Question 6 of 13: Island Bend Dam — Spillway Discharge and Gate Force
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 6: Island Bend Dam — Spillway Discharge and Gate Force (5 marks)
Section through one spillway bay: F.S.L. sits 7.92 m above the crest; the closed radial gate (8.76 m wide × 8.23 m tall) spans crest to just above F.S.L.
Find. (a) Discharge Q over the crest with one gate lifted clear; (b) horizontal force on one closed gate at F.S.L.
Approach. (a) Treat the open bay as a rectangular sharp-crested weir of length equal to the gate width, head = F.S.L. − crest. (b) Treat the closed gate as a submerged vertical plane of that same width, wetted from the crest up to F.S.L., and find the resultant horizontal hydrostatic force.
Head over the crest.
$$H=\text{F.S.L.}-\text{Crest}=1185.67-1177.75=\boxed{7.92\ \text{m}}$$
(a) Weir discharge, one bay (width $L=8.76$ m, the dimension "slightly greater than" the 8.23 m gate height).
$$Q=C_d\left(\tfrac23\right)\sqrt{2g}\,L\,H^{3/2}=0.80\left(\tfrac23\right)\sqrt{19.62}\,(8.76)(7.92)^{1.5}=0.80(0.667)(4.429)(8.76)(22.29)$$
$$Q=\boxed{461\ \text{m}^3/\text{s}}$$
(b) Wetted height of the closed gate at F.S.L. The gate seats on the crest and, per the paper's note, its top (8.23 m above the crest) is slightly above F.S.L., so the water surface — not the gate top — sets the wetted height:
$$h=\text{F.S.L.}-\text{Crest}=7.92\ \text{m (same value as } H\text{)}$$
(b) Horizontal force on one gate. The horizontal component of hydrostatic force on a curved gate equals the force on its vertical projection — a submerged rectangle $8.76\ \text{m}\times7.92\ \text{m}$, centroid depth $y_c=h/2$:
$$F_H=\rho g\,y_c\,A=(1000)(9.81)\left(\frac{7.92}{2}\right)(8.76\times7.92)=(9810)(3.96)(69.38)$$
$$F_H=\boxed{2{,}695{,}000\ \text{N} \approx 2.70\ \text{MN}}$$
