Question 13 of 13: Barge Passing Over an Aqueduct — Change in Pillar Load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2017-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 capillarity (Ch. 2), integral analysis and buoyancy (Ch. 3), viscous flow in ducts, pipe friction and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), open-channel flow and the hydraulic jump (Ch. 10).
Check — assumptions used across this paper:
Q2's figure shows the vertical leaf of the gate rising above the free surface, so water acts over the full depth x = 2.0 m; both "0.3 m" dimensions locate the centre of gravity (0.3 m right of the vertical leaf, 0.3 m above the arm). Reservoir water also fills the space beneath the 1.2 m arm (the vertical lines below O are dimension extension lines, not a wall), so the arm carries uplift at pressure ρgx, and the arm tip bears up against a lip on the spillway crest — the gate can only open by rotating clockwise, vertical leaf toward the spillway.
Q4's flow coefficient K ≈ 0.69 is read from the attached VDI chart on the Do/D1 = 0.50 curve (third from the top, the one that levels off at 0.62) at the approach Reynolds number R ≈ 2550. The printed curve gives 0.695 at R = 2000 and 0.684 at R = 3000; a reading of 0.68–0.70 moves the answer by only ±0.3 kPa.
Q7 and Q8 friction factors are obtained from the Colebrook–White/Haaland equation the Moody chart itself plots (smooth-wall case, as both problems specify or imply smooth surfaces).
Q9's sphere drag coefficient vs. Reynolds number is obtained from the Schiller–Naumann correlation $C_D=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)$, which reproduces the published sphere-drag curve (the same curve reprinted in the attachment) to within a few percent for $Re<1000$ — used here because the resulting Reynolds number (≈19–20) is well above the range where the simple Stokes'-law formula ($C_D=24/Re$) alone is valid.
Question 13: Barge Passing Over an Aqueduct — Change in Pillar Load (5 marks)
Given. Barge $15\ \text{m}\times3\ \text{m}$, draught $1.2\ \text{m}$; canal $5\ \text{m}$ wide, $2\ \text{m}$ deep, carried across a valley on an aqueduct supported by pillars; the canal is a continuous, open channel (not a sealed tank) whose free-surface elevation is maintained by the rest of the waterway.
Canal carried on pillars across a valley: the barge's own weight substitutes for the weight of water it displaces.
Find. The change in compressive force on the aqueduct pillars while the barge is passing over them.
Approach. Compute the barge's displaced-water volume and weight directly, then reason from Archimedes' principle and the open (level-controlled) nature of the canal to determine the net change in load on the pillars.
Displaced volume and equivalent water weight. $V_{disp}=15\times3\times1.2=54\ \text{m}^3$:
$$W_{disp}=\rho g V_{disp}=1000\times9.81\times54=\boxed{529{,}740\ \text{N} = 529.7\ \text{kN}}$$
By Archimedes' principle this is exactly the barge's own weight (hull + cargo), since it floats in equilibrium.
Water removed from the aqueduct span. The canal's free-surface elevation is fixed by the continuous, open waterway beyond the aqueduct (not by a sealed local volume), so the $54\ \text{m}^3$ of water the barge's hull displaces simply flows off the aqueduct span into the adjoining canal reaches — it does not raise the local level. The weight of water resting on the aqueduct therefore decreases by exactly $W_{disp}=529.7\ \text{kN}$.
Net change in load on the pillars. The barge's own weight ($+529.7\ \text{kN}$, now resting on the aqueduct via water pressure) exactly cancels the reduction in water weight ($-529.7\ \text{kN}$) caused by the displaced volume flowing away:
$$\Delta F = W_{barge}-W_{disp} = 529.7\ \text{kN}-529.7\ \text{kN}=\boxed{0\ \text{kN (no change)}}$$
Quantity
Value
Displaced volume
54 m³
Weight of displaced water (= barge weight)
529.7 kN
Net change in compressive force on pillars
0 kN (no change)
This is a direct demonstration of Archimedes' principle in an open system: a floating body always displaces exactly its own weight of fluid, regardless of the shape of the vessel it floats in. Because the canal is open and its level is externally maintained, the "missing" water simply relocates to the rest of the waterway rather than piling up on the aqueduct — so the aqueduct's pillars see no net change in load whether the barge is present or not. (Had the aqueduct instead been a sealed, fixed-volume tank with no outflow, the barge's full weight would have added directly to the load, since none of the displaced water could escape to compensate.)
