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04-BS-7 · May 2013

Question 7 of 13: Propeller Thrust

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

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2013-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.

Reference texts: Crowe, C.T., Elger, D.F. & Roberson, J.A., Engineering Fluid Mechanics (the exam's own Moody chart and drag-coefficient chart are reproduced from this text); Douglas, J.F., Gasiorek, J.M., Swaffield, J.A. & Jack, L.B., Fluid Mechanics; White, F.M., Fluid Mechanics.

Check — assumptions used across this paper:
  • Where a question does not restate an ambient temperature, air is taken at the ISA sea-level standard of 15°C, giving ρair = 1.21 kg/m³ (the Constants table's 15°C value) — used in Q5, Q6, Q7 and Q9.
  • Q8's Moody chart and Q9's sphere drag-coefficient chart are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 (the Moody chart is a graphical solution of Colebrook–White) and the Morrison (2013) curve-fit for sphere drag versus Reynolds number for Q9 (which reproduces the published "Sphere" curve to within a few percent over this Re range).
  • Q1, Q3 and Q11 depend on their figures; the relevant crop of the exam page is reproduced beside each, and every reading used is taken from it.

Question 7: Propeller Thrust (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. Propeller diameter $D=2\ \text{m}$; aircraft (upstream) velocity $V_1=500\ \text{km/hr}=138.9\ \text{m/s}$; downstream jet velocity $V_2=650\ \text{km/hr}=180.6\ \text{m/s}$; air density $\rho_{air}=1.21\ \text{kg/m}^3$.

disk V₁=500 km/hr V₂=650 km/hr V_disk = (V₁+V₂)/2
Actuator-disk streamtube: the disk velocity is the average of the far-upstream and far-downstream velocities.

Find. Thrust developed by the propeller.

Approach. Actuator disk (momentum) theory: the velocity through the disk is the mean of the upstream and fully-developed downstream velocities; the mass flow rate through the disk times the velocity change gives the thrust (aircraft propulsion equation supplied on the reference sheet).

  1. Velocity through the disk. $V_{disk}=\tfrac12(V_1+V_2) = \tfrac12(138.9+180.6) = 159.7\ \text{m/s}$.
  2. Mass flow rate through the disk. $A=\pi D^2/4 = \pi(2)^2/4=3.1416\ \text{m}^2$: $$\dot M = \rho_{air} A V_{disk} = 1.21\times3.1416\times159.7 = 607.2\ \text{kg/s}$$
  3. Thrust. $F_{thrust}=\dot M(V_2-V_1) = 607.2\times(180.6-138.9)=607.2\times41.67$ $$\boxed{F_{thrust} = 25{,}300\ \text{N} = 25.3\ \text{kN}}$$
QuantityValue
Velocity at the disk159.7 m/s
Mass flow rate607.2 kg/s
Thrust25.3 kN