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Power transmission in aircraft propulsion

Short answer

The design requires minimum weight, high reliability, forced lubrication, and ability to carry thrust loads. A reduction gearbox adapts high turbine rpm to low propeller rpm for efficiency (tip speed below sonic). Epicyclic trains are chosen for compactness, power sharing among planets, and balanced forces. Ratio: 12000/1500 = 8, so ring/sun teeth = 7. Propeller torque = 9549 N·m.

Explain the design requirements of power transmission in an aircraft with a turboprop or piston engine. Why is a reduction gearbox used, and why are epicyclic (planetary) gear trains preferred? A turboprop engine delivers 1500 kW at 12000 rpm, and the propeller must run at 1500 rpm using a simple planetary stage with the sun as input, the ring fixed and the carrier as output. Find the required ratio of ring to sun teeth and the torque at the propeller shaft (neglect losses).

Requirements

  • Minimum weight with high power: use of alloy steels, case-hardened ground gears, thin webs and light casings.
  • High reliability and safety: redundancy, high factor on fatigue life, vibration and condition monitoring; the gearbox is flight-critical.
  • Operation at high speed and temperature with forced oil lubrication and cooling.
  • Compact size and the correct shaft geometry, with low noise and vibration.
  • Ability to carry the propeller thrust and gyroscopic loads.

Why a reduction gearbox

A gas turbine runs efficiently at 10000 to 30000 rpm, but a large propeller must turn slowly (about 1000 to 2000 rpm) so that blade tip speed stays below sonic speed for efficiency and low noise. A reduction gearbox gives this speed ratio.

Why epicyclic

Power is shared among several planet gears, so each tooth carries less load; coaxial input and output; compact and light; balanced forces on the sun and carrier give lower bearing loads.

Numerical

Sun input, ring fixed, carrier output: speed ratio =1+zrzs= 1 + \dfrac{z_r}{z_s}.

NinNout=120001500=8=1+zrzs  ⇒  zrzs=7\frac{N_{in}}{N_{out}} = \frac{12000}{1500} = 8 = 1 + \frac{z_r}{z_s} \;\Rightarrow\; \frac{z_r}{z_s} = 7

For example zs=18z_s = 18, zr=126z_r = 126 (planet teeth =(126−18)/2=54= (126-18)/2 = 54).

Input torque Tin=Pω=1 500 00012000×2π/60=1193.7 N mT_{in} = \dfrac{P}{\omega} = \dfrac{1\,500\,000}{12000\times2\pi/60} = 1193.7\ \text{N m}.

Output torque (no loss) =Tin×8=9549 N m= T_{in}\times 8 = 9549\ \text{N m}.

Answer: zr/zs=7z_r/z_s = 7; propeller torque about 9550 N m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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