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Aircraft systems: hydraulics, fuel and landing gear

Aeronautical / Aerospace Engineering (B.E./B.Tech) · Aircraft systems and maintenance · free study notes for exams, placements and interviews

In short

Aircraft hydraulic, fuel and landing gear systems are the "muscle, blood and legs" of an aeroplane: hydraulics transmit power to move flight controls, gear, brakes and flaps; the fuel system stores and delivers fuel to the engines at the right pressure and in an order that keeps the centre of gravity (CG) in limits; the landing gear supports the aircraft on ground and absorbs landing energy. These three are mechanical-system subjects, so they are examined through fluid power, thermodynamics of fuel, energy absorption and structural load paths, and they dominate aircraft-systems papers in B.E./B.Tech aerospace and aeronautical programmes as well as AME (DGCA) Paper on Aircraft Systems. In interviews at HAL, Boeing India, Airbus India, Collins Aerospace, Safran, Liebherr Aerospace, Tata Advanced Systems, Air India Engineering Services, AIESL/IndiGo/Air India maintenance and Indian Air Force technical branch, candidates are asked why hydraulics instead of electrics, what an oleo strut does, and how fuel is sequenced. Design, airworthiness certification (DGCA CAR Series/CAR-21, CAR-145, and FAR/CS-25 as adopted) and line maintenance roles all use this knowledge daily.

Explained

Hydraulic system fundamentals

Aircraft fuel systems

Landing gear: types and construction

Integration, failures and maintenance practice

Key points

Formulas and examples

Pascal / force multiplication: P = F / A ; F₂ = F₁ × (A₂ / A₁) ; x₁A₁ = x₂A₂
P = pressure (Pa or N/m²; 1 bar = 10⁵ Pa, 1 psi = 6,895 Pa), F = force (N), A = piston area (m²), x = stroke (m)
Actuator force (double-acting): F_extend = P·A_p − P_ret·(A_p − A_r) ; F_retract = P·(A_p − A_r)
A_p = π D²/4 (full bore area), A_r = π d²/4 (rod area), D = bore dia., d = rod dia. (m)
Hydraulic power: P_hyd = ΔP × Q (W, with ΔP in Pa and Q in m³/s) ; shaft power = P_hyd / η_overall
Pump flow: Q = V_d × N × η_vol , V_d = displacement per revolution (m³/rev), N = rev/s
Accumulator (isothermal, Boyle's law): P₁V₁ = P₂V₂ ; usable fluid ΔV = V₁(P₁/P₃ − P₁/P₂), P₂ = system pressure, P₃ = minimum useful pressure
Adiabatic (fast discharge): P₁V₁n = P₂V₂n with n ≈ 1.4 for nitrogen
Fuel mass = volume × density: m = V × ρ ; endurance E = usable fuel mass / fuel flow ; range ≈ E × V_TAS
Energy = m × CV ; CV(Jet A-1) ≈ 43.2 MJ/kg ; ρ ≈ 0.80 kg/L nominal
Landing energy absorbed by gear: E = ½ m v² + (W − L)·s and E = η × F_max × s
v = sink rate (m/s), s = shock-strut stroke (m), η = strut efficiency, F_max = peak strut load (N), L = wing lift
Load factor at landing: n = F_max / W ; Ultimate load = 1.5 × limit load
Tyre/pavement: contact pressure ≈ tyre inflation pressure ; load per wheel = static reaction / number of wheels
Static gear reactions (tricycle, CG at distance a aft of nose wheel, wheelbase b):
R_main = W × a / b ; R_nose = W × (b − a) / b

Example 1 — Actuator sizing for landing gear retraction. A gear actuator must supply 60 kN to retract the leg. System pressure is 3,000 psi.

Example 2 — Accumulator usable volume for emergency braking. A 4-litre piston accumulator is pre-charged with nitrogen to 1,500 psi; system pressure is 3,000 psi. Brakes remain usable down to 1,800 psi. Assume isothermal behaviour.

Example 3 — Fuel mass, endurance and tankering. An A320-class aircraft is uplifted 14,500 litres of Jet A-1 at a measured density of 0.795 kg/L. Cruise fuel flow is 2,400 kg/h total.

Example 4 — Oleo strut stroke and peak load. An aircraft of 22,000 kg lands at a sink rate of 3.05 m/s (10 ft/s). Two main gears absorb the energy, strut efficiency 0.85, and wing lift equals the weight during the impact (so the (W − L)·s term is zero).

Example 5 — Static gear reactions and wheel load. Aircraft weight 78,000 kg; wheelbase (nose to main axle) 12.6 m; CG 11.5 m aft of the nose wheel. Four main wheels, two nose wheels.

Shock absorber typeEfficiencyRemarks
Steel coil spring≈ 50%Stores, barely damps; bounces
Rubber / bungee cord≈ 60%Light aircraft, cheap, ages
Steel leaf / spring strut≈ 50–60%Cessna-type; no moving parts, no servicing
Air (pneumatic) only≈ 45–50%Poor damping
Oleo-pneumatic≈ 75–90%Damps and springs; needs oil + nitrogen servicing
FeatureHydraulicPneumaticElectric (EMA)
MediumIncompressible oilCompressible airNone
Stiffness / position holdingVery highPoor (spongy)High
Force-to-weightHighestMediumGood, improving
Leak consequenceFluid loss, mess, fire riskHarmless lossNone
Return line neededYesNo (vent to atmosphere)No
Typical useFlight controls, gear, brakesEmergency gear blow-down, some brakesSpoilers, brakes (B787 EMA), trim
Tank typeWhere usedAdvantageLimitation
Integral (wet wing)All transport aircraftLightest, maximum volumeSealant leaks, hard to repair
Rigid removableLight aircraftEasy to remove/inspectHeavy, wastes volume
Bladder cellHelicopters, fightersCrashworthy, self-sealing optionsAges, chafes, costly
Auxiliary / trim / drop tankLong range, militaryExtra range, CG controlDrag, complexity, transfer system

Where it is used

Interview questions

Common mistakes

Related topics

Avionics basics · Aircraft maintenance and DGCA licensing · Aircraft design process · Drones and UAV basics · Fluid mechanics and fluid power · Aircraft structures and materials

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