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04-BS-7 · Undated paper

Question 4 of 13: Minimum Wind Speed to Lift an Insulation Panel

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

Notes on this paper

04-BS-7 Mechanics of Fluids — undated sitting, identified as May 2019 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs “do seven”; Section B (Analytical) offers 4 questions and instructs “do three.” A complete paper is any 10 of the 13, each worth 5 marks. Every question is answered below (13 of 13) so the set can be used as a full study resource. Constants used throughout (from the paper's own Constants page, p.12): g = 9.81 m/s², patm = 100 kPa, ρwater = 1000 kg/m³, SGbenzene = 0.90, SGmercury = 13.56, SGcarbon tetrachloride = 1.59, ρair = 1.19 kg/m³ (20°C), μwater = 1.0×10⁻³ N·s/m², μair = 1.8×10⁻⁵ N·s/m².

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), hydrostatic forces on plane surfaces (Ch. 2), the Bernoulli/continuity pair and orifice flow (Ch. 3), the linear-momentum equation for moving vanes (Ch. 3), pipe friction and the Moody/Colebrook relation (Ch. 6), boundary layers and drag (Ch. 7), capillary rise (Ch. 1), high-lift devices and aircraft wing aerodynamics (J. D. Anderson, Fundamentals of Aerodynamics, Ch. 4–5).

Question 4: Minimum Wind Speed to Lift an Insulation Panel (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
Panel dimensions2.438 m × 1.219 m × 25 mm
Panel density100 kg/m³
Air condition20°C, ρ = 1.19 kg/m³
Under-panel conditiongrass brings air to stagnation (V ≈ 0)

Find. The minimum free-stream wind speed (km/hr) that lifts the panel off the grass.

wind, V flows freely above panel stagnant air trapped below (grass)
Free-stream air passes over the top of the panel at speed V while the grass beneath brings the trapped air to stagnation pressure; the resulting pressure difference must exceed the panel's weight per unit area.

Approach. Bernoulli between the free stream (above the panel) and the stagnation point (below the panel, brought to rest by the grass) gives the pressure difference across the panel; set that difference, times the panel area, equal to the panel's weight.

  1. Panel weight. Volume V = 2.438(1.219)(0.025) = 0.07430 m³: $$ W = \rho_{panel}\,V\,g = (100)(0.07430)(9.81) \approx 72.9\ \text{N} $$
  2. Pressure difference needed to lift it. Plan area A = 2.438(1.219) = 2.972 m²: $$ \Delta p = \frac{W}{A} = \frac{72.9}{2.972} \approx 24.5\ \text{Pa} $$
  3. Bernoulli: relate Δp to wind speed. The under-panel air is stagnant (V=0), so the entire dynamic pressure of the free stream above the panel becomes the lifting Δp: $$ \Delta p = \tfrac12 \rho_{air} V^2 \quad\Rightarrow\quad V = \sqrt{\frac{2\Delta p}{\rho_{air}}} = \sqrt{\frac{2(24.5)}{1.19}} $$ $$ V \approx 6.42\ \text{m/s} $$
  4. Convert to km/hr. $$ \boxed{V \approx 6.42 \times 3.6 \approx 23.1\ \text{km/hr}} $$
QuantityResult
Panel weight72.9 N
Minimum lift-off wind speed6.42 m/s = 23.1 km/hr