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04-BS-7 · December 2017

Question 13 of 13: Jet Force and Energy Transfer — Flat vs. Curved Plate

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

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2017-Dec. 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 flow measurement (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), turbomachinery and wind turbines (Ch. 11).

Check — assumptions used across this paper:
  • Q1's manometer is read from the printed figure as a benzene(hatched)–mercury(black)–carbon-tetrachloride(clear) chain: benzene fills pipe A up and over the first bend and down the U-tube's left arm to the benzene/mercury interface at the UPPER dimension line (2.0 m + 400 mm = 2.4 m above A); the mercury stands 400 mm lower in the right arm, at the mercury/CCl₄ interface on the LOWER line (2.0 m above A); CCl₄ then fills the rest of the run over the second bend down to B, 3.0 m below A. (The mercury is higher on the A side, so pA < pB.)
  • Q6's spillway/gate width is read from the drawing as the 8.76 m dimension (the question's own hint: "width of each gate is slightly greater than its height" – 8.76 m > 8.23 m); the closed gate's wetted height at F.S.L. is F.S.L. − Crest = 7.92 m (the gate's own 8.23 m height extends slightly above F.S.L., matching the paper's note that "the top of the gate is higher than F.S.L.").
  • Q7 and Q9's friction/drag coefficients come from the Colebrook–White equation and the plotted drag curve — the same relations the attached Moody and drag charts plot.
  • Q9's Reynolds number (≈5.9×107) is beyond the attached drag chart's plotted range (10−1 to 106); CD = 0.3 is taken from the right-hand end of the cylinder curve (minimum ≈0.3 past the drag crisis, ≈0.33 at 106) — the best available reading — and flagged here as an extrapolation.
  • Q8(c)'s fuel density (needed to convert a fuel mass into litres) is not stated in Q8 itself; the paper lists no fuel density on its Constants page, so the gasoline SG = 0.75 given in Q4 of this same paper is adopted (Q2's 0.72 would give 3.50 L/100 km instead of 3.36).

Question 13: Jet Force and Energy Transfer — Flat vs. Curved Plate (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.

FLAT PLATE — 90° deflection jet splits, leaves ⊥ to itself CURVED PLATE — 180° turn jet reverses, exits opposite in direction to entry
The flat plate deflects the jet through 90°; the curved (U-turn) plate reverses it through 180°.

The curved plate is subject to the greater force when stationary, and the curved plate also gives the best transfer of energy when moving in the jet's direction.

Force when stationary — momentum viewpoint. For a jet of density $\rho$, area $A$ and velocity $V$ striking a plate and being deflected through an angle $\theta$ from its original direction, the force on the plate in the original jet direction is

$$F=\rho Q V(1-\cos\theta),\qquad Q=AV$$

A large flat plate splits the jet symmetrically and the flow leaves running parallel to the plate surface — i.e. deflected through $\theta=90^\circ$, so $\cos\theta=0$ and $F_{flat}=\rho QV$. A curved (hemispherical, 180°) plate turns the jet completely around, $\theta=180^\circ$, $\cos\theta=-1$, giving $F_{curved}=\rho QV(1-(-1))=2\rho QV$. Since the flow rate and velocity are stated to be identical for both plates,

$$\frac{F_{curved}}{F_{flat}}=\frac{1-\cos180^\circ}{1-\cos90^\circ}=\frac{2}{1}=\boxed{2}$$

the curved plate feels exactly twice the force of the flat plate: reversing the jet's momentum entirely removes it (a change of $2V$ in the flow direction), while the flat plate only removes the momentum component along the original direction, leaving the deflected streams to carry the rest away sideways with no net effect on that component.

Energy transfer when moving — energy viewpoint. For a plate (or a row of vanes) moving with velocity $U$ in the jet direction, the power delivered to the plate is $P=F_{rel}\,U=\rho A(V-U)^2(1-\cos\theta)\,U$, which is maximized at $U=V/3$ for any deflection angle, giving

$$P_{max}=\frac{4}{27}\rho A V^3(1-\cos\theta)$$

Because $(1-\cos\theta)$ is the same factor that controlled the stationary force, it again doubles going from the flat plate ($\theta=90^\circ$, factor 1) to the curved plate ($\theta=180^\circ$, factor 2). At the same operating speed $U$, the curved plate therefore extracts twice the mechanical power from the jet — and hence converts twice the fraction of the jet's kinetic energy — that the flat plate does. Physically, a vane that reverses the relative flow entirely leaves the water with the least possible kinetic energy in the plate's frame, which is exactly the condition for maximum energy transfer from jet to plate (this is the working principle behind Pelton-wheel buckets, which are curved close to 180° for this reason).

QuantityFlat plate (θ=90°)Curved plate (θ=180°)
Stationary force factor, $(1-\cos\theta)$12 (greater)
Max moving power, relative1×2× (best transfer)
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