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04-BS-7 · December 2016

Question 5 of 13: Propeller Thrust via Actuator-Disk (Momentum) Theory

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — December 2016 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Graphical and Analytical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource. Constants used throughout (from the paper's own Constants page): g = 9.81 m/s², patm = 100 kPa, ρwater = 1000 kg/m³, ρconcrete = 2400 kg/m³, ρair = 1.19 kg/m³ (20°C) / 1.21 kg/m³ (15°C), μwater = 1.0×10⁻³ N·s/m², μair = 1.8×10⁻⁵ N·s/m², Rair = 287 J/kg·K.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and buoyancy/stability (Ch. 2), Bernoulli and control-volume momentum (Ch. 3), potential (inviscid) flow past a cylinder (Ch. 8), pipe friction and the Moody/Colebrook relation (Ch. 6), open-channel flow (Ch. 10), drag on immersed bodies and Stokes' law (Ch. 7); B. R. Munson et al., Fundamentals of Fluid Mechanics — actuator-disk propeller theory and variable-area tank draining.

Question 5: Propeller Thrust via Actuator-Disk (Momentum) Theory (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Propeller diameter D2 m
Aircraft (upstream) velocity V1500 km/h = 138.9 m/s
Far-downstream slipstream velocity V2650 km/h = 180.6 m/s
Air density ρ (assumed 15°C)1.21 kg/m³

Find. The thrust T developed by the propeller.

Check: no ambient air density is stated for this question; 1.21 kg/m³ (15°C, per the Constants page) is assumed, consistent with Question 4.

Approach. Actuator-disk (Rankine–Froude) momentum theory: the velocity actually through the propeller disk is the average of the far-upstream and far-downstream velocities, and thrust equals mass flow rate through the disk times the net velocity change.

  1. Disk area and velocity at the disk. $$A = \frac{\pi D^2}{4} = \frac{\pi(2)^2}{4} = 3.1416\ \text{m}^2, \qquad V_d = \frac{V_1+V_2}{2} = \frac{138.9+180.6}{2} = \boxed{159.7\ \text{m/s}}$$
  2. Mass flow rate through the disk. $$\dot m = \rho A V_d = 1.21(3.1416)(159.7) = \boxed{607.2\ \text{kg/s}}$$
  3. Thrust from the momentum change. $$T = \dot m (V_2-V_1) = 607.2(180.6-138.9) = 607.2(41.7) = \boxed{25{,}298\ \text{N} \approx 25.3\ \text{kN}}$$
QuantityResult
Disk area A3.14 m²
Velocity at the disk Vd159.7 m/s
Mass flow rate607.2 kg/s
Thrust T25.3 kN