulic motor requires 0.02 m³/s but a fixed-displacement pump delivers 0.025 m³/s at a relief setting of 10 MPa. What happens to the excess flow, and what does it cost?Numerical
The excess is 0.025 − 0.02 = 0.005 m³/s. With a fixed pump, pressure rises until the relief valve opens and dumps this flow to tank at the relief pressure, so the wasted power is p × ΔQ = 10 × 10⁶ × 0.005 = 50 kW, all turned into heat in the oil. That is why such circuits need coolers, and why variable-displacement, pressure-compensated or load-sensing pumps are used to deliver only the flow required.
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