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04-BS-7 · December 2017

Question 9 of 13: Wind Loading on a Multi-Flue Chimney

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 9: Wind Loading on a Multi-Flue Chimney (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
Chimney diameter / height20 m / 275 m
Wind velocity, $V$160 km/h = 44.44 m/s
Air density / viscosity (Constants)1.19 kg/m³ / 1.8×10⁻⁵ Ns/m²
H = 275 m D = 20 m V = 44.44 m/s
Uniform 160 km/h wind acting over the full 275 m windshield height, treated as an infinite cylinder for the drag coefficient.

Find. (a) $C_D$; (b) total horizontal wind force on the chimney.

Approach. Compute the Reynolds number for the cylinder in cross-flow, read $C_D$ from the attached drag chart, then apply the total-drag formula over the projected (frontal) area.

  1. (a) Reynolds number. $$Re=\frac{DV\rho}{\mu}=\frac{(20)(44.44)(1.19)}{1.8\times10^{-5}}=\boxed{5.9\times10^{7}}$$ This is beyond the attached chart's plotted range ($10^{-1}$ to $10^6$); the infinite-circular-cylinder curve bottoms out at about 0.3 just past the drag crisis (about $5\times10^5$) and has only climbed to about 0.33 at $10^6$, the chart's right-hand limit, so the chart reading is $C_D\approx\boxed{0.3}$ (see check note above the questions). Full-scale data at $Re\sim10^7$ put a smooth cylinder nearer 0.5–0.7, so a design check would treat this chart value as a lower bound.
  2. (b) Projected (frontal) area. $$A=D\times H=(20)(275)=\boxed{5500\ \text{m}^2}$$
  3. (b) Total horizontal drag force. $$F_D=C_D\,\tfrac12\rho V^2A=(0.3)\left(\tfrac12\right)(1.19)(44.44)^2(5500)$$ $$F_D=(0.3)(1175.1)(5500)=\boxed{1.94\times10^{6}\ \text{N} \approx 1.94\ \text{MN}}$$
QuantityValue
Reynolds number5.9 × 10⁷
Drag coefficient, $C_D$0.3
Frontal area5500 m²
Total horizontal force, $F_D$1.94 MN