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04-BS-7 · December 2016

Question 4 of 13: Wind-Induced Stagnation and Side Pressure on a Chimney

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

Notes on this paper

04-BS-7 Mechanics of Fluids — December 2016 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Graphical and Analytical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource. Constants used throughout (from the paper's own Constants page): g = 9.81 m/s², patm = 100 kPa, ρwater = 1000 kg/m³, ρconcrete = 2400 kg/m³, ρair = 1.19 kg/m³ (20°C) / 1.21 kg/m³ (15°C), μwater = 1.0×10⁻³ N·s/m², μair = 1.8×10⁻⁵ N·s/m², Rair = 287 J/kg·K.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and buoyancy/stability (Ch. 2), Bernoulli and control-volume momentum (Ch. 3), potential (inviscid) flow past a cylinder (Ch. 8), pipe friction and the Moody/Colebrook relation (Ch. 6), open-channel flow (Ch. 10), drag on immersed bodies and Stokes' law (Ch. 7); B. R. Munson et al., Fundamentals of Fluid Mechanics — actuator-disk propeller theory and variable-area tank draining.

Question 4: Wind-Induced Stagnation and Side Pressure on a 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 D20 m
Chimney height275 m
Free-stream wind velocity V0100 km/h = 27.78 m/s
Surface velocity relationV = 2V0 sinθ
Air density ρ (assumed 15°C, per Constants page)1.21 kg/m³

Find. V and gauge pressure at the stagnation point (θ=0°) and at the side (θ=90°).

Check: the paper does not restate an air temperature for this question; the Constants page gives ρair at both 15°C and 20°C, so 15°C (1.21 kg/m³) is assumed, consistent with the ambient conditions used elsewhere in the paper (Question 1). The chimney's height and 3-D shape do not affect this 2-D cross-flow result.

Approach. Evaluate the given surface-velocity relation at θ=0° and θ=90°, then apply the Bernoulli equation between the undisturbed free stream and each surface point to get the gauge pressure.

  1. Surface velocities. $$V_1 = 2V_0\sin(0^\circ) = \boxed{0\ \text{m/s (stagnation)}}, \qquad V_2 = 2V_0\sin(90^\circ) = 2(27.78) = \boxed{55.56\ \text{m/s}}$$
  2. Bernoulli between the free stream and each surface point (same elevation, gauge pressure referenced to the undisturbed atmosphere where p=0, V=V0). $$p + \tfrac12\rho V^2 = 0 + \tfrac12\rho V_0^2 \quad\Rightarrow\quad p = \tfrac12\rho\left(V_0^2 - V^2\right)$$
  3. Stagnation point (front). $$p_1 = \tfrac12(1.21)(27.78^2 - 0^2) = \boxed{466.8\ \text{Pa (gauge)}}$$
  4. Side point (90°). $$p_2 = \tfrac12(1.21)\left(27.78^2 - 55.56^2\right) = \boxed{-1400.5\ \text{Pa (gauge)}}$$
QuantityResult
Velocity at stagnation point (front)0 m/s
Gauge pressure at stagnation point+466.8 Pa
Velocity at side (90°)55.56 m/s
Gauge pressure at side (90°)−1400.5 Pa (suction)