Question 6 of 13: Minimum Wind Velocity 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 — National Examination, 2013-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (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: Crowe, C.T., Elger, D.F. & Roberson, J.A., Engineering Fluid Mechanics (the exam's own Moody chart and drag-coefficient chart are reproduced from this text); Douglas, J.F., Gasiorek, J.M., Swaffield, J.A. & Jack, L.B., Fluid Mechanics; White, F.M., Fluid Mechanics.
Check — assumptions used across this paper:
Where a question does not restate an ambient temperature, air is taken at the ISA sea-level standard of 15°C, giving ρair = 1.21 kg/m³ (the Constants table's 15°C value) — used in Q5, Q6, Q7 and Q9.
Q8's Moody chart and Q9's sphere drag-coefficient chart are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 (the Moody chart is a graphical solution of Colebrook–White) and the Morrison (2013) curve-fit for sphere drag versus Reynolds number for Q9 (which reproduces the published "Sphere" curve to within a few percent over this Re range).
Q1, Q3 and Q11 depend on their figures; the relevant crop of the exam page is reproduced beside each, and every reading used is taken from it.
Question 6: Minimum Wind Velocity to Lift an Insulation Panel (5 marks)
Given. Panel $2.438\ \text{m}\times1.219\ \text{m}\times0.025\ \text{m}$, $\rho_{panel}=100\ \text{kg/m}^3$; grass reduces the under-panel velocity to stagnation ($V\approx0$); air density $\rho_{air}=1.21\ \text{kg/m}^3$.
Wind grazes the top face at full speed V while grass stagnates the flow underneath, creating a net upward (lift) pressure difference.
Find. Minimum wind velocity to lift the panel off the grass.
Approach. At lift-off, the net upward pressure force (Bernoulli difference between stagnation below and free-stream above) equals the panel's weight; solve for V.
Pressure difference at lift-off. Under the panel $V\approx0$ (stagnation), so $p_{under}-p_\infty=\tfrac12\rho_{air}V^2$; above the panel the wind still moves at V, so $p_{over}\approx p_\infty$. Net lift pressure $=\tfrac12\rho_{air}V^2$.
Force balance at the threshold. $\tfrac12\rho_{air}V^2 A = W$, with plan area $A=2.438\times1.219=2.972\ \text{m}^2$:
$$V = \sqrt{\frac{W}{0.5\,\rho_{air}A}} = \sqrt{\frac{72.9}{0.5\times1.21\times2.972}} = 6.37\ \text{m/s}$$
$$\boxed{V = 6.37\ \text{m/s} = 22.9\ \text{km/hr}}$$
