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04-BS-7 · May 2018

Question 7 of 13: Jet Engine Forward and Reverse Thrust

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

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2018-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & 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: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and manometry (Ch. 2), Bernoulli and the energy equation (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), potential flow and the Magnus effect (Ch. 8), open-channel flow and the hydraulic jump (Ch. 10), turbomachinery and jet propulsion (Ch. 11).

Check — assumptions used across this paper:
  • Q1's manometer chain is read off the extraction as a two-stage water–mercury–glycerine–(air)–glycerine–mercury system. The enclosed air pocket between the two glycerine columns is treated as weightless (uniform pressure), so only the one described open end is needed to close the hydrostatic chain back to pipe P; the second "opening" is not load-bearing for this calculation.
  • Q5's wave/hydraulic-jump analysis takes the depth "in front of the wave" (0.15 m, undisturbed, at rest) as the upstream state and "behind the wave" (0.75 m) as the downstream state, per the question's own prose (the raw figure-label ordering in the extraction is a reconstruction and is not used to override the stated text). The classic hydraulic-jump head-loss formula is applied to the given depths, and the swept flow rate uses the measured wave celerity directly — a standard engineering estimate, not a fully momentum-self-consistent bore solution.
  • Q6's air properties are taken at 20°C (domestic ambient, ρ=1.19 kg/m³) since no duct-air temperature is stated.
  • Q7(c)'s "discharged at right angles to the initial direction (20° becomes 0°)" is read as: the reverser's exhaust jet is normally angled 20° forward of the fully-radial (right-angle) direction; part (c) removes that forward lean entirely, leaving a purely radial (90° to the engine axis) discharge with zero axial velocity component.
  • Q8's terminal velocity and Q9's cable drag coefficient are obtained from the Reynolds-number relations the attached charts themselves plot (Morrison's sphere-drag correlation for Q8; the flat subcritical Cd≈1.2 plateau of the smooth-cylinder curve for Q9, since Re≈3×104 falls solidly within it).

Question 7: Jet Engine Forward and Reverse 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.

QuantityValue
Mass flow rate, $\dot M$143 kg/s
Inlet (flight) velocity, $V_{in}$220 m/s
Exhaust velocity, $V_{ex}$650 m/s
Reverser forward-lean angle (from purely radial)20°
engine / fuselage V_in=220 V_ex=650 (normal) reverse jet, 20° off radial radial (90°)
Normal operation: exhaust leaves rearward at 650 m/s. Reverse thrust: cascades redirect the exhaust to 20° forward of the purely radial (right-angle) direction, giving it a small forward-pointing axial component.

Find. (a) Forward thrust, normal operation; (b) reverse thrust, reverser deployed; (c) whether reverse thrust persists with a purely radial (right-angle) discharge.

Approach. Apply the linear-momentum (thrust) equation $F=\dot M(V_{ex,\text{axial}}-V_{in})$ along the flight axis, using the appropriate axial component of the exhaust velocity for each configuration.

  1. (a) Forward thrust, normal operation. Exhaust leaves fully rearward: $$F_{fwd}=\dot M(V_{ex}-V_{in})=143(650-220)=\boxed{61{,}490\ \text{N}\approx61.5\ \text{kN, forward}}$$
  2. (b) Reverse thrust, reverser deployed (20° forward of radial). The exhaust's axial component now points forward (toward the nose) with magnitude $V_{ex}\sin(20^\circ)$, opposing the inlet's forward-facing intake momentum: $$F_{rev}=\dot M\big(V_{ex}\sin20^\circ+V_{in}\big)=143\big(650(0.342)+220\big)=143(222.3+220)=\boxed{63{,}250\ \text{N}\approx63.3\ \text{kN, rearward (braking)}}$$
  3. (c) Purely radial discharge (20°→0° forward lean). With the exhaust leaving exactly at right angles to the engine axis, its axial component is zero — but the engine still ingests air at the flight speed through the inlet: $$F_{(c)}=\dot M(0-V_{in})=143(0-220)=\boxed{-31{,}460\ \text{N}\approx31.5\ \text{kN, still rearward}}$$ Yes, reverse (braking) thrust would still exist, though reduced to about half of (b): the intake alone continuously removes forward momentum from the air stream (ram drag) with no compensating rearward push from the exhaust, so a net retarding force remains even without any forward-directed exhaust component.
CaseForceDirection on aircraft
(a) Normal operation61.5 kNForward (thrust)
(b) Reverser, 20° forward of radial63.3 kNRearward (braking)
(c) Purely radial discharge31.5 kNRearward (still braking)