Question 5 of 13: Ideal Flow Around a Cylinder — Chimney in Wind
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 5: Ideal Flow Around a Cylinder — Chimney in Wind (5 marks)
Given. $V_0=100\ \text{km/hr}=27.78\ \text{m/s}$; $V_t=2V_0\sin\theta$; air density $\rho_{air}=1.21\ \text{kg/m}^3$ (standard atmosphere, no other temperature stated). The 20 m diameter and 275 m height describe the chimney but do not enter the potential-flow surface-velocity formula, which is independent of cylinder radius.
Point 1 is the stagnation point (θ=0); point 2 is at θ=90° where the ideal-flow surface velocity peaks.
Find. Velocity and gauge pressure at point 1 (θ=0) and point 2 (θ=90°).
Approach. Evaluate $V_t$ at each angle, then apply Bernoulli between the free stream and each surface point (same elevation) to get gauge pressure.
Point 1 — stagnation. $V_1=2V_0\sin(0^\circ)=0$. Bernoulli gives the full dynamic pressure recovered:
$$p_1 = \tfrac12\rho_{air}V_0^2 = 0.5\times1.21\times27.78^2 = \boxed{466.8\ \text{Pa gauge}}$$
Point 2 — side (θ=90°). $V_2=2V_0\sin(90^\circ)=2V_0=\boxed{55.56\ \text{m/s}}$.
Pressure at point 2. Bernoulli, same elevation: $p_2 = \tfrac12\rho_{air}(V_0^2-V_2^2) = 0.5\times1.21\times(27.78^2-55.56^2) = \boxed{-1400.5\ \text{Pa gauge}}$ (a suction, since the surface velocity there exceeds the free-stream velocity).
