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04-BS-7 · December 2016

Question 6 of 13: Wind-Induced Suction Ventilating an Outhouse Pit

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 6: Wind-Induced Suction Ventilating an Outhouse Pit (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
Wind speed V25 km/h = 6.944 m/s
Air temperature20°C
Air density ρ (20°C)1.19 kg/m³

Find. The differential pressure Δp between the pit interior and the outside atmosphere that draws air out of the pit.

Approach. Wind blowing across the open top of the vent pipe accelerates from calm (near-zero) conditions at the sheltered pit to the full wind speed at the pipe's exposed outlet; Bernoulli's equation converts this speed-up directly into a pressure drop at the outlet, which is what draws (sucks) air continuously out of the pit.

  1. Bernoulli between the still air in the pit and the wind-swept vent outlet (same elevation). $$p_{pit} + \tfrac12\rho(0)^2 = p_{outlet} + \tfrac12\rho V^2 \quad\Rightarrow\quad p_{pit}-p_{outlet} = \tfrac12\rho V^2$$
  2. Evaluate the dynamic pressure of the wind. $$\Delta p = \tfrac12(1.19)(6.944)^2 = \boxed{28.7\ \text{Pa}}$$ The vent outlet runs below atmospheric by this amount, so air is continuously drawn from the (relatively higher-pressure, still) pit up through the vent pipe and out.
QuantityResult
Wind speed6.944 m/s
Differential pressure (pit above outlet)28.7 Pa